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For my second-order energy correction for a harmonic oscillator in an electric field I have the following:

$$q^2\varepsilon^2\sum_{m\neq n}\frac{|\langle m|x|n\rangle|^2}{E^{(0)}_n-E^{(0)}_m}+\text{ }...$$

  1. My first question is what values of $m$ do we sum over?

  2. Secondly, my textbook says we neglect higher-power terms, does this mean higher values of $m$ or what?

If I have left any information out (which is very likely) just say.

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    The sum over m is over all states n different from m. Higher power terms are higher order in q. – Urgje Aug 04 '14 at 08:00
  • Where do you get terms in higher order q from? –  Aug 04 '14 at 08:07
  • From higher order corrections, from the dots, right? – Luboš Motl Aug 04 '14 at 08:58
  • Oh ok that explains a lot! I am still a bit confused about what values of m we go sum over though please could someone explain this clearer, thanks again –  Aug 04 '14 at 09:03
  • @LubošMotl By the way have you had a chance to have a look at Ben's yesterday post: http://physics.stackexchange.com/q/129324/ it's be interesting to have your view/approach on it. – Ellie Aug 04 '14 at 09:17
  • Hi Phonon, see also e.g. http://www.mathpages.com/home/kmath242/kmath242.htm - the difference between odd and even dimensions may be seen in the Taylor expansions. Effectively, one needs things like $(d/2)!$ in the denominator of the Taylor coefficients, by a recursive relationship, and those behave differently for even and odd $d$. There are other ways to see the difference but the difference is really qualitative. – Luboš Motl Aug 04 '14 at 12:47
  • @LubošMotl alright, very interesting, this is why I asked you, you always have these neat mathematical insights! I will add a new comment to Ben's post regarding the physical interpretation, feel free to join in on the discussion if you have time, it'd be great. – Ellie Aug 04 '14 at 13:39
  • BTW, Phonon, there are huge differences between spinors in even and odd dimensions. In fact, the periodicity of the qualitative behavior of spinors isn't just 2, it is 8. In odd dimensions, there is only one type of spinors. In even dimensions, there are two inequivalent types of spinors. These differences may probably be related to the differences in the wave equation, too. A key difference is that the parity, sign flip of all the coordinates, is a "mirroring operation" (detrminant minus one) in odd dims and a rotation (not mirroring) in even dim. – Luboš Motl Aug 04 '14 at 13:59

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No it doesn't mean higher values of m, as m here represents the set of orthonormal eigenstates of the system under study. In order to better understand where and how one arrives at 1st, 2nd, 3rd, ... order corrections, you need to look at the derivation of time independent perturbation theory that you find in most QM textbooks, there's also a rather detailed derivation on wikipedia.

An overview:

When you write out the time-independent Schrödinger equation for the perturbed system: $$(H_0 + \lambda V)\left|n\right > = E_n \left|n\right > $$ Where $H_0$ is the Hamiltonian of the unperturbed system, $\left|n\right >$ are the eigenstates of the perturbed system, $\lambda$ is a perturbation parameter describing its strength, $E_n$ energy eigenvalues, and $V$ can be an external field seen as perturbative. Now for small perturbations, i.e. small $\lambda$, we can expand the above equation as a power series of $\lambda$: $$E_n = E_n^{0} +\lambda E_n^{1}+\lambda^2 E_n^{2}+\lambda^3...$$ and $$\left|n\right >=\left|n^0\right >+\lambda\left|n^1\right >+\lambda^2\left|n^2\right >+...$$ Substituting the expanded version into Schrödinger's equation again, you get a system of equation with as many equations as the power of $\lambda$ you intend to keep, $\lambda^0,\lambda^1,\lambda^2$..., solving for each one of them brings you to:

First order correction (expectation value of the perturbation Hamiltonian taken in the unperturbed state): $$E_n^1=\left<n^0 \right|V\left|n^0\right >$$ Second order: $$E_n^2= \left<n^0 \right|V\left|n^1\right > = \sum_{m\ne n}\frac{\left|\left<m^0 \right|V\left|n^0\right >\right|^2}{E_n^0-E_m^0}$$ Third order: $$E_n^3= \left<n^0 \right|V\left|n^2\right >=... $$

All this only valid for non-degenerate states, as I didn't not take into account degeneracy.

EDIT: Concerning your ''m'' confusion: Example of m: Take the hydrogen atom, first eigenstate would be $(n,l,m_l)=(1,0,0)$, second state $(2,0,0)$ and so on... (remember the wave-functions...)

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Ellie
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  • Hydrogen atom is degenerate system, so for the energy we must use the secular equation. – Andrew McAddams Aug 04 '14 at 09:32
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    Are there infinity many 'm' values we sum over then? either way why do can we approximate it by just looking at the m=n-1 and m=n+1 terms?? thanks –  Aug 04 '14 at 09:34
  • @Joseph No, it varies from case to case, sometimes because of the parity of the perturbation, only odd or even terms survive, like when we put a delta-function bump in the center of the infinite square well: $V'(x)=\alpha \delta(x-a)$, then only odd terms m=3,5,7,.. survive as the perturbation of the even terms is canceled out in the averaging. Other times you may have to solve the integral directly (expectation value of the perturbation), or if possible you use creation, annihilation operators to avoid the integral...so it's not like you sit and sum over each term one by one. – Ellie Aug 04 '14 at 10:05
  • @AndrewMcAddams should have just used the harmonic oscillator example or the square well with bump...was only trying to clarify for Joseph what the m's are. – Ellie Aug 04 '14 at 10:08