Spins first with computation

I’m teaching quantum mechanics (PHY 3310) at HPU for the first time this semester, and I decided to try a “spins first” approach where the whole framework is developed in the context of two-level systems before doing the standard “wavefunction stuff.” We’re using McIntyre as the textbook, and I’m surprised to report it’s actually going pretty well so far. I used Griffiths when I took the course (the 2nd edition had just come out, to date myself), and I TA’d a course using that edition of Griffiths twice in grad school. Suffice it to say I’ve internalized the flow of Griffiths QM pretty well, and I like it (enough). But taking a hard right turn from the comfortable road seemed like a good idea.

Because adopting a completely new approach to the subject wasn’t enough, I thought introducing multiparticle states on the third week would be fun. McIntyre has a short chapter on EPR/Bell’s inequalities, and I wanted to turn this into a computational activity. I think it’s great to go into this area so early, but you have to either introduce two-particle formalism or just pull the Bell states out of a hat and say “we’ll talk more about these in 300 pages.” McIntyre takes the latter approach, and I thought it could use some more foundation. So here’s a nifty approach I found that makes for some fun numerical calculations.

We’ve introduced the “matrix representation” of a quantum state \left|\psi\right\rangle = \alpha \left|+\right\rangle + \beta\left|-\right\rangle as

\displaystyle \left|\psi\right\rangle \dot{=} \left(\begin{array}{c} \alpha\\ \beta\end{array}\right)

Two spin-\frac{1}{2} particle states can be described by the basis \left|+\right\rangle\otimes \left|+\right\rangle, \left|+\right\rangle\otimes \left|-\right\rangle, \left|-\right\rangle\otimes \left|+\right\rangle, \left|-\right\rangle\otimes \left|-\right\rangle. This tensor product \otimes can be represented by the Kronecker product,

\displaystyle \left|\psi\right\rangle \otimes \left|\phi\right\rangle \dot{=} \left(\begin{array}{c} \alpha \\ \beta\end{array}\right)\otimes\left(\begin{array}{c} \gamma \\ \delta\end{array}\right) \equiv \left(\begin{array}{c} \alpha \gamma \\ \alpha\delta \\ \beta\gamma \\ \beta\delta\end{array}\right)

Using \left|+\right\rangle \dot{=}\left(\begin{array}{c} 1\\ 0 \end{array}\right), \left|-\right\rangle \dot{=}\left(\begin{array}{c} 0\\ 1 \end{array}\right), one gets the “matrix” representations of all the simple basis states:

\displaystyle \left|+\right\rangle \left|+\right\rangle \dot{=} \left(\begin{array}{c} 1\\ 0 \\ 0 \\ 0\end{array}\right),\;\;\;\; \displaystyle \left|+\right\rangle \left|-\right\rangle \dot{=} \left(\begin{array}{c} 0\\ 1 \\ 0 \\ 0\end{array}\right),\;\;\;\; \displaystyle \left|-\right\rangle \left|+\right\rangle \dot{=} \left(\begin{array}{c} 0\\ 0 \\ 1 \\ 0\end{array}\right),\;\;\;\; \displaystyle \left|-\right\rangle \left|-\right\rangle \dot{=} \left(\begin{array}{c} 0\\ 0 \\ 0 \\ 1\end{array}\right).

The book has introduced the total spin operator in almost a throwaway line,

{\bf \hat{S}}^{2}\left|s,m\right\rangle = s(s+1)\hbar^{2}\left|s,m\right\rangle.

I first give the students a Jupyter notebook with functions defined that generate \hat{S}^{x}, \hat{S}^{y}, \hat{S}^{z} in the standard basis for spin s. From this, they can construct {\bf \hat{S}}^{2} = \left[\hat{S}^{x}\right]^{2} + \left[\hat{S}^{y}\right]^{2} +\left[\hat{S}^{z}\right]^{2} and verify that it’s indeed s(s+1) times the identity matrix. We set \hbar = 1 so the numerical coefficients can be seen easily, and there was only been marginal pushback. So at this point, they have some intuition for spin s and they have this basis of two-particle states.

At this point, I should probably emphasize that I’m trying to build up some intuition for why the textbook says “a spin zero particle decays into two spin-\frac{1}{2} particles with total spin zero” and then presents the singlet state. And I’m trying hard to do this in a computationally-friendly way that doesn’t use a lot of fancy calculations or Clebsh-Gordon coefficients. One route to explaining the singlet is to invoke indistinguishability. The total state should exhibit invariance with respect to particle interchange. So we can define an operator \hat{\Pi} which switches the states of the particles,

\hat{\Pi}\left|+\right\rangle\left|+\right\rangle = \left|+\right\rangle\left|+\right\rangle,\;\;\;\;\;\;\;\; \hat{\Pi}\left|+\right\rangle\left|-\right\rangle = \left|-\right\rangle\left|+\right\rangle.

\hat{\Pi}\left|-\right\rangle\left|+\right\rangle = \left|+\right\rangle\left|-\right\rangle,\;\;\;\;\;\;\;\; \hat{\Pi}\left|-\right\rangle\left|-\right\rangle = \left|-\right\rangle\left|-\right\rangle.

If we buy into this new indistinguishability postulate, then \left|+\right\rangle \left|+\right\rangle and \left|-\right\rangle\left|-\right\rangle check out fine, but the two other states get mixed up. We can fix this by writing out the matrix representation for \hat{\Pi} and obtaining its eigenvalues. Regrettably, I was petrified by matrices as an undergraduate and did anything I could to avoid actually thinking about them. I was a little nervous to give this to the class as an exercise, but I was impressed by how well they were able to reason through this. My advice was to start with the identity matrix and shift the ones around until the matrix multiplication with the basis vectors respected the operator action outlined above. Most folks obtained

\displaystyle \hat{\Pi} \dot{=} \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1\end{array}\right).

It’s not actually hard to diagonalize this by hand, but we were working in Jupyter notebooks, so they could just do it numerically, obtaining eigenvectors corresponding to the representations of \left|+\right\rangle\left|+\right\rangle and \left|-\right\rangle\left|-\right\rangle with eigenvalue 1. No surprises. But you also get

\displaystyle \frac{1}{\sqrt{2}}\left(\left|+\right\rangle \left|-\right\rangle + \left|+\right\rangle \left|-\right\rangle \right)

with eigenvalue +1 and

\displaystyle \frac{1}{\sqrt{2}}\left(\left|+\right\rangle \left|-\right\rangle - \left|+\right\rangle \left|-\right\rangle \right)

with eigenvalue -1. The negative eigenvalue doesn’t scream invariance since that state picks up a factor of -1 upon the action of \hat{\Pi}. But we had already discussed the irrelevance of an overall phase e^{i\phi} to probability calculations, and -1 = e^{i\pi}, so most of them seemed to believe that these states were suitably invariant. Now this is all fine, but I don’t find it particularly satisfying because we’ve pulled this notion of indistinguishability out of a hat. And we’ve made zero connection to spin. Since it’s a simple call to a built-in function to diagonalize matrices using Numpy, I decided to have them build the total components of spin by exploiting the (built-in) Kronecker product,

\displaystyle \hat{S}^{x} = \hat{S}^{x}_{1} + \hat{S}^{x}_{2} = \hat{S}^{x}\otimes\hat{I} + \hat{I}\otimes\hat{S}^{x},

with similar expressions for \hat{S}^{y} and \hat{S}^{z}. I should emphasize that I do spin calculations with the Pauli matrices for a living and I wouldn’t really want to do any of this tensor-product hocus pocus by hand. It’s really convenient to have this built into Numpy (and even Sympy for symbolic work). Either way you do it, if you square each of these components and add them up, you find

\displaystyle {\bf \hat{S}}^{2} \dot{=} \hbar^{2}\left(\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 2\end{array}\right).

Here’s where it gets fun: this is an object they’ve already computed for a single spin-s state, where it’s always proportional to the identity matrix and they can read off s from the factor s(s+1) out front. In this system, the same operator is no longer even diagonal. What does that mean?

Well, it means that the middle two states aren’t by themselves states of definite spin. If we “do what we always do” when we encounter a non-diagonal matrix and look for the eigenvalues and eigenvectors, we find exactly the same eigenvectors that we got from diagonalizing \hat{\Pi}. The singlet state has eigenvalue 0 while the other three all come with 2\hbar^{2}. Quite nicely, if we interpret the eigenvalue to be s(s+1)\hbar^{2}, it means there is one state with s = 0 and three with s = 1. So the singlet actually corresponds to a legitimately spinless state. What of the other three? Well, a spin-1 state should have three possible basis vectors and we have precisely three states here. Sure enough, if we print out the total z component of spin,

\displaystyle \hat{S}_{tot}^{z} = \hat{S}^{z}\otimes \hat{I} + \hat{I}\otimes\hat{S}^{z} = \hbar\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1\end{array}\right).

That is, \left|+\right\rangle\left|+\right\rangle has eigenvalue +\hbar and \left|-\right\rangle\left|-\right\rangle has eigenvalue -\hbar. The mixed states are degenerate, and both the singlet and positive linear combination have vanishing projection along the z direction. For the singlet, this has to be the case since s=0 and there is no spin “vector” to project.

This all works out cleanly. A little too cleanly. The next layer of complication might be to consider three spin-\frac{1}{2} particles. This gives eight basis states \left|+\right\rangle\left|+\right\rangle\left|+\right\rangle, \cdots, \left|-\right\rangle\left|-\right\rangle\left|-\right\rangle. The indistinguishability business is a little more complicated here, but maybe we can cut directly to the eigenstates of total spin where now

\displaystyle \hat{S}^{x}_{tot} = \hat{S}^{x}\otimes \hat{I}\otimes\hat{I} + \hat{I}\otimes\hat{S}^{x}\otimes \hat{I} + \hat{I}\otimes\hat{I}\otimes\hat{S}^{x}

\displaystyle \hat{S}^{y}_{tot} = \hat{S}^{y}\otimes \hat{I}\otimes\hat{I}+ \hat{I}\otimes\hat{S}^{y}\otimes \hat{I} + \hat{I}\otimes\hat{I}\otimes\hat{S}^{y}

\displaystyle \hat{S}^{z}_{tot} = \hat{S}^{z}\otimes \hat{I}\otimes\hat{I} + \hat{I}\otimes \hat{S}^{z}\otimes \hat{I}+ \hat{I}\otimes\hat{I}\otimes\hat{S}^{z}

By just blindly building up {\bf \hat{S}}^{2} and looking at the eigenvalues, you find four instances of \lambda = \frac{3}{4}\hbar^{2} and instances of \lambda = \frac{15}{4}\hbar^{2} corresponding to s = \frac{1}{2} and s = \frac{3}{2}, respectively. Now the s = \frac{3}{2} makes sense, because there should be four states corresponding to S^{z} = \pm\frac{3\hbar}{2} and \pm\frac{\hbar}{2}. Looking at the eigenvectors, one can also confirm that \left|+\right\rangle \left|+\right\rangle \left|+\right\rangle is the +\frac{3\hbar}{2} state and \left|-\right\rangle \left|-\right\rangle \left|-\right\rangle is the -\frac{3\hbar}{2} state. The “middle” states are superpositions, and they fall out from the explicit eigenvectors just as they would by defining a lowering operator \hat{J}^{-} and acting repeatedly on the state \left|+\right\rangle\left|+\right\rangle\left|+\right\rangle.

The s = \frac{1}{2} states are a little harder to interpret. Strangely, you get two sets of spin-\frac{1}{2} states. I remember having to calculate all these states using operators in grad school. It was actually an exercise from Ballentine‘s textbook, and he had an ominous note in his solution claiming that at the level of three spins, total spin and total spin projection along one axis are no longer sufficient to fully characterize the states. Here that plays out because at the level of total spin and z projection, you get “two copies.” They aren’t the same, and so you can index them by symmetry properties or projection of one of the spins in addition to the “good” quantum numbers.

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