This was a question on my exam and I don't know how to solve it.

Use the variational principle to prove that the first order perturbation theory always overestimates the energy of the fundamental state. Also prove that the second order term is always negative. Any Ideas?

Assume that the perturbation operator is hermitian.


closed as too localized by David Z Jan 23 '13 at 11:11

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    $\begingroup$ Hi Presbitero - questions like this where you just ask for a solution aren't appropriate on this site. This is a place for conceptual physics questions, so if you'd like to edit your question to focus on the concept that is giving you trouble, someone can reopen it. See our homework policy for more information. $\endgroup$ – David Z Jan 23 '13 at 11:13
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    $\begingroup$ The reason this is not a conceptual physics question is that you're not asking about a concept, you're asking us to solve a problem for you. It has nothing to do with the fact that the problem came from an exam. You can find some examples of good (conceptual) questions in the homework policy I linked to. $\endgroup$ – David Z Jan 23 '13 at 12:02
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    $\begingroup$ First order perturbation theory result $E_0^{(1)} = E_0^{(0)} +V_{00}$ overestimates the value of $E_0$ because the second-order correction is negative (see the corresponding formula). However, it is implied that the perturbation operator $\hat{V}$ is Hermitian. If it is not Hermitian, the second order correction can be positive. $\endgroup$ – Vladimir Kalitvianski Jan 23 '13 at 12:21
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    $\begingroup$ @DavidZaslavsky I'm not so sure I agree it's too localized. First order perturbation theory is going to miss the true energy, but it's mildly interesting that it definitively overshoots not undershoots, and that the second order term goes the other way (I haven't checked that it does!) $\endgroup$ – twistor59 Jan 23 '13 at 12:24
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    $\begingroup$ @twistor59 and others: regardless of the fact that it is an interesting question, it clearly goes against our homework policy, as it is just asking us to solve a problem without narrowing it down to a conceptual issue. So I really don't understand the argument that this is not too localized. $\endgroup$ – David Z Jan 23 '13 at 16:35