I came across this simple proof:
We show that Hermitian operators have real eigenvalues. The definition
of a Hermitian operator is\begin{equation} \langle \phi_i | \hat A | \phi \rangle = \langle
\phi_i | \hat A | \phi \rangle^* \tag{1} \end{equation}Then if $|\psi\rangle$ is an eigenvector of $\hat A$, we have
$$ \hat A|\psi\rangle = \lambda|\psi\rangle \tag{2}$$
and therefore
$$ \langle\psi|\hat A|\psi\rangle = \lambda . \tag{3}$$
If $\hat A$ is hermitian, we my apply (1) so that
$$\langle\psi|\hat A|\psi\rangle = \langle\psi|\hat A|\psi\rangle^*$$
$$\lambda = \lambda^*.$$
What I am not getting, is the step from (2) to (3). Seems to me that would be true only if $\psi$ is normed ($\langle\psi|\psi\rangle = \int \psi^*\psi \text d \tau = 1$).
Is it true in general that eigenvectors/eigenfunctions of operators are normed and orthogonal?
Best Answer
The property of orthogonality can always be imposed, but it is not required at all in the excerpt you've cited.
The normalization of the eigenvectors can always be assured (independently of whether the operator is hermitian or not), by virtue of the fact that if $Av=\lambda v$, then any multiple $w=\alpha v$ of that vector will obey $$ Aw = A\alpha v = \alpha A v = \alpha \lambda v = \lambda w. $$ Thus, given any eigenvector of any operator, you can always assume (for free) that it's been normalized to unity.
However, this is also not necessary for the manipulations you've cited: if you remove that normalization, then your equation $(3)$ becomes $$ \langle\psi|\hat A|\psi\rangle = \lambda \langle\psi|\psi\rangle, \tag{3'}$$ in which $\lambda \langle\psi|\psi\rangle $ is (by the properties of the inner product) a real and positive number. The rest of the manipulations are unaffected: you get to $$ \lambda \langle\psi|\psi\rangle = \lambda^* \langle\psi|\psi\rangle $$ and all you need to do is divide by $\langle\psi|\psi\rangle $.