There is one way to get in trouble: if En(0) = Em(0) for any combination of n and m. In that case |<m|H'|n>|2 / (En(0) - Em(0)) is not ordinarily smaller than <n|H'|n>! So the whole theory (so far) fails if the states of the unperturbed Hamiltonian are degenerate.
The cure is to ensure that <m|H'|n> = 0 whenever En(0) = Em(0). Then there is no problem with infinities. This is possible and legitimate because any basis within the space of degenerate states is a legitimate set of unperturbed states. So anytime you are going to do perturbation theory where the unperturbed states are degenerate, first diagonalize the perturbing Hamiltonian. Note that the diagonalization is necessary even if you are not going beyond first order.
1/2 | 0 | 0 | 0 | 0 | 0 |
0 | -1/2 | sqrt(2)/2 | 0 | 0 | 0 |
0 | sqrt(2)/2 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | sqrt(2)/2 | 0 |
0 | 0 | 0 | sqrt(2)/2 | -1/2 | 0 |
0 | 0 | 0 | 0 | 0 | 1/2 |
Background: Summary of hydrogen atom | |
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m e2 1 En = - [ ------- (-----)2] -- 2 hbar2 4p e0 n2 1 = - a2 m c2 ---- 2 n2 |
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ynlm =
Rnl(r) Ylm(q, f) = Nnl e -2r rl Ln-l-12l+1(r) Ylm(q, f ) |
r = 2r/(na) a = 4 p e0 hbar2 / (m e2) Nnl2 = {[2/(na)]3 (n-l-1)!} / {2n [(n+l)!]3} |
R10
= 2 a -3/2 e - r/a
R20
= [1/sqrt(2)] a -3/2 [1 - r / (2a)]
e - r / (2a) |
Y00 = 1/ sqrt(4p) Y10 = sqrt[3 / (4p) ] cosq Y11 = - sqrt[3 / (8p) ] sinq e imf Y1 -1 = sqrt[3 / (8p) ] sinq e -imf |