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seen Dec 16 at 1:29

Dec
1
comment What is the Difference Between BCS Theory and Ginzburg-Landau Theory?
Ginzburg-Landau theory is the low energy effective theory that describes the macroscopic phenomenology of superconductivity. But it does not say how from microscopic interactions the SC state came to be. This is what BCS theory does for so-called conventional SC's. So the difference is macroscopic vs microscopic. You can essentially derive Ginzburg-Landau theory from BCS theory. However, the Ginzburg-Landau theory is still valid for High T_C SC's. But we don't know a general microscopic theory for those (BCS dosen't work there).
Nov
20
awarded  Good Answer
Nov
12
comment Why isn't the path integral defined for non homotopic paths?
The path integral is defined as a sum over all possible paths. In order to do such a summation, you have to integrate within each homotopy class and then sum up all homotopy classes. This is also what the paper states in Theorem 3.1.
Oct
26
awarded  Revival
Sep
12
comment What is the mathematical reason for topological edge states?
Very nice approach!
Aug
24
awarded  Nice Question
May
9
comment When can we take the Brillouin zone to be a sphere?
@JiaYiyang I have been very busy lately, but I will try to write an answer over the weekend. If I forget write a comment with @ to remind me. But the short answer is (2) but with some subtleties. The BZ is a torus, but if you instead think of it as a sphere you only get what people call strong topological insulators. If you take a torus, you get strong AND weak topological insulators, the reason for that is some complicated algebraic topology. But since the strong one is more interesting, we can for simplicity take the sphere. The weak ones are not interesting since they are not really robust.
Apr
22
comment Number of Grassmann generators for Dirac field?
I had not seen your earlier comment, sorry for not replying. I think its always good to add @ to make sure people notice your comment. I have added an answer with a few more details.
Apr
22
answered Number of Grassmann generators for Dirac field?
Apr
11
revised Topological insulators: why K-theory classification rather than homotopy classification?
added 1 characters in body
Apr
1
comment Lie group Homomorphism $SU(2) \to SO(3)$
In this language, $\phi_\star$ is essentially the differential of $\phi$ (at identity), sometimes also called the pushforward (this is what the star in the index means). en.wikipedia.org/wiki/Pushforward_(differential) This question is probably more appropriate at math.SE.
Mar
20
comment Topological insulators: why K-theory classification rather than homotopy classification?
Yes, just parametrizing the space of gapped Hamiltonians is extremely hard if you go beyond free theories. But this is exactly what Kitaev is doing for free theories, he calls these spaces for classifying spaces (modulo subtleties I will not get into).
Mar
20
comment Topological insulators: why K-theory classification rather than homotopy classification?
Note however that what I describe is in general very hard to do. It can be done if you restrict $\mathcal M$ to be only the space of gapped fermionic Hamiltonians, which are non-interacting and possibly have certain symmetry (time-reversal, charge conservation, particle-hole). This will lead to homotopy classification of topological insulators. But weakening the equivalence notion as described in the answer, you end up with the K-theory classification. The general classification for interacting theories are under active research, but by very different methods.
Mar
20
comment Topological insulators: why K-theory classification rather than homotopy classification?
Homotopies are deformations of the Hamiltonian that do NOT close the gap, because we are interested in the manifold of gapped Hamiltonians. Whenever the gap closes, it can be a critical point/phase transition between two topologically distinct phases. So your reasoning is correct.
Mar
20
comment Topological insulators: why K-theory classification rather than homotopy classification?
(continued) If $\mathcal M$ is path-wise connected, then we can deform (by local perturbations) any gapped Hamiltonian into any other without closing the gap and there is thus only one phase. If the space is not connected, then each connected component of $\mathcal M$ corresponds to a distinct topological phase, because you cannot deform a Hamiltonian from one connected component into one from another component without closing the gap! In other words, on the manifold $\mathcal M$ any perturbation that closes the gap is NOT a continues deformation and thus not homotopy! (continued)
Mar
20
comment Topological insulators: why K-theory classification rather than homotopy classification?
@JiaYiyang Your reasoning is completely correct, but it is taken into account in both the homotopy and K-theory classifications. One way to think about the problem is that we are not interested in the space of all Hamiltonian $\mathcal A$, but only the sub-manifold containing gapped Hamiltonians only $\mathcal M\subset\mathcal A$ (insulators are gapped by definition). Any gapped Hamiltonian is just a point of this manifold $H\in\mathcal M$. The interesting thing is now is the topology of $\mathcal M$, in particular is this space connected or not (as measured by $\pi_0(\mathcal M)$). (cont.)
Mar
20
answered Topological insulators: why K-theory classification rather than homotopy classification?
Jan
19
awarded  Yearling
Dec
28
comment Is there a critical order of the Abelian gauge theory in (2+1)D
This line of thinking seem to imply that for larger $n$, the $\mathbb Z_n$ topological state is less stable against perturbations that induce transition to the trivial (confined) phase.
Dec
28
comment Is there a critical order of the Abelian gauge theory in (2+1)D
Just some random comments. I guess that all $\mathbb Z_n$ gauge theories have both a confined and deconfined phases, one can for example write down toric code like models for all $\mathbb Z_n$ and consider perturbations (like string tension/magnetic field) which will at some point induce a phase transition. So instead of the existence of a critical $n_c$, isn't it more natural to suspect that the size of the deconfined $\mathbb Z_n$ phase (as a function of perturbations), shrinks as a function of $n$? Such that in the limit $n\rightarrow\infty$, the deconfined phase vanishes gradually?