††institutetext: Department of Physics, Brandeis University, Waltham,
Massachusetts 02453, USA††institutetext: Condensed Matter Theory Center and
Joint Quantum Institute, Department of Physics, University of Maryland,
College Park, Maryland 20742, USA

# Phase transition in von Neumann entanglement entropy from replica symmetry
breaking

Shao-Kai Jian  Brian Swingle<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We study the entanglement transition in monitored Brownian SYK chains in the
large-$N$ limit. Without measurement the steady state $n$-th Rényi entropy is
obtained by summing over a class of solutions, and is found to saturate to the
Page value in the $n\rightarrow 1$ limit. In the presence of measurements, the
analytical continuation $n\rightarrow 1$ is performed using the cyclic
symmetric solution. The result shows that as the monitoring rate increases, a
continuous von Neumann entanglement entropy transition from volume-law to
area-law occurs at the point of replica symmetry unbreaking.

##  1 Introduction

Starting from an unentangled product state, a chaotic Hamiltonian or circuit
generates entanglement between different parts of the system, and eventually
leads to a state with volume-law entanglement at late time. On the contrary,
local measurements extract information and diminish the entanglement between
the measured qubits and the rest of the system. The competition between these
two effects leads to a transition in the degree of entanglement in the long-
time steady state, the so-called measurement-induced phase transition
li2018quantum ; li2019measurement ; skinner2019measurement ;
gullans2020dynamical ; chan2019unitary .

Following this realization, there has been a surge of investigations of this
transition and of related phenomena vasseur2019entanglement ; choi2020quantum
; jian2020measurement ; bao2020theory ; zabalo2020critical ;
gullans2020scalable ; li2020conformal ; iaconis2020measurement ;
nahum2021measurement ; li2021statistical ; sang2021measurement ;
lavasani2021measurement ; ippoliti2021entanglement ; bao2021symmetry ;
bentsen2021measurement ; jian2021syk ; li2021entanglement ; jian2021quantum ;
yang2021entanglement . In principle, one can diagnose transitions in
entanglement properties by introducing replicas and using them to compute the
$n$-th Rényi entropy of the long-time steady state. In this context, it was
found that the entanglement transition can be understood as an unbreaking
transition of replica symmetry vasseur2019entanglement ; jian2020measurement ;
bao2020theory ; nahum2021measurement . Nevertheless, as the von Neumann
entanglement entropy is obtained in the $n\rightarrow 1$ limit, where $n$ is
the number of replicas, how the replica symmetry unbreaking defined for $n\geq
2$ interplays with the replica limit $n\rightarrow 1$ is still an outstanding
question. Part of the difficulty lies in the hardness of analytical
continuation. In the context of Haar random hybrid circuits, the transition
can be effectively mapped to a statistical model in the replica space, which
in the limit of infinite local Hilbert space dimension is described by bond
percolation skinner2019measurement ; jian2020measurement ; bao2020theory , and
the replica limit $n\rightarrow 1$ is notoriously subtle even in this limit.
In this paper, we fill this gap by introducing a monitored Brownian SYK model
kitaev2015simple ; sachdev1993gapless ; maldacena2016remarks ;
saad2018semiclassical ; sunderhauf2019quantum ; liu2021non ; jian2020note ;
jian2021syk ; stanford2021subleading , and calculating the von Neumann
entanglement entropy using the replica trick.

This model, which features continuous weak monitoring, admits a path integral
representation, and consequently allows saddle-point analysis in the large-$N$
limit. We show explicitly how summing over a class of replica symmetry broken
solutions can correctly produce the Page value page1993average  of the
entanglement entropy in the absence of monitoring. A crucial step in the
calculation is a mapping of part of the saddle point evaluation to the
evaluation of a transition amplitude of a quantum mechanical model. Turning on
measurement decreases the degree of replica symmetry breaking and eventually
leads to its unbreaking. We then take the $n\rightarrow 1$ limit based on the
cyclic symmetric solution and extract the behavior of the entanglement entropy
per site as a function of measurement strength. The transition in the von
Neumann entropy density in the $n\rightarrow 1$ limit coincides with the
unbreaking transition of replica symmetry for $n\geq 2$.

The paper is organized as follows. In Section 2, we define the quasi entropy
of a monitored system napp2019efficient . Because of the non-unitary nature of
measurements, a proper density matrix needs to be normalized after each
measurement. This leads to difficulties in analytical calculations as the
normalization is a nonlinear procedure that depends on the measurement
outcome. The quasi entropy is the Rényi entropy of the unnormalized density
matrix weighted such that it vanishes for a pure state. The quasi entropy has
two crucial properties: it admits a path integral representation and, like the
Rényi entropy, the $n\rightarrow 1$ limit gives the von Neumann entanglement
entropy averaged over quantum trajectories.

In Section 3, we introduce our model which contains both a chaotic unitary
part and a measurement part. The unitary part is given by two independent
Brownian SYK chains. The Brownian SYK Hamiltonian is able to efficiently
generate entanglement among different parts of the chain. The measurement part
amounts to weakly projecting onto the Fermi parity of the complex fermions
formed by the left-right pair of Majoranas at each site. Importantly, it is a
local measurement that extracts information about a single site of the left-
right coupled system, and its dark state corresponds to a product state
between different sites with definite Fermi parity. Therefore, the competition
between the two parts will result in a measurement-induced entanglement
transition.

In Section 4, we first study a simpler version of our model that only consists
of the unitary part. It is given by two coupled Brownian SYK clusters. After
properly defining the initial state, the entanglement entropy between the two
clusters can be obtained via a large-$N$ path integral in the replica space.
We find a class of saddle points that are described by permutation matrices.
The calculation of the onshell action can be done by decomposing the
permutation into cycles, and then mapping each cycle to a transition amplitude
of a Kitaev chain kitaev2001unpaired . An interesting point arising in the
calculation is that the Pfaffian effectively counts the number of cycles in
the solution given by a general permutation matrix. Then, by summing over this
class of saddle points, we get a result that predicts maximal entanglement at
leading order with subleading corrections consistent with the Page value
page1993average  and the symmetry of the model.

In Section 5, we generalize the calculation into two Brownian SYK chains with
a measurement that couples the two chains. The aforementioned class of saddle
points that saturate the Page value are changed in the presence of the
measurement. When the measurement rate gets larger than a critical value, all
saddle points reduce to a diagonal solution where no correlation between
different replica exists, i.e., the replica symmetry is restored. The
calculation of quasi entropy now involves two coupled Kitaev chains, where the
hopping within each chain is given by the correlation between different
replicas, and the coupling between two chains is proportional to the
measurement rate. After the coupling increases to a critical point, the
hopping within each chain vanishes, and the eigenmodes become local. We take
an analytical continuation of the cyclic symmetric solution to get the von
Neumann entanglement entropy. The entropy density is obtained, and it shows a
continuous phase transition right at the point where the replica symmetry is
restored.

We conclude the paper in Section 6, where we also study the Landau-Ginzburg
theory. We leave the technical details of the permutation operator that is
used to calculate the quasi entropy, the explicit saddle-point solution for
$n=2$, and the product of special trigonometric functions that is used to
evaluate the transition amplitude to the appendices.

##  2 Quasi entropy and trajectory averaged entanglement entropy

We consider time evolution generated by unitary evolution and measurements,
and are interested in the entanglement entropy of the long-time steady state.
The state itself is not steady in a microscopic sense, but its entanglement
structure is expected to stabilize at long times. Importantly, to really
assess the entanglement properties of the resulting state, one needs to keep
track of each measurement record. A specific evolution with given measurement
outcomes forms a quantum trajectory. Ultimately, we will calculate the von
Neumann entanglement entropy averaged over quantum trajectories.

As the unitary evolution part of the evolution is well known, here we focus on
the measurement part. The measurement is described by a set of Kraus operator
$\\{K_{\nu}\\}$, $\sum_{\nu}K_{\nu}^{\dagger}K_{\nu}=1$ wiseman1996quantum ;
jian2020measurement , where $\nu$ numerates possible measurement outcomes.
Starting from an initial density matrix $\rho$, given a sequence of
measurement outcomes $\bm{\nu}=\nu_{1}\nu_{2}...\nu_{m}$, the corresponding
quantum trajectory is

$\displaystyle\tilde{\rho}_{\bm{\nu}}=K_{\nu_{m}}...K_{\nu_{2}}K_{\nu_{1}}\rho
K_{\nu_{1}}^{\dagger}K_{\nu_{2}}^{\dagger}...K_{\nu_{m}}^{\dagger}.$ (1)

Note that $\tilde{\rho}_{\bm{\nu}}$ is an unnormalized density matrix, whose
trace gives the probability of the measurement ourcome $\bm{\nu}$. Summing the
probabilities of all possible measurement outcomes gives unity, i.e.,
$\sum_{\bm{\nu}}\text{Tr}(\tilde{\rho}_{\bm{\nu}})=1$. We are interested in
calculating the quasi entropy of bipartite system $A\bar{A}$ napp2019efficient
,

$\displaystyle S_{A}^{(n)}$ $\displaystyle=$
$\displaystyle\frac{1}{1-n}\log\frac{\sum_{\bm{\nu}}\text{Tr}[(\tilde{\rho}_{\bm{\nu}})^{\otimes
n}M_{\text{cyc}}(A)]}{\sum_{\bm{\nu}}(\text{Tr}[\tilde{\rho}_{\bm{\nu}}])^{n}},$
(2)

where $M_{\text{cyc}}(A)$ is the cyclic permutation operator acting on the
subsystem $A$ of the $n$ replicas. Note that the summation over quantum
trajectories appears both in the numerator and denominator.

Although the quasi entropy is distinct from the Rényi entropy of the
normalized state with measurement outcome $\bm{\nu}$, the limit $n\rightarrow
1$ of the quasi entropy converges to the averaged von Neumann entanglement
entropy of the normalized state. To show this, recall that the Rényi entropy
of subsystem $A$ for a given quantum trajectory $\bm{\nu}$ is

$\displaystyle
S_{A}^{(n)}(\bm{\nu})=\frac{1}{1-n}\log\frac{\text{Tr}[(\tilde{\rho}_{\bm{\nu}})^{\otimes
n}M_{\text{cyc}}(A)]}{(\text{Tr}[\tilde{\rho}_{\bm{\nu}}])^{n}}.$ (3)

The von Neumann entanglement entropy can be obtained by analytically
continuing the Rényi entropy to $n\rightarrow 1$, i.e.,
$S_{A}(\bm{\nu})=\lim_{n\rightarrow 1}S_{A}^{(n)}(\bm{\nu})$. We have the
following relation

$\displaystyle\frac{\text{Tr}[(\tilde{\rho}_{\bm{\nu}})^{\otimes
n}M_{\text{cyc}}(A)]}{(\text{Tr}[\tilde{\rho}_{\bm{\nu}}])^{n}}\approx
1+(1-n)S_{A}(\bm{\nu})+\mathcal{O}\left((n-1)^{2}\right).$ (4)

Using this relation, we can take $n\rightarrow 1$ limit of the quasi entropy
defined in (154),

$\displaystyle\lim_{n\rightarrow 1}S_{A}^{(n)}$ $\displaystyle=$
$\displaystyle\lim_{n\rightarrow
1}\frac{1}{1-n}\log\frac{\sum_{\bm{\nu}}(1+(1-n)S_{A}(\bm{\nu}))(\text{Tr}[\tilde{\rho}_{\bm{\nu}}])^{n}}{\sum_{\bm{\nu}}(\text{Tr}[\tilde{\rho}_{\bm{\nu}}])^{n}}$
(5) $\displaystyle=$ $\displaystyle\lim_{n\rightarrow
1}\frac{1}{1-n}\log\sum_{\bm{\nu}}\left(1+(1-n)S_{A}(\bm{\nu})\right)\text{Tr}[\tilde{\rho}_{\bm{\nu}}]$
(6) $\displaystyle=$ $\displaystyle\lim_{n\rightarrow
1}\frac{1}{1-n}\log\left(1+(1-n)\sum_{\bm{\nu}}\text{Tr}[\tilde{\rho}_{\bm{\nu}}]S_{A}(\bm{\nu})\right)$
(7) $\displaystyle=$
$\displaystyle\sum_{\bm{\nu}}\text{Tr}[\tilde{\rho}_{\bm{\nu}}]S_{A}(\bm{\nu}),$
(8)

where in the first step we use (4) and in the second step and third step we
use $\sum_{\bm{\nu}}\text{Tr}[\tilde{\rho}_{\bm{\nu}}]=1$. Therefore, the
$n\rightarrow 1$ limit of the quasi entropy is the quantum trajectory averaged
von Neumann entanglement entropy. In the following, we will calculate the
quasi entropy for arbitrary $n$, and then take the $n\rightarrow 1$ limit to
get the von Neumann entanglement entropy.

##  3 Model and setup

Our model consists of two coupled Brownian SYK chains, with a unitary part and
non-unitary monitoring part. The unitary evolution is governed by the
following Brownian SYK Hamiltonian describing left ($L$) and right ($R$)
chains,

$\displaystyle H(\psi)$ $\displaystyle=$
$\displaystyle\sum_{x;a=L,R}\Big{(}\sum_{i<j}iJ_{a,ij}^{x}(t)\psi_{x,a,i}\psi_{x,a,j}$
(9)
$\displaystyle+\sum_{j_{1},...,j_{q}}i^{q/2}U_{a,j_{1}...j_{q}}^{x,x+1}(t)\psi_{x,a,j_{1}}...\psi_{x,a,j_{q/2}}\psi_{x+1,a,j_{q/2+1}}...\psi_{x+1,a,j_{q}}\Big{)},$

where $\psi_{x,a,i}$ $i=1,...,N$ denotes $i$-th of $N$ Majorana fermions at
each site $x=1,...,L$ of the $a=L,R$ chains,
$\\{\psi_{x,a,j},\psi_{x^{\prime},a^{\prime},j^{\prime}}\\}=\delta_{xx^{\prime}}\delta_{aa^{\prime}}\delta_{jj^{\prime}}$.
$L$ is the number of sites (it should not be confused with the $L$ chain), and
periodic boundary conditions in real space are assumed in this paper.
$J_{a,ij}^{x}$ is two Majorana coupling within each site, and
$U^{x}_{a,j_{1},...,j_{q}}$ is $q$ Majorana interaction between nearest-
neighbor sites. Here we consider $q=4k,k\in\mathbb{Z}^{+}$ ($\mathbb{Z}^{+}$
denotes positive integers) to conserve local Fermi parity. The couplings in
the left and right chains are independent Gaussian variables with mean zero
and variances

$\displaystyle\overline{J_{a,ij}^{x}(t_{1})J_{a^{\prime},ij}^{x^{\prime}}(t_{2})}$
$\displaystyle=$
$\displaystyle\frac{4J}{N}\delta(t_{12})\delta_{aa^{\prime}}\delta^{x,x^{\prime}},$
(10)
$\displaystyle\overline{U_{a,j_{1}...j_{q}}^{x,x+1}(t_{1})U_{a^{\prime},j_{1}...j_{q}}^{x^{\prime},x^{\prime}+1}(t_{2})}$
$\displaystyle=$
$\displaystyle\frac{2^{q}(q/2)!^{2}U}{qN^{q-1}}\delta(t_{12})\delta_{aa^{\prime}}\delta^{x,x^{\prime}}.$
(11)

where $t_{12}=t_{1}-t_{2}$. The Dirac $\delta$ function $\delta(t_{12})$
indicates the Brownian nature of the couplings. In the unitary evolution, the
left and right chains do not couple, but the evolution does scramble
information over the entire individual chains jian2021quantum .

To describe the monitoring part, we first divide the continuous time into
infinitesimal steps and define proper Kraus operators, then we take the
continuum limit to get an effective action description. In each infinitesimal
time step $\delta t$, we consider the following measurement. The local
measurement operator couples the $L$ annd $R$ fermions at each site, as
described by the Kraus operators jian2021syk

$\displaystyle\\{K_{1}^{x,i},K^{x,i}_{2}\\}=\left\\{\pi^{-}_{x,i}+\sqrt{1-\mathfrak{s}^{2}}\pi^{+}_{x,i},\mathfrak{s}\pi^{+}_{x,i}\right\\},$
(12)

where $\pi^{\pm}_{x,i}=\frac{1}{2}(1\mp i2\psi_{x,L,i}\psi_{x,R,i})$ is the
projection onto one of the Fermi parity eigenstates and $0\leq\mathfrak{s}\leq
1$ is the measurement strength. We will see in the following that the
appropriate continuous monitoring strength should be
$\mathfrak{s}\propto\sqrt{\delta t}$. The Kraus operators satisfy the
nomralization condition $\sum_{j=1}^{2}(K_{j}^{x,i})^{\dagger}K^{x,i}_{j}=1$.

During a single time step in $\delta t$, the evolution of the unnormalized
density matrix for a quantum trajectory with measurement outcome $\nu_{x,i}$
for each flavor at each site is

$\displaystyle\tilde{\rho}_{\bm{\nu}}=(\otimes_{x,i}K_{\nu_{x,i}})\rho(\otimes_{x,i}K_{\nu_{x,i}}^{\dagger}),$
(13)

where on the right-hand side we suppress the superscript for the flavor and
site in Kraus operators $K_{j}^{x,i}$ as they can be inferred from the
measurement outcome. The quasi entropy requires $n$ replicas, i.e.,

$\displaystyle\tilde{\rho}_{\bm{\nu}}^{\otimes
n}=\otimes_{\alpha=1}^{n}(\otimes_{x,i}K_{\nu_{x,i}}^{\alpha})\rho^{\alpha}(\otimes_{x,i}K_{\nu_{x,i}}^{\alpha{\dagger}}),$
(14)

where we have introduced a new superscript Greek letter $\alpha=1,...,n$ to
denote different replicas (it should not be confused with the original
superscript of the Kraus operator that is used to denote site and flavor). We
will use the superscript Greek letter as the replica index throughout the
paper. According to the definition of quasi entropy, we can change the order
of the trajectory average and the trace over the density matrix, and consider
the trajectory average first, namely,
$\sum_{\bm{\nu}}\text{Tr}[(\tilde{\rho}_{\bm{\nu}})^{\otimes
n}\cdots]=\text{Tr}[\sum_{\bm{\nu}}(\tilde{\rho}_{\bm{\nu}})^{\otimes
n}\cdots]$ for any $\cdots$ independent of $\bm{\nu}$. The effect of the
monitoring operator on the replicas is given by the following operator,

$\displaystyle\sum_{\bm{\nu}}\otimes_{\alpha=1}^{n}(\otimes_{x,i}K_{\nu_{x,i}}^{\alpha})\otimes(\otimes_{x,i}K_{\nu_{x,i}}^{\alpha{\dagger}})$
$\displaystyle=$
$\displaystyle\otimes_{x,i}\sum_{\nu_{x,i}=1,2}\otimes_{\alpha=1}^{n}\otimes_{s=\pm}K_{\nu_{x,i},s}^{\alpha},$
(15)

where we have introduced a Keldysh-like contour index $s=\pm$ to denote
$K_{\nu_{x,i},+}^{\alpha}=K_{\nu_{x,i}}^{\alpha}$, and
$K_{\nu_{x,i},-}^{\alpha}=K_{\nu_{x,i}}^{\alpha{\dagger}}$. Finally, the
average over measurement outcomes is jian2021syk

$\displaystyle\otimes_{x,i}\sum_{\nu_{x,i}=1,2}\otimes_{\alpha,s}K_{\nu_{x,i},s}^{\alpha}$
$\displaystyle\approx$
$\displaystyle\otimes_{x,i}\left(1-\frac{\mathfrak{s}^{2}}{2}\sum_{\alpha=1}^{n}\sum_{s=\pm}\pi^{+,\alpha}_{x,i,s}\right)$
(16) $\displaystyle\approx$
$\displaystyle\otimes_{x,i}\exp\left(-\frac{\mathfrak{s}^{2}}{2}\sum_{\alpha,s}\pi^{+,\alpha}_{x,i,s}\right)$
(17) $\displaystyle=$ $\displaystyle\exp\left(\frac{\delta
t\mu}{2}\sum_{x,i,\alpha,s}i\psi_{x,L,i,s}^{\alpha}\psi_{x,R,i,s}^{\alpha}\right),$
(18)

in which we have used the relation $\pi^{+}_{x,a,j}+\pi^{-}_{x,a,j}=1$ and
assumed $\mathfrak{s}\ll 1$ and kept orders up to
$\mathcal{O}(\mathfrak{s}^{2})$. In the last line, we introduce
$\mu=\frac{\mathfrak{s}^{2}}{\delta t}$ as the effective monitoring strength.
When the continuum limit $\delta t\rightarrow 0$ is taken, $\mu$ is kept
fixed. This means that the appropriate measurement strength is
$\mathfrak{s}\propto\sqrt{\delta t}$. Constants are neglected because they
will not affect the dynamics. After taking the continuum limit, we arrive at
the following action description of the monitoring,

$\displaystyle-I_{\mu}=\int
dt\frac{i\mu}{2}\sum_{x,i,\alpha,s}\psi_{x,L,i,s}^{\alpha}(t)\psi_{x,R,i,s}^{\alpha}(t),$
(19)

where the summation over time steps results in an integral in the continuum
limit.

Since our model contains not only trajectory average but also disorder average
due to the Brownian coupling, we generalize the quasi entropy to also include
disorder average,

$\displaystyle S_{A}^{(n)}$ $\displaystyle=$
$\displaystyle\frac{1}{1-n}\log\frac{\mathbb{E}\text{Tr}[(\tilde{\rho}_{\bm{\nu}})^{\otimes
n}M_{\text{cyc}}(A)]}{\mathbb{E}(\text{Tr}[\tilde{\rho}_{\bm{\nu}}])^{n}},$
(20)

where $\mathbb{E}$ denotes both the trajectory average and the disorder
average in (10).

##  4 Entanglement entropy in two coupled Brownian SYK clusters

Before we consider the full calculation of entanglement entropy with
monitoring, let us first consider a simpler case without measurement: two
coupled Brownian SYK clusters, considered as a single chain with only two
sites called $x=1,2$. Because the $R$ chain plays no role in this calculation,
we will suppress the chain index in this section, so the relevant fermions are
$\psi_{x,j}=\psi_{x,L,j}$. We will divide the system into two parts, with all
fermions at site $x=1$ in $A$ and all fermions at site $x=2$ in the complement
$\bar{A}$, and refer to $x=1$ as $x\in A$, or simply $A$, and $x=2$ as
$x\in\bar{A}$ or simply $\bar{A}$.

###  4.1 Rényi entropy of a time-evolved EPR state

We consider the following setup: starting from the tensor product of
thermofield double (TFD) states in each of the clusters ($x=1,2$), we compute
the quasi entropy between the two clusters at late time penington2019replica ;
gu2017spread ; chen2020replica ; jian2020note . Because Brownian random
interactions do not conserve energy, we simply consider an infinite
temperature TFD state or, equivalently, a fermionic EPR state for each
cluster. To describe such a state, we double the Hilbert space by introducing
two copies of the fermions, the original fermions $\psi_{x,j}$ and copies
$\chi_{x,j}$, for both clusters $x=1,2$. The initial density matrix is given
by

$\displaystyle\rho=|\text{EPR}\rangle\langle\text{EPR}|,\quad(\psi_{x,j}+i\chi_{x,j})|\text{EPR}\rangle=0,\quad\forall
x=1,2,\quad\forall j=1,...,N.$ (21)

The time evolution is generated by the sum of $\psi$ and $\chi$ Hamiltonians,
$H(\psi)$ and $H(-i\chi)$. This choice implies that
$H(\psi)|\text{EPR}\rangle=H(-i\chi)|\text{EPR}\rangle$. The time evolved
density matrix becomes

$\displaystyle\rho(T)=U(T)\rho U^{\dagger}(T),\quad
U(T)=\mathcal{T}e^{-i\int_{0}^{T/2}dt(H(\psi)+H(-i\chi))},$ (22)

where $\mathcal{T}$ denotes time ordering. This joint $\psi$-$\chi$ evolution
of the EPR state for time $T/2$ is equivalent to just evolving with
$U_{\psi}(T)=\mathcal{T}e^{-i\int_{0}^{T}dtH(\psi)}$ or
$U_{\chi}(T)=\mathcal{T}e^{-i\int_{0}^{T}dtH(-i\chi)}$ for time $T$ by virtue
of the identity $H(\psi)|\text{EPR}\rangle=H(-i\chi)|\text{EPR}\rangle$.

Without measurement, the quasi entropy is same as annealed average of the
Rényi entropy,

$\displaystyle S_{A}^{(n)}=\frac{1}{1-n}\log\mathbb{E}\text{Tr}[\rho^{\otimes
n}M_{\text{cyc}}(A)].$ (23)

For the EPR state, this quantity is related to the two-point function of
cyclic permutation operators. Notice that the cyclic permutation operator
factorizes, $M_{\text{cyc}}(A)=M_{\text{cyc}}^{\psi}(A)\otimes
M_{\text{cyc}}^{\chi}(A)$ for the doubled Hilbert space, where each of them is
given by (see Appendix A)

$\displaystyle M^{\psi}_{\text{cyc}}(A)=\prod_{x\in
A}\prod_{j=1}^{N}e^{\frac{\pi}{2}\psi^{1}_{x,j}\psi^{2}_{x,j}}e^{\frac{\pi}{2}\psi^{3}_{x,j}\psi^{4}_{x,j}}...e^{\frac{\pi}{2}\psi^{n-3}_{x,j}\psi^{n-2}_{x,j}}e^{\frac{\pi}{2}\psi^{n-1}_{x,j}\psi^{n}_{x,j}},$
(24)

where the superscript denotes replica index, and the same for the $\chi$
fermions. Then the trace can be cast into

$\displaystyle\text{Tr}[\rho(T)^{\otimes n}M_{\text{cyc}}(A)]$
$\displaystyle=$ $\displaystyle\langle\text{EPR}|^{\otimes
n}U^{\dagger}_{\chi}(T)^{\otimes n}[M_{\text{cyc}}^{\psi}(A)\otimes
M_{\text{cyc}}^{\chi}(A)]U_{\chi}(T)^{\otimes n}|\text{EPR}\rangle^{\otimes
n}$ (25) $\displaystyle=$ $\displaystyle\langle\text{EPR}|^{\otimes
n}M_{\text{cyc}}^{\chi{\dagger}}(A)U^{\dagger}_{\chi}(T)^{\otimes
n}M_{\text{cyc}}^{\chi}(A)U_{\chi}(T)^{\otimes n}|\text{EPR}\rangle^{\otimes
n}$ (26) $\displaystyle=$ $\displaystyle
2^{-nN/2}\text{Tr}[M_{\text{cyc}}^{\chi{\dagger}}(A)U^{\dagger}_{\chi}(T)^{\otimes
n}M_{\text{cyc}}^{\chi}(A)U_{\chi}(T)^{\otimes n}],$ (27)

where in the first line we use
$\rho(T)=U_{\chi}(T)|\text{EPR}\rangle\langle\text{EPR}|U_{\chi}^{\dagger}(T)$,
in the second line we note
$M^{\chi}_{\text{cyc}}(A)|\text{EPR}\rangle^{\otimes
n}=M^{\psi{\dagger}}_{\text{cyc}}(A)|\text{EPR}\rangle^{\otimes n}$, and in
the last line because the operators contain only $\psi$ fields, we can trace
over the $\chi$ Hilbert space. As is seen from the last expression, the quasi
entropy of the EPR state at time $T$ is given by the averaged two-point
correlation function of the cyclic permutation operators. If $T=0$, it is easy
to see the trace is one, consistent with the initial state being a pure state.

To get a path integral representation of the trace, it is useful to write it
as

$\displaystyle\text{Tr}[\rho(T)^{\otimes n}M_{\text{cyc}}(A)]$
$\displaystyle=$ $\displaystyle\langle\text{EPR}|^{\otimes
n}M_{\text{cyc}}^{\chi{\dagger}}(A)[U^{\dagger}_{\chi}(T)\otimes
U_{\psi}(T)]^{\otimes n}M_{\text{cyc}}^{\chi}(A)|\text{EPR}\rangle^{\otimes
n}.$ (28)

If there were not permutation operators, then the trace would be one since
these two time evolutions cancel for the EPR state. One then recognizes
$U_{\psi}(T)$ and $U^{\dagger}_{\chi}(T)$ as the forward and backward
evolution in Keldysh contour language. Indeed, the boundary conditions for the
fields in conventional Keldysh field theory are altland2010condensed

$\displaystyle\psi_{x,j,+}^{\alpha}(0)=-\psi^{\alpha}_{x,j,-}(0),\quad\psi_{x,j,+}^{\alpha}(T)=\psi_{x,j,-}^{\alpha}(T),\quad\forall
j=1,...,N,\quad\forall\alpha=1,...,n,$ (29)

where the minus sign at time $t=0$ is from the fermionic coherent state path
integral. Recall that for an EPR state,

$\displaystyle(\psi^{\alpha}_{x,j}+i\chi^{\alpha}_{x,j})|\text{EPR}\rangle^{\otimes
n}=0,\quad\langle\text{EPR}|^{\otimes
n}(\psi^{\alpha}_{x,j}-i\chi^{\alpha}_{x,j})=0,\quad\forall
j=1,...,N,\quad\forall\alpha=1,...,n,$

which suggests that by taking $\psi_{x,j}^{\alpha}\rightarrow\psi_{x,j,+}$ and
$i\chi_{x,j}^{\alpha}\rightarrow\psi_{x,j,-}^{\alpha}$, we can write the trace
as a $n$-replicated Keldysh field theory with a total of $2n$ contours. In the
following, we will use notation like $\psi_{+}$, $\psi_{-}$ for Keldysh
fields, while notation like $\psi$, $\chi$ for operators defined in EPR
states.

The presence of the permutation operator then changes the boundary conditions
for the fermions in subsystem $A$, for $x\in A$,

$\displaystyle\psi^{\alpha}_{x,j}M_{\text{cyc}}^{\chi}(A)|\text{EPR}\rangle^{\otimes
n}$ $\displaystyle=$ $\displaystyle
M_{\text{cyc}}^{\chi}(A)(-i\chi^{\alpha}_{x,j})|\text{EPR}\rangle^{\otimes n}$
(31) $\displaystyle=$ $\displaystyle
M_{\text{cyc}}^{\chi}(A)(-i\chi^{\alpha}_{x,j})M_{\text{cyc}}^{\chi{\dagger}}(A)M_{\text{cyc}}^{\chi}(A)|\text{EPR}\rangle^{\otimes
n}$ (32) $\displaystyle=$
$\displaystyle\sum_{\beta}\text{sgn}(\alpha-\beta)\delta^{\alpha+1,\beta}(-i\chi^{\beta}_{x,j})M_{\text{cyc}}^{\chi}(A)|\text{EPR}\rangle^{\otimes
n}.$ (33)

In the last line, when $\alpha=n$, the symbol means
$\delta^{n+1,\beta}=\delta^{1,\beta}$. The $n$-replicated Keldysh field theory
is coupled through boundary conditions due to the permutation operator. We can
redefine the fields in the backward contours to have the conventional boundary
conditions, and so bring the effect of the twisted boundary conditions to the
bulk action, i.e.,

$\displaystyle\sum_{\beta}\text{sgn}(\alpha-\beta)\delta^{\alpha+1,\beta}\psi_{x,j,-}^{\beta}\rightarrow\psi_{x,j,-}^{\alpha},\quad\forall
x\in A,\quad\forall\alpha=1,...,n,\quad\forall j=1,...N.$ (34)

With this field redefinition, we can derive the path integral,
$\mathbb{E}\text{Tr}[\rho^{\otimes n}M_{\text{cyc}}(A)]=\int DGD\Sigma
e^{-I}$, where the $G$-$\Sigma$ action reads penington2019replica ;
jian2021syk ; jian2020note

$\displaystyle-\frac{I}{N}$ $\displaystyle=$
$\displaystyle\sum_{x=1,2}\Big{[}\log\text{Pf}[\hat{S}\partial_{t}-\hat{\Sigma}_{x}]+\int
dt_{1}dt_{2}\big{(}-\frac{1}{2}\Sigma_{x,ss^{\prime}}^{\alpha\beta}G_{x,ss^{\prime}}^{\alpha\beta}+\frac{J}{8}\delta(t_{12})c_{ss^{\prime}}(2G_{x,ss^{\prime}}^{\alpha\beta}(t_{1},t_{2}))^{2}\big{)}\Big{]}$
(35) $\displaystyle+\frac{U}{2q}\int
dt_{1}dt_{2}\delta(t_{12})c_{ss^{\prime}}(2G_{1,ss^{\prime}}^{\alpha\beta}(t_{1},t_{2}))^{q/2}M^{\alpha\gamma}_{s}M^{\beta\delta}_{s^{\prime}}(2G_{2,ss^{\prime}}^{\gamma\delta}(t_{1},t_{2}))^{q/2},$

where $t_{12}\equiv t_{1}-t_{2}$, and $s=\pm$ is introduced to denote the
forward and backward evolution. $S_{++}=1$, $S_{--}=-1$, $S_{+-}=S_{-+}=0$,
and $c_{++}=c_{--}=-1$, $c_{+-}=c_{-+}=1$ capture the structure of the forward
and backward evolutions. The summation over the replica indices and the
contour indices is implicit. The $M$ matrix is defined as
$M^{\alpha\beta}_{+}=\delta^{\alpha\beta}$,
$M^{\alpha\beta}_{-}=\epsilon^{\alpha\beta}$, where
$\epsilon^{\alpha\beta}\equiv\delta^{\alpha+1,\beta}$ is the cyclic
permutation matrix originating from the redefinition in (34). Because our
model conserves the local Fermi parity, the sign from the redefinition (34)
disappears. The boundary conditions are the same as (29). In the large-$N$
action, there is an emergent $SO(N)$ symmetry among the flavors of Majorana
$\psi_{x,j,s}^{\alpha}$, $j=1,...,N$. For simplicity, we will often talk about
the Majorana fermions simply as $\psi_{x,s}^{\alpha}$ without referring to
each individual flavor.

###  4.2 Saddle-point solutions given by permutations

The saddle-point equations following from (35) are

$\displaystyle\hat{G}^{-1}_{x}$ $\displaystyle=$
$\displaystyle\hat{S}\partial_{t}-\hat{\Sigma}_{x},$ (36)
$\displaystyle\Sigma_{x,ss^{\prime}}^{\alpha\beta}$ $\displaystyle=$
$\displaystyle
c_{ss^{\prime}}\delta(t_{12})\Big{[}J(2G_{x,ss^{\prime}}^{\alpha\beta})+U(2G_{x,ss^{\prime}}^{\alpha\beta})^{q/2-1}[M_{(x)}]^{\alpha\gamma}_{s}[M_{(x)}]^{\beta\delta}_{s^{\prime}}(2G_{\bar{x},ss^{\prime}}^{\gamma\delta})^{q/2}\Big{]},$
(37)

where we have defined $[M_{(1)}]_{s}^{\alpha\beta}=M_{s}^{\alpha\beta}$,
$[M_{(2)}]_{s}^{\alpha\beta}=M_{s}^{\beta\alpha}$, and $\bar{1}=2$,
$\bar{2}=1$. Because we are interested in the steady state in the long-time
limit, we assume the solution depends only on the time difference. Then the
first equation can be solved by Fourier transform,

$\displaystyle\hat{G}^{-1}_{x}(\omega)$ $\displaystyle=$
$\displaystyle-i\omega\hat{S}-\hat{\Sigma}_{x}(\omega),$ (38)

where
$\hat{G}_{x}(t_{1},t_{2})=\int\frac{d\omega}{2\pi}\hat{G}_{x}(\omega)e^{i\omega
t_{12}}$,
$\hat{\Sigma}_{x}(t_{1},t_{2})=\int\frac{d\omega}{2\pi}\hat{\Sigma}_{x}(\omega)e^{i\omega
t_{12}}$, with $t_{12}=t_{1}-t_{2}$. The Fourier transform is over the
continuous frequency as we take $T\rightarrow\infty$. Moreover, because the
self energy is proportional to a Dirac $\delta$ function, $\delta(t_{12})$, in
frequency space it is just a constant,
$\hat{\Sigma}_{x}(\omega)=\hat{\Sigma}_{x}$. In the following, we will always
use $\hat{\Sigma}_{x}$ without argument to denote the constant self energy in
frequency space.

An inspection of the self energy equation of motion,

$\displaystyle\Sigma_{1,-+}^{\alpha\beta}$ $\displaystyle=$ $\displaystyle
J(2G_{1,-+}^{\alpha\beta})+U(2G_{1,-+}^{\alpha\beta})^{q/2-1}\sum_{\gamma}\epsilon^{\alpha\gamma}(2G_{2,-+}^{\gamma\beta})^{q/2},$
(39) $\displaystyle\Sigma_{2,-+}^{\alpha\beta}$ $\displaystyle=$
$\displaystyle
J(2G_{2,-+}^{\alpha\beta})+U(2G_{2,-+}^{\alpha\beta})^{q/2-1}\sum_{\gamma}(\epsilon^{T})^{\alpha\gamma}(2G_{2,-+}^{\gamma\beta})^{q/2},$
(40)

suggests that the solution is given by permutation matrices. Let $\tau$ denote
a general $n$-by-$n$ permutation matrix. It is easy to check for any pairs of
permutation matrices, $(\tau_{A},\tau_{\bar{A}})$, satisfying
$\tau_{A}=\epsilon\tau_{\bar{A}}$, there is a solution given by the Green’s
function,
$\hat{G}_{A}(t_{1},t_{2})=\hat{G}(t_{1},t_{2},\tau_{A}),\hat{\Sigma}_{A}=\hat{\Sigma}(\tau_{A})$,
and
$\hat{G}_{\bar{A}}(t_{1},t_{2})=\hat{G}(t_{1},t_{2},\tau_{\bar{A}}),\hat{\Sigma}_{\bar{A}}=\hat{\Sigma}(\tau_{\bar{A}})$,
where the functions are defined by

$\displaystyle\hat{G}(t_{1},t_{2},\tau)=\frac{e^{-\Lambda|t_{12}|}}{2}\left(\begin{array}[]{cccc}\text{sgn}(t_{12})&-\tau^{T}\\\
\tau&-\text{sgn}(t_{12})\end{array}\right),\quad\hat{\Sigma}(\tau)=\Lambda\left(\begin{array}[]{cccc}0&-\tau^{T}\\\
\tau&0\end{array}\right),$ (45)

where $t_{12}=t_{1}-t_{2}$, $\Lambda=J+U$, the solution is written in the
Keldysh space and $\tau$ is a $n$-by-$n$ permutation matrix in the replica
space.

To show these are the center-of-mass time invariant solutions, we take the
transpose of the first equation and use the antisymmetric property of the
bilocal field $\hat{G}^{T}=-\hat{G}$ to get stanford2021subleading

$\displaystyle\partial_{t_{1}}\hat{S}\hat{G}_{x}(t_{1},t_{2})+\partial_{t_{2}}\hat{G}_{x}(t_{1},t_{2})\hat{S}=[\hat{\Sigma}_{x}(t_{1},t_{2}),\hat{G}_{x}(t_{1},t_{2})].$
(46)

For the Brownian model, we have further
$\hat{\Sigma}_{x}(t_{1},t_{2})=\hat{\Sigma}_{x}(t_{1})\delta(t_{1}-t_{2})$.
Setting $t_{1}=t_{2}$, the differential equation for center-of-mass time is

$\displaystyle\partial_{t}\hat{S}\hat{G}_{x}(t,t)\hat{S}$ $\displaystyle=$
$\displaystyle[\hat{\Sigma}_{x}(t),\hat{G}_{x}(t,t)].$ (47)

The right-hand side vanishes for the saddle-point solutions (45).

The solution given by (45) is onshell in the bulk $0<t<T$, but does not
satisfy the boundary condition. The full dynamics set by the equation of
motion (47) with the boundary condition (29) is a complicated problem for
general $n$, involving multiple-variable non-linear differential equations for
which a general solution does not exist. The solution given by a permutation
matrix (45) is actually a fixed point of the equation of motion in the long-
time limit. We show in Appendix B that the real onshell solution in the long-
time limit is given by (45) in the bulk, with deviations from it near the
boundary to account for the correct boundary conditions. The deviation is
exponentially suppressed as time moves away from the boundary, with a time
scale set by $\frac{1}{\sqrt{U(2J+U)}}$. In the following, we will see that
the solution (45) gives an essential contribution to the entanglement of the
steady state. In Appendix B, we also argue that the solution (45) correctly
captures the time-independent part of the onshell action at the late time.
There are two further remarks on the solution given by (45) being the relevant
one. First, summing over these solutions correctly leads to the Page value in
the long-time limit. If there were a contribution due to the deviation near
the boundary, the Rényi entropy would have involved microscopic parameters
like $U/J$, which is not reasonable for a scrambling system at late time.
Second, it is reasonable to expect that when the $n\rightarrow 1$ limit is
taken, the effect of the boundary conditions on the solutions vanishes as
there is nothing to permute at $n=1$. This is similar to the vanishing of
backreaction in the gravity setup almheiri2019replica ; penington2019replica .

Actually, for any choice of $s^{\alpha\beta}_{x}=\pm$, $\alpha\neq\beta$ (we
define $s^{\alpha\alpha}_{x}=1$ for completeness), there is a center-of-mass
time invariant solution given by the modified permutation matrix
$\tau_{x}^{\alpha\beta}\equiv s^{\alpha\beta}_{x}\tau^{\alpha\beta}$, namely,
$G_{x\in A}=G(t_{1},t_{2},\tau_{A,x})$,
$G_{x\in\bar{A}}=G(t_{1},t_{2},\tau_{\bar{A},x})$, where we have referred to
$x=1$ as $x\in A$, and $x=2$ as $x\in\bar{A}$. The emergence of these
solutions is due to the Fermi parity symmetry. The theory (9) has Fermi parity
conservation at each site, whose transformation law is
$\psi_{x,j}\rightarrow-\psi_{x,j},j=1,...,N$. When we extend the theory to $n$
replicas, the symmetry is also extended to independent transformation in each
replica, namely, $\psi_{x,\pm}^{\alpha}\rightarrow-\psi_{x,\pm}^{\alpha}$ in
terms of the Keldysh field 111Note that the large-$N$ action has an emergent
$SO(N)$ symmetry, and we can talk about $\psi_{x,j,s}^{\psi}$ without
referring to an individual $j$.. The solutions given by
$\tau_{x}^{\alpha\beta}$ can be divided into classes that within each class
they are related to each other via these transformations. There are
inequivalent classes of solutions distinguished by,

$\displaystyle
P_{x}=\prod_{\alpha=1}^{n}(2i\chi_{x}^{\alpha}\psi_{x}^{\alpha}).$ (48)

Because of the $SO(N)$ symmetry, the Fermi parity is conserved for each
individual flavor. Thus, for each site there are two distinct classes of
solutions for each permutation matrix $\tau_{x}$ distinguished by $P_{x}=\pm
1$.

Put it in another way, the saddle-point solution spontaneously breaks the
local Fermi parity symmetry, and the parity transformation can bring one
solution to the other. Those solutions that can be connected by the
transformation is distinguished by $P_{x}=\pm 1$ in (48), i.e., there are two
inequivalent classes for each site. Since the initial state is a product EPR
state between each site where for each site, the Fermi parity is
$P_{x}=1,\forall x$, there is a unique class of solutions that is allowed by
the initial state.

###  4.3 Summing over all saddle points

Let us now compute the onshell action (35) by plugging in the saddle-point
solution. First notice that the forward and backward evolution cancels because
of $c_{ss^{\prime}}$,

$\displaystyle\int
dt\sum_{x}\frac{J}{8}c_{ss^{\prime}}(2G_{x,ss^{\prime}}^{\alpha\beta}(t,t))^{2}+\frac{U}{2q}c_{ss^{\prime}}(2G_{1,ss^{\prime}}^{\alpha\beta}(t,t))^{q/2}M^{\alpha\gamma}_{s}M^{\beta\delta}_{s^{\prime}}(2G_{2,ss^{\prime}}^{\gamma\delta}(t,t))^{q/2}=0.$
(49)

There is also an exponential decaying part,

$\displaystyle\exp\left(-\frac{1}{2}\int
dt\sum_{x}\Sigma_{x,ss^{\prime}}^{\alpha\beta}G_{x,ss^{\prime}}^{\alpha\beta}(t,t)\right)=\exp(-n\Lambda
T).$ (50)

We discuss the Pfaffian in detail. The evaluation of the Pfaffian can be
mapped to a Kitaev chain problem. Because the large-$N$ action (35) has an
$SO(N)$ symmetry, the Pfaffian at each site (we first suppress the site index,
and later restore it when we are familiar with the evaluation for a single
site) is equivalent to a transition amplitude of an EPR state of an effective
single flavor Majorana fermion for two contours of each replica,
$\psi_{\pm}^{\alpha}$. The EPR state is
$(\psi^{\alpha}+i\chi^{\alpha})|\text{EPR}\rangle=0$, $\forall\alpha=1,...,n$,
and the Pfaffian is

$\displaystyle\text{Pf}\Big{[}\hat{S}\partial_{t}-\Sigma\Big{]}=\langle\text{EPR}|e^{TH}|\text{EPR}\rangle,\quad
H=\frac{1}{2}(\psi,\chi)\left(\begin{array}[]{cccc}1&0\\\
0&i\end{array}\right)\Sigma\left(\begin{array}[]{cccc}1&0\\\
0&i\end{array}\right)\left(\begin{array}[]{cccc}\psi\\\
\chi\end{array}\right).$ (57)

Here the self-energy $\Sigma$ is given by some $n$-by-$n$ permutation matrix.

Now, all permutations of $n$ elements can be decomposed into cyclic
permutations (or cycles) with length $n_{\check{\tau}_{(i)}}$ such that
$\sum_{i}n_{\check{\tau}_{(i)}}=n$. We use a check over the letter,
$\check{\tau}$, to denote cycles. Here since the problem is quadratic, we can
indeed simplify the Hamiltonian by taking the permutation matrix $\tau$ into
block diagonal individual cycles $\tau=\oplus_{i}\check{\tau}_{(i)}$, and for
each of these cycles, we can bring them to the following canonical form,

$\displaystyle\check{\tau}_{(i)}^{\alpha\beta}=\begin{cases}\delta^{\alpha+1,\beta},&n_{\check{\tau}_{(i)}}=\text{odd}\\\
\text{sgn}(\beta-\alpha)\delta^{\alpha+1,\beta},&n_{\check{\tau}_{(i)}}=\text{even}\end{cases},$
(58)

where we defined $\delta^{n+1,\beta}=\delta^{1,\beta}$. Note that for the even
length cycles, it has an additional minus sign to be consistent with the even
Fermi parity $P_{x}=1$.

Now the Hamiltonian decomposes into individual cycles. Let us consider a cycle
$\check{\tau}$ in the canonical form with length $n_{\check{\tau}}$. We define

$\displaystyle
H(\check{\tau})=\frac{\Lambda}{2}(\psi,\chi)\left(\begin{array}[]{cccc}0&-i\check{\tau}^{T}\\\
i\check{\tau}&0\end{array}\right)\left(\begin{array}[]{cccc}\psi\\\
\chi\end{array}\right),$ (63)

which is nothing but a Kitaev chain kitaev2001unpaired  with length
$n_{\check{\tau}}$. Using the complex fermion,

$\displaystyle c_{\alpha}=\frac{\psi^{\alpha}+i\chi^{\alpha}}{\sqrt{2}},\quad
c_{\alpha}^{\dagger}=\frac{\psi^{\alpha}-i\chi^{\alpha}}{\sqrt{2}},$ (64)

satisfying $\\{c_{\alpha},c_{\beta}^{\dagger}\\}=\delta_{\alpha\beta}$, we can
bring the Hamiltonian into

$\displaystyle H(\check{\tau})$ $\displaystyle=$
$\displaystyle\frac{\Lambda}{2}\sum_{\alpha=1}^{n_{\check{\tau}}}[(c_{\alpha}^{\dagger}c_{\alpha+1}+c_{\alpha}c_{\alpha+1}+h.c.)+((-1)^{n_{\check{\tau}}+1}c_{n}c_{1}^{\dagger}+h.c.)]$
(65) $\displaystyle=$
$\displaystyle\Lambda\sum_{k}(c_{k}^{\dagger},c_{-k})\left(\begin{array}[]{cccc}-\cos
k&-i\sin k\\\ i\sin k&\cos
k\end{array}\right)\left(\begin{array}[]{cccc}c_{k}\\\
c_{-k}^{\dagger}\end{array}\right)+\text{mod}(n_{\check{\tau}},2)\Lambda(c_{0}^{\dagger}c_{0}+\frac{1}{2}),$
(70)

where we have the BdG Hamiltonian altland2010condensed  in the second line.
The choice of momentum needs some explanation because the odd and even lengths
have different boundary conditions. When $n_{\check{\tau}}$ is an odd integer,
the fermion satisfies periodic boundary condition,
$k=\frac{2j\pi}{n_{\check{\tau}}}$, and the summation is over
$j=1,...,\frac{n_{\check{\tau}}-1}{2}$. When $n_{\check{\tau}}$ is an even
integer, the fermion satisfies anti-periodic boundary condition,
$k=\frac{(2j-1)\pi}{n_{\check{\tau}}}$, and the summation is over
$j=1,...,\frac{n_{\check{\tau}}}{2}$. The last term with zero momentum in the
second line appears only for odd integer $n_{\check{\tau}}$ because it is
allowed under periodic boundary conditions.

It is easy to diagonalize the Hamiltonian with the Bogoliubov quasiparticle
operators,

$\displaystyle d_{-,k}$ $\displaystyle=$
$\displaystyle-i\cos\frac{k}{2}c_{k}+\sin\frac{k}{2}c_{-k}^{\dagger}$ (71)
$\displaystyle d_{+,k}$ $\displaystyle=$ $\displaystyle
i\sin\frac{k}{2}c_{k}+\cos\frac{k}{2}c_{-k}^{\dagger},$ (72)

such that

$\displaystyle H(\check{\tau})$ $\displaystyle=$
$\displaystyle\Lambda\sum_{k}(d_{+,k}^{\dagger}d_{+,k}-d_{-,k}^{\dagger}d_{-,k})+\text{mod}(n_{\check{\tau}},2)\Lambda(c_{0}^{\dagger}c_{0}+\frac{1}{2}).$
(73)

Each of the eigenmodes is independent for the choice of momentum discuss in
above, and the exponential of the Hamiltonian is given by

$\displaystyle e^{TH(\check{\tau})}$ $\displaystyle=$ $\displaystyle
e^{\text{mod}(n_{\check{\tau}},2)\frac{\Lambda}{2}}\left(1+(e^{\Lambda
T}-1)c_{0}^{\dagger}c_{0}\right)$ (75)
$\displaystyle\times\prod_{k}\left(1+(e^{\Lambda
T}-1)d_{+,k}^{\dagger}d_{+,k}+1+(e^{-\Lambda
T}-1)d_{-,k}^{\dagger}d_{-,k}\right)$ $\displaystyle\ni$ $\displaystyle
e^{\text{mod}(n_{\check{\tau}},2)\frac{\Lambda}{2}}\prod_{k}e^{\Lambda
T}\cos^{2}\frac{k}{2}\,c_{k}c_{k}^{\dagger},$ (76)

where because the EPR state is the vacuum of a complex fermion (64), we
express the Bogoliubov quasiparticle using the complex fermion and show the
largest nonvanishing component for each $k$ in the second line. Then it is
straightforward to evaluate the transition amplitude of the EPR state,

$\displaystyle\langle\text{EPR}|e^{TH(\check{\tau})}|\text{EPR}\rangle=e^{\frac{n_{\check{\tau}}}{2}\Lambda
T}\prod_{k}\cos^{2}\frac{k}{2}=e^{\frac{n_{\check{\tau}}}{2}\Lambda
T}2^{-n_{\check{\tau}}+1}.$ (77)

where the range of product of $k$ is the same as the summation we discussed
before.

We now know the contribution for a cycle $\check{\tau}$. It is time to discuss
the full solution including different sites. For the full solution, recall
that the solution in $x\in A$ and $x\in\bar{A}$ are related by the cyclic
permutation $\epsilon$, namely, $\tau_{A}=\epsilon\tau_{\bar{A}}$. Suppose
that they can be decomposed into $m_{\text{cyc}}$ number cycles, i.e., those
two permutation matrices can be built using $\check{\tau}_{(i)}$,
$i=1,2,...,m_{\text{cyc}}$. The sum of the length of each cycles should
satisfy $\sum_{j=1}^{m_{\text{cyc}}}n_{\check{\tau}_{(j)}}=2n$. Then the
Pfaffian becomes

$\displaystyle\prod_{x=1,2}\text{Pf}\Big{[}\hat{S}\partial_{t}-\Sigma_{x}\Big{]}=\langle\text{EPR}|e^{T\sum_{i=1}^{m_{\text{cyc}}}H(\check{\tau}_{(i)})}|\text{EPR}\rangle=\prod_{i=1}^{m_{\text{cyc}}}e^{\frac{n_{\check{\tau}}}{2}\Lambda
T}2^{-n_{\check{\tau}}+1}=e^{n\Lambda T}2^{-2n+m_{\text{cyc}}}.$

Because $n$ is the number of replicas, it is natural to expect the action to
have a factor of $n$. What is interesting is that the Pfaffian depends on the
number of cycles in the solution, $m_{\text{cyc}}$, and does not depend on
other details of the permutation matrix.

For a pair of permutations, $(\sigma,\tau)$, that are related by
$\sigma=\epsilon\tau$, it is known that the maximal possible number of cycles
that can be decomposed from them is $m_{\text{cyc}}=n+1$ jacques1968genre ;
dulucq1998combinatorial . Then we have

$\displaystyle\prod_{x=1,2}\text{Pf}\Big{[}\hat{S}\partial_{t}-\Sigma_{x}\Big{]}=\langle\text{EPR}|e^{T\sum_{i=1}^{n+1}H(\check{\tau}_{(i)})}|\text{EPR}\rangle=e^{n\Lambda
T}2^{-n+1}.$ (79)

The exponential growing in time part cancels the decaying part in (50), and
the onshell action becomes $e^{-I_{\text{onshell}}}=e^{(1-n)N\log 2}$.

For those pairs with less cycles, the contribution will be exponentially
suppressed, i.e., $2^{Nm_{\text{cyc}}}$, so we ignore them. There are $C_{n}$
pairs of permutation matrices, $(\sigma,\tau)$, that are related by
$\sigma=\epsilon\tau$, having the maximal number of cycles, where $C_{n}$ is
the Catalan number jacques1968genre ; dulucq1998combinatorial . In addition to
that, we have also a degeneracy coming from the local Fermi parity symmetry
$\psi^{\alpha}_{j}\rightarrow-\psi^{\alpha}_{j},j=1,...,N$ independently for
each replica $\alpha=1,..,n$, which leads to a number
$(2^{n}/2)^{2}=2^{2(n-1)}$ 222The $1/2$ inside the parentheses is due to the
fact that the Fermi parity of the initial state is even at each site, so that
only half of solutions are allowed, while the $2$ in the power is because we
have two sites.. Taking into account these degenerate saddle points, the
steady state Rényi entropy is given by

$\displaystyle S_{A}^{(n)}=(N-2)\log 2+\frac{1}{1-n}\log C_{n}.$ (80)

To get the von Neumann entanglement entropy, it is useful to represent the
Catalan number as $C_{n}=\frac{1}{2\pi}\int_{0}^{4}dxx^{n-1}\sqrt{(4-x)x}$. We
can analytically continue the number to $n\rightarrow 1$, and this leads to

$\displaystyle\lim_{n\rightarrow 1}\frac{1}{1-n}\log C_{n}$ $\displaystyle=$
$\displaystyle\lim_{n\rightarrow
1}\frac{1}{1-n}\log\left(\frac{1}{2\pi}\int_{0}^{4}dx\sqrt{(4-x)x}-(1-n)\frac{1}{2\pi}\int_{0}^{4}dx\log
x\sqrt{(4-x)x}\right)$ (81) $\displaystyle=$ $\displaystyle\lim_{n\rightarrow
1}\frac{1}{1-n}\log(1-\frac{1}{2}(1-n))=-\frac{1}{2}.$

so the von Neumann entanglement entropy is $S_{A}=N\log 2-2\log
2-\frac{1}{2}$. The first $1/N$ is due to the Fermi parity symmetry, i.e.,
starting from an EPR state the steady state can explore half of the total
Hilbert space at each site jian2020note . The second $1/N$ correction is
consistent with the Page value in the limit of large Hilbert space dimension
page1993average .

We can do better than just calculating the von Neumann entanglement entropy.
This explicit form (80) allows us to calculate the entanglement spectrum
density of states. This is done by considering the resolvant of reduced
density matrix of subsystem $A$ penington2019replica ,

$\displaystyle
R(\lambda)=\frac{1}{\lambda-\rho_{A}}=\frac{1}{\lambda}+\sum_{k=1}^{\infty}\rho_{A}^{k},\quad\rho_{A}=\text{Tr}_{\bar{A}}\rho.$
(82)

The entanglement spectrum density of states is given by the resolvant through
333As we know the steady state can only explore the Hilbert space with even
Fermi parity at each site, we restrict the trace to that subspace, e.g.,
$\text{Tr}[1]=2^{N-2}$.

$\displaystyle D(\lambda)=\frac{1}{2\pi i}\lim_{\delta\rightarrow
0}(\text{Tr}[R(\lambda-i\delta)]-\text{Tr}[R(\lambda+i\delta)]).$ (83)

The trace of resolvant can be obtained by the saddle-point action,

$\displaystyle\text{Tr}[R(\lambda)]$ $\displaystyle=$
$\displaystyle\frac{1}{\lambda}2^{N-2}+\sum_{k=1}^{\infty}\frac{\text{Tr}[\rho_{A}^{k}(\lambda)]}{\lambda^{k+1}}=\frac{1}{\lambda}2^{N-2}+\sum_{k=1}^{\infty}\frac{\text{Tr}[\rho^{\otimes
k}M_{\text{cyc}}(A)]}{\lambda^{k+1}}$ (84) $\displaystyle=$
$\displaystyle\frac{1}{\lambda}2^{N-2}+\sum_{k=1}^{\infty}\frac{C_{k}e^{(1-k)(N-2)\log
2}}{\lambda^{k+1}}=2^{2N-5}\left(1-\sqrt{1-\frac{2^{4-N}}{\lambda}}\right),$
(85)

where in the third equality we used the saddle-point results and in the last
equality we performed the summation. The density of state function becomes

$\displaystyle
D(\lambda)=\frac{2^{2N-5}}{\pi}\frac{\sqrt{\lambda(2^{4-N}-\lambda)}}{\lambda},$
(86)

which is equivalent to the result of a random pure state page1993average  in
the subsector accessible from the initial EPR state. It is expected since
unitary evolution generated from the Brwonian SYK Hamiltonian approaches the
Haar random unitary at late time jian2021quantum ; stanford2021subleading .
This result strongly supports that the saddle-point solutions given by the
permutation matrix (45) play an essential role in late-time entanglement
entropy.

If there is no cyclic permutation operator in the path integral, we are just
calculating the $n$-th power of the trace of a density matrix. In this case,
the solutions described by the permutation matrix also exists. However, the
relation between solution in $A$ and $\bar{A}$ changes to
$\tau_{A}=\tau_{\bar{A}}$, so the maximal decomposition to cycles is given by
the identity matrix, where each diagonal element is a trivial cycle with
length one. Taking $\tau_{A}$ and $\tau_{\bar{A}}$ to be the identity matrix
where the number of cycles is $m_{\text{cyc}}=2n$ in total, we have from (4.3)

$\displaystyle\prod_{x=1,2}\text{Pf}\Big{[}\hat{S}\partial_{t}-\Sigma_{x}\Big{]}=e^{n\Lambda
T}.$ (87)

The exponential growing part gets cancelled exactly by (50), so we get one,
which is $n$-th power of the trace of a density matrix.

Finally, although here we only consider two coupled clusters, it is
straightforward to generalize the calculation to the half-chain quasi entropy
of a chain with $L$ sites. The calculation is the same: the solution in $A$
and $\bar{A}$ are given by pairs of permutation matrices,
$(\tau_{A},\tau_{\bar{A}})$ with $\tau_{A}=\epsilon\tau_{\bar{A}}$, where now
$A$ is half of the chain and $\bar{A}$ is the rest. Taking into account the
length of the chain and of the subregion $A$, the half-chain entanglement is
given by $S_{A}=\frac{NL}{2}\log 2-L\log 2-\frac{1}{2}$, where the first $1/N$
is due to the Fermi parity symmetry at each site, i.e., starting from the EPR
state, the steady state can only explore the subspace of even Fermi parity at
each site.

##  5 Entanglement phase transition in the monitored Brownian SYK chains

In this section, we will consider the monitored Brownian SYK chains introduced
in Section 3. We focus on the effect of the measurement on the saddle-point
solutions and thus the quasi entropy. Then we take analytical continuation
$n\rightarrow 1$ to get the von Neumann entanglement entropy for the cyclic
symmetric solution.

###  5.1 Saddle-point analysis

To generalize the story in the coupled Brownian SYK clusters into monitored
Brownian SYK chains, we double the degrees of freedom into left and right
chains, and add the effective action description (19) for continuous
monitoring. Similar to the unitary case, we will consider EPR initial states.
It is described by two copies of the fermions, denoted by $\psi_{x,a,j}$ and
$\chi_{x,a,j}$ at every site in both chains. The initial density matrix is
given by

$\displaystyle\rho=|\text{EPR}\rangle\langle\text{EPR}|,\quad(\psi_{x,a,j}+i\chi_{x,a,j})|\text{EPR}\rangle=0,$
(88) $\displaystyle\quad\forall x=1,...,L,\quad\forall a=L,R,\quad\forall
j=1,...,N.$ (89)

We divide the chain into $A$ and $\bar{A}$, and compute the quasi entropy of
the steady state (20). The only modification of the quasi entropy is the
presence of monitoring, and we have shown that the effect of monitoring part
is described by (19). Note that the redefinition (34) will not change the form
of the effective action (19). It is then straightforward to get the
$G$-$\Sigma$ action for $\mathbb{E}\text{Tr}[\tilde{\rho}^{\otimes
n}M_{\text{cyc}}(A)]=\int DGD\Sigma e^{-I(A)}$,

$\displaystyle-\frac{I(A)}{N}=\log\text{Pf}[\hat{S}\partial_{t}-\hat{\Sigma}_{x}]+\int
dt_{1}dt_{2}\Big{(}-\frac{1}{2}\Sigma_{x,ab,ss^{\prime}}^{\alpha\beta}G_{x,ab,ss^{\prime}}^{\alpha\beta}+\frac{J}{8}\delta(t_{12})\delta_{ab}c_{ss^{\prime}}(2G_{x,ab,ss^{\prime}}^{\alpha\beta}(t_{1},t_{2}))^{2}\Big{)}$
$\displaystyle+\frac{U}{4q}\int
dt_{1}dt_{2}\delta(t_{12})\delta_{ab}c_{ss^{\prime}}(2G_{x,ab,ss^{\prime}}^{\alpha\beta}(t_{1},t_{2}))^{q/2}M^{\alpha\gamma}_{s}(x,x+1)M^{\beta\delta}_{s^{\prime}}(x,x+1)(2G_{x+1,ab,ss^{\prime}}^{\gamma\delta}(t_{1},t_{2}))^{q/2}$
$\displaystyle+\frac{i\mu}{2}\int
dt_{1}dt_{2}\delta(t_{12})\delta_{ss^{\prime}}\delta^{\alpha\beta}(\delta_{aL}\delta_{bR}-\delta_{aR}\delta_{bL})G_{x,ab,ss^{\prime}}^{\alpha\beta}(t_{1},t_{2}),$
(90)

where the summation over the site indices, the chain indices, the replica
indices, and the contour indices,
$x,a,b,\alpha,\beta,\gamma,\delta,s,s^{\prime}$, is implicit. The third line
is the monitoring part from (19). In the second line, $M_{s}(x,y)$ is the
cyclic permutation operator at the boundary of intervals $A$ and $\bar{A}$,

$\displaystyle M^{\alpha\beta}_{+}(x,y)$ $\displaystyle=$
$\displaystyle\delta^{\alpha\beta},$ (91) $\displaystyle
M^{\alpha\beta}_{-}(x,y)$ $\displaystyle=$
$\displaystyle\begin{cases}\delta^{\alpha\beta}&x,y\in A\text{ or
}x,y\in\bar{A}\\\ \epsilon^{\alpha\beta}&x\in A\text{ and }y\in\bar{A}\\\
\epsilon^{\beta\alpha}&x\in\bar{A}\text{ and }y\in A\end{cases},$ (92)

which is only nontrivial when $x$ and $y$ are located at different subsystems,
and is introduced to calculate the quasi entropy between $A$ and $\bar{A}$. We
assume the length of $A$ and $\bar{A}$ are equal and given by $L/2$. The
boundary conditions of the fields are similar to (29) with a trivial extension
to left and right chains.

The equations of motion from the action read

$\displaystyle\hat{G}^{-1}_{x}$ $\displaystyle=$
$\displaystyle\hat{S}\partial_{t}-\hat{\Sigma}_{x},$ (93)
$\displaystyle\Sigma_{x,ab,ss^{\prime}}^{\alpha\beta}$ $\displaystyle=$
$\displaystyle\delta(t_{12})\Big{[}J\delta_{ab}c_{ss^{\prime}}(2G_{x,ab,ss^{\prime}}^{\alpha\beta})+i\mu\delta_{ss^{\prime}}\delta^{\alpha\beta}(\delta_{aL}\delta_{bR}-\delta_{aR}\delta_{bL})$
(94)
$\displaystyle+\frac{U}{2}\delta_{ab}c_{ss^{\prime}}\Big{(}(2G_{x,ab,ss^{\prime}}^{\alpha\beta})^{q/2-1}M^{\alpha\gamma}_{s}(x,x+1)M^{\beta\delta}_{s^{\prime}}(x,x+1)(2G_{x+1,ab,ss^{\prime}}^{\gamma\delta})^{q/2}$
$\displaystyle+(2G_{x-1,ab,ss^{\prime}}^{\gamma\delta})^{q/2}M^{\gamma\alpha}_{s}(x-1,x)M^{\delta\beta}_{s^{\prime}}(x-1,x)(2G_{x,ab,ss^{\prime}}^{\alpha\beta})^{q/2-1}\Big{)}\Big{]}.$

We assume that the solutions are uniform in $A$ and $\bar{A}$ up to the parity
transformation. The invariant quantity that distinguishes different classes of
solutions now becomes the total Fermi parity of left and right Majorana
fermions at each site, $P_{x,L}P_{x,R}$. In the presence of measurement, only
the total parity of left and right Majorana fermions at each site is
conserved, i.e., the theory (5.1) is invariant under transformation that acts
on both left and right Majorana fermoins,
$\psi_{x,L/R,\pm}^{\alpha}\rightarrow-\psi_{x,L/R,\pm}^{\alpha}$.

The solution is now modified by the presence of the monitoring. Similar to the
discussion in the previous section, it is easy to check that the ansatz
$G_{x\in A}=G(t_{1},t_{2},\tau_{A,x})$, $\Sigma_{x\in
A}(t_{1},t_{2})=\Sigma(\tau_{A,x})$, and
$G_{x\in\bar{A}}(t_{1},t_{2})=G(t_{1},t_{2},\tau_{\bar{A},x})$,
$\Sigma_{x\in\bar{A}}=\Sigma(\tau_{\bar{A},x})$ solves the equation of motion,
with $\tau_{A}=\epsilon\tau_{\bar{A}}$, where $G(t_{1},t_{2},\tau)$ and
$\Sigma(\tau)$ are now given by

$\displaystyle
G(t_{1},t_{2},\tau)=\frac{e^{-\Lambda|t_{12}|}}{2}\left(\begin{array}[]{cccc}\text{sgn}(t_{12})&-i\sin\theta&-\cos\theta\,\tau^{T}&0\\\
i\sin\theta&\text{sgn}(t_{12})&0&-\cos\theta\,\tau^{T}\\\
\cos\theta\,\tau&0&-\text{sgn}(t_{12})&-i\sin\theta\\\
0&\cos\theta\,\tau&i\sin\theta&-\text{sgn}(t_{12})\end{array}\right),$ (99)
$\displaystyle\Sigma(\tau)=\left(\begin{array}[]{cccc}0&i\mu&-\Lambda\tau^{T}&0\\\
-i\mu&0&0&-\Lambda\tau^{T}\\\ \Lambda\tau&0&0&i\mu\\\
0&\Lambda\tau&-i\mu&0\end{array}\right),$ (104)

The solution is written in the
$\left(\psi_{L,+}^{\alpha},\psi_{R,+}^{\alpha},\psi_{L,-}^{\alpha},\psi_{R,-}^{\alpha}\right)$
basis, and for the Green’s function, we have introduced
$\tan\theta=\frac{\mu}{\Lambda}$, such that when $\theta=0$, it reduces to the
unmeasured case. Recall that
$\tau_{x}^{\alpha\beta}=s_{x}^{\alpha\beta}\tau^{\alpha\beta}$ is the
permutation matrix that characterizes the correlation between different
replicas. The EPR initial state allows the solutions with $P_{x,L}P_{x,R}=1$,
but only the permutation solution for $P_{x,L}=P_{x,R}=1$ can smoothly connect
to the zero measurement limit. Note that in the non-measurement case, the
Fermi parity at two chains is conserved separately. So, we consider this
sector of solutions, which includes $(2^{n}/2)^{L}=2^{L(n-1)}$ 444Because the
parity transformation acts simultaneously on left and right Majorana fermions,
we have a similar degeneracy as before. degenerate solutions from the parity
transformation.

In the presence of measurement, the parameter $\Lambda$ satisfies

$\displaystyle\Lambda=U\left(\frac{\Lambda}{\sqrt{\Lambda^{2}+\mu^{2}}}\right)^{q-1}+J\frac{\Lambda}{\sqrt{\Lambda^{2}+\mu^{2}}}.$
(105)

For small interaction strength $U\ll J$, $\Lambda$ is given by

$\displaystyle\Lambda=J\left(1-\frac{\mu^{2}}{J^{2}}\right)^{\frac{1}{2}}+U\left(1-\frac{\mu^{2}}{J^{2}}\right)^{\frac{q-3}{2}}+\mathcal{O}\left(U^{2}\right).$
(106)

$\Lambda$ vanishes at $\mu=J$. At this point $\theta=\frac{\pi}{2}$, and the
correlation between left and right chains is maximal while the correlation
between different replicas vanishes. The solution restores the replica
symmetry. We will see in the following section that this corresponds to an
entanglement transition, namely, when the replica symmetry is broken, the von
Neumann entanglement entropy satisfies a volume law, while when the replica
symmetry is unbroken, the von Neumann entanglement entropy satisfies an area
law. A plot of the phase diagram for $q=4$ as a function of the monitoring
rate is shown in Fig. 1.

Figure 1: The phase diagram as a function of measurement strength for $q=4$.
The phase boundary is determined by (105). The solid line indicates a
continuous transition, while the dashed line indicates a first-order
transition.

###  5.2 Quasi entropy by summing over all saddle points

Now we evaluate the onshell action on the above solution. Due to the presence
of monitoring, the forward and backward evolution no longer cancel,

$\displaystyle\int
dt\Big{[}\frac{U}{4q}\delta_{ab}c_{ss^{\prime}}(2G_{x,ab,ss^{\prime}}^{\alpha\beta}(t,t))^{q/2}M^{\alpha\gamma}_{s}(x,x+1)M^{\beta\delta}_{s^{\prime}}(x,x+1)(2G_{x+1,ab,ss^{\prime}}^{\gamma\delta}(t,t))^{q/2}$
$\displaystyle+\frac{J}{8}\delta_{ab}c_{ss^{\prime}}(2G_{x,ab,ss^{\prime}}^{\alpha\beta}(t,t))^{q}\Big{]}=\left[\frac{J}{2}\Big{[}\Big{(}\frac{\Lambda}{\sqrt{\Lambda^{2}+\mu^{2}}}\Big{)}^{2}-1\Big{]}+\frac{U}{q}\Big{[}\Big{(}\frac{\Lambda}{\sqrt{\Lambda^{2}+\mu^{2}}}\Big{)}^{q}-1\Big{]}\right]nLT.$

From this expression, we can check that when the monitoring strength is zero,
$\mu=0$, the contributions from the forward and backward evolution do cancel.
The monitoring part itself also leads to a contribution,

$\displaystyle\int
dt\frac{i\mu}{2}\delta_{ss^{\prime}}\delta^{\alpha\beta}(\delta_{aL}\delta_{bR}-\delta_{aR}\delta_{bL})G_{x,ab,ss^{\prime}}^{\alpha\beta}(t,t)=-\frac{\mu^{2}}{\sqrt{\Lambda^{2}+\mu^{2}}}nLT.$
(108)

Similar to the unitary case, we have also the following contribution,

$\displaystyle-\frac{1}{2}\int
dt\Sigma_{x,ab,ss^{\prime}}^{\alpha\beta}G_{x,ab,ss^{\prime}}^{\alpha\beta}(t,t)=-\sqrt{\Lambda^{2}+\mu^{2}}nLT.$
(109)

In the unitary case, this exponential decaying part is cancelled by the
Pfaffian. We will see that it is also true for monitoring case. Nevertheless,
the other parts (5.2) and (108) cannot be canceled. This is consistent with
$\tilde{\rho}_{\nu}$ being an unnormalized density matrix.

The evaluation of the Pfaffian can again be mapped to a transition amplitude
of Kitaev chains, but now we have two coupled Kitaev chains. Similarly,
because the large-$N$ action (5.1) has an $SO(N)$ symmetry, the Pfaffian is
equivalent to the transition amplitude of an EPR state of an effective single
flavor Majorana fermion for two contours of each replica,
$\psi_{x,a,s}^{\alpha}$. The EPR state is
$(\psi_{x,a}^{\alpha}+i\chi_{x,a}^{\alpha})|\text{EPR}\rangle=0$, $\forall
x=1,...,L$, $\forall\alpha=1,...,n$, and $\forall a=L,R$. The Pfaffian is
given by the following transition amplitude,

$\displaystyle\prod_{x=1}^{L}\text{Pf}\Big{[}\hat{S}\partial_{t}-\Sigma_{x}\Big{]}$
$\displaystyle=$
$\displaystyle\langle\text{EPR}|e^{T\sum_{x=1}^{L}H_{x}}|\text{EPR}\rangle,$
(110)

with the Hamiltonian given by

$\displaystyle H_{x}$ $\displaystyle=$
$\displaystyle\sum_{a=L,R}\frac{1}{2}(\psi_{x,a},\chi_{x,a})\left(\begin{array}[]{cccc}1&0\\\
0&i\end{array}\right)\Sigma_{x}\left(\begin{array}[]{cccc}1&0\\\
0&i\end{array}\right)\left(\begin{array}[]{cccc}\psi_{x,a}\\\
\chi_{x,a}\end{array}\right).$ (117)

Similar to the unitary case, the self-energy $\Sigma_{x}$ is given by some
permutation matrix which can be decomposed into cycles with length
$n_{\check{\tau}_{(i)}}$ such that $\sum_{i}n_{\check{\tau}_{(i)}}=n$, and for
each of cycles, we can bring them to the canonical form (58).

Consider a cyclic permutation $\check{\tau}$ at one site (so we suppress the
site index), by extending the previous calculation to left and right chains,
the Pfaffian can be mapped to the amplitude of EPR state with the following
Hamiltonian

$\displaystyle
H(\check{\tau})=\frac{1}{2}(\psi_{L},\chi_{L},\psi_{R},\chi_{R})\left(\begin{array}[]{cccc}0&-i\Lambda\check{\tau}^{T}&i\mu&0\\\
i\Lambda\check{\tau}&0&0&-i\mu\\\ -i\mu&0&0&-i\Lambda\check{\tau}^{T}\\\
0&i\mu&i\Lambda\check{\tau}&0\end{array}\right)\left(\begin{array}[]{cccc}\psi_{L}\\\
\chi_{L}\\\ \psi_{R}\\\ \chi_{R}\end{array}\right).$ (126)

Using the complex fermion,

$\displaystyle
c_{a,\alpha}=\frac{\psi^{\alpha}_{a}+i\chi^{\alpha}_{a}}{\sqrt{2}},\quad
c_{a,\alpha}^{\dagger}=\frac{\psi^{\alpha}_{a}-i\chi^{\alpha}_{a}}{\sqrt{2}},$
(127)

satisfying $\\{c_{a,\alpha},c_{b,\beta}\\}=\delta_{ab}\delta_{\alpha\beta}$,
we can bring the Hamiltonian to

$\displaystyle H(\check{\tau})$ $\displaystyle=$
$\displaystyle\frac{\Lambda}{2}\sum_{a,\alpha}(c_{a,\alpha}^{\dagger}c_{a,\alpha+1}+c_{a,\alpha}c_{a,\alpha+1}+h.c.)+\frac{\mu}{2}\sum_{\alpha}(ic_{L,\alpha}c_{R,\alpha}+ic_{L,\alpha}^{\dagger}c_{R,\alpha}^{\dagger}+h.c.)$
(128) $\displaystyle=$
$\displaystyle\sum_{k}(c_{L,k}^{\dagger},c_{L,-k},c_{R,k}^{\dagger},c_{R,-k})\left(\begin{array}[]{cccc}-\Lambda\cos
k&-i\Lambda\sin k&0&i\mu\\\ i\Lambda\sin k&\Lambda\cos k&i\mu&0\\\
0&-i\mu&-\Lambda\cos k&-i\Lambda\sin k\\\ -i\mu&0&i\Lambda\sin k&\Lambda\cos
k\end{array}\right)\left(\begin{array}[]{cccc}c_{L,k}\\\ c_{L,-k}^{\dagger}\\\
c_{R,k}\\\ c_{R,-k}^{\dagger}\end{array}\right).$ (137)

Because the odd and even length have different boundary conditions, the choice
of momentum is: when $n_{\check{\tau}}$ is an odd integer, the fermion
satisfies periodic boundary condition, $k=\frac{2j\pi}{n_{\check{\tau}}}$, and
the summation is over $j=0,1,...,\frac{n_{\check{\tau}}-1}{2}$. When
$n_{\check{\tau}}$ is an even integer, the fermion satisfies anti-periodic
boundary condition, $k=\frac{(2j-1)\pi}{n_{\check{\tau}}}$, and the summation
is over $j=1,...,\frac{n_{\check{\tau}}}{2}$.

The Bogoliubov quasiparticle annihilation operators altland2010condensed  are
given by

$\displaystyle d_{L,-,k}$ $\displaystyle=$
$\displaystyle\cos\frac{\theta}{2}\left(i\cos\frac{k}{2}c_{L,k}-\sin\frac{k}{2}c_{L,-k}^{\dagger}\right)+\sin\frac{\theta}{2}\left(i\sin\frac{k}{2}c_{R,k}+\cos\frac{k}{2}c_{R,-k}^{\dagger}\right),$
(139) $\displaystyle d_{R,-,k}$ $\displaystyle=$
$\displaystyle\sin\frac{\theta}{2}\left(i\sin\frac{k}{2}c_{L,k}+\cos\frac{k}{2}c_{L,-k}^{\dagger}\right)+\cos\frac{\theta}{2}\left(-i\cos\frac{k}{2}c_{R,k}+\sin\frac{k}{2}c_{R,-k}^{\dagger}\right),$
(140) $\displaystyle d_{L,+,k}$ $\displaystyle=$
$\displaystyle\cos\frac{\theta}{2}\left(-i\sin\frac{k}{2}c_{L,k}-\cos\frac{k}{2}c_{L,-k}^{\dagger}\right)+\sin\frac{\theta}{2}\left(-i\cos\frac{k}{2}c_{R,k}+\sin\frac{k}{2}c_{R,-k}^{\dagger}\right),$
(141) $\displaystyle d_{R,+,k}$ $\displaystyle=$
$\displaystyle\sin\frac{\theta}{2}\left(-i\cos\frac{k}{2}c_{L,k}+\sin\frac{k}{2}c_{L,-k}^{\dagger}\right)+\cos\frac{\theta}{2}\left(i\sin\frac{k}{2}c_{R,k}+\cos\frac{k}{2}c_{R,-k}^{\dagger}\right),$
(142)

where $\tan\theta=\frac{\mu}{\Lambda}$, and in terms of these operators, the
Hamiltonian becomes diagonal,

$\displaystyle H(\check{\tau})$ $\displaystyle=$
$\displaystyle\sqrt{\Lambda^{2}+\mu^{2}}\sum_{a=L,R;k>0}(d_{a,+,k}^{\dagger}d_{a,+,k}-d_{a,-,k}^{\dagger}d_{a,-,k})$
(144)
$\displaystyle+\text{mod}(n_{\check{\tau}},2)\sqrt{\Lambda^{2}+\mu^{2}}(d_{L,+,0}^{\dagger}d_{L,+,0}-d_{R,-,0}^{\dagger}d_{R,-,0}).$

Note that only the odd $n_{\check{\tau}}$ can have zero momentum mode because
even $n_{\check{\tau}}$ should satisfies the anti-periodic boundary condition.
With the help of the Bogoliubov quasiparticle operator, the exponential of the
Hamiltonian can be simplified by

$\displaystyle e^{TH(\check{\tau})}$ $\displaystyle=$
$\displaystyle\left(1+(e^{\sqrt{\Lambda^{2}+\mu^{2}}T}-1)d_{L,+,0}^{\dagger}d_{L,+,0}+1+(e^{-\sqrt{\Lambda^{2}+\mu^{2}}T}-1)d_{R,-,0}^{\dagger}d_{R,-,0}\right)^{\text{mod}(n_{\check{\tau}},2)}$
(145)
$\displaystyle\times\prod_{a=L,R;k>0}\left(1+(e^{\sqrt{\Lambda^{2}+\mu^{2}}T}-1)d_{a,+,k}^{\dagger}d_{a,+,k}+1+(e^{-\sqrt{\Lambda^{2}+\mu^{2}}T}-1)d_{a,-,k}^{\dagger}d_{a,-,k}\right)$
$\displaystyle\ni$
$\displaystyle\left(e^{\sqrt{\Lambda^{2}+\mu^{2}}T}\cos^{2}\frac{\theta}{2}c_{L,0}c_{L,0}^{\dagger}\right)^{\text{mod}(n_{\check{\tau}},2)}$
$\displaystyle\times\prod_{a=L,R;k>0}e^{\sqrt{\Lambda^{2}+\mu^{2}}T}\left(\cos^{2}\frac{\theta}{2}\cos^{2}\frac{k}{2}+\sin^{2}\frac{\theta}{2}\sin^{2}\frac{k}{2}\right)c_{a,k}c_{a,k}^{\dagger}.$

where in the first line we explicit show the contribution of $k=0$ sector that
is present for odd $n_{\check{\tau}}$ only. Remember the transition amplitude
is between the vacuum state of the original complex fermion (127), so we only
show the dominant contribution for each $k$ in the second equality. Therefore,
the transition amplitude for a cycle $\check{\tau}$ becomes

$\displaystyle\langle\text{EPR}|e^{TH(\check{\tau})}|\text{EPR}\rangle=e^{n_{\check{\tau}}\sqrt{\Lambda^{2}+\mu^{2}}T}\prod_{k}\left(\cos^{2}\frac{\theta}{2}\cos^{2}\frac{k}{2}+\sin^{2}\frac{\theta}{2}\sin^{2}\frac{k}{2}\right),$
(146)

Here the choice of momentum in the product is that for an odd
$n_{\check{\tau}}$, $k=\frac{2j\pi}{n_{\check{\tau}}}$, with
$j=-\frac{n_{\check{\tau}}-1}{2},...,-1,0,1,...,\frac{n_{\check{\tau}}-1}{2}$,
and for an even $n_{\check{\tau}}$, $k=\frac{(2j-1)\pi}{n_{\check{\tau}}}$,
with
$j=-\frac{n_{\check{\tau}}}{2}+1,...,-1,0,1,...,\frac{n_{\check{\tau}}}{2}$,
by noticing each term in the product is an even function of $k$. The product
is evaluated to be (see Appendix C)

$\displaystyle\prod_{k}\left(\cos^{2}\frac{\theta}{2}\cos^{2}\frac{k}{2}+\sin^{2}\frac{\theta}{2}\sin^{2}\frac{k}{2}\right)=\prod_{k}\frac{1}{2}\left(1+\cos\theta\cos
k\right)$ (147)
$\displaystyle=2^{1-2n_{\check{\tau}}}{\cos^{n_{\check{\tau}}}\theta}\left[{}_{2}F_{1}\left(n_{\check{\tau}},-n_{\check{\tau}},\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right],$
(148)

where ${}_{2}F_{1}(a,b,c;z)$ is the Gaussian hypergeometric function.

We now know the contribution from a cycle $\check{\tau}$, which can be used to
obtain the result of the full solution. Again, the solution in $A$ and
$\bar{A}$ are related by the cyclic permutation $\epsilon$,
$\tau_{A}=\epsilon\tau_{\bar{A}}$. As we discussed before, the maximal number
of cycles is $n+1$, i.e., $\check{\tau}_{(i)}$, $i=1,...,n+1$, where the sum
of the length of each cyclic permutation satisfies
$\sum_{j=1}^{n+1}n_{\check{\tau}_{(j)}}=2n$. Thus we have

$\displaystyle\prod_{x=1}^{L}\text{Pf}\Big{[}\hat{S}\partial_{t}-\Sigma_{x}\Big{]}=\left(\langle\text{EPR}|e^{T\sum_{i=1}^{n+1}H(\check{\tau}_{(i)})}|\text{EPR}\rangle\right)^{L/2}$
(149)
$\displaystyle=\left[e^{2n\sqrt{\Lambda^{2}+\mu^{2}}T}2^{1-3n}\cos^{2n}\theta\prod_{i=1}^{n+1}\left[{}_{2}F_{1}\left(n_{\check{\tau}_{(i)}},-n_{\check{\tau}_{(i)}},\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right]\right]^{L/2}.$
(150)

where in the first line we take a $L/2$ power as there are $L/2$ contributions
for the half-chain quasi entropy, i.e., both $A$ and $\bar{A}$ have $L/2$
number of sites.

Due to the presence of the measurement, we have to consider the denominator in
quasi entropy (20), $\mathbb{E}\text{Tr}[\tilde{\rho}^{\otimes n}]$. The
calculation is paralleled to $\mathbb{E}\text{Tr}[\tilde{\rho}^{\otimes
n}M_{\text{cyc}}(A)]$. The difference is that the relation between $\tau_{A}$
and $\tau_{\bar{A}}$ should be modified from $\tau_{A}=\epsilon\tau_{\bar{A}}$
to $\tau_{A}=\tau_{\bar{A}}$, and as we discussed around (87) in the previous
section, to maximize the number of cycles, it is given by the identity matrix.
So taking $\tau_{A}$ and $\tau_{\bar{A}}$ to be the identity which have $2n$
cycles in total, we have gotten the Pfaffian as follows,

$\displaystyle\prod_{x=1}^{L}\text{Pf}\Big{[}\hat{S}\partial_{t}-\Sigma_{x}\Big{]}=\left(\langle\text{EPR}|e^{T\sum_{i=1}^{2n}H(1)}|\text{EPR}\rangle\right)^{L/2}$
(151)
$\displaystyle=\left(e^{\sqrt{\Lambda^{2}+\mu^{2}}T}2^{-1}{\cos\theta}\left[{}_{2}F_{1}\left(1,-1,\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right]\right)^{nL}$
(152)
$\displaystyle=e^{n\sqrt{\Lambda^{2}+\mu^{2}}TL}\left(\cos\frac{\theta}{2}\right)^{2nL}.$
(153)

where $H(1)$ in the first line denotes (126) with trivial cycle
$\check{\tau}=1$. On the other hand, the calculation of (5.2), (108) and (109)
is the same as before. Thus, all the time dependent contributions cancel when
we take the ratio between these two quantities, namely,
$\frac{\mathbb{E}\text{Tr}[\tilde{\rho}^{\otimes
n}S_{\text{cyc}}(A)]}{\mathbb{E}\text{Tr}[\tilde{\rho}^{\otimes n}]}$. And
finally, we get the quasi entropy,

$\displaystyle
e^{(1-n)S_{A}^{(n)}}=2^{L(n-1)}\sum_{\mu=1}^{C_{n}}\left[\frac{2^{1-3n}\cos^{2n}\theta\prod_{i=1}^{n+1}\left[{}_{2}F_{1}\left(n_{\check{\tau}^{\mu}_{(i)}},-n_{\check{\tau}^{\mu}_{(i)}},\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right]}{\cos^{4n}\frac{\theta}{2}}\right]^{NL/2},$
(154)

where $\\{\check{\tau}^{\mu}_{(i)}\\}$, $i=1,...,n+1$, is the cycle
decomposition of the solution $\tau_{A}^{\mu}$ and $\tau_{\bar{A}}^{\mu}$, and
there are $C_{n}$ number of such solutions, $\mu=1,...,C_{n}$. $C_{n}$ is the
Catalan number. In the prefactor we also include the degeneracy coming from
the local Fermi parity symmetry.

Without measurements, i.e., $\theta=0$, (154) reduces to two decoupled unitary
chains discussed at the end of Section 4. With the measurement, we are not
able to evaluate the product explicitly. It is however clear that as
$\theta\rightarrow\frac{\pi}{2}$, the different saddle-point solutions tend to
the same form with no correlation between replicas, because $\Lambda=0$ in
(99). And according to

$\displaystyle\lim_{\theta\rightarrow\frac{\pi}{2}}\cos^{n}\theta\left[{}_{2}F_{1}\left(n_{\check{\tau}^{\mu}_{(i)}},-n_{\check{\tau}^{\mu}_{(i)}},\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right]=2^{n-1}e^{i2\pi
n},$ (155)

we have $e^{(1-n)S_{A}^{(n)}}=\left(\frac{2^{1-3n}2^{n-1}e^{i2\pi
n}}{(1/\sqrt{2})^{4n}}\right)^{NL/2}=1$ from (154) where all degeneracy
disappears as the replica symmetric solution is unique, for any integer $n\geq
2$. This implies that the quasi entropy of any order vanishes at
$\theta=\frac{\pi}{2}$ in the large-$N$ limit. In the next section, we will
consider a special solution, the cyclic symmetric one, and analytically
continue the solution to $n\rightarrow 1$ to show the von Neumann entanglement
entropy transition.

###  5.3 von Neumann entanglement entropy from replica trick

Figure 2: The plot of von Neumann entanglement entropy per site per flavor
$\sigma(\theta)$ as a function of the monitoring rate
$\theta=\arctan\frac{\mu}{\Lambda}$.

Among the $C_{n}$ different permutation that satisfy
$\tau_{A}=\epsilon\tau_{\bar{A}}$ and have the maximal number of cycles, there
is a cyclic symmetric one, where $\tau_{\bar{A}}$ is identity, and $\tau_{A}$
is itself a cyclic permutation of length $n$ 555In the gravity setup
penington2019replica , similar type of solution is considered in taking the
$n\rightarrow 1$ limit.. So the number of cycles is maximal, $n+1$. This is
the only nontrivial solution when $n=2$, i.e., $C_{2}=1$, and we expect it to
dominate for $n<2$ in the replica limit. Focusing on the contribution of this
special solution, we make the following analytic continuation,

$\displaystyle
e^{(1-x)S^{(x)}_{A}}=\left[\frac{\left(\cos^{2}\frac{\theta}{2}\right)^{x}\left(\frac{\cos^{x}\theta}{2^{2x-1}}\left[{}_{2}F_{1}\left(x,-x,\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right]\right)}{\cos^{4x}\frac{\theta}{2}}\right]^{NL/2},$
(156)

where $x$ is a real number 666We neglect the degeneracy from the Fermi parity
transformation since it is $\mathcal{O}(1)$ contribution, and we are
interested in $\mathcal{O}(N)$ contribution.. In the numerator inside the
parentheses, the first factor is the contribution from identity
$\tau_{\bar{A}}^{\alpha\beta}=\delta^{\alpha\beta}$, and the second factor is
the contribution from the cyclic permutation
$\tau_{A}^{\alpha\beta}=\epsilon^{\alpha\beta}$. We expand the result near
$x=1$ to get the von Neumann entanglement entropy

$\displaystyle S_{A}=\lim_{x\rightarrow
1}\frac{1}{1-x}\log\left[1+(1-x)\sigma(\theta)\right]^{NL/2}=\sigma(\theta)\frac{NL}{2},$
(157)

where the von Neumann entanglement entropy per site per flavor is

$\displaystyle\sigma(\theta)=\left(\log
2(1+\sec\theta)+\tan\frac{\theta}{2}\log(\sec\theta-\tan\theta)\right),\quad\tan\theta=\frac{\mu}{\Lambda}.$
(158)

A plot of this function is shown in Fig. 2. At $\theta=0$, we recover the
maximal von Neumann entanglement entropy $S_{A}=NL\log 2$. The von Neumann
entanglement entropy density vanishes smoothly at $\theta=\frac{\pi}{2}$,
where the von Neumann entanglement entropy becomes area-law. It shows clearly
that the measurement-induced phase transition occurs at the unbreaking point
of replica symmetry. Near the transition, we can expand the von Neumann
entanglement entropy to get

$\displaystyle
S_{A}=\frac{NL}{2}\left[\log\left(\frac{\pi}{2}-\theta\right)-\log
2e\right]\left(\frac{\pi}{2}-\theta\right).$ (159)

Thus, by noticing the relation between $\theta$ and $\Lambda$ in (106), the
entanglement density vanishes as
$(1-\frac{\mu}{J})^{1/2}\log(1-\frac{\mu}{J})$ near the critical point
$\mu=J$. This result differs from the known universality classes.

It is instructive to consider a more general case for $L_{A}\neq L_{\bar{A}}$.
When $L_{A}<L_{\bar{A}}$, the solution is given by the identity in
$L_{\bar{A}}$ and the replica symmetric permutation matrix in $L_{A}$, because
the identity is onshell in the bulk and satisfies the boundary condition (see
Appendix B). Then the von Neumann entanglement entropy for this solution is

$\displaystyle
e^{(1-n)S^{(n)}_{A}}=\frac{(\cos^{2n}\frac{\theta}{2})^{NL_{\bar{A}}}\left(\frac{\cos^{n}\theta}{2^{2n-1}}\left[{}_{2}F_{1}\left(n,-n,\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right]\right)^{NL_{A}}}{(\cos^{2n}\frac{\theta}{2})^{N(L_{A}+L_{\bar{A}})}}.$
(160)

It is clear that the numerator has two factors, the first one from
$L_{\bar{A}}$ and the second one from $L_{\bar{A}}$. Taking a similar analytic
continuation, we get the same result, $S_{A}=\sigma(\theta)NL_{A}$.

##  6 Conclusion and discussion

To conclude, we introduced a large-$N$ model in which the von Neumann entropy
can be calculated analytically. We obtained the Rényi entropy by summing over
a class of solutions related to the permutation group for the non-measured
case, and we analytically established a von Neumann entanglement entropy
transition using the replica trick in the meansured case. The result in
Section 4 strongly suggests that the late-time von Neumann entanglement
entropy of a chaotic unitary system is dominated by saddle points that connect
different replicas. Interestingly, the degeneracy of these replica non-
diagonal solutions play an important role in determining the entanglement
entropy, and a single replica non-diagonal solution can lead to an inaccurate
density spectrum density of state. For instance, if we neglected the
degeneracy given by the Catalan number, we would get
$D(\lambda)=2^{N-2}\delta(\lambda-2^{-N+2})$. This would imply that all Rényi
entropies are given by the same value, similar to stabilizer states generated
by Clifford circuits. It would be interesting to understand if there is any
larger significance to this observation, for example, related to a path
integral representation of Clifford dynamics.

The continuous transition between replica symmetry broken and unbroken
solutions studied in Section 5 should be contrasted with the discontinuous
transition between two similar large-$N$ solutions in the unitary time
evolution of Rényi entropy jian2020note . As the replica non-diagonal solution
is closely related to the replica wormhole observed in the context of black
hole information paradox penington2019replica ; almheiri2019replica , it is
worth speculating about the physics of a monitored black hole. In the current
setup, the monitoring is implemented on the full system and causes the
restoration of replica symmetry. We expect that by increasing the monitoring
of the black hole and its environment at late time, the replica wormhole
continuously disappears. Because the so-called entanglement island is obtained
by continuing the cyclic symmetric wormhole penington2019replica , it should
also disappear with sufficiently frequent measurements. An interesting
question is to monitor the environment, and study the effect of such
measurements on the black hole. Though the measurement destroys the
entanglement between black hole and radiation, the state remaining in black
hole is expected to be maximally scrambled. We leave this exploration as a
future work.

More broadly, the approach of mapping the path integral of multi-replicas to
the transition amplitude of quantum states provides a general tool in
evaluations of entropy related quantities. We expect generalization of such an
approach to other models that admit saddle-point analysis to be
straightforward. For example, it is of value to generalize and utilize this
tool in bosonic models bentsen2021measurement . In this case, the emergent
transition amplitude will governed by a spin model living in the replica
space. The transition amplitude for a general $n$ might be complicated, but we
expect that a similar cyclic symmetric replica non-diagonal solution dominates
near $n\approx 1$, which could be analysed in detail.

Finally, the model defined by the action (5.1) is distinct from the model we
considered previously in Ref. jian2021syk . The major distinction lies in that
the action (9) conserves local Fermi parity, whereas the model studied in Ref.
jian2021syk  breaks the local Fermi parity by two Majorana hopping between
sites. If we consider $n=2$, the Landau-Ginzbug action for $q=4$ would be

$\displaystyle\frac{I_{\text{eff}}}{N}$ $\displaystyle=$ $\displaystyle\int
dtdx\Big{(}\frac{1}{2}\big{[}(\partial_{t}\vec{\phi})^{2}+\phi_{1}^{2}(\partial_{x}\phi_{1})^{2}+\phi_{2}^{2}(\partial_{x}\phi_{2})^{2}\big{]}+r\vec{\phi}^{2}+\lambda|\vec{\phi}|^{4}+\lambda^{\prime}(\phi_{1}^{4}+\phi_{2}^{4})\Big{)},$
(161)

where $\vec{\phi}=(\phi_{1},\phi_{2})$ is a two-dimensional order parameter
related to rotations between two replicas. The space derivative appears in the
forth order term, and it leads to a distinct universality class. It will be
interesting to derive the Landau-Ginzburg action in the limit $n\rightarrow
1$.

## Acknowledgements

We would like to thank Pengfei Zhang and Zhuo-Yu Xian for useful discussions.
This work is supported by the Simons Foundation via the It From Qubit
Collaboration. The work of BGS is also supported in part by the AFOSR under
grant number FA9550-19-1-0360.

##  Appendix A Permutation operator for fermionic state

Consider a Hilbert space consisting of $N$ Majorana fermions,
$\psi_{1},...,\psi_{N}$, $\\{\psi_{i},\psi_{j}\\}=\delta_{ij}$, and $N$ is an
even integer. It is convenient to double the Hilbert space and work in a state
language. To do that, we introduce another set of Majoran fermion called
$\chi_{1},...,\chi_{N}$, $\\{\chi_{i},\chi_{j}\\}=\delta_{ij}$, and the
maximally entangled state between these two sets of Majrana fermions,
$|\text{EPR}\rangle$, defined by

$\displaystyle(\psi_{j}+i\chi_{j})|\text{EPR}\rangle=0,\quad\forall
j=1,...,N.$ (162)

Then any operator acting on the $\psi$ Hilbert space can be uniquely mapped to
a state in the doubled Hilbert space by acting it on the EPR state
$O_{\psi}\rightarrow O_{\psi}|\text{EPR}\rangle$.

Let us first consider a simpler case with $N=2$. By pairing the two Majorana
into a single complex fermion, $c_{\psi}=\frac{\psi_{1}+i\psi_{2}}{\sqrt{2}}$,
$c_{\psi}^{\dagger}=\frac{\psi_{1}-i\psi_{2}}{\sqrt{2}}$ and
$c_{\chi}=i\frac{\chi_{1}-i\chi_{2}}{\sqrt{2}}$,
$c_{\chi}^{\dagger}=-i\frac{\chi_{1}+i\chi_{2}}{\sqrt{2}}$ the Hilbert space
is spanned by

$\displaystyle\\{|00\rangle,|10\rangle=c_{\psi}^{\dagger}|00\rangle,|01\rangle=c_{\chi}^{\dagger}|00\rangle,|11\rangle=c_{\psi}^{\dagger}c_{\chi}^{\dagger}|00\rangle\\},$
(163)

with the EPR state expressed as
$|\text{EPR}\rangle=\frac{|00\rangle+|11\rangle}{\sqrt{2}}$. It is convenient
to label it as
$|\text{EPR}\rangle=\frac{|0\bar{0}\rangle+|1\bar{1}\rangle}{\sqrt{2}}=\sum_{a}\frac{|a\bar{a}\rangle}{\sqrt{2}}$.
For arbitrary even integer $N$, we have the similar expression
$|\text{EPR}\rangle=\sum_{a}\frac{|a\bar{a}\rangle}{2^{N/2}}$.

We take $n$ copies of the doubled Hilbert space, and denote the Majorana
operators as $\psi_{i}^{\alpha},\chi_{i}^{\alpha}$, $i=1,2,...,N$,
$\alpha=1,2,...,n$. The cyclic permutation operator is defined to take

$\displaystyle
M_{\text{cyc}}|a_{1}\bar{a}_{1}a_{2}\bar{a}_{2}a_{3}\bar{a}_{3}...a_{n}\bar{a}_{n}\rangle\rightarrow|a_{1}\bar{a}_{n}a_{2}\bar{a}_{1}a_{3}\bar{a}_{2}...a_{n}\bar{a}_{n-1}\rangle,$
(164)

and it is trivial to show that

$\displaystyle\text{Tr}[\rho^{n}]=\langle\text{EPR}|^{\otimes n}\rho^{\otimes
n}M_{\text{cyc}}|\text{EPR}\rangle^{\otimes
n},\quad|\text{EPR}\rangle^{\otimes
n}=2^{-nN/2}\sum_{a_{1},...,a_{n}}|a_{1}\bar{a}_{1}a_{2}\bar{a}_{2}a_{3}\bar{a}_{3}...a_{n}\bar{a}_{n}\rangle.$

It is also easy to show that the following operator gives (164),

$\displaystyle
M_{\text{cyc}}=\prod_{i=1}^{N}e^{\frac{\pi}{2}\chi^{n}_{i}\chi^{n-1}_{i}}e^{\frac{\pi}{2}\chi^{n-1}_{i}\chi^{n-2}_{i}}...e^{\frac{\pi}{2}\chi^{3}_{i}\chi^{2}_{i}}e^{\frac{\pi}{2}\chi^{2}_{i}\chi^{1}_{i}}.$
(166)

In terms of the operator in the $\psi$ Hilbert space, the cyclic permutation
becomes

$\displaystyle
M_{\text{cyc}}=\prod_{i=1}^{N}e^{\frac{\pi}{2}\psi^{1}_{i}\psi^{2}_{i}}e^{\frac{\pi}{2}\psi^{2}_{i}\psi^{3}_{i}}...e^{\frac{\pi}{2}\psi^{n-2}_{i}\psi^{n-1}_{i}}e^{\frac{\pi}{2}\psi^{n-1}_{i}\psi^{n}_{i}},$
(167)

with the action on the $\psi_{i}^{\alpha}$ field

$\displaystyle
M_{\text{cyc}}\psi_{i}^{\alpha}M_{\text{cyc}}^{\dagger}=\sum_{\beta}\text{sgn}(\alpha-\beta)\delta^{\alpha+1,\beta}\psi_{i}^{\beta},\quad\forall
i=1,...,N,$ (168)

where we define $\delta^{n+1,\beta}=\delta^{1,\beta}$ when $\alpha=n$. Thus,
(168) is the cyclic permutation operator for fermionic states, and is used in
Section 4.

##  Appendix B Saddle-point solutions

The equations of motion and boundary conditions for the saddle-point analysis
given in the following for general $n$ is a complicated dynamical problem,

$\displaystyle\partial_{t}\hat{S}\hat{G}_{x}(t,t)\hat{S}$ $\displaystyle=$
$\displaystyle[\hat{\Sigma}_{x}(t),\hat{G}_{x}(t,t)],$ (169)
$\displaystyle\Sigma_{x,ss^{\prime}}^{\alpha\beta}(t)$ $\displaystyle=$
$\displaystyle
c_{ss^{\prime}}\Big{[}J(2G_{x,ss^{\prime}}^{\alpha\beta})+U(2G_{x,ss^{\prime}}^{\alpha\beta})^{q/2-1}[M_{(x)}]^{\alpha\gamma}_{s}[M_{(x)}]^{\beta\delta}_{s^{\prime}}(2G_{\bar{x},ss^{\prime}}^{\gamma\delta})^{q/2}\Big{]},$
(170) $\displaystyle\psi_{x,+}^{\alpha}(0)$ $\displaystyle=$
$\displaystyle-\psi^{\alpha}_{x,-}(0),\quad\psi_{x,+}^{\alpha}(T)=\psi_{x,-}^{\alpha}(T),\quad\forall\alpha=1,...,n.$
(171)

One can show that there is a conserved quantity for each site, given by

$\displaystyle\text{Tr}(\hat{S}\hat{G}_{x}\hat{S}\hat{G}_{x})=\text{const},$
(172)

where the constant is set by the boundary condition. This follows from

$\displaystyle\frac{d}{dt}\text{Tr}(\hat{S}\hat{G}_{x}\hat{S}\hat{G}_{x})=2\text{Tr}\left(\frac{d(\hat{S}\hat{G}_{x}\hat{S})}{dt}\hat{G}_{x}\right)=2\text{Tr}\left([\Sigma_{x},G_{x}]\hat{G}_{x}\right)=0.$
(173)

Except for this conserved quantity, in general, the problem becomes a coupled
nonlinear differential equation that has no general solution. For $n=2$, we
explicitly write down the equations of motion

$\displaystyle\frac{dx_{1}}{dt}$ $\displaystyle=$ $\displaystyle
4Jx_{2}z_{1}+2Ux_{2}z_{1}(y_{1}^{2}+w^{2}),$ (174)
$\displaystyle\frac{dx_{2}}{dt}$ $\displaystyle=$
$\displaystyle-4Jx_{1}z_{1}-2Ux_{1}z_{1}(y_{2}^{2}+w^{2}),$ (175)
$\displaystyle\frac{dz_{1}}{dt}$ $\displaystyle=$ $\displaystyle
2Ux_{1}x_{2}(y_{1}^{2}-y_{2}^{2}),$ (176) $\displaystyle\frac{dy_{1}}{dt}$
$\displaystyle=$ $\displaystyle 4Jy_{2}w_{1}+2Uy_{2}w_{1}(x_{1}^{2}+z^{2}),$
(177) $\displaystyle\frac{dy_{2}}{dt}$ $\displaystyle=$
$\displaystyle-4Jy_{1}w_{1}-2Uy_{1}w_{1}(x_{2}^{2}+z^{2}),$ (178)
$\displaystyle\frac{dw_{1}}{dt}$ $\displaystyle=$ $\displaystyle
2Uy_{1}y_{2}(x_{1}^{2}-x_{2}^{2}),$ (179)

where the coefficients are defined via

$\displaystyle
G_{A}(t,t)=\frac{1}{2}\left(\sum_{k=1}^{n}x_{k}(t)X_{k}+\sum_{m=1}^{[\frac{n}{2}]}z_{m}(t)Z_{m}\right),\quad
G_{\bar{A}}(t,t)=\frac{1}{2}\left(\sum_{k=1}^{n}y_{k}(t)X_{k}+\sum_{m=1}^{[\frac{n}{2}]}w_{m}(t)Z_{m}\right),$

with

$\displaystyle X_{k}=\left(\begin{array}[]{cccc}0&-\sigma^{-k}\\\
\sigma^{k}&0\end{array}\right),$ $\displaystyle\quad k=1,...,n,$ (183)
$\displaystyle
Z_{m}=\frac{1}{2}\left(\begin{array}[]{cccc}\sigma^{m}-\sigma^{-m}&0\\\
0&\sigma^{-m}-\sigma^{m}\end{array}\right),$ $\displaystyle\quad
m=1,...,\left[\frac{n}{2}\right].$ (186)

Here $\left[\frac{n}{2}\right]$ is the largest integer that is less than
$\frac{n}{2}$, and
$\sigma^{\alpha\beta}=\text{sgn}(\alpha-\beta)\delta^{\alpha+1,\beta}$ is the
permutation matrix with a proper sign to be consistent with the even Fermi
parity. When $\alpha=n$, the symbol means
$\delta^{n+1,\beta}=\delta^{1,\beta}$.

Although we have only explicitly write down the equation for $n=2$, it can be
seen from the definition of the matrices that these type of coefficients form
a closed set of dynamical variables. It reduces the number of variable from
$\sim n^{2}$ to $\sim n$. For $n=2$, this representation actually captures all
independent variables. In terms of this set of variables, the conserved
quantity is given by

$\displaystyle\sum_{k=1}^{n}x_{k}^{2}-\sum_{m=1}^{[\frac{n}{2}]}z_{m}^{2}=\text{const},\quad\sum_{k=1}^{n}y_{k}^{2}-\sum_{m=1}^{[\frac{n}{2}]}w_{m}^{2}=\text{const}.$
(187)

To further simplify the problem, we assume that in $\bar{A}$ the conserved
quantity is given by $y_{1}^{2}+y_{2}^{2}-w_{1}^{2}=1$ (as we discuss in the
following, this is required to have a time-independent action), and since the
boundary condition is $y_{2}=-1$, it uniquely determines the steady solution
$y_{1}=w_{1}=0$ and $y_{2}=-1$. The equations of motion reduce to

$\displaystyle\frac{dx_{1}}{dt}$ $\displaystyle=$ $\displaystyle
4Jx_{2}z_{1},$ (188) $\displaystyle\frac{dx_{2}}{dt}$ $\displaystyle=$
$\displaystyle-(4J+2U)x_{1}z_{1},$ (189) $\displaystyle\frac{dz_{1}}{dt}$
$\displaystyle=$ $\displaystyle-2Ux_{1}x_{2},$ (190)

and $x_{1}^{2}+x_{2}^{2}-z_{1}^{2}$ is conserved. There is a discrete symmetry
given by multiplying any two of the three variables $x_{1},x_{2},z_{1}$ by a
minus one, leading to four-fold degenerate nontrivial solutions. Among these
solutions, half of them are determined by the initial value of $x_{2}$, i.e.,
by the boundary condition, and given the boundary condition, there are still
two solutions related by the Fermi parity transformation discussed in the main
text. In the following, we will discuss one explicit solution.

These differential equations can be solved via Jacobi elliptic functions,

$\displaystyle x_{1}$ $\displaystyle=$
$\displaystyle\sqrt{c_{1}}\text{sn}\left(2\sqrt{U(2J+U)c_{2}}(t-t_{0}),\frac{c_{1}}{c_{2}}\right),$
(191) $\displaystyle x_{2}$ $\displaystyle=$
$\displaystyle-\sqrt{\frac{(2J+U)c_{1}}{2J}}\left(1-\text{sn}\left(2\sqrt{U(2J+U)c_{2}}(t-t_{0}),\frac{c_{1}}{c_{2}}\right)\right)^{1/2},$
(192) $\displaystyle z_{1}$ $\displaystyle=$
$\displaystyle-\sqrt{\frac{Uc_{2}}{2J}}\text{dn}\left(2\sqrt{U(2J+U)c_{2}}(t-t_{0}),\frac{c_{1}}{c_{2}}\right),$
(193)

where $\text{sn}(u,c)$ and $\text{dn}(u,c)$ are the Jacobi elliptic functions.
$c_{1},c_{2},t_{0}$ are integral constants to be determined below. $t_{0}$ is
the shift of time argument, and $c_{1},c_{2}$ are related to the conserved
quantity by $x_{1}^{2}+x_{2}^{2}-z_{1}^{2}=c_{1}+\frac{U}{2J}(c_{1}-c_{2})$.

The conserved quantity is related to the time dependence of the onshell
action. Because the action is dimensionless, it can only depend on $JT$ and
$UT$. Taking the derivative with respect to $T$, we have

$\displaystyle\frac{dI(JT,UT)}{dT}$ $\displaystyle=$
$\displaystyle\frac{J}{T}\frac{\partial I}{\partial
J}+\frac{U}{T}\frac{\partial I}{\partial U}$ (194) $\displaystyle=$
$\displaystyle-c_{ss^{\prime}}\int
dt\Big{[}\frac{J}{8T}\sum_{x}(2G_{x,ss^{\prime}}^{\alpha\beta}(t,t))^{2}+\frac{U}{8T}(2G_{1,ss^{\prime}}^{\alpha\beta}(t,t))^{2}M^{\alpha\gamma}_{s}M^{\beta\delta}_{s^{\prime}}(2G_{2,ss^{\prime}}^{\gamma\delta}(t,t))^{2}\Big{]}$
$\displaystyle=$ $\displaystyle-\int
dt\Big{[}\frac{J}{2T}(x_{1}(t)^{2}+x_{2}(t)^{2}-z_{1}(t)^{2}-1)+\frac{U}{2T}(x_{1}(t)^{2}-1)\Big{]},$

where we have plugged in $y_{1}=w_{1}=0$ and $y_{2}=-1$. In the third line,
the first term in parentheses is given by the conserved quantity. To have a
time-independent result, the conserved quantity should be
$x_{1}(t)^{2}+x_{2}(t)^{2}-z_{1}(t)^{2}=1$. The second term vanishes at long
times when $x_{1}(t)\rightarrow 1$. With these two facts, $c_{1}$, $c_{2}$ can
be determined to be $c_{1}=1$ and $c_{2}=1$ as $T\rightarrow\infty$. The
boundary condition at $t=0$ fixes
$t_{0}=\frac{1}{4\sqrt{U(2J+U)}}\cosh^{-1}\frac{J+U}{J}$ such that
$x_{2}(0)=-1$. An example is shown in Fig. 3, where we allow $c_{2}$ to be
slightly less than $1$ to capture the boundary condition at $t=T$. As $c_{2}$
approaches $1$, the time $T$ approaches $\infty$. In the long-time limit, the
solution becomes,

$\displaystyle x_{1}$ $\displaystyle=$
$\displaystyle\tanh\left(2\sqrt{U(2J+U)}(t-t_{0})\right),$ (195)
$\displaystyle x_{2}$ $\displaystyle=$
$\displaystyle-\sqrt{\frac{2J+U}{2J}}\frac{1}{\cosh\left(2\sqrt{U(2J+U)}(t-t_{0})\right)},$
(196) $\displaystyle z_{1}$ $\displaystyle=$
$\displaystyle-\sqrt{\frac{U}{2J}}\frac{1}{\cosh\left(2\sqrt{U(2J+U)}(t-t_{0})\right)},$
(197)

where indeed the solution in the bulk takes the form of (45), and the effect
of the boundary conditions decreases exponentially fast into the bulk. Now in
the long-time limit, the onshell action tends to a constant determined by the
ratio $U/J$ because it is the only dimensionless quantity when
$T\rightarrow\infty$. The long-time onshell action cannot depend on $U(2J+U)$,
so we can take this advantage to send $U(2J+U)\rightarrow\infty$ without
changing the onshell action. In this case, the solution is given by (45).
Therefore, although (45) does not satisfy the boundary condition in any finite
$U(2J+U)$, the action evaluated via (45) gives the correct answer in the long-
time limit. The argument implicitly states

$\displaystyle\lim_{E\rightarrow\infty}I\left[G_{\text{exact}},\Sigma_{\text{exact}}\right]=I\left[\lim_{E\rightarrow\infty}G_{\text{exact}},\lim_{E\rightarrow\infty}\Sigma_{\text{exact}}\right]=I\left[G,\Sigma\right],$
(198)

where $E$ is a microscopic energy scale in the model, i.e., $U=uE$, $J=jE$
777This is same as taking $T\rightarrow\infty$, because the action is
dimensionless and depends on $TE$ only.. In the above limit, $U/J=u/j$ is kept
fixed. $G_{\text{exact}}$ and $\Sigma_{\text{exact}}$ denote the exact saddle-
point solution satisfying the boundary condition, and $G$ and $\Sigma$ in the
last equality is the solution given in (45). This must be true because the
long-time limit of the Rényi entropy and von Neumann entanglement entropy of
the Brownian SYK clusters saturate to a constant set by the dimension of
Hilbert space and the symmetry of the model independent of any microscopic
energy scale page1993average ; jian2021quantum ; stanford2021subleading . And
the evaluation of the von Neumann entanglement entropy from the permutation
solution (45) correctly gives the Page value (see Section 4).

Figure 3: A plot of the solution of $x_{1},x_{2},z_{1}$. We choose $J=1$,
$U=0.4$, $c_{1}=1$, $c_{2}=1-10^{-7}$.

For general $n$, we expect the same situation happens. The conserved quantity
(187) is given by $1$, and the bulk is dominated by one of the permutation
matrix $X_{k}$ given by (183) to have a time-independent onshell action. When
the measurement is turned on, the addition of (19) is an integration over the
Green’s function, for which we expect (198) still holds. Thus, in Section 5,
we consider a generalization of solutions of the form (45) to the measurement
case, c.f. (99), and study how the measurement affects the solution and causes
the entanglement transition.

##  Appendix C Finite products of trigonometric functions

In this section we first show that the following identity holds and then use
it to derive (148),

$\displaystyle
2(T_{n}(a)+1)=\begin{cases}\prod_{k=-\frac{n}{2}+1}^{\frac{n}{2}}(2a+2\cos\frac{(2k+1)\pi}{n}),&n=\text{even},\\\
\\\
\prod_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}(2a+2\cos\frac{2k\pi}{n}),&n=\text{odd}.\end{cases}$
(199)

for $k$ being integers, where $T_{n}(a)$ is the Chebyshev polynomial of the
first kind, namely, $T_{n}(\cos\phi)=\cos n\phi$. For an even integer $n$,
note that (199) is equivalent to

$\displaystyle x^{n}+x^{-n}+2\cos
n\varphi=\prod_{k=-\frac{n}{2}+1}^{\frac{n}{2}}\left[x+x^{-1}-2\cos\left(\varphi+\frac{(2k-1)\pi}{n}\right)\right],$
(200)

by setting $\varphi=\pi$ and $x=e^{i\phi}$. The left-hand side of (200) has
the roots,

$\displaystyle x=e^{i(\varphi+\frac{(2k-1)\pi}{n})},\quad
k\in\\{-\frac{n}{2}+1,-\frac{n}{2}+2,...,\frac{n}{2}\\}.$ (201)

By noticing

$\displaystyle x+x^{-1}=2\cos\left(\varphi+\frac{(2k-1)\pi}{n}\right),$ (202)

we arrive the right-hand side of (200). The proof for an odd integer $n$ is
similar. (199) is equivalent to

$\displaystyle x^{n}+x^{-n}-2\cos
n\varphi=\prod_{k=-\frac{n-1}{2}}^{-\frac{n-1}{2}}\left[x+x^{-1}-2\cos\left(\varphi+\frac{2k\pi}{n}\right)\right],$
(203)

by setting $\varphi=\pi$ and $x=e^{i\phi}$. The left-hand side of (203) has
the roots,

$\displaystyle x=e^{i(\varphi+\frac{2k\pi}{n})},\quad
k\in\\{-\frac{n-1}{2},-\frac{n-1}{2}+1,...,\frac{n-1}{2}\\}.$ (204)

Then by noticing

$\displaystyle x+x^{-1}=2\cos\left(\varphi+\frac{2k\pi}{n}\right),$ (205)

we get the right-hand side of (200).

Using the identity (199), we can derive (148). The derivation is simply given
here. For even integers $n$,

$\displaystyle\prod_{k=-\frac{n}{2}+1}^{\frac{n}{2}}\frac{1}{2}\left(1+\cos\theta\cos
k\right)=2^{-2n}\cos^{n}\theta\prod_{k=-\frac{n}{2}+1}^{\frac{n}{2}}\left(\frac{2}{\cos\theta}+2\cos
k\right)$ (206)
$\displaystyle=2^{1-2n}\cos^{n}\theta(T_{n}(\sec\theta)+1)=2^{1-2n}{\cos^{n}\theta}\left[{}_{2}F_{1}\left(n,-n,\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right],$
(207)

and for odd integers $n$,

$\displaystyle\prod_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}\frac{1}{2}\left(1+\cos\theta\cos
k\right)=2^{-2n}\cos^{n}\theta\prod_{k=-\frac{n-1}{2}}^{\frac{n-1}{2}}\left(\frac{2}{\cos\theta}+2\cos
k\right)$ (208)
$\displaystyle=2^{1-2n}\cos^{n}\theta(T_{n}(\sec\theta)+1)=2^{1-2n}{\cos^{n}\theta}\left[{}_{2}F_{1}\left(n,-n,\frac{1}{2};\frac{1}{2}(1-\sec\theta)\right)+1\right].$
(209)

where in the last step of the derivation, we use the fact that the Chebyshev
polynomial is equal to the Gaussian hypergeometic function,

$\displaystyle
T_{x}(\cos\phi)={}_{2}F_{1}\left(x,-x,\frac{1}{2};\frac{1}{2}(1-\cos\phi)\right),$
(210)

for integer $x$. Then we use the fact that the Gaussian hypergeometric
function is defined for real number $x$ to make analytical continuations.

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