- #1

andresB

- 629

- 375

One can construct a Hilbert space "over" ##\mathbb{R}^{3}## by considering the set of square integrable functions ##\int_{\mathbb{R}^{3}}\left|\psi(\mathbf{r})\right|^{2}<\infty##. That's what is done in QM, and there, even if they are not normalizable, to every ##\mathbf{r}\in\mathbb{R}^{3}## we can asign a base ket ##\left|\mathbf{r}\right\rangle ## so that vectors in the Hilbert space can be written as ##\left|\psi\right\rangle =\int_{\mathbb{R}^{3}}\left\langle \mathbf{r}\right|\left.\psi\right\rangle \left|\mathbf{r}\right\rangle d\mathbf{r}=\int_{\mathbb{R}^{3}}\psi(\mathbf{r})\left|\mathbf{r}\right\rangle d\mathbf{r}##.

This can be extended trivially to any ##\mathbb{R}^{3n}## for a collection of (non-identical) particles. It can also be done for the tangent bundle of ##\mathbb{R}^{3}## since the tanget bundle can be identified with ##\mathbb{R}^{6}##.

Actual Question:

On a general manifold, I don't see a reason why volume integral can't be used to define square integrability and inner products so a Hilbert space can be build. Moreover, those concepts are coordinate independent.

But what about the base kets? can a non-normalizable ket (modulo a phase) be defined for each point on the manifold so that every element of the Hilbert space can be written as a superposition of the base kets? If so, I can't think on how to write the base kets in a coordinate-free way.

I suppose this have to be a question people have considered (and answered) already. QM in curved space time is a thing, after all. I'm particularly interested on Hilbert spaces "over" tangent bundles.