Electron Spin — Question 9

Question 9

A full spin rotation can change an interferometer output

An ideal two-path interferometer recombines equal amplitudes with zero initial relative phase. Only path A applies $U_z(\theta)=\exp(-i\theta\sigma_z/2)$ to an incident $|+z\rangle$. The probability at its initially bright output is $P=|(1+u)/2|^2$, where $U_z|+z\rangle=u|+z\rangle$.

1. Find P as a function of the rotation angle.

2. Compare $\theta=2\pi$ and $4\pi$, and explain why the phase is observable here.

Question 9 - Solution

A full spin rotation can change an interferometer output

(a) Method and calculation

$$\sigma_z|+z\rangle=|+z\rangle\Rightarrow u=e^{-i\theta/2}.$$$$P=\tfrac14(2+2\cos(\theta/2))=\boxed{\cos^2(\theta/4)}.$$

(b) Interpretation and further calculation

$$\boxed{P(2\pi)=0,\quad P(4\pi)=1}.$$$$2\pi:\quad |A\rangle+|B\rangle\longrightarrow-|A\rangle+|B\rangle.$$

Physical check

A phase applied to only one path is relative to the other path. Rotating both paths identically would produce only a common phase and no intensity change.