# Proving orthogonality of eigenstates of a Hamiltonian

Suppose we have $\Psi_{1}$ and $\Psi_{2}$ which are eigenstates of some (self-adjoint) Hamiltonian $\hat{H}$ with unequal eigenvalues. Could you explain me how can I prove that these arbitrary $\Psi_{1}$ and $\Psi_{2}$ are orthogonal?

P.S. I saw some similar problems here, but I still didn't get it.

• "P.S. I saw some similar problems here, but I still didn't get it." Give a link. – DanielSank Jul 17 '18 at 20:31
• @DanielSank I saw this one. – user201777 Jul 17 '18 at 20:44

Given $H$ self-adjoint and $\psi_1$, $\psi_2$ eigenvector from distinct eigenspaces, we want to prove that $(\psi_1,\psi_2)=0$.

Since $\psi_1$ and $\psi_2$ are eigenvectors from distinct eigenspaces of $H$, and $H$ is self-adjoint, there are two distinct real numbers $E_1$ and $E_2$ such that $H\psi_i = E_i\psi_i$, $i=1,2$. Therefore, if we take the inner product between $\psi_1$ and $H\psi_2$ we get

$$(\psi_1,H\psi_2)=E_2(\psi_1,\psi_2)$$

Since $H$ is self-adjoint, we have the identity

$$(\psi_1,H\psi_2) = (H^*\psi_1,\psi_2) = (H\psi_1,\psi_2).$$

But, since $\psi_1$ is an eigenvector of $H$, we also have

$$(H\psi_1,\psi_2)=\overline{E_1}(\psi_1,\psi_2)=E_1(\psi_1,\psi_2),$$

the last equality following from the fact that $E_1$ is real. We then get to the equality

$$E_1(\psi_1,\psi_2) = E_2(\psi_1,\psi_2),$$

with $E_1\neq E_2$ by hypothesis. The only way this can be true is if $(\psi_1,\psi_2) = 0$, i.e. if $\psi_1$ and $\psi_2$ are orthogonal.

• This seems to hinge on the assumption of non-degeneracy. Could there is a more general proof for degenerate eigenstates? – VVC Apr 2 at 9:39
• All you need is to show that eigenspaces are orthogonal, as done above. Then in each of them you can find an ONS – Phoenix87 Apr 2 at 18:46

$E_1\langle\Psi_1|\Psi_2\rangle=\langle\Psi_1|\hat{H}|\Psi_2\rangle=E_2\langle\Psi_1|\Psi_2\rangle$, so $(E_1-E_2)\langle\Psi_1|\Psi_2\rangle=0$. Cancelling $E_1-E_2\ne 0$ gives $\langle\Psi_1|\Psi_2\rangle=0$ as required.