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.

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.