Talk:Perturbation theory (quantum mechanics)

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Should a link from the deep theory of quantum mechanics not go directly to the subpage for adiabatically specific to quantum theory rather than the disambiguation page?

http://en.wikipedia.org/wiki/Adiabatic

1 Cancellation of Terms
2 Nonperturbative?
3 Intermediate normalization
4 Lambda dropped

[edit] Cancellation of Terms

The article claims:

The first-order equation is

$H_0 |n^{(1)}\rang + V |n^{(0)}\rang = E_n^{(0)} |n^{(1)}\rang + E_n^{(1)} |n^{(0)}\rang$

Multiply through by <n⁽⁰⁾|. The first term on the left-hand side cancels with the first term on the right-hand side. (Recall, the unperturbed Hamiltonian is hermitian). This leads to the first-order energy shift:

(emphasis added) After multiplying through by <n⁽⁰⁾|, it looks like the first term on the left-hand side goes to zero on it's own (H_0 pops out of the bra-ket and turns into an E_0, leaving you with <n⁽⁰⁾| n⁽¹⁾> which is zero (orthogonality). (As opposed to canceling with the term on the right-hand side.) I believe the statement I've italicized is incorrect.--GameGod (talk) 18:22, 7 December 2007 (UTC)

[edit] Nonperturbative?

Could someone highlight Nonperturbative and explain it more explicity ... the wlnk comes here ... but then it's not highlighted nor explained well ... thanks. J. D. Redding 16:24, 10 March 2006 (UTC)

[edit] Intermediate normalization

I get so tired of people getting their fingers into articles, while not knowing what they are doing. Now somebody removed the very well-known intermediate normalization condition and made an error that has the consequence that the first-order correction to the wave function is zero. Namely, (s)he writes (through first order)

$1=\langle n | n \rangle = \langle n^{(0)} + \lambda n^{(1)} | n^{(0)} + \lambda n^{(1)} \rangle = \langle n^{(0)} | n^{(0)} \rangle + \lambda \langle n^{(0)} | n^{(1)} \rangle + \lambda \langle n^{(1)} | n^{(0)} \rangle + \lambda^2 \langle n^{(1)} | n^{(1)} \rangle$

$= 1 + 0 + 0 + \lambda^2 \langle n^{(1)} | n^{(1)} \rangle$

so that, because a Hilbert space inner product is positive definite,

$\langle n^{(1)} | n^{(1)} \rangle = 0 \Longrightarrow |n^{(1)}\rangle = 0$

Hence I reverted--P.wormer 14:41, 9 August 2007 (UTC)

at second order you are missing terms. you should have 2<n⁽⁰⁾|n⁽²⁾>+<n⁽¹⁾|n⁽¹⁾>. you can check this with the second order expressions actually given in the article!

the statement in the text regarding the intermediate normalization condition: "The same choice is made for all higher-order state corrections" is just wrong.

The unit normalization of the exact wave function is not convenient. Already the second-order wave function $|n^{(2)}\rangle$ obtains the non-zero component $\langle n^{(0)}| n^{(2)}\rangle$ along the zeroth-order wave function $|n^{(0)}\rangle$ in unit normalization. This means that the weight (prefactor) of the zeroth-order wave function changes when the second-order correction is included. This continues for higher orders: every order included changes the weight of the zeroth-order wave function. The equations for higher order wave functions become very awkward in unit normalization. An elegant way out is to solve all the higher order contributions in the orthogonal complement of $|n^{(0)}\rangle$ , so that $\langle n^{(k)}| n^{(0)}\rangle = 0$ for all k > 0. As a consequence $\langle n| n^{(0)}\rangle = \langle n^{(0)}| n^{(0)}\rangle = 1$ . --P.wormer 13:05, 13 August 2007 (UTC).

I see that an anonymous deleted again the part on intermediate normalization (including three references to reputable books!). I give up.--P.wormer 15:58, 13 August 2007 (UTC)

i reverted the article because as it stood it was wrong. according to the actual expressions given at second order $\langle n^{(k)}| n^{(0)}\rangle = 0$ does not hold for k=2. feel free to use a more convenient normalization but you will need to correct the expressions given in the article.

[edit] Lambda dropped

$λ$ is lost in the final equation for the energy. HolIgor 20:29, 16 October 2007 (UTC)

Yes, thank you for the catch. But let me put it up for debate: Is it clearer to use the approach begun in the derivation here and used on the Perturbation theory page (which takes a formal mathematical standpoint), or would it be better to take the more utilitarian road and drop the lambda in favor of an alternative notation, such as the substitution $\lambda V \rarr H_1$ (as done by, among others, the standard textbook by Shankar)? Cgd8d (talk) 17:30, 8 May 2008 (UTC)