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I often hear: "The degenerate groundstate subspace of a QFT is often a TQFT".

I'm trying to work out an example of this for, say, superconductors: In the context of condensed matter physics, the spacetime of a 2D superconductor is

$$\Sigma\times \mathbb R$$

Where $\Sigma$ is some compact, oriented 2-manifold. Now let's consider the Ginzburg-Landau model, which is a good effective QFT for a superconductor:

$$S_\text{GL}(\phi,A)\equiv \int_{\Sigma\times \mathbb R} (|\nabla^{2A}\phi|^2+|F_A|^2)\,d^3x$$

Now this model has classical solution space $H^1(\Sigma;\mathbb Z_2)$, based off of recognizing that, classically, $\nabla^{2A}$ is a flat spin connection and thus we are just counting spin structures on $\Sigma$. Moreover, this action is supposed to represent, at $T=0$, the groundstate energy of a superconductor.

Anyways, my questions are, (1), does this mean that the classical solutions of the GL model correspond to quantum mechanical groundstates? (2) how is the GL model a topological quantum field theory (because I see a $U(1)$ Yang-Mills term and a Bochner Laplacian)?

If the GL model is a TQFT, then if we write $$Z_\text{GL}(\Sigma\times [0,t])=\int_{C^\infty(\cdots)} D\phi\,DA\, \exp(iS_\text{GL}/\hbar),$$ This is supposed to represent a linear map $U(t):H^1(\Sigma;\mathbb Z_2)\to H^1(\Sigma;\mathbb Z_2)$. However, since we're on a cylinder, the time-evolution should be trivial, no?

Also: Where do we get all the nice quasiparticle braiding? if we add quasiparticles to the action, they certainly do not affect the topology of $\Sigma$ (they're not massive enough). But then, because of diffeomorphism-invariance, quasiparticle processes do not affect the partition function - so we can't switch between groundstates - and so topological quantum computing is impossible!

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  • $\begingroup$ It is possible to write a TQFT for a superconductor, see Hansson's et al papers. arxiv.org/abs/cond-mat/0404327 and arxiv.org/abs/1105.5031. To obtain an effective TQFT at low energies, you might at least integrate the fermionic part of the action. It is not clear how to do that from the Ginzburg-Landau formalism. The ground state of a superconductor is degenerate, see Greiter's paper arxiv.org/abs/cond-mat/0503400 for instance. I can not help you for the part concerning the braiding. I do not remember if Hansson's et al discuss this point but please refresh my mind :-) $\endgroup$ – FraSchelle Feb 28 '16 at 17:38
  • $\begingroup$ About the braiding, I'm just remembering that they might well be trivial in trivial superconductors, say s-wave, as one could in principle check around vortices (I'm not sure it exists experiments about that), following Caroli-Matricon-deGennes. About p-wave superconductor, perhaps a beginning of an answer for you would be the paper by Ivanov: arxiv.org/abs/cond-mat/0005069 (the paper form Caroli et al. is cited in this one). Have no hesitation to make this interesting point an other separated question. $\endgroup$ – FraSchelle Feb 28 '16 at 17:48
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One can analyze this in a limit where the LG potential is very strong, ie. we study $$\int |\nabla_A \phi|^2 + |dA|^2 + g^2(|\phi|^2 - a^2)^2$$

with $g^2 >> 1$. We separate $\phi$ into amplitude and phase parts separately, and the amplitude fluctuations of $\phi$ are very massive, centered around their vev $|\phi|=a$. Meanwhile, the phase fluctuations are quick. The effective Lagrangian for the phase $\theta$ is

$$a^2|d\theta - 2A|^2 + |dA|^2.$$

Now we study the limit $a^2 >>1$, for which we obtain the constraint

$$2A = d\theta.$$

We see that this forces $dA = 0$, which knocks out the Maxwell term, and also forces $A$ to have holonomy $\int A = 0$ or $\pi$ around all cycles. So we see that the TQFT we get is the (untwisted) $\mathbb{Z}_2$ gauge theory. For more calculations like these, I recommend https://arxiv.org/abs/1307.4793 and https://arxiv.org/abs/1308.2926 .

To actually "see" the nontrivial braiding takes a bit of work. In this case the theory has a fermionic quasiparticle, which is a bound state between the unit charge

$$\exp i \int_\gamma A$$

and the $\pi$-flux, which is a disorder operator along a worldline $\gamma'$ which forces $\int_S dA = \pi$ through every surface $S$ intersecting $\gamma'$, so you can think of it as a narrow flux tube around $\gamma'$ containing $\pi$ units of magnetic flux. It's easy to see that this braids the Wilson loop above with a minus sign, and therefore that their bound state is a fermion (they are both bosons on their own, which can also be checked).

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