# 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?

• Hint: use the fact that the Hamiltonian is Hermitian, and consider the inner product $\langle f | Hg \rangle$. – Brian Bi Jan 6 '14 at 23:11
• To use @BrianBi 's hint you must also assume that the eigenvalues corresponding to the eigenstates are different. If the eigenvalues are the same, then you can arrange for them to be orthogonal by Gramm-Schmidt. – Selene Routley Jan 6 '14 at 23:29
• – Qmechanic Jan 6 '14 at 23:36
• With regards your answer in that second link, Qmechanic, I don't see the problem with non-orthogonal eigenspaces, providing we modify our Born rule to 'the probability of collapse into the state $\vec{e}_1$ is the squared $\vec{e}_1$ component of the vector $\psi$ in the basis $\vec{e}_i$'. For an orthonormal basis, we can get this component by taking the inner product with $\vec{e}_1$ (recovering the Born rule), but in an arbitrary basis we can still extract a component, we just need the reciprocal basis vector. Projections still have meaning in non-orthonormal bases. – gj255 Jan 7 '14 at 2:26

## 1 Answer

1. 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.

2. 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

$\left\langle\alpha |\alpha\right\rangle = 1$

3. 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.
• +1 You probably need to add that if the eigenvalues are equal, then you can arrange a finite set of degenerate eigenvectors to be orthogonal by the Gramm-Schmidt process. – Selene Routley Jan 7 '14 at 3:50