4
$\begingroup$

Consider a particular compact 2D symplectic manifold $\mathcal{M}$ defined as follows:

  • The topology of $\mathcal{M}$ is a 2-torus.
  • Let $\theta$ and $\varphi$ be the coordinate patch on $\mathcal{M}$ with the following identifications: $\theta+ 2 \pi m \sim \theta$ and $\varphi + 2 \pi m \sim \varphi$ for integer $m \in \mathbb{Z}$. In another words, the torus was formed by gluing the opposite edges of a square $[0,2\pi)^2$.
  • The symplectic form $\omega$ is given by $$\omega = d\varphi \wedge d\theta, \quad d\omega = 0.$$

Since the torus is flat, such a form exists globally.

Question: I want to see through the quantization of this symplectic manifold. According to literature, this should result in a finite-dimensional Hilbert space. How can this be shown explicitly? How can the matrices corresponding to $\hat{\theta}$ and $\hat{\varphi}$ be computed? How does this construction depend on $\hbar$?

Comment: I am sure that the Hilbert space should be finite-dimensional, because I've seen this claim made many times by multiple authors. E.g. Witten mentions it in his "QFT and the Jones polynomial" (see pdf, page 18/367, last paragraph before section 3.1).

Update: one can calculate the relevant $C^{*}$ algebra of observables by taking the basis of functions over the phase space:

$$ W_{n,m} = e^{i n \varphi + i m \theta} $$

and computing the Moyal bracket. Using the Kontsevich formula and making use of the fact that $\partial_{\mu} \omega^{\nu \sigma} = 0$ (because the torus is flat), one obtains the exponential:

$$ W_{n,m} * W_{n',m'} = e^{\frac{1}{2} i \hbar (m n' - n m')} W_{n+n', m+m'}. $$

(The involution is given by $W_{n,m}^{*} = W_{-n,-m}$).

One can then hope that this $C^{*}$ algebra acts naturally on some finite-dimensional Hilbert space, but I wasn't able to obtain such a space. I wonder if GNS construction can be employed here. Tried it for the following state: $$ \rho(W_{n, m}) = \delta_{n0} \delta_{m0}, $$

but the resulting Hilbert space seems to be infinite-dimensional. Not sure how to proceed further.

$\endgroup$
  • $\begingroup$ More on torus phase space: physics.stackexchange.com/q/126676/2451 $\endgroup$ – Qmechanic Mar 24 '18 at 9:24
  • $\begingroup$ @Qmechanic there is nothing about quantization there. $\endgroup$ – Prof. Legolasov Mar 24 '18 at 9:47
  • $\begingroup$ @CosmasZachos in Witten's own words, "Since in quantum mechanics there is one quantum state per unit volume in classical phase space, the finiteness of the volume of $\mathcal{M}$ means that the quantum Hubert spaces will be finite dimensional." $\endgroup$ – Prof. Legolasov Mar 24 '18 at 14:26
  • 1
    $\begingroup$ This paper answers all your questions: arxiv.org/abs/1503.00597 (the way you get a finite dimensional rep from your C* algebra above is just like how Verma modules split off finite dimensional pieces in usual rep theory when you have a integral weight) $\endgroup$ – Ryan Thorngren Mar 24 '18 at 15:08
  • 1
    $\begingroup$ @RyanThorngren thanks! The title sounds exactly like smth I'm looking for. How about posting an answer with the link so that I can upvote&accept? $\endgroup$ – Prof. Legolasov Mar 24 '18 at 15:09
2
$\begingroup$

Here is a paper (pdf) which discusses quantization of the torus phase space. In particular in eqn (24), they consider exactly the basis functions you considered. In section 4, in particular eqns 33 and 34, they show that the naive representation of this algebra on L^2(T^2) has a finite dimensional sub-representation that gives the physical quantization.

By the way, and since you had the other question about Chern-Simons, this theory of the quantum torus is exactly the same as in the "orbit method" representation theory of Lie groups actually, where one naively has an infinite dimensional Verma module, but a finite dimensional irrep splits off from it when you have an integral weight, which in the torus case corresponds to a torus with integral symplectic form. A connection between the two is used in Witten's paper on QFT and the Jones Polynomial to insert punctures into the Hilbert space.

In particular, while Heisenberg uncertainty seems to imply that a compact phase space has finitely many states, you also need the phase space volume to be exactly an integer multiple of $\hbar^n$, where $2n$ is the dimension of phase space, otherwise strange, unphysical things happen, like your Verma modules never closing. This phenomenon is exactly why observables end up being quantized.

$\endgroup$
  • $\begingroup$ Thanks for the link! Could you please expand your second paragraph? I've only heard about Verma modules in context of representation theory of the Virasoro algebra, but there's no finite-dimensional representations there afaik. I would greatly appreciate if you could post a real example of how one can obtain a finite-dimensional rep from my $C^{*}$ algebra. (See, I've seen to many people use analogies when it comes to TQFT, and IMO this is not at all useful, in fact, quite the opposite. I don't like the "just like with lowest Landau Level" or "just like Verma modules" explanations). $\endgroup$ – Prof. Legolasov Mar 24 '18 at 15:30
  • $\begingroup$ @CosmasZachos please understand that most of those words don't mean hamburgers to me... If you indeed understand how the finite-dim representation is derived from the algebra (and some ad-hoc assumptions, whatever), please, please, I'm begging you, consider simply posting a draft derivation from the point "we have the algebra" to the point "we have a Hilbert space" with no handwaving, analogies and general words that don't really mean anything. You don't have to carefully work the math, I can do that myself, just draft the key steps... I would be so much grateful, you couldn't imagine. $\endgroup$ – Prof. Legolasov Mar 24 '18 at 17:28
  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Mar 24 '18 at 18:03
  • $\begingroup$ @Cosmas Zachos: Thanks. Actually, I was referring to the Ryan Thorngren's answer. $\endgroup$ – Qmechanic Mar 24 '18 at 18:20
  • $\begingroup$ Virasoro does have finite dimensional irreps for c < 1 and they split off from the Verma modules for highest weight states. This is described in the big yellow book. I'd also recommend Kirillov's book on the orbit method. What I'm saying is totally precise by the way, not an analogy. The torus you're talking about is a coadjoint orbit of the abelian Lie group U(1) x U(1), and the character variety is a kind of generalized coadjoint orbit. $\endgroup$ – Ryan Thorngren Mar 25 '18 at 6:17
2
$\begingroup$

OK, as per intense request, I'll review some standard facts on the Moyal algebra, summarized in our (1990) paper and my 1989 Tahoe conference talk. I'm not sure they do what you are after, and they certainly don't get into subtleties of geometric quantization, Jones polynomials, or Morse theories. Just basic, primitive formal facts.

Starting from your Fourier modes on the torus, for integer m, n, $$ W_{m,n}=e^{im\phi + in\theta}, $$ we use the flat ("Moyal") $\star$-product introduced by Groenewold, with $\hbar\mapsto 2k$, $x\mapsto \phi$, $p \mapsto \theta$, to define the pseudodifferential operator $$ \mathfrak {W}_{m,n}=\frac{1}{2ik}W_{m,n} \star ~~. $$

It provides a realization of the infinite-dimensional Moyal Lie algebra, $$ [\mathfrak {W}_{m,n} \mathfrak {W}_{m',n'}]= \frac{\sin k(mn'-nm')}{k} ~\mathfrak {W}_{m+m,n+n'} ~. $$

If, for whatever reason, one focuses on special ks, $$ k=2\pi/N , $$ where N is an integer, (take it prime to avoid questions and fuss), you note that the structure constants' integer indices identify modulo N; so we can identify the integer indices m, n with their correspondents (differing by an integer multiple of N) inside an integer N×N fundamental lattice around the origin. Moreover, $\mathfrak {W}_{0,0}$ and its correspondents don't appear on the r-h side and commute with the rest, so we drop them.

Folding over corresponding generators, ad infinitum, results in a finite-dimensional subalgebra of this Moyal algebra, $$ {\cal W}_{m,n}\equiv\sum_{s,t} \mathfrak {W}_{m+Ns,n+Nt}~, $$ where we've summed over all integers s and t, ahem!, to produce "lattice average" generators... There are less sleazy ways of arranging that, but the field has been churning for over a quarter century, now... I'm sure Wootters and Vourdas reviews would cover such things adequately.

I've left the physics out, precisely because the topological arguments are subtle, but, then again, I understood the request as the basic "if, then" statements, and the above here is a standard maneuver. N just came out of a hat, somebody's physics hat, strenuously avoided here.

But the point is this algebra of the $\cal{W}$s, with the same structure constants, is now a finite-dimensional algebra of dimension $N^2-1$, and it should not be a surprise that it includes all the classical Lie algebras, best seen in Sylvester's 1867 clock and shift matrix basis. The fundamental rep is N×N matrices acting on N-vectors. The large N limit, with suitable normalizations of the generators, results into the Poisson Bracket algebra, as a contraction of the Moyal one, but that is probably beyond the point.

$\endgroup$
  • $\begingroup$ Thanks, this actually makes sense (though the main idea is pulled out of a hat indeed) – I see how the Hilbert space arises, step by step. If only TQFT authors could do the same, instead of resorting to analogies. $\endgroup$ – Prof. Legolasov Mar 25 '18 at 11:33
  • 1
    $\begingroup$ I suspect it is cultural: they are anxious to appear "cutting edge" and throw in colorful code "you know"s to reassure themselves of their imagined position in the parade... ignore it... $\endgroup$ – Cosmas Zachos Mar 25 '18 at 12:57
  • $\begingroup$ ^^^^^^^^^^^^^lol $\endgroup$ – Ryan Thorngren Mar 26 '18 at 2:16
  • $\begingroup$ Thanks again for this answer. I had some time to rethink this through, and now I think I understand this problem much better because of people like you and Ryan Thorngren. Actually, now I think there's three special cases to this problem: (a) when $N = a b / 2 \pi \hbar$ is an integer (which indeed gives a finite-dim space), (b) when it is rational and (c) when it is irrational. I would expect (b) to still give a finite-dim vector space, and (c) to give an infinite-dim vector space. However, the literature claims that (b) and (c) are inconsistent. Could you clarify this? $\endgroup$ – Prof. Legolasov Mar 31 '18 at 12:42
  • $\begingroup$ No, not really... Rational smells a normalization away from integer... But, no, no insights there.... $\endgroup$ – Cosmas Zachos Mar 31 '18 at 12:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.