The Stern-Gerlach Experiment — Question 6

Section 40-4 · Physics pages 6642–6643

Question 6 A tilted intermediate selector maximizes transmission between opposite filters A beam already prepared in | + z⟩passes a positive-n selector followed by a negative-z selector. Axis n is tilted through 0 ≤θ ≤π from z. Use the spin-half overlap rule P(+n| + z) = cos2(θ/2). z x ˆn θ Schematic; labels, not drawing lengths, specify the data. (a) Find the transmitted fraction as a function of angle. (b) Find the angle and maximum fraction.
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Question 6 - Solution A tilted intermediate selector maximizes transmission between opposite filters z x ˆn θ (a) Method and calculation P(−z| + n) = sin2(θ/2), T = cos2(θ/2) sin2(θ/2). T(θ) = 1 4 sin2 θ . (b) Interpretation and further calculation dT dθ = 1 2 sin θ cos θ = 0. θmax = π/2, Tmax = 1/4 . Physical check This fraction refers to the already prepared input. An additional initial positive-z selection from an unpolarized source would halve the total source-to-detector transmission.

Question 6

A tilted intermediate selector maximizes transmission between opposite filters

A beam already prepared in |+z⟩|+z\rangle passes a positive-n selector followed by a negative-z selector. Axis n is tilted through 0≤θ≤π0\le\theta\le\pi from z. Use the spin-half overlap rule P(+n|+z)=cos⁡2(θ/2)P(+n|+z)=\cos^2(\theta/2).

Diagram shown in the original SVG worksheet.

  1. Find the transmitted fraction as a function of angle.

  2. Find the angle and maximum fraction.

Question 6 - Solution

A tilted intermediate selector maximizes transmission between opposite filters

Diagram shown in the original SVG worksheet.

(a) Method and calculation

P(−z|+n)=sin⁡2(θ/2),T=cos⁡2(θ/2)sin⁡2(θ/2).P(-z|+n)=\sin^2(\theta/2),\quad T=\cos^2(\theta/2)\sin^2(\theta/2).T(θ)=14sin⁡2θ.\boxed{T(\theta)=\tfrac14\sin^2\theta}.

(b) Interpretation and further calculation

dTdθ=12sin⁡θcos⁡θ=0.\frac{dT}{d\theta}=\tfrac12\sin\theta\cos\theta=0.θmax=π/2,Tmax=1/4.\boxed{\theta_{\max}=\pi/2,\quad T_{\max}=1/4}.

Physical check

This fraction refers to the already prepared input. An additional initial positive-z selection from an unpolarized source would halve the total source-to-detector transmission.