QM particle on a ring

Stream: PhysLean

[Joseph Tooby-Smith]

I think it might be a fun and useful exercise to formalize the QM system corresponding to a particle on a ring.

• https://en.wikipedia.org/wiki/Particle_in_a_ring

I think this might be useful since it will give an example of how to do derivatives not on $\mathbb{R}^n$.

I think there are two questions to get it started:

1. What is the correct hilbert space?
2. Given a function $\psi : S^1 \to \mathbb{C}$, what is the correct implementation of the derivative $\frac{\partial \psi}{\partial \theta}$?

[Alex Meiburg]

I think the correct way to do this is to talk about the space as the manifold S1, and then you can use docs#mfderiv for the derivative.

[Alex Meiburg]

I think the correct Hilbert space here is the set of smooth functions.

the Schwartz space, a vector space of smooth functions stable under the Fourier transform.

[Joseph Tooby-Smith]

I'm slightly worried that the Schwartz space of functions doesn't actually form a hilbert space - I'm pretty sure it doesn't for $\mathbb{R}$, but things might be different for $S^1$?

[Michiel Huttener]

Indeed, it does not. Its natural topology (via suprema of derivatives) makes it a Fréchet space, but is it not normed since one norm cannot take into account all derivatives at once. The argument works both on $\R$ and $S^1$.
Also note that the fourier transform will map functions on $S^1$ to sequences on $\Z$, so "stable under the fourier transform" does not make sense, I think. (The sequences belonging to smooth functions are precisely those with polynomial decay.)

[Michiel Huttener]

(Disclaimer: I know barely anything about physics. :upside_down: )

[Michiel Huttener]

Don't people usually start from the operator (here: fourier transform?) on some Sobolev space and then consider its closure?

[Joseph Tooby-Smith]

Physicists aren't too careful about these things, they just tend to talk vaguely about "square integrable functions". The operator I want is really an unbounded operator that represents differentiation, the eigenvalues of which correspond to $e^{ixn}$.
The fourier transform should come in with regard to the completion of the $e^{ixn}$ in the Hilbert space.

To me starting with the operator seems the wrong way around, but I can't claim whether this is something people do or not.

[Michiel Huttener]

Ok, so is the goal more or less to show the spectral theorem for self-adjoint unbounded in the case of the derivative on $L^2(S^1)$, and the interesting part from a physicist's perspective is that the Fourier transform is the intertwiner that lets one view derivation as multiplication (on the Fourier side), so one can then solve "the" Schrödinger equation?

[Joseph Tooby-Smith]

Yes exactly! I think this is the correct interpretation of things.

[Michiel Huttener]

Great! This seems like a fun little project to work on. (I won't claim it yet :wink: but I'll revisit this thread in a couple of weeks if I get some more Lean experience.)