Degenerate Eigenvalues

Every eigenvalue of the eigenvalue equation

$$-\nabla^2\psi = k^2\psi$$

is highly degenerate. In fact, each eigenvalue $k^2$ is infinitely degenerate. This means that for one and the same eigenvalue $k^2$ , there is an infinite set of eigenfunctions, namely,

$$\{ e^{i(k_x x+k_yy)}\colon k^2_x+k^2_y = k^2\}$$

or

$$\{ e^{ikr\cos (\alpha -\theta )}\colon \alpha\textrm{~is~a~constant}\}\,.$$

These solutions form a basis for the subspace of solutions to the Helmholtz equation

$$(\nabla^2+k^2)\psi =0~~.$$

Any solution to this equation is a unique superposition of the basis elements. We shall refer to this subspace as the eigenspace of the (degenerate) eigenvalue $k^2$ .

A matrix, and more generally an operator, is diagonal relative to its eigenvector basis. The Helmholtz operator $-\nabla^2$ can, therefore, be viewed as an infinite diagonal matrix

$$-\nabla^2 = \left[\begin{array}{ccc} \ddots &&\\ &\boxed{\begin{... ...c} k^2 &&0\\ &\ddots &\\ 0 &&k^2\end{array}} &\\ &&\ddots\end{array}\right]$$

with degenerate eigenvalues $k^2$ along the diagonal.

The question now is, how does one tell the difference between the eigenfunctions having the same eigenvalue $k^2$ ? Physically one says that these eigenfunctions are plane waves propagating into different directions. However, one also would like to express the difference algebraically.