# First-Order Perturbation Theory Derivation in Griffiths

1. Jan 17, 2013

### BucketOfFish

1. The problem statement, all variables and given/known data

On page 251 of Griffiths's quantum book, when deriving a result in first-order perturbation theory, the author makes the claim that $$<\psi^0|H^0\psi^1> = <H^0\psi^0|\psi^1>$$ where $H^0$ is the unperturbed Hamiltonian and $\psi^0$ and $\psi^1$ are the unperturbed wavefunction and its first-order correction.

2. Relevant equations

$$H^0\psi^0=E^0\psi^0$$

3. The attempt at a solution

From the derivations I've seen of Hermitian operators, I seem to understand that both sides of the expression in the braket have to be eigenfunctions of the operator for the operator to be symmetric. If this is the case, then I feel that Griffiths's explanation may be lacking something, since there is no guarantee that the first-order wavefunction correction is an eigenfunction of the unperturbed Hamiltonian. In this case, you could not switch the Hamiltonian operator around like this.

I must be misunderstanding something! Can someone help me clear this up?

Last edited: Jan 17, 2013
2. Jan 17, 2013

### vela

Staff Emeritus
That relation must hold for all states, not just eigenstates, if the operator is Hermitian.

3. Jan 17, 2013

### BucketOfFish

Ah yes, I forgot that observables are not just for stationary states. Thanks for your help, vela!