How to Find the Expectation Value in the Stern-Gerlach Experiment?

[bmxicle]

Homework Statement

An electron starts in a spin state $$|\psi(t=0)\rangle = |z \uparrow \rangle$$ and evolves in a magnetic field $$B_0(\hat{x} + \hat{z})$$. The Hamiltonian of the system is $$\hat{H} = \alpha \vec{B}\cdot\vec{S}$$. Evaluate $$\langle \psi (t_{1/2}) | S_x | \psi(t_{1/2}) \rangle$$

Homework Equations

$$S_x = \frac{\hbar}{2}\left( \begin{array}{cc}0 & 1\\ 1 & 0 \end{array} \right)$$
$$S_z = \frac{\hbar}{2}\left( \begin{array}{cc}1 & 0\\ 0 & -1 \end{array} \right)$$

The Attempt at a Solution

From another part of the question I found the period as $$T = \dfrac{\sqrt{2}\pi}{\alpha B_o}$$ since $$\hat{H} = \alpha B_o(\hat{S_x} +\hat{S_z})$$

The energy eigenvalues/eigenvectors are are $$E_+ = \sqrt{2} \rightarrow |E_+\rangle = \frac{1}{A} \left( \begin{array}{cc} 1 + \sqrt{2} \\ 1 \end{array} \right)$$

$$E_- = -\sqrt{2} \rightarrow |E_-\rangle = \frac{1}{B} \left( \begin{array}{cc} 1 - \sqrt{2} \\ 1 \end{array} \right)$$
$$A = (4+2\sqrt{2})^{1/2} \ and \ B = (4-2\sqrt{2})^{1/2}$$

Which when plugging in the value for t_1/2 gives: $$|\psi(t_{1/2})\rangle =Ae^{-i\frac{\pi}{2}}|E_+\rangle + B e^{i\frac{\pi}{2}} |E_-\rangle = |\psi(t_{1/2})\rangle =-iA|E_+\rangle + iB|E_-\rangle$$

I'm confused about changing basis to compute the expectation. Since the electron is initially in $$| z \uparrow \rangle$$ does this mean I want to change the expression I have for energy in terms in the eigenstates of S_x?
ie. Are the the S matrices all in the z-basis since that is what [tex] |\psi(t=0)\rangle[\tex] is given in?

 

[NateD]

So then I would have S_x = \frac{\hbar}{2}\left( \begin{array}{cc}0 & 1\\ 1 & 0 \end{array} \right) = \frac{1}{A}\left( \begin{array}{cc} A^2 & AB \\ AB & B^2 \end{array} \right) And the expectation value would be \langle \psi (t_{1/2}) | S_x | \psi(t_{1/2}) \rangle = \frac{iAB}{A^2B^2}(-iA^2 + iB^2) = \frac{AB}{A^2B^2}(B^2-A^2) = \frac{AB}{A^2B^2}(-2\sqrt{2}) = \frac{-2\sqrt{2}}{4+2\sqrt{2}}(-2\sqrt{2}) = -\frac{4}{4+2\sqrt{2}}