Question

A particle of mass $m$ is in a one-dimensional potential $V(x)=\begin{cases}0,\; 0<x<L \\ \infty,\ \text{otherwise}\end{cases}$. 

At some instant its wave function is given by $\psi(x)=\dfrac{1}{\sqrt{3}}\psi_1(x) + i\sqrt{\dfrac{2}{3}}\,\psi_2(x)$, where $\psi_1,\psi_2$ are the ground and first excited states. Identify the correct statement. 
 

• A. $\langle x\rangle = \dfrac{L}{2}, \langle E\rangle = \dfrac{h^2}{2m}\dfrac{3\pi^2}{L^2}$
• B. $\langle x\rangle = \dfrac{2L}{3}, \langle E\rangle = \dfrac{h^2}{2m}\dfrac{\pi^2}{L^2}$
• C. $\langle x\rangle = \dfrac{L}{2}, \langle E\rangle = \dfrac{h^2}{2m}\dfrac{8\pi^2}{L^2}$
• D. $\langle x\rangle = \dfrac{2L}{3}, \langle E\rangle = \dfrac{h^2}{2m}\dfrac{4\pi^2}{3L^2}$

Hint:

For superpositions of stationary states, energy expectation is always a probability-weighted average of the individual energies.

Answer 1

The Correct Option is A

Step 1: Expectation value of position.
Both $\psi_1$ and $\psi_2$ are symmetric about $x=L/2$. Thus $\langle x \rangle = L/2$, independent of coefficients.

Step 2: Compute expectation value of energy.
Energies of infinite well: $\displaystyle E_n = \dfrac{n^2\pi^2\hbar^2}{2mL^2}.$ Given coefficients: $|c_1|^2 = 1/3, |c_2|^2 = 2/3.$

Step 3: Weighted energy.
$\displaystyle \langle E\rangle = |c_1|^2 E_1 + |c_2|^2 E_2 = \frac{1}{3}E_1 + \frac{2}{3}E_2.$
$E_1 = \dfrac{\pi^2\hbar^2}{2mL^2}, E_2 = \dfrac{4\pi^2\hbar^2}{2mL^2}.$
Thus $\displaystyle \langle E\rangle = \frac{1}{3}\left(\frac{\pi^2\hbar^2}{2mL^2}\right) + \frac{2}{3}\left(\frac{4\pi^2\hbar^2}{2mL^2}\right) = \frac{3\pi^2\hbar^2}{2mL^2}.$

Step 4: Conclusion.
Correct values: $\langle x\rangle = L/2$ and $\langle E\rangle = \dfrac{3\pi^2\hbar^2}{2mL^2}$, matching option (A).