Electron Spin — Question 5

Question 5

A transverse measurement saturates the spin uncertainty bound

An electron is in $|+z\rangle$. Use $S_x^2=S_y^2=S_z^2=\hbar^2 I/4$, $\langle S_x\rangle=\langle S_y\rangle=0$, and $[S_x,S_y]=i\hbar S_z$.

1. Calculate both transverse standard deviations.

2. Compare their product with $\frac12|\langle[S_x,S_y]\rangle|$.

Question 5 - Solution

A transverse measurement saturates the spin uncertainty bound

(a) Method and calculation

$$(\Delta S_x)^2=\langle S_x^2\rangle-\langle S_x\rangle^2=\hbar^2/4.$$$$\boxed{\Delta S_x=\Delta S_y=\hbar/2}.$$

(b) Interpretation and further calculation

$$\Delta S_x\Delta S_y=\hbar^2/4.$$$$\tfrac12|i\hbar\langle S_z\rangle|=\hbar^2/4\Rightarrow\boxed{\text{bound saturated}}.$$

Physical check

A precisely known z projection does not make transverse measurements precise. This uncertainty describes repeated measurements on identically prepared electrons.