Discussion
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 :
\[ \begin{aligned} Y_{0}^{0} &= \sqrt{\frac{1}{4\pi}} , \\ Y_{1}^{0} &= \sqrt{\frac{3}{4\pi}}\cos\theta , \\ Y_{1}^{\pm 1} &= \mp\sqrt{\frac{3}{8\pi}}\sin\theta\, e^{\pm i\phi} \end{aligned} \]
Answer: C
The eigenvalues are \(L^{2} = 2\hbar^{2}\) and \(L_{z} = 0\).
Write \(z = r\cos\theta\), so
\[ \begin{aligned} \psi &= N r\cos\theta\, e^{-\alpha r^{2}} \\ &\propto r e^{-\alpha r^{2}}\, Y_{1}^{0}(\theta) \end{aligned} \]
The angular part is exactly \(Y_{1}^{0}\), so the state has \(\ell = 1\) and \(m = 0\). The eigenvalues are \(\ell(\ell + 1)\hbar^{2} = 2\hbar^{2}\) and \(m\hbar = 0\). The radial factor \(re^{-\alpha r^{2}}\) does not affect the angular momentum.
0 and 0 would need the spherically symmetric \(Y_{0}^{0}\) (no factor of \(z\)). The options with \(\hbar^{2}\) misuse \(\ell^{2}\hbar^{2}\) for \(L^{2}\), and \(\pm\hbar\) for \(L_{z}\) would need the \(x \pm iy\) combinations that go with \(Y_{1}^{\pm 1}\).
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