Eigenvalues dependent on choice of $\vec{A}$?

[CharlieCW]

Homework Statement

A particle with spin s=1/2 moves under the influence of a magnetic field given by:

$$\vec{A}=B(-y,0,0)$$

Find the eigenvalues of the corresponding Pauli hamiltonian. Repeat the same process for:

$$\vec{A}=\frac{B}{2}(-y,x,0)$$

Explain your result by relating the wavefunctions of both cases.

Homework Equations

$$\vec{\nabla}\times \vec{A}$$

$$\hat{H}=H_0+\frac{1}{2m}(-e\vec{A}\cdot\vec{p}+i\hbar\vec{\nabla}\cdot\vec{A}+e^2\vec{A}^2)+e\phi+\mu_b \vec{\sigma}\cdot\vec{B}$$

$$L_z=-ih(x\partial_y-y\partial_x)$$

$$\mu_B=\frac{e\hbar}{2m}$$

$$\hat{H}\Psi(n,l,m)=E\Psi(n,l,m)$$

$$L_z\Psi(n,l,m)=m\hbar\Psi(n,l,m)$$

The Attempt at a Solution

My question is actually punctual as I've solved most of the exercise. My problem arises in the hamiltonian used to calculate the energies, as they should be the same but the calculations say otherwise.

We can easily verify that, for both cases, we have the same magnetic field $\vec{B}=B\hat{z}$, which is not surprising since the vector potential is not unique for a given field.

Let's begin with the second case with $\vec{A}=\frac{B}{2}(-y,x,0)$ since it's easier. Calculating some important products:

$$\vec{A}\cdot\vec{p}=A_xp_x+A_yp_y=-Bi\hbar(y\partial_x-x\partial_y)=B(L_z)$$

$$\vec{\nabla}\cdot\vec{A}=0$$

Substituting in the Pauli hamiltonian (with $\phi=0$ and $e\rightarrow-e$ for an electron, and neglecting second order terms in $\vec{A}$):

$$\hat{H}=H_0+\frac{1}{2m}(eB(L_z))+e\phi+\mu_b \vec{\sigma}\cdot\vec{B}$$

$$\hat{H}=H_0+\frac{\mu_B B}{\hbar}(L_z+\sigma_z)$$

Substituting our hamiltonian in the time-independent Schrodinger equation, we find straightforward the energies by applying the operators to a hydrogen-like wavefunction:

$$(H_0+\frac{\mu_B B}{\hbar}(L_z+\sigma_z))\Psi(n,l,m)=E\Psi(n,l,m)$$

Since the Pauli matrix $\sigma_z=(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array})$, we get 2 equations with only a sign difference on the last term, so we get two pairs of energies:

$$E^{\pm}_{nm}=E_n+\mu_B B (m\pm 1)$$

Now for the first case, with $\vec{A}=\frac{B}{2}(-y,0,0)$, the procedure is similar except for one key difference. Let's calculate the relevant products again:

$$\vec{A}\cdot\vec{p}=A_xp_x=B(i\hbar y\partial_x)=B(L_z+i\hbar x\partial_y)$$

$$\vec{\nabla}\cdot\vec{A}=0$$

Now the hamiltonian has the form:

$$\hat{H}=H_0+\frac{1}{2m}(eB(L_z+i\hbar x\partial_y))+e\phi+\mu_b \vec{\sigma}\cdot\vec{B}$$

Substituting into time-independent Schrodinger, we finally get:

$$H_0\Psi'(n,l,m)+\frac{\mu_B B}{\hbar}(L_z+\sigma_z+i\hbar x\partial_y)\Psi'(n,l,m)=E\Psi'(n,l,m)$$

Which seems to don't give the same eigenvalues (I'm not sure how to evaluate the last term $(i\hbar x\partial_y)\Psi(r)$). However, this doesn't make sense, since the energies shouldn't depend on the choice of the vector potential but on the external magnetic field, which is the same for both cases.

My idea is that the wavefunction should "absorb" this small difference so both energies (eigenvalues) are the same, but I'm not sure how to proceed here. Or if there's a way to act the last term on the eigenfunction.

 

[CharlieCW]

Solved it.