Visualization and Analogies

Like many, my favorite subject in high school mathematics was geometry. This was probably the case because it was one of the few subjects where I was able to directly visualize everything that was going on.

I find that I am prone to thinking in pictures or “visual analogies”, because this enables me to understand and remember concepts better. Solutions to certain problems may then become obvious. I’ve illustrated this kind of thinking on a couple occasions on this blog when addressing plasmons, LO-TO splitting and the long-range Coulomb interaction and also when speaking about the “physicist’s proof”.

Let me give another example, that of measurement probability. Suppose I have a spin-1/2 particle in the following initial state:

$|\psi\rangle{}=\sqrt{\frac{1}{3}}|+\rangle{} + \sqrt{\frac{2}{3}}|-\rangle{}$

In this case, when measuring $S_z$, we find that the probability of finding the particle in the spin-up state will be $P(S_z = +) = 1/3$.

Let us now consider a slightly more thought-provoking problem. Imagine that we have a spin-1 particle in the following state:

$|\psi\rangle{}=\sqrt{\frac{1}{3}}|1\rangle{} + \sqrt{\frac{1}{2}}|0\rangle{} + \sqrt{\frac{1}{6}}|-1\rangle{}$

Suppose now I measure $S_z^2$ and obtain $+1$. This measurement eliminates the possibility of measuring the $|0\rangle{}$ state henceforth. The question is : what is the probability of now measuring $-1$ if I measure $S_z$, i.e. $P(S_z = -1 | S_z^2 = 1)$? (Have a go at solving this problem before reading my solution below.)

My favorite solution to this problem involves a visual interpretation (obviously!). Imagine axes labelled by the $S_z$ kets and the initial state by a vector in this space as so:

Now, the key involves thinking of the measurement of $S_z^2$ as a projection onto the $(1,-1)$ plane as so:

After this projection, though, the wavefunction is unnormalized. Therefore, one needs to normalize (or re-scale) the wavefunction again so that the probabilities still add up to one. This can done quite simply and the new wavefunction is therefore:

$|\psi_{new}\rangle{}=\sqrt{\frac{2}{3}}|1\rangle{} + \sqrt{\frac{1}{3}}|-1\rangle{}$

Hence the probability of measuring $S_z = -1$ is now increased to $1/3$, whereas it was only $1/6$ before measuring $S_z^2$. It is important to note that the ratio of the probabilities, $P(S_z=1)/P(S_z=-1)$, and the relative phase $e^{i\phi}$ between the $|+1\rangle{}$ and the $|-1\rangle{}$ kets does not change after the projection.

I find this solution to the problem particularly illuminating because it permits a visual geometric interpretation and is actually helpful in solving the problem.

Please let me know if you find this kind of visualization as helpful as I do, because I hope to write posts in the future about the Anderson pseudo-spin representation of BCS theory and about the water analogy in electronic circuits.