1
$\begingroup$

A particle is in the one dimensional harmonic potential $V(x)=\frac{1}{2}m\omega^2x^2$ with a small perturbation $V'$. I want to calculate the first- and second order correction to the ground state energy through the matrix elements: \begin{equation} \langle n'|V'|n\rangle =\frac{\hbar\omega\alpha}{4}[\sqrt{(n+1)(n+2)}\delta_{n',n+2}+(1+2n)\delta_{n',n}+\sqrt{(n)(n-1)}\delta_{n',n-2}] \end{equation} From this I got that $V'(x)=\frac{1}{2}m\omega^2\alpha x^2$ and I calculated the first order correction to the ground state energy as $E^1_0=\frac{1}{4}\alpha\omega\hbar$. I would like to calculate the second order correction to the ground state energy through the matrix elements, but I do not know how to tackle that part, and I am hoping that somebody can give me a hint.

$\endgroup$
0
$\begingroup$

We don't have to worry about degeneracy here, so your first and second order energy corrections follow from

$$E_n^{(1)} = \langle n^{(0)}|V|n^{(0)}\rangle \\ \\ E_n^{(2)} = \sum_{i\neq n}\frac{|\langle n^{(0)}|V|i^{(0)}\rangle|^2}{E_n^{(0)}-E_i^{(0)}}.$$

Now use your rewritten form of $V$, which you deduced from raising and lowering operators, to evaluate $|\langle n^{(0)}|V|i^{(0)}\rangle|^2$. Clear from your Kronecker deltas, this is only nonzero for the $i$'s that fall within $n\pm2$, but not necessarily all of them (can you see why?).

$\endgroup$
5
  • $\begingroup$ Thanks for your answer. That puts the limit on the sum, of course. However, I am not quite sure if I am allowed/supposed to choose $n=0$ for the ground state. In that case, I get that $\sum_{i\neq n}|\langle n^{(0)}|V|i^{(0)}\rangle|^2=\frac{1}{2}\hbar\omega\alpha$, since i cannot be smaller than the ground state. $\endgroup$
    – user226866
    Mar 30 '19 at 19:38
  • $\begingroup$ Restriction on the sum is one, so there is no correction from the ground state on the ground state, but I was also getting at there is no correction from neighboring states, only from those that are $2$ energy levels away. So, for your ground state, only a correction from $|2^{(0)}\rangle$, which it looks like you're just about there, just need to check the factors: $(\frac{\sqrt{2}\hbar\omega\alpha}{4})^2/(-2\hbar\omega)$ $\endgroup$
    – dsm
    Mar 30 '19 at 19:52
  • $\begingroup$ Great, then I understand it. Thanks for your time! Edit: oh yes, I have to square the whole term. I forgot to square $\frac{\hbar \omega\alpha}{4}$. Again, thank you. $\endgroup$
    – user226866
    Mar 30 '19 at 20:03
  • $\begingroup$ Sure thing, and welcome to the physics SE :). If you feel someone has sufficiently answered your question, you can click the check mark to indicate that it's complete. Cheers $\endgroup$
    – dsm
    Mar 30 '19 at 20:07
  • 1
    $\begingroup$ Thank you. Just did. :) $\endgroup$
    – user226866
    Mar 30 '19 at 20:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy