Canonical perturbation theory

[noospace]

Homework Statement

An electron is inside a magnetic field oriented in the z-direction. No measurement of the electron has been made. A magnetic field in the x-direction is now switched on. Calculate the first-order change in the energy levels as a result of this perturbation.

The Attempt at a Solution

I got this wrong, claiming that the first order perturbation is zero. I said this because if you evaluate the expectation of the $S_x$ matrix in the states up/down individually, you obviously get zero since $S_x$ is anti-diagonal.

Apparently you are supposed to evaluate the expectation of $S_x$ in the superposition of states $\psi = 1/\sqrt{2}(1,0)^T + 1/\sqrt{2}(0,1)^T$. I have trouble understanding this because $\psi$ is not an eigenstate of the orginial Hamiltonian, which it should be for canonical perturbation theory.

Any help is greatly appreciated.

 

[Physics Monkey]

Hi noospace,

Something is very strange here. Assuming the z-direction magnetic field is large and the x-direction magnetic field is small, your calculation is correct. There is no first order change in the energy levels in such a situation. (One could just solve the complete problem trivially and see this fact.) Using eigenstates of the perturbation to calculate expectation values is certainly wrong. I suspect there is some kind of confusion here. Any ideas?

 

[noospace]

Hi Physics Monkey,

Thanks for your response. My lecturer's approach to this question is to evaluate the expectation of the operator $S_x$ in the superposition of states $\psi = \frac{1}{\sqrt{2}} | 1/2 1/2 \rangle + \frac{1}{\sqrt{2}}}|1/2 \pm 1/2\rangle$. The answer given is

$\pm\frac{B_x\hbar}{2m_e}$

My lecturer claims that since $| 1/2 1/2 \rangle, | 1/2 -1/2 \rangle$ are separately eigenstates of the Hamiltonian then so should their superposition. I claim this is false since the sum of two eigenvectors need not be an eigenvector.

Who is correct?