Let |m〉 and |n〉 denote the energy ei genstates of a one -dimensional simple
harmonic oscillator . The position and momentum operators are X̂ and P̂,
respectively. The matrix element 〈m|P̂X̂|n〉 is non-zero when
Options:
(A) m = n ± 2 only
(B) m = n or m = n ± 2
(C) m = n ± 3 only
(D) m = n ± 1 only
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.43)**
**1. Harmonic Oscillator Ladder Operators**:
- Position and momentum operators expressed in terms of raising ($a^\dagger$) and lowering ($a$) operators:
$\hat{X} = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)$
$\hat{P} = -i\sqrt{\frac{m\hbar\omega}{2}}(a - a^\dagger)$
**2. Step-by-Step Operator Algebra**:
- Form the operator product $\hat{P}\hat{X}$:
$\hat{P}\hat{X} = -i\frac{\hbar}{2}(a - a^\dagger)(a + a^\dagger) = -i\frac{\hbar}{2}(a^2 + a a^\dagger - a^\dagger a - (a^\dagger)^2)$
- Using the canonical bosonic commutation relation $[a, a^\dagger] = a a^\dagger - a^\dagger a = 1$:
$\hat{P}\hat{X} = -i\frac{\hbar}{2}\left(a^2 + 1 - (a^\dagger)^2\right)$
- Matrix element $\langle m | \hat{P}\hat{X} | n \rangle$:
$\langle m | \hat{P}\hat{X} | n \rangle = -i\frac{\hbar}{2}\Big[\langle m | a^2 | n \rangle + \langle m | 1 | n \rangle - \langle m | (a^\dagger)^2 | n \rangle\Big]$
- The identity term $\langle m | 1 | n \rangle = \delta_{m, n} \implies m = n$.
- The lowering term $\langle m | a^2 | n \rangle = \sqrt{n(n-1)}\delta_{m, n-2} \implies m = n - 2$.
- The raising term $\langle m | (a^\dagger)^2 | n \rangle = \sqrt{(n+1)(n+2)}\delta_{m, n+2} \implies m = n + 2$.
- Hence, the matrix element is non-zero exclusively when **$m = n$ or $m = n \pm 2$**.
**3. Official Verification**:
The verified official IIT answer is **Option (B)** (`m = n or m = n ± 2`).
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