Why is the set of eigenfunctions of a Hermitian operator complete? When I was studying quantum chemistry, I was told that given the time-independent Schrodinger equation 
$$\hat H \psi = E \psi$$
since $\hat H$ is Hermitian, the set of eigenfunctions $\{\psi_i \}_{i = 1}^{\infty}$ of $\hat H$ is complete. 
What does it mean that the set is complete? Does it mean this set of functions can replace Fourier series? And how do we prove that the set is complete? 
 A: 
Does it mean this set of functions can replace Fourier series?

Basically, yes. It means that any state $\psi$ can be written as the infinite sum
$$
\psi = \sum_{i=1}^\infty \langle \psi_i,\psi\rangle \:\psi_i,
$$
where $\langle\psi_i,\psi\rangle$ is the inner product on Hilbert space. This result is known as the spectral theorem and it is a foundational result of functional analysis, though its proof is generally too difficult for physics books and courses.
A word of caution, though: the result as stated is valid only for self-adjoint operators, a condition which requires hermiticity (also known as symmetry) in the sense that
$$
\langle \phi,H \psi \rangle = \langle H\phi, \psi \rangle
$$
for any two states $\phi$ and $\psi$ in the domain of $H$, but which also makes additional requirements on the domain of $H$ and its adjoint, and their relationship to the full space.
A: "Complete" means that any state in your space can be written as a superposition of energy eigenstates; that is, energy eigenstates span all the state space. It is standard subject in any textbook on mathematical physics that this is indeed a complete set. Check any source on "spectral theorem" and you will see.
A: The same question has troubled me.
Regarding the statement that there is always such a complete set of eigenvectors for $\hat{H}$, it can be found in Levine's Quantum Chemistry (Levine, I. R., Quantum chemistry 7th edition, Pearson, 2013) the following (Ch. 7, p. 165):
" We now postulate that the set of eigenfunctions of every Hermitian operator that
represents a physical quantity is a complete set. (Completeness of the eigenfunctions can
be proved in many cases, but must be postulated in the general case.) Thus, every wellbehaved function that satisfies the same boundary conditions as the set of eigenfunctions can be expanded according to
$$
f=\sum_{i} a_i g_i
$$
(...)"
(Where the $a_i$'s are constants and the $g_i$'s are a set of functions).
Thus, it can be proved in a case basis, but apparently there is yet no general proof.
