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!

  • $\begingroup$ 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
    Commented Feb 28, 2016 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
    Commented Feb 28, 2016 at 17:48

1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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