2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.