Why does the energy gap of sublattice-symmetric systems never close?

Question

I am studying from this famous site some symmetries useful for topological quantum matter. At some point, talking about the particle-hole symmetry, it states:

You can however notice that, unlike in the case of sublattice symmetry, energy levels do not repel around zero energy, so that crossings at zero energy appear.

However, in the paragraph about sublattice symmetry, this is poorly discussed. It just says that:

[...] if $(\psi_A,\psi_B)^T$ is an eigenvector of the Hamiltonian with energy ε , then $(\psi_A,-\psi_B)^T$ is an eigenvector with energy −ε . A symmetric spectrum is the consequence of sublattice symmetry.

What does this mean for the topological classification? Clearly, the number of states with negative energy is the same as the number of states with positive energy, and that means we don’t ever expect a single level to cross zero energy.

[...] Indeed, we can deform all the Hamiltonians with sublattice symmetry into one another without closing the gap. This means that an extra symmetry may render topological classification trivial.

I really don't understand these last four raws and why if a system presents sublattice symmetry, this prevents ε from being null. This is more diffucult to understand if I think to the Kitaev chain, which present both ph and sublattice symmetry, and still has conditions of gap closing.

Can someone please explain this to me and maybe give me a more accurate definition of sublattice symmetry to help me understand?

Answer 1

Due to the off-diagonal terms which are always present in any Hamiltonian with sublattice symmetry, there is a phenomenon of level repulsion. This means that if two different eigenstates get close to each other in energy, there is a repulsion between the two energy eigenvalues, such that they never cross.

As an exercise, calculate explicitly the eigenvectors and eigenvalues of a Hamiltonian given by: $$H'' = [[E,W],[W^*,E]]$$ where $W$ is any complex number representing the interaction between sublattices and $E$ is a real number representing the energy of the individual sublattices. The eigenvalues will be $ E \pm |W|$. Even setting $E = 0$ would not make the eigenvalues degenerate.

For sublattice symmetry, $U = \sigma_z$. It is easy to verify that for the hamiltonian $$H = [[0, H_{AB}],[H^\dagger_{AB}, 0]]$$ $UHU^\dagger = -H$, if $\psi = [\psi_1, \psi_2]^T$ is an eigenvector with eigen energy $E$, then so is $\psi' = [\psi_1, -\psi_2]^T$ with eigen energy $-E$. We now have two independent set of eigen equations: $$H_{AB} \psi_2 = E \psi_1; \quad H^\dagger_{AB} \psi_1 = E \psi_2$$ and $$H_{AB} \psi_1 = -E \psi_2;\quad - H^\dagger_{AB} \psi_2 = E \psi_1$$

Due to sublattice symmetry, the Hamiltonian is necessarily symmetric around the fermi level, so the energies of the symmetric partners can only be degenerate at the fermi level. Whenever one of the bands approaches the fermi level, so does its symmetric partner. This then creates level repulsion, so that the bands never actually have degenerate energy. Thus, the bands never cross the fermi level.

Contrast this with the case of particle-hole symmetry exhibited by the BdG Hamiltonian $H'$. A similar phenomenon occurs, where the bands are symmetric around the fermi-level. But this time, it is not two different eigenstates! It is rather a single eigenstate. This is due to the fact that the operator that implements the particle-hole symmetry is an antiunitary operator (as opposed to a unitary operator in the case of sublattice symmetry). Let's check that.

Now for particle hole symmetry, $U = i \sigma_z K$. It is easy to verify that since $UH'U^\dagger = -H'$, if $\psi = [\psi_1, \psi_2]^T$ is an eigenvector, then so is $\psi' = [\psi_2^*, \psi_1^*]^T$. We now only have a single independent set of eigen equation: $$H \psi_2^* + \Delta \psi_1^* = -E \psi_2^*; \quad \Delta^* \psi_2^* + H^* \psi_1^* = E \psi_1^*$$ The other eigenequation is the complex conjugate of this one.

So crossings can occur around the Fermi surface since level repulsion only applies for crossings between different eigenstates, and here we only have a single eigenstate.

Answer 2

Well, am I really late to the party!

I think the OP's interpretation of those quoted passages is wrong. (Though neither do I think they convey the intended meaning clearly.)

Using the notation adopted by that famous site, the class of Hamiltonian with S-symmetry is every that satisfies $$ \sigma_z H \sigma_z = - H. $$ It should be easy to come up with any number of examples where a Hamiltonian in the class is actually gapless. So S-symmetry does not prevent gap closing, per se.

What it does, and what the famous site tries to say, is that: if you have two gapped Hamiltonians in the class, then you can deform one into the other, without closing the gap, while respecting the S-symmetry all the way.

There are gapless Hamiltonians in the class. But you can always avoid them as long as they are not your beginning and end points.

One simple example is the Dirac Hamiltonian in 1D: $$ H = \sigma_x (-i\partial_x) + \sigma_z m. $$ Here the sublattice transformation is instead $\sigma_y$. Put the system on a periodic lattice and replace $-i\partial_x$ with a nearest neighbor hopping matrix if you are allergic to models in continuum space. This thing has not only $S = \sigma_y$, but also time reversal $T = \sigma_z$ and charge conjugation $C = \sigma_x$. If you insist on keeping all $S$, $C$ and $T$, the only way to change the sign $m \rightarrow -m$ is to go through $m = 0$.

But if you are fine with breaking $T$ and $C$, you can first deform the Hamiltonian into $$ H \rightarrow \sigma_x \left [(-i\partial_x)^2 + A\right ] + \sigma_z m, $$ where $A \neq 0$ is a constant. Now $m$ can be brought through zero without gap closing.