- #1

bmxicle

- 55

- 0

## Homework Statement

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

## Homework Equations

[tex] S_x = \frac{\hbar}{2}\left( \begin{array}{cc}0 & 1\\ 1 & 0 \end{array} \right) [/tex]

[tex] S_z = \frac{\hbar}{2}\left( \begin{array}{cc}1 & 0\\ 0 & -1 \end{array} \right) [/tex]

## The Attempt at a Solution

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

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

[tex] E_- = -\sqrt{2} \rightarrow |E_-\rangle = \frac{1}{B} \left( \begin{array}{cc} 1 - \sqrt{2} \\ 1 \end{array} \right) [/tex]

[tex] A = (4+2\sqrt{2})^{1/2} \ and \ B = (4-2\sqrt{2})^{1/2} [/tex]

Which when plugging in the value for t_1/2 gives: [tex] |\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 [/tex]

I'm confused about changing basis to compute the expectation. Since the electron is initially in [tex] | z \uparrow \rangle [/tex] 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?