Time independent perturbation theory

[barefeet]

Homework Statement

The following text on the time independent perturbation theory is given in a textbook:

\hat{H} = \hat{H}_0 + \alpha \hat{H'}
We expand its eigenstates \mid n \rangle in the convenient basis of \mid n \rangle^{(0)}
\mid n \rangle = \sum_m c_{nm} \mid m \rangle^{(0)}
The Schrödinger equation in these notations becomes
\left\{ E_n(\alpha) - E_m^{(0)} \right\}c_{nm} = \alpha \sum_p c_{np} M_{mp}
With
M_{nm} = \langle n \mid \hat{H'} \mid m \rangleI don't understand how the second last equation is derived and I don't know how the Schrödinger equation is used

Homework Equations

The Attempt at a Solution

The only thing I can think of is to use the first equation and let both sides be sandwiched between an eigenstate \mid n \rangle of the operator \hat{H}

\langle n \mid \hat{H} \mid n \rangle = \langle n \mid \hat{H_0} \mid n \rangle + \alpha \langle n \mid \hat{H'} \mid n \rangle

\langle n \mid E_n(\alpha) \mid n \rangle = \sum_m c_{nm}^* \langle m \mid^{(0)} \hat{H_0} \mid \sum_k c_{nk} \mid k \rangle^{(0)} + \alpha \langle n \mid \hat{H'} \mid \sum_p c_{np} \mid p \rangle^{(0)}

E_n(\alpha) = \sum_m c_{nm}^*c_{nm} E_m^{(0)} + \alpha \sum_p c_{np} \langle n \mid \hat{H'} \mid p \rangle^{(0)}

E_n(\alpha) - \sum_m |c_{nm}|^2 E_m^{(0)} = \alpha \sum_p c_{np} M_{np}

And here I am stuck:
- E_n(\alpha) doesn't have a factor c_{nm}
- E_m^{(0)} is still a summation and has a factor of |c_{nm}|^2 instead of c_{nm}
- I have M_{np} instead of M_{mp}
- The p's are eigenstates of H_0 and not of H

 

[TSny]

The Schrödinger equation in these notations becomes
\left\{ E_n(\alpha) - E_m^{(0)} \right\}c_{nm} = \alpha \sum_p c_{np} M_{mp}
With
M_{nm} = \langle n \mid \hat{H'} \mid m \rangle

In the last expression for $M_{nm}$, should there be superscripts (0) on the n and m states?

 

[barefeet]

I am assuming that as well, it is not specifically stated in the textbook.
In my derivation it is, but my equation isn't the same anyway, so that doesn't help much
But only on the ket vector not the bra vector

 

[TSny]

The only thing I can think of is to use the first equation and let both sides be sandwiched between an eigenstate \mid n \rangle of the operator \hat{H}

\langle n \mid \hat{H} \mid n \rangle = \langle n \mid \hat{H_0} \mid n \rangle + \alpha \langle n \mid \hat{H'} \mid n \rangle

Try looking at ^{(0)} \! \langle m\mid \hat{H} \mid n \rangle where the m state is an eigenstate of H⁽⁰⁾ and the n state is an eigenstate of H.

 

[barefeet]

Yes I got it, but only if there are superscripts for the matrix elements. I tried this before but the absence of superscripts threw me off. I guess they must be eigenstates of the unperturbed state otherwise it wouldn't make sense. Thanks

 

[TSny]

OK. Sounds good.