Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've asked this on Math.SE, but with no avail. So, I decided to ask it here.

I was wondering about the following after reading the Wikipedia article on TQFTs.

It is said that TQFTs have vanishing Hamiltonians $\hat{\mathcal{H}}$. Firstly, I would like to ask:

Why is this so?

Secondly, consider the Schroedinger equation: $$i\hbar\dfrac{\partial}{\partial t}\Psi=\hat{\mathcal{H}}\Psi=0$$ So, there are no dynamics or propagation. This leads to my second question:

What does one mean by "nontrivial dynamics/propagation" (as mentioned in the Wikipedia article)?

I've understood the TQFT must be invariant under time reparameterizations, the reason being pretty obvious.

Thank you!

share|cite|improve this question
Cross-posted from – Qmechanic Mar 7 '14 at 0:11
@Qmechanic Yes, I'll add the link in the post. – Sanath K. Devalapurkar Mar 7 '14 at 0:11
Note that the preferred procedure is to not cross-post, but to flag for migration. And if one does cross-post, one should as a minimum link to the other post (on both sides!) in order not to waste potential answerer's time. – Qmechanic Mar 7 '14 at 0:16
Since you originally posted it on MathSE, I reckon that you might be interested in a mathematical perspective, in which case you might want to check out Atiyah's introduction of the categorical approach to TQFT, for starters. from there, you could go onto something like this paper by Baez and Dolan using higher categories, and ultimately something like Lurie's classification of $n$-dim TQFTs. – Ralph Mellish Mar 7 '14 at 2:02
Well, the reason I put a comment instead of an answer is since I didn't really give an answer, so much as a few links that possibly contain an answer. If I have time later, I will try to type up a legitimate answer (although you might want to hold out for a bit to see if Urs Schreiber shows up -- he is much, much more qualified to give such an answer than me :) ). – Ralph Mellish Mar 7 '14 at 2:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.