# What's wrong with my argument about entanglement entropy in QFT being time-independent?

Let's say we need to compute the entanglement entropy (EE) of a subsystem $$A$$ ($$A=[0,L]$$, $$L>0$$) in a 2D CFT.

The density matrix of the total system (i.e., the real axis) is given by $$\rho(t)=\text{e}^{-iHt}\rho_0~\text{e}^{iHt},$$ where $$\rho_0$$ is certian time-independent positive semi-definite matrix with unit trace.

Then, I shall argue that the EE of $$A$$ should be time-independent.

#### My argument:

\begin{align} \because&~\text{The Hamiltonian } H=\int_{\mathbb{R}^1}T_{00}dx=\int_{A}T_{00}dx+\int_{\bar{A}}T_{00}dx=H_A+H_{\bar A},\\ \therefore&~\text{the evolution operator}~\text{e}^{\pm iHt}=\text{e}^{\pm i(H_A+H_{\bar A})~t}=\text{e}^{\pm i H_A~t}\otimes\text{e}^{\pm iH_{\bar A}~t}.\\ \end{align} Therefore the reduced density matrix of $$A$$ reads: $$\rho_A(t)=\text{Tr}_{\bar A}[\rho(t)]=e^{-iH_A~t}\text{Tr}_{\bar A}[e^{-iH_\bar A~t}\rho_0e^{iH_\bar A~t}]e^{iH_A~t}=e^{-iH_A~t}\text{Tr}_{\bar A}[\rho_0]e^{iH_A~t}.$$ Since $$\rho_A(t)$$ and $$\text{Tr}_{\bar A}[\rho_0]$$ differ by only a unitary transformation, the EE obtained by $$\rho_A(t)$$ should be the same for the EE obtained by $$\text{Tr}_{\bar A}[\rho_0]$$. What's more, $$S_A\big(\text{Tr}_{\bar A}[\rho_0]\big)=-\text{Tr}_A\big[\text{Tr}_{\bar A}[\rho_0]\log \text{Tr}_{\bar A}[\rho_0]\big]$$ is time-independent, which finally leads to a time-indepentent $$S_A\big(\rho_A(t)\big)$$. $$\square$$

### Contradiction with known results:

The result derived by the above argument is contradict with many known results.

For exapmle, in 2014, T. Takayanagi et al found that the $$S_A$$ for a locally excited state (A state generated by acting a local operator $$O(-x)~(x>0)$$ on the vaccum) behaves like $$S_A(t)= \begin{cases} S_{A,vacuum},&tx+L,\\ S_{A,vacuum}+\log d_O,& x where $$S_{A,vacuum}=\frac{c}{3}\log\frac{L}{\epsilon}$$ stands for the EE of $$A$$ when the total system is in the vacuum, $$d_O$$ stands for the quantum dimension of $$O$$. Obviously their result is time dependent.

### My question:

Physically, I agree that in some states the entanglement entropy of the subsystem should be time dependent, but what is the problem with my argument?

• You’re assuming that $\rho_0$ and the region A is time independent. But neither of those has to be true. Commented Aug 4, 2022 at 9:08
• @Prahar Thanks for the comment, but sorry i can't get your point. What do you mean by "the region A is time independent" Commented Aug 4, 2022 at 9:31

You are implicitly making a very strong assumption in your argument. Given $$H=H_A+H_{\bar A},$$ $$e^{\pm iHt}=e^{\pm i(H_A+H_{\bar A})~t}$$ is true, but your next step $$e^{\pm i(H_A+H_{\bar A})~t} =e^{\pm i H_A~t} e^{\pm iH_{\bar A}~t}.$$ only holds if $$\left[ H_A, H_\bar{A}\right]=0$$, recall the Zassenhaus formula.
• Thanks for the answer! : ) You're probably right, and I need to take some time to think about it. BTW, why do you say that "$[H_A,H_{\bar{A}}]=0$ is a very strong assumption"? Could you please comment more on that? Thanks again! ^_^ Commented Aug 7, 2022 at 1:15
• Well, saying that $H_A$ and $H_\bar{A}$ can be simultaneously diagonalized rather constrains what models you are considering. It is certainly not generally true. In particular, you're ruling out many interacting lattice models with CFT low-energy descriptions. Commented Aug 7, 2022 at 15:50