Litle help with perturbation theory

[baranas]

Why when we analyse time dependent perturbation theory, we take that the diagonal elements of matrix <i|W(t)|j> are equal to zero?
Why in degenerate perturbation theory we assume that perturbed wavefunctions of degenerate states can be expressed in the base of unperturbed wavefunctions of degenerate states and deny other functions?



P.S. Sorry for my English

 

[strangerep]

Why when we analyse time dependent perturbation theory, we take that the diagonal elements of matrix <i|W(t)|j> are equal to zero?
Why in degenerate perturbation theory we assume that perturbed wavefunctions of degenerate states can be expressed in the base of unperturbed wavefunctions of degenerate states and deny other functions?

I'm not sure I understand your question, but I think the answer is:
"because we're assuming that the set of all eigenstates of the
unperturbed Hamiltonian is a basis for the Hilbert space".

(If that doesn't make sense, then I guess I didn't understand
the question properly.)

 

[baranas]

I'm not sure I understand your question, but I think the answer is:
"because we're assuming that the set of all eigenstates of the
unperturbed Hamiltonian is a basis for the Hilbert space".

No, the problem is that in degenerate perturbation theory we take the basis for our perturbed state function only the wave-functions of degenerate states from unperturbed system and skip others, which belong to nendegenerate states.

H₀\phi⁰_n=E⁰ⁿ\phi⁰_n

Assume that we have r degenerate states E⁰_f\rightarrow\phi⁰_fa, with a=1,2,3...r. The question is why do we assume, that a wave-function after perturbation will be in the form
\Psi_na=\sumc_b\phi⁰_fb and skip the part \sumc_n\phi⁰_n of nondegenerate functions

 

[strangerep]

No, the problem is that in degenerate perturbation theory we take the basis for our perturbed state function only the wave-functions of degenerate states from unperturbed system and skip others, which belong to nondegenerate states.

In Ballentine, "QM - A Modern Development", sect 10.5, pp284-285, one starts with a full set of (unperturbed) states, with an additional label to represent any degeneracies. Then one derives an equation involving only the degenerate unperturbed states belonging to a
particular unperturbed energy. [See Ballentine's eq(10.92) and discussion leading to it.]

So we haven't "skipped" the others. They're still there, but we derived an extra
condition that allows the perturbation procedure to go ahead.