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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!

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  • $\begingroup$ Nope! $\delta E_n$ and $E^{(1)}_n$ is the same thing in the leading approximation. Both of them are "infinitesimal" relatively to $E_n$. $\endgroup$ – Luboš Motl Jan 7 '13 at 16:57
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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 $(\delta E_n)^{(0)}=E_n^{(1)}$.

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

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