2nd order pertubation theory for harmonic oscillator

I'm having some trouble calculating the 2nd order energy shift in a problem. I am given the pertubation:

$\hat{H}'=\alpha \hat{p}$,

where $\alpha$ is a constant, and $\hat{p}$ is given by:

$p=i\sqrt{\frac{\hbar m\omega }{2}}\left( {{a}_{+}}-{{a}_{-}} \right)$,

where ${a}_{+}$ and ${a}_{-}$ are the usual ladder operators.

Now, according to my book, the 2nd order energy shift is given by:

$E_{n}^{2}=\sum\limits_{m\ne n}{\frac{{{\left| \left\langle \psi _{m}^{0} \right|H'\left| \psi _{n}^{0} \right\rangle \right|}^{2}}}{E_{n}^{0}-E_{m}^{0}}}$

Now, what I have tried to do is to calculate the term inside the power of 2. And so far I have done this:

\begin{align} & E_{n}^{1}=\alpha i\sqrt{\frac{\hbar m\omega }{2}}\int{\psi _{m}^{*}\left( {{{\hat{a}}}_{+}}-{{{\hat{a}}}_{-}} \right)}\,{{\psi }_{n}}\,dx=\alpha i\sqrt{\frac{\hbar m\omega }{2}}\left( \int{\psi _{m}^{*}\,{{{\hat{a}}}_{+}}{{\psi }_{n}}\,dx-\int{\psi _{m}^{*}\,{{{\hat{a}}}_{-}}{{\psi }_{n}}\,dx}} \right) \\ & =\alpha i\sqrt{\frac{\hbar m\omega }{2}}\left( \sqrt{n+1}\int{\psi _{m}^{*}\,{{\psi }_{n+1}}\,dx-\sqrt{n}\int{\psi _{m}^{*}\,{{\psi }_{n-1}}\,dx}} \right) \end{align}

As you can see, I end up with the two integrals. But I don't know what to do next ? 'Cause if $m > n$, and only by 1, then the first integral will be 1, and the other will be zero. And if $n > m$, only by 1, then the second integral will be 1, and the first will be zero. Otherwise both will be zero. And it seems wrong to have to make two expressions for the energy shift for $n > m$ and $m > n$.

So am I on the right track, or doing it totally wrong ?

Regards

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Well, you end up with integrals but those are very, very easy to solve for the harmonic oscillator!

Since your problem is already formulated in terms of the raising and lowering operators $a_+$ and $a_-$. Recall that $$a_+ | n \rangle = \sqrt{n+1} | n+1 \rangle$$ $$a_- | n \rangle = \sqrt{n} | n - 1 \rangle$$ $$a_+ a_- | n \rangle = n | n \rangle$$

where $|n\rangle$ is short-hand for $|\psi_n^0 \rangle$.

These relations make it almost trivial to compute matrix elements involving eigenstates of the harmonic oscillator and those operators.

Just as an example, let's prove that all eigenstates have zero expectation value for $x$: We know $x$ is proportional to $a_+ + a_-$. Inserting that into the matrix element gives us $$\langle n | x | n \rangle \propto \langle n | a_+ | n \rangle + \langle n | a_- | n \rangle = \sqrt{n+1} \langle n | n+1\rangle + \sqrt{n} \langle n | n- 1 \rangle = 0$$ because eigenstates are orthogonal.

EDIT: Continuing with the derivation where you left off, you see that you get a non-zero contribution only if $m = n+1$ or $m = n-1$. So in the infinite sum over all states $m \not= n$, only two terms will contribute, making it possible to easily carry out that sum: Just add those two non-zero terms.

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But in my case I have n and m eigenstates. So in some cases, won't the inner product of the to states be one, if they end out with the same index, and for the most part be zero ? If what you have written is true, then what I wrote should also be zero, right ? But according to the problem, the answer is a non-zero answer. –  Denver Dang May 10 '13 at 12:02
Yes, the inner product will be finite if they end out with the same index after you applied the operators. For example $\langle 2 | a_+ | 1 \rangle = \sqrt{2} \langle 2 | 2 \rangle = \sqrt{2}$. –  Lagerbaer May 10 '13 at 14:10
But haven't I already done that in my equation above ? When I used the ladder operators I end out with two integrals where the first one is $\left\langle m | n+1 \right\rangle$ and the second one $\left\langle m | n-1 \right\rangle$ So there is no way, as far as I can see, that both of them can equal 1 at the same time ? If $n = m$ then it is 0 in both cases. But if $n > m$ by 1 only, then the second integral will be 1, and the first will be 0. And if $m > n$ by 1 only, then the first integral will be 1, and the second will be 0. So from this, won't I end up with two expressions? –  Denver Dang May 10 '13 at 14:38
Why would it bother you that they can't both be non-zero at the same time? To compute the energy shift of level $n$, you'd sum over all states $m \not= n$ anyway, but instead of an infinite sum, you only get contributions when $m = n-1$ and when $m = n+1$, i.e. you get two terms, that you add, to get the final result. –  Lagerbaer May 10 '13 at 15:06
Ok, I see now :) Thank you very much. –  Denver Dang May 10 '13 at 21:27