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*〉C* _{1}* + |

*2*〉C

*+ |*

_{2}*3*〉C

*+|*

_{3}*4*〉C

_{4 }with C

_{i }= 〈

*i*|ψ〉

We refer to |ψ〉 as the *spin state *of the system, and so it’s *determined *by those four C_{i }amplitudes. Now, we know that those C_{i }amplitudes 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* = −μ**σ**·**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}^{e}σ_{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}^{e}σ_{z}^{p}-like combinations, because σ_{x}^{e}σ_{z}^{p }= σ_{z}^{p}σ_{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}^{e}σ_{z}^{p }= σ_{z}^{p}σ_{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 E_{0} constant is, of course, to be multiplied by the unit matrix, so we should actually write E_{0}δ_{ij} instead of E_{0}, but… Well… Quantum physicists always want to confuse you. 🙂 Second, the **σ**^{e}**σ**^{p }is 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}^{p}^{ }matrices as being the three *components* of two new (matrix) vectors, which we write as **σ**^{e }and **σ**^{p }respectively. 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

*functions should have the following functional form:*

_{i}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:

*a*_{1} = 1 and *a*_{2} = *a*_{3} = *a*_{4} = 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:

*a*_{4} = 1 and *a*_{1} = *a*_{2} = *a*_{3} = 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 !

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