3rd order energy perturbation correction

  • Thread starter valtorEN
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Homework Statement



Derive the general expression of 3rd-order perturbation energy for a

non-degenerate quantum system.


Homework Equations



for nth order we have

(Ho-Eo)|n>+(H'-E1)|n-1> -E2\n-2>-En|0>=0 (given)

also,

<0|0>=1,

<1|0> = <0|1>=0,

<0|2>=<2|0>=-1/2<1|1>

<0|n>=<n|0>=-1/2*(<n-1|1>+<n-2|2>+....+<2|n-2>+<1|n-1>),

The Attempt at a Solution



for n=3 we get

(Ho-Eo)|3>+(H'-E1)|2>-E2|1>-E3|0>=0

now, multiply by <0| (this

we get <0|(H0-E0)|3> +<0|(H'-E1)|2>-<0|E2|1>-<0|E3|0>=0

by the rules above, the third term is = 0, the fourth is just =-E3 and the 2nd
is -1/2<1|1>

this gives E3=<0|(Ho-Eo)|3>-1/2*<1|(H'-E1)|1>

for <0|n>=<n|0>=-1/2(<n-1|1>+<n-2|2>+....+<2|n-2>+<1|n-1>)

for n=3

i get

E3=-1/2(<2|(H0-E0)|1>+<1|(H0-E0)|2>+<0|(H0-E0)|3>+<2|(H0-E0)|1>

+<1|(H0-E0)|2>)-1/2*(<1|(H'-E1)|1>) (phew!)

I am not sure what do do next

what is <1|2>=<2|1> = to?

i can see all the terms only have states 1 and 2 involved in them except the

3rd term <0|3> and the last term <1|1>, are these important?

does <1|1>=0?
 

Answers and Replies

  • #2
Dr Transport
Science Advisor
Gold Member
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you have to find the third order perturbation expansion for the energy correction, & not just the interaction between the n = 0,1,2 & 3 states.
 

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