I am finding that the eigenfunctions of my Hermitian differential operator are not orthogonal and I do not know why. Consider the differential operator
$$ \mathcal{L} = x^2 \frac{d^2}{dx^2} + 2x \frac{d}{dx} + c $$
where $c$ is a constant. This is an Hermitian operator with respect to the inner product
$$ \langle \psi , \phi \rangle =\int_{- \infty}^\infty dx\psi^*\phi$$
We have
$$ \langle \psi , \mathcal{L} \phi \rangle =\int_{- \infty}^\infty dx\psi^*\left( x^2 \frac{d^2 \phi}{dx^2} + 2x \frac{d \phi}{dx} + c \phi\right) \\
= \int_{- \infty}^\infty dx \left(\frac{d^2}{dx^2} \left( x^2 \psi^* \right) \phi – \frac{d}{dx}\left( 2x \psi^* \right) \phi + c \psi^* \phi \right) \\
= \int_{- \infty}^\infty dx\left( 2\psi^* + 4x \frac{d \psi^*}{dx} + x^2 \frac{d^2 \psi^*}{dx^2} – 2\psi^* – 2x \frac{d \psi^*}{dx} + c\psi^* \phi\right) \\
= \langle \mathcal{L} \psi , \phi \rangle $$
where I have assumed my solutions vanish at $\pm \infty$ so the boundary terms vanish when I integrate by parts. So my operator is Hermitian and I expect my eigenfunctions to be orthogonal. Consdider the eigenvalue equation $\mathcal{L} \psi = \lambda \psi$, this yields the differential equation
$$ \quad x^2 \psi''(x) + 2x \psi'(x) + (c – \lambda)\psi = 0$$
The eigenvalue equation is therefore an Euler differential equation. If we take a trial solution $\psi(x) = x^n$, then substituting this in yields the quadratic equation
$$ n^2 + n + (c- \lambda) = 0 \quad \Rightarrow \quad n=-\frac{1}{2} \pm \frac{1}{2} \sqrt{1- 4(c-\lambda)}$$
Suppose we took the special case where the eigenvalues are negative and of the form $\lambda = -E^2$, for some $E$, and let $ c = \frac{1}{4}$, then we have $ n = -\frac{1}{2} \pm i E$ and the solutions will be given by
$$ \psi_\pm(x) = \frac{1}{\sqrt{x}} x^{\pm i E}$$
My problem is that these solutions do not appear to be orthogonal for different eigenvalues. If we take the solutions whose eigenvalues are $\lambda $ and $\lambda'$, then the inner product would be
$$ \langle \psi , \psi' \rangle = \int_{-\infty}^\infty dx \frac{1}{x} x^{\pm i (E'-E)} $$
which according to Wolfram is divergent. I am not sure why my solutions for different eigenvalues are not orthogonal. Any hints would be greatly appreciated.
Best Answer
The function of yours that you have named $\psi_{\pm}$ are not both in $L^2(0,\infty)$ or both in $L^2(-\infty,0)$. Are they? Orthogonality has no meaning when it comes to functions that are not in the same inner product space.