Simple eigenvalue for eigenfunctions

Question

I know what it means that an eigenvalue of a matrix is simple. My question is about the eigenvalues of operators as the Schröndiger one. Here we work with eigenfunctions rather than with eigenvectors so I can not see what the algebraic multiplicity is.

Answer 1

When you solve the Schrodinger equation, you solve for the possible energies of the system. These are the eigenvalues of the Hamiltonian operator. Just like a single eigenvalue can correspond to multiple eigenvectors, a single energy can correspond to multiple eigenfunctions of the Hamiltonian. This is called degeneracy. An eigenvalue (energy) is called simple if it corresponds to a single eigenfunction of the Hamiltonian.

Take, for example, the one dimensional harmonic oscillator. The Schrodinger equation is - $$\left (-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+\frac{1}{2}m\omega x^2 \right )\psi=E\psi$$

And the corresponding energies are - $$E_n=\hbar \omega\left (n+\frac{1}{2}\right ), n\in\mathbb N$$

In this case, since each value of $n$ corresponds to a distinct eigenfunction, we know that each energy (eigenvalue) has only one corresponding eigenfunction, and the eigenvalues are thus all simple. This is unlike the case of the $2$ or $3$ dimensional harmonic oscillator, which can have degenerate states - energies with certain multiplicity.