# What is the topological number of 1D Haldane phase?

According to the topological classification work [e.g. Chen et al. Science 338, 1604 (2012)], the 1D Haldane phase should have a topological number $Z_2$, which has close relationship with the edge states. But how can we understand this $Z_2$ number only from its bulk state?

• This question may be ill defined. It seems that you're asking "why a change of topological invariant between two bulks is the hallmark of edge state at the interface between the two" which has been answered many times. In fact you answer your own question : what is the topological number = it's Z2 ! – FraSchelle Apr 19 '16 at 6:27

Suppose you have an infinitely long spin chain in the Haldane phase. Now imagine partitioning it in two, so you have a left and a right half. These two halves will be entangled with one another, and by Schmidt decomposition the total system can be decomposed as follows: $\boxed{| \psi \rangle }= \sum_\alpha \boxed{\Lambda_\alpha \; |\psi_L,\alpha \rangle \otimes |\psi_R,\alpha \rangle }$. The $\Lambda_\alpha$ coefficients characterize the entanglement between the two halves (these are called the Schmidt values). Now it turns out that for gapped topological phases in 1D, these Schmidt values are degenerate. For example, the AKLT state (which is an exactly solvable model in the Haldane phase) has only two non-zero Schmidt values: $\Lambda_\alpha = (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$. In other words in this case if you partition the ground state in two, the two halves form a Bell state!
So this already shows that topological phases have an interesting entanglement structure. Then the natural question arises: why? What is ensuring this degeneracy? That is where the symmetry comes in. It turns out that in general if you have a symmetry of your ground state (which is not broken), then this symmetry acts on the above entanglement decomposition. For example in the case of the Haldane phase, it turns out that if you act with the $\pi$-rotation matrices on the physical spin-$1$ particles, that the above Schmidt states transform under the Pauli matrices, which as we know form a spin-$\frac{1}{2}$ representation of the rotation group. But since we know the ground state in the Haldane phase is invariant under these rotations, the Schmidt values must be degenerate in this two-dimensional subspace of the spin-$\frac{1}{2}$ representation. This ensures the two-fold degeneracy.
More generally, in the case of spin rotation symmetry, the topological invariant is defined as $0$ if the rotation acts with an integer representation on the entanglement decomposition, and is defined as $1$ if it acts as a half-integer representation. This gives a $\mathbb Z_2$ invariant in the bulk. This extends naturally to other symmetry groups (that is in fact how the classification is obtained: group cohomology labels the projective representations that the bulk symmetry can have on the entanglement states).
Note that the above reasoning also naturally explains why these states have edges states: imagine removing these two halves far away from each other. Since our Hamiltonian is local, we know that the energy of the left half is unaffected by anything we do on the right half. But since these two states form a Bell state, we know that depending on a measurement we do on the right half, our left half can collapse into two different states. This proves that the left half has a twofold degenerate ground state. Moreover since the entanglement in 1D is always short range for gapped phases, this degeneracy must live on the edge of the left half. Also, since this degeneracy comes from the entanglement structure, and we knew the entanglement states transform as a spin-$\frac{1}{2}$, we expect our edge states to behave as spin-$\frac{1}{2}$ particles. So you see that entanglement is the most fundamental way to understand these topological phases in 1D, naturally giving rise to all the associated properties :)