Quantum Mechanics: Angular Momentum Eigenvalues

Problem Setup

We are given that a particle's state function is an eigenfunction of the angular momentum squared operator, $\\hat{L}^2$, with a specific eigenvalue. We need to find the possible eigenvalues of the angular momentum z-component squared operator, $\\hat{L}_z^2$, for this same state.

The given eigenvalue for $\\hat{L}^2$ is $30\hbar^2$.

Solution Steps

1. Relate the given $\\hat{L}^2$ eigenvalue to the quantum number $l$. The eigenvalue of $\\hat{L}^2$ is given by the formula $l(l+1)\hbar^2$, where $l$ is the angular momentum quantum number ($l \ge 0$).

   Set the given eigenvalue equal to the formula:

   $ l(l+1)\hbar^2 = 30\hbar^2 $

   Solve for $l$:

   $ l(l+1) = 30 $

   By inspection or solving the quadratic equation $l^2 + l - 30 = 0$, we find the positive integer solution $l=5$. ($ (l+6)(l-5) = 0 \implies l=5 $ since $l \ge 0$).

2. Determine the possible values of the magnetic quantum number, $m$. For a given $l$, the possible values of $m$ (the magnetic quantum number) range from $-l$ to $+l$ in integer steps.

   Since $l=5$, the possible values for $m$ are:

   $ m = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 $

3. Find the eigenvalues of $\\hat{L}_z^2$. The eigenvalue of $\\hat{L}_z$ is $m\hbar$. Therefore, the eigenvalue of $\\hat{L}_z^2$ is $(m\hbar)^2 = m^2\hbar^2$.

   Calculate $m^2\hbar^2$ for each possible value of $m$:

   • For $m=0$, eigenvalue $= 0^2\hbar^2 = 0$.
   • For $m=\pm 1$, eigenvalue $= (\pm 1)^2\hbar^2 = 1\hbar^2$.
   • For $m=\pm 2$, eigenvalue $= (\pm 2)^2\hbar^2 = 4\hbar^2$.
   • For $m=\pm 3$, eigenvalue $= (\pm 3)^2\hbar^2 = 9\hbar^2$.
   • For $m=\pm 4$, eigenvalue $= (\pm 4)^2\hbar^2 = 16\hbar^2$.
   • For $m=\pm 5$, eigenvalue $= (\pm 5)^2\hbar^2 = 25\hbar^2$.

4. Identify the possible eigenvalues from the given options. The possible eigenvalues for $\\hat{L}_z^2$ are $0, \hbar^2, 4\hbar^2, 9\hbar^2, 16\hbar^2, 25\hbar^2$. Comparing these with the options provided:

   • $10\hbar^2$: Not possible, as $m^2=10$ has no integer solution for $m$.
   • $16\hbar^2$: Possible (when $m=\pm 4$).
   • $25\hbar^2$: Possible (when $m=\pm 5$).
   • $0$: Possible (when $m=0$).

Therefore, the possible eigenvalues of $\\hat{L}_z^2$ are $0$, $16\hbar^2$, and $25\hbar^2$.