# Constant magnetic field applied to a quantum harmonic oscillator

I have a spinless particle of mass $m$ and charge $q$ which is an isotropic harmonic oscillator of frequency $\omega_0$, then I apply a constant magnetic field in the $z$ direction. We can show the Hamiltonian operator in this case is:

$$\hat {\mathcal H} = \frac{\hat {\mathbf p} ^2}{2m}-\frac{qB}{2m} \hat L_z + \frac{q^2B^2}{8m} (\hat x^2 + \hat y^2) + \frac{1}{2}m\omega_0^2 (\hat x^2+\hat y^2+\hat z^2)$$

Then the exercise asks me to immediately give the exact energy levels without doing any calculation, but I don't see how that's possible.

• The Hamiltonian is not separable in rectangular coordinates because of the $\hat L_z = i\hbar (\hat x \hat p_y - \hat y \hat p_x)$ term. My friend told me to just use the fact that $\left \{\hat {\mathcal H},\hat L^2, \hat L_z \right \}$ form a CSCO (complete set of commuting observables) and thus have the same eigenvectors, but I don't think that's true; they form a CSCO only if the potential is central.
• Changing the Hamiltonian to cylindrical or spherical coordinates doesn't seem to help.
• I could try via a perturbation method:

$$\hat {\mathcal H} = \hat {\mathcal H}_{0x} + \hat {\mathcal H}_{0y} + \hat {\mathcal H}_{0z} + \hat {\mathcal H}_{V}$$

with:

\begin{align*} \hat {\mathcal H}_{0x} &= \frac{\hat p_x ^2}{2m} + \frac{1}{2}m \left [ \left ( \frac{qB}{2m} \right )^2 + \omega_0 ^2 \right ]\hat x^2\\ \hat {\mathcal H}_{0y} &= \frac{\hat p_y ^2}{2m} + \frac{1}{2}m \left [ \left ( \frac{qB}{2m} \right )^2 + \omega_0 ^2 \right ]\hat y^2\\ \hat {\mathcal H}_{0x} &= \frac{\hat p_z ^2}{2m} + \frac{1}{2}m \omega_0 ^2 \hat z^2\\ \hat {\mathcal H}_{V} &= -\frac{qB}{2m} \hat L_z \end{align*}

But this requires doing some calculation and doesn't give the exact energy levels.

Thanks for the help!

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You should be using cylindrical coordinates because the symmetry of the problem demands it. Doing anything else is just playing hard-headed.

You should split the hamiltonian into a $z$ component and a cylindrical radial component: \begin{align*} \hat {\mathcal H}_{\rho} &= \frac{\hat p_x ^2+\hat p_y^2}{2m} + \frac{1}{2}m \left ( \omega_\text{L}^2 + \omega_0 ^2 \right)\left(\hat x^2+\hat y^2\right)-\omega_\text{L} \hat L_z,\\ \hat {\mathcal H}_{x} &= \frac{\hat p_z ^2}{2m} + \frac{1}{2}m \omega_0 ^2 \hat z^2, \end{align*} where $\omega_\text{L}=\frac{qB}{2m}$ is the relevant Larmor frequency. The $z$ component is immediate. The radial component has a 2D central potential which commutes with $\hat L_z$, so you can simply add the energies. The $\hat L_z$-less radial hamiltonian can then be decomposed into trivial $x$ and $y$ components to find the eigenenergies.

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So let me see, I could write $\hat {\mathcal H}_\rho = \hat {\mathcal H}_\text{harm} - \omega_L \hat L_z$. Then $\hat {\mathcal H}_\text{harm}$ and $\hat L_z$ form a CSCO. So $\hat {\mathcal H}_\rho |\psi\rangle = E|\psi\rangle \implies (\hat {\mathcal H}_\text{harm} - \omega_L \hat L_z) |\psi\rangle = E|\psi\rangle \implies \hbar \omega(n+1) -\omega_L \hbar m = E$, where $\omega$ is the mixture of the Larmor frequency and the natural frequency. Does this work? – jasmine Feb 22 '13 at 21:40
Yes. Keep in mind, though, that in that notation you must have $m\leq n$ (or some similar condition). This is essentially because $\hat T\geq L_z^2/2mr^2$. You can work out the precise condition by working out the degree of polynomial needed to match the factor of $(x\pm i y)^{|m|}$ in the wavefunction required by having $L_z=\hbar m$. – Emilio Pisanty Feb 22 '13 at 21:59
@EmilioPisanty I'm doing a similar problem myself. Do you mind elaborating on the last point? – Pricklebush Tickletush Sep 7 '13 at 3:27
@AlecS the last comment or the end of the answer? – Emilio Pisanty Sep 7 '13 at 12:52
@EmilioPisanty Basically, what are the restrictions on $m$? – Pricklebush Tickletush Sep 7 '13 at 14:08

I don't think there is any need to change coordinates, or anything like that. Just note that $\hat L_z$ commutes with the Hamiltonian, so you can simultaneously diagonalise them. Now you can replace $\hat L_z$ with its eigenvalue, so that term becomes a constant, and the rest of the Hamiltonian just represents three decoupled harmonic oscillators.

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The problem is that the $x$ and $y$ oscillators don't individually commute with $\hat{L}_z$, only their combination does. So you can't solve for them in terms of independent $n_x,\ n_y$ (there is some nontrivial constraint condition on the expansion coefficients). – Michael Brown Feb 23 '13 at 0:31
Thanks Michael; I was too hasty. – Rhys Feb 25 '13 at 17:08

Just to add a different perspective, let me comment the following, with the punch line being that $L_z$ is proportional to a $J_y$ momentum operator. If you write $L_z$ in terms of the ladder operators, you get $L_z=-i\hbar (a_x^\dagger a_y-a_x a_y^\dagger)$. Then, taking into account the Schwinger oscillator model of angular momentum (where for example $J_+=a_x^\dagger a_y$) $L_z$ reads $L_z=2\hbar J_y$. On the other hand, $J^2=\frac{N}{2}(\frac{N}{2}+1)$, where $N$ is the sum of the occupation numbers $N_x$ and $N_y$, so $j=N/2$. With this in mind, the eigeinstates are $|j m_y\rangle$, the states with definite (Schwinger) total angular momentum $j$ and $m_y$. These can be obtained from the standard ones $|j m\rangle=|n_x n_y\rangle$ by a rotation of $-\pi/2$ around $\hat{x}$.

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Actually, combining the oscillators in $x$ and $y$ one can have, schematically, $H=N_++N_-+1$ and $L_z=N_+-N_-$. See this link link – Alan Garbarz Oct 11 '15 at 21:36