9
$\begingroup$

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!

$\endgroup$
4
$\begingroup$

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: $$ \begin{align*} \hat {\mathcal H}_{\rho} &= \frac{\hat p_x ^2+\hat p_y^2}{2m} + \frac{1}{2}m \left ( \omega_\text{c}^2 + \omega_0 ^2 \right)\left(\hat x^2+\hat y^2\right)-\omega_\text{c} \hat L_z,\\ \hat {\mathcal H}_{z} &= \frac{\hat p_z ^2}{2m} + \frac{1}{2}m \omega_0 ^2 \hat z^2, \end{align*} $$ where $\omega_\text{c}=\frac{qB}{2m}$ is the relevant cyclotron frequency. To solve this without any calculation:

  • The $z$ component is immediate.
  • The radial component has a 2D central potential $\mathcal H_{\rho,0}=\frac{\hat p_x ^2+\hat p_y^2}{2m} + \frac{1}{2}m \left ( \omega_\text{c}^2 + \omega_0 ^2 \right)\left(\hat x^2+\hat y^2\right)$ which commutes with $\hat L_z$, so you can simply add the energies: that is, $\mathcal H_{\rho,0}$ has a common eigenbasis with $\omega_\text{c} \hat L_z$, and you can diagonalize them simultaneously; this 2D harmonic oscillator is relatively standard but you can still solve it explicitly.

    To do that, you can find the eigenenergies of $\mathcal H_{\rho,0}$ by splitting it into two 1D harmonic oscillators, but the product basis of those oscillators does not give you an angular-momentum eigenbasis. Instead, the thing to do is to note that the set of eigenfunctions with a fixed total excitation number $n$, i.e. the subspace $$\mathcal L_n=\mathrm{span} \left\{|n_x\rangle \otimes |n_y\rangle : n_x+n_y=n\right\}$$ is invariant under rotations (because the hamiltonian is) so you can diagonalize $\hat L_z$ in this subspace.

    How do you do that, then? Well, you notice that your basis functions, of the form $H_{n_x}(x)H_{n_y}(y)$ once you remove the gaussian, are polynomials of degree $n$ in $x,y$ and with definite parity, and you're looking to decompose them as linear combinations of $(x\pm iy)^m$ for $m=\ldots,n-4,n-2,n$, which are the eigenfunctions of $\hat L_z$.

    That then tells you what eigenvalues $m$ of $\hat L_z$ are allowed for each eigenspace of $\mathcal H_{\rho,0}$ with excitation number $n$, i.e. $m\leq n$ with both of the same parity. That's sufficient to fix the spectrum.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Michael Brown Feb 23 '13 at 0:31
0
$\begingroup$

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 $\hat{x}$.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Alan Garbarz Oct 11 '15 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.