How can I show the symmetry of an operator?

Question

Given the operator $L$ and $p, q, V$ smooth functions: \begin{equation} L=\frac{1}{p(x)}\left[\frac{d}{dx}\left(q(x)\frac{d}{dx}\right)+V(x)\right] \end{equation} I should show that, whenever p is not constant, the operator $L$ is not symmetric on $L^2(\mathbb{R})$ but on $L^2(\mathbb{R}; p(x)dx)$. My problem is that I do not actually understand what does it mean to be symmetric on $L^2(\mathbb{R}; p(x)dx)$: is $p(x)dx$ a measure? And if yes, what does it matter it with the calculation of hermiticity and consequently of symmetry of the operator? Thank you in advance for any help!

Answer 1

Note: I'm going to just assume everything is real-valued, for easy of notation (you'll need that $p,q, V$ are real-valued, and throwing a complex conjugate in the inner product won't add in any difficulty).

First off, yes, $p(x) dx$ is a measure, as we require that $p>0.$ Next, if we take $u,v$ to live in an appropriate domain of $L$ (say $C_0^\infty$), then we need to compute $(Lu,v)_{L^2(\mathbb{R},p(x) dx)}$ as $(u,L^*v)_{L^2(\mathbb{R},p(x) dx)}$ (and, for symmetry, we need that $L=L^*$ here). You'll find, via integrating by parts twice, that

\begin{align*} (Lu,v)_{L^2(\mathbb{R},p(x) dx)}&=\int\limits_{-\infty}^\infty \left(\frac{1}{p(x)}\left[\frac{d}{dx}\left(q(x)\frac{d}{dx}\right)+V(x)\right]\right)u(x)v(x)p(x)\, dx\\ &=\int\limits_{-\infty}^\infty\left[\frac{d}{dx}\left(q(x)\frac{d}{dx}\right)u(x)+V(x)u(x)\right] v(x)\, dx\\ &=\int\limits_{-\infty}^\infty \left(-q(x)\frac{du}{dx}\frac{dv}{dx}+ V(x)u(x)v(x)\right)\, dx\\ &=\int\limits_{-\infty}^\infty u(x)\left[\frac{d}{dx}\left(q(x)\frac{d}{dx}\right)+V(x)\right]v(x)\, dx\\ &=\int\limits_{-\infty}^\infty u(x)\frac{1}{p(x)}\left[\frac{d}{dx}\left(q(x)\frac{d}{dx}\right)+V(x)\right]v(x)p(x)\, dx\\ &=(u,Lv)_{L^2(\mathbb{R},p(x) dx)}, \end{align*} which shows symmetry. If you didn't weight the inner product, the derivatives wouldn't move over as nicely, due to the $1/p(x).$