GATE 2026 PH – Question 26
If the perturbation $V=\lambda x^3$ is added to the Hamiltonian of a one-dimensional harmonic oscillator, the matrix element $\langle m|V|0\rangle$ is/are non-zero for which of the following states? Here, the eigenstates of the harmonic oscillator are denoted by $|n\rangle$.
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Correct answer: (A) $|m=3\rangle$; (B) $|m=1\rangle$
Explanation
The position operator is $\hat x=\sqrt{\dfrac{\hbar}{2m\omega}}\,(a+a^\dagger)$. The perturbation $V=\lambda x^3$ contains $(a+a^\dagger)^3$.
**Acting on the ground state** $|0\rangle$, with $a|0\rangle=0$ and $a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$:
- $(a+a^\dagger)|0\rangle=|1\rangle$,
- $(a+a^\dagger)^2|0\rangle=(a+a^\dagger)|1\rangle=|0\rangle+\sqrt2|2\rangle$,
- $(a+a^\dagger)^3|0\rangle=(a+a^\dagger)\left(|0\rangle+\sqrt2|2\rangle\right)=|1\rangle+\sqrt2\left(\sqrt2|1\rangle+\sqrt3|3\rangle\right)=3|1\rangle+\sqrt6|3\rangle$.
So $x^3|0\rangle$ has components only on $|1\rangle$ and $|3\rangle$, and the matrix elements $\langle m|V|0\rangle$ are non-zero only for $m=1$ and $m=3$.
Answer **A and B**.