QM: position and spin dependent potential

[Heirot]

Homework Statement

A spin 1/2 particle of mass m is described by the Hamiltonian: H = p^2/(2m) + 1/2 mw^2 x^2 + g * x * sigma_x, where sigma_x is one of the Pauli matrices.

Homework Equations

The Attempt at a Solution

I have no idea where to start. It's obvious that the harmonic oscillator part of H should be multiplied by unit matrix. Does this mean that I need to diagonalize the whole matrix H? That would give me two eigenvalues containing operators p and x. When I solve the two equations, I should get the energies for spin up and spin down? I'm not certain is it ok to mix the spin and position variables in such a way.

 

[fzero]

Think of the wavefunction as a two-component vector

$$\begin{pmatrix} \psi_+(x) \\ \psi_-(x) \end{pmatrix}.$$

 

[Heirot]

Yes, that's exactly what I did and it leads to the following equations:
$$(\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2-E_+)\Psi_+ + gx\Psi_-=0 \qquad<br /> (\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2-E_-)\Psi_- + gx\Psi_+=0$$

Now, for this system to be consistent, it's necessary for the determinant to vanish. But here the coefficients are not numbers but operators! I'm not sure whether I'm allowed to calculate the determinant of an operator?

 

[fzero]

Yes, that's exactly what I did and it leads to the following equations:
$$(\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2-E_+)\Psi_+ + gx\Psi_-=0 \qquad<br /> (\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2-E_-)\Psi_- + gx\Psi_+=0$$

Now, for this system to be consistent, it's necessary for the determinant to vanish. But here the coefficients are not numbers but operators! I'm not sure whether I'm allowed to calculate the determinant of an operator?

You still want to solve

$$H \psi = E \psi,$$

so don't introduce $$E_\pm$$ at this point. I was a bit hasty to introduce $$\psi_\pm$$. Things will be cleaner if you use $$(\Psi_1, \Psi_2)$$, since we're not working in the basis that diagonalizes $$\sigma_x$$. All together, we'll have

$$(\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2-E)\Psi_1 + gx\Psi_2=0 \qquad<br /> (\frac{p^2}{2m}+\frac{m\omega^2}{2}x^2-E)\Psi_2 + gx\Psi_1=0.$$

Don't worry about any determinants yet, just consider the sum and difference of these equations and find the spectrum. The determinant of those equations is zero whenever $$E=E_\pm$$, the energy eigenvalues, so it's consistent.