For a hermitian operator A, which does NOT commute with the Hamiltonian H, let $\psi_{1}$ be an eigenfunction of A and $\psi_{2}$ be an eigenfunction of H. The correct statement regarding the average of A and value of the commutator of A with H ([A,H]) is
(a)both $\psi_{1}[\mathrm{A}, \mathrm{H}] \psi_{1}$ and $\psi_{2}[\mathrm{A}, \mathrm{H}] \psi_{2}$ are non-zero
(b)only $\psi_{1}[\mathrm{A}, \mathrm{H}] \psi_{1}$ is zero , but $\psi_{2}[\mathrm{A}, \mathrm{H}] \psi_{2}$ is non-zero
(c)only $\psi_{2}[\mathrm{A}, \mathrm{H}] \psi_{2}$ is zero , but $\psi_{1}[\mathrm{A}, \mathrm{H}] \psi_{1}$ is non-zero
(d)both $\psi_{1}[\mathrm{A}, \mathrm{H}] \psi_{1}$ and $\psi_{2}[\mathrm{A}, \mathrm{H}] \psi_{2}$ are zero
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
For an eigenfunction of A: ⟨ψ₁|AH − HA|ψ₁⟩ = a⟨H⟩ − a⟨H⟩ = 0; for an eigenfunction of H likewise ⟨ψ₂|[A,H]|ψ₂⟩ = E⟨A⟩ − E⟨A⟩ = 0. Both vanish even though the operators do not commute.