For what finite values of $p,q$ is the following inner product, defined in the usual sense of $L_2\left(\mathbb{R}\right),$
$$\left<\phi_m^p(x)\Big|\phi_n^q(x)\right>$$ equal to $0$? $\phi(x)$ are the $m^{th}$ and $n^{th}$ eigenfunctions of some Sturm-Liouville operator, and $p,q$ are integer powers greater than $0$. I used $\sin\left(\frac{n\pi x}{L}\right)$ to get a feel for this, but I was wondering if there was a proof someone could provide for certainty.
In general, if it's for all $m\neq n$, then I doubt such an identity is true for any $p,q>1$. As a quick example, if $\phi_n(x)=\sin(\frac{n\pi x}{2})$ on $[0,2]$, then $\langle\phi_m^p,\phi_n^q\rangle>0$ if $p$ and $q$ are even. Of course, for odd $p,q$, we have $\langle\phi_m^p,\phi_n^q\rangle=0$ if the sets $\{km:k=1,\dots,p\}$ and $\{\ell n:\ell=1,\dots,q\}$ are disjoint, but this need not be true for all $m\neq n$.
This difficulty has less to do with the eigenvalues being integers and more to do with exponentiation $f(x)\mapsto f(x)^p$ "raising" the energy of the eigenfunctions. For example, consider the bounded eigenfunctions of $-\nabla^2$ on $\mathbb R$, namely $\phi_\xi(x)=e^{2\pi i\xi x}$. In this case, the eigenfunctions behave nicely under exponentiation: $\phi_\xi^p=\phi_{p\xi }$. In the sense of distributions, we have $\langle \phi_\xi,\phi_\eta\rangle=\delta(\xi-\eta)$, so $$ \langle \phi_\xi ^p,\phi_\eta^q\rangle=\langle \phi_{p\xi},\phi_{q\eta}\rangle=\delta(p\xi-q\eta). $$ If $p,q>1$, then this cannot be zero for all $\xi\neq \eta$.