Discussion
A particle of mass m is in a one dimensional potential
\[ V(x) = \begin{cases} 0, & 0 \lt x \lt L \\ \infty, & \text{otherwise} \end{cases} \]
At some instant its wave function is given by
\[ \begin{aligned} \psi(x) &= \frac{1}{\sqrt{3}}\psi_{1}(x) \\ &\quad + i\sqrt{\frac{2}{3}}\psi_{2}(x), \end{aligned} \]
where \(\psi_{1}(x)\) and \(\psi_{2}(x)\) are the ground and the first excited states, respectively. Identify the correct statement.
Answer: A
The answer is \(\langle x \rangle = \frac{L}{2}\); \(\langle E \rangle = \frac{\hbar^{2}}{2m}\frac{3\pi^{2}}{L^{2}}\).
Energy: the states have \(E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}\), and the probabilities are \(\frac{1}{3}\) and \(\frac{2}{3}\):
\[ \begin{aligned} \langle E \rangle &= \tfrac{1}{3}E_{1} + \tfrac{2}{3}(4E_{1}) = 3E_{1} \\ &= \frac{\hbar^{2}}{2m}\frac{3\pi^{2}}{L^{2}} \end{aligned} \]
Position: each stationary state alone has \(\langle x \rangle = L/2\). The cross term is proportional to \(2\,\text{Re}(c_{1}^{*}c_{2})\int \psi_{1}x\psi_{2}\,dx\), and since \(c_{1}^{*}c_{2} = i\sqrt{2}/3\) is purely imaginary, it vanishes at this instant. So \(\langle x \rangle = L/2\).
The \(\frac{2L}{3}\) options mistake the probability weights for positions, and \(8\pi^{2}\) is \(2E_{2}\), double counting the excited state.
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