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 $\epsilon^{12} = 1$.
- Calculate the Hamiltonian of the system and quantize the system by imposing the
canonical commutation relations. Solve the Heisenberg equations of the quantum system. - 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:
Best Answer
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 $N=n+m$.