3rd-order Energy Correction Derivation

[cscott]

Homework Statement

Time-independent, non-degenerate perturbation theory:

- Let |k> represent k-th order wave function correction.
- Let E^{(k)} represent k-th order energy correction
- H' is the perturbed Hamiltonian.

Want the third order energy correction:
E^{(3)} = <1|H' - E^{(1)}|1> - 2E^{(2)}<0|1>

Homework Equations

Equating coefficients of equal powers of the parameter:
H_0|3> + H'|2> = E^{(0)}|3> + E^{(1)}|2> + E^{(2)}|1> + E^{(3)}|0>

The Attempt at a Solution

Tried multiplying <0| on the left like when deriving the first order correction, E^{(1)}.

I get,
E^{(3)} = &lt;0|H&#039; - E^{(1)}|2&gt;

I don't know what I'm missing to proceed... I'm in the same situation with a few other similar problems. I don't see how multiplying by anything else on the left would make sense either.

Hint on what I'm missing? Thanks

 

[cscott]

did I state my question poorly? :\

This is problem 8.1 from B&J's book.

 

[cscott]

One last bump... haven't gotten anywhere with this.

 

[bjorklund]

Recently had to solve the same problem, and since no one has commented, I figured I'd leave a hint or two for the next person who comes along.

Hint 1:
The attempted solution in the opening post is the correct way to proceed, except for a term which has been forgotten:
E^{(3)}=&lt;0|H&#039;-E^{(1)}\mathbb{I}|2&gt;-E^{(2)}&lt;0|1&gt;

(0)​

The secret now is to look at the equations of the coefficients for the lower orders
H_0|1&gt;+H&#039;|0&gt;=E^{(0)}|1&gt;+E^{(1)}|0&gt;

(1)​

and
H_0|2&gt;+H&#039;|1&gt;=E^{(0)}|2&gt;+E^{(1)}|1&gt;+E^{(2)}|0&gt;.

(2)​

Using these, the third order energy correction may be expressed in the form
E^{(3)}=&lt;1|H&#039;-E^{(1)}\mathbb{I}|1&gt;-2E^{(2)}\textt{Re}&lt;0|1&gt;.

(3)​

Hint 2:
The "dual equation" to equation (1) may be written as
&lt;1|H_0=E^{(0)}&lt;1|+E^{(1)}&lt;0|-&lt;0|H&#039;.

(4)​

Spoiler

Multiply <1| with equation (2).
Insert (4) in the first term of this and rearrange.
Insert this in (0).