GATE 2025 PH – Question 43

Quantum Mechanics · one dimensional potentials: step potential, finite rectangular well, tunnelling from a potential barrier, particle in 1,2,3-dimensional box, particle in single and double delta function potentials, 1,2,3 dimensional harmonic oscillator: concept of degeneracy · 2 marks · Multiple choice

Let $|m\rangle,|n\rangle$ denote energy eigenstates of a one-dimensional simple harmonic oscillator, with position and momentum operators $\hat X,\hat P$. The matrix element $\langle m|\hat P\hat X|n\rangle$ is nonzero when

$m=n\pm2$ only
$m=n$ or $m=n\pm2$
$m=n\pm3$ only
$m=n\pm1$ only

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Correct answer: (B) $m=n$ or $m=n\pm2$

Explanation

Write the position and momentum operators in terms of the ladder operators $a$ and $a^\dagger$:
$$\hat X=\sqrt{\frac{\hbar}{2m\omega}}\,(a+a^\dagger),\qquad\hat P=i\sqrt{\frac{m\hbar\omega}{2}}\,(a^\dagger-a).$$

**The product:**
$$\hat P\hat X=\frac{i\hbar}{2}\,(a^\dagger-a)(a+a^\dagger)=\frac{i\hbar}{2}\left(a^\dagger a+a^{\dagger2}-a^2-aa^\dagger\right).$$

$a^{\dagger2}$ changes $n\to n+2$.
$a^2$ changes $n\to n-2$.
$a^\dagger a-aa^\dagger=-1$, a constant, which gives a nonzero **diagonal** term ($m=n$).

So $\langle m|\hat P\hat X|n\rangle\neq0$ for $m=n$ and $m=n\pm2$ (option B). This is consistent with $\hat P\hat X=\tfrac12(\hat P\hat X+\hat X\hat P)-\tfrac{i\hbar}{2}$, where the symmetric part connects $n\to n\pm2$ and the commutator is a constant.