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,
$$E^{(3)} = <0|H' - E^{(1)}|2>$$

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:
$$E^{(3)}=<0|H'-E^{(1)}\mathbb{I}|2>-E^{(2)}<0|1>$$

(0)​

The secret now is to look at the equations of the coefficients for the lower orders
$$H_0|1>+H'|0>=E^{(0)}|1>+E^{(1)}|0>$$

(1)​

and
$$H_0|2>+H'|1>=E^{(0)}|2>+E^{(1)}|1>+E^{(2)}|0>.$$

(2)​

Using these, the third order energy correction may be expressed in the form
$$E^{(3)}=<1|H'-E^{(1)}\mathbb{I}|1>-2E^{(2)}\textt{Re}<0|1>.$$

(3)​

Hint 2:
The "dual equation" to equation (1) may be written as
$$<1|H_0=E^{(0)}<1|+E^{(1)}<0|-<0|H'.$$

(4)​

Spoiler

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