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A quantum particle moves in 2 dimensions with Hamiltonian H:

$ H = \frac1{2m} ((P_1 + \frac12 eBX_2)^2 + (P_2 - \frac12 eBX_1)^2) $

For constants $e,B,m$ with $e$ and $B$ nonzero.

Show that the energy levels are of the form $ (n + \frac12)\bar h |eB|\frac{1}{m}$

The hint given is to define $\bar P$ and $\bar X$ as proportional to $P_1 + \frac12 eBX_2$ and $P_2 - \frac12 eBX_1$ and show that the original Hamiltonian has the form

$\frac1{2m} P^2 + \frac12m\omega^2X^2$ for some $\omega$, where

$P_j = -i\bar h \frac{\partial}{\partial x_j}$ and $X_j = x_j$

We are given that this has energy levels $(n+\frac12)\bar h \omega$.

2 Answers2

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Not an answer, but some advice: you may have more luck if you help us translate the physics problem into a question about mathematics. For example, my bumbling attempt, knowing no quantum mechanics:

$H$ is an operator on complex-valued functions $\Psi$ over the real plane given by $$H\Psi = \frac{1}{m}\left[\left(-i \hbar\frac{\partial}{\partial x} + \frac{1}{2}eBy \right)^2+\left(-i \hbar\frac{\partial }{\partial y} - \frac{1}{2}eBx \right)^2\right]\Psi$$ where $\hbar, e, B$ and $m$ are (real?) constants.

How do I show that the eigenvalues of $H$ are of the form $\left(n+\frac{1}{2}\right)\hbar |eB|\frac{1}{m}$ for $n\in\mathbb{Z}$?

user7530
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Depending on how you want to approach this problem I've seen the following done: \begin{eqnarray} \hat{H} &=& \frac{1}{2m}[\hat{p}_x ^2+\hat{p}_x \hat{y}eB+\frac{1}{4}\hat{y}^2e^2B^2-\hat{p}_y ^2+\hat{p}_y \hat{x}eB+\frac{1}{4}\hat{x}^2e^2B^2]\\ &=&\frac{1}{2m}[\hat{p}_x^2+\hat{p}_y^2]+\hat{L}_z \frac{eB}{2m}+\frac{e^2 B^2}{8m}(\hat{x}^2+\hat{y}^2) \end{eqnarray} It's provable that $\hat{L}_z$ commutes with $\hat{p}_x ^2 + \hat{p}_y ^2$ and $\hat{x}^2 + \hat{y}^2$. You can thus form a complete set of commuting operators for $L_z$ and what appears to be a spring Hamiltonian.

You could probably calculate the levels of $L_z$ and the levels of a spring hamiltonian, so I'll leave the rest to you.

  • Don't know if you got the point you were at through the book telling you to go that route/telling you the answer, or if you derived it from the definition of canonical momentum and appreciation of azimuthal symmetry, but if it were the latter then I really admire your concise setup! – Kitter Catter Nov 27 '16 at 20:37