# Why surface and bulk states of topological insulators (TI) have different topological invariants and how it leads to conductive states?

I am trying to understand how TIs work. Right now, the main thing I understand is that the surface states and bulk states have different topological invariants. This leads to spin orbit coupling being higher on the surface, allowing for band inversion which creates dirac cones and if these obey time reversal symmetry, we get conducting states.

I know it has something to do with bulk-boundary correspondence but I don't quite understand that argument (as it feels like it should be applied to any insulator then).

I also do not quite understand the connection between time-reversal symmetry and conductive states- is it just that TRS is what protects the surface states?

Perhaps I am missing something though...

In the standard sense, when we think about topological classification/topological order - we think of dividing ground states of local and gapped Hamiltonians (with the symmetry eigenvalue if we are looking at symmetry-protected topological (SPT) order) into equivalence classes distinguished by the topological invariants. The relation is defined such that two states are in the same phase/equivalence class if they can be adiabatically connected.

To gain some intuition, we can think about it less formally, as some number characterizing a band in our system (sum over all bands is equal to 0). This number is invariant - any deformation/perturbation will not change its value as long as it is not strong enough to close the gap in the energy spectrum (or, in other words - changing of the invariant occurs through closing and reopening the gap between bands). Here, we also usually think about infinite-size systems (with periodic boundary conditions) to avoid any surface states that make it more tricky to define the invariant.

Right now, the main thing I understand is that the surface states and bulk states have different topological invariants

This is not true. Edge/surface states appear when you have an interface between two different phases (phases characterized by different invariants, e.g., topological phase - trivial phase or two topological phases with different invariants). So, it is not that a surface state has an invariant of its own but that the number/existence of edge states is a consequence of the difference in the topological invariant across the interface - this is the famous bulk-boundary correspondence. (As a side note, this is also why formal proofs of bulk-boundary correspondence are rather involved - bulk topological invariants are necessarily defined in infinite systems without a boundary, while the edge states exist in finite systems, so connecting the two concepts requires heavy mathematical tools)

For example, in a Haldane model, in which we can construct phases with non-zero Chern numbers - the number of edge states at the interface between a trivial and topological insulator is $$N_R - N_L = \Delta C,$$ where $$N_{R/L}$$ is the number of right/left moving modes, and $$\Delta C$$ is the difference of the Chern number across the interface.

I know it has something to do with bulk-boundary correspondence, but I don't quite understand that argument (as it feels like it should be applied to any insulator).

It applies to any insulator - it's just that non-topological insulators are all topologically trivial, and you don't get any non-zero difference in the invariant across an interface - so you don't get any topological edge modes on the boundary.

This leads to spin orbit coupling being higher on the surface, allowing for band inversion which creates dirac cones and if these obey time reversal symmetry, we get conducting states.

I'm not sure, what exactly you mean here, perhaps you can clarify in a comment. Hopefully the explanation below will also help though.

I also do not quite understand the connection between time-reversal symmetry and conductive states- is it just that TRS is what protects the surface states?

Seems like you are talking about some particular model. In general, there are systems with intrinsic topological order where you don't need any special symmetries to protect the surface states (e.g., Quantum Hall systems). In fact, you need to actually break the time-reversal symmetry (e.g., by applying a magnetic field) to get any topological behavior in such systems. The conducting behavior is then insensitive to disorder/or perturbations, as there is no state that the edge mode could scatter into (only on the opposite edge do we have a state that flows in the opposite direction, so they are spatially separated).

I assume you are then talking about Symmetry Protected topological phases like $$\mathbb{Z}_2$$ Topological Insulators. There, indeed, it is the Time Reversal Symmetry $$\mathcal{T}$$ that protects the edge states. This is due to Kramer's theorem stating that for systems with half-integer spin, for every eigenstate $$\vert n\rangle$$, $$\mathcal{T}\vert n\rangle$$ is also an eigenstate with the same energy.

Thus, if we look at momenta invariant under time-reversal symmetry ($$\vec{k} \leftrightarrow -\vec{k}$$) in the BZ such as $$(0,0)$$ or $$(\pi,\pi)$$ in 2D $$\mathbb{Z}_2$$ TI - any energy crossing at those momenta has to be doubly degenerate (and the degeneracy is protected by TRS). Away from the TR invariant momenta, the spin-orbit coupling can lift the degeneracy - and so if you have an edge state (that cannot be trivially contracted - connecting the two conductance and valence band), it will be protected by TRS - see Fig 11 in [1].

So, to answer your question - in $$\mathbb{Z}_2$$ Topological Insulators, the main connection between the TRS and edge modes is that the symmetry forces degeneracy at TR invariant moment through Kramers Theorem and protects the edge modes from perturbations (which don't break TRS).

References:

• thank you! this clarifies a lot. I was confusing different models and different ways of explaining the phenomena. The references are also helpful so essentially (for Z2) "the bulk has a non-trivial topological invariant. However, with a boundary it meets (like vacuum), bulk/boundary correspondence forces the surface to have a special state with a closed the bandgap and is protected by TRS?" I was talking about SOC because a lot of sources described it was the mechanism closing the bandgap, but I guess more so, it is one of the influences on the topological invariant being non-trivial? Commented Jun 25 at 18:08
• @Lynn0903 Yes, the boundary forces the surface to have a special conducting state within the bandgap, and in the case of Z2TI, these surface states are protected by TRS. SOC in the answer was used to show that the edge modes are stable against perturbations. That even if you add SOC to the system - which oftentimes lifts the spin degeneracy, the degeneracy at TR invariant momenta remains unchanged (due to the protection by TRS). and thus SOC is insufficient to destroy the topological surface modes.