# Applying Schrodinger equation to find the energies of a free electron model in a metal [closed]

The one-particle Hamiltonian is given by $$\hat{H}=\frac{1}{2m}\left(p+\frac{e}{c}A\right)$$

with $e > 0$ and vector potential $A=(0,x,0)B$, such $B=\triangledown \times A=(0,0,B)$

Question:

"I was asked to show that the degrees of freedom in the x-y plane are those of a harmonic oscillator. To determine the one-particle energies and the degeneracy, $\mathrm{g}$, of the corrsponding one-particle states".

Hint:

Use the ansatz $\psi(x.y.z)=u(x)\exp(iky)\exp(ikz)$ with $p=\hbar\vec{k}$ for the Schrodinger equation $\hat{H}\Psi(r)=E\Psi(r)$, and find an equation for $u(x)$ by multiplying out the Hamiltonian operator $\hat{H}$. Separate the motion in $z$-direction from the $xy$-plane by using $\varepsilon=E+\frac{\hbar^2 k_z^2}{2m}$, where E is th energy from motion in the xy-plane. You will find a harmonic oscillator equation for $u(x)$ with equilibrium point $x_0(k_y)$. Determine the degeneracy factor, $\mathrm{g}$, of the corresponding energy levels, $E(n)$, $n=0,1,2,...$ by using $0<x_0(k_y)<L$ and $k_y=\frac{2\pi}{L}\ell$ with integer $\ell$.

## closed as off-topic by John Rennie, ACuriousMind♦, Danu, WetSavannaAnimal, Kyle KanosApr 30 '15 at 13:43

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, ACuriousMind, Danu, WetSavannaAnimal, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

• What did you get when you worked through the procedure suggested in the hint? – WetSavannaAnimal Apr 30 '15 at 11:22

First I think that the hamiltonian is $\hat{H}=\frac{1}{2m}\left(p+\frac{e}{c}A\right)^2$, not $\hat{H}=\frac{1}{2m}\left(p+\frac{e}{c}A\right)$.
Then the remaining is just calculation. Using $[\hat{p},A]=0$, expanding hamiltonian, we get $$\hat{H}=\frac{1}{2m}(-\hbar^2 \frac{\partial^2}{\partial x^2}+\frac{e^2B^2}{c^2}x^2+\frac{2 \hbar e B}{ci}x\frac{\partial}{\partial y} - \hbar^2 \frac{\partial^2}{\partial y^2} - \hbar^2 \frac{\partial^2}{\partial z^2})$$
The $z$ part of hamiltonian is just free particle, and if you use ansatz $\psi(x.y.z)=u(x)\exp(iky)\exp(ikz)$, you can also remove the $\partial_y$ term. Then the hamiltonian becomes
$$\hat{H}=\frac{1}{2m}(-\hbar^2 \frac{\partial^2}{\partial x^2}+\frac{e^2B^2}{c^2}x^2+\frac{2 \hbar e B k_y}{c}x + \hbar^2k_y^2+\hbar^2k_z^2)$$