First-Order Perturbation Theory Derivation in Griffiths

[BucketOfFish]

Homework Statement

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.

Homework Equations

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

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?

 

[vela]

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

 

[BucketOfFish]

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