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 $$S_A(t)= \begin{cases} S_{A,vacuum},&t<x~\text{or}~t>x+L,\\ S_{A,vacuum}+\log d_O,& x<t<x+L, \end{cases} $$ 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?

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

1 Answer 1


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.

  • $\begingroup$ 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! ^_^ $\endgroup$
    – Hezaraki
    Commented Aug 7, 2022 at 1:15
  • $\begingroup$ 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. $\endgroup$
    – Anyon
    Commented Aug 7, 2022 at 15:50

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