Hilbert spaces and kets "over" manifolds

  • A
  • Thread starter andresB
  • Start date
  • #1
385
136
Background:
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.
 

Answers and Replies

  • #2
martinbn
Science Advisor
2,282
782
Even for ##\mathbb R^3##, to make it rigorous needs more work. The functions ## |r\rangle## are not square integrable. You need a bigger space of distributions for that, or the usual Gelfand's triples. All that extends to manifolds.
 
  • #3
385
136
Indeed. I'm aware of the construction of Rigged Hilbert spaces, though at the moment I'm ok with a less rigorous treatment.

Another question: It is even possible to define a self-adjoint position operator over general manifolds?
 
  • #4
martinbn
Science Advisor
2,282
782
Indeed. I'm aware of the construction of Rigged Hilbert spaces, though at the moment I'm ok with a less rigorous treatment.

Another question: It is even possible to define a self-adjoint position operator over general manifolds?
How is it defined in ##\mathbb R^3##?
 
  • #5
385
136
How is it defined in ##\mathbb R^3##?
For a standard construction you can check Quantum Mechanics by Albert Messiah, sections V.8, VII.13 and VIII.6. But I think you are already familiar with it.
 
  • #6
strangerep
Science Advisor
3,243
1,156
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.
Try Birrell & Davies, "Quantum Fields on Curved Space". The real problem is that the Hilbert space you might construct at one spacetime point is unitarily inequivalent to another at an infinitesimally nearby point. (If you're not familiar with the notion of unitary inequivalence, Bogoliubov transformations, and all that, try Umezawa's "Advanced Field Theory".)
 
  • #7
martinbn
Science Advisor
2,282
782
For a standard construction you can check Quantum Mechanics by Albert Messiah, sections V.8, VII.13 and VIII.6. But I think you are already familiar with it.
I have seen it, but it is usually very non rigorus coordinate and basis dependent, so it isn't clear to me what the construction is.
 
  • #8
385
136
I have seen it, but it is usually very non rigorus coordinate and basis dependent, so it isn't clear to me what the construction is.
Well, you are correct. Mathematical rigour is most usually handwaved away. Though textbooks tend to comfort the reader by stating that rigorous treatments do exist. Messiah's book has a small section on distribution theory.

Not sure what you mean by basis dependent, though.
 
  • #9
385
136
Try Birrell & Davies, "Quantum Fields on Curved Space". The real problem is that the Hilbert space you might construct at one spacetime point is unitarily inequivalent to another at an infinitesimally nearby point. (If you're not familiar with the notion of unitary inequivalence, Bogoliubov transformations, and all that, try Umezawa's "Advanced Field Theory".)
Thanks for the reference.

It seems that relativistic considerations create a lot of complications. After a first quick read I think, but I'm not sure, that not having time as a coordinate would make things easier.

I mention it because my interest arises after reading about non-relativistic wave mechanics on curved surfaces https://arxiv.org/abs/1602.00528. And while the wave mechanics formulation has no problem, I wonder if the entire Dirac formalism can be constructed on the surface.
 
  • #10
martinbn
Science Advisor
2,282
782
Well, you are correct. Mathematical rigour is most usually handwaved away. Though textbooks tend to comfort the reader by stating that rigorous treatments do exist. Messiah's book has a small section on distribution theory.

Not sure what you mean by basis dependent, though.
What I have seen about the position operator is that one first defines the position kets ##|x\rangle##, and then the operator through that "basis", defining it to be the operator having these as eigenvectors with the coressponding eignevalues. By the way I couldn't find the part in Messiah, that is specific for the position operator.

ps I am asking because your question is interesting and I would like to learn more, not because I will be able to give you an answer.
 
  • #11
385
136
What I have seen about the position operator is that one first defines the position kets ##|x\rangle##, and then the operator through that "basis", defining it to be the operator having these as eigenvectors with the coressponding eignevalues. By the way I couldn't find the part in Messiah, that is specific for the position operator.

ps I am asking because your question is interesting and I would like to learn more, not because I will be able to give you an answer.

There are several sections dedicated to the formalism of QM in Messiah. Section VIII.6 (of my Dover version, at least) deals with the position operator.

But it is as you say. The process is more or less the following
(1) Define/postulate a complete set of commuting observables. Ignoring spin, the components of the position operator are a complete set.
(2) Define commuting relations with the other observables of the theory. Messiah only deals with momentum.
(3) Find the spectrum of the commuting observables
(4) The Hilbert space is defined as the space of all the linear combination of the eigenvectors of the commuting observables.
(5) Use commutation relations to find the spectrum of all the observables of the theory
(5) check internal consistency


==============
Now, on a general manifold I'm unsure on how to proceed from the very first step. But assuming I can define a position operator on a local chart just as usual, then the second step seems problematic. Usually, the spectrum of the position operator is found using its commuting relation with a self-adjoint momentum operator. But in general coordinates this seems complicated https://aapt.scitation.org/doi/abs/10.1119/1.12675?journalCode=ajp
 

Related Threads on Hilbert spaces and kets "over" manifolds

  • Last Post
Replies
4
Views
3K
Replies
4
Views
2K
Replies
9
Views
3K
Replies
8
Views
1K
  • Last Post
Replies
5
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
11
Views
5K
Replies
63
Views
5K
  • Last Post
Replies
11
Views
5K
Replies
5
Views
2K
Top