Why are eigenspaces of a Hermitian operator mutually orthogonal? In Quantum Mechanics, from the properties of the solution of Schrodinger's Equation inside the infinite well, is that they are:


*

*Mutually orthogonal for different eigenvalues.

*Orthonormal.

*Complete.


What is the proof of their orthogonality? What is the proof of the orthogonality property in the solution of Schrodinger's equation inside the infinite well?
 A: Any observables on the wavefunction have to have Hermitian form, which gives that their eigenvalues must be real, and you can always find an orthogonal basis from the eigenstates related to these values even if they are degenerate. 
A hermitian matrix is one which is equal to its hermitian conjugate:
$$A=A^\dagger=\overline{A^T}$$
We can prove orthogonality of an hermitian matrix's eigenvectors for distinct eigenvalues:
We have two eigenvectors related to different eigenvalues
$$Av^i=\lambda_i v^i$$
$$Av^j=\lambda_j v^j$$
We take the hermitian conjugate of the $\lambda_i$ case, and operate on the right of $v^j$:
$$(v^i)^\dagger A^\dagger v^j = \overline{\lambda_i }(v^i)^\dagger v^j=\lambda_i (v^i)^\dagger v^j$$
Since the eigenvalues are real.
We can also multiply the $\lambda_j$ case on the left by $(v^i)^\dagger$:
$$(v^i)^\dagger A v^j=\lambda_j (v^i)^\dagger v^j$$
Since $A=A^\dagger$ we have:
$$0=(\lambda_i-\lambda_j)(v^i)^\dagger v^j$$
Since the eigenvectors are for different values, $\lambda_i≠\lambda_j$ then $(v^i)^\dagger v^j=0$ which implies orthogonality. In the case of degenerate eigenvalues, we can use Gram-Schmidt orthogonalisation to produce an orthogonal basis set since any linear sum of eigenvectors of an eigenvalue is also an eigenvector of that value.
Normality follows from the fact that if $v^i$ is an eigenvector of $A$ with eigenvalue $\lambda_i$ then $\alpha v^i$ is as well, so you can normalise these values.
