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.
It’s obvious these two equations are usually not easy to solve: the C1 and C2 functions 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 C2 functions 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·tvI + 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:
- PI = |a1e−(i/ħ)·EI·t|2 = |a1|2·|e−(i/ħ)·EI·t|2 = |a1|2
- PII = |a2e−(i/ħ)·EII·t|2 = |a2|2·|e−(i/ħ)·EII·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)·EI·t and e−(i/ħ)·EII·t as:
- CI = 〈 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 〉 CI + | 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 〉 CI + | 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 C1 and 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 C1 and 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·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 λI and 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·CI + a2·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)|2 = 1 = |a1·vI1·CI(0) + a2·vII1·CII(0)|2 and |C2|2 = 0 = |a1·vI1·CI(0) + a2·vII1·CII(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·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/a2 = 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/a2 ratio! And… Well… That just doesn’t make sense. The system must come with some specific value for a1 and 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 a1 and a2 coefficients as he start his chapter on ‘other two-state systems’. It’s the symbols he’s using: ‘his’ a1 and 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 a1 and 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! 🙂