The question arises from an exercise but tackles deeper understanding of angular momentum operators.

Suppose we have a 2D harmonic oscillator and an infinite square well in the third dimension:

\begin{cases} \frac{1}{2}m\omega^2 (x^2+y^2), & 0 < z < a \\ \infty & \text{elsewhere} \end{cases}

Consider the degenerate subspace of energy $2\hbar\omega + \frac{\pi^2 \hbar^2}{2ma^2}$. There are two vectors in this subspace, $v_1 = |n_x=1, n_y=0, n_z=1\rangle$ and $v_2 = |n_x=0, n_y=1, n_z=1\rangle,$ assuming $ \frac{\pi^2 \hbar^2}{2ma^2} \neq \hbar \omega$.

We can easily find eigenstates of $L_z$ by expressing it in the form of ladder operators, $L_z = i \hbar (a_x a_y^{\dagger} - a_x^{\dagger}a_y)$, then acting with that on $v_1$ and $v_2$ and diagonalizing the resulting matrix. We find that the eingestates of $L_z$ are $\frac{1}{\sqrt{2}} (v_1 + i v_2)$ and $\frac{1}{\sqrt{2}} (v_1 - i v_2)$, with eigenvalues $\hbar$ and $-\hbar$. We can even explicitly check that by expressing them by the appropriate Hermite polynomials and evaluating $L_z = -i\hbar \frac{\partial}{\partial \phi}$.

The question is - can we find eigenstates of $L^2$ in this subspace? Intuition tells me that the infinite well in the $z$ direction doesn't allow it, but I can't exactly see why. Maybe it has to do with dimensions? The subspace has dimension 2, so $L^2$ has to be a square matrix. But we already found that the projections of $L$ on $z$ are $m=\pm1$, so $l$ would have to be at least $1$ and such an $L^2$ would have at least dimension 3.

In the case of a 3D harmonic oscillator we could find in the subspace $E = \frac{5}{2}\hbar\omega$ eigenstates of $L_z$ with $m=1,0,-1$. I know that they correspond to eigenstates of $L^2$: $|l=1, m=1\rangle$, $|l=1, m=0\rangle$ and $|l=1, m=-1\rangle$, but I'm also not sure how to prove $l=1$ in this case, besides explicitly writing down those states with Hermites and identifying spherical harmonics.

So generally I think I'm missing some intuition involving the relation between energy, $L_z$ and $L^2$ eigenstates.

  • $\begingroup$ $L_z$ is the only symmetry of your system: the full spherical symmetry 𝐿⃗, and hence $𝐿^2$ simply do not commute with the potential, so you could probably never consider them profitably. That is, the generators 𝐿𝑥, 𝐿𝑦 are badly broken in your system, much unlike in the 3D oscillator. The representation structure of SO(2) and SO(3) are very different, so it is not apparent what puzzles you... $\endgroup$ – Cosmas Zachos Jun 16 '19 at 20:29

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