Whenever one talks about topological insulators, the breaking of time reversal symmetry is always mentioned.
Is there an intuitive reason as to why one need time reversal symmetry to be broken in order for topological effects to appear?
$\begingroup$
$\endgroup$
2
asked Aug 14, 2018 at 23:11
-
$\begingroup$ There are time-reversal invariant topological insulators (for example the $\mathbb{Z}_2$ insulator in 2D protected by time-reversal or the Chern insulator in 4D). Insulators with chiral edge states (such as the Chern insulator in 2D), however, must break time reversal symmetry in the non-trivial phase (since the edge state flips chirality on time reversal). $\endgroup$– Sebastian RieseAug 14, 2018 at 23:28
-
$\begingroup$ The standard topological insulator has symmetry protected topological states, where one of those symmetries is time reverse invariance. On the boundary there can be edge states which have fractional quantum Hall physics. Because of that time reversal invariance is broken. This can be seen in elementary terms with the Lorentz equation with velocity and magnetic field that both flip with time reverse. $\endgroup$– Lawrence B. CrowellAug 14, 2018 at 23:34
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The easiest way to understand this is as follows: Those systems that you hear people talking about are most usually either the IQHE or analogs of it. In that case, you have two space dimensions and the (bulk) invariant is the Hall conductivity. If $P$ is your Fermi projection (derived from a Hamiltonian $H$), then the Hall conductivity is given (up to a constant) by \begin{align} \sigma_H(P) &= \mathrm{index}(PUP+P^{\perp}) \end{align} where $U:=\exp(i \arg(X_1+i X_2))$ and $X_i$ are the position operators, and $\mathrm{index}(F) \equiv\dim\ker F - \dim\ker F^\ast$. Note this formula is equivalent to the usual integral over the Berry curvature and is associated with a magnetic flux insertion at the origin to implement the electric field.
Now if your system has time reversal symmetry, then there is some anti-unitary operator $\Theta$ which squares to $-1$ (for Fermions) which commutes with $H$ and hence also with $P$: $[P,\Theta]=0$. These two facts imply $$P = -\Theta P \Theta\,.$$ Since $\Theta$ commutes with the position operator and it is anti-unitary, we have $\Theta U \Theta = -U^\ast$. Hence with $F:=PUP+P^{\perp}$ we find \begin{align}F &\equiv PUP+P^{\perp} \\ &= (-\Theta P \Theta)U(-\Theta P \Theta)-\Theta P^{\perp}\Theta \\ &= -\Theta(P(-\Theta U \Theta) P + P^\perp)\Theta \\ &= -\Theta(P U^\ast P + P^\perp)\Theta \\ &= -\Theta(P U P + P^\perp)^\ast\Theta \\ &= -\Theta F^\ast \Theta\,.\end{align}
But now one can use a property of $\mathrm{index}$ which is that: (1) It is logarithmic $\mathrm{index}(AB) = \mathrm{index}(A) + \mathrm{index}(B)$; (2) It is zero on isomorphisms $\mathrm{index}(A) = 0 $ if $A$ is an isomorphism; (3) Clearly from $\mathrm{index}(F) \equiv\dim\ker F - \dim\ker F^\ast$ it is obvious that $\mathrm{index}(F^\ast) = -\mathrm{index}(F)$. Since $\Theta$ is an isomorphism as well as $-\mathrm{Id}$ (multiplication by $-1$), we conclude that $$\mathrm{index}(PUP+P^{\perp}) = 0$$ for all time-reversal invariant $P$! That's why people try so hard to break time-reversal. Either that, or look for different (i.e. not the Hall conductivity) invariants.
