How do I show that the eigenstates of a Hamiltonian can be made orthonormal? I've been tearing my hair out over this all evening. It should be simple but I must be missing something somewhere. Can someone show me how to prove that the eigenstates of a Hamiltonian can be made orthonormal, please?
 A: *

*We first prove orthogonality of non-degenerate eigenvectors of the
Hamiltonian. Consider the braket and act with the Hamiltonian in
both directions,
$ \left\langle\alpha | H |\beta\right\rangle =
    E_\alpha\left\langle\alpha  |\beta\right\rangle = E
    _\beta\left\langle\alpha  |\beta\right\rangle $
If the states are not orthogonal ($\left\langle\alpha 
    |\beta\right\rangle \neq 0 $) then we would get a contradiction
since we assume the states are non-degenerate ($E_\alpha\neq E_\beta
    $). So we must have 
$\left\langle\alpha  |\beta\right\rangle = 0 $
for distinct states.

*Now we need to prove that the braket of two eigenstates is equal to
$1$ up to a phase. Consider the braket:
$ \left\langle\alpha  |\alpha\right\rangle = \sum_n
    \left\langle\alpha  |n\right\rangle \left\langle n 
    |\alpha\right\rangle = \left\langle\alpha  |\alpha\right\rangle
    \left\langle\alpha  |\alpha\right\rangle $
where we have inserted a sum over the states of the Hamiltonian and
then used the orthogonality relation that we proved above. Now we
can divide both sides by $\left\langle\alpha  |\alpha\right\rangle $
to get 
$\left\langle\alpha  |\alpha\right\rangle = 1 $

*Thus for we have only considered non-degenerate eigenvectors.
Degenerate eigenvectors can't be distinguished and they don't need
to be orthogonal to each other. However, for any set of linearly
independent vectors (all wavefunctions of a Hamiltonian are linearly
independent) there exists linear combinations of them that are orthogonal which can be found through the Gram–Schmidt procedure. Thus one can choose the vectors to be linearly independent.

