The Stern-Gerlach Experiment — Question 10

Question 10

Finite detector widths set a minimum useful magnetic gradient

A two-spot Stern-Gerlach apparatus has $z_\pm=\pm\mu GL(D+L/2)/(Mv^2)$. Each spot has standard deviation $\sigma=.100\,\mathrm{mm}$. Adopt the design criterion $\Delta z\ge4\sigma$, with $\mu=9.27\times10^{-24}\,\mathrm{J/T}$, $M=1.80\times10^{-25}\,\mathrm{kg}$, $v=500\,\mathrm{m/s}$, $L=.100\,\mathrm m$, $D=.400\,\mathrm m$.

1. Derive the minimum gradient.

2. Find its numerical value and explain how doubling v affects it if the spot width is unchanged.

Question 10 - Solution

Finite detector widths set a minimum useful magnetic gradient

(a) Method and calculation

$$\frac{2\mu G L(D+L/2)}{Mv^2}\ge4\sigma.$$$$\boxed{G_{\min}=\frac{2\sigma Mv^2}{\mu L(D+L/2)}}.$$

(b) Interpretation and further calculation

$$\boxed{G_{\min}=21.57\,\mathrm{T/m}}.$$$$G_{\min}\propto v^2\Rightarrow\boxed{G_{\min}(2v)=4G_{\min}(v)}.$$

Physical check

Four standard deviations is a specified engineering criterion, not a universal quantum boundary between resolved and unresolved spots. Other widths may themselves depend on velocity.