# Time independent perturbation theory

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

I don't understand how the second last equation is derived and I don't know how the Schrodinger equation is used

## 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
Homework Helper
Gold Member
The Schrodinger 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?

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
Homework Helper
Gold Member
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.

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
Homework Helper
Gold Member
OK. Sounds good.