# Why is this the eigenvalue?

The equation I am looking at (the wave equation for the electric field) has the following form:

$$\nabla_t^2\mathcal{E} + [(\frac{\omega}{c})^2n^2(x,y) - \beta^2]\mathcal{E} = 0$$

where $$\nabla_t^2 = \nabla^2 - \partial^2/\partial z^2$$ (the subscript $$t$$ denotes transverse components of the Laplacian).

The author now says that this equation can be viewed as an eigenvalue problem with $$\mathcal{E}$$ being the eigenfunctions and $$\beta^2$$ the eigenvalues.

Shouldn't the eigenvalues be $$\beta^2 - (\frac{\omega}{c})^2n^2(x,y)$$ ?

Notice that $$n(x,y)$$ is not constant, and the eigenvalues should be constant. The eigenvalue problem might be clearer if it is written in the form $$\left[\nabla_t^2 + \left(\frac{\omega}{c}\right)^2n^2(x,y)\right]\mathcal{E} = \beta^2 \mathcal{E}.$$
The operator whose eigenvalue problem is being considered is $$\nabla_t^2 + \left(\frac{\omega}{c}\right)^2n^2(x,y)$$, with $$\beta^2$$ playing the role of the eigenvalue. $$n(x,y)$$ plays the role of an operator, just as $$\frac{\partial}{\partial x}$$ and $$\frac{\partial}{\partial y}$$ do.