Question

A particle has wavefunction \[ \psi(x, y, z) = N z e^{-\alpha(x^2 + y^2 + z^2)}, \] where \(N\) is a normalization constant and \(\alpha\) is a positive constant. In this state, which one of the following options represents the eigenvalues of \(L^2\) and \(L_z\) respectively?
Some values of \(Y_\ell^m\) are: \[ Y_0^0 = \frac{1}{\sqrt{4\pi}}, \ Y_1^0 = \frac{\sqrt{3}}{\sqrt{4\pi}} \cos \theta, \ Y_1^{\pm 1} = \mp \frac{\sqrt{3}}{8\pi} \sin \theta e^{\pm i \phi} \]

• A. 0 and 0
• B. \(\hbar^2\) and \(-\hbar\)
• C. \(2\hbar^2\) and \(\hbar\)
• D. \(\hbar^2\) and \(\hbar\)

Hint:

For spherical harmonics, the eigenvalues of \(L^2\) are given by \(\ell(\ell + 1) \hbar^2\), and the eigenvalues of \(L_z\) are \(m\hbar\), where \(\ell\) and \(m\) are the angular momentum quantum numbers.

Answer 1

The Correct Option is C

- The wavefunction is a function of \(z\), which means the system has spherical symmetry. The angular momentum eigenvalues are given by the quantum numbers associated with the spherical harmonics \(Y_\ell^m\). In this case, the eigenvalues for \(L^2\) and \(L_z\) are determined by the \(Y_1^0\) state, which corresponds to \(\ell = 1\) and \(m = 0\). Thus, the eigenvalue of \(L^2\) is \(\ell(\ell+1)\hbar^2 = 2\hbar^2\) and the eigenvalue of \(L_z\) is \(m\hbar = \hbar\).
- Hence, the correct answer is (C) \(2\hbar^2\) and \(\hbar\).