Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi guys I have a quick question about perturbation theory in quantum mechanics, particularly about energy shifts.

We write: $E_n = E_n^{(0)} + \delta E_n$ where $E_n^{(0)}$ is the unperturbed hamiltonian. If I now have to calculate the first order energy i.e. $E_n^ {(1)}$ I have to calculate $\delta E_n$ to first order, which is (from writing down TISE for perturbed hamiltonian and keeping all the terms with only one $\delta$ in it) $\delta E_n = \int \psi_n^* \delta H \psi_n dx$

Is it correct to say now that $E_n^{(1)} = E_n - \delta E_n$ ? I know this is probably a really basic question but I'm just a little confused about the whole thing.

I hope you can clear things up for me!

Thx in advance!

share|cite|improve this question
Nope! $\delta E_n$ and $E^{(1)}_n$ is the same thing in the leading approximation. Both of them are "infinitesimal" relatively to $E_n$. – Luboš Motl Jan 7 '13 at 16:57

No, it is better to write: $$E_n=E_n^{(0)}+E_n^{(1)}+E_n^{(2)}...$$

Then $$E_n^{(1)}=\int \psi_n^{*(0)} \delta H \psi_n^{0)}dx.$$

Your $\delta E_n$ depends on how you define it. You may want to define it as $\delta E_n = E_n - E_n^{(0)}\;$ and $(\delta E_n)^{(0)}=E_n^{(1)}$.

share|cite|improve this answer
Aaaah of course, $\delta E_n$ depends on which order terms we include. Thanks for clearing this up for me:) – user17574 Jan 7 '13 at 17:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.