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I was wondering about the following 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)?

Thank you!

Note that I have cross-posted this to Phys.SE.

Andrews
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    All I've managed to get is this: the TQFT must be invariant under time reparameterizations, the reason being pretty obvious. –  Mar 06 '14 at 01:09
  • Cross-posted to http://physics.stackexchange.com/q/102371/2451 – Qmechanic Mar 07 '14 at 00:11
  • @SanathDevalapurkar, time reparametrization invariance forces the Hamiltonian to be a constraint; a great review of this can be found in Henneaux and Teitelboim's Quantization of Gauge Systems, viz. chapter 4. – Alex Nelson Mar 08 '14 at 01:41
  • @AlexNelson True. A vanishing Hamiltonian imposes time reparametrization invariance. –  Mar 08 '14 at 01:42

1 Answers1

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For your first question:

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

Why is this so?

It's because in a Schwarz-type topological field theory, there is no preferred notion of time...some you have time re-parametrization invariance as a symmetry. Well, to be strictly honest, the metric doesn't even appear in their Lagrangians...usually it's just several differential forms (not their time derivatives) wedged together.

Hence they don't depend on any choice of time coordinate.

Really, they boil down to computing topological invariants of some sort. That makes them easier to handle :)

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

Great question! I asked that myself when I started learning this stuff.

The Wikipedia article on TQFT is misleading a bit, they are handwavy-ily implying "Since $H=0$, and quantum theory has its time evolution determined by the operator $\exp(-\mathrm{i}\delta t\,H)|\psi(t,x)\rangle = |\psi(t+\delta t, x)\rangle$, then obviously trying to do that here won't work."

(Note that $\exp(h\mathrm{d}/\mathrm{d}x)f(x)=f(x+h)$ is precisely the Taylor series about $x$.)

That's kind of vacuous. For topological quantum field theories, we aren't interested in that question. We're interested in what sort of topological invariants the field theory gives us.

Basically, the result from the partition function for a topological field theory doesn't depend on anything other than the topology of the domain of integration.

References

  1. R. K. Kaul, T. R. Govindarajan, P. Ramadevi, "Schwarz Type Topological Quantum Field Theories." Eprint arXiv:hep-th/0504100
  2. Danny Birmingham, et al., "Geometry and Quantization of Topological Field Theories". Preprint
  3. Albert Schwarz, "Topological Quantum Field Theory." Eprint arXiv:hep-th/0011260

Addendum

I would also like to note when considering a parametrization-invariant system, $t$ acts (usually) as a coordinate and not as "proper time".

(Caveat: for the parametrized Newtonian particle, or any parametrized Newtonian system, these two coincide usually. Also, when time transforms "like a connection" instead of a scalar, the Hamiltonian constraint is whack.)

When $\tau$ is the parameter describing the position of the particle along its trajectory, the Hamiltonian operator is

$$\hat{H}\psi = i\hbar\frac{\partial}{\partial\tau}\psi=0.$$

Exercise. Consider the (special) relativistic particle. Its action is $$S = -m\int^{\tau_{2}}_{\tau_{1}}\sqrt{-\dot{x}^{2}}\,\mathrm{d}\tau$$ where $$\dot{x}^{2}=\eta_{\alpha\beta}\frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\tau}\frac{\mathrm{d}x^{\beta}}{\mathrm{d}\tau}.$$ Prove:

  1. This action is reparametrization-invariant: $\tau\to f(\tau)$

  2. Find the canonical momenta. Prove it obeys the mass-shell constraint.

  3. Find the Hamiltonian. Prove it is a constraint. Also prove it is a multiple of the mass-shell constraint.

  4. Prove the Dirac quantized Hamiltonian constraint is precisely the Klein-Gordon equation.

Remark. Not all systems with a Hamiltonian constraint can be "decoupled" like this, i.e., are equivalent to an unconstrained system with interesting dynamics. It "seems like" we can't do this trick with General Relativity. See, e.g., C.G. Torre, "Is General Relativity an 'Already Parametrized' Theory?" Phys.Rev.D 46 (1993) 3231-3234; arXiv:hep-th/9204014. (End of Remark)

Alex Nelson
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  • +1! But, what if one is interested in the dynamics (I'm not - I'm just curious) of the TQFT? What then? Also, I'm curious - where do you study? –  Mar 08 '14 at 03:23
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    In, e.g., 1+1 dimensional TQFT, dynamics is done by specifying the number of loops you begin with at time $t=0$, and how many you have at $t=1$, as well as the topology of the world-sheet for $0<t<1$. BUT the partition function (controlling dynamics) then becomes a function of the topological invariants (which invariants depends on the TQFT). This is good for, e.g., BF-theory since computing topological invariants is simpler than, say, solving the Wheeler-DeWitt equation :) – Alex Nelson Mar 08 '14 at 22:13
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    @SanathDevalapurkar, also, where I'm studying -- I studied at UC Davis as an undergraduate. My current situation is rather strange (not that I'm private about it, I just cannot describe it in 140 characters!). I still study quantum gravity, though :) – Alex Nelson Mar 08 '14 at 22:14
  • Thanks a lot for your answers! –  Mar 08 '14 at 22:28