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Quantum Mechanics}Perturbation Theory

The energy for first-order perturbation theory of H=H_0+\Delta H (H_0 is the known Hamiltonian and \Delta H is the perturbed Hamiltonian) is given by E_1=\langle \psi_0 | \Delta H \psi \rangle_0, where the wave-functions are the unperturbed ones.

Thus, the problem amounts to calculating E_1=\langle n | V(a+a^\dag)^2 | n \rangle. This is just raising and lowering operator mechanics.

(a+a^\dag)^2=a^2+a^{\dag^2}+aa^\dag + a^\dag a. But, after bra-ketting, one finds that the expectation value of a^2 and a^{\dag^2} are 0, since n and n+2, n-2 are orthogonal. Thus, the problem becomes,

E_1=\langle n | (a a^\dag + a^\dag a) | n \rangle. Applying the given eigen-equations, one finds that E_1=\langle n | (a a^\dag + a^\dag a) | n \rangle = (2n+1)V. For n=2, one finds E_1=5V, as in choice (E).

(Note that: a^\dag a |n\rangle = a^\dag \sqrt{n} | n-1 \rangle = n | n \rangle and a a^\dag |n\rangle = a \sqrt{n+1} | n+1 \rangle = (n+1)|n\rangle and a^2 = a \sqrt{n}|n-1\rangle = \sqrt{n}\sqrt{n-1}|n-2\rangle.)

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