Zeeman effect - eigenstates and its degeneracy - with and without magnetic field

Question

Consider a hydrogen atom in a homogeneous magnetic field $\vec{B}=B\vec{e_z}$. Using the coulomb gauge ($\nabla \vec{A}=0$) we can take $\vec{A}=\frac{1}{2}\vec{B}\times \vec{r}$ as a vector potential. The spin of the electrons should not be considered in this exercise. Work with gaussian units.

a) Show that by neglecting any terms ~$B^2$ the hamiltonian can be written as

$H=\frac{\vec{p}^2}{2m}+V(r)-\frac{q}{2mc}\vec{L}\cdot \vec{B}$

Hint: Use $H=\frac{1}{2m}(\vec{p}-\frac{q}{c}\vec{A})^2 +q\phi$.

b) Show further that the known eigenstates $\left|nlm\right>$ of the hydrogen atom without the field are also eigenstates of the atom in the magnetic field. Determine the eigenvalues and its degeneracy.

I'm doing some exercises on the side to keep me fresh and this is one I'm kind of stuck at.

Here is my work so far:

a) Basically I plugged in the definition of A (I won't bother to write it with the vector arrow) into the hint

$\begin{eqnarray}H&=&\frac{1}{2m}(p^2-\frac{2q}{c}pA+\frac{q^2}{c^2}A^2)+q\phi \\ &=&\frac{1}{2m}(p^2-\frac{q}{c}p\cdot(B\times r)+\frac{q^2}{4c^2}(B\times r)^2)+q\phi\end{eqnarray}$

From the exercise I can already forget about the term $\frac{q^2}{4c^2}(B\times r)^2$ since it's proportional to $B^2$. All I have left is

$H=\frac{1}{2m}(p^2-\frac{q}{c}L\cdot B)+q\phi$, now I assume that $q\phi=V(r)$ for some reason (would be nice if anybody told me why, although I assume that $\phi$ is a scalar potential and thus a legitimate assumption).

Part a) was very straightforward actually. The real problem lies within b).

I don't even know how to approach it and further I'm not really used to the notation of eigenstates in that way $\left|nlm\right>$.

I would appreciate any help I can get on this.

Answer 1

Your answer for part (a) looks correct.

For part (b), you are looking at the energy eigenfunctions of the Hydrogen atom, the derivation of which can be found in any elementary text on quantum mechanics. They are described by three quantum numbers $l$, $m$, and $n$, such that in the position representation, the eigenfunctions are

$$\left< \mathbf{x'} | n, l, m \right> = R_{nl}(r) ~Y_l^{m}(\theta, \phi)$$

where $R$ is the radial component of the wavefunction, which depends only on position, and the $Y$s are functions called spherical harmonics, which only have an angular dependence.

Note that these states are simultaneous eigenstates of the Hamiltonian and the angular momentum operator, $L_z$, with eigenvalue $m \hbar$.

To demonstrate that the $\left| n,l,m \right>$ states are eigenstates of the Zeeman Hamiltonian, express the Hamiltonian as

$$H_{zeeman} = H_{hydrogen} - \frac{qB_z}{2mc}L_z$$

where I've used the fact that $\mathbf{B}$ only has a $z$ component. I've already told you how $L_z$ acts on the $\left| n,l,m \right>$ states. I'll leave the rest to you.