Discussion
The electron in a hydrogen atom is in a superposition state described by the wave function
\(\psi(\vec{r}) = \frac{1}{6}[4\psi_{100}(\vec{r}) - 2\psi_{211}(\vec{r})\) \(+ \sqrt{6}\psi_{210}(\vec{r}) - \sqrt{10}\psi_{21-1}(\vec{r})]\),
where \(\psi_{nem}(\vec{r})\) is a normalized eigenstate. The expectation value of \(L_{z}\) is :
Answer: A
The expectation value is \(-\hbar/6\).
Each \(\psi_{nlm}\) is an eigenstate of \(L_{z}\) with eigenvalue \(m\hbar\), and the probability of each is the squared coefficient. The coefficients are 4/6, -2/6, \(\sqrt{6}/6\) and \(-\sqrt{10}/6\), giving probabilities 16/36, 4/36, 6/36 and 10/36, which add to 1, so the state is normalized.
\[ \begin{aligned} \langle L_{z}\rangle &= \hbar\Big[0\cdot\tfrac{16}{36} + 1\cdot\tfrac{4}{36} \\ &\quad + 0\cdot\tfrac{6}{36} - 1\cdot\tfrac{10}{36}\Big] \\ &= -\frac{6\hbar}{36} = -\frac{\hbar}{6} \end{aligned} \]
Only the m = 1 and m = -1 components contribute. The value \(+\hbar/6\) has the sign reversed, and \(\pm 6\hbar\) forget to divide by the normalization 36.
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