Neither Example 4.4 nor Problem 4.73 actually solved the Schrödinger equation for the Stern–Gerlach experiment. In this problem we will see how to set up that calculation. The Hamiltonian for a neutral, spin 1/2 particle traveling through a Stern–Gerlach device is H = p^2 /2m - γB.S. where B is

Question

Neither Example 4.4 nor Problem 4.73 actually solved the Schrödinger equation for the Stern–Gerlach experiment. In this problem we will see how to set up that calculation. The Hamiltonian for a neutral, spin 1/2 particle traveling through a Stern–Gerlach device is

[latex] H=\frac{p^{2}}{2 m}-\gamma B \cdot S [/latex].

where B is given by Equation 4.169. The most general wave function for a spin- particle—including both spatial and spin degrees of freedom—is

[latex] B (x, y, z)=-\alpha x \hat{\imath}+\left(B_{0}+\alpha z ight) \hat{k} [/latex] (4.169).

[latex] \Psi( r , t)=\Psi_{+}( r , t) \chi_{+}+\Psi_{-}( r , t) \chi_{-} [/latex].

(a) Put [latex] \Psi ( r , t) [/latex] into the Schrödinger equation

[latex] H \Psi=i \hbar \frac{\partial}{\partial t} \Psi [/latex].

to obtain a pair of coupled equations for [latex] \Psi_{\pm} [/latex]. Partial answer:

[latex] -\frac{\hbar^{2}}{2 m} abla^{2} \Psi_{+}-\frac{\hbar}{2} \gamma\left(B_{0}+\alpha z ight) \Psi_{+}+\frac{\hbar}{2} \gamma \alpha x \Psi_{-}=i \hbar \frac{\partial}{\partial t} \Psi_{+} [/latex].

(b) We know from Example 4.3 that the spin will precess in a uniform field [latex] B_{0} \hat{k} [/latex]. We can factor this behavior out of our solution—with no loss of generality—by writing

[latex] \Psi_{\pm}( r , t)=e^{\pm i \gamma B_{0} t / 2} ilde{\Psi}( r , t) [/latex].

Find the coupled equations for [latex] ilde{\Psi}_{\pm} [/latex]. Partial answer:

[latex] -\frac{\hbar^{2}}{2 m} abla^{2} ilde{\Psi}_{+}-\frac{\hbar}{2} \gamma \alpha z ilde{\Psi}_{+}+\frac{\hbar}{2} \gamma \alpha x e^{-i \gamma B_{0} t} ilde{\Psi}_{-}=i \hbar \frac{\partial}{\partial t} ilde{\Psi}_{+} [/latex].

(c) If one ignores the oscillatory term in the solution to (b)—on the grounds that it averages to zero (see discussion in Example 4.4)—one obtains uncoupled equations of the form

[latex] -\frac{\hbar^{2}}{2 m} abla^{2} ilde{\Psi}_{\pm}+V_{\pm} ilde{\Psi}_{\pm}=i \hbar \frac{\partial}{\partial t} ilde{\Psi}_{\pm} [/latex].

Based upon the motion you would expect for a particle in the “potential” [latex] V_{\pm} [/latex], explain the Stern–Gerlach experiment.

Accepted Answer

(a) Write S in terms of raising and lowering operators:

[latex] H=\frac{p^{2}}{2 m}-\gamma\left[B_{x} \frac{S_{+}+S_{-}}{2}+B_{y} \frac{S_{+}-S_{-}}{2 i}+B_{z} S_{z} ight]=\frac{p^{2}}{2 m}+\frac{1}{2} \gamma \alpha x\left(S_{+}+S_{-} ight)-\gamma\left(B_{0}+\alpha z ight) S_{z} [/latex].

Then

[latex] H \Psi=\frac{p^{2}}{2 m} \Psi_{+}( r , t) \chi_{+}+\frac{p^{2}}{2 m} \Psi_{-}( r , t) \chi_{-}+\frac{1}{2} \gamma \alpha x \Psi_{+}( r , t) \hbar \chi_{-}+\frac{1}{2} \gamma \alpha x \Psi_{-}( r , t) \hbar \chi_{+} [/latex]

[latex] -\gamma\left(B_{0}+\alpha z ight) \Psi_{+}( r , t) \frac{\hbar}{2} \chi_{+}-\gamma\left(B_{0}+\alpha z ight) \Psi_{-}( r , t)\left(-\frac{\hbar}{2} \chi_{-} ight) [/latex].

[latex] =\left[\frac{p^{2}}{2 m} \Psi_{+}+\frac{\hbar}{2} \gamma \alpha x \Psi_{-}-\frac{\hbar}{2} \gamma\left(B_{0}+\alpha z ight) \Psi_{+} ight] \chi_{+}+\left[\frac{p^{2}}{2 m} \Psi_{-}+\frac{\hbar}{2} \gamma \alpha x \Psi_{+}+\frac{\hbar}{2} \gamma\left(B_{0}+\alpha z ight) \Psi_{-} ight] \chi_{-} [/latex],

while

[latex] i \hbar \frac{\partial}{\partial t} \Psi=i \hbar \frac{\partial \Psi_{+}}{\partial t} \chi_{+}+i \hbar \frac{\partial \Psi_{-}}{\partial t} \chi_{-} [/latex].

Since [latex] \chi_{+} ext {and } \chi_{-} [/latex] are orthogonal, the coefficients on either side of Schrödinger's equation must match and we have

[latex] \frac{p^{2}}{2 m} \Psi_{+}+\frac{\hbar}{2} \gamma \alpha x \Psi_{-}-\frac{\hbar}{2} \gamma\left(B_{0}+\alpha z ight) \Psi_{+}=i \hbar \frac{\partial \Psi_{+}}{\partial t} [/latex].

[latex] \frac{p^{2}}{2 m} \Psi_{-}+\frac{\hbar}{2} \gamma \alpha x \Psi_{+}+\frac{\hbar}{2} \gamma\left(B_{0}+\alpha z ight) \Psi_{-}=i \hbar \frac{\partial \Psi_{-}}{\partial t} [/latex].

(b)

[latex] \frac{\partial \Psi_{\pm}}{\partial t}=\pm \frac{i}{2} \gamma B_{0} \Psi_{\pm}+e^{\pm i \gamma B_{0} t / 2} \frac{\partial}{\partial t} ilde{\Psi}_{\pm} [/latex].

Plugging into the results from (a), and cancelling the exponential factor:

[latex] \frac{p^{2}}{2 m} ilde{\Psi}_{+}+\frac{\hbar}{2} \gamma \alpha x e^{-i \gamma B_{0} t} ilde{\Psi}_{-}-\frac{\hbar}{2} \gamma \alpha z ilde{\Psi}_{+}=i \hbar \frac{\partial ilde{\Psi}_{+}}{\partial t} [/latex].

[latex] \frac{p^{2}}{2 m} ilde{\Psi}_{-}+\frac{\hbar}{2} \gamma \alpha x e^{i \gamma B_{0} t} ilde{\Psi}_{+}+\frac{\hbar}{2} \gamma \alpha z ilde{\Psi}_{-}=i \hbar \frac{\partial ilde{\Psi}_{-}}{\partial t} [/latex].

(c) The potential [latex] V_{\pm}=\mp \gamma \alpha \hbar z / 2 [/latex] is a linear potential that decreases in the positive-z direction for spin up states, and in the negative-z direction for spin down. Therefore if the particle starts in a state where [latex] \Psi_{+}=\Psi_{-} [/latex], once it enters the magnet [latex] \Psi_{+} [/latex] will propagate in the positive-z direction and [latex] \Psi_{-} [/latex] will propagate in the negative-z.

Question 4.74: Neither Example 4.4 nor Problem 4.73 actually solved the Sch...

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