The analysis of a two-state system in my post on the ammonia molecule ‘flipping’ its spin direction from ‘up’ to ‘down’, or vice versa) 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 P_{1} (i.e. the probability to be in state 1) and P_{2} (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_{μ}x_{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 = E_{0}/ħ. What’s E_{0}? 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 = *a*^{2}*, always*. The ‘normalization’ condition (so that’s the condition that probabilities have to add up to one) implies that P_{1} = P_{2} = *a*^{2}* *= 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, *no*where, 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 that post on the quantum math involved in analyzing the energy levels of a molecule: we had the ammonia molecule, i.e. an NH_{3} 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 E_{0 }+ A and E_{0 }− 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:

- C
_{1 }= (1/2)·*e*^{−(i/ħ)·(E0 − A)·t }+ (1/2)·*e*^{−(i/ħ)·(E0 + A)·t} - C
_{2 }= (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 ω_{1 }= (E_{0 }− A)/ħ and ω_{1 }= (E_{0 }+ A)/ħ respectively. [As for the minus sign in front of the second term in the wave equation for C_{2}, −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 C_{1}(t) and C_{2}(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 E_{0 }but, 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:

- |C
_{1}(t)|^{2 }=*cos*^{2}[(A/ħ)·t], and - |C
_{2}(t)|^{2 }=*sin*^{2}[(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 E_{0 }– 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 = *c*^{2}/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 that method. 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*newton·**meter*(force over distance, remember?) or*electronvolts*(1 eV = 1.6×10^{−19 }J = 1.6×10^{−19 }N·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_{μ}x_{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 =*c*^{2}/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 = E_{0 }± σ_{E}, and p = p_{0 }± 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. 🙂

**Addendum**:

Frankly, I tried to fool you above. If the functional form of the wavefunction is *a*·e^{−(}^{i/ħ)·(E·t − p·x)}, then we can measure E and **p** in whatever unit we want, including h or ħ, but we can*not* re-scale the argument of the function, i.e. the *phase *θ, without *changing *the functional form itself. I explained that in that post for my kids on wavefunctions:, in which I explained we may represent the same electromagnetic* *wave by two different functional forms:

F(*c*t−x) = G(t−x/*c*)

So F and G represent the same wave, but they are different wavefunctions. In this regard, you should note that the argument of F is expressed in *distance units*, as we multiply t with the speed of light (so it’s like our time unit is 299,792,458 m now), while the argument of G is expressed in *time units*, as we divide x by the distance traveled in one second). But F and G are different functional forms. Just do an example and take a simple sine function: you’ll agree that sin(θ) ≠ sin(θ/*c*) for all values of θ, except 0. Re-scaling changes the frequency, or the wavelength, and it does so quite drastically in this case. Likewise, you can see that *a·e*^{−i(φ/E) }= [*a·e*^{−iφ}]^{1/E}, so that’s a *very *different function. In short, we were a bit too adventurous above. But I hope it helped you to think things through *yourself*. 🙂