Are eigenfunctions always normed and orthogonal?

I came across this simple proof:

We show that Hermitian operators have real eigenvalues. The definition of a Hermitian operator is

$$\langle \phi_i | \hat A | \phi \rangle = \langle \phi_i | \hat A | \phi \rangle^* \tag{1}$$

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?

• – John Rennie Jun 22 '18 at 14:59
• @JohnRennie The duplicate question is asking about the general consequences of observables corresponding to Hermitian operators. This question seems to be asking a minor technical clarification about eigenvector normalisation. – gj255 Jun 22 '18 at 15:12
• @SeanBone Eigenvectors are defined merely as non-zero vectors $|\psi\rangle$ such that $A |\psi\rangle = \lambda |\psi\rangle$. There is no constraint on normalisation, and you can easily check that if $|\psi\rangle$ is an eigenvector, then so is $c|\psi\rangle$ for any non-zero $c$. Hence in going from (2) to (3), there is an implicit assumption as you suspect. Commonly we take states in QM to be of unit norm, but not always. – gj255 Jun 22 '18 at 15:15
• If the wave function is not normalised then predictions based on it will be wrong. There is no theoretical basis for normalisation. – my2cts Jun 22 '18 at 18:23

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 $\langle\psi|\psi\rangle$.
Note that if $\hat{A}|\psi\rangle=\lambda|\psi\rangle,\,\hat{A}^\dagger|\phi\rangle=\mu|\phi\rangle$ then $\langle \phi|\hat{A}|\psi\rangle$ is equal to both $\lambda\langle\phi|\psi\rangle$ and $\mu^\ast\langle\phi|\psi\rangle$, so either the vectors are orthogonal or $\lambda=\mu^\ast$. We can even ensure orthogonality in this special case with a basis change called the Gram-Schmidt process. Finally, we can rescale eigenvectors to have unit norm. This allows such convenient results as $\operatorname{id}=\sum_i |i\rangle\langle i|$ so that $|\Psi\rangle=\sum_i \langle i|\Psi\rangle|i\rangle$.