Symmetries of the Helmholtz Equations

It is easy to see that if

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

then

$$X_{a\ast}\psi\,,~~Y_{b\ast}\psi~~\textrm{and}~~R_{\gamma\ast}\psi$$

are also solutions to the Hermholtz equation. In other words,

$$(\nabla^2+k^2)(X_a\psi )=0\,,~~\textrm{etc.}$$

This is because the partial derivative can be interchanged and the coefficient of $\nabla^2$ are independent of $x$ , $y$ , and $\theta$ . One refers to this independence by saying that $x$ , $y$ and $\theta$ are cyclic coordinates, or equivalently, that $X_{a\ast}$ , $Y_{b\ast}$ , and $R_{\gamma\ast}$ are symmetries of $\nabla^2$ .

This independence implies that the eigenspace of $\nabla^2$ is invariant under $X_{a*}$ , $Y_{b*}$ , and also $R_{\gamma*}$ . This is a very powerful result. It says that if $\psi$ is a solution, then one obtains the additional solutions

$$X_{a*}\psi,~~Y_{b*}\psi, ~~\textrm{and}~~R_{\gamma*}\psi~~,$$

which are parametrized by the translation parameter $a$ and $b$ , and by the angle $\gamma$ respectively.