# 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?

• This is a standard proof in textbooks Oct 2, 2018 at 14:00
• @Elio Fabri I thought they are the same, thank you for pointing this out. Oct 2, 2018 at 14:09
• @meTchaikovsky Sorry, I had deleted my comment because I recognized that the matter is actually much more complicated and cannot dealt with within a comment. But you succeeded in reading it anyhow. Self-adjoint is not enough, but I'm afraid I cannot find the time to write a correct and useful answer. Oct 2, 2018 at 14:39

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.

• Self-adjointness is not enough to have a basis of eigenvectors, one needs that either the operator or its resolvent (the "inverse", more precisely the inverse of $H+z$, $\Im z\neq0$) is a compact operator. If else, there may not be a basis of eigenvectors (typical examples are the position operator, and the Laplacian on $L^2(\mathbb{R}^d)$: since they have purely continuous spectrum their resolvent cannot be compact, and there is no basis of eigenvectors - actually there are no eigenvectors at all belonging to $L^2$). Oct 2, 2018 at 15:27
• Let me remark that an example of a physically relevant self-adjoint operator that is unbounded but has compact resolvent is the harmonic oscillator on $L^2$: $H=-\Delta+x^2$. In fact it admits an orthonormal basis of eigenvectors, given by the well-known Hermite functions. Oct 2, 2018 at 15:33
• @yuggib Yes, and then the discrete sum needs to be replaced with a spectral measure and all sorts of technical complications - which sum up to very minor changes to the core of this answer. You're welcome to provide your own answer going into as much depth into those topics as you want, but I find them pretty irrelevant here. Oct 2, 2018 at 15:49

"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.

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.