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 $B=\triangledown \times A=(0,0,B)$


"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".


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 Kanos Apr 30 '15 at 13:43

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    $\begingroup$ What did you get when you worked through the procedure suggested in the hint? $\endgroup$ – 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

$$\hat{H}=\frac{1}{2m}(-\hbar^2 \frac{\partial^2}{\partial x^2}+\frac{e^2B^2}{c^2}x^2+\frac{2 \hbar e B k_y}{c}x + \hbar^2k_y^2+\hbar^2k_z^2)$$

This hamiltonian resmbles that of hamonic oscillator, so if you change variables properly, you can change it to hamonic oscillator problem.


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