The Stern-Gerlach Experiment — Question 3

Question 3

Time-of-flight selection removes speed-induced Stern-Gerlach broadening

For a fixed apparatus one positive branch has $z=C/v^2$. Its mean speed is $v_0$ with small standard deviation $\sigma_v$. Ignore all other sources of width. A time-of-flight selector reduces $\sigma_v/v_0$ from .060 to .010.

1. Derive the leading fractional width $\sigma_z/z_0$.

2. If $z_0=1.50\,\mathrm{mm}$, calculate the width before and after selection.

Question 3 - Solution

Time-of-flight selection removes speed-induced Stern-Gerlach broadening

(a) Method and calculation

$$z(v)\simeq z_0+z\prime(v_0)(v-v_0),\quad z\prime(v_0)=-2C/v_0^3.$$$$\boxed{\sigma_z/z_0\simeq2\sigma_v/v_0}.$$

(b) Interpretation and further calculation

$$\sigma_{z,\mathrm{before}}=2(.060)(1.50\,\mathrm{mm})=\boxed{.180\,\mathrm{mm}}.$$$$\sigma_{z,\mathrm{after}}=2(.010)(1.50\,\mathrm{mm})=\boxed{.0300\,\mathrm{mm}}.$$

Physical check

This linear uncertainty propagation requires a narrow speed distribution. Slower atoms deflect farther, so a broad speed distribution can produce asymmetric spot shapes.