Discussion
The wavefunction of a particle in an infinite one-dimensional potential well at time \(t\) is
\[ \begin{aligned} \Psi(x,t) &= \sqrt{\tfrac{2}{3}}\,e^{-iE_{1}t/\hbar}\psi_{1}(x) \\ &+ \tfrac{1}{\sqrt{6}}\,e^{i\pi/6}e^{-iE_{2}t/\hbar}\psi_{2}(x) \\ &+ \tfrac{1}{\sqrt{6}}\,e^{i\pi/4}e^{-iE_{3}t/\hbar}\psi_{3}(x) \end{aligned} \]
where \(\psi_{1}\), \(\psi_{2}\) and \(\psi_{3}\) are the normalized ground state, the normalized first excited state and the normalized second excited state, respectively. \(E_{1}\), \(E_{2}\) and \(E_{3}\) are the eigen-energies corresponding to \(\psi_{1}\), \(\psi_{2}\) and \(\psi_{3}\), respectively. The expectation value of energy of the particle in state \(\Psi(x,t)\) is
Answer: A
The expectation value of energy is \(\dfrac{17}{6}E_{1}\).
For a stationary-state expansion, ⟨E⟩ = Σ |cn|2En; the phase factors drop out of the squared moduli. For an infinite well En = n2E1, so E2 = 4E1 and E3 = 9E1:
\[ \begin{aligned} \langle E \rangle &= \tfrac{2}{3}E_{1} + \tfrac{1}{6}(4E_{1}) + \tfrac{1}{6}(9E_{1}) \\ &= \tfrac{4 + 4 + 9}{6}E_{1} = \tfrac{17}{6}E_{1} \end{aligned} \]
The probabilities 2/3 + 1/6 + 1/6 add to 1, so the state is normalized. 2/3 E1 keeps only the ground-state term, 3/2 E1 treats the levels as En = nE1, and 14E1 adds the energies without weighting them.
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