# 2D Harmonic oscillator with angular momentum term

Problem

Consider the quantum dynamics of a particle of mass m in the real plane $$\mathbb{R}^2$$ with coordinates $$q = (q_1, q_2)$$ and classical Lagrangian $$L = \frac{m}{2} \sum_{i=1, 2} \dot{q}_i \dot{q}_i + \frac{B}{2} \sum_{i,j=1,2} \epsilon^{ij} q_i \dot{q}_j − \frac{k}{2} \sum_{i=1,2} q_i q_i$$ where B and k are two real positive constants and $$\epsilon^{ij}$$ the antisymmetric tensor in two dimensions with component $$\epsilon^{12} = 1$$.

1. Calculate the Hamiltonian of the system and quantize the system by imposing the canonical commutation relations. Solve the Heisenberg equations of the quantum system.
2. Calculate the simultaneous eigenstates of the Hamiltonian and the angular momentum operator $$M = \sum_{i,j=1,2} \epsilon^{ij} q_i p_j$$ and their spectrum of eigenvalues. [Hint: write the total Hamiltonian as $$H = H_{red} − \frac{B}{2m} M$$. Note that eigenfunctions of $$M$$ with eigenvalue $$+\lambda$$ and $$−\lambda$$ are degenerate in the spectrum of $$H_{red}$$ and use this fact to write the eigenfunctions of $$H$$ in the form $$\psi(q) = \chi(q)\psi(q)$$ where $$\chi(q)$$ are eigenfunctions of $$M$$ and $$\psi(q)$$ eigenfunctions of $$H_{red}$$ with zero eigenvalue w.r.t. $$M$$]

Solution attempt

My problem is with part 2 of the exercise.

The canonical momenta can be calculated from the classical Lagrangian by

$$p_k = \frac{\partial L}{\partial \dot{q}_k} = m \dot{q}_k + \frac{B}{2} \sum_{i=1,2} \epsilon^{ik} q_i.$$

The Hamiltonian can be written in an expanded form as

$$H = \frac{p_1^2 + p_2^2}{2m} - \frac{B}{2m}M + \left (\frac{B^2}{8m} + \frac{k}{2} \right ) (q_1^2 + q_2^2)$$

where $$M = q_1 p_2 - q_2 p_1$$.

The exercise instructs to calculate the simultaneous eigenstates of $$H$$ and $$M$$ and their eigenvalues.

The hint says to write $$H = H_{red} - \frac{B}{2m}M$$. Then $$H_{red}$$ is just a simple harmonic oscillator, and $$M$$ is just the angular momentum.

One can verify that $$[H_{red}, M] = 0$$ since $$[p_1^2 + p_2^2, M] = 0$$ and $$[q_1^2 + q_2^2, M] = 0$$. Also, $$H_{red}$$ eigenfunctions are just occupation numbers $$|n,m\rangle$$ where $$n, m \in \mathbb{N}$$.

I wrote $$M$$ in terms of ladder operators as $$M = q_1 p_2 - q_2 p_1 = \frac{a_1 + a_1^\dagger}{2} \frac{a_2 - a_2^\dagger}{2i} - \frac{a_2 + a_2^\dagger}{2} \frac{a_1 - a_1^\dagger}{2i} = \frac{a_1^\dagger a_2 - a_1 a_2^\dagger}{2i},$$

but the eigenstates $$|n,m\rangle$$ generally are not the eigenstates if $$M$$, since

$$M |n, m\rangle = \frac{a_1^\dagger a_2 - a_1 a_2^\dagger}{2i} | n, m \rangle = \frac{\sqrt{(n + 1) m}| n+1, m-1 \rangle - \sqrt{n (m + 1)} | n-1, m+1 \rangle}{2i} \qquad n, m \geq 1.$$

Suppose that $$\psi$$ is a simultaneous eigenstate, that is $$M \psi = m \psi$$ and $$H_{red} \psi = E_{red} \psi$$. Then $$H_{red} \psi$$ is also an eigenstate of $$M$$ and $$M \psi$$ is also an eigenstate of $$H_{red}$$, since

$$H_{red} M \psi = m H_{red} \psi = M H_{red} \psi = E_{red} M \psi$$

Obviously $$M \psi$$ is an eigenstate of $$H_{red}$$ in general, but the condition $$M \psi = m \psi$$ only really works for $$\psi = | 0, 0 \rangle$$ with $$m = 0$$ as we have seen previously, and $$| 0, 0 \rangle$$ is not considered to be an eigenstate of $$M$$ (or is it?). I'm sure that this is not the solution, and I'm really puzzled right now.

Question: Does this mean that there are no simultaneous eigenstates? If so, why does the exercise prompt me to consider eigenfunctions of $$H$$ in the form $$\psi(q) = \chi(q)\psi(q)$$? Can I construct eigenstates of $$M$$ somehow?

Source:

You should be able to verify that $$L_z$$ preserve $$n+m$$, which means that you can find the common eigenstates of $$L_z$$ and $$H_{red}$$ inside each subspace of constant $$N=n+m$$.
• I couldn't find the common eigenstates apparently, sorry. Let's suppose that $N = 1$, and that $| \psi \rangle = \alpha | 0, 1\rangle + \beta | 1, 0\rangle$ is an eigenstate. Then $L_z | \psi \rangle = \alpha | 1, 0 \rangle - \beta | 0, 1 \rangle = k (\alpha | 0, 1\rangle + \beta | 1, 0\rangle)$, but this is only possible if $\alpha = \beta = 0$. Basically, my issue with this problem is that $H_{red}$ and $L_z$ should share a common eigenbasis since they commute, but they don't seem to share one here. Commented Jan 31, 2022 at 20:41
• You should first construct the matrix representation of $L_z$ in the 2-dimensional space, find first the eigenvalues and then find the eigenvectors. Basically with $k=1$ and $\alpha=\beta=\frac{1}{\sqrt{2}}$ you got an eigenstate right there, so your conclusion $\alpha=\beta=0$ is demonstrably false. Commented Jan 31, 2022 at 21:17
• It works, thank you! My previous conclusion is indeed ill-considered: I should've written this properly as an eigenvalue problem. To be more precise, for $N=1$ eigenvectors are e.g. $\frac{\pm | 0, 1\rangle + i | 1, 0 \rangle}{\sqrt{2}}$. Commented Jan 31, 2022 at 22:33
• FYI you might care to define $L_z=M/2$ and then you (should?) have the usual setup where eigenvalues of $L_z$ differ by $1$ there are $N+1$ of them. Commented Jan 31, 2022 at 22:36