First order perturbation correction $\Delta \varepsilon_n^{(1)}$ to energy level $\varepsilon_n$ of a simple harmonic oscillator due to the anharmonicity perturbation $\lambda x^3$ is given by
(a)$\Delta \varepsilon_n^{(1)}=\gamma$
(b)$\Delta \varepsilon_n^{(1)}=\gamma^2$
(c)$\Delta \varepsilon_n^{(1)}=\gamma^{-1}$
(d)$\Delta \varepsilon_n^{(1)}=0$
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
E_n⁽¹⁾ = ⟨n|x³|n⟩ = 0 because x³ is odd while |ψ_n|² is even for every oscillator state.