Linear perturbation to harmonic oscillator

[vbrasic]

Homework Statement

Find the first-order corrections to energy and the wavefunction, for a 1D harmonic oscillator which is linearly perturbed by $H'=ax$.

Homework Equations

First-order correction to the energy is given by, $E^{(1)}=\langle n|H'|n\rangle$, while first-order correction to the wave-function is, $$|n^{(1)}\rangle=\Sigma_{m\neq n}\frac{\langle m|H'|n\rangle}{E_n-E_m}|m\rangle.$$

The Attempt at a Solution

The energy correction is just $0$ by orthogonality arguments. For the corrections to the wavefunction, I have, $$|n^{(1)}\rangle=\frac{a\sqrt{\frac{\hbar}{2m\omega}}}{\hbar\omega}\Sigma_{m\neq n}\frac{\langle m|a+a^+|n\rangle}{n-m}|m\rangle.$$ The first term within the summation is only exists for $m=n-1$, such that the term becomes, $\sqrt{n}|n-1\rangle$. Similarly, the second term in the summation only exists for $m=n+1$, such that the term becomes,$-\sqrt{n+1}|n+1\rangle$. Hence, the first order correction should be, $$\frac{a\sqrt{\frac{\hbar}{2m\omega}}}{\hbar\omega}(\sqrt{n}|n-1\rangle-\sqrt{n+1}|n+1\rangle).$$ I am not sure however, as I'm still getting my feet wet with perturbation theory and would like to know if I'm on the correct track.

 

[blue_leaf77]

Looks fine, except that the last expression is applicable only for $n>0$. What about the one for $n=0$?

 

[vbrasic]

Looks fine, except that the last expression is applicable only for $n>0$. What about the one for $n=0$?

For, $n=0$, I'm thinking that the term involving the $|n-1\rangle$ would just go to $0$ by definition of the raising/lowering operators. Would that be correct?

 

[blue_leaf77]

For, $n=0$, I'm thinking that the term involving the $|n-1\rangle$ would just go to $0$ by definition of the raising/lowering operators. Would that be correct?

Yes.