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.
$E_1\langle\Psi_1|\Psi_2\rangle=\langle\Psi_1|\hat{H}|\Psi_2\rangle=E_2\langle\Psi_1|\Psi_2\rangle$, 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.
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.
