Understanding entanglement entropy in the transverse field Ising model

Question

I'm working on a project studying the transverse field Ising model in 1D with periodic boundary conditions, given by $$H = - J \sum_{i=1}^{N} \sigma^z_i \sigma^z_{i+1} - h \sum_{i=1}^N \sigma^x_i$$ I have written a code to construct and diagonalize the Hamiltonian. Taking the ground state eigenvector, I have written some code to compute the entanglement entropy between two halves of system $S(x)$, where in the left half of the system (subsystem A) there are $x$ spins and in the right half of the system (subsystem B) there are $N-x$ spins. I've plotted the results for various values of $h$ here:

The fit is the expected result from conformal field theory at the critical point $h_c=1$, $$S(x) = A + \frac{c}{3}\log\left[\frac{N}{\pi} \sin \frac{\pi x}{N}\right]$$ where $c$ is the central charge. From my fit I got $c \approx 0.510$, pretty close to the exact value of $c = 0.5$.

First I note that when $h=0$ we have a purely classical theory, the groundstate is simply all spins up or all spins down, and there is no entanglement entropy. In the limit $h\to\infty$, we again have a classical theory and all spins point along the field direction. Again the entanglement entropy is zero.

My questions is in the interpretation of these results in the regions $h<1$ and $h>1$. How do I understand why for $h<1$ the entanglement is large and relatively uniform, while for $h>1$ the entanglement is small and relatively uniform? And how do I understand why at $h=1$ the entaglement is "least flat", displaying the largest variation (small regions display low entanglement, larger regions display high entanglement)?

Answer 1

Let's first consider the two extremal cases, $h=0$ and $h=\infty$. In both cases, the ground state is a product state - either $|0\cdots0\rangle$ or $|+\cdots+\rangle$, so the entropy is zero.

Intuitively, as we change $h$ towards $h=1$, the system acquires a finite length scale, which (crudely speaking) leads to entanglement over this scale. This is why the entanglement goes up as we change $h$ towards $h=1$. Its value is slightly smaller towards the boundary since there are no degrees of freedom beyond the boundary - the entanglement is cut off.

So how come the entanglement for $h<1$ is larger? First, at $h=0$ there are in fact two degenerate ground states, $|0\cdots0\rangle$ and $|1\cdots1\rangle$. As soon as $h>0$ (but $<1$), these two states are not perfectly degenerate, but there is a splitting of order $h^N$ (from $N$th order perturbation theory) between $$ |\psi_\pm\rangle = |0\cdots0\rangle + |1\cdots1\rangle\ . $$ Thus, the only ground states is $|\psi_+\rangle$. This is a "cat state", i.e., a macroscopic superposition, which is unstable against arbitrarily small perturbations as $N\to\infty$ and is therefore not "physical". The fact that we have this macroscopic superposition gives roughly an extra entanglement of $\log 2$, which is what you see in your simulation.

At the criticial point, the system is scale invariant, this is, degrees of freedom at all length scales are equally entangled. A chain of length $L$ has $\log L$ length scales, and thus, we expect the entanglement to scale like $\log L$. This is exactly what you see again (and CFT predicts); moreover, if you cut the chain not in the middle, but $L$ away from the boundary, the shorter piece is what matters for the entanglement (as this piece limits the available degrees of freedom), and thus, you get the $\log L$ scaling as a function of the cut.