# Transforming amplitudes for spin-1/2 particles

Some say it is not possible to fully understand quantum-mechanical spin. Now, I do agree it is difficult, but I do not believe it is impossible. That’s why I wrote so many posts on it. Most of these focused on elaborating how the classical view of how a rotating charge precesses in a magnetic field might translate into the weird world of quantum mechanics. Others were more focused on the corollary of the quantization of the angular momentum, which is that, in the quantum-mechanical world, the angular momentum is never quite all in one direction only—so that explains some of the seemingly inexplicable randomness in particle behavior.

Frankly, I think those explanations help us quite a bit already but… Well… We need to go the extra mile, right? In fact, that’s drives my search for a geometric (or physical) interpretation of the wavefunction: the extra mile. 🙂

Now, in one of these many posts on spin and angular momentum, I advise my readers – you, that is – to try to work yourself through Feynman’s 6th Lecture on quantum mechanics, which is highly abstract and, therefore, usually skipped. [Feynman himself told his students to skip it, so I am sure that’s what they did.] However, if we believe the physical (or geometric) interpretation of the wavefunction that we presented in previous posts is, somehow, true, then we need to relate it to the abstract math of these so-called transformations between representations. That’s what we’re going to try to do here. It’s going to be just a start, and I will probably end up doing several posts on this but… Well… We do have to start somewhere, right? So let’s see where we get today. 🙂

The thought experiment that Feynman uses throughout his Lecture makes use of what Feynman’s refers to as modified or improved Stern-Gerlach apparatuses. They allow us to prepare a pure state or, alternatively, as Feynman puts it, to analyze a state. In theory, that is. The illustration below present a side and top view of such apparatus. We may already note that the apparatus itself—or, to be precise, our perspective of it—gives us two directions: (1) the up direction, so that’s the positive direction of the z-axis, and (2) the direction of travel of our particle, which coincides with the positive direction of the y-axis. [This is obvious and, at the same time, not so obvious, but I’ll talk about that in my next post. In this one, we basically need to work ourselves through the math, so we don’t want to think too much about philosophical stuff.]

The kind of questions we want to answer in this post are variants of the following basic one: if a spin-1/2 particle (let’s think of an electron here, even if the Stern-Gerlach experiment is usually done with an atomic beam) was prepared in a given condition by one apparatus S, say the +S state, what is the probability (or the amplitude) that it will get through a second apparatus T if that was set to filter out the +T state?

The result will, of course, depend on the angles between the two apparatuses S and T, as illustrated below. [Just to respect copyright, I should explicitly note here that all illustrations are taken from the mentioned Lecture, and that the line of reasoning sticks close to Feynman’s treatment of the matter too.]

We should make a few remarks here. First, this thought experiment assumes our particle doesn’t get lost. That’s obvious but… Well… If you haven’t thought about this possibility, I suspect you will at some point in time. So we do assume that, somehow, this particle makes a turn. It’s an important point because… Well… Feynman’s argument—who, remember, represents mainstream physics—somehow assumes that doesn’t really matter. It’s the same particle, right? It just took a turn, so it’s going in some other direction. That’s all, right? Hmm… That’s where I part ways with mainstream physics: the transformation matrices for the amplitudes that we’ll find here describe something real, I think. It’s not just perspective: something happened to the electron. That something does not only change the amplitudes but… Well… It describes a different electron. It describes an electron that goes in a different direction now. But… Well… As said, these are reflections I will further develop in my next post. 🙂 Let’s focus on the math here. The philosophy will follow later. 🙂 Next remark.

Second, we assume the (a) and (b) illustrations above represent the same physical reality because the relative orientation between the two apparatuses, as measured by the angle α, is the same. Now that is obvious, you’ll say, but, as Feynman notes, we can only make that assumption because experiments effectively confirm that spacetime is, effectively, isotropic. In other words, there is no aether allowing us to establish some sense of absolute direction. Directions are relativerelative to the observer, that is… But… Well… Again, in my next post, I’ll argue that it’s not because directions are relative that they are, somehow, not real. Indeed, in my humble opinion, it does matter whether an electron goes here or, alternatively, there. These two different directions are not just two different coordinate frames. But… Well… Again. The philosophy will follow later. We need to stay focused on the math here.

Third and final remark. This one is actually very tricky. In his argument, Feynman also assumes the two set-ups below are, somehow, equivalent.

You’ll say: Huh? If not, say it! Huh? 🙂 Yes. Good. Huh? Feynman writes equivalentnot the same because… Well… They’re not the same, obviously:

1. In the first set-up (a), T is wide open, so the apparatus is not supposed to do anything with the beam: it just splits and re-combines it.
2. In set-up (b) the T apparatus is, quite simply, not there, so… Well… Again. Nothing is supposed to happen with our particles as they come out of S and travel to U.

The fundamental idea here is that our spin-1/2 particle (again, think of an electron here) enters apparatus U in the same state as it left apparatus S. In both set-ups, that is! Now that is a very tricky assumption, because… Well… While the net turn of our electron is the same, it is quite obvious it has to take two turns to get to U in (a), while it only takes one turn in (b). And so… Well… You can probably think of other differences too. So… Yes. And no. Same-same but different, right? 🙂

Right. That is why Feynman goes out of his way to explain the nitty-gritty behind: he actually devotes a full page in small print on this, which I’ll try to summarize in just a few paragraphs here. [And, yes, you should check my summary against Feynman’s actual writing on this.] It’s like this. While traveling through apparatus T in set-up (a), time goes by and, therefore, the amplitude would be different by some phase factor δ. [Feynman doesn’t say anything about this, but… Well… In the particle’s own frame of reference, this phase factor depend on the energy, the momentum and the time and distance traveled. Think of the argument of the elementary wavefunction here: θ = (E∙t – px)/ħ).] Now, if we believe that the amplitude is just some mathematical construct—so that’s what mainstream physicists (not me!) believe—then we could effectively say that the physics of (a) and (b) are the same, as Feynman does. In fact, let me quote him here:

“The physics of set-up (a) and (b) should be the same but the amplitudes could be different by some phase factor without changing the result of any calculation about the real world.”

Hmm… It’s one of those mysterious short passages where we’d all like geniuses like Feynman (or Einstein, or whomever) to be more explicit on their world view: if the amplitudes are different, can the physics really be the same? I mean… Exactly the same? It all boils down to that unfathomable belief that, somehow, the particle is real but the wavefunction that describes it, is not. Of course, I admit that it’s true that choosing another zero point for the time variable would also change all amplitudes by a common phase factor and… Well… That’s something that I consider to be not real. But… Well… The time and distance traveled in the apparatus is the time and distance traveled in the apparatus, right?

Bon… I have to stay away from these questions as for now—we need to move on with the math here—but I will come back to it later. But… Well… Talking math, I should note a very interesting mathematical point here. We have these transformation matrices for amplitudes, right? Well… Not yet. In fact, the coefficient of these matrices are exactly what we’re going to try to derive in this post, but… Well… Let’s assume we know them already. 🙂 So we have a 2-by-2 matrix to go from S to T, from T to U, and then one to go from S to U without going through T, which we can write as RSTRTU,  and RSU respectively. Adding the subscripts for the base states in each representation, the equivalence between the (a) and (b) situations can then be captured by the following formula:

So we have that phase factor here: the left- and right-hand side of this equation is, effectively, same-same but different, as they would say in Asia. 🙂 Now, Feynman develops a beautiful mathematical argument to show that the eiδ factor effectively disappears if we convert our rotation matrices to some rather special form that is defined as follows:

I won’t copy his argument here, but I’d recommend you go over it because it is wonderfully easy to follow and very intriguing at the same time. [Yes. Simple things can be very intriguing.] Indeed, the calculation below shows that the determinant of these special rotation matrices will be equal to 1.

So… Well… So what? You’re right. I am being sidetracked here. The point is that, if we put all of our rotation matrices in this special form, the eiδ factor vanishes and the formula above reduces to:

So… Yes. End of excursion. Let us remind ourselves of what it is that we are trying to do here. As mentioned above, the kind of questions we want to answer will be variants of the following basic one: if a spin-1/2 particle was prepared in a given condition by one apparatus (S), say the +S state, what is the probability (or the amplitude) that it will get through a second apparatus (T) if that was set to filter out the +T state?

We said the result would depend on the angles between the two apparatuses S and T. I wrote: angles—plural. Why? Because a rotation will generally be described by the three so-called Euler angles:  α, β and γ. Now, it is easy to make a mistake here, because there is a sequence to these so-called elemental rotations—and right-hand rules, of course—but I will let you figure that out. 🙂

The basic idea is the following: if we can work out the transformation matrices for each of these elemental rotations, then we can combine them and find the transformation matrix for any rotation. So… Well… That fills most of Feynman’s Lecture on this, so we don’t want to copy all that. We’ll limit ourselves to the logic for a rotation about the z-axis, and then… Well… You’ll see. 🙂

So… The z-axis… We take that to be the direction along which we are measuring the angular momentum of our electron, so that’s the direction of the (magnetic) field gradient, so that’s the up-axis of the apparatus. In the illustration below, that direction points out of the page, so to speak, because it is perpendicular to the direction of the x– and the y-axis that are shown. Note that the y-axis is the initial direction of our beam.

Now, because the (physical) orientation of the fields and the field gradients of S and T is the same, Feynman says that—despite the angle—the probability for a particle to be up or down with regard to and T respectively should be the same. Well… Let’s be fair. He does not only say that: experiment shows it to be true. [Again, I am tempted to interject here that it is not because the probabilities for (a) and (b) are the same, that the reality of (a) and (b) is the same, but… Well… You get me. That’s for the next post. Let’s get back to the lesson here.] The probability is, of course, the square of the absolute value of the amplitude, which we will denote as C+C, C’+, and C’ respectively. Hence, we can write the following:

Now, the absolute values (or the magnitudes) are the same, but the amplitudes may differ. In fact, they must be different by some phase factor because, otherwise, we would not be able to distinguish the two situations, which are obviously different. As Feynman, finally, admits himself—jokingly or seriously: “There must be some way for a particle to know that it has turned the corner at P1.” [P1 is the midway point between and in the illustration, of course—not some probability.]

So… Well… We write:

C’+ = eiλ ·C+ and C’ = eiμ ·C

Now, Feynman notes that an equal phase change in all amplitudes has no physical consequence (think of re-defining our t0 = 0 point), so we can add some arbitrary amount to both λ and μ without changing any of the physics. So then we can choose this amount as −(λ + μ)/2. We write:

Now, it shouldn’t you too long to figure out that λ’ is equal to λ’ = λ/2 + μ/2 = −μ’. So… Well… Then we can just adopt the convention that λ = −μ. So our C’+ = eiλ ·C+ and C’ = eiμ ·C equations can now be written as:

C’+ = eiλ ·C+ and C’ = eiλ·C

The absolute values are the same, but the phases are different. Right. OK. Good move. What’s next?

Well… The next assumption is that the phase shift λ is proportional to the angle (α) between the two apparatuses. Hence, λ is equal to λ = m·α, and we can re-write the above as:

C’+ = ei·C+ and C’ = ei·C

Now, this assumption may or may not seem reasonable. Feynman justifies it with a continuity argument, arguing any rotation can be built up as a sequence of infinitesimal rotations and… Well… Let’s not get into the nitty-gritty here. [If you want it, check Feynman’s Lecture itself.] Back to the main line of reasoning. So we’ll assume we can write λ as λ = m·α. The next question then is: what is the value for m? Now, we obviously do get exactly the same physics if we rotate by 360°, or 2π radians. So we might conclude that the amplitudes should be the same and, therefore, that ei = eim·2π has to be equal to one, so C’+ = C+ and C’ = C . That’s the case if m is equal to 1. But… Well… No. It’s the same thing again: the probabilities (or the magnitudes) have to be the same, but the amplitudes may be different because of some phase factor. In fact, they should be different. If m = 1/2, then we also get the same physics, even if the amplitudes are not the same. They will be each other’s opposite:

Huh? Yes. Think of it. The coefficient of proportionality (m) cannot be equal to 1. If it would be equal to 1, and we’d rotate by 180° only, then we’d also get those C’+ = −C+ and C’ = −C equations, and so these coefficients would, therefore, also describe the same physical situation. Now, you will understand, intuitively, that a rotation of the apparatus by 180° will not give us the same physical situation… So… Well… In case you’d want a more formal argument proving a rotation by 180° does not give us the same physical situation, Feynman has one for you. 🙂

I know that, by now, you’re totally tired and bored, and so you only want the grand conclusion at this point. Well… All of what I wrote above should, hopefully, help you to understand that conclusion, which – I quote Feynman here – is the following:

If we know the amplitudes C+ and C of spin one-half particles with respect to a reference frame S, and we then use new base states, defined with respect to a reference frame T which is obtained from S by a rotation α around the z-axis, the new amplitudes are given in terms of the old by the following formulas:

[Feynman denotes our angle α by phi (φ) because… He uses the Euler angles a bit differently. But don’t worry: it’s the same angle.]

What about the amplitude to go from the C to the C’+ state, and from the C+ to the C’ state? Well… That amplitude is zero. So the transformation matrix is this one:

Let’s take a moment and think about this. Feynman notes the following, among other things: “It is very curious to say that if you turn the apparatus 360° you get new amplitudes. [They aren’t really new, though, because the common change of sign doesn’t give any different physics.] But if something has been rotated by a sequence of small rotations whose net result is to return it to the original orientation, then it is possible to define the idea that it has been rotated 360°—as distinct from zero net rotation—if you have kept track of the whole history.”

This is very deep. It connects space and time into one single geometric space, so to speak. But… Well… I’ll try to explain this rather sweeping statement later. Feynman also notes that a net rotation of 720° does give us the same amplitudes and, therefore, cannot be distinguished from the original orientation. Feynman finds that intriguing but… Well… I am not sure if it’s very significant. I do note some symmetries in quantum physics involve 720° rotations but… Well… I’ll let you think about this. 🙂

Note that the determinant of our matrix is equal to a·b·ceiφ/2·eiφ/2 = 1. So… Well… Our rotation matrix is, effectively, in that special form! How comes? Well… When equating λ = −μ, we are effectively putting the transformation into that special form.  Let us also, just for fun, quickly check the normalization condition. It requires that the probabilities, in any given representation, add to up to one. So… Well… Do they? When they come out of S, our electrons are equally likely to be in the up or down state. So the amplitudes are 1/√2. [To be precise, they are ±1/√2 but… Well… It’s the phase factor story once again.] That’s normalized: |1/√2|2 + |1/√2|2 = 1. The amplitudes to come out of the apparatus in the up or down state are eiφ/2/√2 and eiφ/2/√2 respectively, so the probabilities add up to |eiφ/2/√2|2 + |eiφ/2/√2|2 = … Well… It’s 1. Check it. 🙂

Let me add an extra remark here. The normalization condition will result in matrices whose determinant will be equal to some pure imaginary exponential, like eiα. So is that what we have here? Yes. We can re-write 1 as 1 = ei·0 = e0, so α = 0. 🙂 Capito? Probably not, but… Well… Don’t worry about it. Just think about the grand results. As Feynman puts it, this Lecture is really “a sort of cultural excursion.” 🙂

Let’s do a practical calculation here. Let’s suppose the angle is, effectively, 180°. So the eiφ/2 and eiφ/2/√2 factors are equal to eiπ/2 = +i and eiπ/2 = −i, so… Well… What does that mean—in terms of the geometry of the wavefunction? Hmm… We need to do some more thinking about the implications of all this transformation business for our geometric interpretation of he wavefunction, but so we’ll do that in our next post. Let us first work our way out of this rather hellish transformation logic. 🙂 [See? I do admit it is all quite difficult and abstruse, but… Well… We can do this, right?]

So what’s next? Well… Feynman develops a similar argument (I should say same-same but different once more) to derive the coefficients for a rotation of ±90° around the y-axis. Why 90° only? Well… Let me quote Feynman here, as I can’t sum it up more succinctly than he does: “With just two transformations—90° about the y-axis, and an arbitrary angle about the z-axis [which we described above]—we can generate any rotation at all.”

So how does that work? Check the illustration below. In Feynman’s words again: “Suppose that we want the angle α around x. We know how to deal with the angle α α around z, but now we want it around x. How do we get it? First, we turn the axis down onto x—which is a rotation of +90°. Then we turn through the angle α around z’. Then we rotate 90° about y”. The net result of the three rotations is the same as turning around x by the angle α. It is a property of space.”

Besides helping us greatly to derive the transformation matrix for any rotation, the mentioned property of space is rather mysterious and deep. It sort of reduces the degrees of freedom, so to speak. Feynman writes the following about this:

“These facts of the combinations of rotations, and what they produce, are hard to grasp intuitively. It is rather strange, because we live in three dimensions, but it is hard for us to appreciate what happens if we turn this way and then that way. Perhaps, if we were fish or birds and had a real appreciation of what happens when we turn somersaults in space, we could more easily appreciate such things.”

In any case, I should limit the number of philosophical interjections. If you go through the motions, then you’ll find the following elemental rotation matrices:

What about the determinants of the Rx(φ) and Ry(φ) matrices? They’re also equal to one, so… Yes. A pure imaginary exponential, right? 1 = ei·0 = e0. 🙂

What’s next? Well… We’re done. We can now combine the elemental transformations above in a more general format, using the standardized Euler angles. Again, just go through the motions. The Grand Result is:

Does it give us normalized amplitudes? It should, but it looks like our determinant is going to be a much more complicated complex exponential. 🙂 Hmm… Let’s take some time to mull over this. As promised, I’ll be back with more reflections in my next post.

# 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 Cand Cfunctions 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 Hij may also vary in space and in time, and so that‘s what makes things quite nightmarish to solve. [Note that, while H11 and H22 represent some energy level and, hence, are usually real numbers, H12 and H21 may 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 Cfunctions 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 = a1eλI·tv+ a2eλII·tvII

The λI and λII 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)Hij elements. The vI and vII 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 λI and λII values. The two eigenvalues of our system above can be written as:

λI = −(i/ħ)·EI and λII = −(i/ħ)·EII.

EI and EII 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 a1eλI·t and a2eλII·t expressions gives us a1e−(i/ħ)·EI·t and a2e−(i/ħ)·EII·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= |a1e−(i/ħ)·EI·t|= |a1|2·|e−(i/ħ)·EI·t|= |a1|2
• PII = |a2e−(i/ħ)·EII·t|= |a2|2·|e−(i/ħ)·EII·t|= |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)·EI·t and e−(i/ħ)·EII·t as:

• C= 〈 I | ψ 〉 =  e−(i/ħ)·EI·t
• CII = 〈 II | ψ 〉 = e−(i/ħ)·EII·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| II 〉 CII = | 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| II 〉 CII.

Of course, we should also be able to revert back to the base states we started out with so, once we’ve calculated Cand 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 Cand C2, we can also write:

| ψ 〉 = | 1 〉 C| 2 〉 C= | 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·CI·vI + a2·CII·vII

Hmm… So what’s that? We’ve seen something like C = a1·CI + a2·CII , as we wrote something like C1 = (a/2)·CI + (b/2)·CII b in our previous posts, for example—but what are those eigenvectors vI and vII? 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−λII)vI = 0 and (A−λIII)vII = 0 equations, so they come with the eigenvalues, and their components will depend on the eigenvalues λand λI 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 EI and EII 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 vI and vII. 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 CI and CII 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 vI and vII as vI1 and vI2, and vII1 and vII2 respectively, then the C = a1·CI·vI + a2·CII·vII expression is equivalent to:

• C1 = a1·vI1·Ca2·vII1·CII
• C2 = a1·vI2·CI + a2·vII2 ·CII

Hence, a1·vI1 = a2·vII2 = cos(α/2) and a2·vII1 = −a1·vI2 = 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)|= 1 = |a1·vI1·CI(0) + a2·vII1·CII(0)|and |C2|= 0 = |a1·vI1·CI(0) + a2·vII1·CII(0)|

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·vI1 = a2·vII2 = cos(α/2) and a2·vII1 = −a1·vI2 = sin(α/2) conditions only make sense when we equate α with −π/2, so we can write:

• a1·vI1 = a2·vII2 = cos(−π/4) = 1/√2
• a2·vII1 = −a1·vI2 = sin(−π/4) = –1/√2

It’s only then that we get a unique ratio for a1/a= vI1/vII2 = −vII1/vI2. [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 vI1/vII2 and −vII1/vI2 ratios would be different, we’d have a huge problem, because we’d have two different values for the a1/aratio! And… Well… That just doesn’t make sense. The system must come with some specific value for aand 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 aand a2 coefficients as he start his chapter on ‘other two-state systems’. It’s the symbols he’s using: ‘his’ aand a2, and the other doublet with the primes, i.e. a1‘ and a2‘, are the transformation amplitudesnot 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 aand 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! 🙂