# Understanding entanglement entropy in the transverse field Ising model

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)?

• Did you take the exact ground state for h<0, or the symmetry-broken states? (It seems you took the former, which for N->infty could be argued to be unphysical.) BTW, I don't understand why you claim "more disorder = more entanglement". Why is this a "principle"? Apr 21, 2018 at 11:35
• Hmm I hadn't thought too hard about it but generally was considering that a thermal state is maximally entangled, and in these notes it is said that "In a random state...we expect subsystem A to be almost maximally entangled with B". I suppose the statement needs mroe consideration and was too cavalier.
– Kai
Apr 21, 2018 at 15:41
• Regarding the ground state, I simply took the lowest energy state. Are you are saying that the state I am using is actually a superposition of the two symmetry broken states? How would I take the two lowest eigenvectors and recover the symmetry broken states? This is an interesting point I had not considered but would explain some other things I was confused about. Why would this state be unphysical in the $N\to\infty$ limit?
– Kai
Apr 21, 2018 at 15:44
• Thermal states at T->infty are not entangled at all. (Where did you get this from?) Also, ground states of local Hamiltonians are not random at all (as can be seen by parameter counting). Yes, the state you are using is a superposition of the sym. broken states (there is a splitting of $e^{-N}$). For a finite system, getting a sym.-broken state is tricky (though if you understand the model, as for the Ising, less so), but you could e.g. add a small field. They are unphysical because they are not stable against (arbitrarily) small fields (arb small as N->inf). Apr 21, 2018 at 18:40
• Note that you have chosen an unfortunate system size: It makes it look like the maximum at h=1 has sth. to do with the value at h<1, which is not at all the case: The former grows as log(N), the latter is constant = log 2. Apr 22, 2018 at 10:50

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.
• Thank you that makes a lot more sense now. Although note I was using periodic boundary conditions, but I don't think it makes a difference. Could you point me to a source for the "roughly an extra $\log 2$ of entanglement? I have very little experience with cat states though I have seen them before.