Question

The normalized wave functions \(\psi_1\) and \(\psi_2\) correspond to the ground state and the first excited state of a particle in a potential. The operator \(\hat{A}\) acts on the wave functions as \(\hat{A}\psi_1 = \psi_2\) and \(\hat{A}\psi_2 = \psi_1\). The expectation value of the operator \(\hat{A}\) for the state \(\psi = (3\psi_1 + 4\psi_2)/5\) is:

• A. 0.96
• B. −0.32
• C. 0.75
• D. 0

Hint:

Apply \(\hat{A}\) (which swaps the states) to \(\psi\), then take the overlap with \(\psi\); the cross terms give \(2\cdot 3\cdot 4/25\).

Answer 1

The Correct Option is A

Step 1: Write the state as a coefficient vector in the $\{\psi_1,\psi_2\}$ basis, $c=\tfrac{1}{5}(3,4)^{T}$, which is already normalized since $\tfrac{1}{25}(9+16)=1$.

Step 2: In this basis the operator that swaps the two states has the matrix
\[\hat{A} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\]
because it sends $\psi_1\to\psi_2$ and $\psi_2\to\psi_1$.

Step 3: Compute $\hat{A}c=\tfrac{1}{5}(4,3)^{T}$.

Step 4: Take the dot product with $c$:
\[\langle A\rangle = c^{T}\hat{A}c = \tfrac{1}{25}\big(3\cdot 4 + 4\cdot 3\big)=\tfrac{24}{25}.\]
Step 5: Convert to decimal form.
\[\boxed{\langle A\rangle = 0.96}\]