Computation of two-site entanglement entropy of the critical transverse-field Ising chain

The paper Osborne and Nielsen, Phys. Rev. A 66, 032110 gives an exact solution for the two-site reduced density matrix for the transverse-field Ising model with periodic boundary conditions (Eq. 26): $$\rho_{0r} =\frac{I_{0r}+\left<\sigma^z\right>\left(\sigma_0^z+\sigma_r^z\right)+\sum_{k=1}^3\left<\sigma_0^k\sigma_r^k\right>\sigma_0^k\sigma_r^k}{4}$$ where $$0r$$ indicates that site 0 and $$r$$ are in subsystem A and the rest of the infinite chain in subsystem B.

Question: When I evaluate this at criticality and compute the entanglement entropy $$S_{0r}=Tr(\rho_{0r}\log{\rho_{0r}})$$, I obtain a complex number... How does one set up this computation correctly? (see below)

The correlation functions at criticality are known (Eqs. 49-52): \begin{align} \left<\sigma_0^x\sigma_r^x\right>&=\left(\frac{2}{\pi}\right)^r 2^{2r(r-1)}\frac{H(r)^4}{H(2r)}\\ \left<\sigma_0^y\sigma_r^y\right>&=-\frac{\left<\sigma_0^x\sigma_r^x\right>}{4r^2-1}\\ \left<\sigma_0^z\sigma_r^z\right>&=\frac{4}{\pi^2}\frac{1}{4r^2-1}\\ \left<\sigma^z\right>&=\frac{2}{\pi} \end{align} where $$H(r)=1^{r-1}2^{r-2}\dots (r-1)$$.

Evaluating at $$r=1$$: \begin{align} \left<\sigma_0^x\sigma_1^x\right>&=\frac{2}{\pi}\approx0.63662\\ \left<\sigma_0^y\sigma_1^y\right>&=-\frac{2}{3\pi}\approx -0.212207\\ \left<\sigma_0^z\sigma_1^z\right>&=\frac{4}{3\pi^2}\approx 0.135095\\ \left<\sigma^z\right>&=\frac{2}{\pi} \end{align} These values match Pfeuty (1970) (where $$n=r$$):

So far so good. Using the usual $$2\times 2$$ identity matrix and Pauli matrices $$\sigma^k$$, I use the Kronecker product to construct the full matrices: \begin{align} \sigma_0^k &= \sigma^k \otimes I\\ \sigma_r^k &= I \otimes \sigma^k\\ I_{0r} &= I\otimes I \end{align}

Finally, the reduced density matrix is: \begin{align} \rho_{01} &=\frac{I_{01}+\left<\sigma^z\right>\left(\sigma_0^z+\sigma_1^z\right)+\sum_{k=1}^3\left<\sigma_0^k\sigma_1^k\right>\sigma_0^k\sigma_1^k}{4}\\ &=\begin{pmatrix} \frac{1}{4}+\frac{1}{3\pi^2}+\frac{1}{\pi} & 0 & 0 & \frac{2}{3\pi}\\ 0 & \frac{1}{4}-\frac{1}{3\pi^2} & \frac{1}{3\pi} & 0\\ 0 & \frac{1}{3\pi} & \frac{1}{4}-\frac{1}{3\pi^2} & 0\\ \frac{2}{3\pi} & 0 & 0 & \frac{4+3(\pi-4)\pi}{12\pi^2} \end{pmatrix} \end{align} Note that $$\rho^*=\rho$$ and $$Tr(\rho)=1$$, so that is good. However, the two-site entanglement entropy is now $$S_{01}=Tr(\rho_{01}\log{\rho_{01}})\approx0.468674 + 0.223869 i.$$

What is going wrong here?

I noticed that the references Pfeuty and Osborne/Nielsen differ by a factor of $$1/\pi$$ for $$\left<\sigma_0^z\sigma_r^z\right>$$, which appears to be a typo, but both these values lead to a complex number for $$S_{01}$$, so the problem is not caused by that.

This is a follow-up question on Entanglement entropy of the infinite transverse field Ising chain at the critical point, but it is a separate technical problem, hence the new post here.

The problem arises because of an incorrect expression for $$\langle\sigma_0^z\sigma_1^z\rangle$$. The correct expression is $$\langle\sigma_0^z\sigma_1^z\rangle = \langle\sigma_0^z\rangle \langle\sigma_1^z\rangle + \frac4{3\pi^2}.$$
• Thank you! Now I get $S_{01}\approx 0.428054$ which matches the exact diagonalization results. Jul 30, 2022 at 12:32