# Why are the eigenfunctions of my Hermitian operator not orthogonal?

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.

• Are you sure you're supposed to work on the whole real line? – Cameron Williams Oct 20 '20 at 21:16
• The spectral theorem only applies to self-adjoint operators. Being Hermitian is necessary but not sufficient to be self-adjoint. – eyeballfrog Oct 20 '20 at 21:28
• @CameronWilliams Yes – Matt0410 Oct 20 '20 at 21:48
• It's only for eigenfunctions in $L^2$ that the usual proof that eigenfunctions with different eigenvalues are orthogonal goes through. The eigenfunctions you wrote down aren't in $L^2$ for any value of $E$. Also you need to be a bit more careful about what $\sqrt{x}$ means if $x$ is negative. – Qiaochu Yuan Oct 20 '20 at 21:52
• @QiaochuYuan's comment is exactly why I asked if you're supposed to be working on $\mathbb{R}$, OP. These are generalized eigenfunctions in a sense, but even then it's messy. Presumably a sort of quantization on the eigenvalues comes in to make sure things are "well" defined re: making sense of powers of $x$. – Cameron Williams Oct 20 '20 at 23:37

## 1 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.

• I don’t understand why they both aren’t valid in $(0,\infty)$? They can be written as $\psi_\pm (x) = \frac{1}{\sqrt{x}} e^{\pm i E \ln x}$ and that is valid for both $\pm$ surely? – Matt0410 Oct 23 '20 at 8:47
• @Matt0410 : Are these functions square-integrable? If not, they are not in the space, and you can't form inner products with them. $e^{isx}$ is not in $L^2(\mathbb{R})$, for example. – Disintegrating By Parts Oct 24 '20 at 1:50