# Entanglement entropy of 1D chiral Fermion

I was told that the entanglement entropy $S_E$ on the ground state of a (1+1)D conformal field theory (CFT) follows the logarithmic behavior $S_E=\frac{c}{12}\ln L$ where $L$ is the length scale between the entanglement cuts. I do not know how the CFT works, so I would like to convince myself by starting from a special case, say the chiral (and free) fermion.

My question is how to calculate the entanglement entropy of 1D chiral fermion using the 2nd quantization language without referring to bosonization or mapping to CFT?

Here is my attempt to approach the problem. Suppose we have a chiral fermion chain described by the Hamiltonian $H=\sum_k k c_k^\dagger c_k$. Consider ground state (at zero temperature), it would be $|\psi\rangle=\prod_{k<0}c_k^\dagger |0\rangle$. The density matrix can be constructed from the ground state as $\rho=|\psi\rangle\langle\psi|$. Then I should make entanglement cuts to separate the system into sectors A and B. Tracing out the fermion degrees of freedoms in B to obtain the reduced density matrix $\rho_A=\mathrm{Tr}_B\rho$. Then I was suppose to diagonalize $\rho_A$ to find the entanglement spectrum and evaluate the entanglement entropy.

But when I tried to work out the details, I was stuck in the last several steps. Let me illustrate what I have obtained so far. First to understand the structure of $\rho$, I started from the correlation function, and found $$\begin{split} \mathrm{Tr}\rho c_{x_1}^\dagger c_{x_2} &= \langle c_{x_1}^\dagger c_{x_2} \rangle\\ &=\sum_{k_1 k_2}\langle c_{k_1}^\dagger c_{k_2}\rangle e^{i(k_2 x_2-k_1x_1)}\\ &=\sum_{k<0} e^{ik(x_2-x_1)}\\ &\simeq \frac{i}{x_1-x_2},\end{split}$$ where $x_1$ and $x_2$ are two spacial coordinates restricted in the sector A. So on the one-particle subspace, the density matrix should be $$\rho_{A1}=\int_0^L\mathrm{d}x_1\int_0^L\mathrm{d}x_2\;c_{x_1}^\dagger\frac{i}{x_1-x_2}c_{x_2}.$$ I also noticed that in the zero particle subspace, the density matrix is simply the identity $\rho_{A0}=1$. So I would generalize that in the $n$ particle space, the density matrix should be $\rho_{An}=\rho_{A1}^n$. (Let me know if I am wrong here.) Then the reduced density matrix would be $$\rho_A=\sum_{n=0}^{\infty}\rho_{An}=(1-\rho_{A1})^{-1}.$$ So here is where I stopped. I can not figure out how to diagonalize the reduced density matrix $\rho_A$. Even for $\rho_{A1}$, I do not know how to deal with it. The entanglement cuts breaks the space translational symmetry, and I can not do diagonalization by Fourier transform to the momentum space. Even if I tried some numerics by discretization, the eigen values varies from negative to positive, and I can not figure out a clue. I would appreciate so much if anyone could help me out from here.

## 1 Answer

A useful reference is http://arxiv.org/pdf/0906.1663.pdf, by Peschel and Eisler. A common approach is to make use of the fact that the two point function you calculated is independent of whether one uses the full density matrix or the reduced density matrix, provided one looks at operators that are local to the region that one is not tracing over. If one then assumes that the reduced density matrix is also Gaussian, there is a simple relationship that can be constructed between the eigenvalues of the reduced density matrix and the eigenvalues of the two point function, treated as a matrix indexed by position. (See equation (17) in the reference above.)

To actually use this approach, I would put the fermion on a lattice, which introduces the usual fermion doubling problem. So I would not be able to study chiral fermions...