Dirac’s delta function and Schrödinger’s equation in three dimensions

Feynman’s rather informal derivation of Schrödinger’s equation – following Schrödinger’s own logic when he published his famous paper on it back in 1926 – is wonderfully simple but, as I mentioned in my post on it, does lack some mathematical rigor here and there. Hence, Feynman hastens to dot all of the i‘s and cross all of the t‘s in the subsequent Lectures. We’ll look at two things here:

Dirac’s delta function, which ensures proper ‘normalization’. In fact, as you’ll see in a moment, it’s more about ‘orthogonalization’ than normalization. 🙂
The generalization of Schrödinger’s equation to three dimensions (in space) and also including the presence of external force fields (as opposed to the usual ‘free space’ assumption).

The second topic is the most interesting, of course, and also the easiest, really. However, let’s first use our energy to grind through the first topic. 🙂

Dirac’s delta function

When working with a finite set of discrete states, a fundamental condition is that the base states be ‘orthogonal’, i.e. they must satisfy the following equation:

〈 i | j 〉 = δ_ij, with δ_ij= 1 if i = j and δ_ij= 0 if i ≠ j

Needless to say, the base states i and j are rather special vectors in a rather special mathematical space (a so-called Hilbert space) and so it’s rather tricky to interpret their ‘orthogonality’ in any geometric way, although such geometric interpretation is often actually possible in simple quantum-mechanical systems: you’ll just notice a ‘right’ angle may actually be 45°, or 180° angles, or whatever. 🙂 In any case, that’s not the point here. The question is: if we move an infinite number of base states – like we did when we introduced the ψ(x) and φ(p) wavefunctions – what happens to that condition?

Your first reaction is going to be: nothing. Because… Well… Remember that, for a two-state system, in which we have two base states only, we’d fully describe some state | φ 〉 as a linear combination of the base states, so we’d write:

| φ 〉 =| I 〉 C_I + | II 〉 C_II

Now, while saying we were talking a Hilbert space here, I did add we could use the same expression to define the base states themselves, so I wrote the following triviality:

Trivial but sensible. So we’d associate the base state | I 〉 with the base vector (1, 0) and, likewise, base state | II 〉 with the base vector (0, 1). When explaining this, I added that we could easily extend to an N-state system and so there’s a perfect analogy between the 〈 i | j 〉 bra-ket expression in quantum math and the e_i·e_j product in the run-of-the-mill coordinate spaces that you’re used to. So why can’t we just extend the concept to an infinite-state system and move to base vectors with an infinite number of elements, which we could write as e_i =(…, 0, e_i = 1, 0, 0,,…) and e_j =(…, 0, 0, e_j = 1, 0,…), thereby ensuring 〈 i | j 〉 = e_i·e_j = δ_ij—always! The ‘orthogonality’ condition looks simple enough indeed, and so we could re-write it as:

〈 x | x’ 〉 = δ_xx’, with δ_xx’= 1 if x = x’ and δ_xx’= 0 if if x ≠ x’

However, when moving from a space with a finite number of dimensions to a space with an infinite number of dimensions, there are some issues. They pop up, for example, when we insert that 〈 x | x’ 〉 = δ_xx’ function (note that we’re talking some function here of x and x’, indeed, so we’ll write it as f(x, x’) in the next step) in that 〈φ|ψ〉 = ∫〈φ|x〉〈x|ψ〉dx integral.

Huh? What integral? Relax: that 〈φ|ψ〉 = ∫〈φ|x〉〈x|ψ〉dx integral just generalizes our 〈φ|ψ〉 = ∑〈φ|x〉〈x|ψ〉 expression for discrete settings for the continuous case. Just look at it. When substituting φ for x’, we get:

〈x’|ψ〉 = ψ(x’) = ∫ 〈x’|x〉 〈x|ψ〉 dx ⇔ ψ(x’) = ∫ 〈x’|x〉 ψ(x) dx

You’ll say: what’s the problem? Well… From a mathematical point of view, it’s a bit difficult to find a function 〈x’|x〉 = f(x, x’) which, when multiplied with a wavefunction ψ(x), and integrated over all x, will just give us ψ(x’). A bit difficult? Well… It’s worse than that: it’s actually impossible!

Huh? Yes. Feynman illustrates the difficulty for x’ = 0, but he could have picked whatever value, really. In any case, if x’ = 0, we can write f(x, 0) = f(x), and our integral now reduces to:

ψ(0) = ∫ f(x) ψ(0) dx

This is a weird expression: the value of the integral (i.e. the right-hand side of the expression) does not depend on x: it is just some non-zero value ψ(0). However, we know that the f(x) in the integrand is zero for all x ≠ 0. Hence, this integral will be zero. So we have an impossible situation: we wish a function to be zero everywhere but for one point, and, at the same time, we also want it to give us a finite integral when using it in that integral above.

You’re likely to shake your head now and say: what the hell? Does it matter? It does: it is an actual problem in quantum math. Well… I should say: it was an actual problem in quantum math. Dirac solved it. He invented a new function which looks a bit less simple than our suggested generalization of Kronecker’s delta for the continuous case (i.e. that 〈 x | x’ 〉 = δ_xx’ conjecture above). Dirac’s function is – quite logically – referred to as the Dirac delta function, and it’s actually defined by that integral above, in the sense that we impose the following two conditions on it:

δ(x − x‘) = 0 if x ≠ x’ (so that’s just like the first of our two conditions for that 〈 x | x’ 〉 = δ_xx’ function)
∫ δ(x − x‘)ψ(x) dx = ψ(x’) (so that’s not like the second of our two condition for that 〈 x | x’ 〉 = δ_xx’ function)

Indeed, that second condition is much more sophisticated than our 〈 x | x’ 〉 = 1 if x = x’ condition. In fact, one can show that the second condition amounts to finding some function satisfying this condition:

∫ δ(x)dx = 1

We get this by equating x’ to zero once more and, additionally, by equating ψ(x) to 1. [Please do double-check yourself.] Of course, this ‘normalization’ (or ‘orthogonalization’) problem all sounds like a lot of hocus-pocus and, in many ways, it is. In fact, we’re actually talking a mathematical problem here which had been lying around for centuries (for a brief overview, see the Wikipedia article on it). So… Well… Without further ado, I’ll just give you the mathematical expression now—and please don’t stop reading now, as I’ll explain it in a moment:

I will also credit Wikipedia with the following animation, which shows that the expression above is just the normal distribution function, and which shows what happens when that a, i.e. its standard deviation, goes to zero: Dirac’s delta function is just the limit of a sequence of (zero-centered) normal distributions. That’s all. Nothing more, nothing less.

But how do we interpret it? Well… I can’t do better than Feynman as he describes what’s going on really:

“Dirac’s δ(x) function has the property that it is zero everywhere except at x = 0 but, at the same time, it has a finite integral equal to unity. [See the ∫ δ(x)dx = 1 equation.] One should imagine that the δ(x) function has such fantastic infinity at one point that the total area comes out equal to one.”

Well… That says it all, I guess. 🙂 Don’t you love the way he puts it? It’s not an ‘ordinary’ infinity. No. It’s fantastic. Frankly, I think these guys were all fantastic. 🙂 The point is: that special function, Dirac’s delta function, solves our problem. The equivalent expression for the 〈 i | j 〉 = δ_ij condition for a finite and discrete set of base states is the following one for the continuous case:

〈 x | x’ 〉 = δ(x − x’)

The only thing left now is to generalize this result to three dimensions. Now that’s fairly straightforward. The ‘normalization’ condition above is all that’s needed in terms of modifying the equations for dealing with the continuum of base states corresponding to the points along a line. Extending the analysis to three dimensions goes as follows:

First, we replace the x coordinate by the vector r = (x, y, z)
As a result, integrals over x, become integrals over x, y and z. In other words, they become volume integrals.
Finally, the one-dimensional δ-function must be replaced by the product of three δ-functions: one in x, one in y and one in z. We write:

〈 r | r‘ 〉 = δ(x − x’) δ(y − y’)δ(z − z’)

Feynman summarizes it all together as follows:

What if we have two particles, or more? Well… Once again, I won’t bother to try to re-phrase the Grand Master as he explains it. I’ll just italicize or boldface the key points:

Suppose there are two particles, which we can call particle 1 and particle 2. What shall we use for the base states? One perfectly good set can be described by saying that particle 1 is at x₁and particle 2 is at x₂, which we can write as | x₁x₂〉. Notice that describing the position of only one particle does not define a base state. Each base state must define the condition of the entire system, so you must not think that each particle moves independently as a wave in three dimensions. Any physical state | ψ 〉 can be defined by giving all of the amplitudes 〈 x₁x₂| ψ 〉 to find the two particles at x₁and x₂. This generalized amplitude is therefore a function of the two sets of coordinates x₁and x₁. You see that such a function is not a wave in the sense of an oscillation that moves along in three dimensions. Neither is it generally simply a product of two individual waves, one for each particle. It is, in general, some kind of a wave in the six dimensions defined by x₁and x₁. Hence, if there are two particles in Nature which are interacting, there is no way of describing what happens to one of the particles by trying to write down a wave function for it alone. The famous paradoxes that we considered in earlier chapters—where the measurements made on one particle were claimed to be able to tell what was going to happen to another particle, or were able to destroy an interference—have caused people all sorts of trouble because they have tried to think of the wave function of one particle alone, rather than the correct wave function in the coordinates of both particles. The complete description can be given correctly only in terms of functions of the coordinates of both particles.

Now we really know it all, don’t we? 🙂

Well… Almost. I promised to tackle another topic as well. So here it is:

Schrödinger’s equation in three dimensions

Let me start by jotting down what we had found already, i.e. Schrödinger’s equation when only one coordinate in space is involved. It’s written as:

Now, the extension to three dimensions is remarkably simple: we just substitute the ∂/∂x²operator by the ∇²operator, i.e. ∇²= ∂/∂x²+ ∂/∂y²+ ∂/∂z². We get:

Finally, we can also put forces on the particle, so now we are not looking at a particle moving in free space: we’ve got some force field working on it. It turns out the required modification is equally simple. The grand result is Schrödinger’s original equation in three dimensions:

V = V(x, y, z) is, of course, just the potential here. Remarkably simple equations but… How do we get these? Well… Sorry. The math is not too difficult, but you’re well equipped now to look at Feynman’s Lecture on it yourself now. You really are. Trust me. I really dealt with all of the ‘serious’ stuff you need to understand how he’s going about it in my previous posts so, yes, now I’ll just sit back and relax. Or go biking. Or whatever. 🙂

The Uncertainty Principle

In my previous post, I showed how Feynman derives Schrödinger’s equation using a historical and, therefore, quite intuitive approach. The approach was intuitive because the argument used a discrete model, so that’s stuff we are well acquainted with—like a crystal lattice, for example. However, now we’re now going to think continuity from the start. Let’s first see what changes in terms of notation.

New notations

Our C(x_n, t) = 〈x_n|ψ〉 now becomes C(x) = 〈x|ψ〉. This notation does not explicitly show the time dependence but then you know amplitudes like this do vary in space as well as in time. Having said that, the analysis below focuses mainly on their behavior in space, so it does make sense to not explicitly mention the time variable. It’s the usual trick: we look at how stuff behaves in space or, alternatively, in time. So we temporarily ‘forget’ about the other variable. That’s just how we work: it’s hard for our mind to think about these wavefunctions in both dimensions simultaneously although, ideally, we should do that.

Now, you also know that quantum physicists prefer to denote the wavefunction C(x) with some Greek letter: ψ (psi) or φ (phi). Feynman think it’s somewhat confusing because we use the same to denote a state itself, but I don’t agree. I think it’s pretty straightforward. In any case, we write:

ψ(x) = C_ψ(x) = C(x) = 〈x|ψ〉

The next thing is the associated probabilities. From your high school math course, you’ll surely remember that we have two types of probability distributions: they are either discrete or, else, continuous. If they’re continuous, then our probability distribution becomes a probability density function (PDF) and, strictly speaking, we should no longer say that the probability of finding our particle at any particular point x at some time t is this or that. That probability is, strictly speaking, zero: if our variable is continuous, then our probability is defined for an interval only, and the P[x] value itself is referred to as a probability density. So we’ll look at little intervals Δx, and we can write the associated probability as:

prob (x, Δx) = |〈x|ψ〉|²Δx = |ψ(x)|²Δx

The idea is illustrated below. We just re-divide our continuous scale in little intervals and calculate the surface of some tiny elongated rectangle now. 🙂

It is also easy to see that, when moving to an infinite set of states, our 〈φ|ψ〉 = ∑〈φ|x〉〈x|ψ〉 (over all x) formula for calculating the amplitude for a particle to go from state ψ to state φ should now be written as an infinite sum, i.e. as the following integral:

Now, we know that 〈φ|x〉 = 〈x|φ〉* and, therefore, this integral can also be written as:

For example, if φ(x) = 〈x|φ〉 is equal to a simple exponential, so we can write φ(x) = a·e⁻ⁱ^θ, then φ*(x) = 〈φ|x〉 = a·e⁺ⁱ^θ.

With that, we’re ready for the plat de résistance, except for one thing, perhaps: we don’t look at spin here. If we’d do that, we’d have to take two sets of base sets: one for up and one for down spin—but we don’t worry about this, for the time being, that is. 🙂

The momentum wavefunction

Our wavefunction 〈x|ψ〉 varies in time as well as in space. That’s obvious. How exactly depends on the energy and the momentum: both are related and, hence, if there’s uncertainty in the momentum, there will be uncertainty in the momentum, and vice versa. Uncertainty in the momentum changes the behavior of the wavefunction in space—through the p = ħk factor in the argument of the wavefunction (θ = ω·t − k·x)—while uncertainty in the energy changes the behavior of the wavefunction in time—through the E = ħω relation. As mentioned above, we focus on the variation in space here. We’ll do so y defining a new state, which is referred to as a state of definite momentum. We’ll write it as mom p, and so now we can use the Dirac notation to write the amplitude for an electron to have a definite momentum equal to p as:

φ(p) = 〈 mom p | ψ 〉

Now, you may think that the 〈x|ψ〉 and 〈mom p|ψ〉 amplitudes should be the same because, surely, we do associate the x state with a definite momentum p, don’t we? Well… No! If we want to localize our wave ‘packet’, i.e. localize our particle, then we’re actually not going to associate it with a definite momentum. See my previous posts: we’re going to introduce some uncertainty so our wavefunction is actually a superposition of more elementary waves with slightly different (spatial) frequencies. So we should just go through the motions here and apply our integral formula to ‘unpack’ this amplitude. That goes as follows:

So, as usual, when seeing a formula like this, we should remind ourselves of what we need to solve. Here, we assume we somehow know the ψ(x) = 〈x|ψ〉 wavefunction, so the question is: what do we use for 〈 mom p | x 〉? At this point, Feynman wanders off to start a digression on normalization, which really confuses the picture. When everything is said and done, the easiest thing to do is to just jot down the formula for that 〈mom p | x〉 in the integrand and think about it for a while:

〈mom p | x〉 = e^{−i(p/ħ)∙x}

I mean… What else could it be? This formula is very fundamental, and I am not going to try to explain it. As mentioned above, Feynman tries to ‘explain’ it by some story about probabilities and normalization, but I think his ‘explanation’ just confuses things even more. Really, what else would it be? The formula above really encapsulates what it means if we say that p and x are conjugate variables. [I can already note, of course, that symmetry implies that we can write something similar for energy and time. Indeed, we can define a state of definite energy as 〈E | ψ〉, and then ‘unpack’ it in the same way, and see that one of the two factors in the integrand would be equal to 〈E | t〉 and, of course, we’d associate a similar formula with it:

〈E | t〉 = e^i(E/ħ)∙t

But let me get back to the lesson here. We’re analyzing stuff in space now, not in time. Feynman gives a simple example here. He suggests a wavefunction which has the following form:

ψ(x) = K·e^−x²/4σ²

The example is somewhat disingenuous because this is not a complex– but real-valued function. In fact, squaring it, and then calculating K applying the normalization condition (all probabilities have to add up to one), yields the normal probability distribution:

prob (x, Δx) = P(x)dx = (2πσ²)^−1/2e^−x²/2σ²dx

So that’s just the normal distribution for μ = 0, as illustrated below.

In any case, the integral we have to solve now is:

Now, I hate integrals as much as you do (probably more) and so I assume you’re also only interested in the result (if you want the detail: check it in Feynman), which we can write as:

φ(p) = (2πη²)^−1/4·e^−p²/4η², with η = ħ/2σ

This formula is totally identical to the ψ(x) = (2πσ²)^−1/4·e^−x²/4σ²distribution we started with, except that it’s got another sigma value, which we denoted by η (and that’s not nu but eta), with

η = ħ/2σ

Just for the record, Feynman refers to η and σ as the ‘half-width’ of the respective distributions. Mathematicians would say they’re the standard deviation. The concept are nearly the same, but not quite. In any case, that’s another thing I’ll let you find our for yourself. 🙂 The point is: η and σ are inversely proportional to each other, and the constant of proportionality is equal to ħ/2.

Now, if we take η and σ as measures of the uncertainty in p and x respectively – which is what they are, obviously ! – then we can re-write that η = ħ/2σ as ησ = ħ/2 or, better still, as the Uncertainty Principle itself:

ΔpΔx = ħ/2

You’ll say: that’s great, but we usually see the Uncertainty Principle written as:

ΔpΔx ≥ ħ/2

So where does that come from? Well… We choose a normal distribution (or the Gaussian distribution, as physicists call it), and so that yields the ΔpΔx = ħ/2 identity. If we’d chosen another one, we’d find a slightly different relation and so… Well… Let me quote Feynman here: “Interestingly enough, it is possible to prove that for any other form of a distribution in x or p, the product ΔpΔx cannot be smaller than the one we have found here, so the Gaussian distribution gives the smallest possible value for the ΔpΔx product.”

This is great. So what about the even more approximate ΔpΔx ≥ ħ formula? Where does that come from? Well… That’s more like a qualitative version of it: it basically says the minimum value of the same product is of the same order as ħ which, as you know, is pretty tiny: it’s about 0.0000000000000000000000000000000006626 J·s. 🙂 The last thing to note is its dimension: momentum is expressed in newton-second and position in meter, obviously. So the uncertainties in them are expressed in the same unit, and so the dimension of the product is N·m·s = J·s. So this dimension combines force, distance and time. That’s quite appropriate, I’d say. The ΔEΔt product obviously does the same. But… Well… That’s it, folks! I enjoyed writing this – and I cannot always say the same of other posts! So I hope you enjoyed reading it. 🙂

Schrödinger’s equation: the original approach

Of course, your first question when seeing the title of this post is: what’s original, really? Well… The answer is simple: it’s the historical approach, and it’s original because it’s actually quite intuitive. Indeed, Lecture no. 16 in Feynman’s third Volume of Lectures on Physics is like a trip down memory lane as Feynman himself acknowledges, after presenting Schrödinger’s equation using that very rudimentary model we developed in our previous post:

So… Well… Let’s have a look at it. 🙂 We were looking at some electron we described in terms of its location at one or the other atom in a linear array (think of it as a line). We did so by defining base states |n〉 = |x_n〉, noting that the state of the electron at any point in time could then be written as:

|φ〉 = ∑ |x_n〉C_n(t) = ∑ |x_n〉〈x_n|φ〉 over all n

The C_n(t) = 〈x_n|φ〉 coefficient is the amplitude for the electron to be at x_nat t. Hence, the C_n(t) amplitudes vary with t as well as with x. We’ll re-write them as C_n(t) = C(x_n, t) = C(x_n). Note that the latter notation does not explicitly show the time dependence. The Hamiltonian equation we derived in our previous post is now written as:

iħ·(∂C(x_n)/∂t) = E₀C(x_n) − AC(x_n+b) − AC(x_n−b)

Note that, as part of our move from the C_n(t) to the C(x_n) notation, we write the time derivative dC_n(t)/dt now as ∂C(x_n)/∂t, so we use the partial derivative symbol now (∂). Of course, the other partial derivative will be ∂C(x)/∂x) as we move from the count variable x_nto the continuous variable x, but let’s not get ahead of ourselves here. The solution we found for our C(x_n) functions was the following wavefunction:

C(x_n) = a·e^{i(k∙x_n−ω·t)}= a·e^−i∙ω·t·eⁱ^∙k∙x_n= a·e^{−i·(E/ħ)·t}·e^i·k∙x_n

We also found the following relationship between E and k:

E = E₀ − 2A·cos(kb)

Now, even Feynman struggles a bit with the definition of E₀ and k here, and their relationship with E, which is graphed below.

Indeed, he first writes, as he starts developing the model, that E₀ is, physically, the energy the electron would have if it couldn’t leak away from one of the atoms, but then he also adds: “It represents really nothing but our choice of the zero of energy.”

This is all quite enigmatic because we cannot just do whatever we want when discussing the energy of a particle. As I pointed out in one of my previous posts, when discussing the energy of a particle in the context of the wavefunction, we generally consider it to be the sum of three different energy concepts:

The particle’s rest energy m₀c², which de Broglie referred to as internal energy (E_int), and which includes the rest mass of the ‘internal pieces’, as Feynman puts it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’ interaction energy).
Any potential energy it may have because of some field (i.e. if it is not traveling in free space), which we usually denote by U. This field can be anything—gravitational, electromagnetic: it’s whatever changes the energy of the particle because of its position in space.
The particle’s kinetic energy, which we write in terms of its momentum p: m·v²/2 = m²·v²/(2m) = (m·v)²/(2m) = p²/(2m).

It’s obvious that we cannot just “choose” the zero point here: the particle’s rest energy is its rest energy, and its velocity is its velocity. So it’s not quite clear what the E₀ in our model really is. As far as I am concerned, it represents the average energy of the system really, so it’s just like the E₀ for our ammonia molecule, or the E₀ for whatever two-state system we’ve seen so far. In fact, when Feynman writes that we can “choose our zero of energy so that E₀ − 2A = 0″ (so the minimum of that curve above is at the zero of energy), he actually makes some assumption in regard to the relative magnitude of the various amplitudes involved.

We should probably think about it in this way: −(i/ħ)·E₀ is the amplitude for the electron to just stay where it is, while i·A/ħ is the amplitude to go somewhere else—and note we’ve got two possibilities here: the electron can go to |x_n+1〉, or, alternatively, it can go to |x_n−1〉. Now, amplitudes can be associated with probabilities by taking the absolute square, so I’d re-write the E₀ − 2A = 0 assumption as:

E₀= 2A ⇔ |−(i/ħ)·E₀|²= |(i/ħ)·2A|²

Hence, in my humble opinion, Feynman’s assumption that E₀ − 2A = 0 has nothing to do with ‘choosing the zero of energy’. It’s more like a symmetry assumption: we’re basically saying it’s as likely for the electron to stay where it is as it is to move to the next position. It’s an idea I need to develop somewhat further, as Feynman seems to just gloss over these little things. For example, I am sure it is not a coincidence that the E_I, E_II, E_III and E_IVenergy levels we found when discussing the hyperfine splitting of the hydrogen ground state also add up to 0. In fact, you’ll remember we could actually measure those energy levels (E_I= E_II= E_III= A ≈ 9.23×10⁻⁶ eV, and E_IV= −3A ≈ −27.7×10⁻⁶ eV), so saying that we can “choose” some zero energy point is plain nonsense. The question just doesn’t arise. In any case, as I have to continue the development here, I’ll leave this point for further analysis in the future. So… Well… Just note this E₀ − 2A = 0 assumption, as we’ll need it in a moment.

The second assumption we’ll need concerns the variation in k. As you know, we can only get a wave packet if we allow for uncertainty in k which, in turn, translates into uncertainty for E. We write:

ΔE = Δ[E₀ − 2A·cos(kb)]

Of course, we’d need to interpret the Δ as a variance (σ²) or a standard deviation (σ) so we can apply the usual rules – i.e. var(a) = 0, var(aX) = a²·var(X), and var(aX ± bY) = a²·var(X) + b²·var(Y) ± 2ab·cov(X, Y) – to be a bit more precise about what we’re writing here, but you get the idea. In fact, let me quickly write it out:

var[E₀ − 2A·cos(kb)] = var(E₀) + 4A²·var[cos(kb)] ⇔ var(E) = 4A²·var[cos(kb)]

Now, you should check my post scriptum to my page on the Essentials, to see how the probability density function of the cosine of a randomly distributed variable looks like, and then you should go online to find a formula for its variance, and then you can work it all out yourself, because… Well… I am not going to do it for you. What I want to do here is just show how Feynman gets Schrödinger’s equation out of all of these simplifications.

So what’s the second assumption? Well… As the graph shows, our k can take any value between −π/b and +π/b, and therefore, the kb argument in our cosine function can take on any value between −π and +π. In other words, kb could be any angle. However, as Feynman puts it—we’ll be assuming that kb is ‘small enough’, so we can use the small-angle approximations whenever we see the cos(kb) and/or sin(kb) functions. So we write: sin(kb) ≈ kb and cos(kb) ≈ 1 − (kb)²/2 = 1 − k²b²/2. Now, that assumption led to another grand result, which we also derived in our previous post. It had to do with the group velocity of our wave packet, which we calculated as:

v = dω/dk = (2Ab²/ħ)·k

Of course, we should interpret our k here as “the typical k“. Huh? Yes… That’s how Feynman refers to it, and I have no better term for it. It’s some kind of ‘average’ of the Δk interval, obviously, but… Well… Feynman does not give us any exact definition here. Of course, if you look at the graph once more, you’ll say that, if the typical kb has to be “small enough”, then its expected value should be zero. Well… Yes and no. If the typical kb is zero, or if k is zero, then v is zero, and then we’ve got a stationary electron, i.e. an electron with zero momentum. However, because we’re doing what we’re doing (that is, we’re studying “stuff that moves”—as I put it unrespectfully in a few of my posts, so as to distinguish from our analyses of “stuff that doesn’t move”, like our two-state systems, for example), our “typical k” should not be zero here. OK… We can now calculate what’s referred to as the effective mass of the electron, i.e. the mass that appears in the classical kinetic energy formula: K.E. = m·v²/2. Now, there are two ways to do that, and both are somewhat tricky in their interpretation:

1. Using both the E₀ − 2A = 0 as well as the “small kb” assumption, we find that E = E₀ − 2A·(1 − k²b²/2) = A·k²b². Using that for the K.E. in our formula yields:

m_eff= 2A·k²b²/v²= 2A·k²b²/[(2Ab²/ħ)·k]²= ħ²/(2Ab²)

2. We can use the classical momentum formula (p = m·v), and then the 2nd de Broglie equation, which tells us that each wavenumber (k) is to be associated with a value for the momentum (p) using the p = ħk (so p is proportional to k, with ħ as the factor of proportionality). So we can now calculate m_eff as m_eff = ħk/v. Substituting v again for what we’ve found above, gives us the same:

m_eff= 2A·k²b²/v= ħ·k/[(2Ab²/ħ)·k]= ħ²/(2Ab²)

Of course, we’re not supposed to know the de Broglie relations at this point in time. 🙂 But, now that you’ve seen them anyway, note how we have two formulas for the momentum:

The classical formula (p = m·v) tells us that the momentum is proportional to the classical velocity of our particle, and m is then the factor of proportionality.
The quantum-mechanical formula (p = ħk) tells us that the (typical) momentum is proportional to the (typical) wavenumber, with Planck’s constant (ħ) as the factor of proportionality. Combining both combines the classical and quantum-mechanical perspective of a moving particle:

m·v = ħk

I know… It’s an obvious equation but… Well… Think of it. It’s time to get back to the main story now. Remember we were trying to find Schrödinger’s equation? So let’s get on with it. 🙂

To do so, we need one more assumption. It’s the third major simplification and, just like the others, the assumption is obvious on first, but not on second thought. 😦 So… What is it? Well… It’s easy to see that, in our m_eff= ħ²/(2Ab²) formula, all depends on the value of 2Ab². So, just like we should wonder what happens with that kb factor in the argument of our sine or cosine function if b goes to zero—i.e. if we’re letting the lattice spacing go to zero, so we’re moving from a discrete to a continuous analysis now—we should also wonder what happens with that 2Ab² factor! Well… Think about it. Wouldn’t it be reasonable to assume that the effective mass of our electron is determined by some property of the material, or the medium (so that’s the silicon in our previous post) and, hence, that it’s constant really. Think of it: we’re not changing the fundamentals really—we just have some electron roaming around in some medium and all that we’re doing now is bringing those x_ncloser together. Much closer. It’s only logical, then, that our amplitude to jump from x_n_±1to x_nwould also increase, no? So what we’re saying is that 2Ab² is some constant which we write as ħ²/m_effor, what amounts to the same, that Ab²= ħ²/2·m_eff.

Of course, you may raise two objections here:

The Ab²= ħ²/2·m_eff assumption establishes a very particular relation between A and b, as we can write A as A = [ħ²/(2m_eff)]·b⁻² now. So we’ve got like an y = 1/x² relation here. Where the hell does that come from?
We were talking some real stuff here: a crystal lattice with atoms that, in reality, do have some spacing, so that corresponds to some real value for b. So that spacing gives some actual physical significance to those x_nvalues.

Well… What can I say? I think you should re-read that quote of Feynman when I started this post. We’re going to get Schrödinger’s equation – i.e. the ultimate prize for all of the hard work that we’ve been doing so far – but… Yes. It’s really very heuristic, indeed! 🙂 But let’s get on with it now! We can re-write our Hamiltonian equation as:

iħ·(∂C(x_n)/∂t) = E₀C(x_n) − AC(x_n+b) − AC(x_n−b)]

= (E₀−2A)C(x_n) + A[2C(x_n) − C(x_n+b) − C(x_n−b) = A[2C(x_n) − C(x_n+b) − C(x_n−b)]

Now, I know your brain is about to melt down but, fiddling with this equation as we’re doing right now, Schrödinger recognized a formula for the second-order derivative of a function. I’ll just jot it down, and you can google it so as to double-check where it comes from:

Just substitute f(x) for C(x_n) in the second part of our equation above, and you’ll see we can effectively write that 2C(x_n) − C(x_n+b) − C(x_n−b) factor as:

We’re done. We just iħ·(∂C(x_n)/∂t) on the left-hand side now and multiply the expression above with A, to get what we wanted to get, and that’s – YES! – Schrödinger’s equation:

Whatever your objections to this ‘derivation’, it is the correct equation. For a particle in free space, we just write m instead of m_eff, but it’s exactly the same. I’ll now give you Feynman’s full quote, which is quite enlightening:

“We do not intend to have you think we have derived the Schrödinger equation but only wish to show you one way of thinking about it. When Schrödinger first wrote it down, he gave a kind of derivation based on some heuristic arguments and some brilliant intuitive guesses. Some of the arguments he used were even false, but that does not matter; the only important thing is that the ultimate equation gives a correct description of nature. The purpose of our discussion is then simply to show you that the correct fundamental quantum mechanical equation [i.e. Schrödinger’s equation] has the same form you get for the limiting case of an electron moving along a line of atoms. We can think of it as describing the diffusion of a probability amplitude from one point to the next along the line. That is, if an electron has a certain amplitude to be at one point, it will, a little time later, have some amplitude to be at neighboring points. In fact, the equation looks something like the diffusion equations which we have used in Volume I. But there is one main difference: the imaginary coefficient in front of the time derivative makes the behavior completely different from the ordinary diffusion such as you would have for a gas spreading out along a thin tube. Ordinary diffusion gives rise to real exponential solutions, whereas the solutions of Schrödinger’s equation are complex waves.”

So… That says it all, I guess. Isn’t it great to be where we are? We’ve really climbed a mountain here. And I think the view is gorgeous. 🙂

Oh—just in case you’d think I did not give you Schrödinger’s equation, let me write it in the form you’ll usually see it:

schrodinger 3

Done! 🙂

Quantum math in solid-state physics

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

Original post:

I’ve said it a couple of times already: so far, we’ve only studied stuff that does not move in space. Hence, till now, time was the only variable in our wavefunctions. So now it’s time to… Well… Study stuff that does move in space. 🙂

Is that compatible with the title of this post? Solid-state physics? Solid-state stuff doesn’t move, does it? Well… No. But what we’re going to look at is how an electron travels through a solid crystal or, more generally, how an atomic excitation can travel through. In fact, what we’re really going to look at is how the wavefunction itself travels through space. However, that’s a rather bold statement, and so you should just read this post and judge for yourself. To be specific, we’re going to look at what happens in semiconductor material, like the silicon that’s used in microelectronic components like transistors and integrated circuits (ICs). You surely know the classical idea of that, which involves imagining an electron can be situated in a kind of ‘pit’ at one particular atom (or an electron hole, as it’s usually referred to), and it just moves from pit to pit. The Wikipedia article on it defines an electron hole as follows: an electron hole is the absence of an electron from a full valence band: the concept is used to conceptualize the interactions of the electrons within a nearly full system, i.e. a system which is missing just a few electrons. But here we’re going to forget about the classical picture. We’ll try to model it using the wavefunction concept. So how does that work? Feynman approaches it as follows.

If we look at a (one-dimensional) line of atoms – we can extend to a two- and three-dimensional analysis later – we may define an infinite number of base states for the extra electron that we think of as moving through the crystal. If the electron is with the n-th atom, then we’ll say it’s in a base state which we shall write as |n〉. Likewise, if it’s at atom n+1 or n−1, then we’ll associate that with base state |n+1〉 and |n−1〉 respectively. That’s what visualized below, and you should just along with the story here: don’t think classically, i.e. in terms of the electron is either here or, else, somewhere else. No. It’s got an amplitude to be anywhere. If you can’t take that… Well… I am sorry but that’s what QM is all about!

As usual, we write the amplitude for the electron to be in one of those states |n〉 as C_n(t) = 〈n|φ〉, and so we can the write the electron’s state at any point in time t by superposing all base states, so that’s the weighted sum of all base states, with the weights being equal to the associated amplitudes. So we write:

|φ〉 = ∑ |n〉C_n(t) = ∑ |n〉〈n|φ〉 over all n

Now we add some assumptions. One assumption is that an electron cannot directly jump to its next nearest neighbor: if it goes to the next nearest one, it will first have to go to nearest one. So two steps are needed to go from state |n−1〉 to state |n+1〉. This assumption simplifies the analysis: we can discuss more general cases later. To be specific, we’ll assume the amplitude to go from one base state to another, e.g. from |n〉 to |n+1〉, or |n−1〉 to state |n〉, is equal to i·A/ħ. You may wonder where this comes from, but it’s totally in line with equating our Hamiltonian non-diagonal elements to –A. Let me quickly insert a small digression here—for those who do really wonder where this comes from. 🙂

START OF DIGRESSION

Just check out those two-state systems we described, or that post of mine in which I explained why the following formulas are actually quite intuitive and easy to understand:

U₁₂(t + Δt, t) = − (i/ħ)·H₁₂(t)·Δt = (i/ħ)·A·Δt and
U₂₁(t + Δt, t) = − (i/ħ)·H₂₁(t)·Δt = (i/ħ)·A·Δt

More generally, you’ll remember that we wrote U_ij(t + Δt, t) as:

U_ij(t + Δt, t) = U_ij(t, t) + K_ij·Δt = δ_ij(t, t) + K_ij·Δt = δ_ij(t, t) − (i/ħ)·H_ij(t)·Δt

That looks monstrous but, frankly, what we have here is just a formula like this:

f(x+Δx) = f(x) + [df(x)/dt]·Δx

In case you didn’t notice, the formula is just the definition of the derivative if we write it as Δy/Δx = df(x)/dt for Δx → 0. Hence, the K_ijcoefficient in this formula is to be interpreted as a time derivative. Now, we re-wrote that K_ijcoefficient as the amplitude −(i/ħ)·H_ij(t) and, therefore, that amplitude – i.e. the i·A/ħ factor (for i ≠ j) I introduced above – is to be interpreted as a time derivative. [Now that we’re here, let me quickly add that a time derivative gives the time rate of change of some quantity per unit time. So that i·A/ħ factor is also expressed per unit time.] We’d then just move the − (i/ħ) factor in that −(i/ħ)·H_ij(t) coefficient to the other side to get the grand result we got for two-state systems, i.e. the Hamiltonian equations, which we could write in a number of ways, as shown below:

So… Well… That’s all there is to it, basically. Quantum math is not easy but, if anything, it is logical. You just have to get used to that imaginary unit (i) in front of stuff. That makes it always look very mysterious. 🙂 However, it should never scare you. You can just move it in or out of the differential operator, for example: i·df(x)/dt = d[i·f(x)]/dt. [Of course, i·f(x) ≠ f(i·x)!] So just think of i as a reminder that the number that follows it points in a different direction. To be precise: its angle with the other number is 90°. It doesn’t matter what we call those two numbers. The convention is to say that one is the real part of the wavefunction, while the other is the imaginary part but, frankly, in quantum math, both numbers are just as real. 🙂

END OF DIGRESSION

Yes. Let me get back to the lesson here. The assumption is that the Hamiltonian equations for our system here, i.e. the electron traveling from hole to hole, look like the following equation:

It’s really like those iħ·(dC₁/dt) = E₀C₁− AC₂and iħ·(dC₂/dt) = − AC₁+ E₀C₂ equations above, except that we’ve got three terms here:

−(i/ħ)·E₀ is the amplitude for the electron to just stay where it is, so we multiply that with the amplitude of the electron to be there at that time, i.e. the amplitude C_n(t), and bingo! That’s the first contribution to the time rate of change of the C_n amplitude (i.e. dC_n/dt). [Note that all I brought that iħ factor in front to the other side: 1/(iħ) = −(i/ħ).] Of course, you also need to know what E₀is now: that’s just the (average) energy of our electron. So it’s really like the E₀of our ammonia molecule—or the average energy of any two-state system, really.
−(i/ħ)·(−A) = i·A/ħ is the amplitude to go from one base state to another, i.e. from |n+1〉 to |n〉, for example. In fact, the second term models exactly that: i·A/ħ times the amplitude to be in state |n+1〉 is the second contribution to to the time rate of change of the C_n amplitude.
Finally, the electron may also be in state |n−1〉 and go to |n〉 from there, so i·A/ħ times the amplitude to be in state |n−1〉 is yet another contribution to to the time rate of change of the C_n amplitude.

Now, we don’t want to think about what happens at the start and the end of our line of atoms, so we’ll just assume we’ve got an infinite number of them. As a result, we get an infinite number of equations, which Feynman summarizes as:

Holy cow! How do we solve that? We know that the general solution for those C_n amplitudes is likely to be some function like this:

C_n(t) = a_n·e^{−(i/ħ)·E·t}

In case you wonder where this comes from, check my post on the general solution for N-state systems. If we substitute that trial solution in that iħ·(dC_n/dt) = E₀C_n− AC_n+1− AC_n−1, we get:

Ea_n= E₀a_n− Aa_n+1− Aa_n−1

[Just do that derivative, and you’ll see the iħ can be scrapped. Also, the exponentials on both sides of the equation cancel each other out.] Now, that doesn’t look too bad, and we can also write it as (E − E₀)·a_n= − A(a_n+1+ a_n−1 ), but… Well… What’s the next step? We’ve got an infinite number of coefficients a_nhere, so we can’t use the usual methods to solve this set of equations. Feynman tries something completely different here. It looks weird but… Well… He gets a sensible result, so… Well… Let’s go for it.

He first writes these coefficients a_nas a function of a distance, which he defines as x_n= x_n−1 + b, with b the atomic spacing, i.e. the distance between two atoms (see the illustration). So now we write a_n= f(x_n) = a(x_n). Note that we don’t write a_n= f_n(x_n) = a_n(x_n). No. It’s just one function f = a, not an infinite number of functions f_n= a_n. Of course, once you see what comes of it, you’ll say: sure! The (complex) a_ncoefficient in that function is the non-time-varying part of our function, and it’s about time we insert some part that’s varying in space and so… Well… Yes, of course! Our a_ncoefficients don’t vary in time, so they must vary in space. Well… Yes. I guess so. 🙂 Our Ea_n= E₀a_n− Aa_n+1− Aa_n−1equation becomes:

E·a(x_n) = E₀·a(x_n) − A·a(x_n+1)− A·a(x_n+1) = E₀·a(x_n) − A·a(x_n+b)− A·a(x_n−b)

We can write this, once again, as (E − E₀)·a(x_n)= − A·[a(x_n+b)+ a(x_n−b)]. Feynman notes this equation is like a differential equation, in the sense that it relates the value of some function (i.e. our a function, of course) at some point x to the values of the same function at nearby points, i.e. x ± b here. Frankly, I struggle a bit to see how it works exactly but Feynman now offers the following trial solution:

a(x_n) = e^ikx_n

Huh? Why? And what’s k? Be patient. Just go along with this for a while. Let’s first do a graph. Think of x_nas a nearly continuous variable representing position in space. We then know that this parameter k is then equal to the spatial frequency of our wavefunction: larger values for k give the wavefunction a higher density in space, as shown below.

In fact, I shouldn’t confuse you here, but you’ll surely think of the wavefunction you saw so many times already:

ψ(x, t) = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} = a·e^{−i·(ω·t − k∙x)}= a·e^{i(k∙x−ω·t)}= a·e^i∙k∙x·e^−i∙ω·t

This was the elementary wavefunction we’d associate with any particle, and so k would be equal to p/ħ, which is just the second of the two de Broglie relations: E = ħω and p = ħk (or, what amounts to the same: E = hf and λ = h/p). But you shouldn’t get confused. Not at this point. Or… Well… Not yet. 🙂

Let’s just take this proposed solution and plug it in. We get:

E·e^ikx_n = E₀·e^ikx_n − A·e^ik(x_n+b)− A·e^ik(x_n−b)⇔ E = E₀ − A·e^ikb− A·e^−ikb⇔ E = E₀ − 2A·cos(kb)

[In case you wonder what happens here: we just divide both sides by the common factor e^ikx_nand then we also know that e^iθ+e^−iθ= 2·cosθ property.] So each k is associated with some energy E. In fact, to be precise, that E = E₀ − 2A·cos(kb) function is a periodic function: it’s depicted below, and it reaches a maximum at k = ± π/b. [It’s easy to see why: E₀ − 2A·cos(kb) reaches a maximum if cos(kb) = −1, i.e. if kb = ± π.]

Of course, we still don’t really know what k or E are really supposed to represent, but think of it: it’s obvious that E can never be larger or smaller than E ± 2A, whatever the value of k. Note that, once again, it doesn’t matter if we used +A or −A in our equations: the energy band remains the same. And… Well… We’ve dropped the term now: the energy band of a semiconductor. That’s what it’s all about. What we’re saying here is that our electron, as it moves about, can have no other energies than the values in this band. Having said, that still doesn’t determine its energy: any energy level within that energy band is possible. So what does that mean? Hmm… Let’s take a break and not bother too much about k for the moment. Let’s look at our C_n(t) equations once more. We can now write them as:

C_n(t) = e^ikx_n·e^{−(i/ħ)·E·t} = e^ikx_n·e^{−(i/ħ)·[E₀ − 2A·cos(kb)]·t}

You have enough experience now to sort of visualize what happens here. We can look at a certain x_nvalue – read: a certain position in the lattice and watch, as time goes by, how the real and imaginary part of our little C_nwavefunction varies sinusoidally. We can also do it the other way around, and take a snapshot of the lattice at a certain point in time, and then we see how the amplitudes vary from point to point. That’s easy enough.

The thing is: we’re interested in probabilities in the end, and our wavefunction does not satisfy us in that regard: if we take the absolute square, its phase vanishes, and so we get the same probability everywhere! [Note that we didn’t normalize our wavefunctions here. It doesn’t matter. We can always do that later.] Now that’s not great. So what can we do about that? Now that’s where that k comes back in the game. Let’s have a look.

The effective mass of an electron

We’d like to find a solution which sort of ‘localizes’ our electron in space. Now, we know that we can do, in general, by superposing wavefunctions having different frequencies. There are a number of ways to go about, but the general idea is illustrated below.

The first animation (for which credit must go to Wikipedia once more) is, obviously, the most sophisticated one. It shows how a new function – in red, and denoted by s₆(x) – is constructed by summing six sine functions of different amplitudes and with harmonically related frequencies. This particular sum is referred to as a Fourier series, and the so-called Fourier transform, i.e. the S(f) function (in blue), depicts the six frequencies and their amplitudes.

We’re more interested in the second animation here (for which credit goes to another nice site), which shows how a pattern of beats is created by just mixing two slightly different cosine waves. We want to do something similar here: we want to get a ‘wave packet‘ like the one below, which shows the real part only—but you can imagine the imaginary part 🙂 of course. [That’s exactly the same but with a phase shift, cf. the sine and cosine bit in Euler’s formula: e^iθ= cosθ + i·sinθ.]

As you know, we must know make a distinction between the group velocity of the wave, and its phase velocity. That’s got to do with the dispersion relation, but we’re not going to get into the nitty-gritty here. Just remember that the group velocity corresponds to the classical velocity of our particle – so that must be the classical velocity of our electron here – and, equally important, also remember the following formula for that group velocity:

Let’s see how that plays out. The ω in this equation is equal to E/ħ = [E₀ − 2A·cos(kb)]/ħ, so dω/dk = d[− (2A/ħ)·cos(kb)]/dk = (2Ab/ħ)·sin(kb). However, we’ll usually assume k is fairly small, so the variation of the amplitude from one x_n to the other is fairly small. In that case, kb will be fairly small, and then we can use the so-called small angle approximation formula sin(ε) ≈ ε. [Note the reasoning here is a bit tricky, though, because – theoretically – k may vary between −π/b and +π/b and, hence, kb can take any value between −π and +π.] Using the small angle approximation, we get:

So we’ve got a quantum-mechanical calculation here that yields a classical velocity. Now, we can do something interesting now: we can calculate what is known as the effective mass of the electron, i.e. the mass that appears in the classical kinetic energy formula: K.E. = m·v²/2. Or in the classical momentum formula: p = m·v. So we can now write: K.E. = m_eff·v²/2 and p = m_eff·v. But… Well… The second de Broglie equation tells us that p = ħk, so we find that m_eff = ħk/v. Substituting v for what we’ve found above, gives us:

Unsurprisingly, we find that the value of m_eff is inversely proportional to A. It’s usually stated in units of the true mass of the electron, i.e. its mass in free space (m_e≈ 9.11×10⁻³¹ kg) and, in these units, it’s usually in the range of 0.01 to 10. You’ll say: 0.01, i.e. one percent of its actual mass? Yes. An electron may travel more freely in matter than it does in free space. 🙂 That’s weird but… Well… Quantum mechanics is weird.

In any case, I’ll wrap this post up now. You’ ve got a nice model here. As Feynman puts it:

“We have now explained a remarkable mystery—how an electron in a crystal (like an extra electron put into germanium) can ride right through the crystal and flow perfectly freely even though it has to hit all the atoms. It does so by having its amplitudes going pip-pip-pip from one atom to the next, working its way through the crystal. That is how a solid can conduct electricity.”

Well… There you go. 🙂

Systems with 2 spin-1/2 particles (II)

Original post:

In our previous post, we noted the Hamiltonian for a simple system of two spin-1/2 particles—a proton and an electron (i.e. a hydrogen atom, in other words):

After noting that this Hamiltonian is “the only thing that it can be, by the symmetry of space, i.e. so long as there is no external field,” Feynman also notes the constant term (A) depends on the level we choose to measure energies from, so one might just as well take E₀= 0, in which case the formula reduces to H = Aσ^e·σ^p. Feynman analyzes this term as follows:

If there are two magnets near each other with magnetic moments μ_e and μ_p, the mutual energy will depend on μ_e·μ_p= |μ_e||μ_p|cosα = μ_eμ_pcosα — among other things. Now, the classical thing that we call μ_e or μ_p appears in quantum mechanics as μ_eσ^eand μ_pσ^prespectively (where μ_pis the magnetic moment of the proton, which is about 1000 times smaller than μ_e, and has the opposite sign). So the H = Aσ^e·σ^p equation says that the interaction energy is like the interaction between two magnets—only not quite, because the interaction of the two magnets depends on the radial distance between them. But the equation could be—and, in fact, is—some kind of an average interaction. The electron is moving all around inside the atom, and our Hamiltonian gives only the average interaction energy. All it says is that for a prescribed arrangement in space for the electron and proton there is an energy proportional to the cosine of the angle between the two magnetic moments, speaking classically. Such a classical qualitative picture may help you to understand where the H = Aσ^e·σ^pequation comes from.

That’s loud and clear, I guess. The next step is to introduce an external field. The formula for the Hamiltonian (we don’t distinguish between the matrix and the operator here) then becomes:

H = Aσ^e·σ^p − μ_eσ^e·B − μ_pσ^p·B

The first term is the term we already had. The second term is the energy the electron would have in the magnetic field if it were there alone. Likewise, the third term is the energy the proton would have in the magnetic field if it were there alone. When reading this, you should remember the following convention: classically, we write the energy U as U = −μ·B, because the energy is lowest when the moment is along the field. Hence, for positive particles, the magnetic moment is parallel to the spin, while for negative particles it’s opposite. In other words, μ_p is a positive number, while μ_e is negative. Feynman sums it all up as follows:

Classically, the energy of the electron and the proton together, would be the sum of the two, and that works also quantum mechanically. In a magnetic field, the energy of interaction due to the magnetic field is just the sum of the energy of interaction of the electron with the external field, and of the proton with the field—both expressed in terms of the sigma operators. In quantum mechanics these terms are not really the energies, but thinking of the classical formulas for the energy is a way of remembering the rules for writing down the Hamiltonian.

That’s also loud and clear. So now we need to solve those Hamiltonian equations once again. Feynman does so first assuming B is constant and in the z-direction. I’ll refer you to him for the nitty-gritty. The important thing is the results here:

He visualizes these – as a function of μB/A – as follows:

The illustration shows how the four energy levels have a different B-dependence:

E_I, E_II, E_III start at (0, 1) but E_I increases linearly with B—with slope μ, to be precise (cf. the E_I = A + μB expression);
In contrast, E_II decreases linearly with B—again, with slope μ (cf. the E_II = A − μB expression);
We then have the E_III and E_IV curves, which start out horizontally, to then curve and approach straight lines for large B, with slopes equal to μ’.

Oh—I realize I forget to define μ and μ’. Let me do that now: μ = −(μ_e+μ_p) and μ’ = −(μ_e−μ_p). And remember what we said above: μ_pis about 1000 times smaller than μ_e, and has opposite sign. OK. The point is: the magnetic field shifts the energy levels of our hydrogen atom. This is referred to as the Zeeman effect. Feynman describes it as follows:

The curves show the Zeeman splitting of the ground state of hydrogen. When there is no magnetic field, we get just one spectral line from the hyperfine structure of hydrogen. The transitions between state IV and any one of the others occurs with the absorption or emission of a photon whose (angular) frequency is 1/ħ times the energy difference 4A. [See my previous post for the calculation.] However, when the atom is in a magnetic field B, there are many more lines, and there can be transitions between any two of the four states. So if we have atoms in all four states, energy can be absorbed—or emitted—in any one of the six transitions shown by the vertical arrows in the illustration above.

The last question is: what makes the transitions go? Let me also quote Feynman’s answer to that:

The transitions will occur if you apply a small disturbing magnetic field that varies with time (in addition to the steady strong field B). It’s just as we saw for a varying electric field on the ammonia molecule. Only here, it is the magnetic field which couples with the magnetic moments and does the trick. But the theory follows through in the same way that we worked it out for the ammonia. The theory is the simplest if you take a perturbing magnetic field that rotates in the xy-plane—although any horizontal oscillating field will do. When you put in this perturbing field as an additional term in the Hamiltonian, you get solutions in which the amplitudes vary with time—as we found for the ammonia molecule. So you can calculate easily and accurately the probability of a transition from one state to another. And you find that it all agrees with experiment.

Alright! All loud and clear. 🙂

The magnetic quantum number

At very low magnetic fields, we still have the Zeeman splitting, but we can now approximate it as follows:

This simplified representation of things explains an older concept you may still see mentioned: the magnetic quantum number, which is usually denoted by m. Feynman’s explanation of it is quite straightforward, and so I’ll just copy it as is:

As he notes: the concept of the magnetic quantum number has nothing to do with new physics. It’s all just a matter of notation. 🙂

Well… This concludes our short study of four-state systems. On to the next! 🙂

Systems with 2 spin-1/2 particles (I)

Original post:

I agree: this is probably the most boring title of a post ever. However, it should be interesting, as we’re going to apply what we’ve learned so far – i.e. the quantum-mechanical model of two-state systems – to a much more complicated problem—the solution of which can then be generalized to describe even more complicated situations.

Two spin-1/2 particles? Let’s recall the most obvious example. In the ground state of a hydrogen atom (H), we have one electron that’s bound to one proton. The electron occupies the lowest energy state in its ground state, which – as Feynman shows in one of his first quantum-mechanical calculations – is equal to −13.6 eV. More or less, that is. 🙂 You’ll remember the reason for the minus sign: the electron has more energy when it’s unbound, which it releases as radiation when it joins an ionized hydrogen atom or, to put it simply, when a proton and an electron come together. In-between being bound and unbound, there are other discrete energy states – illustrated below – and we’ll learn how to describe the patterns of motion of the electron in each of those states soon enough.

Not in this post, however. 😦 In this post, we want to focus on the ground state only. Why? Just because. That’s today’s topic. 🙂 The proton and the electron can be in either of two spin states. As a result, the so-called ground state is not really a single definite-energy state. The spin states cause the so-called hyperfine structure in the energy levels: it splits them into several nearly equal energy levels, so that’s what referred to as hyperfine splitting.

[…] OK. Let’s go for it. As Feynman points out, the whole model is reduced to a set of four base states:

State 1: |++〉 = |1〉 (the electron and proton are both ‘up’)
State 2: |+−〉 = |2〉 (the electron is ‘up’ and the proton is ‘down’)
State 3: |−+〉 = |3〉 (the electron is ‘down’ and the proton is ‘up’)
State 4: |−−〉 = |4〉 (the electron and proton are both ‘down’)

The simplification is huge. As you know, the spin of electrically charged elementary particles is related to their motion in space, but so we don’t care about exact spatial relationships here: the direction of spin can be in any direction, but all that matters here is the relative orientation, and so all is simplified to some direction as defined by the proton and the electron itself. Full stop.

You know that the whole problem is to find the Hamiltonian coefficients, i.e. the energy matrix. Let me give them to you straight away. The energy levels involved are the following:

E_I= E_II= E_III= A ≈ 9.23×10⁻⁶ eV
E_IV= −3A ≈ 27.7×10⁻⁶ eV

So the difference in energy levels is measured in ten-millionths of an electron-volt and, hence, the hyperfine splitting is really hyper-fine. The question is: how do we get these values? So that is what this post is about. Let’s start by reminding ourselves of what we learned so far.

The Hamiltonian operator

We know that, in quantum mechanics, we describe any state in terms of the base states. In this particular case, we’d do so as follows:

|ψ〉 = |1〉C₁ + |2〉C₂ + |3〉C₃ +|4〉C₄with C_i= 〈i|ψ〉

We refer to |ψ〉 as the spin state of the system, and so it’s determined by those four C_iamplitudes. Now, we know that those C_iamplitudes are functions of time, and they are, in turn, determined by the Hamiltonian matrix. To be precise, we find them by solving a set of linear differential equations that we referred to as Hamiltonian equations. To be precise, we’d describe the behavior of |ψ〉 in time by the following equation:

In case you forgot, the expression above is a short-hand for the following expression:

The index j would range over all base states and, therefore, this expression gives us everything we want: it really does describe the behavior, in time, of an N-state system. You’ll also remember that, when we’d use the Hamiltonian matrix in the way it’s used above (i.e. as an operator on a state), we’d put a little hat over it, so we defined the Hamiltonian operator as:

So far, so good—but this does not solve our problem: how do we find the Hamiltonian for this four-state system? What is it?

Well… There’s no one-size-fits-all answer to that: the analysis of two different two-state systems, like an ammonia molecule, or one spin-1/2 particle in a magnetic field, was different. Having said that, we did find we could generalize some of the solutions we’d find. For example, we’d write the Hamiltonian for a spin-1/2 particle, with a magnetic moment that’s assumed to be equal to μ, in a magnetic field B = (B_x, B_y, B_z) as:

In this equation, we’ve got a set of 4 two-by-two matrices (three so-called sigma matrices (σ_x, σ_y, σ_z), and then the unit matrix δ_ij = 1) which we referred to as the Pauli spin matrices, and which we wrote as:

You’ll remember that expression – which we further abbreviated, even more elegantly, to H = −μσ·B – covered all two-state systems involving a magnetic moment in a magnetic field. In fact, you’ll remember we could actually easily adapt the model to cover two-state systems in electric fields as well.

In short, these sigma matrices made our life very easy—as they covered a whole range of two-state models. So… Well… To make a long story short, what we want to do here is find some similar sigma matrices for four-state problems. So… Well… Let’s do that.

First, you should remind yourself of the fact that we could also use these sigma matrices as little operators themselves. To be specific, we’d let them ‘operate’ on the base states, and we’d find they’d do the following:

You need to read this carefully. What it says that the σ_z matrix, as an operator, acting on the ‘up’ base state, yields the same base state (i.e. ‘up’), and that the same operator, acting on the ‘down’ state, gives us the same but with a minus sign in front. Likewise, the σ_y matrix operating on the ‘up’ and ‘down’ states respectively, will give us i·|down〉 and −i·|up〉 respectively.

The trick to solve our problem here (i.e. our four-state system) is to apply those sigma matrices to the electron and the proton separately. Feynman introduces a new notation here by distinguishing the electron and proton sigma operators: the electron sigma operators (σ_x^e, σ_y^e, and σ_z^e) operate on the electron spin only, while – you guessed it – the proton sigma operator ((σ_x^p, σ_y^p, and σ_z^p) acts on the proton spin only. Applying it to the four states we’re looking at (i.e. |++〉, |+−〉, |−+〉 and |−−〉), we get the following bifurcation for our σ_x operator:

σ_x^e|++〉 = |−+〉
σ_x^e|+−〉 = |−−〉
σ_x^e|−+〉 = |++〉
σ_x^e|−−〉 = |+−〉
σ_x^p|++〉 = |+−〉
σ_x^p|+−〉 = |++〉
σ_x^p|−+〉 = |−−〉
σ_x^p|−−〉 = |−+〉

You get the idea. We had three operators acting on two states, i.e. 6 possibilities. Now we combine these three operators with two different particles, so we have six operators now, and we let them act on four possible system states, so we have 24 possibilities now. Now, we can, of course, let these operators act one after another. Check the following for example:

σ_x^eσ_z^p|+−〉 = σ_x^e[σ_z^p|+−〉] = –σ_x^e|+−〉 = –|–−〉

[I now realize that I should have used the ↑ and ↓ symbols for the ‘up’ and ‘down’ states, as the minus sign is used to denote two very different things here, but… Well… So be it.]

Note that we only have nine possible σ_x^eσ_z^p-like combinations, because σ_x^eσ_z^p= σ_z^pσ_x^e, and then we have the 2×3 = six σ^e and σ^p operators themselves, so that makes for 15 new operators. [Note that the commutativity of these operators (σ_x^eσ_z^p= σ_z^pσ_x^e) is not some general property of quantum-mechanical operators.] If we include the unit operator (δ_ij = 1) – i.e. an operator that leaves all unchanged – we’ve got 16 in total. Now, we mentioned that we could write the Hamiltonian for a two-state system – i.e. a two-by-two matrix – as a linear combination of the four Pauli spin matrices. Likewise, one can demonstrate that the Hamiltonian for a four-state system can always be written as some linear combination of those sixteen ‘double-spin’ matrices. To be specific, we can write it as:

We should note a few things here. First, the E₀ constant is, of course, to be multiplied by the unit matrix, so we should actually write E₀δ_ij instead of E₀, but… Well… Quantum physicists always want to confuse you. 🙂 Second, the σ^eσ^pis like the σ·B notation: we can look at the σ_x^e, σ_y^e, σ_z^e and σ_x^p, σ_y^p, σ_z^pmatrices as being the three components of two new (matrix) vectors, which we write as σ^eand σ^prespectively. Thirdly, and most importantly, you’ll want proof of that equation above. Well… I am sorry but I am going to refer you to Feynman here: he shows that the expression above “is the only thing that the Hamiltonian can be.” The proof is based on the fundamental symmetry of space. He also adds that space is symmetrical only so long as there is no external field. 🙂

Final question: what’s A? Well… Feynman is quite honest here as he says the following: “ $A$ can be calculated accurately once you understand the complete quantum theory of the hydrogen atom—which we so far do not. It has, in fact, been calculated to an accuracy of about $30$ parts in one million. So, unlike the flip-flop constant $A$ of the ammonia molecule, which couldn’t be calculated at all well by a theory, our constant $A$ for the hydrogen can be calculated from a more detailed theory. But never mind, we will for our present purposes think of the $A$ as a number which could be determined by experiment, and analyze the physics of the situation.”

So… Well… So far so good. We’ve got the Hamiltonian. That’s all we wanted, actually. But, now that we have come so far, let’s write it all out now.

Solving the equations

If that expression above is the Hamiltonian – and we assume it is, of course! – then our system of Hamiltonian equations can be written as:

[Note that we’ve switched to Newton’s ‘over-dot’ notation to denote time derivatives here.] Now, I could walk you through Feynman’s exposé but I guess you’ll trust the result. The equation above is equivalent to the following set of four equations:

We know that, because the Hamiltonian looks like this:

How do we know that? Well… Sorry: just check Feynman. 🙂 He just writes it all out. Now, we want to find those C_i functions. [When studying physics, the most important thing is to remember what it is that you’re trying to do. 🙂 ] Now, from my previous post (i.e. my post on the general solution for N-state systems), you’ll remember that those C_i functions should have the following functional form:

C_i(t) = a_i·e^{−i·(E/ħ)·t}

If we substituting C_i(t) for that functional form in our set of Hamiltonian equations, we can cancel the exponentials so we get the following delightfully simple set of new equations:

The trivial solution, of course, is that all of the a_i coefficients are zero, but – as mentioned in my previous post – we’re looking for non-trivial solutions here. Well… From what you see above, it’s easy to appreciate that one non-trivial but simple solution is:

a₁ = 1 and a₂ = a₃ = a₄ = 0

So we’ve got one set of a_i coefficients here, and we’ll associate it with the first eigenvalue, or energy level, really—which we’ll denote as E_I. [I am just being consistent here with what I wrote in my previous post, which explained how general solutions to N-state systems look like.] So we find the following:

E_I= A

[Another thing you learn when studying physics is that the most amazing things are often summarized in super-terse equations, like this one here. 🙂 ]

But – Hey! Look at the symmetry between the first and last equation!

We immediately get another simple – but non-trivial! – solution:

a₄ = 1 and a₁ = a₂ = a₃ = 0

We’ll associate the second energy level with that, so we write:

E_II= A

We’ve got two left. I’ll leave that to Feynman to solve:

Done! Four energy levels E_n(n = I, II, III, IV), and four associated energy state vectors – |n〉 – that describe their configuration (and which, as Feynman puts it, have the time dependence “factored out”). Perfect!

Now, we mentioned the experimental values:

E_I= E_II= E_III= A ≈ 9.23×10⁻⁶ eV
E_IV= −3A ≈ 27.7×10⁻⁶ eV

How can scientists measure these values? The theoretical analysis gives us the A and −3A values, but what about the empirical measurements? Well… We should find those values as the hydrogen atoms in state I, II or III should get rid of the energy by emitting some radiation. Now, the frequency of that radiation will give us the information we need, as illustrated below. The difference between E_I= E_II= E_III= A and E_IV= −3A (i.e. 4A) should correspond to the (angular) frequency of the radiation that’s being emitted or absorbed as atoms go from one energy state to the other. Now, hydrogen atoms do absorb and emit microwave radiation with a frequency that’s equal to 1,420,405,751.8 Hz. More or less, that is. 🙂 The standard error in the measurement is about two parts in 100 billion—and I am quoting some measurement done in the early 1960s here!]

Bingo! If f = ω/2π = (4A/ħ)/2π = 1,420,405,751.8 Hz, then A = f·2π·ħ/4 ≈ 9.23×10⁻⁶ eV.

So… Well… We’re done! I’ll see you tomorrow. 🙂 Tomorrow, we’re going to look at what happens when space is not symmetric, i.e. when we would have some external field! C u ! Cheers !

N-state systems

Original post:

On the 10th of December, last year, I wrote that my next post would generalize the results we got for two-state systems. That didn’t happen: I didn’t write the ‘next post’—not till now, that is. No. Instead, I started digging—as you can see from all the posts in-between this one and the 10 December piece. And you may also want to take a look at my new Essentials page. 🙂 In any case, it is now time to get back to Feynman’s Lectures on quantum mechanics. Remember where we are: halfway, really. The first half was all about stuff that doesn’t move in space. The second half, i.e. all that we’re going to study now, is about… Well… You guessed it. 🙂 That’s going to be about stuff that does move in space. To see how that works, we first need to generalize the two-state model to an N-state model. Let’s do it.

You’ll remember that, in quantum mechanics, we describe stuff by saying it’s in some state which, as long as we don’t measure in what state exactly, is written as some linear combination of a set of base states. [And please do think about what I highlight here: some state, measure, exactly. It all matters. Think about it!] The coefficients in that linear combination are complex-valued functions, which we referred to as wavefunctions, or (probability) amplitudes. To make a long story short, we wrote:

These C_i coefficients are a shorthand for 〈 i | ψ(t) 〉 amplitudes. As such, they give us the amplitude of the system to be in state i as a function of time. Their dynamics (i.e. the way they evolve in time) are governed by the Hamiltonian equations, i.e.:

The H_ij coefficients in this set of equations are organized in the Hamiltonian matrix, which Feynman refers to as the energy matrix, because these coefficients do represent energies indeed. So we applied all of this to two-state systems and, hence, things should not be too hard now, because it’s all the same, except that we have N base states now, instead of just two.

So we have a N×N matrix whose diagonal elements H_ijare real numbers. The non-diagonal elements may be complex numbers but, if they are, the following rule applies: H_ij* = H_ji. [In case you wonder: that’s got to do with the fact that we can write any final 〈χ| or 〈φ| state as the conjugate transpose of the initial |χ〉 or |φ〉 state, so we can write: 〈χ| = |χ〉*, or 〈φ| = |φ〉*.]

As usual, the trick is to find those N C_i(t) functions: we do so by solving that set of N equations, assuming we know those Hamiltonian coefficients. [As you may suspect, the real challenge is to determine the Hamiltonian, which we assume to be given here. But… Well… You first need to learn how to model stuff. Once you get your degree, you’ll be paid to actually solve problems using those models. 🙂 ] We know the complex exponential is a functional form that usually does that trick. Hence, generalizing the results from our analysis of two-state systems once more, the following general solution is suggested:

C_i(t) = a_i·e^{−i·(E/ħ)·t}

Note that we introduce only one E variable here, but N a_i coefficients, which may be real- or complex-valued. Indeed, my examples – see my previous posts – often involved real coefficients, but that’s not necessarily the case. Think of the C₂(t) = i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t] function describing one of the two base state amplitudes for the ammonia molecule—for example. 🙂

Now, that proposed general solution allows us to calculate the derivatives in our Hamiltonian equations (i.e. the d[C_i(t)]/dt functions) as follows:

d[C_i(t)]/dt = −i·(E/ħ)·a_i·e^{−i·(E/ħ)·t}

You can now double-check that the set of equations reduces to the following:

Please do write it out: because we have one E only, the e^{−i·(E/ħ)·t}factor is common to all terms, and so we can cancel it. The other stuff is plain arithmetic: −i·i = −i² = 1, and the ħ constants cancel out too. So there we are: we’ve got a very simple set of N equations here, with N unknowns (i.e. these a₁, a₂,…, a_N coefficients, to be specific). We can re-write this system as:

The δ_ij here is the Kronecker delta, of course (it’s one for i = j and zero for i ≠ j), and we are now looking at a homogeneous system of equations here, i.e. a set of linear equations in which all the constant terms are zero. You should remember it from your high school math course. To be specific, you’d write it as Ax = 0, with A the coefficient matrix. The trivial solution is the zero solution, of course: all a₁, a₂,…, a_N coefficients are zero. But we don’t want the trivial solution. Now, as Feynman points out – tongue-in-cheek, really – we actually have to be lucky to have a non-trivial solution. Indeed, you may or may not remember that the zero solution was actually the only solution if the determinant of the coefficient matrix was not equal to zero. So we only had a non-trivial solution if the determinant of A was equal to zero, i.e. if Det[A] = 0. So A has to be some so-called singular matrix. You’ll also remember that, in that case, we got an infinite number of solutions, to which we could apply the so-called superposition principle: if x and y are two solutions to the homogeneous set of equations Ax = 0, then any linear combination of x and y is also a solution. I wrote an addendum to this post (just scroll down and you’ll find it), which explains what systems of linear equations are all about, so I’ll refer you to that in case you’d need more detail here. I need to continue our story here. The bottom line is: the [H_ij–δ_ijE] matrix needs to be singular for the system to have meaningful solutions, so we will only have a non-trivial solution for those values of E for which

Det[H_ij–δ_ijE] = 0

Let’s spell it out. The condition above is the same as writing:

So far, so good. What’s next? Well… The formula for the determinant is the following:

That looks like a monster, and it is, but, in essence, what we’ve got here is an expression for the determinant in terms of the permutations of the matrix elements. This is not a math course so I’ll just refer you Wikipedia for a detailed explanation of this formula for the determinant. The bottom line is: if we write it all out, then Det[H_ij–δ_ijE] is just an Nth order polynomial in E. In other words: it’s just a sum of products with powers of E up to E^N, and so our Det[H_ij–δ_ijE] = 0 condition amounts to equating it with zero.

In general, we’ll have N roots, but – sorry you need to remember so much from your high school math classes here – some of them may be multiple roots (i.e. two or more roots may be equal). We’ll call those roots—you guessed it:

E_I, E_II,…, E_n,…, E_N

Note I am following Feynman’s exposé, and so he uses n, rather than k, to denote the n^th Roman numeral (as opposed to Latin numerals). Now, I know your brain is near the melting point… But… Well… We’re not done yet. Just hang on. For each of these values E = E_I, E_II,…, E_n,…, E_N, we have an associated set of solutions a_i. As Feynman puts it: you get a set which belongs to E_n. In order to not forget that, for each E_n, we’re talking a set of N coefficients a_i (i = 1, 2,…, N), we denote that set not by a_i(n) but by a_i(n). So that’s why we use boldface for our index n: it’s special—and not only because it denotes a Roman numeral! It’s just one of Feynman’s many meaningful conventions.

Now remember that C_i(t) = a_i·e^{−i·(E/ħ)·t}formula. For each set of a_i(n) coefficients, we’ll have a set of C_i(n) functions which, naturally, we can write as:

C_i(n) = a_i(n)·e^{−i·(E_n/ħ)·t}

So far, so good. We have N a_i(n) coefficients and N C_i(n) functions. That’s easy enough to understand. Now we’ll define also define a set of N new vectors, which we’ll write as |n〉, and which we’ll refer to as the state vectors that describe the configuration of the definite energy states E_n(n = I, II,… N). [Just breathe right now: I’ll (try to) explain this in a moment.] Moreover, we’ll write our set of coefficients a_i(n) as 〈i|n〉. Again, the boldface n reminds us we’re talking a set of N complex numbers here. So we re-write that set of N C_i(n) functions as follows:

C_i(n) = 〈i|n〉·e^{−i·(E_n/ħ)·t}

We can expand this as follows:

C_i(n) = 〈 i | ψ_n(t) 〉 = 〈 i | n 〉·e^{−i·(E_n/ħ)·t}

which, of course, implies that:

| ψ_n(t) 〉 = |n〉·e^{−i·(E_n/ħ)·t}

So now you may understand Feynman’s description of those |n〉 vectors somewhat better. As he puts it:

“The |n〉 vectors – of which there are N – are the state vectors that describe the configuration of the definite energy states E_n(n = I, II,… N), but have the time dependence factored out.”

Hmm… I know. This stuff is hard to swallow, but we’re not done yet: if your brain hasn’t melted yet, it may do so now. You’ll remember we talked about eigenvalues and eigenvectors in our post on the math behind the quantum-mechanical model of our ammonia molecule. Well… We can generalize the results we got there:

The energies E_I, E_II,…, E_n,…, E_Nare the eigenvalues of the Hamiltonian matrix H.
The state vectors |n〉 that are associated with each energy E_n, i.e. the set of vectors |n〉, are the corresponding eigenstates.

So… Well… That’s it! We’re done! This is all there is to it. I know it’s a lot but… Well… We’ve got a general description of N-state systems here, and so that’s great!

Let me make some concluding remarks though.

First, note the following property: if we let the Hamiltonian matrix act on one of those state vectors |n〉, the result is just E_n times the same state. We write:

We’re writing nothing new here really: it’s just a consequence of the definition of eigenstates and eigenvalues. The more interesting thing is the following. When describing our two-state systems, we saw we could use the states that we associated with the E_Iand E_IIas a new base set. The same is true for N-state systems: the state vectors |n〉 can also be used as a base set. Of course, for that to be the case, all of the states must be orthogonal, meaning that for any two of them, say |n〉 and |m〉, the following equation must hold:

〈n|m〉 = 0

Feynman shows this will be true automatically if all the energies are different. If they’re not – i.e. if our polynomial in E would accidentally have two (or more) roots with the same energy – then things are more complicated. However, as Feynman points out, this problem can be solved by ‘cooking up’ two new states that do have the same energy but are also orthogonal. I’ll refer you to him for the detail, as well as for the proof of that 〈n|m〉 = 0 equation.

Finally, you should also note that – because of the homogeneity principle – it’s possible to multiply the N a_i(n) coefficients by a suitable factor so that all the states are normalized, by which we mean:

〈n|n〉 = 1

Well… We’re done! For today, at least! 🙂

Addendum on Systems of Linear Equations

It’s probably good to briefly remind you of your high school math class on systems of linear equations. First note the difference between homogeneous and non-homogeneous equations. Non-homogeneous equations have a non-zero constant term. The following three equations are an example of a non-homogeneous set of equations:

3x + 2y − z = 1
2x − 2y + 4z = −2
−x + y/2 − z = 0

We have a point solution here: (x, y, z) = (1, −2, −2). The geometry of the situation is something like this:

One of the equations may be a linear combination of the two others. In that case, that equation can be removed without affecting the solution set. For the three-dimensional case, we get a line solution, as illustrated below.

Homogeneous and non-homogeneous sets of linear equations are closely related. If we write a homogeneous set as Ax = 0, then a non-homogeneous set of equations can be written as Ax = b. They are related. More in particular, the solution set for Ax = b is going to be a translation of the solution set for Ax = 0. We can write that more formally as follows:

If p is any specific solution to the linear system Ax = b, then the entire solution set can be described as {p + v|v is any solution to Ax = 0}

The solution set for a homogeneous system is a linear subspace. In the example above, which had three variables and, hence, for which the vector space was three-dimensional, there were three possibilities: a point, line or plane solution. All are (linear) subspaces—although you’d want to drop the term ‘linear’ for the point solution, of course. 🙂 Formally, a subspace is defined as follows: if V is a vector space, then W is a subspace if and only if:

The zero vector (i.e. 0) is in W.
If x is an element of W, then any scalar multiple ax will be an element of W too (this is often referred to as the property of homogeneity).
If x and y are elements of W, then the sum of x and y (i.e. x + y) will be an element of W too (this is referred to as the property of additivity).

As you can see, the superposition principle actually combines the properties of homogeneity and additivity: if x and y are solutions, then any linear combination of them will be a solution too.

The solution set for a non-homogeneous system of equations is referred to as a flat. It’s a subset too, so it’s like a subspace, except that it need not pass through the origin. Again, the flats in two-dimensional space are points and lines, while in three-dimensional space we have points, lines and planes. In general, we’ll have flats, and subspaces, of every dimension from 0 to n−1 in n-dimensional space.

OK. That’s clear enough, but what is all that talk about eigenstates and eigenvalues about? Mathematically, we define eigenvectors, aka as characteristic vectors, as follows:

The non-zero vector v is an eigenvector of a square matrix A if Av is a scalar multiple of v, i.e. Av = λv.
The associated scalar λ is known as the eigenvalue (or characteristic value) associated with the eigenvector v.

Now, in physics, we talk states, rather than vectors—although our states are vectors, of course. So we’ll call them eigenstates, rather than eigenvectors. But the principle is the same, really. Now, I won’t copy what you can find elsewhere—especially not in an addendum to a post, like this one. So let me just refer you elswhere. Paul’s Online Math Notes, for example, are quite good on this—especially in the context of solving a set of differential equations, which is what we are doing here. And you can also find a more general treatment in the Wikipedia article on eigenvalues and eigenstates which, while being general, highlights their particular use in quantum math.

The Hamiltonian revisited

I want to come back to something I mentioned in a previous post: when looking at that formula for those U_ij amplitudes—which I’ll jot down once more:

U_ij(t + Δt, t) = δ_ij + ΔU_ij(t + Δt, t) = δ_ij + K_ij(t)·Δt ⇔ U_ij(t + Δt, t) = δ_ij − (i/ħ)·H_ij(t)·Δt

—I noted that it resembles the general y(t + Δt) = y(t) + Δy = y(t) + (dy/dt)·Δt formula. So we can look at our K_ij(t) function as being equal to the time derivative of the U_ij(t + Δt, t) function. I want to re-visit that here, as it triggers a whole range of questions, which may or may not help to understand quantum math somewhat more intuitively. Let’s quickly sum up what we’ve learned so far: it’s basically all about quantum-mechanical stuff that does not move in space. Hence, the x in our wavefunction ψ(x, t) is some fixed point in space and, therefore, our elementary wavefunction—which we wrote as:

ψ(x, t) = a·e^−i·θ= a·e^{−i·(ω·t − k∙x)} = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]}

—reduces to ψ(t) = a·e^−i·ω·t = a·e^{−i·[(E/ħ)·t}.

Unlike what you might think, we’re not equating x with zero here. No. It’s the p = m·v factor that becomes zero, because our reference frame is that of the system that we’re looking at, so its velocity is zero: it doesn’t move in our reference frame. That immediately answers an obvious question: does our wavefunction look any different when choosing another reference frame? The answer is obviously: yes! It surely matters if the system moves or not, and it also matters how fast it moves, because it changes the energy and momentum values from E and p to some E’ and p’. However, we’ll not consider such complications here: that’s the realm of relativistic quantum mechanics. Let’s start with the simplest of situations.

A simple two-state system

One of the simplest examples of a quantum-mechanical system that does not move in space, is the textbook example of the ammonia molecule. The picture was as simple as the one below: an ammonia molecule consists of one nitrogen atom and three hydrogen atoms, and the nitrogen atom could be ‘up’ or ‘down’ with regard to the motion of the NH₃molecule around its axis of symmetry, as shown below.

It’s important to note that this ‘up’ or ‘down’ direction is, once again, defined with respect to the reference frame of the system itself. The motion of the molecule around its axis of symmetry is referred to as its spin—a term that’s used in a variety of contexts and, therefore, is annoyingly ambiguous. When we use the term ‘spin’ (up or down) to describe an electron state, for example, we’d associate it with the direction of its magnetic moment. Such magnetic moment arises from the fact that, for all practical purposes, we can think of an electron as a spinning electric charge. Now, while our ammonia molecule is electrically neutral, as a whole, the two states are actually associated with opposite electric dipole moments, as illustrated below. Hence, when we’d apply an electric field (denoted as ε) below, the two states are effectively associated with different energy levels, which we wrote as E₀ ± εμ.

But we’re getting ahead of ourselves here. Let’s revert to the system in free space, i.e. without an electromagnetic force field—or, what amounts to saying the same, without potential. Now, the ammonia molecule is a quantum-mechanical system, and so there is some amplitude for the nitrogen atom to tunnel through the plane of hydrogens. I told you before that this is the key to understanding quantum mechanics really: there is an energy barrier there and, classically, the nitrogen atom should not sneak across. But it does. It’s like it can borrow some energy – which we denote by A – to penetrate the energy barrier.

In quantum mechanics, the dynamics of this system are modeled using a set of two differential equations. These differential equations are really the equivalent of Newton’s classical Law of Motion (I am referring to the F = m·(dv/dt) = m·a equation here) in quantum mechanics, so I’ll have to explain them—which is not so easy as explaining Newton’s Law, because we’re talking complex-valued functions, but… Well… Let me first insert the solution of that set of differential equations:

This graph shows how the probability of the nitrogen atom (or the ammonia molecule itself) being in state 1 (i.e. ‘up’) or, else, in state 2 (i.e. ‘down’), varies sinusoidally in time. Let me also give you the equations for the amplitudes to be in state 1 or 2 respectively:

C₁(t) = 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}+ (1/2)·e^{−(i/ħ)·(E₀+ A)·t}= e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t]
C₂(t) = 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}– (1/2)·e^{−(i/ħ)·(E₀+ A)·t}= i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t]

So the P₁(t) and P₂(t) probabilities above are just the absolute square of these C₁(t) and C₂(t) functions. So as to help you understand what’s going on here, let me quickly insert the following technical remarks:

In case you wonder how we go from those exponentials to a simple sine and cosine factor, remember that the sum of complex conjugates, i.e e^iθ+ e^−iθreduces to 2·cosθ, while e^iθ− e^−iθreduces to 2·i·sinθ.
As for how to take the absolute square… Well… I shouldn’t be explaining that here, but you should be able to work that out remembering that (i) |a·b·c|² = |a|²·|b|²·|c|²; (ii) |e^iθ|² = |e^−iθ|²= 1² = 1 (for any value of θ); and (iii) |i|² = 1.
As for the periodicity of both probability functions, note that the period of the squared sine and cosine functions is equal to π. Hence, the argument of our sine and cosine function will be equal to 0, π, 2π, 3π etcetera if (A/ħ)·t = 0, π, 2π, 3π etcetera, i.e. if t = 0·ħ/A, π·ħ/A, 2π·ħ/A, 3π·ħ/A etcetera. So that’s why we measure time in units of ħ/A above.

The graph above is actually tricky to interpret, as it assumes that we know in what state the molecule starts out with at t = 0. This assumption is tricky because we usually do not know that: we have to make some observation which, curiously enough, will always yield one of the two states—nothing in-between. Or, else, we can use a state selector—an inhomogeneous electric field which will separate the ammonia molecules according to their state. It’s a weird thing really, and it summarizes all of the ‘craziness’ of quantum-mechanics: as long as we don’t measure anything – by applying that force field – our molecule is in some kind of abstract state, which mixes the two base states. But when we do make the measurement, always along some specific direction (which we usually take to be the z-direction in our reference frame), we’ll always find the molecule is either ‘up’ or, else, ‘down’. We never measure it as something in-between. Personally, I like to think the measurement apparatus – I am talking the electric field here – causes the nitrogen atom to sort of ‘snap into place’. However, physicists use more precise language here: they would say that the electric field does result in the two positions having very different energy levels (E₀ + εμ and E₀ – εμ, to be precise) and that, as a result, the amplitude for the nitrogen atom to flip back and forth has little effect. Now how do we model that?

The Hamiltonian equations

I shouldn’t be using the term above, as it usually refers to a set of differential equations describing classical systems. However, I’ll also use it for the quantum-mechanical analog, which amounts to the following for our simple two-state example above:

Don’t panic. We’ll explain. The equations above are all the same but use different formats: the first block writes them as a set of equations, while the second uses the matrix notation, which involves the use of that rather infamous Hamiltonian matrix, which we denote by H = [H_ij]. Now, we’ve postponed a lot of technical stuff, so… Well… We can’t avoid it any longer. Let’s look at those Hamiltonian coefficients H_ij first. Where do they come from?

You’ll remember we thought of time as some kind of apparatus, with particles entering in some initial state φ and coming out in some final state χ. Both are to be described in terms of our base states. To be precise, we associated the (complex) coefficients C₁ and C₂ with |φ〉 and D₁ and D₂ with |χ〉. However, the χ state is a final state, so we have to write it as 〈χ| = |χ〉† (read: chi dagger). The dagger symbol tells us we need to take the conjugate transpose of |χ〉, so the column vector becomes a row vector, and its coefficients are the complex conjugate of D₁ and D₂, which we denote as D₁* and D₂*. We combined this with Dirac’s bra-ket notation for the amplitude to go from one base state to another, as a function in time (or a function of time, I should say):

U_ij(t + Δt, t) = 〈i|U(t + Δt, t)|j〉

This allowed us to write the following matrix equation:

To see what it means, you should write it all out:

〈χ|U(t + Δt, t)|φ〉 = D₁*·(U₁₁(t + Δt, t)·C₁ + U₁₂(t + Δt, t)·C₂) + D₂*·(U₂₁(t + Δt, t)·C₁ + U₂₂(t + Δt, t)·C₂)

= D₁*·U₁₁(t + Δt, t)·C₁+ D₁*·U₁₂(t + Δt, t)·C₂+ D₂*·U₂₁(t + Δt, t)·C₁+ D₂*·U₂₂(t + Δt, t)·C₂

It’s a horrendous expression, but it’s a complex-valued amplitude or, quite simply, a complex number. So this is not nonsensical. We can now take the next step, and that’s to go from those U_ij amplitudes to the H_ij amplitudes of the Hamiltonian matrix. The key is to consider the following: if Δt goes to zero, nothing happens, so we write: U_ij = 〈i|U|j〉 → 〈i|j〉 = δ_ij for Δt → 0, with δ_ij = 1 if i = j, and δ_ij = 0 if i ≠ j. We then assume that, for small t, those U_ij amplitudes should differ from δ_ij (i.e. from 1 or 0) by amounts that are proportional to Δt. So we write:

U_ij(t + Δt, t) = δ_ij + ΔU_ij(t + Δt, t) = δ_ij + K_ij(t)·Δt

We then equated those K_ij(t) factors with − (i/ħ)·H_ij(t), and we were done: U_ij(t + Δt, t) = δ_ij − (i/ħ)·H_ij(t)·Δt. […] Well… I show you how we get those differential equations in a moment. Let’s pause here for a while to see what’s going on really. You’ll probably remember how one can mathematically ‘construct’ the complex exponential e^iθby using the linear approximation eⁱ^ε = 1 + iε near θ = 0 and for infinitesimally small values of ε. In case you forgot, we basically used the definition of the derivative of the real exponential e^εfor ε going to zero:

So we’ve got something similar here for U₁₁(t + Δt, t) = 1 − i·[H₁₁(t)/ħ]·Δt and U₂₂(t + Δt, t) = 1 − i·[H₂₂(t)/ħ]·Δt. Just replace the ε in eⁱ^ε = 1 + iε by ε = − (E₀/ħ)·Δt. Indeed, we know that H₁₁ = H₂₂ = E₀, and E₀/ħ is, of course, just the energy measured in (reduced) Planck units, i.e. in its natural unit. Hence, if our ammonia molecule is in one of the two base states, we start at θ = 0 and then we just start moving on the unit circle, clockwise, because of the minus sign in e^−iθ. Let’s write it out:

U₁₁(t + Δt, t) = 1 − i·[H₁₁(t)/ħ]·Δt = 1 − i·[E₀/ħ]·Δt and

U₂₂(t + Δt, t) = 1 − i·[H₂₂(t)/ħ]·Δt = 1 − i·[E₀/ħ]·Δt

But what about U₁₂and U₂₁? Is there a similar interpretation? Let’s write those equations down and think about them:

U₁₂(t + Δt, t) = 0 − i·[H₁₂(t)/ħ]·Δt = 0 + i·[A/ħ]·Δt and

U₂₁(t + Δt, t) = 0 − i·[H₂₁(t)/ħ]·Δt = 0 + i·[A/ħ]·Δt

We can visualize this as follows:

Let’s remind ourselves of the definition of the derivative of a function by looking at the illustration below:The f(x₀) value in this illustration corresponds to the U_ij(t, t), obviously. So now things make somewhat more sense: U₁₁(t, t) = U₁₁(t, t) = 1, obviously, and U₁₂(t, t) = U₂₁(t, t) = 0. We then add the ΔU_ij(t + Δt, t) to U_ij(t, t). Hence, we can, and probably should, think of those K_ij(t) coefficients as the derivative of the U_ij(t, t) functions with respect to time. So we can write something like this:

These derivatives are pure imaginary numbers. That does not mean that the U_ij(t + Δt, t) functions are purely imaginary: U₁₁(t + Δt, t) and U₂₂(t + Δt, t) can be approximated by 1 − i·[E₀/ħ]·Δt for small Δt, so they do have a real part. In contrast, U₁₂(t + Δt, t) and U₂₁(t + Δt, t) are, effectively, purely imaginary (for small Δt, that is).

I can’t help thinking these formulas reflect a deep and beautiful geometry, but its meaning escapes me so far. 😦 When everything is said and done, none of the reflections above makes things somewhat more intuitive: these wavefunctions remain as mysterious as ever.

I keep staring at those P₁(t) and P₂(t) functions, and the C₁(t) and C₂(t) functions that ‘generate’ them, so to speak. They’re not independent, obviously. In fact, they’re exactly the same, except for a phase difference, which corresponds to the phase difference between the sine and cosine. So it’s all one reality, really: all can be described in one single functional form, so to speak. I hope things become more obvious as I move forward.

Post scriptum: I promised I’d show you how to get those differential equations but… Well… I’ve done that in other posts, so I’ll refer you to one of those. Sorry for not repeating myself. 🙂

The de Broglie relations, the wave equation, and relativistic length contraction

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. So no use to read this. Read my recent papers instead. 🙂

Original post:

You know the two de Broglie relations, also known as matter-wave equations:

f = E/h and λ = h/p

You’ll find them in almost any popular account of quantum mechanics, and the writers of those popular books will tell you that f is the frequency of the ‘matter-wave’, and λ is its wavelength. In fact, to add some more weight to their narrative, they’ll usually write them in a somewhat more sophisticated form: they’ll write them using ω and k. The omega symbol (using a Greek letter always makes a big impression, doesn’t it?) denotes the angular frequency, while k is the so-called wavenumber. Now, k = 2π/λ and ω = 2π·f and, therefore, using the definition of the reduced Planck constant, i.e. ħ = h/2π, they’ll write the same relations as:

λ = h/p = 2π/k ⇔ k = 2π·p/h
f = E/h = (ω/2π)

⇒ k = p/ħ and ω = E/ħ

They’re the same thing: it’s just that working with angular frequencies and wavenumbers is more convenient, from a mathematical point of view that is: it’s why we prefer expressing angles in radians rather than in degrees (k is expressed in radians per meter, while ω is expressed in radians per second). In any case, the ‘matter wave’ – even Wikipedia uses that term now – is, of course, the amplitude, i.e. the wave-function ψ(x, t), which has a frequency and a wavelength, indeed. In fact, as I’ll show in a moment, it’s got two frequencies: one temporal, and one spatial. I am modest and, hence, I’ll admit it took me quite a while to fully distinguish the two frequencies, and so that’s why I always had trouble connecting these two ‘matter wave’ equations.

Indeed, if they represent the same thing, they must be related, right? But how exactly? It should be easy enough. The wavelength and the frequency must be related through the wave velocity, so we can write: f·λ = v, with v the velocity of the wave, which must be equal to the classical particle velocity, right? And then momentum and energy are also related. To be precise, we have the relativistic energy-momentum relationship: p·c = m_v·v·c = m_v·c²·v/c = E·v/c. So it’s just a matter of substitution. We should be able to go from one equation to the other, and vice versa. Right?

Well… No. It’s not that simple. We can start with either of the two equations but it doesn’t work. Try it. Whatever substitution you try, there’s no way you can derive one of the two equations above from the other. The fact that it’s impossible is evidenced by what we get when we’d multiply both equations. We get:

f·λ = (E/h)·(h/p) = E/p
v = f·λ ⇒ f·λ = v = E/p ⇔ E = v·p = v·(m·v)

⇒ E = m·v²

Huh? What kind of formula is that? E = m·v²? That’s a formula you’ve never ever seen, have you? It reminds you of the kinetic energy formula of course—K.E. = m·v²/2—but… That factor 1/2 should not be there. Let’s think about it for a while. First note that this E = m·v²relation makes perfectly sense if v = c. In that case, we get Einstein’s mass-energy equivalence (E = m·c²), but that’s besides the point here. The point is: if v = c, then our ‘particle’ is a photon, really, and then the E = h·f is referred to as the Planck-Einstein relation. The wave velocity is then equal to c and, therefore, f·λ = c, and so we can effectively substitute to find what we’re looking for:

E/p = (h·f)/(h/λ) = f·λ = c ⇒ E = p·c

So that’s fine: we just showed that the de Broglie relations are correct for photons. [You remember that E = p·c relation, no? If not, check out my post on it.] However, while that’s all nice, it is not what the de Broglie equations are about: we’re talking the matter-wave here, and so we want to do something more than just re-confirm that Planck-Einstein relation, which you can interpret as the limit of the de Broglie relations for v = c. In short, we’re doing something wrong here! Of course, we are. I’ll tell you what exactly in a moment: it’s got to do with the fact we’ve got two frequencies really.

Let’s first try something else. We’ve been using the relativistic E = m_v·c² equation above. Let’s try some other energy concept: let’s substitute the E in the f = E/h by the kinetic energy and then see where we get—if anywhere at all. So we’ll use the E_kinetic = m∙v²/2 equation. We can then use the definition of momentum (p = m∙v) to write E = p²/(2m), and then we can relate the frequency f to the wavelength λ using the v = λ∙f formula once again. That should work, no? Let’s do it. We write:

E = p²/(2m)
E = h∙f = h·v/λ

⇒ λ = h·v/E = h·v/(p²/(2m)) = h·v/[m²·v²/(2m)] = h/[m·v/2] = 2∙h/p

So we find λ = 2∙h/p. That is almost right, but not quite: that factor 2 should not be there. Well… Of course you’re smart enough to see it’s just that factor 1/2 popping up once more—but as a reciprocal, this time around. 🙂 So what’s going on? The honest answer is: you can try anything but it will never work, because the f = E/h and λ = h/p equations cannot be related—or at least not so easily. The substitutions above only work if we use that E = m·v² energy concept which, you’ll agree, doesn’t make much sense—at first, at least. Again: what’s going on? Well… Same honest answer: the f = E/h and λ = h/p equations cannot be related—or at least not so easily—because the wave equation itself is not so easy.

Let’s review the basics once again.

The wavefunction

The amplitude of a particle is represented by a wavefunction. If we have no information whatsoever on its position, then we usually write that wavefunction as the following complex-valued exponential:

ψ(x, t) = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} = a·e^{−i·(ω·t − k∙x)}= a·e^{i(k∙x−ω·t)}= a·e^iθ= a·(cosθ + i·sinθ)

θ is the so-called phase of our wavefunction and, as you can see, it’s the argument of a wavefunction indeed, with temporal frequency ω and spatial frequency k (if we choose our x-axis so its direction is the same as the direction of k, then we can substitute the k and x vectors for the k and x scalars, so that’s what we’re doing here). Now, we know we shouldn’t worry too much about a, because that’s just some normalization constant (remember: all probabilities have to add up to one). However, let’s quickly develop some logic here. Taking the absolute square of this wavefunction gives us the probability of our particle being somewhere in space at some point in time. So we get the probability as a function of x and t. We write:

P(x ,t) = |a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]}|²= a²

As all probabilities have to add up to one, we must assume we’re looking at some box in spacetime here. So, if the length of our box is Δx = x₂ − x₁, then (Δx)·a² = (x₂−x₁)·a²= 1 ⇔ Δx = 1/a². [We obviously simplify the analysis by assuming a one-dimensional space only here, but the gist of the argument is essentially correct.] So, freezing time (i.e. equating t to some point t = t₀), we get the following probability density function:

That’s simple enough. The point is: the two de Broglie equations f = E/h and λ = h/p give us the temporal and spatial frequencies in that ψ(x, t) = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} relation. As you can see, that’s an equation that implies a much more complicated relationship between E/ħ = ω and p/ħ = k. Or… Well… Much more complicated than what one would think of at first.

To appreciate what’s being represented here, it’s good to play a bit. We’ll continue with our simple exponential above, which also illustrates how we usually analyze those wavefunctions: we either assume we’re looking at the wavefunction in space at some fixed point in time (t = t₀) or, else, at how the wavefunction changes in time at some fixed point in space (x = x₀). Of course, we know that Einstein told us we shouldn’t do that: space and time are related and, hence, we should try to think of spacetime, i.e. some ‘kind of union’ of space and time—as Minkowski famously put it. However, when everything is said and done, mere mortals like us are not so good at that, and so we’re sort of condemned to try to imagine things using the classical cut-up of things. 🙂 So we’ll just an online graphing tool to play with that a·e^{i(k∙x−ω·t)}= a·e^iθ= a·(cosθ + i·sinθ) formula.

Compare the following two graps, for example. Just imagine we either look at how the wavefunction behaves at some point in space, with the time fixed at some point t = t₀, or, alternatively, that we look at how the wavefunction behaves in time at some point in space x = x₀. As you can see, increasing k = p/ħ or increasing ω = E/ħ gives the wavefunction a higher ‘density’ in space or, alternatively, in time.

That makes sense, intuitively. In fact, when thinking about how the energy, or the momentum, affects the shape of the wavefunction, I am reminded of an airplane propeller: as it spins, faster and faster, it gives the propeller some ‘density’, in space as well as in time, as its blades cover more space in less time. It’s an interesting analogy: it helps—me, at least—to think through what that wavefunction might actually represent.

So as to stimulate your imagination even more, you should also think of representing the real and complex part of that ψ = a·e^{i(k∙x−ω·t)}= a·e^iθ= a·(cosθ + i·sinθ) formula in a different way. In the graphs above, we just showed the sine and cosine in the same plane but, as you know, the real and the imaginary axis are orthogonal, so Euler’s formula a·e^iθ= a·(cosθ + i·sinθ) = a·cosθ + i·a·sinθ = Re(ψ) + i·Im(ψ) may also be graphed as follows:

The illustration above should make you think of yet another illustration you’ve probably seen like a hundred times before: the electromagnetic wave, propagating through space as the magnetic and electric field induce each other, as illustrated below. However, there’s a big difference: Euler’s formula incorporates a phase shift—remember: sinθ = cos(θ − π/2)—and you don’t have that in the graph below. The difference is much more fundamental, however: it’s really hard to see how one could possibly relate the magnetic and electric field to the real and imaginary part of the wavefunction respectively. Having said that, the mathematical similarity makes one think!

Of course, you should remind yourself of what E and B stand for: they represent the strength of the electric (E) and magnetic (B) field at some point x at some time t. So you shouldn’t think of those wavefunctions above as occupying some three-dimensional space. They don’t. Likewise, our wavefunction ψ(x, t) does not occupy some physical space: it’s some complex number—an amplitude that’s associated with each and every point in spacetime. Nevertheless, as mentioned above, the visuals make one think and, as such, do help us as we try to understand all of this in a more intuitive way.

Let’s now look at that energy-momentum relationship once again, but using the wavefunction, rather than those two de Broglie relations.

Energy and momentum in the wavefunction

I am not going to talk about uncertainty here. You know that Spiel. If there’s uncertainty, it’s in the energy or the momentum, or in both. The uncertainty determines the size of that ‘box’ (in spacetime) in which we hope to find our particle, and it’s modeled by a splitting of the energy levels. We’ll say the energy of the particle may be E₀, but it might also be some other value, which we’ll write as E_n = E₀ ± n·ħ. The thing to note is that energy levels will always be separated by some integer multiple of ħ, so ħ is, effectively , the quantum of energy for all practical—and theoretical—purposes. We then super-impose the various wave equations to get a wave function that might—or might not—resemble something like this:

Who knows? 🙂 In any case, that’s not what I want to talk about here. Let’s repeat the basics once more: if we write our wavefunction a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} as a·e^{−i·[ω·t − k∙x]}, we refer to ω = E/ħ as the temporal frequency, i.e. the frequency of our wavefunction in time (i.e. the frequency it has if we keep the position fixed), and to k = p/ħ as the spatial frequency (i.e. the frequency of our wavefunction in space (so now we stop the clock and just look at the wave in space). Now, let’s think about the energy concept first. The energy of a particle is generally thought of to consist of three parts:

The particle’s rest energy m₀c², which de Broglie referred to as internal energy (E_int): it includes the rest mass of the ‘internal pieces’, as Feynman puts it (now we call those ‘internal pieces’ quarks), as well as their binding energy (i.e. the quarks’ interaction energy);
Any potential energy it may have because of some field (so de Broglie was not assuming the particle was traveling in free space), which we’ll denote by U, and note that the field can be anything—gravitational, electromagnetic: it’s whatever changes the energy because of the position of the particle;
The particle’s kinetic energy, which we write in terms of its momentum p: m·v²/2 = m²·v²/(2m) = (m·v)²/(2m) = p²/(2m).

So we have one energy concept here (the rest energy) that does not depend on the particle’s position in spacetime, and two energy concepts that do depend on position (potential energy) and/or how that position changes because of its velocity and/or momentum (kinetic energy). The two last bits are related through the energy conservation principle. The total energy is E = m_vc², of course—with the little subscript (v) ensuring the mass incorporates the equivalent mass of the particle’s kinetic energy.

So what? Well… In my post on quantum tunneling, I drew attention to the fact that different potentials , so different potential energies (indeed, as our particle travels one region to another, the field is likely to vary) have no impact on the temporal frequency. Let me re-visit the argument, because it’s an important one. Imagine two different regions in space that differ in potential—because the field has a larger or smaller magnitude there, or points in a different direction, or whatever: just different fields, which corresponds to different values for U₁ and U₂, i.e. the potential in region 1 versus region 2. Now, the different potential will change the momentum: the particle will accelerate or decelerate as it moves from one region to the other, so we also have a different p₁ and p₂. Having said that, the internal energy doesn’t change, so we can write the corresponding amplitudes, or wavefunctions, as:

ψ₁(θ₁) = Ψ₁(x, t) = a·e⁻ⁱ^θ^₁ = a·e^{−i[(E_int+ p₁²/(2m) + U₁)·t − p₁∙x]/ħ}
ψ₂(θ₂) = Ψ₂(x, t) = a·e⁻ⁱ^θ^₂ = a·e^{−i[(E_int+ p₂²/(2m) + U₂)·t − p₂∙x]/ħ}

Now how should we think about these two equations? We are definitely talking different wavefunctions. However, their temporal frequencies ω₁= E_int+ p₁²/(2m) + U₁ and ω₁= E_int+ p₂²/(2m) + U₂must be the same. Why? Because of the energy conservation principle—or its equivalent in quantum mechanics, I should say: the temporal frequency f or ω, i.e. the time-rate of change of the phase of the wavefunction, does not change: all of the change in potential, and the corresponding change in kinetic energy, goes into changing the spatial frequency, i.e. the wave number k or the wavelength λ, as potential energy becomes kinetic or vice versa. The sum of the potential and kinetic energy doesn’t change, indeed. So the energy remains the same and, therefore, the temporal frequency does not change. In fact, we need this quantum-mechanical equivalent of the energy conservation principle to calculate how the momentum and, hence, the spatial frequency of our wavefunction, changes. We do so by boldly equating ω₁= E_int+ p₁²/(2m) + U₁and ω₂ = E_int+ p₂²/(2m) + U₂, and so we write:

ω₁= ω₂ ⇔ E_int+ p₁²/(2m) + U₁= E_int+ p₂²/(2m) + U₂

⇔ p₁²/(2m) − p₂²/(2m) = U₂– U₁⇔ p₂²= (2m)·[p₁²/(2m) – (U₂– U₁)]

⇔ p₂= (p₁² – 2m·ΔU)^1/2

We played with this in a previous post, assuming that p₁² is larger than 2m·ΔU, so as to get a positive number on the right-hand side of the equation for p₂², so then we can confidently take the positive square root of that (p₁² – 2m·ΔU) expression to calculate p₂. For example, when the potential difference ΔU = U₂– U₁ was negative, so ΔU < 0, then we’re safe and sure to get some real positive value for p₂.

Having said that, we also contemplated the possibility that p₂²= p₁² – 2m·ΔU was negative, in which case p₂ has to be some pure imaginary number, which we wrote as p₂= i·p’ (so p’ (read: p prime) is a real positive number here). We could work with that: it resulted in an exponentially decreasing factor e^{−p’·x/ħ} that ended up ‘killing’ the wavefunction in space. However, its limited existence still allowed particles to ‘tunnel’ through potential energy barriers, thereby explaining the quantum-mechanical tunneling phenomenon.

This is rather weird—at first, at least. Indeed, one would think that, because of the E/ħ = ω equation, any change in energy would lead to some change in ω. But no! The total energy doesn’t change, and the potential and kinetic energy are like communicating vessels: any change in potential energy is associated with a change in p, and vice versa. It’s a really funny thing. It helps to think it’s because the potential depends on position only, and so it should not have an impact on the temporal frequency of our wavefunction. Of course, it’s equally obvious that the story would change drastically if the potential would change with time, but… Well… We’re not looking at that right now. In short, we’re assuming energy is being conserved in our quantum-mechanical system too, and so that implies what’s described above: no change in ω, but we obviously do have changes in p whenever our particle goes from one region in space to another, and the potentials differ. So… Well… Just remember: the energy conservation principle implies that the temporal frequency of our wave function doesn’t change. Any change in potential, as our particle travels from one place to another, plays out through the momentum.

Now that we know that, let’s look at those de Broglie relations once again.

**Re-visiting the de Broglie relations**

As mentioned above, we usually think in one dimension only: we either freeze time or, else, we freeze space. If we do that, we can derive some funny new relationships. Let’s first simplify the analysis by re-writing the argument of the wavefunction as:

θ = E·t − p·x

Of course, you’ll say: the argument of the wavefunction is not equal to E·t − p·x: it’s (E/ħ)·t − (p/ħ)∙x. Moreover, θ should have a minus sign in front. Well… Yes, you’re right. We should put that 1/ħ factor in front, but we can change units, and so let’s just measure both E as well as p in units of ħ here. We can do that. No worries. And, yes, the minus sign should be there—Nature choose a clockwise direction for θ—but that doesn’t matter for the analysis hereunder.

The E·t − p·x expression reminds one of those invariant quantities in relativity theory. But let’s be precise here. We’re thinking about those so-called four-vectors here, which we wrote as p_μ = (E, p_x, p_y, p_z) = (E, p) and x_μ = (t, x, y, z) = (t, x) respectively. [Well… OK… You’re right. We wrote those four-vectors as p_μ = (E, p_x·c , p_y·c, p_z·c) = (E, p·c) and x_μ = (c·t, x, y, z) = (t, x). So what we write is true only if we measure time and distance in equivalent units so we have c = 1. So… Well… Let’s do that and move on.] In any case, what was invariant was not E·t − p·x·c or c·t − x (that’s a nonsensical expression anyway: you cannot subtract a vector from a scalar), but p_μ² = p_μp_μ= E² − (p·c)² = E² − p²·c²= E² − (p_x² + p_y² + p_z²)·c² and x_μ² = x_μx_μ= (c·t)² − x² = c²·t² − (x² + y² + z²) respectively. [Remember p_μp_μand x_μx_μare four-vector dot products, so they have that +— signature, unlike the p² and x² or a·b dot products, which are just a simple sum of the squared components.] So… Well… E·t − p·x is not an invariant quantity. Let’s try something else.

Let’s re-simplify by equating ħ as well as c to one again, so we write: ħ = c = 1. [You may wonder if it is possible to ‘normalize’ both physical constants simultaneously, but the answer is yes. The Planck unit system is an example.] then our relativistic energy-momentum relationship can be re-written as E/p = 1/v. [If c would not be one, we’d write: E·β = p·c, with β = v/c. So we got E/p = c/β. We referred to β as the relative velocity of our particle: it was the velocity, but measured as a ratio of the speed of light. So here it’s the same, except that we use the velocity symbol v now for that ratio.]

Now think of a particle moving in free space, i.e. without any fields acting on it, so we don’t have any potential changing the spatial frequency of the wavefunction of our particle, and let’s also assume we choose our x-axis such that it’s the direction of travel, so the position vector (x) can be replaced by a simple scalar (x). Finally, we will also choose the origin of our x-axis such that x = 0 zero when t = 0, so we write: x(t = 0) = 0. It’s obvious then that, if our particle is traveling in spacetime with some velocity v, then the ratio of its position x and the time t that it’s been traveling will always be equal to v = x/t. Hence, for that very special position in spacetime (t, x = v·t) – so we’re talking the actual position of the particle in spacetime here – we get: θ = E·t − p·x = E·t − p·v·t = E·t − m·v·v·t= (E − m∙v²)·t. So… Well… There we have the m∙v² factor.

The question is: what does it mean? How do we interpret this? I am not sure. When I first jotted this thing down, I thought of choosing a different reference potential: some negative value such that it ensures that the sum of kinetic, rest and potential energy is zero, so I could write E = 0 and then the wavefunction would reduce to ψ(t) = e^{−i·m∙v²·t}. Feynman refers to that as ‘choosing the zero of our energy scale such that E = 0’, and you’ll find this in many other works too. However, it’s not that simple. Free space is free space: if there’s no change in potential from one region to another, then the concept of some reference point for the potential becomes meaningless. There is only rest energy and kinetic energy, then. The total energy reduces to E = m (because we chose our units such that c = 1 and, therefore, E = mc² = m·1² = m) and so our wavefunction reduces to:

ψ(t) = a·e^{−i·m·(1 − v²)·t}

We can’t reduce this any further. The mass is the mass: it’s a measure for inertia, as measured in our inertial frame of reference. And the velocity is the velocity, of course—also as measured in our frame of reference. We can re-write it, of course, by substituting t for t = x/v, so we get:

ψ(x) = a·e^{−i·m·(1/v − v)·x}

For both functions, we get constant probabilities, but a wavefunction that’s ‘denser’ for higher values of m. The (1 − v²) and (1/v − v) factors are different, however: these factors becomes smaller for higher v, so our wavefunction becomes less dense for higher v. In fact, for v = 1 (so for travel at the speed of light, i.e. for photons), we get that ψ(t) = ψ(x) = e⁰ = 1. [You should use the graphing tool once more, and you’ll see the imaginary part, i.e. the sine of the a·(cosθ + i·sinθ) expression, just vanishes, as sinθ = 0 for θ = 0.]

The wavefunction and relativistic length contraction

Are exercises like this useful? As mentioned above, these constant probability wavefunctions are a bit nonsensical, so you may wonder why I wrote what I wrote. There may be no real conclusion, indeed: I was just fiddling around a bit, and playing with equations and functions. I feel stuff like this helps me to understand what that wavefunction actually is somewhat better. If anything, it does illustrate that idea of the ‘density’ of a wavefunction, in space or in time. What we’ve been doing by substituting x for x = v·t or t for t = x/v is showing how, when everything is said and done, the mass and the velocity of a particle are the actual variables determining that ‘density’ and, frankly, I really like that ‘airplane propeller’ idea as a pedagogic device. In fact, I feel it may be more than just a pedagogic device, and so I’ll surely re-visit it—once I’ve gone through the rest of Feynman’s Lectures, that is. 🙂

That brings me to what I added in the title of this post: relativistic length contraction. You’ll wonder why I am bringing that into a discussion like this. Well… Just play a bit with those (1 − v²) and (1/v − v) factors. As mentioned above, they decrease the density of the wavefunction. In other words, it’s like space is being ‘stretched out’. Also, it can’t be a coincidence we find the same (1 − v²) factor in the relativistic length contraction formula: L = L₀·√(1 − v²), in which L₀ is the so-called proper length (i.e. the length in the stationary frame of reference) and v is the (relative) velocity of the moving frame of reference. Of course, we also find it in the relativistic mass formula: m = m_v = m₀/√(1−v²). In fact, things become much more obvious when substituting m for m₀/√(1−v²) in that ψ(t) = e^{−i·m·(1 − v²)·t}function. We get:

ψ(t) = a·e^{−i·m·(1 − v²)·t}= a·e^{−i·m₀·√(1−v²)·t}

Well… We’re surely getting somewhere here. What if we go back to our original ψ(x, t) = a·e^{−i·[(E/ħ)·t − (p/ħ)∙x]} function? Using natural units once again, that’s equivalent to:

ψ(x, t) = a·e^{−i·(m·t − p∙x)}= a·e^{−i·[(m₀/√(1−v²))·t − (m₀·v/√(1−v²)∙x)}

= a·e^{−i·[m₀/√(1−v²)]·(t − v∙x)}

Interesting! We’ve got a wavefunction that’s a function of x and t, but with the rest mass (or rest energy) and velocity as parameters! Now that really starts to make sense. Look at the (blue) graph for that 1/√(1−v²) factor: it goes from one (1) to infinity (∞) as v goes from 0 to 1 (remember we ‘normalized’ v: it’s a ratio between 0 and 1 now). So that’s the factor that comes into play for t. For x, it’s the red graph, which has the same shape but goes from zero (0) to infinity (∞) as v goes from 0 to 1.

Now that makes sense: the ‘density’ of the wavefunction, in time and in space, increases as the velocity v increases. In space, that should correspond to the relativistic length contraction effect: it’s like space is contracting, as the velocity increases and, therefore, the length of the object we’re watching contracts too. For time, the reasoning is a bit more complicated: it’s our time that becomes more dense and, therefore, our clock that seems to tick faster.

[…]

I know I need to explore this further—if only so as to assure you I have not gone crazy. Unfortunately, I have no time to do that right now. Indeed, from time to time, I need to work on other stuff besides this physics ‘hobby’ of mine.

Post scriptum 1: As for the E = m·v²formula, I also have a funny feeling that it might be related to the fact that, in quantum mechanics, both the real and imaginary part of the oscillation actually matter. You’ll remember that we’d represent any oscillator in physics by a complex exponential, because it eased our calculations. So instead of writing A = A₀·cos(ωt + Δ), we’d write: A = A₀·e^{i(ωt + Δ)}= A₀·cos(ωt + Δ) + i·A₀·sin(ωt + Δ). When calculating the energy or intensity of a wave, however, we couldn’t just take the square of the complex amplitude of the wave – remembering that E ∼ A². No! We had to get back to the real part only, i.e. the cosine or the sine only. Now the mean (or average) value of the squared cosine function (or a squared sine function), over one or more cycles, is 1/2, so the mean of A² is equal to 1/2 = A₀². cos(ωt + Δ). I am not sure, and it’s probably a long shot, but one must be able to show that, if the imaginary part of the oscillation would actually matter – which is obviously the case for our matter-wave – then 1/2 + 1/2 is obviously equal to 1. I mean: try to think of an image with a mass attached to two springs, rather than one only. Does that make sense? 🙂 […] I know: I am just freewheeling here. 🙂

Post scriptum 2: The other thing that this E = m·v²equation makes me think of is – curiously enough – an eternally expanding spring. Indeed, the kinetic energy of a mass on a spring and the potential energy that’s stored in the spring always add up to some constant, and the average potential and kinetic energy are equal to each other. To be precise: 〈K.E.〉 + 〈P.E.〉 = (1/4)·k·A² + (1/4)·k·A²= k·A²/2. It means that, on average, the total energy of the system is twice the average kinetic energy (or potential energy). You’ll say: so what? Well… I don’t know. Can we think of a spring that expands eternally, with the mass on its end not gaining or losing any speed? In that case, v is constant, and the total energy of the system would, effectively, be equal to E_total = 2·〈K.E.〉 = (1/2)·m·v²/2 = m·v².

Post scriptum 3: That substitution I made above – substituting x for x = v·t – is kinda weird. Indeed, if that E = m∙v² equation makes any sense, then E − m∙v² = 0, of course, and, therefore, θ = E·t − p·x = E·t − p·v·t = E·t − m·v·v·t= (E − m∙v²)·t = 0·t = 0. So the argument of our wavefunction is 0 and, therefore, we get a·e⁰= a for our wavefunction. It basically means our particle is where it is. 🙂

Post scriptum 4: This post scriptum – no. 4 – was added later—much later. On 29 February 2016, to be precise. The solution to the ‘riddle’ above is actually quite simple. We just need to make a distinction between the group and the phase velocity of our complex-valued wave. The solution came to me when I was writing a little piece on Schrödinger’s equation. I noticed that we do not find that weird E = m∙v² formula when substituting ψ for ψ = e^{i(kx − ωt)}in Schrödinger’s equation, i.e. in:

Let me quickly go over the logic. To keep things simple, we’ll just assume one-dimensional space, so ∇²ψ = ∂²ψ/∂x². The time derivative on the left-hand side is ∂ψ/∂t = −iω·e^{i(kx − ωt)}. The second-order derivative on the right-hand side is ∂²ψ/∂x² = (ik)·(ik)·e^{i(kx − ωt)}= −k²·e^{i(kx − ωt)} . The e^{i(kx − ωt)} factor on both sides cancels out and, hence, equating both sides gives us the following condition:

−iω = −(iħ/2m)·k² ⇔ ω = (ħ/2m)·k²

Substituting ω = E/ħ and k = p/ħ yields:

E/ħ = (ħ/2m)·p²/ħ²= m²·v²/(2m·ħ) = m·v²/(2ħ) ⇔ E = m·v²/2

In short: the E = m·v²/2 is the correct formula. It must be, because… Well… Because Schrödinger’s equation is a formula we surely shouldn’t doubt, right? So the only logical conclusion is that we must be doing something wrong when multiplying the two de Broglie equations. To be precise: our v = f·λ equation must be wrong. Why? Well… It’s just something one shouldn’t apply to our complex-valued wavefunction. The ‘correct’ velocity formula for the complex-valued wavefunction should have that 1/2 factor, so we’d write 2·f·λ = v to make things come out alright. But where would this formula come from? The period of cosθ + isinθ is the period of the sine and cosine function: cos(θ+2π) + isin(θ+2π) = cosθ + isinθ, so T = 2π and f = 1/T = 1/2π do not change.

But so that’s a mathematical point of view. From a physical point of view, it’s clear we got two oscillations for the price of one: one ‘real’ and one ‘imaginary’—but both are equally essential and, hence, equally ‘real’. So the answer must lie in the distinction between the group and the phase velocity when we’re combining waves. Indeed, the group velocity of a sum of waves is equal to v_g = dω/dk. In this case, we have:

v_g = d[E/ħ]/d[p/ħ] = dE/dp

We can now use the kinetic energy formula to write E as E = m·v²/2 = p·v/2. Now, v and p are related through m (p = m·v, so v = p/m). So we should write this as E = m·v²/2 = p²/(2m). Substituting E and p = m·v in the equation above then gives us the following:

dω/dk = d[p²/(2m)]/dp = 2p/(2m) = v_g= v

However, for the phase velocity, we can just use the v_p= ω/k formula, which gives us that 1/2 factor:

v_p= ω/k = (E/ħ)/(p/ħ) = E/p = (m·v²/2)/(m·v) = v/2

Bingo! Riddle solved! 🙂 Isn’t it nice that our formula for the group velocity also applies to our complex-valued wavefunction? I think that’s amazing, really! But I’ll let you think about it. 🙂

The hydrogen molecule as a two-state system

My posts on the state transitions of an ammonia molecule weren’t easy, were they? So let’s try another two-state system. The illustration below shows an ionized hydrogen molecule in two possible states which, as usual, we’ll denote as |1〉 and |2〉. An ionized hydrogen molecule is an H₂ molecule which lost an electron, so it’s two protons with one electron only, so we denote it as H₂⁺. The difference between the two states is obvious: the electron is either with the first proton or with the second.

It’s an example taken from Feynman’s Lecture on two-state systems. The illustration itself raises a lot of questions, of course. The most obvious question is: how do we know which proton is which? We’re talking identical particles, right? Right. We should think of the proton spins! However, protons are fermions and, hence, they can’t be in the same state, so they must have opposite spins. Of course, now you’ll say: they’re not in the same state because they’re at different locations. Well… Now you’ve answered your own question. 🙂 However you want to look at this, the point is: we can distinguish both protons. Having said that, the reflections above raise other questions: what reference frame are we using? The answer is: it’s the reference frame of the system. We can mirror or rotate this image however we want – as I am doing below – but state |1〉 is state |1〉, and state |2〉 is state |2〉.

The other obvious question is more difficult. If you’ve read anything at all about quantum mechanics, you’ll ask: what about the in-between states? The electron is actually being shared by the two protons, isn’t it? That’s what chemical bonds are all about, no? Molecular orbitals rather than atomic orbitals, right? Right. That’s actually what this post is all about. We know that, in quantum mechanics, the actual state – or what we think is the actual state – is always expressed as some linear combination of so-called base states. We wrote:

|ψ〉 = |1〉C₁+ |2〉C₂= |1〉〈1|ψ〉 + |2 〉〈2|ψ 〉

In terms of representing what’s actually going on, we only have these probability functions: they say that, if we would take a measurement, the probability of finding the electron near the first or the second proton varies as shown below:

If the |1〉 and |2〉 states were actually representing two dual physical realities, the actual state of our H₂⁺molecule would be represented by some square or some pulse wave, as illustrated below. [We should be calling it a square function but that term has been reserved for a function like y = x².]

Of course, the symmetry of the situation implies that the average pulse duration τ would be one-half of the (average) period T, so we’d be talking a square wavefunction indeed. The two wavefunctions both qualify as probability density functions: the system is always in one state or the other, and the probabilities add up to one. But you’ll agree we prefer the smooth squared sine and cosine functions. To be precise, these smooth functions are:

P₁(t) = |C₁(t)|² = cos²[(A/ħ)·t]
P₂(t) = |C₂(t)|²= sin²[(A/ħ)·t]

So now we only need to explain A here (you know ħ already). But… Well… Why would we actually prefer those smooth functions? An irregular pulse function would seem to be doing a better job when it comes to modeling reality, doesn’t it? The electron should be either here, or there. Isn’t it?

Well… No. At least that’s why I am slowly starting to understand. These pure base states |1〉 and |2〉 are real and not real at the same time. They’re real, because it’s what we’ll get when we verify, or measure, the state, so our measurement will tell us that it’s here or there. There’s no in-between. [I still need to study weak measurement theory.] But then they are not real, because our molecule will never ever be in those two states, except for those ephemeral moments when (A/ħ)·t = n·π (n = 0, 1, 2,…). So we’re really modeling uncertainty here and, while I am still exploring what that actually means, you should think of the electron as being everywhere really, but with an unequal density in space—sort of. 🙂

Now, we’ve learned we can describe the state of a system in terms of an alternative set of base states. We wrote: |ψ〉 = |I〉C_I+ |II〉C_II= |I〉〈I|ψ〉 + |II〉〈II|ψ〉, with the C_{I, II}and C_{1, 2}coefficients being related to each other in exactly the same way as the associated base states, i.e. through a transformation matrix, which we summarized as:

To be specific, the two sets of base states we’ve been working with so far were related as follows:

So we’d write: |ψ〉 = |I〉C_I+ |II〉C_II= |I〉〈I|ψ〉 + |II〉〈II|ψ〉 = |1〉C₁+ |2〉C₂= |1〉〈1|ψ〉 + |2 〉〈2|ψ 〉, and the C_{I, II}and C_{1, 2}coefficients would be related in exactly the same way as the base states:

[In case you’d want to review how that works, see my post on the Hamiltonian and base states.] Now, we cautioned that it’s difficult to try to interpret such base transformations – often referred to as a change in the representation or a different projection – geometrically. Indeed, we acknowledged that (base) states were very much like (base) vectors – from a mathematical point of view, that is – but, at the same time, we said that they were ‘objects’, really: elements in some Hilbert space, which means you can do the operations we’re doing here, i.e. adding and multiplying. Something like |I〉C_I doesn’t mean all that much: C_Iis a complex number – and so we can work with numbers, of course, because we can visualize them – but |I〉 is a ‘base state’, and so what’s the meaning of that, and what’s the meaning of the |I〉C_I or C_I|I〉 product? I could babble about that, but it’s no use: a base state is a base state. It’s some state of the system that makes sense to us. In fact, it may be some state that does not make sense to us—in terms of the physics of the situation, that is – but then there will always be some mathematical sense to it because of that transformation matrix, which establishes a one-to-one relationship between all sets of base states.

You’ll say: why don’t you try to give it some kind of geometrical or whatever meaning? OK. Let’s try. State |1〉 is obviously like minus state |2〉 in space, so let’s see what happens when we equate |1〉 to 1 on the real axis, and |2〉 to −1. Geometrically, that corresponds to the (1, 0) and (−1, 0) points on the unit circle. So let’s multiply those points with (1/√2, −1/√2) and (1/√2, 1/√2) respectively. What do we get? Well… What product should we take? The dot product, the cross product, or the ordinary complex-number product? The dot product gives us a number, so we don’t want that. [If we’re going to represent base states by vectors, we want all states to be vectors.] A cross product will give us a vector that’s orthogonal to both vectors, so it’s a vector in ‘outer space’, so to say. We don’t want that, I must assume, and so we’re left with the complex-number product, which projects our (1, 0) and (−1, 0) vectors into the (1/√2, −1/√2)·(1, 0) = (1/√2−i/√2)·(1+0·i) = √2−i/√2 = (1/√2, −i/√2) and (1/√2, 1/√2)·(−1, 0) = (1/√2+i/√2)·(−1+0·i) = −√2−i/√2 = (−1/√2, −i/√2) respectively.

What does this say? Nothing. Stuff like this only causes confusion. We had two base states that were ‘180 degrees’ apart, and now our new base states are only ’90 degrees’ apart. If we’d ‘transform’ the two new base states once more, they collapse into each other: (1/√2, −1/√2)·(1/√2, −1/√2) = (1/√2−i/√2)² = −i = (0, −1) = (1/√2, 1/√2)·(−1/√2, −1/√2) = −i. This is nonsense, of course. It’s got nothing to do with the angle we picked for our original set of base states: we could have separated our original set of base states by 90 degrees, or 45 degrees. It doesn’t matter. It’s the transformation itself: multiplying by (+1/√2, −1/√2) amounts to a clockwise rotation by 45 degrees, while multiplying by (+1/√2, +1/√2) amounts to the same, but counter-clockwise. So… Well… We should not try to think of our base vectors in any geometric way, because it just doesn’t make any sense. So Let’s not waste time on this: the ‘base states’ are a bit of a mystery, in the sense that they just are what they are: we can’t ‘reduce’ them any further, and trying to interpret them geometrically leads to contradictions, as evidenced by what I tried to do above. Base states are ‘vectors’ in a so-called Hilbert space, and… Well… That’s not your standard vector space. [If you think you can make more sense of it, please do let me know!]

Onwards!

Let’s take our transformation again:

|I〉 = (1/√2)|1〉 − (1/√2)|2〉 = (1/√2)[|1〉 − |2〉]
|II〉 = (1/√2)|1〉 + (1/√2)|2〉 = (1/√2)[|1〉 + |2〉]

Again, trying to geometrically interpret what it means to add or subtract two base states is not what you should be trying to do. In a way, the two expressions above only make sense when combining them with a final state, so when writing:

〈ψ|I〉 = (1/√2)〈ψ|1〉 − (1/√2)〈ψ|2〉 = (1/√2)[〈ψ|1〉 − 〈ψ|2〉]
〈ψ|II〉 = (1/√2)〈ψ|1〉 + (1/√2)〈ψ|2〉 = (1/√2)[〈ψ|1〉 + 〈ψ|2〉]

Taking the complex conjugate of this gives us the amplitudes of the system to be in state I or state II:

〈I|ψ〉 = 〈ψ|I〉* = (1/√2)[〈ψ|1〉* − 〈ψ|2〉*] = (1/√2)[〈1|ψ〉 − 〈2|ψ〉]
〈II|ψ〉 = 〈ψ|II〉* = (1/√2)[〈ψ|1〉* + 〈ψ|2〉*] = (1/√2)[〈1|ψ〉 + 〈2|ψ〉]

That still doesn’t tell us much, because we’d need to know the 〈1|ψ〉 and 〈2|ψ〉 functions, i.e. the amplitudes of the system to be in state 1 and state 2 respectively. What we do know, however, is that the 〈1|ψ〉 and 〈2|ψ〉 functions will have some rather special amplitudes. We wrote:

C_I= 〈 I | ψ 〉 = e^{−(i/ħ)·E_I·t}
C_II= 〈 II | ψ 〉 = e^{−(i/ħ)·E_II·t}

These are amplitudes of so-called stationary states: the associated probabilities – i.e. the absolute square of these functions – do not vary in time: |e^{−(i/ħ)·E_I·t}|² = |e^{−(i/ħ)·E_II·t}|² = 1. For our ionized hydrogen molecule, it means that, if it would happen to be in state I, it will stay in state I, and the same goes for state II. We write:

〈 I | I 〉 = 〈 II | II 〉 = 1 and 〈 I | II 〉 = 〈 II | I 〉 = 0

That’s actually just the so-called ‘orthogonality’ condition for base states, which we wrote as 〈i|j〉 = 〈j|i〉 = δ_ij, but, in light of the fact that we can’t interpret them geometrically, we shouldn’t be calling it like that. The point is: we had those differential equations describing a system like this. If the amplitude to go from state 1 to state 2 was equal to some real- or complex-valued constant A, then we could write those equations either in terms of C₁and C₂, or in terms of C_Iand C_II:

So the two sets of equations are equivalent. However, what we want to do here is look at it in terms of C_Iand C_II. Let’s first analyze those two energy levels E_I= E₀+ A and E_II= E₀− A. Feynman graphs them as follows:

Let me explain. In the first graph, we have E_I= E₀+ A and E_II= E₀− A, and they are depicted as being symmetric, with A depending on the distance between the two protons. As for E₀, that’s the energy of a hydrogen atom, i.e. a proton with a bound electron, and a separate proton. So it’s the energy of a system consisting of a hydrogen atom and a proton, which is obviously not the same as that of an ionized hydrogen molecule. The concept of a molecule assumes the protons are closely together. We assume E₀= 0 if the interproton distance is relatively large but, of course, as the protons come closer, we shouldn’t forget the repulsive electrostatic force between the two protons, which is represented by the dashed line in the first graph. Indeed, unlike the electron and the proton, the two protons will want to push apart, rather than pull together, so the potential energy of the system increases as the interproton distance decreases. So E₀is not constant either: it also depends on the interproton distance. But let’s forget about E₀for a while. Let’s look at the two curves for A now.

A is not varying in time, but its value does depend on the distance between the two protons. We’ll use this in a moment to calculate the approximate size of the hydrogen nucleus in a calculation that closely resembles Feynman’s calculation of the size of a hydrogen atom. That A should be some function of the interproton distance makes sense: the transition probability, and therefore A, will exponentially decrease with distance. There are a few things to reflect on here:

1. In the mentioned calculation of the size of a hydrogen atom, which is based on the Uncertainty Principle, Feynman shows that the energy of the system decreases when an electron is bound to the proton. The reasoning is that, if the potential energy of the electron is zero when it is not bound, then its potential energy will be negative when bound. Think of it: the electron and the proton attract each other, so it requires force to separate them, and force over a distance is energy. From our course in electromagnetics, we know that the potential energy, when bound, should be equal to −e²/a₀, with e²the squared charge of the electron divided by 4πε₀, and a₀ the so-called Bohr radius of the atom. Of course, the electron also has kinetic energy. It can’t just sit on top of the proton because that would violate the Uncertainty Principle: we’d know where it was. Combining the two, Feynman calculates both a₀ as well as the so-called Rydberg energy, i.e. the total energy of the bound electron, which is equal to −13.6 eV. So, yes, the bound state has less energy, so the electron will want to be bound, i.e. it will want to be close to one of the two protons.

2. Now, while that’s not what’s depicted above, it’s clear the magnitude of A will be related to that Rydberg energy which − please note − is quite high. Just compare it with the A for the ammonia molecule, which we calculated in our post on the maser: we found an A of about 0.5×10⁻⁴eV there, so that’s like 270,000 times less! Nevertheless, the possibility is there, and what happens when the electron flips over amounts to tunneling: it penetrates and crosses a potential barrier. We did a post on that, and so you may want to look at how that works. One of the weird things we had to consider when a particle crosses such potential barrier, is that the momentum factor p in its wavefunction was some pure imaginary number, which we wrote as p = i·p’. We then re-wrote that wavefunction as a·e⁻ⁱ^θ = a·e⁻ⁱ^[^{(E/ħ)∙t − (i·p’/ħ)x]} = a·e⁻ⁱ^(E/ħ)∙t·e^{i²·p’·x/ħ} = a·e⁻ⁱ^(E/ħ)∙t·e^{−p’·x/ħ}. Now, it’s easy to see that the e^{−p’·x/ħ} factor in this formula is a real-valued exponential function, with the same shape as the general e^−x function, which I depict below.

This e^{−p’·x/ħ} basically ‘kills’ our wavefunction as we move in the positive x-direction, across the potential barrier, which is what is illustrated below: if the distance is too large, then the amplitude for tunneling goes to zero.

So that’s what depicted in those graphs of E_I= E₀+ A and E_II= E₀− A: A goes to zero when the interproton distance becomes too large. We also recognize the exponential shape for A in those graphs, which can also be derived from the same tunneling story.

Now we can calculate E₀+ A and E₀− A taking into account that both terms vary with the interproton distance as explained, and so that gives us the final curves on the right-hand side, which tell us that the equilibrium configuration of the ionized hydrogen molecule is state II, i.e. the lowest energy state, and the interproton distance there is approximately one Ångstrom, i.e. 1×10⁻¹⁰ m. [You can compare this with the Bohr radius, which we calculated as a₀ = 0.528×10⁻¹⁰ m, so that all makes sense.] Also note the energy scale: ΔE is the excess energy over a proton plus a hydrogen atom, so that’s the energy when the two protons are far apart. Because it’s the excess energy, we have a zero point. That zero point is, obviously, the energy of a hydrogen atom and a proton. [Read this carefully, and please refer back to what I wrote above. The energy of a system consisting of a hydrogen atom and a proton is not the same as that of an ionized hydrogen molecule: the concept of a molecule assumes the protons are closely together.] We then re-scale by dividing by the Rydberg energy E_H= 13.6 eV. So ΔE/E_H≈ −0.2 ⇔ ΔE ≈ −0.2×13.6 = –2.72 eV. That basically says that the energy of our ionized hydrogen molecule is 2.72 eV lower than the energy of a hydrogen atom and a proton.

Why is it lower? We need to think about our model of the hydrogen atom once more: the energy of the electron was minimized by striking a balance between (1) being close to the proton and, therefore, having a low potential energy (or a low coulomb energy, as Feynman calls it) and (2) being further away from the proton and, therefore, lowering its kinetic energy according to the Uncertainty Principle ΔxΔp ≥ ħ/2, which Feynman boldly re-wrote as p = ħ/a₀. Now, a molecular orbital, i.e. the electron being around two protons, results in “more space where the electron can have a low potential energy”, as Feynman puts it, so “the electron can spread out—lowering its kinetic energy—without increasing its potential energy.”

The whole discussion here actually amounts to an explanation for the mechanism by which an electron shared by two protons provides, in effect, an attractive force between the two protons. So we’ve got a single electron actually holding two protons together, which chemists refer to as a “one-electron bond.”

So… Well… That explains why the energy E_II= E₀− A is what it is, so that’s smaller than E₀indeed, with the difference equal to the value A for an interproton distance of 1 Å. But how should we interpret E_I= E₀+ A? What is that higher energy level? What does it mean?

That’s a rather tricky question. There’s no easy interpretation here, like we had for our ammonia molecule: the higher energy level had an obvious physical meaning in an electromagnetic field, as it was related to the electric dipole moment of the molecule. That’s not the case here: we have no magnetic or electric dipole moment here. So, once again, what’s the physical meaning of E_I= E₀+ A? Let me quote Feynman’s enigmatic answer here:

“Notice that this state is the difference of the states |1⟩ and |2⟩. Because of the symmetry of |1⟩ and |2⟩, the difference must have zero amplitude to find the electron half-way between the two protons. This means that the electron is somewhat more confined, which leads to a larger energy.”

What does he mean with that? It seems he’s actually trying to do what I said we shouldn’t try to do, and that is to interpret what adding versus subtracting states actually means. But let’s give it a fair look. We said that the |I〉 = (1/√2)[|1〉 − |2〉] expression didn’t mean much: we should add a final state and write: 〈ψ|I〉 = (1/√2)[〈ψ|1〉 − 〈ψ|2〉], which is equivalent to 〈I|ψ〉 = (1/√2)[〈1|ψ〉 − 〈2|ψ〉]. That still doesn’t tell us anything: we’re still adding amplitudes, and so we should allow for interference, and saying that |1⟩ and |2⟩ are symmetric simply means that 〈1|ψ〉 − 〈2|ψ〉 = 〈2|ψ〉 − 〈1|ψ〉 ⇔ 2·〈1|ψ〉 = 2·〈2|ψ〉 ⇔ 〈1|ψ〉 = 〈2|ψ〉. Wait a moment! That’s an interesting reflection. Following the same reasoning for |II〉 = (1/√2)[|1〉 + |2〉], we get 〈1|ψ〉 + 〈2|ψ〉 = 〈2|ψ〉 + 〈1|ψ〉 ⇔ … Huh? No, that’s trivial: 0 = 0.

Hmm… What to say? I must admit I don’t quite ‘get’ Feynman here: state I, with energy E_I= E₀+ A, seems to be both meaningless as well as impossible. The only energy levels that would seem to make sense here are the energy of a hydrogen atom and a proton and the (lower) energy of an ionized hydrogen molecule, which you get when you bring a hydrogen atom and a proton together. 🙂

But let’s move to the next thing: we’ve added only one electron to the two protons, and that was it, and so we had an ionized hydrogen molecule, i.e. an H₂⁺molecule. Why don’t we do a full-blown H₂molecule now? Two protons. Two electrons. It’s easy to do. The set of base states is quite predictable, and illustrated below: electron a can be either one of the two protons, and the same goes for electron b.

We can then go through the same as for the ion: the molecule’s stability is shown in the graph below, which is very similar to the graph of the energy levels of the ionized hydrogen molecule, i.e. the H₂⁺molecule. The shape is the same, but the values are different: the equilibrium state is at an interproton distance of 0.74 Å, and the energy of the equilibrium state is like 5 eV (ΔE/E_H≈ −0.375) lower than the energy of two separate hydrogen atoms.

The explanation for the lower energy is the same: state II is associated with some kind of molecular orbital for both electrons, resulting in “more space where the electron can have a low potential energy”, as Feynman puts it, so “the electron can spread out—lowering its kinetic energy—without increasing its potential energy.”

However, there’s one extra thing here: the two electrons must have opposite spins. That’s the only way to actually distinguish the two electrons. But there is more to it: if the two electrons would not have opposite spin, we’d violate Fermi’s rule: when identical fermions are involved, and we’re adding amplitudes, then we should do so with a negative sign for the exchanged case. So our transformation would be problematic:

〈II|ψ〉 = (1/√2)[〈1|ψ〉 + 〈2|ψ〉] = (1/√2)[〈2|ψ〉 + 〈1|ψ〉]

When we switch the electrons, we should get a minus sign. The weird thing is: we do get that minus sign for state I:

〈I|ψ〉 = (1/√2)[〈1|ψ〉 − 〈2|ψ〉] = −(1/√2)[〈2|ψ〉 − 〈1|ψ〉]

So… Well… We’ve got a bit of an answer there as to what that the ‘other’ (upper) energy level of E_I= E₀+ A actually means, in physical terms, that is. It models two hydrogens coming together with parallel electron spins. Applying Fermi’s rules – i.e. the exclusion principle, basically – we find that state II is, quite simply, not allowed for parallel electron spins: state I is, and it’s the only one. There’s something deep here, so let me quote the Master himself on it:

“We find that the lowest energy state—the only bound state—of the H₂ molecule has the two electrons with spins opposite. The total spin angular momentum of the electrons is zero. On the other hand, two nearby hydrogen atoms with spins parallel—and so with a total angular momentum ℏ—must be in a higher (unbound) energy state; the atoms repel each other. There is an interesting correlation between the spins and the energies. It gives another illustration of something we mentioned before, which is that there appears to be an “interaction” energy between two spins because the case of parallel spins has a higher energy than the opposite case. In a certain sense you could say that the spins try to reach an antiparallel condition and, in doing so, have the potential to liberate energy—not because there is a large magnetic force, but because of the exclusion principle.”

You should read this a couple of times. It’s an important principle. We’ll discuss it again in the next posts, when we’ll be talking spin in much more detail once again. 🙂 The bottom line is: if the electrons are parallel, then they won’t ‘share’ any space at all and, hence, they are really much more confined in space, and the associated energy level is, therefore, much higher.

Post scriptum: I said we’d ‘calculate’ the equilibrium interproton distance. We didn’t do that. We just gave them through the graphs, which are based on the results of a ‘detailed quantum-mechanical calculation’—or that’s what Feynman claims, at least. I am not sure if they correspond to experimentally determined values, or what calculations are behind, exactly. Feynman notes that “this approximate treatment of the H₂⁺molecule as a two-state system breaks down pretty badly once the protons get as close together as they are at the minimum in the curve and, therefore, it will not give a good value for the actual binding energy. For small separations, the energies of the two “states” we imagined are not really equal to E₀, and a more refined quantum mechanical treatment is needed.”

So… Well… That says it all, I guess.

Two-state systems: the math versus the physics, and vice versa.

I think my previous post, on the math behind the maser, was a bit of a brain racker. However, the results were important and, hence, it is useful to generalize them so we can apply it to other two-state systems. 🙂 Indeed, we’ll use the very same two-state framework to analyze things like the stability of neutral and ionized hydrogen molecules and the binding of diatomic molecules in general – and lots of other stuff that can be analyzed as a two-state system. However, let’s first have look at the math once more. More importantly, let’s analyze the physics behind.

At the center of our little Universe here 🙂 is the fact that the dynamics of a two-state system are described by a set of two differential equations, which we wrote as:

It’s obvious these two equations are usually not easy to solve: the C₁and C₂functions are complex-valued amplitudes which vary not only in time but also in space, obviously, but, in fact, that’s not the problem. The issue is that the Hamiltonian coefficients H_ijmay also vary in space and in time, and so that‘s what makes things quite nightmarish to solve. [Note that, while H₁₁and H₂₂represent some energy level and, hence, are usually real numbers, H₁₂and H₂₁may be complex-valued. However, in the cases we’ll be analyzing, they will be real numbers too, as they will usually also represent some energy. Having noted that, being real- or complex-valued is not the problem: we can work with complex numbers and, as you can see from the matrix equation above, the i/ħ factor in front of our differential equations results in a complex-valued coefficient matrix anyway.]

So… Yes. It’s those non-constant Hamiltonian coefficients that caused us so much trouble when trying to analyze how a maser works or, more generally, how induced transitions work. [The same equations apply to blackbody radiation indeed, or other phenomena involved induced transitions.] In any case, so we won’t do that again – not now, at least – and so we’ll just go back to analyzing ‘simple’ two-state systems, i.e. systems with constant Hamiltonian coefficients.

Now, even for such simple systems, Feynman made life super-easy for us – too easy, I think – because he didn’t use the general mathematical approach to solve the issue on hand. That more general approach would be based on a technique you may or may not remember from your high school or university days: it’s based on finding the so-called eigenvalues and eigenvectors of the coefficient matrix. I won’t say too much about that, as there’s excellent online coverage of that, but… Well… We do need to relate the two approaches, and so that’s where math and physics meet. So let’s have a look at it all.

If we would write the first-order time derivative of those C₁ and C₂functions as C₁‘ and C₂‘ respectively (so we just put a prime instead of writing dC₁/dt and dC₂/dt), and we put them in a two-by-one column matrix, which I’ll write as C‘, and then, likewise, we also put the functions themselves, i.e. C₁ and C₂, in a column matrix, which I’ll write as C, then the system of equations can be written as the following simple expression:

C‘ = AC

One can then show that the general solution will be equal to:

C = a₁e^λ_I·tv_I+ a₂e^λ_II·tv_II

The λ_I and λ_II in the exponential functions are the eigenvalues of A, so that’s that two-by-two matrix in the equation, i.e. the coefficient matrix with the −(i/ħ)H_ijelements. The v_I and v_II column matrices in the solution are the associated eigenvectors. As for a₁ and a₂, these are coefficients that depend on the initial conditions of the system as well as, in our case at least, the normalization condition: the probabilities we’ll calculate have to add up to one. So… Well… It all comes with the system, as we’ll see in a moment.

Let’s first look at those eigenvalues. We get them by calculating the determinant of the A−λI matrix, and equating it to zero, so we write det(A−λI) = 0. If A is a two-by-two matrix (which it is for the two-state systems that we are looking at), then we get a quadratic equation, and its two solutions will be those λ_I and λ_II values. The two eigenvalues of our system above can be written as:

λ_I = −(i/ħ)·E_I and λ_II = −(i/ħ)·E_II.

E_I and E_II are two possible values for the energy of our system, which are referred to as the upper and the lower energy level respectively. We can calculate them as:

Note that we use the Roman numerals I and II for these two energy levels, rather than the usual Arabic numbers 1 and 2. That’s in line with Feynman’s notation: it relates to a special set of base states that we will introduce shortly. Indeed, plugging them into the a₁e^λ_I·t and a₂e^λ_II·t expressions gives us a₁e^{−(i/ħ)·E_I·t} and a₂e^{−(i/ħ)·E_II·t} and…

Well… It’s time to go back to the physics class now. What are we writing here, really? These two functions are amplitudes for so-called stationary states, i.e. states that are associated with probabilities that do not change in time. Indeed, it’s easy to see that their absolute square is equal to:

P_I= |a₁e^{−(i/ħ)·E_I·t}|²= |a₁|²·|e^{−(i/ħ)·E_I·t}|²= |a₁|²
P_II= |a₂e^{−(i/ħ)·E_II·t}|²= |a₂|²·|e^{−(i/ħ)·E_II·t}|²= |a₂|²

Now, the a₁ and a₂ coefficients depend on the initial and/or normalization conditions of the system, so let’s leave those out for the moment and write the rather special amplitudes e^{−(i/ħ)·E_I·t} and e^{−(i/ħ)·E_II·t} as:

C_I= 〈 I | ψ 〉 = e^{−(i/ħ)·E_I·t}
C_II= 〈 II | ψ 〉 = e^{−(i/ħ)·E_II·t}

As you can see, there’s two base states that go with these amplitudes, which we denote as state | I 〉 and | II 〉 respectively, so we can write the state vector of our two-state system – like our ammonia molecule, or whatever – as:

| ψ 〉 = | I 〉 C_I+ | II 〉 C_II= | I 〉〈 I | ψ 〉 + | II 〉〈 II | ψ 〉

In case you forgot, you can apply the magical | = ∑ | i 〉 〈 i | formula to see this makes sense: | ψ 〉 = ∑ | i 〉 〈 i | ψ 〉 = | I 〉 〈 I | ψ 〉 + | II 〉 〈 II | ψ 〉 = | I 〉 C_I+ | II 〉 C_II.

Of course, we should also be able to revert back to the base states we started out with so, once we’ve calculated C₁and C₂, we can also write the state of our system in terms of state | 1 〉 and | 2 〉, which are the states as we defined them when we first looked at the problem. 🙂 In short, once we’ve got C₁and C₂, we can also write:

| ψ 〉 = | 1 〉 C₁+ | 2 〉 C₂= | 1 〉〈 1 | ψ 〉 + | 2 〉〈 2 | ψ 〉

So… Well… I guess you can sort of see how this is coming together. If we substitute what we’ve got so far, we get:

C = a₁·C_I·v_I + a₂·C_II·v_II

Hmm… So what’s that? We’ve seen something like C = a₁·C_I + a₂·C_II, as we wrote something like C₁ = (a/2)·C_I + (b/2)·C_II b in our previous posts, for example—but what are those eigenvectors v_I and v_II? Why do we need them?

Well… They just pop up because we’re solving the system as mathematicians would do it, i.e. not as Feynman-the-Great-Physicist-and-Teacher-cum-Simplifier does it. 🙂 From a mathematical point of view, they’re the vectors that solve the (A−λ_II)v_I = 0 and (A−λ_III)v_II = 0 equations, so they come with the eigenvalues, and their components will depend on the eigenvalues λ_Iand λ_I as well as the Hamiltonian coefficients. [I is the identity matrix in these matrix equations.] In fact, because the eigenvalues are written in terms of the Hamiltonian coefficients, they depend on the Hamiltonian coefficients only, but then it will be convenient to use the E_I and E_II values as a shorthand.

Of course, one can also look at them as base vectors that uniquely specify the solution C as a linear combination of v_I and v_II. Indeed, just ask your math teacher, or google, and you’ll find that eigenvectors can serve as a set of base vectors themselves. In fact, the transformations you need to do to relate them to the so-called natural basis are the ones you’d do when diagonalizing the coefficient matrix A, which you did when solving systems of equations back in high school or whatever you were doing at university. But then you probably forgot, right? 🙂 Well… It’s all rather advanced mathematical stuff, and so let’s cut some corners here. 🙂

We know, from the physics of the situations, that the C₁ and C₂ functions and the C_I and C_II functions are related in the same way as the associated base states. To be precise, we wrote:

This two-by-two matrix here is the transformation matrix for a rotation of state filtering apparatus about the y-axis, over an angle equal to α, when only two states are involved. You’ve seen it before, but we wrote it differently:

In fact, we can be more precise: the angle that we chose was equal to minus 90 degrees. Indeed, we wrote our transformation as:

[Check the values against α = −π/2.] However, let’s keep our analysis somewhat more general for the moment, so as to see if we really need to specify that angle. After all, we’re looking for a general solution here, so… Well… Remembering the definition of the inverse of a matrix (and the fact that cos²α + sin²α = 1), we can write:

Now, if we write the components of v_I and v_II as v_I1 and v_I2, and v_II1 and v_II2 respectively, then the C = a₁·C_I·v_I + a₂·C_II·v_IIexpression is equivalent to:

C₁ = a₁·v_I1·C_I+ a₂·v_II1·C_II
C₂ = a₁·v_I2·C_I + a₂·v_II2·C_II

Hence, a₁·v_I1= a₂·v_II2= cos(α/2) and a₂·v_II1 = −a₁·v_I2= sin(α/2). What can we do with this? Can we solve this? Not really: we’ve got two equations and four variables. So we need to look at the normalization and starting conditions now. For example, we can choose our t = 0 point such that our two-state system is in state 1, or in state I. And then we know it will not be in state 2, or state II. In short, we can impose conditions like:

|C₁(0)|²= 1 = |a₁·v_I1·C_I(0) + a₂·v_II1·C_II(0)|²and |C₂|²= 0 = |a₁·v_I1·C_I(0) + a₂·v_II1·C_II(0)|²

However, as Feynman puts it: “These conditions do not uniquely specify the coefficients. They are still undetermined by an arbitrary phase.”

Hmm… He means the α, of course. So… What to do? Well… It’s simple. What he’s saying here is that we do need to specify that transformation angle. Just look at it: the a₁·v_I1= a₂·v_II2= cos(α/2) and a₂·v_II1 = −a₁·v_I2= sin(α/2) conditions only make sense when we equate α with −π/2, so we can write:

a₁·v_I1= a₂·v_II2= cos(−π/4) = 1/√2
a₂·v_II1 = −a₁·v_I2= sin(−π/4) = –1/√2

It’s only then that we get a unique ratio for a₁/a₂= v_I1/v_II2= −v_II1/v_I2. [In case you think there are two angles in the circle for which the cosine equals minus the sine – or, what amounts to the same, for which the sine equals minus the cosine – then… Well… You’re right, but we’ve got α divided by two in the argument. So if α/2 is equal to the ‘other’ angle, i.e. 3π/4, then α itself will be equal to 6π/4 = 3π/2. And so that’s the same −π/2 angle as above: 3π/2 − 2π = −π/2, indeed. So… Yes. It all makes sense.]

What are we doing here? Well… We’re sort of imposing a ‘common-sense’ condition here. Think of it: if the v_I1/v_II2and −v_II1/v_I2ratios would be different, we’d have a huge problem, because we’d have two different values for the a₁/a₂ratio! And… Well… That just doesn’t make sense. The system must come with some specific value for a₁and a₂. We can’t just invent two ‘new’ ones!

So… Well… We are alright now, and we can analyze whatever two-state system we want now. One example was our ammonia molecule in an electric field, for which we found that the following systems of equations were fully equivalent:

So, the upshot is that you should always remember that everything we’re doing is subject to the condition that the ‘1’ and ‘2’ base states and the ‘I’ and ‘II’ base states (Feynman suggests to read I and II as ‘Eins’ and ‘Zwei’ – or try ‘Uno‘ and ‘Duo‘ instead 🙂 – so as to make a difference with ‘one’ and ‘two’) are ‘separated’ by an angle of (minus) 90 degrees. [Of course, I am not using the ‘right’ language here, obviously. I should say ‘projected’, or ‘orthogonal’, perhaps, but then that’s hard to say for base states: the [1/√2, 1/√2] and [1/√2, −1/√2] vectors are obviously orthogonal, because their dot product is zero, but, as you know, the base states themselves do not have such geometrical interpretation: they’re just ‘objects’ in what’s referred to as a Hilbert space. But… Well… I shouldn’t dwell on that here.]

So… There we are. We’re all set. Good to go! Please note that, in the absence of an electric field, the two Hamiltonians are even simpler:

In fact, they’ll usually do the trick in what we’re going to deal with now.

[…] So… Well… That’s is really! 🙂 We’re now going to apply all this in the next posts, so as to analyze things like the stability of neutral and ionized hydrogen molecules and the binding of diatomic molecules. More interestingly, we’re going to talk about virtual particles. 🙂

Addendum: I started writing this post because Feynman actually does give the impression there’s some kind of ‘doublet’ of a₁and a₂ coefficients as he start his chapter on ‘other two-state systems’. It’s the symbols he’s using: ‘his’ a₁and a₂, and the other doublet with the primes, i.e. a₁‘ and a₂‘, are the transformation amplitudes, not the coefficients that I am calculating above, and that he was calculating (in the previous chapter) too. So… Well… Again, the only thing you should remember from this post is that 90 degree angle as a sort of physical ‘common sense condition’ on the system.

Having criticized the Great Teacher for not being consistent in his use of symbols, I should add that the interesting thing is that, while confusing, his summary in that chapter does give us precise formulas for those transformation amplitudes, which he didn’t do before. Indeed, if we write them as a, b, c and d respectively (so as to avoid that confusing a₁and a₂, and then a₁‘ and a₂‘ notation), so if we have:

then one can show that:

That’s, of course, fully consistent with the ratios we introduced above, as well as with the orthogonality condition that comes with those eigenvectors. Indeed, if a/b = −1 and c/d = +1, then a/b = −c/d and, therefore, a·d + b·c = 0. [I’ll leave it to you to compare the coefficients so as to check that’s the orthogonality condition indeed.]

In short, it all shows everything does come out of the system in a mathematical way too, so the math does match the physics once again—as it should, of course! 🙂

The math behind the maser

Pre-script (dated 26 June 2020): I have come to the conclusion one does not need all this hocus-pocus to explain masers or lasers (and two-state systems in general): classical physics will do. So no use to read this. Read my papers instead. 🙂

Original post:

As I skipped the mathematical arguments in my previous post so as to focus on the essential results only, I thought it would be good to complement that post by looking at the math once again, so as to ensure we understand what it is that we’re doing. So let’s do that now. We start with the easy situation: free space.

The two-state system in free space

We started with an ammonia molecule in free space, i.e. we assumed there were no external force fields, like a gravitational or an electromagnetic force field. Hence, the picture was as simple as the one below: the nitrogen atom could be ‘up’ or ‘down’ with regard to its spin around its axis of symmetry.

It’s important to note that this ‘up’ or ‘down’ direction is defined in regard to the molecule itself, i.e. not in regard to some external reference frame. In other words, the reference frame is that of the molecule itself. For example, if I flip the illustration above – like below – then we’re still talking the same states, i.e. the molecule is still in state 1 in the image on the left-hand side and it’s still in state 2 in the image on the right-hand side.

We then modeled the uncertainty about its state by associating two different energy levels with the molecule: E₀+ A and E₀− A. The idea is that the nitrogen atom needs to tunnel through a potential barrier to get to the other side of the plane of the hydrogens, and that requires energy. At the same time, we’ll show the two energy levels are effectively associated with an ‘up’ or ‘down’ direction of the electric dipole moment of the molecule. So that resembles the two spin states of an electron, which we associated with the +ħ/2 and −ħ/2 energies respectively. So if E₀would be zero (we can always take another reference point, remember?), then we’ve got the same thing: two energy levels that are separated by some definite amount: that amount is 2A for the ammonia molecule, and ħ when we’re talking quantum-mechanical spin. I should make a last note here, before I move on: note that these energies only make sense in the presence of some external field, because the + and − signs in the E₀+ A and E₀− A and +ħ/2 and −ħ/2 expressions make sense only with regard to some external direction defining what’s ‘up’ and what’s ‘down’ really. But I am getting ahead of myself here. Let’s go back to free space: no external fields, so what’s ‘up’ or ‘down’ is completely random here. 🙂

Now, we also know an energy level can be associated with a complex-valued wavefunction, or an amplitude as we call it. To be precise, we can associate it with the generic a·e^{−(i/ħ)·(E·t − p∙x)}expression which you know so well by now. Of course, as the reference frame is that of the molecule itself, its momentum is zero, so the p∙x term in the a·e^{−(i/ħ)·(E·t − p∙x)}expression vanishes and the wavefunction reduces to a·e^−i·ω·t= a·e^{−(i/ħ)·E·t}, with ω = E/ħ. In other words, the energy level determines the temporal frequency, or the temporal variation (as opposed to the spatial frequency or variation), of the amplitude.

We then had to find the amplitudes C₁(t) = 〈 1 | ψ 〉 and C₂(t) =〈 2 | ψ 〉, so that’s the amplitude to be in state 1 or state 2 respectively. In my post on the Hamiltonian, I explained why the dynamics of a situation like this can be represented by the following set of differential equations:

As mentioned, the C₁and C₂functions evolve in time, and so we should write them as C₁= C₁(t) and C₂= C₂(t) respectively. In fact, our Hamiltonian coefficients may also evolve in time, which is why it may be very difficult to solve those differential equations! However, as I’ll show below, one usually assumes they are constant, and then one makes informed guesses about them so as to find a solution that makes sense.

Now, I should remind you here of something you surely know: if C₁and C₂are solutions to this set of differential equations, then the superposition principle tells us that any linear combination a·C₁+ b·C₂will also be a solution. So we need one or more extra conditions, usually some starting condition, which we can combine with a normalization condition, so we can get some unique solution that makes sense.

The H_ijcoefficients are referred to as Hamiltonian coefficients and, as shown in the mentioned post, the H₁₁and H₂₂coefficients are related to the amplitude of the molecule staying in state 1 and state 2 respectively, while the H₁₂and H₂₁coefficients are related to the amplitude of the molecule going from state 1 to state 2 and vice versa. Because of the perfect symmetry of the situation here, it’s easy to see that H₁₁should equal H₂₂, and that H₁₂and H₂₁should also be equal to each other. Indeed, Nature doesn’t care what we call state 1 or 2 here: as mentioned above, we did not define the ‘up’ and ‘down’ direction with respect to some external direction in space, so the molecule can have any orientation and, hence, switching the i an j indices should not make any difference. So that’s one clue, at least, that we can use to solve those equations: the perfect symmetry of the situation and, hence, the perfect symmetry of the Hamiltonian coefficients—in this case, at least!

The other clue is to think about the solution if we’d not have two states but one state only. In that case, we’d need to solve iħ·[dC₁(t)/dt] = H₁₁·C₁(t). That’s simple enough, because you’ll remember that the exponential function is its own derivative. To be precise, we write: d(a·e^iωt)/dt = a·d(e^iωt)/dt = a·iω·e^iωt, and please note that a can be any complex number: we’re not necessarily talking a real number here! In fact, we’re likely to talk complex coefficients, and we multiply with some other complex number (iω) anyway here! So if we write iħ·[dC₁/dt] = H₁₁·C₁ as dC₁/dt = −(i/ħ)·H₁₁·C₁ (remember: i⁻¹ = 1/i = −i), then it’s easy to see that the C₁= a·e^{–(i/ħ)·H₁₁·t}function is the general solution for this differential equation. Let me write it out for you, just to make sure:

dC₁/dt = d[a·e^{–(i/ħ)H₁₁t}]/dt = a·d[e^{–(i/ħ)H₁₁t}]/dt = –a·(i/ħ)·H₁₁·e^{–(i/ħ)H₁₁t}

= –(i/ħ)·H₁₁·a·e^{–(i/ħ)H₁₁t}= −(i/ħ)·H₁₁·C₁

Of course, that reminds us of our generic wavefunction a·e^{−(i/ħ)·E₀·t} wavefunction: we only need to equate H₁₁ with E₀ and we’re done! Hence, in a one-state system, the Hamiltonian coefficient is, quite simply, equal to the energy of the system. In fact, that’s a result can be generalized, as we’ll see below, and so that’s why Feynman says the Hamiltonian ought to be called the energy matrix.

In fact, we actually may have two states that are entirely uncoupled, i.e. a system in which there is no dependence of C₁ on C₂and vice versa. In that case, the two equations reduce to:

iħ·[dC₁/dt] = H₁₁·C₁ and iħ·[dC₂/dt] = H₂₂·C₂

These do not form a coupled system and, hence, their solutions are independent:

C₁(t) = a·e^{–(i/ħ)·H₁₁·t}and C₂(t)= b·e^{–(i/ħ)·H₂₂·t}

The symmetry of the situation suggests we should equate a and b, and then the normalization condition says that the probabilities have to add up to one, so |C₁(t)|²+ |C₂(t)|²= 1, so we’ll find that a = b = 1/√2.

OK. That’s simple enough, and this story has become quite long, so we should wrap it up. The two ‘clues’ – about symmetry and about the Hamiltonian coefficients being energy levels – lead Feynman to suggest that the Hamiltonian matrix for this particular case should be equal to:

Why? Well… It’s just one of Feynman’s clever guesses, and it yields probability functions that makes sense, i.e. they actually describe something real. That’s all. 🙂 I am only half-joking, because it’s a trial-and-error process indeed and, as I’ll explain in a separate section in this post, one needs to be aware of the various approximations involved when doing this stuff. So let’s be explicit about the reasoning here:

We know that H₁₁= H₂₂= E₀if the two states would be identical. In other words, if we’d have only one state, rather than two – i.e. if H₁₂and H₂₁would be zero – then we’d just plug that in. So that’s what Feynman does. So that’s what we do here too! 🙂
However, H₁₂and H₂₁are not zero, of course, and so assume there’s some amplitude to go from one position to the other by tunneling through the energy barrier and flipping to the other side. Now, we need to assign some value to that amplitude and so we’ll just assume that the energy that’s needed for the nitrogen atom to tunnel through the energy barrier and flip to the other side is equal to A. So we equate H₁₂and H₂₁ with −A.

Of course, you’ll wonder: why minus A? Why wouldn’t we try H₁₂= H₂₁ = A? Well… I could say that a particle usually loses potential energy as it moves from one place to another, but… Well… Think about it. Once it’s through, it’s through, isn’t it? And so then the energy is just E₀again. Indeed, if there’s no external field, the + or − sign is quite arbitrary. So what do we choose? The answer is: when considering our molecule in free space, it doesn’t matter. Using +A or −A yields the same probabilities. Indeed, let me give you the amplitudes we get for H₁₁= H₂₂= E₀and H₁₂and H₂₁ = −A:

C₁(t) = 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}+ (1/2)·e^{−(i/ħ)·(E₀+ A)·t}= e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t]
C₂(t) = 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}– (1/2)·e^{−(i/ħ)·(E₀+ A)·t}= i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t]

[In case you wonder how we go from those exponentials to a simple sine and cosine factor, remember that the sum of complex conjugates, i.e e^iθ+ e^−iθreduces to 2·cosθ, while e^iθ− e^−iθreduces to 2·i·sinθ.]

Now, it’s easy to see that, if we’d have used +A rather than −A, we would have gotten something very similar:

C₁(t) = 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀+ A)·t}+ (1/2)·e^{−(i/ħ)·(E₀− A)·t}= e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t]
C₂(t) = 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀+ A)·t}– (1/2)·e^{−(i/ħ)·(E₀− A)·t}= −i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t]

So we get a minus sign in front of our C₂(t) function, because cos(α) = cos(–α) but sin(α) = −sin(α). However, the associated probabilities are exactly the same. For both, we get the same P₁(t) and P₂(t) functions:

P₁(t) = |C₁(t)|² = cos²[(A/ħ)·t]
P₂(t) = |C₂(t)|²= sin²[(A/ħ)·t]

[Remember: the absolute square of i and −i is |i|²= +√1²= +1 and |−i|²= (−1)²|i|²= +1 respectively, so the i and −i in the two C₂(t) formulas disappear.]

You’ll remember the graph:

Of course, you’ll say: that plus or minus sign in front of C₂(t) should matter somehow, doesn’t it? Well… Think about it. Taking the absolute square of some complex number – or some complex function , in this case! – amounts to multiplying it with its complex conjugate. Because the complex conjugate of a product is the product of the complex conjugates, it’s easy to see what happens: the e^{−(i/ħ)·E₀·t} factor in C₁(t) = e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t] and C₂(t) = ±i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t] gets multiplied by e^{+(i/ħ)·E₀·t} and, hence, doesn’t matter: e^{−(i/ħ)·E₀·t}·e^{+(i/ħ)·E₀·t} = e⁰= 1. The cosine factor in C₁(t) = e^{−(i/ħ)·E₀·t}·cos[(A/ħ)·t] is real, and so its complex conjugate is the same. Now, the ±i·sin[(A/ħ)·t] factor in C₂(t) = ±i·e^{−(i/ħ)·E₀·t}·sin[(A/ħ)·t] is a pure imaginary number, and so its complex conjugate is its opposite. For some reason, we’ll find similar solutions for all of the situations we’ll describe below: the factor determining the probability will either be real or, else, a pure imaginary number. Hence, from a math point of view, it really doesn’t matter if we take +A or −A for or real factor for those H₁₂and H₂₁ coefficients. We just need to be consistent in our choice, and I must assume that, in order to be consistent, Feynman likes to think of our nitrogen atom borrowing some energy from the system and, hence, temporarily reducing its energy by an amount that’s equal to −A. If you have a better interpretation, please do let me know! 🙂

OK. We’re done with this section… Except… Well… I have to show you how we got those C₁(t) and C₁(t) functions, no? Let me copy Feynman here:

Note that the ‘trick’ involving the addition and subtraction of the differential equations is a trick we’ll use quite often, so please do have a look at it. As for the value of the a and b coefficients – which, as you can see, we’ve equated to 1 in our solutions for C₁(t) and C₁(t) – we get those because of the following starting condition: we assume that at t = 0, the molecule will be in state 1. Hence, we assume C₁(0) = 1 and C₂(0) = 0. In other words: we assume that we start out on that P₁(t) curve in that graph with the probability functions above, so the C₁(0) = 1 and C₂(0) = 0 starting condition is equivalent to P₁(0) = 1 and P₁(0) = 0. Plugging that in gives us a/2 + b/2 = 1 and a/2 − b/2 = 0, which is possible only if a = b = 1.

Of course, you’ll say: what if we’d choose to start out with state 2, so our starting condition is P₁(0) = 0 and P₁(0) = 1? Then a = 1 and b = −1, and we get the solution we got when equating H₁₂and H₂₁ with +A, rather than with −A. So you can think about that symmetry once again: when we’re in free space, then it’s quite arbitrary what we call ‘up’ or ‘down’.

So… Well… That’s all great. I should, perhaps, just add one more note, and that’s on that A/ħ value. We calculated it in the previous post, because we wanted to actually calculate the period of those P₁(t) and P₂(t) functions. Because we’re talking the square of a cosine and a sine respectively, the period is equal to π, rather than 2π, so we wrote: (A/ħ)·T = π ⇔ T = π·ħ/A. Now, the separation between the two energy levels E₀+ A and E₀− A, so that’s 2A, has been measured as being equal, more or less, to 2A ≈ 10⁻⁴eV.

How does one measure that? As mentioned above, I’ll show you, in a moment, that, when applying some external field, the plus and minus sign do matter, and the separation between those two energy levels E₀+ A and E₀− A will effectively represent something physical. More in particular, we’ll have transitions from one energy level to another and that corresponds to electromagnetic radiation being emitted or absorbed, and so there’s a relation between the energy and the frequency of that radiation. To be precise, we can write 2A = h·f₀. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, which corresponds to microwave radiation with a wavelength of λ = c/f₀ = 1.26 cm. Hence, 2·A ≈ 25×10⁹ Hz times 4×10⁻¹⁵ eV·s = 10⁻⁴eV, indeed, and, therefore, we can write: T = π·ħ/A ≈ 3.14 × 6.6×10⁻¹⁶eV·s divided by 0.5×10⁻⁴eV, so that’s 40×10⁻¹²seconds = 40 picoseconds. That’s 40 trillionths of a seconds. So that’s very short, and surely much shorter than the time that’s associated with, say, a freely emitting sodium atom, which is of the order of 3.2×10⁻⁸seconds. You may think that makes sense, because the photon energy is so much lower: a sodium light photon is associated with an energy equal to E = h·f = 500×10¹² Hz times 4×10⁻¹⁵ eV·s = 2 eV, so that’s 20,000 times 10⁻⁴eV.

There’s a funny thing, however. An oscillation of a frequency of 500 tera-hertz that lasts 3.2×10⁻⁸seconds is equivalent to 500×10¹² Hz times 3.2×10⁻⁸s ≈ 16 million cycles. However, an oscillation of a frequency of 23.97 giga-hertz that only lasts 40×10⁻¹²seconds is equivalent to 23.97×10⁹ Hz times 40×10⁻¹²s ≈ 1000×10⁻³= 1 ! One cycle only? We’re surely not talking resonance here!

So… Well… I am just flagging it here. We’ll have to do some more thinking about that later. [I’ve added an addendum that may or may not help us in this regard. :-)]

The two-state system in a field

As mentioned above, when there is no external force field, we define the ‘up’ or ‘down’ direction of the nitrogen atom was defined with regard to its its spin around its axis of symmetry, so with regard to the molecule itself. However, when we apply an external electromagnetic field, as shown below, we do have some external reference frame.

Now, the external reference frame – i.e. the physics of the situation, really – may make it more convenient to define the whole system using another set of base states, which we’ll refer to as I and II, rather than 1 and 2. Indeed, you’ve seen the picture below: it shows a state selector, or a filter as we called it. In this case, there’s a filtering according to whether our ammonia molecule is in state I or, alternatively, state II. It’s like a Stern-Gerlach apparatus splitting an electron beam according to the spin state of the electrons, which is ‘up’ or ‘down’ too, but in a totally different way than our ammonia molecule. Indeed, the ‘up’ and ‘down’ spin of an electron has to do with its magnetic moment and its angular momentum. However, there are a lot of similarities here, and so you may want to compare the two situations indeed, i.e. the electron beam in an inhomogeneous magnetic field versus the ammonia beam in an inhomogeneous electric field.

Now, when reading Feynman, as he walks us through the relevant Lecture on all of this, you get the impression that it’s the I and II states only that have some kind of physical or geometric interpretation. That’s not the case. Of course, the diagram of the state selector above makes it very obvious that these new I and II base states make very much sense in regard to the orientation of the field, i.e. with regard to external space, rather than with respect to the position of our nitrogen atom vis-á-vis the hydrogens. But… Well… Look at the image below: the direction of the field (which we denote by ε because we’ve been using the E for energy) obviously matters when defining the old ‘up’ and ‘down’ states of our nitrogen atom too!

In other words, our previous | 1 〉 and | 2 〉 base states acquire a new meaning too: it obviously matters whether or not the electric dipole moment of the molecule is in the same or, conversely, in the opposite direction of the field. To be precise, the presence of the electromagnetic field suddenly gives the energy levels that we’d associate with these two states a very different physical interpretation.

Indeed, from the illustration above, it’s easy to see that the electric dipole moment of this particular molecule in state 1 is in the opposite direction and, therefore, temporarily ignoring the amplitude to flip over (so we do not think of A for just a brief little moment), the energy that we’d associate with state 1 would be equal to E₀+ με. Likewise, the energy we’d associate with state 2 is equal to E₀− με. Indeed, you’ll remember that the (potential) energy of an electric dipole is equal to the vector dot product of the electric dipole moment μ and the field vector ε, but with a minus sign in front so as to get the sign for the energy righ. So the energy is equal to −μ·ε = −|μ|·|ε|·cosθ, with θ the angle between both vectors. Now, the illustration above makes it clear that state 1 and 2 are defined for θ = π and θ = 0 respectively. [And, yes! Please do note that state 1 is the highest energy level, because it’s associated with the highest potential energy: the electric dipole moment μ of our ammonia molecule will – obviously! – want to align itself with the electric field ε ! Just think of what it would imply to turn the molecule in the field!]

Therefore, using the same hunches as the ones we used in the free space example, Feynman suggests that, when some external electric field is involved, we should use the following Hamiltonian matrix:

So we’ll need to solve a similar set of differential equations with this Hamiltonian now. We’ll do that later and, as mentioned above, it will be more convenient to switch to another set of base states, or another ‘representation’ as it’s referred to. But… Well… Let’s not get too much ahead of ourselves: I’ll say something about that before we’ll start solving the thing, but let’s first look at that Hamiltonian once more.

When I say that Feynman uses the same clues here, then… Well.. That’s true and not true. You should note that the diagonal elements in the Hamiltonian above are not the same: E₀+ με ≠ E₀+ με. So we’ve lost that symmetry of free space which, from a math point of view, was reflected in those identical H₁₁= H₂₂= E₀coefficients.

That should be obvious from what I write above: state 1 and state 2 are no longer those 1 and 2 states we described when looking at the molecule in free space. Indeed, the | 1 〉 and | 2 〉 states are still ‘up’ or ‘down’, but the illustration above also makes it clear we’re defining state 1 and state 2 not only with respect to the molecule’s spin around its own axis of symmetry but also vis-á-vis some direction in space. To be precise, we’re defining state 1 and state 2 here with respect to the direction of the electric field ε. Now that makes a really big difference in terms of interpreting what’s going on.

In fact, the ‘splitting’ of the energy levels because of that amplitude A is now something physical too, i.e. something that goes beyond just modeling the uncertainty involved. In fact, we’ll find it convenient to distinguish two new energy levels, which we’ll write as E_I= E₀+ A and E_II= E₀− A respectively. They are, of course, related to those new base states | I 〉 and | II 〉 that we’ll want to use. So the E₀+ A and E₀− A energy levels themselves will acquire some physical meaning, and especially the separation between them, i.e. the value of 2A. Indeed, E_I= E₀+ A and E_II= E₀− A will effectively represent an ‘upper’ and a ‘lower’ energy level respectively.

But, again, I am getting ahead of myself. Let’s first, as part of working towards a solution for our equations, look at what happens if and when we’d switch to another representation indeed.

Switching to another representation

Let me remind you of what I wrote in my post on quantum math in this regard. The actual state of our ammonia molecule – or any quantum-mechanical system really – is always to be described in terms of a set of base states. For example, if we have two possible base states only, we’ll write:

| φ 〉 = | 1 〉 C₁ + | 2 〉 C₂

You’ll say: why? Our molecule is obviously always in either state 1 or state 2, isn’t it? Well… Yes and no. That’s the mystery of quantum mechanics: it is and it isn’t. As long as we don’t measure it, there is an amplitude for it to be in state 1 and an amplitude for it to be in state 2. So we can only make sense of its state by actually calculating 〈 1 | φ 〉 and 〈 2 | φ 〉 which, unsurprisingly are equal to 〈 1 | φ 〉 = 〈 1 | 1 〉 C₁ + 〈 1 | 2 〉 C₂ = C₁(t) and 〈 2 | φ 〉 = 〈 2 | 1 〉 C₁ + 〈 2 | 2 〉 C₂ = C₂(t) respectively, and so these two functions give us the probabilities P₁(t) and P₂(t) respectively. So that’s Schrödinger’s cat really: the cat is dead or alive, but we don’t know until we open the box, and we only have a probability function – so we can say that it’s probably dead or probably alive, depending on the odds – as long as we do not open the box. It’s as simple as that.

Now, the ‘dead’ and ‘alive’ condition are, obviously, the ‘base states’ in Schrödinger’s rather famous example, and we can write them as | DEAD 〉 and | ALIVE 〉 you’d agree it would be difficult to find another representation. For example, it doesn’t make much sense to say that we’ve rotated the two base states over 90 degrees and we now have two new states equal to (1/√2)·| DEAD 〉 – (1/√2)·| ALIVE 〉 and (1/√2)·| DEAD 〉 + (1/√2)·| ALIVE 〉 respectively. There’s no direction in space in regard to which we’re defining those two base states: dead is dead, and alive is alive.

The situation really resembles our ammonia molecule in free space: there’s no external reference against which to define the base states. However, as soon as some external field is involved, we do have a direction in space and, as mentioned above, our base states are now defined with respect to a particular orientation in space. That implies two things. The first is that we should no longer say that our molecule will always be in either state 1 or state 2. There’s no reason for it to be perfectly aligned with or against the field. Its orientation can be anything really, and so its state is likely to be some combination of those two pure base states | 1 〉 and | 2 〉.

The second thing is that we may choose another set of base states, and specify the very same state in terms of the new base states. So, assuming we choose some other set of base states | I 〉 and | II 〉, we can write the very same state | φ 〉 = | 1 〉 C₁ + | 2 〉 C₂as:

| φ 〉 = | I 〉 C_I + | II 〉 C_II

It’s really like what you learned about vectors in high school: one can go from one set of base vectors to another by a transformation, such as, for example, a rotation, or a translation. It’s just that, just like in high school, we need some direction in regard to which we define our rotation or our translation.

For state vectors, I showed how a rotation of base states worked in one of my posts on two-state systems. To be specific, we had the following relation between the two representations:

The (1/√2) factor is there because of the normalization condition, and the two-by-two matrix equals the transformation matrix for a rotation of a state filtering apparatus about the y-axis, over an angle equal to (minus) 90 degrees, which we wrote as:

The y-axis? What y-axis? What state filtering apparatus? Just relax. Think about what you’ve learned already. The orientations are shown below: the S apparatus separates ‘up’ and ‘down’ states along the z-axis, while the T-apparatus does so along an axis that is tilted, about the y-axis, over an angle equal to α, or φ, as it’s written in the table above.

Of course, we don’t really introduce an apparatus at this or that angle. We just introduced an electromagnetic field, which re-defined our | 1 〉 and | 2 〉 base states and, therefore, through the rotational transformation matrix, also defines our | I 〉 and | II 〉 base states.

[…] You may have lost me by now, and so then you’ll want to skip to the next section. That’s fine. Just remember that the representations in terms of | I 〉 and | II 〉 base states or in terms of | 1 〉 and | 2 〉 base states are mathematically equivalent. Having said that, if you’re reading this post, and you want to understand it, truly (because you want to truly understand quantum mechanics), then you should try to stick with me here. 🙂 Indeed, there’s a zillion things you could think about right now, but you should stick to the math now. Using that transformation matrix, we can relate the C_Iand C_IIcoefficients in the | φ 〉 = | I 〉 C_I + | II 〉 C_II expression to the C_Iand C_IIcoefficients in the | φ 〉 = | 1 〉 C₁ + | 2 〉 C₂ expression. Indeed, we wrote:

C_I= 〈 I | ψ 〉 = (1/√2)·(C₁− C₂)
C_II= 〈 II | ψ 〉 = (1/√2)·(C₁+ C₂)

That’s exactly the same as writing:

OK. […] Waw! You just took a huge leap, because we can now compare the two sets of differential equations:

They’re mathematically equivalent, but the mathematical behavior of the functions involved is very different. Indeed, unlike the C₁(t) and C₂(t) amplitudes, we find that the C_I(t) and C_II(t) amplitudes are stationary, i.e. the associated probabilities – which we find by taking the absolute square of the amplitudes, as usual – do not vary in time. To be precise, if you write it all out and simplify, you’ll find that the C_I(t) and C_II(t) amplitudes are equal to:

C_I(t) = 〈 I | ψ 〉 = (1/√2)·(C₁− C₂) = (1/√2)·e^{−(i/ħ)·(E₀+ A)·t} = (1/√2)·e^{−(i/ħ)·E_I·t}
C_II(t) = 〈 II | ψ 〉 = (1/√2)·(C₁+ C₂) = (1/√2)·e^{−(i/ħ)·(E₀− A)·t}= (1/√2)·e^{−(i/ħ)·E_II·t}

As the absolute square of the exponential is equal to one, the associated probabilities, i.e. |C_I(t)|² and |C_II(t)|², are, quite simply, equal to |1/√2|² = 1/2. Now, it is very tempting to say that this means that our ammonia molecule has an equal chance to be in state I or state II. In fact, while I may have said something like that in my previous posts, that’s not how one should interpret this. The chance of our molecule being exactly in state I or state II, or in state 1 or state 2 is varying with time, with the probability being ‘dumped’ from one state to the other all of the time.

I mean… The electric dipole moment can point in any direction, really. So saying that our molecule has a 50/50 chance of being in state 1 or state 2 makes no sense. Likewise, saying that our molecule has a 50/50 chance of being in state I or state II makes no sense either. Indeed, the state of our molecule is specified by the | φ 〉 = | I 〉 C_I + | II 〉 C_II= | 1 〉 C₁ + | 2 〉 C₂equations, and neither of these two expressions is a stationary state. They mix two frequencies, because they mix two energy levels.

Having said that, we’re talking quantum mechanics here and, therefore, an external inhomogeneous electric field will effectively split the ammonia molecules according to their state. The situation is really like what a Stern-Gerlach apparatus does to a beam of electrons: it will split the beam according to the electron’s spin, which is either ‘up’ or, else, ‘down’, as shown in the graph below:

The graph for our ammonia molecule, shown below, is very similar. The vertical axis measures the same: energy. And the horizontal axis measures με, which increases with the strength of the electric field ε. So we see a similar ‘splitting’ of the energy of the molecule in an external electric field.

How should we explain this? It is very tempting to think that the presence of an external force field causes the electrons, or the ammonia molecule, to ‘snap into’ one of the two possible states, which are referred to as state I and state II respectively in the illustration of the ammonia state selector below. But… Well… Here we’re entering the murky waters of actually interpreting quantum mechanics, for which (a) we have no time, and (b) we are not qualified. So you should just believe, or take for granted, what’s being shown here: an inhomogeneous electric field will split our ammonia beam according to their state, which we define as I and II respectively, and which are associated with the energy E₀+ A and E₀− A respectively.

electric field

As mentioned above, you should note that these two states are stationary. The Hamiltonian equations which, as they always do, describe the dynamics of this system, imply that the amplitude to go from state I to state II, or vice versa, is zero. To make sure you ‘get’ that, I reproduce the associated Hamiltonian matrix once again:

Of course, that will change when we start our analysis of what’s happening in the maser. Indeed, we will have some non-zero H_I,II and H_II,Iamplitudes in the resonant cavity of our ammonia maser, in which we’ll have an oscillating electric field and, as a result, induced transitions from state I to II and vice versa. However, that’s for later. While I’ll quickly insert the full picture diagram below, you should, for the moment, just think about those two stationary states and those two zeroes. 🙂

Capito? If not… Well… Start reading this post again, I’d say. 🙂

Intermezzo: on approximations

At this point, I need to say a few things about all of the approximations involved, because it can be quite confusing indeed. So let’s take a closer look at those energy levels and the related Hamiltonian coefficients. In fact, in his Lectures, Feynman shows us that we can always have a general solution for the Hamiltonian equations describing a two-state system whenever we have constant Hamiltonian coefficients. That general solution – which, mind you, is derived assuming Hamiltonian coefficients that do not depend on time – can always be written in terms of two stationary base states, i.e. states with a definite energy and, hence, a constant probability. The equations, and the two definite energy levels are:

That yields the following values for the energy levels for the stationary states:

Now, that’s very different from the E_I= E₀+ A and E_II= E₀− A energy levels for those stationary states we had defined in the previous section: those stationary states had no square root, and no μ²ε², in their energy. In fact, that sort of answers the question: if there’s no external field, then that μ²ε² factor is zero, and the square root in the expression becomes ±√A²= ±A. So then we’re back to our E_I= E₀+ A and E_II= E₀− A formulas. The whole point, however, is that we will actually have an electric field in that cavity. Moreover, it’s going to be a field that varies in time, which we’ll write:

Now, part of the confusion in Feynman’s approach is that he constantly switches between representing the system in terms of the I and II base states and the 1 and 2 base states respectively. For a good understanding, we should compare with our original representation of the dynamics in free space, for which the Hamiltonian was the following one:

That matrix can easily be related to the new one we’re going to have to solve, which is equal to:

The interpretation is easy if we look at that illustration again:

If the direction of the electric dipole moment is opposite to the direction ε, then the associated energy is equal to −μ·ε = −μ·ε = −|μ|·|ε|·cosθ = −μ·ε·cos(π) = +με. Conversely, for state 2, we find −μ·ε·cos(0) = −με for the energy that’s associated with the dipole moment. You can and should think about the physics involved here, because they make sense! Thinking of amplitudes, you should note that the +με and −με terms effectively change the H₁₁ and H₂₂ coefficients, so they change the amplitude to stay in state 1 or state 2 respectively. That, of course, will have an impact on the associated probabilities, and so that’s why we’re talking of induced transitions now.

Having said that, the Hamiltonian matrix above keeps the −A for H₁₂ and H₂₁, so the matrix captures spontaneous transitions too!

Still… You may wonder why Feynman doesn’t use those E_Iand E_IIformulas with the square root because that would give us some exact solution, wouldn’t it? The answer to that question is: maybe it would, but would you know how to solve those equations? We’ll have a varying field, remember? So our Hamiltonian H₁₁ and H₂₂ coefficients will no longer be constant, but time-dependent. As you’re going to see, it takes Feynman three pages to solve the whole thing using the +με and −με approximation. So just imagine how complicated it would be using that square root expression! [By the way, do have a look at those asymptotic curves in that illustration showing the splitting of energy levels above, so you see how that approximation looks like.]

So that’s the real answer: we need to simplify somehow, so as to get any solutions at all!

Of course, it’s all quite confusing because, after Feynman first notes that, for strong fields, the A² in that square root is small as compared to μ²ε², thereby justifying the use of the simplified E_I= E₀+ με = H₁₁ and E_II= E₀− με = H₂₂ coefficients, he continues and bluntly uses the very same square root expression to explain how that state selector works, saying that the electric field in the state selector will be rather weak and, hence, that με will be much smaller than A, so one can use the following approximation for the square root in the expressions above:

The energy expressions then reduce to:

And then we can calculate the force on the molecules as:

So the electric field in the state selector is weak, but the electric field in the cavity is supposed to be strong, and so… Well… That’s it, really. The bottom line is that we’ve a beam of ammonia molecules that are all in state I, and it’s what happens with that beam then, that is being described by our new set of differential equations:

Solving the equations

As all molecules in our ammonia beam are described in terms of the | I 〉 and | II 〉 base states – as evidenced by the fact that we say all molecules that enter the cavity are state I – we need to switch to that representation. We do that by using that transformation above, so we write:

C_I= 〈 I | ψ 〉 = (1/√2)·(C₁− C₂)
C_II= 〈 II | ψ 〉 = (1/√2)·(C₁+ C₂)

Keeping these ‘definitions’ of C_Iand C_IIin mind, you should then add the two differential equations, divide the result by the square root of 2, and you should get the following new equation:

Please! Do it and verify the result! You want to learn something here, no? 🙂

Likewise, subtracting the two differential equations, we get:

We can re-write this as:

Now, the problem is that the Hamiltonian constants here are not constant. To be precise, the electric field ε varies in time. We wrote:

So H_I,IIand H_II,I, which are equal to με, are not constant: we’ve got Hamiltonian coefficients that are a function of time themselves. […] So… Well… We just need to get on with it and try to finally solve this thing. Let me just copy Feynman as he grinds through this:

This is only the first step in the process. Feynman just takes two trial functions, which are really similar to the very general C₁= a·e^{–(i/ħ)·H₁₁·t}function we presented when only one equation was involved, or – if you prefer a set of two equations – those C_I(t) = a·e^{−(i/ħ)·E_I·t}and C_I(t) = b·e^{−(i/ħ)·E_II·}^tequations above. The difference is that the coefficients in front, i.e. γ_I and γ_II are not some (complex) constant, but functions of time themselves. The next step in the derivation is as follows:

One needs to do a bit of gymnastics here as well to follow what’s going on, but please do check and you’ll see it works. Feynman derives another set of differential equations here, and they specify these γ_I = γ_I(t) and γ_II = γ_II(t) functions. These equations are written in terms of the frequency of the field, i.e. ω, and the resonant frequency ω₀, which we mentioned above when calculating that 23.79 GHz frequency from the 2A = h·f₀ equation. So ω₀ is the same molecular resonance frequency but expressed as an angular frequency, so ω₀ = f₀/2π = ħ/2A. He then proceeds to simplify, using assumptions one should check. He then continues:

That gives us what we presented in the previous post:

So… Well… What to say? I explained those probability functions in my previous post, indeed. We’ve got two probabilities here:

P_I= cos²[(με₀/ħ)·t]
P_II= sin²[(με₀/ħ)·t]

So that’s just like the P₁= cos²[(A/ħ)·t] and P₂= sin²[(A/ħ)·t] probabilities we found for spontaneous transitions. But so here we are talking induced transitions.

As you can see, the frequency and, hence, the period, depend on the strength, or magnitude, of the electric field, i.e. the ε₀constant in the ε = 2ε₀cos(ω·t) expression. The natural unit for measuring time would be the period once again, which we can easily calculate as (με₀/ħ)·T = π ⇔ T = π·ħ/με₀.

Now, we had that T = (π·ħ)/(2A) expression above, which allowed us to calculate the period of the spontaneous transition frequency, which we found was like 40 picoseconds, i.e. 40×10⁻¹²seconds. Now, the T = (π·ħ)/(2με₀) is very similar, it allows us to calculate the expected, average, or mean time for an induced transition. In fact, if we write T_induced = (π·ħ)/(2με₀) and T_spontaneous = (π·ħ)/(2A), then we can take ratio to find:

T_induced/T_spontaneous = [(π·ħ)/(2με₀)]/[(π·ħ)/(2A)] = A/με₀

This A/με₀ratio is greater than one, so T_induced/T_spontaneous is greater than one, which, in turn, means that the presence of our electric field – which, let me remind you, dances to the beat of the resonant frequency – causes a slower transition than we would have had if the oscillating electric field were not present.

But – Hey! – that’s the wrong comparison! Remember all molecules enter in a stationary state, as they’ve been selected so as to ensure they’re in state I. So there is no such thing as a spontaneous transition frequency here! They’re all polarized, so to speak, and they would remain that way if there was no field in the cavity. So if there was no oscillating electric field, they would never transition. Nothing would happen! Well… In terms of our particular set of base states, of course! Why? Well… Look at the Hamiltonian coefficients H_I,II= H_II,I= με: these coefficients are zero if ε is zero. So… Well… That says it all.

So that‘s what it’s all about: induced emission and, as I explained in my previous post, because all molecules enter in state I, i.e. the upper energy state, literally, they all ‘dump’ a net amount of energy equal to 2A into the cavity at the occasion of their first transition. The molecules then keep dancing, of course, and so they absorb and emit the same amount as they go through the cavity, but… Well… We’ve got a net contribution here, which is not only enough to maintain the cavity oscillations, but actually also provides a small excess of power that can be drawn from the cavity as microwave radiation of the same frequency.

As Feynman notes, an exact description of what actually happens requires an understanding of the quantum mechanics of the field in the cavity, i.e. quantum field theory, which I haven’t studied yet. But… Well… That’s for later, I guess. 🙂

Post scriptum: The sheer length of this post shows we’re not doing something that’s easy here. Frankly, I feel the whole analysis is still quite obscure, in the sense that – despite looking at this thing again and again – it’s hard to sort of interpret what’s going on, in a physical sense that is. But perhaps one shouldn’t try that. I’ve quoted Feynman’s view on how easy or how difficult it is to ‘understand’ quantum mechanics a couple of times already, so let me do it once more:

So… Well… I’ll grind through the remaining Lectures now – I am halfway through Volume III now – and then re-visit all of this. Despite Feynman’s warning, I want to understand it the way I like to, even if I don’t quite know what way that is right now. 🙂

Addendum: As for those cycles and periods, I noted a couple of times already that the Planck-Einstein equation E = h·f can usefully be re-written as E/f = h, as it gives a physical interpretation to the value of the Planck constant. In fact, I said h is the energy that’s associated with one cycle, regardless of the frequency of the radiation involved. Indeed, the energy of a photon divided by the number of cycles per second, should give us the energy per cycle, no?

Well… Yes and no. Planck’s constant h and the frequency f are both expressed referencing the time unit. However, if we say that a sodium atom emits one photon only as its electron transitions from a higher energy level to a lower one, and if we say that involves a decay time of the order of 3.2×10⁻⁸seconds, then what we’re saying really is that a sodium light photon will ‘pack’ like 16 million cycles, which is what we get when we multiply the number of cycles per second (i.e. the mentioned frequency of 500×10¹²Hz) by the decay time (i.e. 3.2×10⁻⁸seconds): (500×10¹²Hz)·(3.2×10⁻⁸s) = 16 ×10⁶cycles, indeed. So the energy per cycle is 2.068 eV (i.e. the photon energy) divided by 16×10⁶, so that’s 0.129×10⁻⁶eV. Unsurprisingly, that’s what we get when we we divide h by 3.2×10⁻⁸s: (4.13567×10⁻¹⁵)/(3.2×10⁻⁸s) = 1.29×10⁻⁷eV. We’re just putting some values in to the E/(f·T) = h/T equation here.

The logic for that 2A = h·f₀ is the same. The frequency of the radiation that’s being absorbed or emitted is 23.79 GHz, so the photon energy is (23.97×10⁹ Hz)·(4.13567×10⁻¹⁵ eV·s) ≈ 1×10⁻⁴eV. Now, we calculated the transition period T as T = π·ħ/A ≈ (π·6.626×10⁻¹⁶eV·s)/(0.5×10⁻⁴eV) ≈ 41.6×10⁻¹²seconds. Now, an oscillation of a frequency of 23.97 giga-hertz that only lasts 41.6×10⁻¹²seconds is an oscillation of one cycle only. The consequence is that, when we continue this style of reasoning, we’d have a photon that packs all of its energy into one cycle!

Let’s think about what this implies in terms of the density in space. The wavelength of our microwave radiation is 1.25×10⁻²m, so we’ve got a ‘density’ of 1×10⁻⁴eV/1.25×10⁻²m = 0.8×10⁻²eV/m = 0.008 eV/m. The wavelength of our sodium light is 0.6×10⁻⁶m, so we get a ‘density’ of 1.29×10⁻⁷eV/0.6×10⁻⁶m = 2.15×10⁻¹eV/m = 0.215 eV/m. So the energy ‘density’ of our sodium light is 26.875 times that of our microwave radiation. 🙂

Frankly, I am not quite sure if calculations like this make much sense. In fact, when talking about energy densities, I should review my posts on the Poynting vector. However, they may help you think things through. 🙂

Quantum math revisited

It’s probably good to review the concepts we’ve learned so far. Let’s start with the foundation of all of our math, i.e. the concept of the state, or the state vector. [The difference between the two concepts is subtle but real. I’ll come back to it.]

State vectors and base states

We used Dirac’s bra-ket notation to denote a state vector, in general, as | ψ 〉. The obvious question is: what is this thing? We called it a vector because we use it like a vector: we multiply it with some number, and then add it to some other vector. So that’s just what you did in high school, when you learned about real vector spaces. In this regard, it is good to remind you of the definition of a vector space. To put it simply, it is is a collection of objects called vectors, which may be added together, and multiplied by numbers. So we have two things here: the ‘objects’, and the ‘numbers’. That’s why we’d say that we have some vector space over a field of numbers. [The term ‘field’ just refers to an algebraic structure, so we can add and multiply and what have you.] Of course, what it means to ‘add’ two ‘objects’, and what it means to ‘multiply’ an object with a number, depends on the type of objects and, unsurprisingly, the type of numbers.

Huh? The type of number?! A number is a number, no?

No, hombre, no! We’ve got natural numbers, rational numbers, real numbers, complex numbers—and you’ve probably heard of quaternions too – and, hence, ‘multiplying’ a ‘number’ with ‘something else’ can mean very different things. At the same time, the general idea is the general idea, so that’s the same, indeed. 🙂 When using real numbers and the kind of vectors you are used to (i.e. Euclidean vectors), then the multiplication amounts to a re-scaling of the vector, and so that’s why a real number is often referred to as a scalar. At the same time, anything that can be used to multiply a vector is often referred to as a scalar in math so… Well… Terminology is often quite confusing. In fact, I’ll give you some more examples of confusing terminology in a moment. But let’s first look at our ‘objects’ here, i.e. our ‘vectors’.

I did a post on Euclidean and non-Euclidean vector spaces two years ago, when I started this blog, but state vectors are obviously very different ‘objects’. They don’t resemble the vectors we’re used to. We’re used to so-called polar vectors, aka as real vectors, like the position vector (x or r), or the momentum vector (p = m·v), or the electric field vector (E). We are also familiar with the so-called pseudo-vectors, aka as axial vectors, like angular momentum (L = r×p), or the magnetic dipole moment. [Unlike what you might think, not all vector cross products yield a pseudo-vector. For example, the cross-product of a polar and an axial vector yields a polar vector.] But here we are talking some very different ‘object’. In math, we say that state vectors are elements in a Hilbert space. So a Hilbert space is a vector space but… Well… With special vectors. 🙂

The key to understanding why we’d refer to states as state vectors is the fact that, just like Euclidean vectors, we can uniquely specify any element in a Hilbert space with respect to a set of base states. So it’s really like using Cartesian coordinates in a two- or three-dimensional Euclidean space. The analogy is complete because, even in the absence of a geometrical interpretation, we’ll require those base states to be orthonormal. Let me be explicit on that by reminding you of your high-school classes on vector analysis: you’d choose a set of orthonormal base vectors e₁, e₂, and e₃, and you’d write any vector A as:

A = (A_x, A_y, A_z) = A_x·e₁ + A_y·e₂+ A_z·e₃ with e_i·e_j = 1 if i = j, and e_i·e_j = 0 if i ≠ j

The e_i·e_j = 1 if i = j and e_i·e_j = 0 if i ≠ j condition expresses the orthonormality condition: the base vectors need to be orthogonal unit vectors. We wrote it as e_i·e_j = δ_ijusing the Kronecker delta (δ_ij= 1 if i = j and 0 if i ≠ j). Now, base states in quantum mechanics do not necessarily have a geometrical interpretation. Indeed, although one often can actually associate them with some position or direction in space, the condition of orthonormality applies in the mathematical sense of the word only. Denoting the base states by i = 1, 2,… – or by Roman numerals, like I and II – so as to distinguish them from the Greek ψ or φ symbols we use to denote state vectors in general, we write the orthonormality condition as follows:

〈 i | j 〉 = δ_ij, with δ_ij= δ_jiis equal to 1 if i = j, and zero if i ≠ j

Now, you may grumble and say: that 〈 i | j 〉 bra-ket does not resemble the e_i·e_j product. Well… It does and it doesn’t. I’ll show why in a moment. First note how we uniquely specify state vectors in general in terms of a set of base states. For example, if we have two possible base states only, we’ll write:

| φ 〉 = | 1 〉 C₁ + | 2 〉 C₂

Or, if we chose some other set of base states | I 〉 and | II 〉, we’ll write:

| φ 〉 = | I 〉 C_I + | II 〉 C_II

You should note that the | 1 〉 C₁ term in the | φ 〉 = | 1 〉 C₁ + | 2 〉 C₂ sum is really like the A_x·e₁ product in the A = A_x·e₁ + A_y·e₂+ A_z·e₃ expression. In fact, you may actually write it as C₁·| 1 〉, or just reverse the order and write C₁| 1 〉. However, that’s not common practice and so I won’t do that, except occasionally. So you should look at | 1 〉 C₁ as a product indeed: it’s the product of a base state and a complex number, so it’s really like m·v, or whatever other product of some scalar and some vector, except that we’ve got a complex scalar here. […] Yes, I know the term ‘complex scalar’ doesn’t make sense, but I hope you know what I mean. 🙂

More generally, we write:

Writing our state vector | ψ 〉, | φ 〉 or | χ 〉 like this also defines these coefficients or coordinates C_i. Unlike our state vectors, or our base states, C_iis an actual number. It has to be, of course: it’s the complex number that makes sense of the whole expression. To be precise, C_iis an amplitude, or a wavefunction, i.e. a function depending on both space and time. In our previous posts, we limited the analysis to amplitudes varying in time only, and we’ll continue to do so for a while. However, at some point, you’ll get the full picture.

Now, what about the supposed similarity between the 〈 i | j〉 bra-ket and the e_i·e_j product? Let me invoke what Feynman, tongue-in-cheek as usual, refers to as the Great Law of Quantum Mechanics:

You get this by taking | ψ 〉 out of the | ψ 〉 = ∑| i 〉〈 i | ψ 〉 expression. And, no, don’t say: what nonsense! Because… Well… Dirac’s notation really is that simple and powerful! You just have to read it from right to left. There’s an order to the symbols, unlike what you’re used to in math, because you’re used to operations that are commutative. But I need to move on. The upshot is that we can specify our base states in terms of the base states too. For example, if we have only two base states, let’s say I and II, then we can write:

| I 〉 = ∑| i 〉〈 i | 1 〉 = 1·| I 〉 + 0·| II 〉 and | II 〉 = ∑| i 〉〈 i | II 〉 = 0·| 1 〉 + 0·| II 〉

We can write this using a matrix notation:

Now that is silly, you’ll say. What’s the use of this? It doesn’t tell us anything new, and it also does not show us why we should think of the 〈 i | j 〉 bra-ket and the e_i·e_j product as being similar! Well… Yes and no. Let me show you something else. Let’s assume we’ve got some state χ and φ, which we specify in terms of our chosen set of base states as | χ 〉 = ∑ | i 〉 D_i and | φ 〉 = ∑ | i 〉 C_i respectively. Now, from our post on quantum math, you’ll remember that 〈 χ | i 〉 and 〈 i | χ 〉 are each other’s complex conjugates, so we know that 〈 χ | i 〉 = 〈 i | χ 〉* = D_i*. So if we have all C_i = 〈 i | φ 〉 and all D_i = 〈 i | χ 〉, i.e. the ‘components’ of both states in terms of our base states, then we can calculate 〈 χ | φ 〉 – i.e. the amplitude to go from some state φ to some state χ as:

〈 χ | φ 〉 = ∑〈 χ | i 〉〈 i | φ 〉 =∑ D_i*C_i= ∑ D_i*〈 i | φ 〉

We can now scrap | φ 〉 in this expression – yes, it’s the power of Dirac’s notation once more! – so we get:

Now, we can re-write this using a matrix notation:

[I assumed that we have three base states now, so as to make the example somewhat less obvious. Please note we can never leave one of the base states out when specifying a state vector, so it’s not like the previous example was not complete. I’ll switch from two-state to three-state systems and back again all the time, so as to show the analysis is pretty general. To visualize things, think of the ammonia molecule as an example of a two-state system versus the spin of a proton or an electron as a three-state system, respectively. OK. Let’s get back to the lesson.]

You’ll say: so what? Well… Look at this:

I just combined the notations for 〈 I | and | III 〉. Can you now see the similarity between the the 〈 i | j〉 bra-ket and the e_i·e_j product? It really is the same: you just need to respect the subtleties in regard to writing the 〈 i | and | j 〉 vector, or the e_iand e_j vectors, as a row vector or a column vector respectively.

It doesn’t stop here, of course. When learning about vectors in high school, we also learned that we could go from one set of base vectors to another by a transformation, such as, for example, a rotation, or a translation. We showed how a rotation worked in one of our post on two-state systems, where we wrote:

So we’ve got that transformation matrix, which, of course, isn’t random. To be precise, we got the matrix equation above (note that we’re back to two states only, so as to simplify) because we defined the C_Iand C_IIcoefficients in the | φ 〉 = | I 〉 C_I + | II 〉 C_II = | 1 〉 C₁ + | 2 〉 C₂ expression as follows:

C_I= 〈 I | ψ 〉 = (1/√2)·(C₁− C₂)
C_II= 〈 II | ψ 〉 = (1/√2)·(C₁+ C₂)

The (1/√2) factor is there because of the normalization condition, and the two-by-two matrix equals the transformation matrix for a rotation of a state filtering apparatus about the y-axis, over an angle equal to (minus) 90 degrees, which we wrote as:

I promised I’d say something more about confusing terminology so let me do that here. We call a set of base states a ‘representation‘, and writing a state vector in terms of a set of base states is often referred to as a ‘projection‘ of that state into the base set. Again, we can see it’s sort of a mathematical projection, rather than a geometrical one. But it makes sense. In any case, that’s enough on state vectors and base states.

Let me wrap it up by inserting one more matrix equation, which you should be able to reconstruct yourself:

The only thing we’re doing here is to substitute 〈 χ | and | φ 〉 for ∑ D_j*〈 j | and ∑ | i 〉 C_i respectively. All the rest follows. Finally, I promised I’d tell you the difference between a state and a state vector. It’s subtle and, in practice, the two concepts refer to the same. However, we write a state as a state, like ψ or, if it’s a base state, like I, or ‘up’, or whatever. When we say a state vector, then we think of a set of numbers. It may be a row vector, like the 〈 χ | row vector with the D_i* coefficients, or a column vector, like the | φ 〉 column vector with the D_i* coefficients. But so if we say vector, then we think of a one-dimensional array of numbers, while the state itself is… Well… The state. So that’s some reality in physics. So you might define the state vector as the set of numbers that describes the state. While the difference is subtle, it’s important. It’s also important to note that the 〈 χ | and | χ 〉 state vectors are different too. The former appears as the final state in an amplitude, while the latter describes the starting condition. The former is referred to as a bra in the 〈 χ | φ 〉 bra-ket, while the latter is a ket in the 〈 φ | χ 〉 = 〈 χ | φ 〉* amplitude. 〈 χ | is a row vector equal to ∑ D_i*〈 i |, while | χ 〉 = ∑ D_i| χ 〉. So it’s quite different. More in general, we’d define bras and kets as row and column vectors respectively, so we write:

That makes it clear that a bra next to a ket is to be understood as a matrix multiplication. From what I wrote, it is also obvious that the conjugate transpose (which is also known as the Hermitian conjugate) of a bra is the corresponding ket and vice versa, so we write:

Let me formally define the conjugate or Hermitian transpose here: the conjugate transpose of an m-by-n matrix $A$ $with complex elements is the$ n-by-m matrix A† obtained from $A$ by taking the transpose (so we write the rows as columns and vice versa) and then taking the complex conjugate of each element (i.e. we switch the sign of the imaginary part of the complex number). A† is read as ‘A dagger’, but mathematicians will usually denote it by $A*$ . In fact, there are a lot of equivalent notations, as we can write:

OK. That’s it on this.

One more thing, perhaps. We’ll often have states, or base states, that make sense, in a physical sense, that is. But it’s not always the case: we’ll sometimes use base states that may not represent some situation we’re likely to encounter, but that make sense mathematically. We gave the example of the ‘mathematical’ | I 〉 and | II 〉 base states, versus the ‘physical’ | 1 〉 and | 2 〉 base states, in our post on the ammonia molecule, so I won’t say more about this here. Do keep it in mind though. Sometimes it may feel like nothing makes sense, physically, but it usually does mathematically and, therefore, all usually comes out alright in the end. 🙂 To be precise, what we did there, was to choose base states with a unambiguous, i.e. a definite, energy level. That made our calculations much easier, and the end result was the same, indeed!

So… Well… I’ll let this sink in, and move on to the next topic.

The Hamiltonian operator

In my post on the post on the Hamiltonian, I explained that those C_i and D_i coefficients are usually a function of time, and how they can be determined. To be precise, they’re determined by a set of differential equations (i.e. equations involving a function and the derivative of that function) which we wrote as:

If we have two base states only, then this set of equations can be written as:

Two equations and two functions – C₁= C₁(t) and C₂= C₂(t) – so we should be able to solve this thing, right? Well… No. We don’t know those H_ijcoefficients. As I explained in that post, they also evolve in time, so we should write them as H_ij(t) instead of H_ijtout court, and so it messes the whole thing up. We have two equations and six functions really. Of course, there’s always a way out, but I won’t dwell on that here—not now at least. What I want to do here is look at the Hamiltonian as an operator.

We introduced operators – but not very rigorously – when explaining the Hamiltonian. We did so by ‘expanding’ our 〈 χ | φ 〉 amplitude as follows. We’d say the amplitude to find a ‘thing’ – like a particle, for example, or some system, of particles or other things – in some state χ at the time t = t₂, when it was in some state φ state at the time t = t₁ was equal to:

Now, a formula like this only makes sense because we’re ‘abstracting away’ from the base states, which we need to describe any state. Hence, to actually describe what’s going on, we have to choose some representation and expand this expression as follows:

That looks pretty monstrous, so we should write it all out. Using the matrix notation I introduced above, we can do that – let’s take a practical example with three base states once again – as follows:

Now, this still looks pretty monstrous, but just think of it. We’re just applying that ‘Great Law of Quantum Physics’ here, i.e. | = ∑ | i 〉〈 i | over all base states i. To be precise, we apply it to an 〈 χ | A | φ 〉 expression, and we do so twice, so we get:

Nothing more, nothing less. 🙂 Now, the idea of an operator is the result of being creative: we just drop the 〈 χ | state from the expression above to write:

Yes. I know. That’s a lot to swallow, but you’ll see it makes sense because of the Great Law of Quantum Mechanics:

Just think about it and continue reading when you’re ready. 🙂 The upshot is: we now think of the particle entering some ‘apparatus’ A in the state ϕ and coming out of A in some state ψ or, looking at A as an operator, we can generalize this. As Feynman puts it:

“The symbol A is neither an amplitude, nor a vector; it is a new kind of thing called an operator. It is something which “operates on” a state to produce a new state.”

Back to our Hamiltonian. Let’s go through the same process of ‘abstraction’. Let’s first re-write that ‘Hamiltonian equation’ as follows:

The H_ij(t) are amplitudes indeed, and we can represent them in a 〈 i | H_ij(t) | j 〉 matrix indeed! Now let’s take the first step in our ‘abstraction process’: let’s scrap the 〈 i | bit. We get:

We can, of course, also abstract away from the | j 〉 bit, so we get:

Look at this! The right-hand side of this expression is exactly the same as that A | χ 〉 format we presented when introducing the concept of an operator. [In fact, when I say you should ‘abstract away’ from the | j 〉 bit, then you should think of the ‘Great Law’ and that matrix notation above.] So H is an operator and, therefore, it’s something which operates on a state to produce a new state.

OK. Clear enough. But what’s that ‘state’ on the left-hand side? I’ll just paraphrase Feynman here, who says we should think of it as follows: “The time derivative of the state vector |ψ〉 times iℏ is equal to what you get by (1) operating with the Hamiltonian operator H on each base state, (2) multiplying by the amplitude that ψ is in the state j (i.e. 〈j|ψ〉), and (1) summing over all j.” Alternatively, you can also say: “The time derivative, times iħ, of a state |ψ〉 is equal to what you get if you operate on it with the Hamiltonian.” Of course, that’s true for any state, so we can ‘abstract away’ the |ψ〉 bit too and, putting a little hat (^) over the operator to remind ourselves that it’s an operator (rather than just any matrix), we get the Hamiltonian operator equation:

Now, that’s all nice and great, but the key question, of course, is: what can you do with this? Well… It turns out his Hamiltonian operators is useful to calculate lots of stuff. In the first place, of course, it’s a useful operator in the context of those differential equations describing the dynamics of a quantum-mechanical system. When everything is said and done, those equations are the equivalent, in quantum physics, of the law of motion in classical physics. [And I am not joking here.]

In addition, the Hamiltonian operator also has other uses. The one I should really mention here is that you can calculate the average or expected value (EV[X]) of the energy of a state ψ (i.e. any state, really) by first operating on | ψ 〉 with the Hamiltonian, and then multiplying 〈 ψ | with the result. That sounds a bit complicated, but you’ll understand it when seeing the mathematical expression, which we can write as:

The formula is pretty straightforward. [If you don’t think so, then just write it all out using the matrix notation.] But you may wonder how it works exactly… Well… Sorry. I don’t want to copy all of Feynman here, so I’ll refer you to him on this. In fact, the proof of this formula is actually very straightforward, and so you should be able to get through it with the math you got here. You may even understand Feynman’s illustration of it for the ‘special case’ when base states are, indeed, those mathematically convenient base states with definite energy levels.

Have fun with it! 🙂

Post scriptum on Hilbert spaces:

As mentioned above, our state vectors are actually functions. To be specific, they are wavefunctions, i.e. periodic functions, evolving in space and time, so we usually write them as ψ = ψ(x, t). Our ‘Hilbert space’, i.e. our collection of state vectors, is, therefore, often referred to as a function space. So it’s a set of functions. At the same time, it is a vector space too, because we have those addition and multiplication operations, so our function space has the algebraic structure of a vector space. As you can imagine, there are some mathematical conditions for a space or a set of objects to ‘qualify’ as a Hilbert space, and the epithet itself comes with a lot of interesting properties. One of them is completeness, which is a property that allows us to jot down those differential equations that describe the dynamics of a quantum-mechanical system. However, as you can find whatever you’d need or want to know about those mathematical properties on the Web, I won’t get into it. The important thing here is to understand the concept of a Hilbert space intuitively. I hope this post has helped you in that regard, at least. 🙂

Occam’s Razor

The analysis of a two-state system (i.e. the rather famous example of an ammonia molecule ‘flipping’ its spin direction from ‘up’ to ‘down’, or vice versa) in my previous post is a good opportunity to think about Occam’s Razor once more. What are we doing? What does the math tell us?

In the example we chose, we didn’t need to worry about space. It was all about time: an evolving state over time. We also knew the answers we wanted to get: if there is some probability for the system to ‘flip’ from one state to another, we know it will, at some point in time. We also want probabilities to add up to one, so we knew the graph below had to be the result we would find: if our molecule can be in two states only, and it starts of in one, then the probability that it will remain in that state will gradually decline, while the probability that it flips into the other state will gradually increase, which is what is depicted below.

However, the graph above is only a Platonic idea: we don’t bother to actually verify what state the molecule is in. If we did, we’d have to ‘re-set’ our t = 0 point, and start all over again. The wavefunction would collapse, as they say, because we’ve made a measurement. However, having said that, yes, in the physicist’s Platonic world of ideas, the probability functions above make perfect sense. They are beautiful. You should note, for example, that P₁ (i.e. the probability to be in state 1) and P₂ (i.e. the probability to be in state 2) add up to 1 all of the time, so we don’t need to integrate over a cycle or something: so it’s all perfect!

These probability functions are based on ideas that are even more Platonic: interfering amplitudes. Let me explain.

Quantum physics is based on the idea that these probabilities are determined by some wavefunction, a complex-valued amplitude that varies in time and space. It’s a two-dimensional thing, and then it’s not. It’s two-dimensional because it combines a sine and cosine, i.e. a real and an imaginary part, but the argument of the sine and the cosine is the same, and the sine and cosine are the same function, except for a phase shift equal to π. We write:

a·e^−iθ= a·cos(θ) – a·sin(−θ) = a·cosθ – a·sinθ

The minus sign is there because it turns out that Nature measures angles, i.e. our phase, clockwise, rather than counterclockwise, so that’s not as per our mathematical convention. But that’s a minor detail, really. [It should give you some food for thought, though.] For the rest, the related graph is as simple as the formula:

Now, the phase of this wavefunction is written as θ = (ω·t − k ∙x). Hence, ω determines how this wavefunction varies in time, and the wavevector k tells us how this wave varies in space. The young Frenchman Comte Louis de Broglie noted the mathematical similarity between the ω·t − k ∙x expression and Einstein’s four-vector product p_μx_μ= E·t − p∙x, which remains invariant under a Lorentz transformation. He also understood that the Planck-Einstein relation E = ħ·ω actually defines the energy unit and, therefore, that any frequency, any oscillation really, in space or in time, is to be expressed in terms of ħ.

[To be precise, the fundamental quantum of energy is h = ħ·2π, because that’s the energy of one cycle. To illustrate the point, think of the Planck-Einstein relation. It gives us the energy of a photon with frequency f: E_γ = h·f. If we re-write this equation as E_γ/f = h, and we do a dimensional analysis, we get: h = E_γ/f ⇔ 6.626×10⁻³⁴ joule·second = [x joule]/[f cycles per second] ⇔ h = 6.626×10⁻³⁴ joule per cycle. It’s only because we are expressing ω and k as angular frequencies (i.e. in radians per second or per meter, rather than in cycles per second or per meter) that we have to think of ħ = h/2π rather than h.]

Louis de Broglie connected the dots between some other equations too. He was fully familiar with the equations determining the phase and group velocity of composite waves, or a wavetrain that actually might represent a wavicle traveling through spacetime. In short, he boldly equated ω with ω = E/ħ and k with k = p/ħ, and all came out alright. It made perfect sense!

I’ve written enough about this. What I want to write about here is how this also makes for the situation on hand: a simple two-state system that depends on time only. So its phase is θ = ω·t = E₀/ħ. What’s E₀? It is the total energy of the system, including the equivalent energy of the particle’s rest mass and any potential energy that may be there because of the presence of one or the other force field. What about kinetic energy? Well… We said it: in this case, there is no translational or linear momentum, so p = 0. So our Platonic wavefunction reduces to:

a·e^−iθ= ae⁻⁽^{i/ħ)·(E₀·t)}

Great! […] But… Well… No! The problem with this wavefunction is that it yields a constant probability. To be precise, when we take the absolute square of this wavefunction – which is what we do when calculating a probability from a wavefunction − we get P = a², always. The ‘normalization’ condition (so that’s the condition that probabilities have to add up to one) implies that P₁ = P₂ = a² = 1/2. Makes sense, you’ll say, but the problem is that this doesn’t reflect reality: these probabilities do not evolve over time and, hence, our ammonia molecule never ‘flips’ its spin direction from ‘up’ to ‘down’, or vice versa. In short, our wavefunction does not explain reality.

The problem is not unlike the problem we’d had with a similar function relating the momentum and the position of a particle. You’ll remember it: we wrote it as a·e^−iθ= ae⁽^{i/ħ)·(p·x)}. [Note that we can write a·e^−iθ= a·e⁻⁽^{i/ħ)·(E₀·t − p·x)}= a·e⁻⁽ⁱ^/ħ)·(E^₀·t)·e⁽^{i/ħ)·(p·x)}, so we can always split our wavefunction in a ‘time’ and a ‘space’ part.] But then we found that this wavefunction also yielded a constant and equal probability all over space, which implies our particle is everywhere (and, therefore, nowhere, really).

In quantum physics, this problem is solved by introducing uncertainty. Introducing some uncertainty about the energy, or about the momentum, is mathematically equivalent to saying that we’re actually looking at a composite wave, i.e. the sum of a finite or infinite set of component waves. So we have the same ω = E/ħ and k = p/ħ relations, but we apply them to n energy levels, or to some continuous range of energy levels ΔE. It amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ.

We know what that does: it ensures our wavefunction is being ‘contained’ in some ‘envelope’. It becomes a wavetrain, or a kind of beat note, as illustrated below:

[The animation also shows the difference between the group and phase velocity: the green dot shows the group velocity, while the red dot travels at the phase velocity.]

This begs the following question: what’s the uncertainty really? Is it an uncertainty in the energy, or is it an uncertainty in the wavefunction? I mean: we have a function relating the energy to a frequency. Introducing some uncertainty about the energy is mathematically equivalent to introducing uncertainty about the frequency. Of course, the answer is: the uncertainty is in both, so it’s in the frequency and in the energy and both are related through the wavefunction. So… Well… Yes. In some way, we’re chasing our own tail. 🙂

However, the trick does the job, and perfectly so. Let me summarize what we did in the previous post: we had the ammonia molecule, i.e. an NH₃ molecule, with the nitrogen ‘flipping’ across the hydrogens from time to time, as illustrated below:

This ‘flip’ requires energy, which is why we associate two energy levels with the molecule, rather than just one. We wrote these two energy levels as E₀+ A and E₀− A. That assumption solved all of our problems. [Note that we don’t specify what the energy barrier really consists of: moving the center of mass obviously requires some energy, but it is likely that a ‘flip’ also involves overcoming some electrostatic forces, as shown by the reversal of the electric dipole moment in the illustration above.] To be specific, it gave us the following wavefunctions for the amplitude to be in the ‘up’ or ‘1’ state versus the ‘down’ or ‘2’ state respectivelly:

C₁= (1/2)·e^{−(i/ħ)·(E₀− A)·t}+ (1/2)·e^{−(i/ħ)·(E₀+ A)·t}
C₂= (1/2)·e^{−(i/ħ)·(E₀− A)·t}– (1/2)·e^{−(i/ħ)·(E₀+ A)·t}

Both are composite waves. To be precise, they are the sum of two component waves with a temporal frequency equal to ω₁= (E₀− A)/ħ and ω₁= (E₀+ A)/ħ respectively. [As for the minus sign in front of the second term in the wave equation for C₂, −1 = e^±iπ, so + (1/2)·e^{−(i/ħ)·(E₀+ A)·t}and – (1/2)·e^{−(i/ħ)·(E₀+ A)·t} are the same wavefunction: they only differ because their relative phase is shifted by ±π.] So the so-called base states of the molecule themselves are associated with two different energy levels: it’s not like one state has more energy than the other.

You’ll say: so what?

Well… Nothing. That’s it really. That’s all I wanted to say here. The absolute square of those two wavefunctions gives us those time-dependent probabilities above, i.e. the graph we started this post with. So… Well… Done!

You’ll say: where’s the ‘envelope’? Oh! Yes! Let me tell you. The C₁(t) and C₂(t) equations can be re-written as:

Now, remembering our rules for adding and subtracting complex conjugates (e^iθ + e^–iθ = 2cosθ and e^iθ − e^–iθ = 2sinθ), we can re-write this as:

So there we are! We’ve got wave equations whose temporal variation is basically defined by E₀but, on top of that, we have an envelope here: the cos(A·t/ħ) and sin(A·t/ħ) factor respectively. So their magnitude is no longer time-independent: both the phase as well as the amplitude now vary with time. The associated probabilities are the ones we plotted:

|C₁(t)|²= cos²[(A/ħ)·t], and
|C₂(t)|²= sin²[(A/ħ)·t].

So, to summarize it all once more, allowing the nitrogen atom to push its way through the three hydrogens, so as to flip to the other side, thereby breaking the energy barrier, is equivalent to associating two energy levels to the ammonia molecule as a whole, thereby introducing some uncertainty, or indefiniteness as to its energy, and that, in turn, gives us the amplitudes and probabilities that we’ve just calculated. [And you may want to note here that the probabilities “sloshing back and forth”, or “dumping into each other” – as Feynman puts it – is the result of the varying magnitudes of our amplitudes, so that’s the ‘envelope’ effect. It’s only because the magnitudes vary in time that their absolute square, i.e. the associated probability, varies too.

So… Well… That’s it. I think this and all of the previous posts served as a nice introduction to quantum physics. More in particular, I hope this post made you appreciate the mathematical framework is not as horrendous as it often seems to be.

When thinking about it, it’s actually all quite straightforward, and it surely respects Occam’s principle of parsimony in philosophical and scientific thought, also know as Occam’s Razor: “When trying to explain something, it is vain to do with more what can be done with less.” So the math we need is the math we need, really: nothing more, nothing less. As I’ve said a couple of times already, Occam would have loved the math behind QM: the physics call for the math, and the math becomes the physics.

That’s what makes it beautiful. 🙂

Post scriptum:

One might think that the addition of a term in the argument in itself would lead to a beat note and, hence, a varying probability but, no! We may look at e^{−(i/ħ)·(E₀+ A)·t}as a product of two amplitudes:

e^{−(i/ħ)·(E₀+ A)·t}= e^{−(i/ħ)·E₀·t}·e^{−(i/ħ)·A·t}

But, when writing this all out, one just gets a cos(α·t+β·t)–sin(α·t+β·t), whose absolute square |cos(α·t+β·t)–sin(α·t+β·t)|²= 1. However, writing e^{−(i/ħ)·(E₀+ A)·t}as a product of two amplitudes in itself is interesting. We multiply amplitudes when an event consists of two sub-events. For example, the amplitude for some particle to go from s to x via some point a is written as:

〈 x | s 〉_{via a} = 〈 x | a 〉〈 a | s 〉

Having said that, the graph of the product is uninteresting: the real and imaginary part of the wavefunction are a simple sine and cosine function, and their absolute square is constant, as shown below.

Adding two waves with very different frequencies – A is a fraction of E₀– gives a much more interesting pattern, like the one below, which shows an e^−iαt+e^−iβt= cos(αt)−i·sin(αt)+cos(βt)−i·sin(βt) = cos(αt)+cos(βt)−i·[sin(αt)+sin(βt)] pattern for α = 1 and β = 0.1.

That doesn’t look a beat note, does it? The graphs below, which use 0.5 and 0.01 for β respectively, are not typical beat notes either.

We get our typical ‘beat note’ only when we’re looking at a wave traveling in space, so then we involve the space variable x again, and the relations that come with in, i.e. a phase velocity v_p= ω/k = (E/ħ)/(p/ħ) = E/p = c²/v (read: all component waves travel at the same speed), and a group velocity v_g= dω/dk = v (read: the composite wave or wavetrain travels at the classical speed of our particle, so it travels with the particle, so to speak). That’s what’s I’ve shown numerous times already, but I’ll insert one more animation here, just to make sure you see what we’re talking about. [Credit for the animation goes to another site, one on acoustics, actually!]

So what’s left? Nothing much. The only thing you may want to do is to continue thinking about that wavefunction. It’s tempting to think it actually is the particle, somehow. But it isn’t. So what is it then? Well… Nobody knows, really, but I like to think it does travel with the particle. So it’s like a fundamental property of the particle. We need it every time when we try to measure something: its position, its momentum, its spin (i.e. angular momentum) or, in the example of our ammonia molecule, its orientation in space. So the funny thing is that, in quantum mechanics,

We can measure probabilities only, so there’s always some randomness. That’s how Nature works: we don’t really know what’s happening. We don’t know the internal wheels and gears, so to speak, or the ‘hidden variables’, as one interpretation of quantum mechanics would say. In fact, the most commonly accepted interpretation of quantum mechanics says there are no ‘hidden variables’.
But then, as Polonius famously put, there is a method in this madness, and the pioneers – I mean Werner Heisenberg, Louis de Broglie, Niels Bohr, Paul Dirac, etcetera – discovered. All probabilities can be found by taking the square of the absolute value of a complex-valued wavefunction (often denoted by Ψ), whose argument, or phase (θ), is given by the de Broglie relations ω = E/ħ and k = p/ħ:

θ = (ω·t − k ∙x) = (E/ħ)·t − (p/ħ)·x

That should be obvious by now, as I’ve written dozens of posts on this by now. 🙂 I still have trouble interpreting this, however—and I am not ashamed, because the Great Ones I just mentioned have trouble with that too. But let’s try to go as far as we can by making a few remarks:

Adding two terms in math implies the two terms should have the same dimension: we can only add apples to apples, and oranges to oranges. We shouldn’t mix them. Now, the (E/ħ)·t and (p/ħ)·x terms are actually dimensionless: they are pure numbers. So that’s even better. Just check it: energy is expressed in newton·meter (force over distance, remember?) or electronvolts (1 eV = 1.6×10⁻¹⁹J = 1.6×10⁻¹⁹N·m); Planck’s constant, as the quantum of action, is expressed in J·s or eV·s; and the unit of (linear) momentum is 1 N·s = 1 kg·m/s = 1 N·s. E/ħ gives a number expressed per second, and p/ħ a number expressed per meter. Therefore, multiplying it by t and x respectively gives us a dimensionless number indeed.
It’s also an invariant number, which means we’ll always get the same value for it. As mentioned above, that’s because the four-vector product p_μx_μ= E·t − p∙x is invariant: it doesn’t change when analyzing a phenomenon in one reference frame (e.g. our inertial reference frame) or another (i.e. in a moving frame).
Now, Planck’s quantum of action h or ħ (they only differ in their dimension: h is measured in cycles per second and ħ is measured in radians per second) is the quantum of energy really. Indeed, if “energy is the currency of the Universe”, and it’s real and/or virtual photons who are exchanging it, then it’s good to know the currency unit is h, i.e. the energy that’s associated with one cycle of a photon.
It’s not only time and space that are related, as evidenced by the fact that t − x itself is an invariant four-vector, E and p are related too, of course! They are related through the classical velocity of the particle that we’re looking at: E/p = c²/v and, therefore, we can write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light. Now, I should add that the t − x four-vector is invariant only if we measure time and space in equivalent units. Otherwise, we have to write c·t − x. If we do that, so our unit of distance becomes c meter, rather than one meter, or our unit of time becomes the time that is needed for light to travel one meter, then c = 1, and the E·β = p·c becomes E·β = p, which we also write as β = p/E: the ratio of the energy and the momentum of our particle is its (relative) velocity.

Combining all of the above, we may want to assume that we are measuring energy and momentum in terms of the Planck constant, i.e. the ‘natural’ unit for both. In addition, we may also want to assume that we’re measuring time and distance in equivalent units. Then the equation for the phase of our wavefunctions reduces to:

θ = (ω·t − k ∙x) = E·t − p·x

Now, θ is the argument of a wavefunction, and we can always re-scale such argument by multiplying or dividing it by some constant. It’s just like writing the argument of a wavefunction as v·t–x or (v·t–x)/v = t –x/v with v the velocity of the waveform that we happen to be looking at. [In case you have trouble following this argument, please check the post I did for my kids on waves and wavefunctions.] Now, the energy conservation principle tells us the energy of a free particle won’t change. [Just to remind you, a ‘free particle’ means it is present in a ‘field-free’ space, so our particle is in a region of uniform potential.] You see what I am going to do now: we can, in this case, treat E as a constant, and divide E·t − p·x by E, so we get a re-scaled phase for our wavefunction, which I’ll write as:

φ = (E·t − p·x)/E = t − (p/E)·x = t − β·x

Now that’s the argument of a wavefunction with the argument expressed in distance units. Alternatively, we could also look at p as some constant, as there is no variation in potential energy that will cause a change in momentum, i.e. in kinetic energy. We’d then divide by p and we’d get (E·t − p·x)/p = (E/p)·t − x) = t/β − x, which amounts to the same, as we can always re-scale by multiplying it with β, which would then yield the same t − β·x argument.

The point is, if we measure energy and momentum in terms of the Planck unit (I mean: in terms of the Planck constant, i.e. the quantum of energy), and if we measure time and distance in ‘natural’ units too, i.e. we take the speed of light to be unity, then our Platonic wavefunction becomes as simple as:

Φ(φ) = a·e^−iφ= a·e^{−i(t − β·x)}

This is a wonderful formula, but let me first answer your most likely question: why would we use a relative velocity?Well… Just think of it: when everything is said and done, the whole theory of relativity and, hence, the whole of physics, is based on one fundamental and experimentally verified fact: the speed of light is absolute. In whatever reference frame, we will always measure it as 299,792,458 m/s. That’s obvious, you’ll say, but it’s actually the weirdest thing ever if you start thinking about it, and it explains why those Lorentz transformations look so damn complicated. In any case, this fact legitimately establishes c as some kind of absolute measure against which all speeds can be measured. Therefore, it is only natural indeed to express a velocity as some number between 0 and 1. Now that amounts to expressing it as the β = v/c ratio.

Let’s now go back to that Φ(φ) = a·e^−iφ= a·e^{−i(t − β·x)}wavefunction. Its temporal frequency ω is equal to one, and its spatial frequency k is equal to β = v/c. It couldn’t be simpler but, of course, we’ve got this remarkably simple result because we re-scaled the argument of our wavefunction using the energy and momentum itself as the scale factor. So, yes, we can re-write the wavefunction of our particle in a particular elegant and simple form using the only information that we have when looking at quantum-mechanical stuff: energy and momentum, because that’s what everything reduces to at that level.

Of course, the analysis above does not include uncertainty. Our information on the energy and the momentum of our particle will be incomplete: we’ll write E = E₀± σ_E, and p = p₀± σ_p. [I am a bit tired of using the Δ symbol, so I am using the σ symbol here, which denotes a standard deviation of some density function. It underlines the probabilistic, or statistical, nature of our approach.] But, including that, we’ve pretty much explained what quantum physics is about here.

You just need to get used to that complex exponential: e^−iφ = cos(−φ) + i·sin(−φ) = cos(φ) − i·sin(φ). Of course, it would have been nice if Nature would have given us a simple sine or cosine function. [Remember the sine and cosine function are actually the same, except for a phase difference of 90 degrees: sin(φ) = cos(π/2−φ) = cos(φ+π/2). So we can go always from one to the other by shifting the origin of our axis.] But… Well… As we’ve shown so many times already, a real-valued wavefunction doesn’t explain the interference we observe, be it interference of electrons or whatever other particles or, for that matter, the interference of electromagnetic waves itself, which, as you know, we also need to look at as a stream of photons , i.e. light quanta, rather than as some kind of infinitely flexible aether that’s undulating, like water or air.

So… Well… Just accept that e^−iφ is a very simple periodic function, consisting of two sine waves rather than just one, as illustrated below.

And then you need to think of stuff like this (the animation is taken from Wikipedia), but then with a projection of the sine of those phasors too. It’s all great fun, so I’ll let you play with it now. 🙂

The Hamiltonian for a two-state system: the ammonia example

Ammonia, i.e. NH₃, is a colorless gas with a strong smell. Its serves as a precursor in the production of fertilizer, but we also know it as a cleaning product, ammonium hydroxide, which is NH₃ dissolved in water. It has a lot of other uses too. For example, its use in this post, is to illustrate a two-state system. 🙂 We’ll apply everything we learned in our previous posts and, as I mentioned when finishing the last of those rather mathematical pieces, I think the example really feels like a reward after all of the tough work on all of those abstract concepts – like that Hamiltonian matrix indeed – so I hope you enjoy it. So… Here we go!

The geometry of the NH₃ molecule can be described by thinking of it as a trigonal pyramid, with the nitrogen atom (N) at its apex, and the three hydrogen atoms (H) at the base, as illustrated below. [Feynman’s illustration is slightly misleading, though, because it may give the impression that the hydrogen atoms are bonded together somehow. That’s not the case: the hydrogen atoms share their electron with the nitrogen, thereby completing the outer shell of both atoms. This is referred to as a covalent bond. You may want to look it up, but it is of no particular relevance to what follows here.]

Here, we will only worry about the spin of the molecule about its axis of symmetry, as shown above, which is either in one direction or in the other, obviously. So we’ll discuss the molecule as a two-state system. So we don’t care about its translational (i.e. linear) momentum, its internal vibrations, or whatever else that might be going on. It is one of those situations illustrating that the spin vector, i.e. the vector representing angular momentum, is an axial vector: the first state, which is denoted by | 1 〉 is not the mirror image of state | 2 〉. In fact, there is a more sophisticated version of the illustration above, which usefully reminds us of the physics involved.

It should be noted, however, that we don’t need to specify what the energy barrier really consists of: moving the center of mass obviously requires some energy, but it is likely that a ‘flip’ also involves overcoming some electrostatic forces, as shown by the reversal of the electric dipole moment in the illustration above. In fact, the illustration may confuse you, because we’re usually thinking about some net electric charge that’s spinning, and so the angular momentum results in a magnetic dipole moment, that’s either ‘up’ or ‘down’, and it’s usually also denoted by the very same μ symbol that’s used below. As I explained in my post on angular momentum and the magnetic moment, it’s related to the angular momentum J through the so-called g-number. In the illustration above, however, the μ symbol is used to denote an electric dipole moment, so that’s different. Don’t rack your brain over it: just accept there’s an energy barrier, and it requires energy to get through it. Don’t worry about its details!

Indeed, in quantum mechanics, we abstract away from such nitty-gritty, and so we just say that we have base states | i 〉 here, with i equal to 1 or 2. One or the other. Now, in our post on quantum math, we introduced what Feynman only half-jokingly refers to as the Great Law of Quantum Physics: | = ∑ | i 〉〈 i | over all base states i. It basically means that we should always describe our initial and end states in terms of base states. Applying that principle to the state of our ammonia molecule, which we’ll denote by | ψ 〉, we can write:

You may – in fact, you should – mechanically apply that | = ∑ | i 〉〈 i | substitution to | ψ 〉 to get what you get here, but you should also think about what you’re writing. It’s not an easy thing to interpret, but it may help you to think of the similarity of the formula above with the description of a vector in terms of its base vectors, which we write as A = A_x·e₁+ A_y·e₂ + A_z·e₃. Just substitute the A_icoefficients for C_i and the e_ibase vectors for the | i 〉 base states, and you may understand this formula somewhat better. It also explains why the | ψ 〉 state is often referred to as the | ψ 〉 state vector: unlike our A = ∑ A_i·e_isum of base vectors, our | 1 〉 C₁ + | 2 〉 C₂sum does not have any geometrical interpretation but… Well… Not all ‘vectors’ in math have a geometric interpretation, and so this is a case in point.

It may also help you to think of the time-dependency. Indeed, this formula makes a lot more sense when realizing that the state of our ammonia molecule, and those coefficients C_i, depend on time, so we write: ψ = ψ(t) and C_i= C_i(t). Hence, if we would know, for sure, that our molecule is always in state | 1 〉, then C₁ = 1 and C₂ = 0, and we’d write: | ψ 〉 = | 1 〉 = | 1 〉 1 + | 2 〉 0. [I am always tempted to insert a little dot (·), and change the order of the factors, so as to show we’re talking some kind of product indeed – so I am tempted to write | ψ 〉 = C₁·| 1 〉 C₁ + C₂·| 2 〉 C₂, but I note that’s not done conventionally, so I won’t do it either.]

Why this time dependency? It’s because we’ll allow for the possibility of the nitrogen to push its way through the pyramid – through the three hydrogens, really – and flip to the other side. It’s unlikely, because it requires a lot of energy to get half-way through (we’ve got what we referred to as an energy barrier here), but it may happen and, as we’ll see shortly, it results in us having to think of the the ammonia molecule as having two separate energy levels, rather than just one. We’ll denote those energy levels as E₀ ± A. However, I am getting ahead of myself here, so let me get back to the main story.

To fully understand the story, you should really read my previous post on the Hamiltonian, which explains how those C_i coefficients, as a function of time, can be determined. They’re determined by a set of differential equations (i.e. equations involving a function and the derivative of that function) which we wrote as:

If we have two base states only – which is the case here – then this set of equations is:

Two equations and two functions – C₁= C₁(t) and C₂= C₂(t) – so we should be able to solve this thing, right? Well… No. We don’t know those H_ijcoefficients. As I explained in my previous post, they also evolve in time, so we should write them as H_ij(t) instead of H_ijtout court, and so it messes the whole thing up. We have two equations and six functions really. There is no way we can solve this! So how do we get out of this mess?

Well… By trial and error, I guess. 🙂 Let us just assume the molecule would behave nicely—which we know it doesn’t, but so let’s push the ‘classical’ analysis as far as we can, so we might get some clues as to how to solve this problem. In fact, our analysis isn’t ‘classical’ at all, because we’re still talking amplitudes here! However, you’ll agree the ‘simple’ solution would be that our ammonia molecule doesn’t ‘tunnel’. It just stays in the same spin direction forever. Then H₁₂and H₂₁must be zero (think of the U₁₂(t + Δt, t) and U₂₁(t + Δt, t) functions) and H₁₁and H₂₂are equal to… Well… I’d love to say they’re equal to 1 but… Well… You should go through my previous posts: these Hamiltonian coefficients are related to probabilities but… Well… Same-same but different, as they say in Asia. 🙂 They’re amplitudes, which are things you use to calculate probabilities. But calculating probabilities involve normalization and other stuff, like allowing for interference of amplitudes, and so… Well… To make a long story short, if our ammonia molecule would stay in the same spin direction forever, then H₁₁and H₂₂are not one but some constant. In any case, the point is that they would not change in time (so H₁₁(t) = H₁₁ and H₂₂(t ) = H₂₂), and, therefore, our two equations would reduce to:

So the coefficients are now proper coefficients, in the sense that they’ve got some definite value, and so we have two equations and two functions only now, and so we can solve this. Indeed, remembering all of the stuff we wrote on the magic of exponential functions (more in particular, remembering that d[e^x]/dx), we can understand the proposed solution:

As Feynman notes: “These are just the amplitudes for stationary states with the energies E₁= H₁₁ and E₂= H₂₂.” Now let’s think about that. Indeed, I find the term ‘stationary’ state quite confusing, as it’s ill-defined. In this context, it basically means that we have a wavefunction that is determined by (i) a definite (i.e. unambiguous, or precise) energy level and (ii) that there is no spatial variation. Let me refer you to my post on the basics of quantum math here. We often use a sort of ‘Platonic’ example of the wavefunction indeed:

a·e^−i·θ= e^{−i·(ω·t − k ∙x)} = a·e^{−(i/ħ)·(E·t − p∙x)}

So that’s a wavefunction assuming the particle we’re looking at has some well-defined energy E and some equally well-defined momentum p. Now, that’s kind of ‘Platonic’ indeed, because it’s more like an idea, rather than something real. Indeed, a wavefunction like that means that the particle is everywhere and nowhere, really—because its wavefunction is spread out all of over space. Of course, we may think of the ‘space’ as some kind of confined space, like a box, and then we can think of this particle as being ‘somewhere’ in that box, and then we look at the temporal variation of this function only – which is what we’re doing now: we don’t consider the space variable x at all. So then the equation reduces to a·e^{–(i/ħ)·(E·t)}, and so… Well… Yes. We do find that our Hamiltonian coefficient H_iiis like the energy of the | i 〉 state of our NH₃ molecule, so we write: H₁₁= E₁, and H₂₂= E₂, and the ‘wavefunctions’ of our C₁and C₂coefficients can be written as:

C₁= a·e^{−(i/ħ)·(H₁₁·t)}= a·e^{−(i/ħ)·(E₁·t)}, with H₁₁= E₁, and
C₂= a·e^{−(i/ħ)·(H₂₂·t)}= a·e^{−(i/ħ)·(E₂·t)}, with H₂₂= E₂.

But can we interpret C₁and C₂as proper amplitudes? They are just coefficients in these equations, aren’t they? Well… Yes and no. From what we wrote in previous posts, you should remember that these C_icoefficients are equal to 〈 i | ψ 〉, so they are the amplitude to find our ammonia molecule in one state or the other.

Back to Feynman now. He adds, logically but brilliantly:

“We note, however, that for the ammonia molecule the two states |1〉 and |2〉 have a definite symmetry. If nature is at all reasonable, the matrix elements H₁₁ and H₂₂ must be equal. We’ll call them both E₀, because they correspond to the energy the states would have if H₁₁ and H₂₂ were zero.”

So our C₁and C₂amplitudes then reduce to:

C₁= 〈 1 | ψ 〉 = a·e^{−(i/ħ)·(E₀·t)}
C₂=〈 2 | ψ 〉 = a·e^{−(i/ħ)·(E₀·t)}

We can now take the absolute square of both to find the probability for the molecule to be in state 1 or in state 2:

|〈 1 | ψ 〉|²= |a·e^{−(i/ħ)·(E₀·t)}|²= a²
|〈 2 | ψ 〉|²= |a·e^{−(i/ħ)·(E₀·t)}|²= a²

Now, the probabilities have to add up to 1, so a²+ a²= 1 and, therefore, the probability to be in either in state 1 or state 2 is 0.5, which is what we’d expect.

Note: At this point, it is probably good to get back to our | ψ 〉 = | 1 〉 C₁ + | 2 〉 C₂equation, so as to try to understand what it really says. Substituting the a·e^{−(i/ħ)·(E₀·t)} expression for C₁ and C₂yields:

| ψ 〉 = | 1 〉 a·e^{−(i/ħ)·(E₀·t)} + | 2 〉 a·e^{−(i/ħ)·(E₀·t)}= [| 1 〉 + | 2 〉] a·e^{−(i/ħ)·(E₀·t)}

Now, what is this saying, really? In our previous post, we explained this is an ‘open’ equation, so it actually doesn’t mean all that much: we need to ‘close’ or ‘complete’ it by adding a ‘bra’, i.e. a state like 〈 χ |, so we get a 〈 χ | ψ〉 type of amplitude that we can actually do something with. Now, in this case, our final 〈 χ | state is either 〈 1 | or 〈 2 |, so we write:

〈 1 | ψ 〉 = [〈 1 | 1 〉 + 〈 1 | 2 〉]·a·e^{−(i/ħ)·(E₀·t)}= [1 + 0]·a·e^{−(i/ħ)·(E₀·t)}· = a·e^{−(i/ħ)·(E₀·t)}
〈 2 | ψ 〉 = [〈 2 | 1 〉 + 〈 2 | 2 〉]·a·e^{−(i/ħ)·(E₀·t)}= [0 + 1]·a·e^{−(i/ħ)·(E₀·t)}· = a·e^{−(i/ħ)·(E₀·t)}

Note that I finally added the multiplication dot (·) because we’re talking proper amplitudes now and, therefore, we’ve got a proper product too: we multiply one complex number with another. We can now take the absolute square of both to find the probability for the molecule to be in state 1 or in state 2:

|〈 1 | ψ 〉|²= |a·e^{−(i/ħ)·(E₀·t)}|²= a²
|〈 2 | ψ 〉|²= |a·e^{−(i/ħ)·(E₀·t)}|²= a²

Unsurprisingly, we find the same thing: these probabilities have to add up to 1, so a²+ a²= 1 and, therefore, the probability to be in state 1 or state 2 is 0.5. So the notation and the logic behind makes perfect sense. But let me get back to the lesson now.

The point is: the true meaning of a ‘stationary’ state here, is that we have non-fluctuating probabilities. So they are and remain equal to some constant, i.e. 1/2 in this case. This implies that the state of the molecule does not change: there is no way to go from state 1 to state 2 and vice versa. Indeed, if we know the molecule is in state 1, it will stay in that state. [Think about what normalization of probabilities means when we’re looking at one state only.]

You should note that these non-varying probabilities are related to the fact that the amplitudes have a non-varying magnitude. The phase of these amplitudes varies in time, of course, but their magnitude is and remains a, always. The amplitude is not being ‘enveloped’ by another curve, so to speak.

OK. That should be clear enough. Sorry I spent so much time on this, but this stuff on ‘stationary’ states comes back again and again and so I just wanted to clear that up as much as I can. Let’s get back to the story.

So we know that, what we’re describing above, is not what ammonia does really. As Feynman puts it: “The equations [i.e. the C₁and C₂equations above] don’t tell us what what ammonia really does. It turns out that it is possible for the nitrogen to push its way through the three hydrogens and flip to the other side. It is quite difficult; to get half-way through requires a lot of energy. How can it get through if it hasn’t got enough energy? There is some amplitude that it will penetrate the energy barrier. It is possible in quantum mechanics to sneak quickly across a region which is illegal energetically. There is, therefore, some [small] amplitude that a molecule which starts in |1〉 will get to the state |2〉. The coefficients H₁₂ and H₂₁ are not really zero.”

He adds: “Again, by symmetry, they should both be the same—at least in magnitude. In fact, we already know that, in general, H_ijmust be equal to the complex conjugate of H_ji.”

His next step, then, is to interpreted as either a stroke of genius or, else, as unexplained. 🙂 He invokes the symmetry of the situation to boldly state that H₁₂is some real negative number, which he denotes as −A, which – because it’s a real number (so the imaginary part is zero) – must be equal to its complex conjugate H₂₁. So then Feynman does this fantastic jump in logic. First, he keeps using the E₀ value for H₁₁ and H₂₂, motivating that as follows: “If nature is at all reasonable, the matrix elements H₁₁ and H₂₂ must be equal, and we’ll call them both E₀, because they correspond to the energy the states would have if H₁₁ and H₂₂ were zero.” Second, he uses that minus A value for H₁₂and H₂₁. In short, the two equations and six functions are now reduced to:

Solving these equations is rather boring. Feynman does it as follows:

solution

Now, what does these equations actually mean? It depends on those a and b coefficients. Looking at the solutions, the most obvious question to ask is: what if a or b are zero? If b is zero, then the second terms in both equations is zero, and so C₁ and C₂ are exactly the same: two amplitudes with the same temporal frequency ω = (E₀− A)/ħ. If a is zero, then C₁ and C₂ are the same too, but with opposite sign: two amplitudes with the same temporal frequency ω = (E₀+ A)/ħ. Squaring them – in both cases (i.e. for a = 0 or b = 0) – yields, once again, an equal and constant probability for the spin of the ammonia molecule to in the ‘up’ or ‘down’ or ‘down’. To be precise, we We can now take the absolute square of both to find the probability for the molecule to be in state 1 or in state 2:

For b = 0: |〈 1 | ψ 〉|²= |(a/2)·e^{−(i/ħ)·(E₀− A)·t}|²= a²/4 = |〈 2 | ψ 〉|²
For a = 0: |〈 1 | ψ 〉|²=|(b/2)·e^{−(i/ħ)·(E₀+ A)·t}|²= b²/4 = |〈 2 | ψ 〉|²(the minus sign in front of b/2 is squared away)

So we get two stationary states now. Why two instead of one? Well… You need to use your imagination a bit here. They actually reflect each other: they’re the same as the one stationary state we found when assuming our nitrogen atom could not ‘flip’ from one position to the other. It’s just that the introduction of that possibility now results in a sort of ‘doublet’ of energy levels. But so we shouldn’t waste our time on this, as we want to analyze the general case, for which the probabilities to be in state 1 or state 2 do vary in time. So that’s when a and b are non-zero.

To analyze it all, we may want to start with equating t to zero. We then get:

This leads us to conclude that a = b = 1, so our equations for C₁(t) and C₂(t) can now be written as:

Remembering our rules for adding and subtracting complex conjugates (e^iθ + e^–iθ = 2cosθ and e^iθ − e^–iθ = 2sinθ), we can re-write this as:

Now these amplitudes are much more interesting. Their temporal variation is defined by E₀but, on top of that, we have an envelope here: the cos(A·t/ħ) and sin(A·t/ħ) factor respectively. So their magnitude is no longer time-independent: both the phase as well as the amplitude now vary with time. What’s going on here becomes quite obvious when calculating and plotting the associated probabilities, which are

|C₁(t)|²= cos²(A·t/ħ), and
|C₂(t)|²= sin²(A·t/ħ)

respectively (note that the absolute square of i is equal to 1, not −1). The graph of these functions is depicted below.

As Feynman puts it: “The probability sloshes back and forth.” Indeed, the way to think about this is that, if our ammonia molecule is in state 1, then it will not stay in that state. In fact, one can be sure the nitrogen atom is going to flip at some point in time, with the probabilities being defined by that fluctuating probability density function above. Indeed, as time goes by, the probability to be in state 2 increases, until it will effectively be in state 2. And then the cycle reverses.

Our | ψ 〉 = | 1 〉 C₁ + | 2 〉 C₂equation is a lot more interesting now, as we do have a proper mix of pure states now: we never really know in what state our molecule will be, as we have these ‘oscillating’ probabilities now, which we should interpret carefully.

The point to note is that the a = 0 and b = 0 solutions came with precise temporal frequencies: (E₀− A)/ħ and (E₀+ A)/ħ respectively, which correspond to two separate energy levels: E₀− A and E₀+ A respectively, with |A| = H₁₂= H₂₁. So everything is related to everything once again: allowing the nitrogen atom to push its way through the three hydrogens, so as to flip to the other side, thereby breaking the energy barrier, is equivalent to associating two energy levels to the ammonia molecule as a whole, thereby introducing some uncertainty, or indefiniteness as to its energy, and that, in turn, gives us the amplitudes and probabilities that we’ve just calculated.

Note that the probabilities “sloshing back and forth”, or “dumping into each other” – as Feynman puts it – is the result of the varying magnitudes of our amplitudes, going up and down and, therefore, their absolute square varies too.

So… Well… That’s it as an introduction to a two-state system. There’s more to come. Ammonia is used in the ammonia maser. Now that is something that’s interesting to analyze—both from a classical as well as from a quantum-mechanical perspective. Feynman devotes a full chapter to it, so I’d say… Well… Have a look. 🙂

Post scriptum: I must assume this analysis of the NH₃ molecule, with the nitrogen ‘flipping’ across the hydrogens, triggers a lot of questions, so let me try to answer some. Let me first insert the illustration once more, so you don’t have to scroll up:

The first thing that you should note is that the ‘flip’ involves a change in the center of mass position. So that requires energy, which is why we associate two different energy levels with the molecule: E₀+ A and E₀− A. However, as mentioned above, we don’t care about the nitty-gritty here: the energy barrier is likely to combine a number of factors, including electrostatic forces, as evidenced by the flip in the electric dipole moment, which is what the μ symbol here represents! Just note that the two energy levels are separated by an amount that’s equal to 2·A, rather than A and that, once again, it becomes obvious now why Feynman would prefer the Hamiltonian to be called the ‘energy matrix’, as its coefficients do represent specific energy levels, or differences between them! Now, that assumption yielded the following wavefunctions for C₁= 〈 1 | ψ 〉 and C₁= 〈 2 | ψ 〉:

C₁= 〈 1 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}+ (1/2)·e^{−(i/ħ)·(E₀+ A)·t}
C₂= 〈 2 | ψ 〉 = (1/2)·e^{−(i/ħ)·(E₀− A)·t}– (1/2)·e^{−(i/ħ)·(E₀+ A)·t}

Now, writing things this way, rather than in terms of probabilities, makes it clear that the two base states of the molecule themselves are associated with two different energy levels, so it is not like one state has more energy than the other. It’s just that the possibility of going from one state to the other requires an uncertainty about the energy, which is reflected by the energy doublet E₀± A in the wavefunction of the base states. Now, if the wavefunction of the base states incorporates that energy doublet, then it is obvious that the state of the ammonia molecule, at any point in time, will also incorporate that energy doublet.

This triggers the following remark: what’s the uncertainty really? Is it an uncertainty in the energy, or is it an uncertainty in the wavefunction? I mean: we have a function relating the energy to a frequency. Introducing some uncertainty about the energy is mathematically equivalent to introducing uncertainty about the frequency. Think of it: two energy levels implies two frequencies, and vice versa. More in general, introducing n energy levels, or some continuous range of energy levels ΔE, amounts to saying that our wave function doesn’t have a specific frequency: it now has n frequencies, or a range of frequencies Δω = ΔE/ħ. Of course, the answer is: the uncertainty is in both, so it’s in the frequency and in the energy and both are related through the wavefunction. So… In a way, we’re chasing our own tail.

Having said that, the energy may be uncertain, but it is real. It’s there, as evidenced by the fact that the ammonia molecule behaves like an atomic oscillator: we can excite it in exactly the same way as we can excite an electron inside an atom, i.e. by shining light on it. The only difference is the photon energies: to cause a transition in an atom, we use photons in the optical or ultraviolet range, and they give us the same radiation back. To cause a transition in an ammonia molecule, we only need photons with energies in the microwave range. Here, I should quickly remind you of the frequencies and energies involved. visible light is radiation in the 400–800 terahertz range and, using the E = h·f equation, we can calculate the associated energies of a photon as 1.6 to 3.2 eV. Microwave radiation – as produced in your microwave oven – is typically in the range of 1 to 2.5 gigahertz, and the associated photon energy is 4 to 10 millionths of an eV. Having illustrated the difference in terms of the energies involved, I should add that masers and lasers are based on the same physical principle: LASER and MASER stand for Light/Micro-wave Amplification by Stimulated Emission of Radiation, respectively.

So… How shall I phrase this? There’s uncertainty, but the way we are modeling that uncertainty matters. So yes, the uncertainty in the frequency of our wavefunction and the uncertainty in the energy are mathematically equivalent, but the wavefunction has a meaning that goes much beyond that. [You may want to reflect on that yourself.]

Finally, another question you may have is why would Feynman take minus A (i.e. −A) for H₁₂ and H₂₁. Frankly, my first thought on this was that it should have something to do with the original equation for these Hamiltonian coefficients, which also has a minus sign: U_ij(t + Δt, t) = δ_ij + K_ij(t)·Δt = δ_ij − (i/ħ)·H_ij(t)·Δt. For i ≠ j, this reduces to:

U_ij(t + Δt, t) = + K_ij(t)·Δt = − (i/ħ)·H_ij(t)·Δt

However, the answer is: it really doesn’t matter. One could write: H₁₂ and H₂₁ = +A, and we’d find the same equations. We’d just switch the indices 1 and 2, and the coefficients a and b. But we get the same solutions. You can figure that out yourself. Have fun with it !

Oh ! And please do let me know if some of the stuff above would trigger other questions. I am not sure if I’ll be able to answer them, but I’ll surely try, and good question always help to ensure we sort of ‘get’ this stuff in a more intuitive way. Indeed, when everything is said and done, the goal of this blog is not simply re-produce stuff, but to truly ‘get’ it, as good as we can. 🙂

Quantum math: the Hamiltonian

Pre-script (dated 26 June 2020): I have come to the conclusion one does not need all this hocus-pocus to explain quantum-mechanical systems: classical physics will do. So no use to read this. Read my papers instead. 🙂

Original post:

After all of the ‘rules’ and ‘laws’ we’ve introduced in our previous post, you might think we’re done but, of course, we aren’t. Things change. As Feynman puts it: “One convenient, delightful ‘apparatus’ to consider is merely a wait of a few minutes; During the delay, various things could be going on—external forces applied or other shenanigans—so that something is happening. At the end of the delay, the amplitude to find the thing in some state χ is no longer exactly the same as it would have been without the delay.”

In short, the picture we presented in the previous posts was a static one. Time was frozen. In reality, time passes, and so we now need to look at how amplitudes change over time. That’s where the Hamiltonian kicks in. So let’s have a look at that now.

[If you happen to understand the Hamiltonian already, you may want to have a look at how we apply it to a real situation: we’ll explain the basics involving state transitions of the ammonia molecule, which are a prerequisite to understanding how a maser works, which is not unlike a laser. But that’s for later. First we need to get the basics.]

Using Dirac’s bra-ket notation, which we introduced in the previous posts, we can write the amplitude to find a ‘thing’ – i.e. a particle, for example, or some system, of particles or other things – in some state χ at the time t = t₂, when it was in some state φ state at the time t = t₁ as follows:

Don’t be scared of this thing. If you’re unfamiliar with the notation, just check out my previous posts: we’re just replacing A by U, and the only thing that we’ve modified is that the amplitudes to go from φ to χ now depend on t₁ and t₂. Of course, we’ll describe all states in terms of base states, so we have to choose some representation and expand this expression, so we write:

I’ve explained the point a couple of time already, but let me note it once more: in quantum physics, we always measure some (vector) quantity – like angular momentum, or spin – in some direction, let’s say the z-direction, or the x-direction, or whatever direction really. Now we can do that in classical mechanics too, of course, and then we find the component of that vector quantity (vector quantities are defined by their magnitude and, importantly, their direction). However, in classical mechanics, we know the components in the x-, y- and z-direction will unambiguously determine that vector quantity. In quantum physics, it doesn’t work that way. The magnitude is never all in one direction only, so we can always some of it in some other direction. (see my post on transformations, or on quantum math in general). So there is an ambiguity in quantum physics has no parallel in classical mechanics. So the concept of a component of a vector needs to be carefully interpreted. There’s nothing definite there, like in classical mechanics: all we have is amplitudes, and all we can do is calculate probabilities, i.e. expected values based on those amplitudes.

In any case, I can’t keep repeating this, so let me move on. In regard to that 〈 χ | U | φ 〉 expression, I should, perhaps, add a few remarks. First, why U instead of A? The answer: no special reason, but it’s true that the use of U reminds us of energy, like potential energy, for example. We might as well have used W. The point is: energy and momentum do appear in the argument of our wavefunctions, and so we might as well remind ourselves of that by choosing symbols like W or U here. Second, we may, of course, want to choose our time scale such that t₁ = 0. However, it’s fine to develop the more general case. Third, it’s probably good to remind ourselves we can think of matrices to model it all. More in particular, if we have three base states, say ‘plus‘, ‘zero, or ‘minus‘, and denoting 〈 i | φ 〉 and 〈 i | χ 〉 as C_i and D_i respectively (so 〈 χ | i 〉 = 〈 i | χ 〉* = D_i*), then we can re-write the expanded expression above as:

Fourth, you may have heard of the S-matrix, which is also known as the scattering matrix—which explains the S in front but it’s actually a more general thing. Feynman defines the S-matrix as the U(t₁, t₂) matrix for t₁→ −∞ and t₂→ +∞, so as some kind of limiting case of U. That’s true in the sense that the S-matrix is used to relate initial and final states, indeed. However, the relation between the S-matrix and the so-called evolution operators U is slightly more complex than he wants us to believe. I can’t say too much about this now, so I’ll just refer you to the Wikipedia article on that, as I have to move on.

The key to the analysis is to break things up once more. More in particular, one should appreciate that we could look at three successive points in time, t₁, t₂, t₃, and write U(t₁, t₃) as:

U(t₃, t₁) = U(t₃, t₂)·U(t₂, t₁)

It’s just like adding another apparatus in series, so it’s just like what did in our previous post, when we wrote:

So we just put a | bar between B and A and wrote it all out. That | bar is really like a factor 1 in multiplication but – let me caution you – you really need to watch the order of the various factors in your product, and read symbols in the right order, which is often from right to left, like in Hebrew or Arab, rather than from left to right. In that regard, you should note that we wrote U(t₃, t₁) rather than U(t₁, t₃): you need to keep your wits about you here! So as to make sure we can all appreciate that point, let me show you what that U(t₃, t₁) = U(t₃, t₂)·U(t₂, t₁) actually says by spelling it out if we have two base states only (like ‘up‘ or ‘down‘, which I’ll note as ‘+’ and ‘−’ again) :

So now you appreciate why we try to simplify our notation as much as we can! But let me get back to the lesson. To explain the Hamiltonian, which we need to describe how states change over time, Feynman embarks on a rather spectacular differential analysis. Now, we’ve done such exercises before, so don’t be too afraid. He substitutes t₁ for t tout court, and t₂for t + Δt, with Δt the infinitesimal you know from Δy = (dy/dx)·Δx, with the derivative dy/dx being defined as the Δy/Δx ratio for Δx → 0. So we write U(t₂, t₁) = U(t + Δt, t). Now, we also explained the idea of an operator in our previous post. It came up when we’re being creative, and so we dropped the 〈 χ | state from the 〈 χ | A | φ〉 expression and just wrote:

If you ‘get’ that, you’ll also understand what I am writing now:

This is quite abstract, however. It is an ‘open’ equation, really: one needs to ‘complete’ it with a ‘bra’, i.e. a state like 〈 χ |, so as to give a 〈 χ | ψ〉 = 〈 χ | A | φ〉 type of amplitude that actually means something. What we’re saying is that our operator (or our ‘apparatus’ if it helps you to think that way) does not mean all that much as long as we don’t measure what comes out, so we have to choose some set of base states, i.e. a representation, which allows us to describe the final state, which we write as 〈 χ |. In fact, what we’re interested in is the following amplitudes:

So now we’re in business, really. 🙂 If we can find those amplitudes, for each of our base states i, we know what’s going on. Of course, we’ll want to express our ψ(t) state in terms of our base states too, so the expression we should be thinking of is:

Phew! That looks rather unwieldy, doesn’t it? You’re right. It does. So let’s simplify. We can do the following substitutions:

〈 i | ψ(t + Δt)〉 = C_i(t + Δt) or, more generally, 〈 j | ψ(t)〉 = C_j(t)
〈 i | U(t₂, t₁) | j〉 = U_ij(t₂, t₁) or, more specifically, 〈 i | U(t + Δt, t) | j〉 = U_ij(t + Δt, t)

As Feynman notes, that’s how the dynamics of quantum mechanics really look like. But, of course, we do need something in terms of derivatives rather than in terms of differentials. That’s where the Δy = (dy/dx)·Δx equation comes in. The analysis looks kinda dicey because it’s like doing some kind of first-order linear approximation of things – rather than an exact kinda thing – but that’s how it is. Let me remind you of the following formula: if we write our function y as y = f(x), and we’re evaluating the function near some point a, then our Δy = (dy/dx)·Δx equation can be used to write:

y = f(x) ≈ f(a) + f'(a)·(x − a) = f(a) + (dy/dx)·Δx

To remind yourself of how this works, you can complete the drawing below with the actual y = f(x) as opposed to the f(a) + Δy approximation, remembering that the (dy/dx) derivative gives you the slope of the tangent to the curve, but it’s all kids’ stuff really and so we shouldn’t waste too much spacetime on this. 🙂

The point is: our U_ij(t + Δt, t) is a function too, not only of time, but also of i and j. It’s just a rather special function, because we know that, for Δt → 0, U_ijwill be equal to 1 if i = j (in plain language: if Δt → 0 goes to zero, nothing happens and we’re just in state i), and equal to 0 if i = j. That’s just as per the definition of our base states. Indeed, remember the first ‘rule’ of quantum math:

〈 i | j〉 = 〈 j | i〉 = δ_ij, with δ_ij= δ_jiis equal to 1 if i = j, and zero if i ≠ j

So we can write our f(x) ≈ f(a) + (dy/dx)·Δx expression for U_ijas:

So K_ij is also some kind of derivative and the Kronecker delta, i.e. δ_ij, serves as the reference point around which we’re evaluating U_ij. However, that’s about as far as the comparison goes. We need to remind ourselves that we’re talking complex-valued amplitudes here. In that regard, it’s probably also good to remind ourselves once more that we need to watch the order of stuff: U_ij = 〈 i | U | j〉, so that’s the amplitude to go from base state j to base state i, rather than the other way around. Of course, we have the 〈 χ | φ 〉 = 〈 φ | χ 〉* rule, but we still need to see how that plays out with an expression like 〈 i | U(t + Δt, t) | j〉. So, in short, we should be careful here!

Having said that, we can actually play a bit with that expression, and so that’s what we’re going to do now. The first thing we’ll do is to write K_ij as a function of time indeed:

K_ij = K_ij(t)

So we don’t have that Δt in the argument. It’s just like dy/dx = f'(x): a derivative is a derivative—a function which we derive from some other function. However, we’ll do something weird now: just like any function, we can multiply or divide it by some constant, so we can write something like G(x) = c·F(x), which is equivalent to saying that F(x) = G(x)/c. I know that sound silly but it is how is, and we can also do it with complex-valued functions: we can define some other function by multiplying or dividing by some complex-valued constant, like a + b·i, or ξ or whatever other constant. Just note we’re no longer talking the base state i but the imaginary unit i. So it’s all done so as to confuse you even more. 🙂

So let’s take −i/ħ as our constant and re-write our K_ij(t) function as −i/ħ times some other function, which we’ll denote by H_ij(t), so K_ij(t) = –(i/ħ)·H_ij(t). You guess it, of course: H_ij(t) is the infamous Hamiltonian, and it’s written the way it’s written both for historical as well as for practical reasons, which you’ll soon discover. Of course, we’re talking one coefficient only and we’ll have nine if we have three base states i and j, or four if we have only two. So we’ve got a n-by-n matrix once more. As for its name… Well… As Feynman notes: “How Hamilton, who worked in the 1830s, got his name on a quantum mechanical matrix is a tale of history. It would be much better called the energy matrix, for reasons that will become apparent as we work with it.”

OK. So we’ll just have to acknowledge that and move on. Our U_ij(t + Δt, t) = δ_ij + K_ij(t)·Δt expression becomes:

U_ij(t + Δt, t) = δ_ij –(i/ħ)·H_ij(t)·Δt

[Isn’t it great you actually start to understand those Chinese-looking formulas? :-)] We’re not there yet, however. In fact, we’ve still got quite a bit of ground to cover. We now need to take that other monster:

So let’s substitute now, so we get:

We can get this in the form we want to get – so that’s the form you’ll find in textbooks 🙂 – by noting that the ∑δ_ij·C_j(t) sum, taking over all j is, quite simply, equal to C_i(t). [Think about the indexes here: we’re looking at some i, and so it’s only the j that’s taking on whatever value it can possibly have.] So we can move that to the other side, which gives us C_i(t + Δt) – C_i(t). We can then divide both sides of our expression by Δt, which gives us an expression like [f(x + Δx) – f(x)]/Δx = Δy//Δx, which is actually the definition of the derivative for Δx going to zero. Now, that allows us to re-write the whole thing in terms of a proper derivative, rather than having to work with this rather unwieldy differential stuff. So, if we substitute [C_i(t + Δt) – C_i(t)]/Δx for d[C_i(t)]/dt, and then also move –(i/ħ) to the left-hand side, remembering that 1/i = –i (and, hence, [–(i/ħ)]⁻¹= i/ħ), we get the formula in the shape we wanted it in:

Done ! Of course, this is a set of differential equations and… Well… Yes. Yet another set of differential equations. 🙂 It seems like we can’t solve anything without involving differential equations in physics, isn’t it? But… Well… I guess that’s the way it is. So, before we turn to some example, let’s note a few things.

First, we know that a particle, or a system, must be in some state at any point of time. That’s equivalent to stating that the sum of the probabilities |C_i(t)|²= |〈 i | ψ(t)〉|²is some constant. In fact, we’d like to say it’s equal to one, but then we haven’t normalized anything here. You can fiddle with the formulas but it’s probably easier to just acknowledge that, if we’d measure anything – think of the angular momentum along the z-direction, or some other direction, if you’d want an example – then we’ll find it’s either ‘up’ or ‘down’ for a spin-1/2 particle, or ‘plus’, ‘zero’, or ‘minus’ for a spin-1 particle.

Now, we know that the complex conjugate of a sum is equal to the sum of the complex conjugates: [∑ z_i]* = ∑ z_i*, and that the complex conjugate of a product is the product of the complex conjugates, so we have [∑ z_iz_j]* = ∑ z_i*z_j*. Now, some fiddling with the formulas above should allow you to prove that H_ij= H_ij*, and the associated matrix is usually referred to as the Hermitian or conjugate transpose. If if the original Hamiltonian matrix is denoted as H, then its conjugate transpose will be denoted by H*, H^† or even H^H(so the H in the superscript stands for Hermitian, instead of Hamiltonean). So… Yes. There’s competing notations around. 🙂

The simplest situation, of course, is when the Hamiltonian do not depend on time. In that case, we’re back in the static case, and all H_ijcoefficients are just constants. For a system with two base states, we’d have the following set of equations:

This set of two equations can be easily solved by remembering the solution for one equation only. Indeed, if we assume there’s only base state – which is like saying: the particle is at rest somewhere (yes: it’s that stupid!) – our set of equations reduces to only one:

This is a differential equation which is easily solved to give:

[As for being ‘easily solved’, just remember the exponential function is its own derivative and, therefore, d[a·e^{–(i/ħ)H_ijt}]/dt = a·d[e^{–(i/ħ)H_ijt}]/dt = –a·(i/ħ)·H_ij·e^{–(i/ħ)H_ijt}, which gives you the differential equation, so… Well… That’s the solution.]

This should, of course, remind you of the equation that inspired Louis de Broglie to write down his now famous matter-wave equation (see my post on the basics of quantum math):

a·e^−i·θ= e^{−i·(ω·t − k ∙x)} = a·e^{−(i/ħ)·(E·t − p∙x)}

Indeed, if we look at the temporal variation of this function only – so we don’t consider the space variable x – then this equation reduces to a·e^{–(i/ħ)·(E·t)}, and so find that our Hamiltonian coefficient H₁₁is equal to the energy of our particle, so we write: H₁₁= E, which, of course, explains why Feynman thinks the Hamiltonian matrix should be referred to as the energy matrix. As he puts it: “The Hamiltonian is the generalization of the energy for more complex situations.”

Now, I’ll conclude this post by giving you the answer to Feynman’s remark on why the Irish 19th century mathematician William Rowan Hamilton should be associated with the Hamiltonian. The truth is: the term ‘Hamiltonian matrix’ may also refer to a more general notion. Let me copy Wikipedia here: “In mathematics, a Hamiltonian matrix is a $2 n$ -by- $2 n$ matrix $A$ such that $JA$ is symmetric, where $J$ is the skew-symmetric matrix

J= \begin{bmatrix} 0 & I_n \\ -I_n & 0 \\ \end{bmatrix}

and $I n$ is the $n$ -by- $n$ identity matrix. In other words, $A$ is Hamiltonian if and only if $(JA) T = JA$ where $() T$ denotes the transpose. So… That’s the answer. 🙂 And there’s another reason too: Hamilton invented the quaternions and… Well… I’ll leave it to you to check out what these have got to do with quantum physics. 🙂

[…] Oh ! And what about the maser example? Well… I am a bit tired now, so I’ll just refer you to Feynman’s exposé on it. It’s not that difficult if you understood all of the above. In fact, it’s actually quite straightforward, and so I really recommend you work your way through the example, as it will give you a much better ‘feel’ for the quantum-mechanical framework we’ve developed so far. In fact, walking through the whole thing is like a kind of ‘reward’ for having worked so hard on the more abstract stuff in this and my previous posts. So… Yes. Just go for it! 🙂 [And, just in case you don’t want to go for it, I did write a little introduction to in the following post. :-)]

Quantum math: transformations

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. 🙂

Original post:

We’ve come a very long way. Now we’re ready for the Big Stuff. We’ll look at the rules for transforming amplitudes from one ‘base’ to ‘another’. [In quantum mechanics, however, we’ll talk about a ‘representation’, rather than a ‘base’, as we’ll reserve the latter term for a ‘base’ state.] In addition, we’ll look at how physicists model how amplitudes evolve over time using the so-called Hamiltonian matrix. So let’s go for it.

Transformations: how should we think about them?

In my previous post, I presented the following hypothetical set-up: we have an S-filter and a T-filter in series, but the T-filter at the angle α with respect to the first. In case you forgot: these ‘filters’ are modified Stern-Gerlach apparatuses, designed to split a particle beam according to the angular momentum in the direction of the gradient of the magnetic field, in which we may place masks to filter out one or more states.

The idea is illustrated in the hypothetical example below. The unpolarized beam goes through S, but we have masks blocking all particles with zero or negative spin in the z-direction, i.e. with respect to S. Hence, all particles entering the T-filter are in the +S state. Now, we assume the set-up of the T-filter is such that it filters out all particles with positive or negative spin. Hence, only particles with zero spin go through. So we’ve got something like this:

However, we need to be careful as what we are saying here. The T-apparatus is tilted, so the gradient of the magnetic field is different. To be precise, it’s got the same tilt as the T-filter itself (α). Hence, it will be filtering out all particles with positive or negative spin with respect to T. So, unlike what you might think at first, some fraction of the particles in the +S state will get through the T-filter, and come out in the 0T state. In fact, we know how many, because we have formulas for situations like this. To be precise, in this case, we should apply the following formula:

〈 0T | +S 〉 = −(1/√2)·sinα

This is a real-valued amplitude. As usual, we get the following probability by taking the absolute square, so P = |−(1/√2)·sinα|²= (1/2)·sin²α, which gives us the following graph of P:

The probability varies between 0 (for α = 0 or π) and 1/2 = 0.5 (for α = π/2 or 3π/2). Now, this graph may or may not make sense to you, so you should think about it. You’ll admit it makes sense to find P = 0 for α = 0, but what about the non-zero values?

Think about what this would mean in classical terms: we’ve got a beam of particles whose angular momentum is ‘up’ in the z-direction. To be precise, this means that J_z = +ħ. [Angular momentum and the quantum of action have the same dimension: the joule·second.] So that’s the maximum value out of the three permitted values, which are +ħ, 0 and –ħ. Note that the particles here must be bosons. So you may think we’re talking photons, in practice but… Well… No. As I’ll explain in a later post, the photon is a spin-one particle but it’s quite particular, because it has no ‘zero spin’-state. Don’t worry about it here – but it’s really quite remarkable. So, instead of thinking of a photon, you should think of some composite matter particle obeying Bose-Einstein statistics. These are not so rare as you may think: all matter-particles that contain an even number of fermions – like elementary particles – have integer spin – but… Well… Their spin number is usually zero – not one. So… Well… Feynman’s particle here is somewhat theoretical – but it doesn’t matter. Let’s move on. 🙂

Let’s look at another transformation formula. More in particular, let’s look at the formula we (should) get for 〈 0T | −S 〉 as a function of α. So we change the set-up of the S-filter to ensure all particles entering T have negative spin. The formula is:

〈 0T | −S 〉 = +(1/√2)·sinα

That gives the same probabilities: |+(1/√2)·sinα|²= (1/2)·sin²α. Adding |〈 0T | +S 〉|² and |〈 0T | −S 〉|²gives us a total probability equal to sin²α, which is equal to 1 if α = π/2 or 3π/2. We may be tempted to interpret this as follows: if a particle is in the +S or −S state before entering the T-apparatus, and the T-apparatus is tilted at an angle α = π/2 or 3π/2 with respect to the S-apparatus, then this particle will come out of the T-apparatus in the 0T-state. No ambiguity here: P = 1.

Is this strange? Well… Let’s think about what it means to tilt the T-apparatus. You’ll have to admit that, if the apparatus is tilted at the angle π/2 or 3π/2, it’s going to measure the angular momentum in the x-direction. [The y-direction is the common axis of both apparatuses here.] So… Well… It’s pretty plausible, isn’t it? If all of the angular momentum is in the positive or negative z-direction, then it’s not going to have any angular momentum in the x-direction, right? And not having any angular momentum in the x-direction effectively corresponds to being in the 0T-state, right?

Oh ! Is it that easy?

Well… No! Not at all! The reasoning above shows how easy it is to be led astray. We forgot to normalize. Remember, if we integrate the probability density function over its domain, i.e. α ∈ [0, 2π], then we have to get one, as all probabilities have to add up to one. The definite integral of (1/2)·sin²α over [0, 2π] is equal to π/2 (the definite integral of the sine or cosine squared over a full cycle is equal to π), so we need to multiply this function by 2/π to get the actual probability density function, i.e. (1/π)·sin²α. It’s got the same shape, obviously, but it gives us maximum probabilities equal to 1/π ≈ 0.32 for α = π/2 or 3π/2, instead of 1/2 = 0.5.

Likewise, the sin²α function we got when adding |〈 0T | +S 〉|² and |〈 0T | −S 〉|²should also be normalized. One really needs to keep one’s wits about oneself here. What we’re saying here is that we have a particle that is either in the +S or the −S state, so let’s say that the chance is 50/50 to be in either of the two states. We then have these probabilities |〈 0T | +S 〉|² and |〈 0T | −S 〉|², which we calculated as (1/π)·sin²α. So the total combined probability is equal to 0.5·(1/π)·sin²α + 0.5·(1/π)·sin²α = (1/π)·sin²α. So we’re now weighing the two (1/π)·sin²α functions – and it doesn’t matter if the weights are 50/50 or 75/25 or whatever, as long as the two weights add up to one. The bottom line is: we get the same (1/π)·sin²α function for P, and the same maximum probability 1/π ≈ 0.32 for α = π/2 or 3π/2.

So we don’t get unity: P ≠ 1 for α = π/2 or 3π/2. Why not? Think about it. The classical analysis made sense, didn’t it? If the angular momentum is all in the z-direction (or in one of the two z-directions, I should say), then we cannot have any of it in the x-direction, can it? Well… The surprising answer is: yes, we can. The remarkable thing is that, in quantum physics, we actually never have all of the angular momentum in one direction. As I explained in my post on spin and angular momentum, the classical concepts of angular momentum, and the related magnetic moment, have their limits in quantum mechanics. In quantum physics, we find that the magnitude of a vector quantity, like angular momentum, or the related magnetic moment, is generally not equal to the maximum value of the component of that quantity in any direction. The general rule is that the maximum value of any component of J in whatever direction – i.e. +ħ in the example we’re discussing here – is smaller than the magnitude of J – which I calculated in the mentioned post as |J| = J = +√2·ħ ≈ 1.414·ħ, so that’s almost 1.5 times ħ! So it’s quite a bit smaller! The upshot is that we cannot associate any precise and unambiguous direction with quantities like the angular momentum J or the magnetic moment μ. So the answer is: the angular momentum can never be all in the z-direction, so we can always have some of it in the x-direction, and so that explains the amplitudes and probabilities we’re having here.

Huh?

Yep. I know. We never seem to get out of this ‘weirdness’, but then that’s how quantum physics is like. Feynman warned us upfront:

“Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to, and it appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct, human experience and of human intuition applies to large objects. We know how large objects will act, but things on a small scale just do not act that way. So we have to learn about them in a sort of abstract or imaginative fashion and not by connection with our direct experience.”

As I see it, quantum physics is about explaining all sorts of weird stuff, like electron interference and tunneling and what have you, so it shouldn’t surprise us that the framework is as weird as the stuff it’s trying to explain. 🙂 So… Well… All we can do is to try to go along with it, isn’t it? And so that’s what we’ll do here. 🙂

Transformations: the formulas

We need to distinguish various cases here. The first case is the case explained above: the T-apparatus shares the same y-axis – along which the particles move – but it’s tilted. To be precise, we should say that it’s rotated about the common y-axis by the angle α. That implies we can relate the x’, y’, z’ coordinate system of T to the x, y, z coordinate system of S through the following equations: z′ = z·cosα + x·sinα, x′ = x·cosα − z·sinα, and y′ = y. Then the transformation amplitudes are:

We used the formula for 〈 0T | +S 〉 and 〈 0T | −S 〉 above, and you can play with the formulas above by imagining the related set-up of the S and T filters, such as the one below:

If you do your homework (just check what formula and what set-up this corresponds to), you should find the following graph for the amplitude and the probability as a function of α: the graph is zero for α = π, but is non-zero everywhere else. As with the other example, you should think about this. It makes sense—sort of, that is. 🙂

OK. Next case. Now we’re going to rotate the T-apparatus around the z-axis by some angle β. To illustrate what we’re doing here, we need to take a ‘top view’ of our apparatus, as shown below, which shows a rotation over 90°. More in general, for any angle β, the coordinate transformation is given by z′ = z, x′ = x·cosβ + y·sinβ, y′ = y·cosβ − x·sinβ. [So it’s quite similar to case 1: we’re only rotating the thing in a different plane.]

The transformation amplitudes are now given by:

As you can see, we get complex-valued transformation amplitudes, unlike our first case, which yielded real-valued transformation amplitudes. That’s just the way it is. Nobody says transformation amplitudes have to be real-valued. On the contrary, one would expect them to be complex numbers. 🙂 Having said that, the combined set of transformation formulas is, obviously, rather remarkable. The amplitude to go from the +S state to, say, the 0T state is zero. Also, when our particle has zero spin when coming out of S, it will always have zero spin when and if it goes through T. In fact, the absolute value of those e^±iβ functions is also equal to one, so they are also associated with probabilities that are equal to one: |e^±iβ|² = 1² = 1. So… Well… Those formulas are simple and weird at the same time, aren’t they? They sure give us plenty of stuff to think about, I’d say.

So what’s next? Well… Not all that much. We’re sort of done, really. Indeed, it’s just a property of space that we can get any rotation of T by combining the two rotations above. As I only want to introduce the basic concepts here, I’ll refer you to Feynman for the details of how exactly that’s being done. [He illustrates it for spin-1/2 particles in particular.] I’ll just wrap up here by generalizing our results from base states to any state.

Transformations: generalization

We mentioned a couple of times already that the base states are like a particular coordinate system: we will usually describe a state in terms of base states indeed. More in particular, choosing S as our representation, we’ll say:

The state φ is defined by the three numbers:

C₊ = 〈 +S | φ 〉,

C₀ = 〈 0S | φ 〉,

C₋ = 〈 −S | φ 〉.

Now, the very same state can, of course, also be described in the ‘T system’, so then our numbers – i.e. the ‘components’ of φ – would be equal to:

C’₊ = 〈 +T | φ 〉, C’₀ = 〈 0T | φ 〉, and C’₋ = 〈 −T | φ 〉.

So how can we go from the unprimed ‘coordinates’ to the primed ones? The trick is to use the second of the three quantum math ‘Laws’ which I introduced in my previous post:

Capture

Just replace χ in [II] by +T, 0T and/or –T. More in general, if we denote +T, 0T or –T by jT, we can re-write this ‘Law’ as:

So the 〈 jT | iS 〉 amplitudes are those nine transformation amplitudes. Now, we can represent those nine amplitudes in a nice three-by-three matrix and, yes, we’ll call that matrix the transformation matrix. So now you know what that is.

To conclude, I should note that it’s only because we’re talking spin-one particles here that we have three base states here and, hence, three ‘components’, which we denoted by C₊, C₋ and C₀, which transform the way they do when going from one representation to another, and so that is very much like what vectors do when we move to a different coordinate system, which is why spin-one particles are often referred to as ‘vector particles‘. [I am just mentioning this in case you’d come across the term and wonder why they’re being called that way. Now you know.] In fact, if we have three base states, in respect to whatever representation, and we define some state φ in terms of them, then we can always re-define that state in terms of the following ‘special’ set of components:

The set is ‘special’ because one can show (you can do that yourself that by using those transformation laws) that these components transform exactly the way as x, y, z transform to x′, y′, z′. But so I’ll leave at this.

[…]

Oh… What about the Hamiltonian? Well… I’ll save that for my next posts, as my posts have become longer and longer, and so it’s probably a good idea to separate them out. 🙂

Post scriptum: transformations for spin-1/2 particles

You should actually really check out that chapter of Feynman. The transformation matrices for spin-1/2 particles look different because… Well… Because there’s only two base states for spin-1/2 particles. It’s a pretty technical chapter, but then spin-1/2 particles are the ones that make up the world. 🙂

Quantum math: the rules – all of them! :-)

Original post:

In my previous post, I made no compromise, and used all of the rules one needs to calculate quantum-mechanical stuff:

However, I didn’t explain them. These rules look simple enough, but let’s analyze them now. They’re simple and not at the same time, indeed.

[I] The first equation uses the Kronecker delta, which sounds fancy but it’s just a simple shorthand: δ_ij= δ_jiis equal to 1 if i = j, and zero if i ≠ j, with i and j representing base states. Equation (I) basically says that base states are all different. For example, the angular momentum in the x-direction of a spin-1/2 particle – think of an electron or a proton – is either +ħ/2 or −ħ/2, not something in-between, or some mixture. So 〈 +x | +x 〉 = 〈 −x | −x 〉 = 1 and 〈 +x | −x 〉 = 〈 −x | +x 〉 = 0.

We’re talking base states here, of course. Base states are like a coordinate system: we settle on an x-, y- and z-axis, and a unit, and any point is defined in terms of an x-, y– and z-number. It’s the same here, except we’re talking ‘points’ in four-dimensional spacetime. To be precise, we’re talking constructs evolving in spacetime. To be even more precise, we’re talking amplitudes with a temporal as well as a spatial frequency, which we’ll often represent as:

a·e^−i·θ= e^{−i·(ω·t − k ∙x)} = a·e^{−(i/ħ)·(E·t − p∙x)}

The coefficient in front (a) is just a normalization constant, ensuring all probabilities add up to one. It may not be a constant, actually: perhaps it just ensure our amplitude stays within some kind of envelope, as illustrated below.

As for the ω = E/ħ and k = p/ħ identities, these are the de Broglie equations for a matter-wave, which the young Comte jotted down as part of his 1924 PhD thesis. He was inspired by the fact that the E·t − p∙x factor is an invariant four-vector product (E·t − p∙x = p_μx_μ) in relativity theory, and noted the striking similarity with the argument of any wave function in space and time (ω·t − k ∙x) and, hence, couldn’t resist equating both. Louis de Broglie was inspired, of course, by the solution to the blackbody radiation problem, which Max Planck and Einstein had convincingly solved by accepting that the ω = E/ħ equation holds for photons. As he wrote it:

“When I conceived the first basic ideas of wave mechanics in 1923–24, I was guided by the aim to perform a real physical synthesis, valid for all particles, of the coexistence of the wave and of the corpuscular aspects that Einstein had introduced for photons in his theory of light quanta in 1905.” (Louis de Broglie, quoted in Wikipedia)

Looking back, you’d of course want the phase of a wavefunction to be some invariant quantity, and the examples we gave our previous post illustrate how one would expect energy and momentum to impact its temporal and spatial frequency. But I am digressing. Let’s look at the second equation. However, before we move on, note that minus sign in the exponent of our wavefunction: a·e^−i·θ. The phase turns counter-clockwise. That’s just the way it is. I’ll come back to this.

[II] The φ and χ symbols do not necessarily represent base states. In fact, Feynman illustrates this law using a variety of examples including both polarized as well as unpolarized beams, or ‘filtered’ as well as ‘unfiltered’ states, as he calls it in the context of the Stern-Gerlach apparatuses he uses to explain what’s going on. Let me summarize his argument here.

I discussed the Stern-Gerlach experiment in my post on spin and angular momentum, but the Wikipedia article on it is very good too. The principle is illustrated below: a inhomogeneous magnetic field – note the direction of the gradient ∇B = (∂B/∂x, ∂B/∂y, ∂B/∂z) – will split a beam of spin-one particles into three beams. [Matter-particles with spin one are rather rare (Lithium-6 is an example), but three states (rather than two only, as we’d have when analyzing spin-1/2 particles, such as electrons or protons) allow for more play in the analysis. 🙂 In any case, the analysis is easily generalized.]

The splitting of the beam is based, of course, on the quantized angular momentum in the z-direction (i.e. the direction of the gradient): its value is either ħ, 0, or −ħ. We’ll denote these base states as +, 0 or −, and we should note they are defined in regard to an apparatus with a specific orientation. If we call this apparatus S, then we can denote these base states as +S, 0S and −S respectively.

The interesting thing in Feynman’s analysis is the imagined modified Stern-Gerlach apparatus, which – I am using Feynman‘s words here 🙂 – “puts Humpty Dumpty back together.” It looks a bit monstruous, but it’s easy enough to understand. Quoting Feynman once more: “It consists of a sequence of three high-gradient magnets. The first one (on the left) is just the usual Stern-Gerlach magnet and splits the incoming beam of spin-one particles into three separate beams. The second magnet has the same cross section as the first, but is twice as long and the polarity of its magnetic field is opposite the field in magnet $1$ . The second magnet pushes in the opposite direction on the atomic magnets and bends their paths back toward the axis, as shown in the trajectories drawn in the lower part of the figure. The third magnet is just like the first, and brings the three beams back together again, so that leaves the exit hole along the axis.”

Now, we can use this apparatus as a filter by inserting blocking masks, as illustrated below.

But let’s get back to the lesson. What about the second ‘Law’ of quantum math? Well… You need to be able to imagine all kinds of situations now. The rather simple set-up below is one of them: we’ve got two of these apparatuses in series now, S and T, with T tilted at the angle α with respect to the first.

I know: you’re getting impatient. What about it? Well… We’re finally ready now. Let’s suppose we’ve got three apparatuses in series, with the first and the last one having the very same orientation, and the one in the middle being tilted. We’ll denote them by S, T and S’ respectively. We’ll also use masks: we’ll block the 0 and − state in the S-filter, like in that illustration above. In addition, we’ll block the + and − state in the T apparatus and, finally, the 0 and − state in the S’ apparatus. Now try to imagine what happens: how many particles will get through?

[…]

Just try to think about it. Make some drawing or something. Please!

[…]

OK… The answer is shown below. Despite the filtering in S, the +S particles that come out do have an amplitude to go through the 0T-filter, and so the number of atoms that come out will be some fraction (α) of the number of atoms (N) that came out of the +S-filter. Likewise, some other fraction (β) will make it through the +S’-filter, so we end up with βαN particles.

Now, I am sure that, if you’d tried to guess the answer yourself, you’d have said zero rather than βαN but, thinking about it, it makes sense: it’s not because we’ve got some angular momentum in one direction that we have none in the other. When everything is said and done, we’re talking components of the total angular momentum here, don’t we? Well… Yes and no. Let’s remove the masks from T. What do we get?

[…]

Come on: what’s your guess? N?

[…] You’re right. It’s N. Perfect. It’s what’s shown below.

Now, that should boost your confidence. Let’s try the next scenario. We block the 0 and − state in the S-filter once again, and the + and − state in the T apparatus, so the first two apparatuses are the same as in our first example. But let’s change the S’ apparatus: let’s close the + and − state there now. Now try to imagine what happens: how many particles will get through?

[…]

Come on! You think it’s a trap, isn’t it? It’s not. It’s perfectly similar: we’ve got some other fraction here, which we’ll write as γαN, as shown below.

Next scenario: S has the 0 and − gate closed once more, and T is fully open, so it has no masks. But, this time, we set S’ so it filters the 0-state with respect to it. What do we get? Come on! Think! Please!

[…]

The answer is zero, as shown below.

Does that make sense to you? Yes? Great! Because many think it’s weird: they think the T apparatus must ‘re-orient’ the angular momentum of the particles. It doesn’t: if the filter is wide open, then “no information is lost”, as Feynman puts it. Still… Have a look at it. It looks like we’re opening ‘more channels’ in the last example: the S and S’ filter are the same, indeed, and T is fully open, while it selected for 0-state particles before. But no particles come through now, while with the 0-channel, we had γαN.

Hmm… It actually is kinda weird, won’t you agree? Sorry I had to talk about this, but it will make you appreciate that second ‘Law’ now: we can always insert a ‘wide-open’ filter and, hence, split the beams into a complete set of base states − with respect to the filter, that is − and bring them back together provided our filter does not produce any unequal disturbances on the three beams. In short, the passage through the wide-open filter should not result in a change of the amplitudes. Again, as Feynman puts it: the wide-open filter should really put Humpty-Dumpty back together again. If it does, we can effectively apply our ‘Law’:

For an example, I’ll refer you to my previous post. This brings me to the third and final ‘Law’.

[III] The amplitude to go from state φ to state χ is the complex conjugate of the amplitude to to go from state χ to state φ:

〈 χ | φ 〉 = 〈 φ | χ 〉*

This is probably the weirdest ‘Law’ of all, even if I should say, straight from the start, we can actually derive it from the second ‘Law’, and the fact that all probabilities have to add up to one. Indeed, a probability is the absolute square of an amplitude and, as we know, the absolute square of a complex number is also equal to the product of itself and its complex conjugate:

|z|²= |z|·|z| = z·z*

[You should go through the trouble of reviewing the difference between the square and the absolute square of a complex number. Just write z as a + ib and calculate (a + ib)²= a² + 2abi + b², as opposed to |z|²= a² + b². Also check what it means when writing z as r·e^iθ= r·(cosθ + i·sinθ).]

Let’s applying the probability rule to a two-filter set-up, i.e. the situation with the S and the tilted T filter which we described above, and let’s assume we’ve got a pure beam of +S particles entering the wide-open T filter, so our particles can come out in either of the three base states with respect to T. We can then write:

〈 +T | +S 〉²+ 〈 0T | +S 〉²+ 〈 −T | +S 〉²= 1

⇔ 〈 +T | +S 〉〈 +T | +S 〉* + 〈 0T | +S 〉〈 0T | +S 〉* + 〈 −T | +S 〉〈 −T | +S 〉* = 1

Of course, we’ve got two other such equations if we start with a 0S or a −S state. Now, we take the 〈 χ | φ 〉 = ∑ 〈 χ | i 〉〈 i | φ 〉 ‘Law’, and substitute χ and φ for +S, and all i states for the base states with regard to T. We get:

〈 +S | +S 〉 = 1 = 〈 +S | +T 〉〈 +T | +S 〉 + 〈 +S | 0T 〉〈 0T | +S 〉 + 〈 +S | –T 〉〈 −T | +S 〉

These equations are consistent only if:

〈 +S | +T 〉 = 〈 +T | +S 〉*,

〈 +S | 0T 〉 = 〈 0T | +S 〉*,

〈 +S | −T 〉 = 〈 −T | +S 〉*,

which is what we wanted to prove. One can then generalize to any state φ and χ. However, proving the result is one thing. Understanding it is something else. One can write down a number of strange consequences, which all point to Feynman‘s rather enigmatic comment on this ‘Law’: “If this Law were not true, probability would not be ‘conserved’, and particles would get ‘lost’.” So what does that mean? Well… You may want to think about the following, perhaps. It’s obvious that we can write:

|〈 φ | χ 〉|²= 〈 φ | χ 〉〈 φ | χ 〉* = 〈 χ | φ 〉*〈 χ | φ 〉 = |〈 χ | φ 〉|²

This says that the probability to go from the φ-state to the χ-state is the same as the probability to go from the χ-state to the φ-state.

Now, when we’re talking base states, that’s rather obvious, because the probabilities involved are either 0 or 1. However, if we substitute for +S and −T, or some more complicated states, then it’s a different thing. My guts instinct tells me this third ‘Law’ – which, as mentioned, can be derived from the other ‘Laws’ – reflects the principle of reversibility in spacetime, which you may also interpret as a causality principle, in the sense that, in theory at least (i.e. not thinking about entropy and/or statistical mechanics), we can reverse what’s happening: we can go back in spacetime.

In this regard, we should also remember that the complex conjugate of a complex number in polar form, i.e. a complex number written as r·e^iθ, is equal to r·e^−iθ, so the argument in the exponent gets a minus sign. Think about what this means for our a·e^−i·θ= e^{−i·(ω·t − k ∙x)} = a·e^{−(i/ħ)·(E·t − p∙x)}function. Taking the complex conjugate of this function amounts to reversing the direction of t and x which, once again, evokes that idea of going back in spacetime.

I feel there’s some more fundamental principle here at work, on which I’ll try to reflect a bit more. Perhaps we can also do something with that relationship between the multiplicative inverse of a complex number and its complex conjugate, i.e. z⁻¹= z*/|z|². I’ll check it out. As for now, however, I’ll leave you to do that, and please let me know if you’ve got any inspirational ideas on this. 🙂

So… Well… Goodbye as for now. I’ll probably talk about the Hamiltonian in my next post. I think we really did a good job in laying the groundwork for the really hardcore stuff, so let’s go for that now. 🙂

Post Scriptum: On the Uncertainty Principle and other rules

After writing all of the above, I realized I should add some remarks to make this post somewhat more readable. First thing: not all of the rules are there—obviously! Most notably, I didn’t say anything about the rules for adding or multiplying amplitudes, but that’s because I wrote extensively about that already, and so I assume you’re familiar with that. [If not, see my page on the essentials.]

Second, I didn’t talk about the Uncertainty Principle. That’s because I didn’t have to. In fact, we don’t need it here. In general, all popular accounts of quantum mechanics have an excessive focus on the position and momentum of a particle, while the approach in this and my previous post is quite different. Of course, it’s Feynman’s approach to QM really. Not ‘mine’. 🙂 All of the examples and all of the theory he presents in his introductory chapters in the Third Volume of Lectures, i.e. the volume on QM, are related to things like:

What is the amplitude for a particle to go from spin state +S to spin state −T?
What is the amplitude for a particle to be scattered, by a crystal, or from some collision with another particle, in the θ direction?
What is the amplitude for two identical particles to be scattered in the same direction?
What is the amplitude for an atom to absorb or emit a photon? [See, for example, Feynman’s approach to the blackbody radiation problem.]
What is the amplitude to go from one place to another?

In short, you read Feynman, and it’s only at the very end of his exposé, that he starts talking about the things popular books start with, such as the amplitude of a particle to be at point (x, t) in spacetime, or the Schrödinger equation, which describes the orbital of an electron in an atom. That’s where the Uncertainty Principle comes in and, hence, one can really avoid it for quite a while. In fact, one should avoid it for quite a while, because it’s now become clear to me that simply presenting the Uncertainty Principle doesn’t help all that much to truly understand quantum mechanics.

Truly understanding quantum mechanics involves understanding all of these weird rules above. To some extent, that involves dissociating the idea of the wavefunction with our conventional ideas of time and position. From the questions above, it should be obvious that ‘the’ wavefunction does actually not exist: we’ve got a wavefunction for anything we can and possibly want to measure. That brings us to the question of the base states: what are they?

Feynman addresses this question in a rather verbose section of his Lectures titled: What are the base states of the world? I won’t copy it here, but I strongly recommend you have a look at it. 🙂

I’ll end here with a final equation that we’ll need frequently: the amplitude for a particle to go from one place (r₁) to another (r₂). It’s referred to as a propagator function, for obvious reasons—one of them being that physicists like fancy terminology!—and it looks like this:

The shape of the e^{(i/ħ)·(p∙r₁₂)}function is now familiar to you. Note the r₁₂in the argument, i.e. the vector pointing from r₁ to r₂. The p∙r₁₂ dot product equals |p|∙|r₁₂|·cosθ = p∙r₁₂·cosθ, with θ the angle between p and r₁₂. If the angle is the same, then cosθ is equal to 1. If the angle is π/2, then it’s 0, and the function reduces to 1/r₁₂. So the angle θ, through the cosθ factor, sort of scales the spatial frequency. Let me try to give you some idea of how this looks like by assuming the angle between p and r₁₂ is the same, so we’re looking at the space in the direction of the momentum only and |p|∙|r₁₂|·cosθ = p∙r₁₂. Now, we can look at the p/ħ factor as a scaling factor, and measure the distance x in units defined by that scale, so we write: x = p∙r₁₂/ħ. The function then reduces to (ħ/p)·e^i∙x/x = (ħ/p)·cos(x)/x + i·(ħ/p)·sin(x)/x, and we just need to square this to get the probability. All of the graphs are drawn hereunder: I’ll let you analyze them. [Note that the graphs do not include the ħ/p factor, which you may look at as yet another scaling factor.] You’ll see – I hope! – that it all makes perfect sense: the probability quickly drops off with distance, both in the positive as well as in the negative x-direction, while it’s going to infinity when very near. [Note that the absolute square, using cos(x)/x and sin(x)/xyields the same graph as squaring 1/x—obviously!]

Taking the magic out of God’s number: some additional reflections

Note: I have published a paper that is very coherent and fully explains this so-called God-given number. There is nothing magical about it. It is just a scaling constant. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

In my previous post, I explained why the fine-structure constant α is not a ‘magical’ number, even if it relates all fundamental properties of the electron: its mass, its energy, its charge, its radius, its photon scattering cross-section (i.e. the Bohr radius, or the size of the atom really) and, finally, the coupling constant for photon-electron interactions. The key to such understanding of α was the model of an electron as a tiny ball of charge. As such, we have two energy formulas for it. One is the energy that’s needed to assemble the charge from infinitely dispersed infinitesimal charges, which we denoted as U_elec. The other formula is the energy of the field of the tiny ball of charge, which we denoted as E_elec.

The formula for E_elec is calculated using the formula for the field momentum of a moving charge and, using the m = E/c²mas-energy equivalence relationship, is equivalent to the electromagnetic mass. We went through the derivation in our previous post, so let me just jot down the result:

The second formula depends on what ball of charge we’re thinking of, because the formulas for a charged sphere and a spherical shell of charge are different: both have the same structure as the relationship above (so the energy is also proportional to the square of the electron charge and inversely proportional to the radius a), but the constant of proportionality is different. For a sphere of charge, we write:

For a spherical shell of charge we write:

shell

To compare the formulas, you need to note that the square of the electron charge e in the formula for the field energy is equal to e²= q_e²/4πε₀= k_e·q_e². So we multiply the square of the actual electron charge by the Coulomb constant k_e= 1/4πε₀. As you can see, the three formulas have exactly the same form then. It’s just the proportionality constant that’s different: it’s 2/3, 3/5 and 1/2 respectively. It’s interesting to quickly reflect on the dimensions here: [k_e] ≈ 9×10⁹N·m²/C², so e² is expressed in N·m². That makes the units come out alright, as we divide by a (so that’s in meter) and so we get the energy in joule (which is newton·meter). In fact, now that we’re here, let’s quickly calculate the value of e²: it’s that k_e·q_e² product, so it’s equal to 2.3×10⁻²⁸N·m². We can quickly check this value because we know that the classical electron radius is equal to:

So we divide 2.3×10⁻²⁸N·m²by m_ec_²≈ 8.2×10⁻¹⁴J, so we get r₀≈ 2.82×10⁻¹⁵m. So we’re spot on! Why did I do this check? Not really to check what I wrote. It’s more to show what’s going on. We’ve got yet another formula relating the energy and the radius of an electron here, so now we have three. In fact we have more because the formula for U_elec depends on the finer details of our model for the electron (sphere versus shell, uniform versus non-uniform distribution):

E_elec= (2/3)·(e²/a): This is the formula for the energy of the field, so we may all it is external energy.
U_elec= (3/5)·(e²/a), or U_elec= (1/2)·(e²/a): This is the energy needed to assemble our electron, so we might, perhaps, call it its internal energy. The first formula assumes our electron is a uniformly charged sphere. The second assumes all charges sit on the surface of the sphere. If we drop the assumption of the charge having to be uniformly distributed, we’ll find yet another formula.
m_ec²= e²/r₀: This is the energy associated with the so-called classical electron radius (r₀) and the electron’s rest mass (m_e).

In our previous posts, we assumed the last equation was the right one. Why? Because it’s the one that’s been verified experimentally. The discrepancies between the various proportionality coefficients – i.e. the difference between 2/3 and 1, basically – are to be explained because of the binding forces within the electron, without which the electron would just ‘explode’, as the French physicist and polymath Henri Poincaré famously put it. Indeed, if the electron is a little ball of negative charge, the repulsive forces between its parts should rip it apart. So we will not say anything more about this. You can have fun yourself by googling all the various theories that try to model these binding forces. [I may do the same some day, but now I’ve got other priorities: I want to move to Feynman’s third volume of Lectures, which is devoted to quantum physics only, so I look very much forward to that.]

In this post, I just wanted to reflect once more on what constants are really fundamental and what constants are somewhat less fundamental. From all what I wrote in my previous post, I said there were three:

The fine-structure constant α, which is a dimensionless number.
Planck’s constant h, whose dimension is joule·second, so that’s the dimension of action.
The speed of light c, whose dimension is that of a velocity.

The three are related through the following expression:

This is an interesting expression. Let’s first check its dimension. We already explained that e_² is expressed in N·m². That’s rather strange, because it means the dimension of e itself is N^1/2·m: what’s the square root of a force of one newton? In fact, to interpret the formula above, it’s probably better to re-write e²as e²= q_e²/4πε₀= k_e·q_e². That shows you how the electron charge and Coulomb’s constant are related. Of course, they are part and parcel of one and the same force law: Coulomb’s law. We don’t need anything else, except for relativity theory, because we need to explain the magnetic force as well—and that we can do because magnetism is just a relativistic effect. Think of the field momentum indeed: the magnetic field comes into play only when we start to move our electron. The relativity effect is captured by c in that formula for α above. As for ħ, ħ = h/2π comes with the E = h·f equation, which links us to the electron’s Compton wavelength λ through the de Broglie relation λ = h/p.

The point is: we should probably not look at α as a ‘fundamental physical constant’. It’s e² that’s the third fundamental constant, besides h and c. Indeed, it’s from e² that all the rest follows: the electron’s internal energy, its external energy, and its radius, and then all the rest by combining stuff with other stuff.

Now, we took the magic out of α by doing what we did in the previous posts, and that’s to combine stuff with other stuff, and so now you may think I am putting the magic back in with that formula for α, which seems to define α in terms of the three mentioned ‘fundamental’ constants. That’s not the case: this relation comes out of all of the other relationships we found, and so it’s nothing new really. It’s actually not a definition of α: it just does what it does, and that’s to relate α to the ‘fundamental’ physical constants behind.

So… No new magic. In fact, I want to close this post by taking away even more of the magic. If you read my previous post, I said that α was ‘God’s cut-off factor’ 🙂 ensuring our energy functions do not blow up, but I also said it was impossible to say why he chose 0.00729735256 as the cut-off factor. The question is actually easily answered by thinking about those two formulas we had for the internal and external energy respectively. Let’s re-write them in natural units and, temporarily, two different subscripts for α, so we write:

E_elec= α_e/r₀: This is the formula for the energy of the field.
U_elec= α_u/r₀: This is the energy needed to assemble our electron.

Both energies are determined by the above-mentioned laws, i.e. Coulomb’s Law and the theory of relativity, so α has got nothing to do what that. However, both energies have to be the same, and so α_ehas to be equal to α_u. In that sense, α is, quite simply, a proportionality constant that achieves that equality. Now that explains why we can derive α from the three other constants which, as mentioned above, are probably more fundamental. In fact, we’ve got only three degrees of freedom here, so if we chose c, h and e as ‘fundamental’, then α isn’t any more.

The underlying deep question behind it all is why those two energies should be equal. Why would our electron have some internal energy if it’s elementary? The answer to that question is: because it has some non-zero radius, and it has some non-zero radius because we don’t want our formula for the field energy (or the field momentum) to blow up. Now, if it has some radius, then it has to have some internal energy.

You’ll say: that makes sense, but it doesn’t answer the question. Why would it have internal energy, with or without a zero radius? If an electron is an elementary particle, then it’s really elementary, isn’t? And so then we shouldn’t try to ‘assemble’ it from an infinite number of infinitesimally small charges. You’re right, and here we can also note that the fact that the electron doesn’t blow up is firm evidence it’s very elementary indeed.

I should also note that Feynman actually doesn’t talk about the energy that’s needed to assemble a charge: he gets his U_elec= (1/2)·(e²/a) by calculating the external field energy for a spherical shell of charge, and he sticks to it—presumably because it’s the same field for a uniform or non-uniform sphere of charge. He only notes there has to be some radius because, if not, the formula he uses blows up, indeed. So – who knows? – perhaps he doesn’t quite believe that formula for the internal energy is relevant either.

So perhaps there is no internal energy indeed. Perhaps there’s just the energy of the field. So… Well… I can’t say much about this… Except… Well… Perhaps just one more thing. Let me note something that, I hope, you noticed as well: the k_e·q_e²is the numerator in Coulomb’s Law itself. You also know that energy equals force times distance. So if we divide both sides by r₀, we get Coulomb’s Law itself F_elec= k_e·q_e²/r₀². The only thing is: what’s the distance? It’s one charge only, and there is no distance between one charge, is there? Well… Yes and no. I have been thinking that the requirement of the internal and external energies being equal resembles the statement that the forces between two charges are equal and opposite. That ties in with the idea of the internal energy itself: remember we were basically talking forces between infinitesimally small elements of charge within the electron itself? So r₀ is, perhaps, some average distance or so. There must be some way of thinking of it like that. But… Well… Which one exactly?

This kind of reflection may not make sense. Who knows? I obviously need to think all of this through and so this post is, indeed, just a bunch of reflections for which I will have more time later—hopefully. 🙂 Perhaps we’re all just pushing the matter too far. Perhaps we should just accept that the external energy has that 2/3 factor but that the actual energy of the electron should also include the equivalent energy of some binding force that holds the electron together. Well… In any case. That’s all I am going to do on this extremely complicated matter. It’s time to move indeed! So the point to take home here is probably just this:

When calculating the radius of an electron using classical theory, we get in trouble: not only do we find different radii, but the radii that we find do not respect the E = m_ec²law. It’s only the m_ec²= e²/r₀ that’s relativistically correct.
That suggests the electron also has some non-electric mass, which are referred to as ‘binding forces’ or ‘Poincaré stresses’, but which remain to be explained convincingly.
All of this shouldn’t surprise us: for all we know, the electron is something fuzzy. 🙂

So my next posts will focus on the ‘essentials’ preparing for Feynman’s Volume on quantum mechanics. Those ‘essentials’ will still involve some classical stuff but, as you will see, even more contradictions, that – hopefully! – will then be solved in the quantum-mechanical picture of it all. 🙂

Taking the magic out of God’s number

Jean Louis Van Belle, 23 December 2018

Original post:

I think the post scriptum to my previous post is interesting enough to separate it out as a piece of its own, so let me do that here. You’ll remember that we were trying to find some kind of a model for the electron, picturing it like a tiny little ball of charge, and then we just applied the classical energy formulas to it to see what comes out of it. The key formula is the integral that gives us the energy that goes into assembling a charge. It was the following thing:

This is a double integral which we simplified in two stages, so we’re looking at an integral within an integral really, but we can substitute the integral over the ρ(2)·dV₂product by the formula we got for the potential, so we write that as Φ(1), and so the integral above becomes:

Now, this integral integrates the ρ(1)·Φ(1)·dV₁product over all of space, so that’s over all points in space, and so we just dropped the index and wrote the whole thing as the integral of ρ·Φ·dV over all of space:

We then established that this integral was mathematically equivalent to the following equation:

So this integral is actually quite simple: it just integrates E•E = E² over all of space. The illustration below shows E as a function of the distance r for a sphere of radius R filled uniformly with charge.

So the field (E) goes as r for r ≤ R and as 1/r²for r ≥ R. So, for r ≥ R, the integral will have (1/r²)² = 1/r⁴in it. Now, you know that the integral of some function is the surface under the graph of that function. Look at the 1/r⁴function below: it blows up between 1 and 0. That’s where the problem is: there needs to be some kind of cut-off, because that integral will effectively blow up when the radius of our little sphere of charge gets ‘too small’. So that makes it clear why it doesn’t make sense to use this formula to try to calculate the energy of a point charge. It just doesn’t make sense to do that.

In fact, the need for a ‘cut-off factor’ so as to ensure our energy function doesn’t ‘blow up’ is not because of the exponent in the 1/r⁴expression: the need is also there for any 1/r relation, as illustrated below. All 1/rⁿfunction have the same pivot point, as you can see from the simple illustration below. So, yes, we cannot go all the way to zero from there when integrating: we have to stop somewhere.

So what’s the ‘cut-off point’? What’s ‘too small’ a radius? Let’s look at the formula we got for our electron as a shell of charge (so the assumption here is that the charge is uniformly distributed on the surface of a sphere with radius a):

So we’ve got an even simpler formula here: it’s just a 1/r relation (a is r in this formula), not 1/r⁴. Why is that? Well… It’s just the way the math turns out: we’re integrating over volumes and so that involves an r³ factor and so it all simplifies to 1/r, and so that gives us this simple inversely proportional relationship between U and r, i.e a, in this case. 🙂 I copied the detail of Feynman’s calculation in my previous post, so you can double-check it. It’s quite wonderful, really. Look at it again: we have a very simple inversely proportional relationship between the radius of our electron and its energy as a sphere of charge. We could write it as:

U_elect = α/a, with α = e²/2

Still… We need the cut-off point’. Also note that, as I pointed out, we don’t necessarily need to assume that the charge in our little ball of charge (i.e. our electron) sits on the surface only: if we’d assume it’s a uniformly charged sphere of charge, we’d just get another constant of proportionality: our 1/2 factor would become a 3/5 factor, so we’d write: U_elect = (3/5)·e²/a. But we’re not interested in finding the right model here. We know the U_elect = (3/5)·e²/a gives us a value for a that differs with a 2/5 factor as the classical electron radius. That’s not so bad and so let’s go along with it. 🙂

We’re going to look at the simple structure of this relation, and all of its implications. The simple equation above says that the energy of our electron is (a) proportional to the square of its charge and (b) inversely proportional to its radius. Now, that is a very remarkable result. In fact, we’ve seen something like this before, and we were astonished. We saw it when we were discussing the wonderful properties of that magical number, the fine-structure constant, which we also denoted by α. However, because we used α already, I’ll denote the fine-structure constant as α_ehere, so you don’t get confused. You’ll remember that the fine-structure constant is a God-like number indeed: it links all of the fundamental properties of the electron, i.e. its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, its mass (and, hence, its energy), its de Broglie wavelength. Whatever: all these physical constants are all related through the fine-structure constant.

In my various posts on this topic, I’ve repeatedly said that, but I never showed why it’s true, and so it was a very magical number indeed. I am going to take some of the magic out now. Not too much but… Well… You can judge for yourself how much of the magic remains after I am done here. 🙂

So, at this stage of the argument, α can be anything, and α_ecannot, of course. It’s just that magical number out there, which relates everything to everything: it’s the God-given number we don’t understand, or didn’t understand, I should say. Past tense. Indeed, we’re going to get some understanding here because we know that one of the many expressions involving α_ewas the following one:

m_e = α_e/r_e

This says that the mass of the electron is equal to the ratio of the fine-structure constant and the electron radius. [Note that we express everything in natural units here, so that’s Planck units. For the detail of the conversion, please see the relevant section on that in my one of my posts on this and other stuff.] In fact, the U = (3/5)·e²/a and m_e = α_e/r_erelations looks exactly the same, because one of the other equations involving the fine-structure constant was: α_e = e_P². So we’ve got the square of the charge here as well! Indeed, as I’ll explain in a moment, the difference between the two formulas is just a matter of units.

Now, mass is equivalent to energy, of course: it’s just a matter of units, so we can equate m_e with E_e (this amounts to expressing the energy of the electron in a kg unit—bit weird, but OK) and so we get:

E_e = α_e/r_e

So there we have: the fine-structure constant α_e is Nature’s ‘cut-off’ factor, so to speak. Why? Only God knows. 🙂 But it’s now (fairly) easy to see why all the relations involving α_e are what they are. As I mentioned already, we also know that α_e is the square of the electron charge expressed in Planck units, so we have:

α_e = e_P² and, therefore, E_e = e_P²/r_e

Now, you can check for yourself: it’s just a matter of re-expressing everything in standard SI units, and relating e_P² to e², and it should all work: you should get the E_elect = (2/3)·e²/a expression. So… Well… At least this takes some of the magic out the fine-structure constant. It’s still a wonderful thing, but so you see that the fundamental relationship between (a) the energy (and, hence, the mass), (b) the radius and (c) the charge of an electron is not something God-given. What’s God-given are Maxwell’s equations, and so the E_e = α_e/r_e= e_P²/r_e is just one of the many wonderful things that you can get out of them.

So we found God’s ‘cut-off factor’ 🙂 It’s equal to α_e ≈ 0.0073 = 7.3×10⁻³. So 7.3 thousands of… What? Well… Nothing. It’s just a pure ratio between the energy and the radius of an electron (if both are expressed in Planck units, of course). And so it determines the electron charge (again, expressed in Planck units). Indeed, we write:

e_P = √α_e

Really? Yes. Just do all these formulas:

e_P = √α_e≈ √0.0073·1.9×10⁻¹⁸coulomb ≈ 1.6 ×10⁻¹⁹C

Just re-check it with all the known decimals: you’ll see it’s bang on. Let’s look at the E_e= m_e = α_e/r_eratio once again. What’s the meaning of it? Let’s first calculate the value of r_e and m_e, i.e. the electron radius and electron mass expressed in Planck units. It’s equal to the classical electron radius divided by the Planck length, and then the same for the mass, so we get the following thing:

r_e ≈ (2.81794×10⁻¹⁵m)/(1.6162×10⁻³⁵m) = 1.7435×10²⁰

m_e ≈ (9.1×10⁻³¹kg)/(2.17651×10⁻⁸kg) = 4.18×10⁻²³

α_e = (4.18×10⁻²³)·(1.7435×10²⁰) ≈ 0.0073

It works like a charm, but what does it mean? Well… It’s just a ratio between two physical quantities, and the scale you use to measure those quantities matters very much. We’ve explained that the Planck mass is a rather large unit at the atomic scale and, therefore, it’s perhaps not quite appropriate to use it here. In fact, out of the many interesting expressions for α_e, I should highlight the following one:

α_e = e²/(ħ·c) ≈ (1.60217662×10⁻¹⁹C)²/(4πε₀·[(1.054572×10⁻³⁴N·m·s)·(2.998×10⁸m/s)]) ≈ 0.0073 once more 🙂

Note that the elementary charge e is actually equal to q_e/4πε₀, which is what I am using in the formula. I know that’s confusing, but it what it is. As for the units, it’s a bit tedious to write it all out, but you’ll get there. Note that ε₀≈ 8.8542×10⁻¹²C²/(N·m²) so… Well… All the units do cancel out, and we get a dimensionless number indeed, which is what α_e is.

The point is: this expression links α_e to the the de Broglie relation (p = h/λ), with λ the wavelength that’s associated with the electron. Of course, because of the Uncertainty Principle, we know we’re talking some wavelength range really, so we should write the de Broglie relation as Δp = h·Δ(1/λ). Now, that, in turn, allows us to try to work out the Bohr radius, which is the other ‘dimension’ we associate with an electron. Of course, now you’ll say: why would you do that. Why would you bring in the de Broglie relation here?

Well… We’re talking energy, and so we have the Planck-Einstein relation first: the energy of some particle can always be written as the product of h and some frequency f: E = h·f. The only thing that de Broglie relation adds is the Uncertainty Principle indeed: the frequency f will be some frequency range, associated with some momentum range, and so that’s what the Uncertainty Principle really says. I can’t dwell too much on that here, because otherwise this post would become a book. 🙂 For more detail, you can check out one of my many posts on the Uncertainty Principle. In fact, the one I am referring to here has Feynman’s calculation of the Bohr radius, so I warmly recommend you check it out. The thrust of the argument is as follows:

If we assume that (a) an electron takes some space – which I’ll denote by r 🙂 – and (b) that it has some momentum p because of its mass m and its velocity v, then the ΔxΔp = ħ relation (i.e. the Uncertainty Principle in its roughest form) suggests that the order of magnitude of r and p should be related in the very same way. Hence, let’s just boldly write r ≈ ħ/p and see what we can do with that.
We know that the kinetic energy of our electron equals mv²/2, which we can write as p²/2m so we get rid of the velocity factor.Well… Substituting our p ≈ ħ/r conjecture, we get K.E. = ħ²/2mr². So that’s a formula for the kinetic energy. Next is potential.
The formula for the potential energy is U = q₁q₂/4πε₀r₁₂. Now, we’re actually talking about the size of an atom here, so one charge is the proton (+e) and the other is the electron (–e), so the potential energy is U = P.E. = –e²/4πε₀r, with r the ‘distance’ between the proton and the electron—so that’s the Bohr radius we’re looking for!
We can now write the total energy (which I’ll denote by E, but don’t confuse it with the electric field vector!) as E = K.E. + P.E. = ħ²/2mr²– e²/4πε₀r. Now, the electron (whatever it is) is, obviously, in some kind of equilibrium state. Why is that obvious? Well… Otherwise our hydrogen atom wouldn’t or couldn’t exist. 🙂 Hence, it’s in some kind of energy ‘well’ indeed, at the bottom. Such equilibrium point ‘at the bottom’ is characterized by its derivative (in respect to whatever variable) being equal to zero. Now, the only ‘variable’ here is r (all the other symbols are physical constants), so we have to solve for dE/dr = 0. Writing it all out yields: dE/dr = –ħ²/mr³+ e²/4πε₀r²= 0 ⇔ r = 4πε₀ħ²/me²
We can now put the values in: r = 4πε₀h²/me²= [(1/(9×10⁹) C²/N·m²)·(1.055×10^–34J·s)²]/[(9.1×10^–31kg)·(1.6×10^–19C)²] = 53×10^–12m = 53 pico-meter (pm)

Done. We’re right on the spot. The Bohr radius is, effectively, about 53 trillionths of a meter indeed!

Phew!

Yes… I know… Relax. We’re almost done. You should now be able to figure out why the classical electron radius and the Bohr radius can also be related to each other through the fine-structure constant. We write:

m_e = α/r_e= α/α²r = 1/αr

So we get that α/r_e= 1/αr and, therefore, we get r_e/r = α², which explains why α is also equal to the so-called junction number, or the coupling constant, for an electron-photon coupling (see my post on the quantum-mechanical aspects of the photon-electron interaction). It gives a physical meaning to the probability (which, as you know, is the absolute square of the probability amplitude) in terms of the chance of a photon actually ‘hitting’ the electron as it goes through the atom. Indeed, the ratio of the Thomson scattering cross-section and the Bohr size of the atom should be of the same order as r_e/r, and so that’s α².

[Note: To be fully correct and complete, I should add that the coupling constant itself is not α² but √α = e_P. Why do we have this square root? You’re right: the fact that the probability is the absolute square of the amplitude explains one square root (√α² = α), but not two. The thing is: the photon-electron interaction consists of two things. First, the electron sort of ‘absorbs’ the photon, and then it emits another one, that has the same or a different frequency depending on whether or not the ‘collision’ was elastic or not. So if we denote the coupling constant as j, then the whole interaction will have a probability amplitude equal to j². In fact, the value which Feynman uses in his wonderful popular presentation of quantum mechanics (The Strange Theory of Light and Matter), is −α ≈ −0.0073. I am not quite sure why the minus sign is there. It must be something with the angles involved (the emitted photon will not be following the trajectory of the incoming photon) or, else, with the special arithmetic involved in boson-fermion interactions (we add amplitudes when bosons are involved, but subtract amplitudes when it’s fermions interacting. I’ll probably find out once I am true through Feynman’s third volume of Lectures, which focus on quantum mechanics only.]

Finally, the last bit of unexplained ‘magic’ in the fine-structure constant is that the fine-structure constant (which I’ve started to write as α again, instead of α_e) also gives us the (classical) relative speed of an electron, so that’s its speed as it orbits around the nucleus (according to the classical theory, that is), so we write

α = v/c = β

I should go through the motions here – I’ll probably do so in the coming days – but you can see we must be able to get it out somehow from all what we wrote above. See how powerful our U_elect ∼ e²/a relation really is? It links the electron, charge, its radius and its energy, and it’s all we need to all the rest out of it: its mass, its momentum, its speed and – through the Uncertainty Principle – the Bohr radius, which is the size of the atom.

We’ve come a long way. This is truly a milestone. We’ve taken the magic out of God’s number—to some extent at least. 🙂

You’ll have one last question, of course: if proportionality constants are all about the scale in which we measure the physical quantities on either side of an equation, is there some way the fine-structure constant would come out differently? That’s the same as asking: what if we’d measure energy in units that are equivalent to the energy of an electron, and the radius of our electron just as… Well… What if we’d equate our unit of distance with the radius of the electron, so we’d write r_e= 1? What would happen to α? Well… I’ll let you figure that one out yourself. I am tired and so I should go to bed now. 🙂

[…] OK. OK. Let me tell you. It’s not that simple here. All those relationships involving α, in one form or the other, are very deep. They relate a lot of stuff to a lot of stuff, and we can appreciate that only when doing a dimensional analysis. A dimensional analysis of the E_e = α_e/r_e= e_P²/r yields [e_P²/r] = C²/m on the right-hand side and [E_e] = J = N·mon the left-hand side. How can we reconcile both? The coulomb is an SI base unit , so we can’t ‘translate’ it into something with N and m. [To be fully correct, for some reason, the ampère (i.e. coulomb per second) was chosen as an SI base unit, but they’re interchangeable in regard to their place in the international system of units: they can’t be reduced.] So we’ve got a problem. Yes. That’s where we sort of ‘smuggled’ the 4πε₀factor in when doing our calculations above. That ε₀constant is, obviously, not ‘as fundamental’ as c or α (just think of the c⁻²= ε₀μ₀relationship to understand what I mean here) but, still, it was necessary to make the dimensions come out alright: we need the reciprocal dimension of ε₀, i.e. (N·m²)/C², to make the dimensional analysis work. We get: (C²/m)·(N·m²)/C² = N·m = J, i.e. joule, so that’s the unit in which we measure energy or – using the E = mc² equivalence – mass, which is the aspect of energy emphasizing its inertia.

So the answer is: no. Changing units won’t change alpha. So all that’s left is to play with it now. Let’s try to do that. Let me first plot that E_e= m_e = α_e/r_e= 0.00729735256/r_e:

Unsurprisingly, we find the pivot point of this curve is at the intersection of the diagonal and the curve itself, so that’s at the (0.00729735256, 0.00729735256) point, where slopes are ± 1, i.e. plus or minus unity. What does this show? Nothing much. What? I can hear you: I should be excited because… Well… Yes! Think of it. If you would have to chose a cut-off point, you’d chose this one, wouldn’t you? 🙂 Sure, you’re right. How exciting! Let me show you. Look at it! It proves that God thinks in terms of logarithms. He has chosen α such that ln(E) = ln(α/r) = lnα – lnr = lnα – lnr = 0, so ln α = lnr and, therefore, α = r. 🙂

Huh? Excuse me?

I am sorry. […] Well… I am not, of course… 🙂 I just wanted to illustrate the kind of exercise some people are tempted to do. It’s no use. The fine-structure constant is what it is: it sort of summarizes an awful lot of formulas. It basically shows what Maxwell’s equation imply in terms of the structure of an atom defined as a negative charge orbiting around some positive charge. It shows we can get calculate everything as a function of something else, and that’s what the fine-structure constant tells us: it relates everything to everything. However, when everything is said and done, the fine-structure constant shows us two things:

Maxwell’s equations are complete: we can construct a complete model of the electron and the atom, which includes: the electron’s energy and mass, its velocity, its own radius, and the radius of the atom. [I might have forgotten one of the dimensions here, but you’ll add it. :-)]
God doesn’t want our equations to blow up. Our equations are all correct but, in reality, there’s a cut-off factor that ensures we don’t go to the limit with them.

So the fine-structure constant anchors our world, so to speak. In other words: of all the worlds that are possible, we live in this one.

[…] It’s pretty good as far as I am concerned. Isn’t it amazing that our mind is able to just grasp things like that? I know my approach here is pretty intuitive, and with ‘intuitive’, I mean ‘not scientific’ here. 🙂 Frankly, I don’t like the talk about physicists “looking into God’s mind.” I don’t think that’s what they’re trying to do. I think they’re just trying to understand the fundamental unity behind it all. And that’s religion enough for me. 🙂

So… What’s the conclusion? Nothing much. We’ve sort of concluded our description of the classical world… Well… Of its ‘electromagnetic sector’ at least. 🙂 That sector can be summarized in Maxwell’s equations, which describe an infinite world of possible worlds. However, God fixed three constants: h, c and α. So we live in a world that’s defined by this Trinity of fundamental physical constants. Why is it not two, or four?

My guts instinct tells me it’s because we live in three dimensions, and so there’s three degrees of freedom really. But what about time? Time is the fourth dimension, isn’t it? Yes. But time is symmetric in the ‘electromagnetic’ sector: we can reverse the arrow of time in our equations and everything still works. The arrow of time involves other theories: statistics (physicists refer to it as ‘statistical mechanics‘) and the ‘weak force’ sector, which I discussed when talking about symmetries in physics. So… Well… We’re not done. God gave us plenty of other stuff to try to understand. 🙂