So far I have learned about topological quantum material, my understanding is that topological order in a quantum material is the way the eigenvectors of the Hamiltonian of the system aligned. So if I am right, I need to know how this topological vector space differs from the eigenvector space of Hamiltonian which describes a normal material that is not topological.
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There are different ways for a material to be called topological, but let's take the example of a two-band Chern insulator. While in the non-topological case, we have no problem writing down a valence band wavefunction $\psi_k$ which is well-behaved over the whole Brillouin zone. In the topological case, it turns out that the non-zero Chern number indicates that we cannot write down a single wavefunction which behaves nicely over the entire Brillouin zone. There is a topological obstruction.
The more precise statement is that a non-zero Chern number indicates an obstruction to a globally smooth choice of periodic gauge. For a recap on how gauge plays a role, see below.
First recall that a Bloch wavefunction must satisfy the Bloch condition: $$ \psi_{nk}(r-R) = e^{-ik\cdot R }\psi_{nk}(r) $$ where $R$ is a lattice vector. We can write down a generic form which satisfies this: $ \psi_{nk}(r) = e^{ik\cdot r}u_{nk}(r) $ where $u_{nk}(r-R)=u_{nk}(r)$. However there is still freedom in this. We can always make the gauge transformation: $$\psi_{nk}\mapsto e^{i\phi(k)}\psi_{nk}$$ for some $\phi(k)$ and the Bloch condition will still be satisfied. Typically we also require $\psi_{n,k+G}=\psi_{nk}$ so we want $\phi(k+G)=\phi(k)$. A non-zero Chern number means that we cannot use a single $\phi(k)$ for the entire BZ. Rather we need two $\phi_1,\phi_2$ which are valid on different patches of the BZ which we glue together at their boundary.
