###### Abstract

By considering the generalized uncertainty principle, the degrees of freedom
near the apparent horizon of Vaidya black hole are calculated with thin film
model. The result shows that a cut-off can be introduced naturally rather than
taking by hand. Furthermore, if the minimal length is chosen to be a specific
value, the statistical entropy will satisfy the conventional area law at the
horizon, which might reveal some deep things of the minimal length.

Entropy of Vaidya Black Hole on Apparent Horizon with Minimal length Revisited
***Supported by National Natural Science Foundation of China under Grant Nos.
11675139, 11605137, 11435006, 11405130 and the Double First-Class University
Construction Project of Northwest University. Bin Wu is also supported by the
China Postdoctoral Science Foundation (No.2017M623219).

Hao Tang,1,3 Bin Wu,1,2,3 †††Corresponding author. Email<EMAIL_ADDRESS>Cheng-yi Sun,1,3

Yu Song,1,3 Rui-hong Yue1,4

1 Institute of Modern Physics, Northwest University, Xi’an 710069, China
2 Shaanxi Key Laboratory for Theoretical Physics Frontiers, Northwest
University, Xi’an 710069, China
3 School of Physics, Northwest University, Xi’an 710069, China
4 College of Physical and Technology, Yangzhou University , Yangzhou 225009,
China

Keywords: entropy; black hole; generalized uncertainty principle.

PACS numbers: 04.70.Dy, 97.60.Lf

Black hole is one of the most intriguing objects which possibly provide a
bridge between an underlying quantum gravity theory and the classical gravity.
Pioneering by the Bekenstein’s [1] and Hawking’s [2] works, the black holes
could have the temperature and entropy as an ordinary thermodynamical
macroscopic system. Moreover, the entropy of a black hole is argued to be
proportional to the area $A$ of the horizon. Since the origin of the entropy
of the black hole at the microscopic level becomes an active and fascinating
field of researches, many efforts have been made and several different methods
have been developed to calculate the entropy [3, 4, 5] of black holes at
statistical level. In Ref.[6], ’t Hooft proposed a widely-used method called
“brick-wall” by supposing that the entropy of the black hole comes from the
contributions of the quantum gases in a “wall" outside the horizon. However,
there exists divergence of the entropy if no further constraint has been
introduced on the brick-wall. So that a cut-off of the brick-wall must be
introduced by hand, which is unnatural in some sense[6, 7, 8, 9, 10, 11, 12,
13, 14, 15].

Recently, the generalized uncertainty principle (GUP), which is developed from
the Heisenberg’s uncertainty principle (HUP), is used in the brick-wall model
to calculate the free energy and the entropy. It shows that the cut-off is
naturally given by a minimal length indicated by the GUP [16, 17, 19, 18, 20,
21, 22, 23, 24, 25, 26, 27]. The bound of the entropy has also been obtained
in the framework of the GUP. However, all those works are concentrated on the
static spherically symmetric black holes. In this paper, we are interested in
the dynamical spacetime. We will follow the approach of brick-wall method with
the GUP to calculate the statistical entropy of Vaidya black hole. Planck unit
$c=G=\hbar=k_{B}=1$ is used throughout this paper for convenience.

Vaidya spacetime is a solution of the Einstein’s equation, which describes the
formation of a Schwarzschild-like black hole out of a collapsing shell of null
dust. The metric of the Vaidya spacetime in the advanced Eddington coordinates
reads [31, 32]

$\displaystyle ds^{2}=-[1-\frac{2m(v)}{r}]{\rm d}v^{2}+2{\rm d}v{\rm
d}r+r^{2}{\rm d}\theta^{2}+r^{2}\sin^{2}\theta{\rm d}\phi^{2},$ (1)

where $m(v)$ is the dynamical mass of the black hole, $v$ is the advanced
Eddington time coordinate and

$\displaystyle f(r)=1-\frac{2m(v)}{r}.$ (2)

In order to characterize the formation of a dynamical black hole, there is a
generalized notion called apparent horizon which is distinct from the event
horizon. To demonstrate, we just consider the apparent horizon which is a
closed space-like hypersurface of co-dimension two and is located at[32]

$\displaystyle r_{AH}=2m(v).$ (3)

The surface gravity of the apparent horizon is

$\displaystyle\kappa_{AH}=\frac{1}{2r_{AH}}=\frac{1}{4m(v)}.$ (4)

The corresponding Hawking temperature is

$\displaystyle T_{AH}=\frac{\kappa_{AH}}{2\pi}=\frac{1}{8\pi
m(v)}=\frac{1}{\beta_{AH}},$ (5)

thus we have

$\displaystyle\beta_{AH}=\frac{2\pi}{\kappa_{AH}}=4\pi r_{AH}.$ (6)

The metric function $f(r)$ can be expanded near the horizon $r_{AH}$ as

$\displaystyle f(r)\approx
f^{\prime}(r_{AH})(r-r_{AH})=2\kappa_{AH}(r-r_{AH}).$ (7)

There are several ways to generalize the Heisenberg’s uncertainty principle,
and the simplest one among of them is [16, 17, 19, 18, 20, 23]

$\Delta x\Delta p\geq\hbar+\frac{\lambda}{\hbar}(\Delta p)^{2},$ (8)

where $\hbar$ is the Planck constant, $\Delta x$ and $\Delta p$ are the
spatial coordinates and momentum respectively, and $\lambda$ is a constant at
order of the Planck length[16]. The Eq.(8) indicate that there exists a
minimal length [21, 22]

$\Delta x\geq 2\sqrt{\lambda}.$ (9)

Since we are considering a system of quantum boson gas in a “wall” outside the
horizon, the Klein-Gordon equation for a scalar field is introduced

$(\Box-\mu^{2})\Phi=0.$ (10)

Decompose the wave function into the form of

$\Phi=e^{-i\omega v}\psi(r,\theta,\phi),$ (11)

and substitute it into Eq.(10), yields

$\begin{split}\frac{\partial^{2}\psi}{\partial
r^{2}}+(\frac{2}{r}+\frac{df(r)}{f(r)dr})\frac{\partial\psi}{\partial
r}+\frac{1}{f(r)}[\frac{\omega^{2}}{f(r)}-\mu^{2}+\\\
\frac{1}{r^{2}}(\frac{\partial^{2}}{\partial\theta^{2}}+\frac{\partial}{\tan\theta\partial\theta}+\frac{\partial}{\cos^{2}\theta\partial\phi^{2}})]\psi=0.\end{split}$
(12)

By using the Wenzel-Kramers-Brillouin (WKB) approximation, i.e., take
$\psi\sim e^{i\mathbb{S}(r,\theta,\phi)}$, one could get [28]

$p_{r}^{2}=\frac{1}{f(r)}[\frac{\omega^{2}}{f(r)}-\mu^{2}-\frac{p_{\theta}^{2}}{r^{2}}-\frac{p_{\phi}^{2}}{r^{2}\cos^{2}\theta}],$
(13)

where

$p_{r}=\frac{\partial\mathbb{S}}{\partial r},\ \
p_{\theta}=\frac{\partial\mathbb{S}}{\partial\theta},\ \
p_{\phi}=\frac{\partial\mathbb{S}}{\partial\phi}.$ (14)

And the square module momentum is

$p^{2}=p_{i}p^{i}=g^{rr}p_{r}^{2}+g^{\theta\theta}p_{\theta}^{2}+g^{\phi\phi}p_{\phi}^{2}=\frac{\omega^{2}}{f(r)}-\mu^{2}.$
(15)

Eq.(15) implies that $\omega\geqslant\mu\sqrt{f(r)}$.

In the framework of the GUP, the number density now becomes [17]

$\frac{d^{3}xd^{3}p}{(2\pi)^{3}(1+\lambda p^{2})^{3}}.$ (16)

Consequently, we obtain the number of quantum states as

$\displaystyle n(\omega)$ $\displaystyle=\frac{1}{(2\pi)^{3}}\int drd\theta
d\phi dp_{r}dp_{\theta}dp_{\phi}\frac{1}{(1+\lambda p^{2})^{3}}$
$\displaystyle=\frac{2}{3\pi}\int
dr\frac{r^{2}(\omega^{2}/f(r)-\mu^{2})^{3/2}}{\sqrt{f(r)}[1+\lambda(\omega^{2}/f(r)-\mu^{2})]^{3}}.$
(17)

One could see that it becomes finite even at the horizons. So that a cut-off
by hand is needless with the help of GUP.

The quantum boson gases in a “wall” outside the horizon are regarded as the
eigenmodes of the scalar field in the vicinity of the black hole. Every wave
solution may be occupied by any integer number of quanta, that the total
number of the quantum boson particles may be not conserved. This is somewhat
analogous to the arguments on the statistics of the photon gas which leads to
the blackbody spectrum. In this case, the parameter $\alpha=-\mu/kT$ in the
normal Bose-Einstein distribution disappears, thus the chemical potential
vanishes. Now the free energy of the scalar field reads

$F=-\int_{\mu\sqrt{f(r)}}^{\infty}d\omega\frac{n(\omega)}{e^{\beta\omega}-1}.$
(18)

Therefore, the entropy can be calculated as

$\displaystyle S$ $\displaystyle=\beta^{2}\frac{\partial F}{\partial\beta}$
$\displaystyle=\frac{2\beta^{2}}{3\pi}\int
dr\frac{r^{2}}{\sqrt{f(r)}}\int_{\mu\sqrt{f(r)}}^{\infty}d\omega\frac{\omega(\omega^{2}/f(r)-\mu^{2})^{3/2}}{4\sinh^{2}(\beta\omega/2)\,[1+\lambda(\omega^{2}/f(r)-\mu^{2})]^{3}}$
$\displaystyle=\frac{\beta^{3}}{12\pi\lambda^{3}}\int
r^{2}f(r)dr\int_{x_{0}}^{\infty}dx\frac{x(x^{2}-x_{0}^{2})^{3/2}}{\sinh^{2}x\,[x^{2}-x_{0}^{2}+\beta^{2}f(r)/(4\lambda)]^{3}},$
(19)

where $x\equiv\beta\omega/2$ and $x_{0}=\beta\mu\sqrt{f(r)}/2$.

It is supposed that the main contribution of entropy comes from a small region
just near the horizon [30], whose thickness corresponds to a proper distance
at the order of the minimal length indicated by the GUP. Now we consider a
thin layer located between $r_{AH}$ and $r_{AH}+\epsilon_{AH}$. Then the range
of integration in Eq.(S0.Ex1) becomes $r_{AH}$ to $r_{AH}+\epsilon_{AH}$ [33].
This method is usually called as the thin film model, which is also developed
from the brick-wall model with GUP. Notice that $x_{0}$ in the above equation
goes to zero at the horizon, the entropy becomes

$\displaystyle
S=\frac{\beta^{3}}{12\pi\lambda^{3}}\int_{0}^{\infty}dx\frac{x^{4}}{\sinh^{2}x}I(x)_{AH},$
(20)

where

$\displaystyle
I(x)_{AH}=\int_{r_{AH}}^{r_{AH}+\epsilon_{AH}}dr\frac{r^{2}f(r)}{[x^{2}+\beta^{2}f(r)/(4\lambda)]^{3}}.$
(21)

On the other hand, the minimal length can be extracted from the relation as
below

$\displaystyle
2\sqrt{\lambda}=\int_{r_{AH}}^{r_{AH}+\epsilon_{AH}}\frac{dr}{\sqrt{f}}\approx\int_{r_{AH}}^{r_{AH}+\epsilon_{AH}}\frac{dr}{\sqrt{2\kappa_{AH}(r-r_{AH})}}=\sqrt{\frac{2\epsilon_{AH}}{\kappa_{AH}}},$
(22)

which gives

$\displaystyle\epsilon_{AH}=2\lambda\kappa_{AH}.$ (23)

Eq.(21) on apparent horizon of Vaidya black hole now becomes

$\displaystyle I(x)_{AH}$
$\displaystyle=\int_{r_{AH}}^{r_{AH}+\epsilon_{AH}}dr\frac{f(r)r}{[x^{2}+\frac{f(r)\beta_{AH}}{4\lambda}]^{3}}$
$\displaystyle\approx\frac{f^{\prime}(r_{AH})\epsilon_{AH}^{2}r_{AH}^{2}}{2x^{2}[x^{2}+\frac{\beta_{AH}^{2}f^{\prime}(r_{AH})\epsilon_{AH}}{4\lambda}]^{2}}$
$\displaystyle=\frac{4\kappa_{AH}^{3}r_{AH}^{2}\lambda^{2}}{(4\pi^{2}x+x^{3})^{2}}.$
(24)

where $x=\frac{\beta_{AH}\omega}{2}$. Then the entropy is

$\displaystyle S_{AH}$
$\displaystyle=\frac{\beta_{AH}^{3}}{12\pi\lambda^{3}}\int_{0}^{\infty}dx\frac{x^{4}}{\sinh^{2}x}I(x)_{AH}$
$\displaystyle=\frac{A_{AH}}{4}\frac{8\pi}{3\lambda}\int_{0}^{\infty}dx\frac{x^{4}}{\sinh^{2}x(4\pi^{2}x+x^{3})^{2}},$
(25)

where $A_{AH}=4\pi r_{AH}^{2}$ is the surface area of the apparent horizon.
The integrand in Eq.(S0.Ex6) could be regarded as a complex function
$h(z)=\frac{z^{4}}{\sinh^{2}z(4\pi^{2}z+z^{3})^{2}}$. With the help of residue
theorem, Eq.(S0.Ex6) turns out to be

$\displaystyle S_{AH}$
$\displaystyle=\frac{A_{AH}}{4}\frac{8}{3\lambda}(-\frac{1}{24}+\frac{1}{\pi^{2}}\sum_{n=1,n\neq
2,n\in\mathbb{N_{+}}}^{\infty}\frac{n(n^{2}+4)}{(n^{2}-4)^{3}})$
$\displaystyle=\frac{A_{AH}}{4}\frac{8}{3\lambda}(-\frac{1}{24}-\frac{25}{32\pi}+\frac{\zeta(3)}{\pi})$
$\displaystyle=\frac{A_{AH}}{4}\frac{\sigma}{\lambda},$ (26)

where
$\sigma=\frac{8}{3}(-\frac{1}{24}-\frac{25}{32\pi}+\frac{\zeta(3)}{\pi})$. We
could see that the entropy of the Vaidya black hole on apparent horizon is
proportional to the surface area at the horizon.

If we further require the entropy satisfy the area law of black hole, one need
to make $\sigma=\lambda$. Then the minimal length could be obtained as
$2\sqrt{\lambda}=2\sqrt{\sigma}l_{p}\approx 0.254969l_{p}$, where $l_{p}$ is
the Planck length. This requirement might provide some deep insight into the
minimal length indicated by GUP. Since the entropy is consider as the semi-
classical property of Hawking radiation from the horizon within a thin layer
of Planck order, so that the range of integration in our method is more
reasonable than brick-wall model with a man-made cut-off.

In summary, we have calculated the entropy of apparent horizon with thin film
model based on generalized uncertainty principle. We have found that the
entropy is proportional to the surface area at the horizon, which is
consistent with the results of Bekenstein and Hawking. Our result is more
reasonable and reliable because there is no unnatural cut-off in this method.
Furthermore, the area law of the black hole requires the minimal length
$\lambda$ in GUP to be a specific value, which deserves future studies to
provide some deep insight into microscopic structure of the black hole.

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# Charged Rényi negativity of massless free bosons

Hui-Huang<EMAIL_ADDRESS>
College of Physics and Communication Electronics, Jiangxi Normal University,
Nanchang 330022, China

In this paper, we consider the computation of charged moments of the reduced
density matrix of two disjoint intervals in the $1+1$ dimensional free
compactified boson conformal field theory (CFT) by studying the four-point
function of the fluxed twist fields. We obtained the exact scaling function of
this four-point function and discussed its decompactification limit. This
scaling function was used to obtain the charged moments of the partial
transpose which we refer as charged Rényi negativity. These charged moments
and the charged moments of the partial transpose are essential for the problem
of symmetry decomposition of the corresponding entanglement measures. We test
our analytic formula against exact numerical computation in the complex
harmonic chain, finding perfect agreements.

###### Contents

  1. 1 Introduction
  2. 2 Entanglement negativity in CFT
    1. 2.1 Moments of reduced density matrix
    2. 2.2 The repica trick for negativity
  3. 3 Symmetry decomposition of negativity
  4. 4 Charged moments of two disjoint interval in free boson CFT
    1. 4.1 Fluxed twist fields and charged moments
    2. 4.2 Small $x$ regime
    3. 4.3 Decompactification regime
  5. 5 Charged Rényi negativity
  6. 6 Numerical results
    1. 6.1 Charged moments
    2. 6.2 Charged Rényi negativity
  7. 7 Conclusion and discussion
  8. A Correlation function of fluxed twist fields
  9. B Normalisation of $\mathcal{F}_{n}(\mu,x)$

##  1 Introduction

In recent years, concepts and ideas coming from quantum information theory
have played crucial roles in connecting condensed matter theory and high
energy physics [1, 2, 3]. In quantum many-body systems, one of the most
important results on entanglement is the so-called area law (see [4] for a
review). In high energy physics, Ryu-Takayanagi formula [5] firstly recovers
the amazing relationship between entanglement and spacetime. People believe
that ideas from quantum information theory would help us solve the information
paradox of black holes [6] ultimately (see [7] as a review for recent
progress). Among all these progress, the entanglement entropies are the most
successful entanglement measures to characterize the bipartite entanglement of
a pure state. Given a system in a pure state $\ket{\Psi}$, and taking the
bipartition of the system $A$ and its complement $B$, the Hilbert space of the
full system factorizes as $\mathcal{H}=\mathcal{H}_{A}\otimes\mathcal{H}_{B}$.
The reduced density matrix (RDM) of subsystem $A$ is defined as
$\rho_{A}={\textrm{Tr}}_{B}\ket{\Psi}\bra{\Psi}$. From the moments of
$\rho_{A}$, i.e. ${\textrm{Tr}}\rho_{A}^{n}$, one can obtain the Von Neumann
entropy through the replica trick [8]

$S_{A}\equiv-{\textrm{Tr}}(\rho_{A}\log\rho_{A})=-\frac{\partial}{\partial
n}{\textrm{Tr}}\rho_{A}^{n}\Big{|}_{n=1}$ (1.1)

For a mixed state, the entanglement entropies are no longer useful measures of
entanglement. Instead, entanglement negativity is one of the most commonly
used measures of quantum entanglement between two subsystems $A_{1}$ and
$A_{2}$ in a generally mixed state [9, 10]. Consider the situation presented
in Fig.1, with the whole system $(A_{1}\cup A_{2})\cup B$ being in a pure
state $\ket{\Psi}$. Let $\rho_{A}$ be the reduced density matrix of subsystem
$A=A_{1}\cup A_{2}$ obtained by tracing out the “environment” $B$ as before,
and consider the “partial transpose” of this density matrix, denoted by
$\rho_{A}^{T_{2}}$, with respect to the Hilbert space corresponding to
$A_{2}$. Then the entanglement negativity (or negativity for short) is defined
as

$\mathcal{N}=\frac{1}{2}({\textrm{Tr}}|\rho_{A}^{T_{2}}|-1)=\sum_{\lambda_{i}<0}|\lambda_{i}|,$
(1.2)

where ${\textrm{Tr}}|O|={\textrm{Tr}}\sqrt{O^{\dagger}O}$ denotes the trace
norm of the operator $O$ and $\lambda_{i}$ are the eigenvalues of
$\rho_{A}^{T_{2}}$. Logarithmic negativity is another equivalent measure which
is defined as

$\mathcal{E}\equiv\log{\textrm{Tr}}|\rho_{A}^{T_{2}}|.$ (1.3)

To apply to the replica trick on entanglement negativity, it is also useful to
introduce the so-called Rényi negativity

$R_{n}={\textrm{Tr}}(\rho_{A}^{T_{2}})^{n}.$ (1.4)

In quantum field theories (QFT), entanglement negativity has been studied
extensively[11, 12, 13, 14, 15, 16]. In recent years, the holographic dual of
entanglement negativity has also been proposed and widely investigated[17, 18,
19, 20, 21, 22, 23].

If our system presents an internal symmetry, the entanglement measures will
split into different sectors corresponding to different representations of the
symmetry group. After the pioneering work [24], a lot of progress has been
made in this area, including symmetry resolved entanglement entropy [25, 26,
27, 28, 29, 30, 31], symmetry resolved relative entropy and trace distance
[32, 33], and symmetry decomposition of entanglement negativity [34, 35],
symmetry resolved entanglement entropy in holographic settings [36, 37, 38,
39]. All these results are mainly focused on $U(1)$ symmetry. Very recently,
symmetry resolved entanglement entropies in Wess-Zumino-Witten (WZW) models
were also discussed [40, 41].

In this manuscript, we will first consider the charged moments of the reduced
density matrix of two disjoint intervals in 1+1 dimensional free boson
conformal field theory (CFT) by the method of fluxed twist fields. As a
result, this problem is related to the four-point function of the fluxed twist
fields, which is determined by a scaling function. We obtained the exact form
of this scaling function and tried to apply this result to studying the
problem that how entanglement negativity decomposes under internal symmetry.
We finally obtained the charged Rényi negativity in this theory and make
various numerical checks of our CFT results. The charge imbalance resolved
negativity can be obtained by Fourier transforming the charged Rényi
negativity and taking the replica limit which has not worked out in this paper
due to technical complications.

The remaining part of this paper is organized as follows. In section 2, we
briefly review the path integral representation of negativity in 1+1
dimensional CFT. In section 3, we will discuss how entanglement negativity
decomposes under internal symmetry. In section 4, we will focus on the
computation of charged moments of two disjoint intervals and discuss two
important regions: the small $x$ region and the decompactification region. In
section 5, we consider the charged Rényi negativity by applying the results
obtained in the last section. In section 6, we test our analytic formula
against exact numerical computation. Finally, we make a conclusion and
discussion of our results in section 7 and some technical details are present
in two appendices.

##  2 Entanglement negativity in CFT

###  2.1 Moments of reduced density matrix

Figure 1: In 1+1 dimensional spacetime, the time slice is one dimensional, we
will consider the density matrix for the field theory on the subsystem
$A=A_{1}\cup A_{2}$ upon tracing out the local degree of freedom in
environment $B$. The endpoints of sub subsystem $A_{1},A_{2}$ is denoted by
$u_{1},v_{1}$ and $u_{2},v_{2}$ respectively.

Let us briefly review the calculation of Rényi entanglement negativity via
path integrals in two-dimensional CFTs. We denote $\phi$ as the basic field in
the theory. The matrix elements of the density matrix $\rho$ of vacuum are
$\rho(\phi,\phi^{\prime})=\braket{\phi}{0}\braket{0}{\phi^{\prime}}$. Now we
consider the subsystem $A$ consists $N$ disjoint interval on the real axis,
$A=\cup_{i=1}^{N}[u_{i},v_{i}]$. The matrix elements of RDM $\rho_{A}$ are
given by tracing out the degree of freedom not in $A$

$\rho_{A}(\phi,\phi^{\prime})=\frac{1}{Z}\int\mathcal{D}\phi_{B}\braket{\phi_{A}=\phi,\phi_{B}}{0}\braket{0}{\phi_{A}=\phi^{\prime},\phi_{B}}.$
(2.1)

For integer $n$, the moments of RDM $\rho_{A}$ are

${\textrm{Tr}}\rho_{A}^{n}=\frac{1}{Z_{1}^{n}}\int\prod_{i=1}^{n}\mathcal{D}\phi_{i}\rho_{A}(\phi_{i},\phi_{i+1}),\quad\phi_{n+1}\equiv\phi_{1},$
(2.2)

where $Z_{1}$ is the partition function on the sphere and from the path
integral above we should glue the cuts cyclicly along with $A$, which turns
out to be the partition function on an $n$-sheeted Riemann surface
$\mathcal{R}_{n,N}$ with genus $(n-1)(N-1)$. The simplest and most important
example is the entanglement entropy of an interval $A$ of length $l$ in an
infinite line. Since the genus of the corresponding Riemann surface for $N=1$
is zero, the entanglement entropy $S_{A}$ has the universal form

$S_{A}=\frac{c}{3}\log\frac{l}{\epsilon}+\mathrm{const},$ (2.3)

where $c$ is the central charge of the CFT, and $\epsilon$ is the UV cutoff.

For general $N>1$, it’s convenient to introduce branch point twist fields in
the complex plane to avoid from studying complicated Riemann surfaces. As a
result, the partition function on Riemann surface $\mathcal{R}_{n,N}$ can be
written as $2N$ point function of branch point twist fields in the complex
plane [8]

${\textrm{Tr}}\rho_{A}^{n}=\langle\prod_{i=1}^{N}\mathcal{T}_{n}(u_{i})\tilde{\mathcal{T}}_{n}(v_{i})\rangle$
(2.4)

The twist fields $\mathcal{T}_{n}$ and anti-twist field
$\tilde{\mathcal{T}_{n}}$ are primary operators and they have the same
dimension

$\Delta_{n}=\frac{c}{12}\left(n-\frac{1}{n}\right).$ (2.5)

By conformal map $w(z)=\frac{(u_{1}-z)(u_{N}-v_{N})}{(u_{1}-u_{N})(z-v_{N})}$,
the above $2N$ point function only depends on $2N-3$ cross-ratios
$x_{1}=w(v_{1}),x_{2}=w(u_{2}),\cdots,x_{2N-3}=w(v_{N-1})$, then we can write

${\textrm{Tr}}\rho_{A}^{n}=c_{n}^{N}\Big{|}\frac{\prod_{i<j}(u_{j}-u_{i})(v_{j}-v_{i})}{\prod_{i,j}(v_{j}-u_{i})}\Big{|}^{2\Delta_{n}}\mathcal{F}_{N,n}(x_{1},\cdots,x_{2N-3}),$
(2.6)

where $\mathcal{F}_{N,n}(x_{1},\cdots,x_{2N-3})$ is sensitive to the full
operator content of the theory and usually is very complicated. Only very few
results are known [42, 43, 44, 45, 46, 47]. For the free real boson
compactified on a circle of radius $\mathrm{R}$, analytic expressions are
available for $N=2$ and for integer $n>1$. In this case, there is only one
independent cross ration
$x=\frac{(u_{1}-v_{1})(u_{2}-v_{2})}{(u_{1}-u_{2})(v_{1}-v_{2})}$. Then eq.
(2.6) specialised to the case $N=2$ reads

${\textrm{Tr}}\rho_{A}^{n}=c_{n}^{2}\left(\frac{(u_{2}-u_{1})(v_{2}-v_{1})}{(v_{1}-u_{1})(v_{2}-u_{2})(v_{2}-u_{1})(u_{2}-v_{1})}\right)^{2\Delta_{n}}\mathcal{F}_{n}(x),$
(2.7)

where $\mathcal{F}_{n}(x)\equiv\mathcal{F}_{2,n}(x)$. The function
$\mathcal{F}_{n}(x)$ is [43](parametrized in terms of $\eta=\mathrm{R}^{2}/2$)

$\mathcal{F}_{n}(x)=\frac{\Theta(\mathbb{0}|\eta\tau)\Theta(\mathbb{0}|\tau/\eta)}{\Theta(\mathbb{0}|\tau)^{2}},$
(2.8)

where $\tau$ is $(n-1)\times(n-1)$ matrix with the following entries:

$\tau_{rs}=\frac{2\mathrm{i}}{n}\sum_{k=1}^{n-1}\sin(\frac{\pi
k}{n})\frac{{}_{2}F_{1}(k/n,1-k/n;1;1-x)}{{}_{2}F_{1}(k/n,1-k/n;1;x)}\cos[\frac{2\pi
k}{n}(r-s)].$ (2.9)

While $\Theta$ is the Siegel theta function defined by

$\Theta(\mathbf{z}|M)=\sum_{\mathbf{m}\in\mathbb{Z}^{n-1}}e^{\mathrm{i}\pi\mathbf{m}^{t}\cdot
M\cdot\mathbf{m}+2\pi\mathrm{i}\mathbf{m}^{t}\cdot\mathbf{z}},$ (2.10)

which is a function of the $(n-1)$ dimensional vector $\mathbf{z}$ and of the
$(n-1)\times(n-1)$ matrix $M$ which must be symmetric and with positive
imaginary part. In the remaining part of this paper, we will mainly focus on
the case $N=2$.

###  2.2 The repica trick for negativity

In this section, we briefly review the computation of entanglement negativity
in 1+1 dimensional CFT using twist fields. Let $\ket{e_{i}^{(1)}}$ and
$\ket{e_{j}^{(2)}}$ be two arbitrary bases of the Hilbert spaces associated to
the degree of freedom on $A_{1}$ and $A_{2}$ respectively. The partial
transpose (with respect to the second space) of $\rho_{A}$ is defined as

$\bra{e_{i}^{(1)}e_{j}^{(2)}}\rho_{A}^{T_{2}}\ket{e_{k}^{(1)}e_{l}^{(2)}}=\bra{e_{i}^{(1)}e_{l}^{(2)}}\rho_{A}\ket{e_{k}^{(1)}e_{j}^{(2)}}.$
(2.11)

The Rényi negativity is defined as

$R_{n}={\textrm{Tr}}\\{(\rho_{A}^{T_{2}})^{n}\\}=\sum_{\lambda_{i}}\lambda_{i}^{n},$
(2.12)

and entanglement negativity is obtained by taking the limit

$\mathcal{N}=\lim_{n\rightarrow\frac{1}{2}}\frac{1}{2}(R_{2n}-1).$ (2.13)

As mentioned in the last subsection, ${\textrm{Tr}}\rho_{A}^{n}$ for the union
of two disjoint interval $A=[u_{1},v_{1}]\cup[u_{2},v_{2}]$ is given by the
correlator
$\langle\mathcal{T}_{n}(u_{1})\tilde{\mathcal{T}}_{n}(v_{1})\mathcal{T}_{n}(u_{2})\tilde{\mathcal{T}}_{n}(v_{2})\rangle$.Taking
partial transpose with respect to the interval $A_{2}$ means the endpoints of
$A_{2}$ are exchanged while $A_{1}$ stay untouched. According to standard
procedure, the path integral representation of
${\textrm{Tr}}\\{(\rho_{A}^{T_{2}})^{n}\\}$ is proportional to the partition
function on the Riemann surface $\tilde{\mathcal{R}}_{n,2}$.
$\tilde{\mathcal{R}}_{n,2}$ differs from $\mathcal{R}_{n,2}$ only by the
exchange $u_{2}\leftrightarrow v_{2}$, and this immediately translates in
terms of correlation function of twist fields as
${\textrm{Tr}}(\rho_{A}^{T_{2}})^{n}=\langle\mathcal{T}_{n}(u_{1})\tilde{\mathcal{T}}_{n}(v_{1})\tilde{\mathcal{T}}_{n}(u_{2})\mathcal{T}_{n}(v_{2})\rangle$.
Since the exchange $u_{2}\leftrightarrow v_{2}$ sends $x\rightarrow x/(x-1)$,
the period matrix $\tilde{\tau}$ of $\tilde{\mathcal{R}}_{n,2}$ for
$x\in(0,1)$ is given by

$\tilde{\tau}(x)=\tau(x/(x-1))=\mathcal{R}+i\mathcal{I},$ (2.14)

where $\tau$ is the period matrix of $\mathcal{R}_{n,2}$ whose elements have
been given in eq. (2.9) and we denote the real and imaginary part of
$\tilde{\tau}(x)$ by $\mathcal{R}$ and $\mathcal{I}$ respectively.

Since the moments of the partial transposed RDM can be written as four-point
functions of twist fields. This four-point function still have the same form
with equation eq. (2.7)

${\textrm{Tr}}(\rho_{A}^{T_{2}})^{n}=c_{n}^{2}\left(\frac{(u_{2}-u_{1})(v_{2}-v_{1})}{(v_{1}-u_{1})(v_{2}-u_{2})(v_{2}-u_{1})(u_{2}-v_{1})}\right)^{2\Delta_{n}}\mathcal{G}_{n}(x),$
(2.15)

and we have the following relation

$\mathcal{G}_{n}(x)=(1-x)^{4\Delta_{n}}\mathcal{F}_{n}\left(\frac{x}{x-1}\right).$
(2.16)

The result eq. (2.8) is valid only for $0<x<1$. For generic $x\in\mathbb{C}$,
it can be written as

$\mathcal{F}_{n}(x,\bar{x})=\frac{\Theta(\mathbf{0}|T)}{|\Theta(\mathbf{0}|\tau)|^{2}},$
(2.17)

where the $2(n-1)\times 2(n-1)$ symmetric matrix $T$ is

$T=\begin{pmatrix}i\eta\mathcal{I}&\mathcal{R}\\\
\mathcal{R}&i\mathcal{I}/\eta\end{pmatrix},$ (2.18)

When $x\in(0,1)$, the period matrix $\tau(x)$ is purely imaginary. Then
$\mathcal{R}=0$ and the Siegel theta function in eq. (2.17) factorizes, giving
back the result eq. (2.8).

##  3 Symmetry decomposition of negativity

Let us first review some basic facts about the symmetry resolution of the
entanglement entropy. In this section and the remaining part of this paper, we
will assume our system presents a global $U(1)$ symmetry $Q$ and assuming that
$[\rho,Q]=0$, which implies $[\rho_{A},Q_{A}]=0$. We take the subsystem $A$
consist of two disjoint region $A=A_{1}\cup A_{2}$. Then $\rho_{A}$ admits
charge decomposition according to eigenvalues $q$ of local charge $Q_{A}$

$\rho_{A}=\oplus_{q}\Pi_{q}\rho_{A}=\oplus_{q}p(q)\rho_{A}(q),\quad
p(q)={\textrm{Tr}}(\Pi_{q}\rho_{A}).$ (3.1)

The symmetry resolved Rényi entropies are then defined as

$S_{n}(q)=\frac{1}{1-n}\log{\textrm{Tr}}[\rho_{A}(q)]^{n}.$ (3.2)

It’s very difficult to compute the symmetry resolved entropy using the above
definition due to the non-local feature of the projector $\Pi_{q}$. However,
we can bypass this difficulty by using the Fourier representation of the
projector and focusing on the charged moments of $\rho_{A}$,

$Z_{n}(\mu)={\textrm{Tr}}(e^{i\mu Q_{A}}\rho_{A}^{n}).$ (3.3)

Then it’s sufficient to compute the Fourier transform

$Z_{n}(q)=\int_{0}^{2\pi}\frac{d\mu}{2\pi}e^{-iq\mu}Z_{n}(\mu)$ (3.4)

to obtain the entropies of the sector of charge $q$ as

$S_{n}(q)=\frac{1}{1-n}\log\left[\frac{Z_{n}(q)}{Z_{1}(q)^{n}}\right].$ (3.5)

Now we turn to the symmetry decomposition of entanglement negativity under
$U(1)$ charge $Q$. Since the charge $Q$ is local, we can write
$Q_{A}=Q_{1}+Q_{2}$. From the relation $[\rho_{A},Q_{A}]=0$, performing a
partial transposition with respect to the second region $A_{2}$ of subsystem
$A$, we obtain

$[\rho_{A}^{T_{2}},\mathcal{Q}]=0,\quad\mathcal{Q}\equiv Q_{1}-Q_{2}^{T_{2}},$
(3.6)

where we have introduced the charge imbalance operator $\mathcal{Q}$ and we
will denote it’s eigenvalues as $\mathrm{q}$ to make a distinction with the
eigenvalues of $Q_{A}$. Then $\rho_{A}^{T_{2}}$ has a block matrix form, each
block was characterized by different eigenvalues $\mathrm{q}$ of the imbalance
operator $\mathcal{Q}$

$\rho_{A}^{T_{2}}=\oplus_{\mathrm{q}}\Pi_{\mathrm{q}}\rho_{A}^{T_{2}}.$ (3.7)

we define the normalized charge imbalance partially transposed density matrix

$\rho_{A}^{T_{2}}({\mathrm{q}})=\frac{\Pi_{\mathrm{q}}\rho_{A}^{T_{2}}}{{\textrm{Tr}}(\Pi_{\mathrm{q}}\rho_{A}^{T_{2}})}.$
(3.8)

Then we can write

$\rho_{A}^{T_{2}}({\mathrm{q}})=\oplus_{\mathrm{q}}p({\mathrm{q}})\rho_{A}^{T_{2}}({\mathrm{q}}),$
(3.9)

where $p({\mathrm{q}})={\textrm{Tr}}(\Pi_{\mathrm{q}}\rho_{A}^{T_{2}})$ is the
probability of finding ${\mathrm{q}}$ as the outcome of measurement of
$\mathcal{Q}$.

The charge imbalance resolved negativity is defined as

$\mathcal{N}({\mathrm{q}})=\frac{1}{2}({\textrm{Tr}}|\rho_{A}^{T_{2}}({\mathrm{q}})|-1).$
(3.10)

The total negativity is given by the sum of charge imbalance resolved
negativity weighted by the corresponding probability

$\mathcal{N}=\sum_{\mathrm{q}}p({\mathrm{q}})\mathcal{N}({\mathrm{q}}).$
(3.11)

The charge imbalance resolved Rényi negativity is defined as

$R_{n}({\mathrm{q}})={\textrm{Tr}}[(\rho_{A}^{T_{2}}({\mathrm{q}}))^{n}]=\frac{1}{p({\mathrm{q}})^{n}}{\textrm{Tr}}[\Pi_{\mathrm{q}}(\rho_{A}^{T_{2}})^{n}].$
(3.12)

Then the charge imbalance entanglement negativity can be obtained by taking
the limit

$\mathcal{N}({\mathrm{q}})=\lim_{n\rightarrow\frac{1}{2}}\frac{1}{2}(R_{2n}({\mathrm{q}})-1).$
(3.13)

The projection operator $\Pi_{\mathrm{q}}$ has the following integral
representation

$\Pi_{\mathrm{q}}=\int_{0}^{2\pi}\frac{d\mu}{2\pi}e^{-i\mu{\mathrm{q}}}e^{i\mu(Q_{1}-Q_{2}^{T_{2}})}.$
(3.14)

We can first compute the charged Rényi negativity

$R_{n}(\mu)={\textrm{Tr}}[(\rho_{A}^{T_{2}})^{n}e^{i\mu(Q_{1}-Q_{2}^{T_{2}})}],$
(3.15)

then the charge imbalance resolved Rényi negativity are related by Fourier
transform

$R_{n}({\mathrm{q}})=\frac{\int_{0}^{2\pi}\frac{d\mu}{2\pi}e^{-i\mu{\mathrm{q}}}R_{n}(\mu)}{[\int_{0}^{2\pi}\frac{d\mu}{2\pi}e^{-i\mu{\mathrm{q}}}R_{1}(\mu)]^{n}}.$
(3.16)

##  4 Charged moments of two disjoint interval in free boson CFT

###  4.1 Fluxed twist fields and charged moments

In a generic two-dimensional QFT, according to the path integral
representation of the charged moments $Z_{n}(\mu)$ defined in eq. (3.3), we
can treat it as the partition function on the Riemann surface
$\mathcal{R}_{n,N}$ pierced by an Aharonov-Bohm flux, such that the total
phase accumulated by the field upon going through the entire surface is $\mu$.
We denote this fluxed Riemann surface by $\mathcal{R}_{n,N}^{(\mu)}$. The
presence of the flux corresponds to impose twist boundary condition. This
boundary condition fuses with twist field at the endpoints of subsystem $A$
can be implemented by two local fields $\mathcal{T}_{n,\mu}$ and
$\tilde{\mathcal{T}}_{n,\mu}$ termed as fluxed twist fields and fluxed anti-
twist fields. These fluxed twist fields take into account not only the
internal permutational symmetry among the replicas but also the presence of
the flux. The partition function on $\mathcal{R}_{n,N}^{(\mu)}$ is related to
the $2N$-point function of the fluxed twist operators.

Studying field theory on the complicated fluxed Riemann surface usually turns
out to be very hard. It’s more convenient to move the topology of the world-
sheet to the target space where the fields lie. In other words, we can work on
a theory defined on a single complex plane formed by $n$ independent copies of
the original theory. In this multi-copy theory, we have to deal with fields
with $n$-component and we denote $\phi_{j}$ as the field on the $j$-th copy.
For subsystem $A$ made of two disjoint intervals, we have

$Z_{n}(\mu)=\langle\mathcal{T}_{n,\mu}(u_{1})\tilde{\mathcal{T}}_{n,\mu}(v_{1})\mathcal{T}_{n,\mu}(u_{2})\tilde{\mathcal{T}}_{n,\mu}(v_{2})\rangle.$
(4.1)

Upon crossing the cut $A$, the field $\phi_{i}$ transforms as
$\phi_{i}\rightarrow(T_{\mu})_{ij}\phi_{j}$, where the matrix elements of the
transformation matrix $T_{\mu}$ are given by
$(T_{\mu})_{ij}=e^{\mathrm{i}\mu/n}\delta_{i+1,j}$ with boundary condition
$n+1\equiv 1$. The eigenvectors of the matrix $T_{\mu}$ are

$\varphi_{k}=\sum_{j=1}^{n}e^{2\pi i\frac{k}{n}j}\phi_{j},\qquad
k=0,1,2,\cdots,n-1,$ (4.2)

with corresponding eigenvalues given by
$\lambda_{k}=e^{\mathrm{i}\frac{\mu}{n}+2\pi\mathrm{i}\frac{k}{n}}$. In other
words, in the space span by the fields $\varphi_{k}$, the action of the fluxed
twist field decoupled, i.e. they act on each $k$ independently. Thus we can
write

$\mathcal{T}_{n,\mu}=\prod_{k=0}^{n-1}\mathcal{T}_{n,\mu,k},\qquad\tilde{\mathcal{T}}_{n,\mu}=\prod_{k=0}^{n-1}\tilde{\mathcal{T}}_{n,\mu,k}.$
(4.3)

Thus, the partition function $Z_{n}(\mu)$ also factorize as

$Z_{n}(\mu)=\prod_{k=0}^{n-1}Z_{n,k}(\mu)=\prod_{k=0}^{n-1}\langle\mathcal{T}_{n,\mu,k}(u_{1})\tilde{\mathcal{T}}_{n,\mu,k}(v_{1})\mathcal{T}_{n,\mu,k}(u_{2})\tilde{\mathcal{T}}_{n,\mu,k}(v_{2})\rangle.$
(4.4)

Now let us consider a concrete model. Namely, we will consider the complex
bosonic free compactified field with Euclidean action

$\mathcal{S}[\phi]=\frac{g}{4\pi}\int
dzd\bar{z}\partial_{z}\phi\partial_{\bar{z}}\phi.$ (4.5)

In the multi-copy theory, we have the $n$-component field
$\Phi=(\phi_{1},\phi_{2},\cdots,\phi_{n})^{T}$, each component $\phi_{j}$ are
free but compactified on a circle. Since the field is complex, the target
space is a torus with radius $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$. Taking the
non-trivial winding into account, encircling the endpoints of $A$ will lead to

$\begin{split}\phi_{j}(e^{2\pi\mathrm{i}}z_{i},e^{-2\pi\mathrm{i}}\bar{z}_{i})=e^{\mathrm{i}\mu/n}\phi_{j-1}(z_{i},\bar{z}_{i})+m_{j,1}\mathrm{R}_{1}+\mathrm{i}m_{j,2}\mathrm{R}_{2},\qquad
m_{j,1},m_{j,2}\in\mathbb{Z}\\\
\phi_{j}(e^{2\pi\mathrm{i}}\tilde{z}_{i},e^{-2\pi\mathrm{i}}\bar{\tilde{z}}_{i})=e^{-\mathrm{i}\mu/n}\phi_{j+1}(\tilde{z}_{i},\bar{\tilde{z}}_{i})+\tilde{m}_{j,1}\mathrm{R}_{1}+\mathrm{i}\tilde{m}_{j,2}\mathrm{R}_{2},\qquad\tilde{m}_{j,1},\tilde{m}_{j,2}\in\mathbb{Z},\end{split}$
(4.6)

where $z_{i}\equiv z-u_{i},\tilde{z}_{i}\equiv z-v_{i}$ for $i=1,2$. In this
theory, $\mathcal{T}_{n,\mu,k}$ and $\tilde{\mathcal{T}}_{n,\mu,k}$ are
bosonic $U(1)$ twist field with dimension (see appendix A for details)

$\Delta_{n,k,\mu}=\left(\frac{k}{n}+\frac{\mu}{2\pi
n}\right)\left(1-\frac{k}{n}-\frac{\mu}{2\pi n}\right).$ (4.7)

Then, the dimension of the fluxed twist fields $\mathcal{T}_{n,\mu}$ can be
obtained by

$\Delta_{n,\mu}=\sum_{k=0}^{n-1}\Delta_{n,k,\mu}=\frac{1}{6}\left(n-\frac{1}{n}\right)-\frac{\mu^{2}}{4\pi^{2}n}+\frac{\mu}{2\pi
n}.$ (4.8)

We can simplify this non-trivial monodromy a bit by introducing
$\varphi_{k}=\sum_{j=1}^{n}e^{2\pi\mathrm{i}jk/n}\phi_{j}$. Then eq. (4.6)
becomes

$\begin{split}\varphi_{k}(e^{2\pi\mathrm{i}}z_{i},e^{-2\pi\mathrm{i}}\bar{z}_{i})=e^{\mathrm{i}\mu/n+2\pi\mathrm{i}k/n}\varphi_{k}(z_{i},\bar{z}_{i})+\mathrm{R}\sum_{j=1}^{n}e^{2\pi\mathrm{i}jk/n}m_{j},\\\
\varphi_{k}(e^{2\pi\mathrm{i}}\tilde{z}_{i},e^{-2\pi\mathrm{i}}\bar{\tilde{z}}_{i})=e^{-\mathrm{i}\mu/n-2\pi\mathrm{i}k/n}\varphi_{k}(\tilde{z}_{i},\bar{\tilde{z}}_{i})+\mathrm{R}\sum_{j=1}^{n}e^{2\pi\mathrm{i}jk/n}\tilde{m}_{j}.\end{split}$
(4.9)

Here, for simplicity we set $\mathrm{R}_{1}=\mathrm{R}_{2}=\mathrm{R}$ and
$m_{j},\tilde{m}_{j}\in\mathbb{Z}+\mathrm{i}\mathbb{Z}$. Consequently, the
target space of the field $\varphi_{k}$ is a complicated two dimensional
lattice $\Lambda$

$\Lambda=\left\\{\mathrm{R}\sum_{j=1}^{n}e^{2\pi\mathrm{i}jk/n}m_{j}\Big{|}m_{j}\in\mathbb{Z}+\mathrm{i}\mathbb{Z}\right\\}.$
(4.10)

In order to calculate the partition function, we can split $\varphi_{k}$ into
a classical and a quantum part
$\varphi_{k}=\varphi^{\mathrm{cl}}_{k}+\varphi^{\mathrm{qu}}_{k}$. The
classical part takes into account the nontrivial structure of the target space

$\varphi^{\mathrm{cl}}_{k}(e^{2\pi\mathrm{i}}z_{i},e^{-2\pi\mathrm{i}}\bar{z}_{i})=e^{2\pi\mathrm{i}a}\varphi^{\mathrm{cl}}_{k}(z_{i},\bar{z}_{i})+\lambda,$
(4.11)

where $\lambda\in\Lambda$, while the quantum part is transparent to it

$\varphi^{\mathrm{qu}}_{k}(e^{2\pi\mathrm{i}}z_{i},e^{-2\pi\mathrm{i}}\bar{z}_{i})=e^{2\pi\mathrm{i}a}\varphi^{\mathrm{qu}}_{k}(z_{i},\bar{z}_{i}),$
(4.12)

where we have defined $a\equiv\frac{k}{n}+\frac{\mu}{2\pi n}$ and similar
relation holds when encircling around $\tilde{z}_{i}$. The computation of
quantum part is already done in [48] and we report its derivation in the
appendix A

$Z^{\mathrm{qu}}_{n,k}(\mu)=\mathrm{const}\left(\frac{(u_{2}-u_{1})(v_{2}-v_{1})}{(v_{1}-u_{1})(v_{2}-u_{2})(v_{2}-u_{1})(u_{2}-v_{1})}\right)^{2\Delta_{n,k,\mu}}\frac{1}{I_{n,k,\mu}(x)},$
(4.13)

where

$I_{n,k,\mu}(x)=2F_{a}(x)F_{a}(1-x)=2\beta_{a}[F_{a}(x)]^{2},$ (4.14)

and

$F_{y}(x)\equiv{{}_{2}F_{1}(y,1-y;1;x)},\quad\beta_{y}(x)\equiv\frac{F_{y}(1-x)}{F_{y}(x)}.$
(4.15)

The action for a given classical configuration is given by (eq. (A.27) with
$\alpha_{a}=0$)

$\mathcal{S}^{\mathrm{cl}}_{a}=\frac{2g\pi\sin\pi
a}{n\beta_{a}}[|\xi_{1}|^{2}+\beta_{a}^{2}|\xi_{2}|^{2}],$ (4.16)

where $\xi_{1},\xi_{2}\in\Lambda$. Then the partition function can be written
as

$Z_{n}(\mu)=\sum_{m\in\mathbb{Z}^{2n}}\prod_{k=0}^{n-1}Z_{a}^{\mathrm{qu}}Z_{a}^{\mathrm{cl}}.$
(4.17)

The classical part of the partition function reads

$Z^{\mathrm{cl}}_{n}(\mu)=\sum_{m\in\mathbb{Z}^{2n}}\prod_{k=0}^{n-1}Z_{a}^{\mathrm{cl}}=\sum_{\mathbf{m}\in\mathbb{Z}^{2n}}\prod_{k=0}^{n-1}\exp\Big{\\{}-\frac{2g\pi\sin\pi
a}{n}[|\xi_{1}|^{2}\beta_{a}+\beta_{a}^{-1}|\xi_{2}|^{2}]\Big{\\}}.$ (4.18)

Since the quantum part does not depend on $\mathbf{m}$, the function
$\mathcal{F}_{n}(\mu,x)$ can be written as

$\mathcal{F}_{n}(\mu,x)=\prod_{k=0}^{n-1}\frac{\mathrm{const}}{\beta_{a}[F_{a}(x)]^{2}}\sum_{\mathbf{m}\in\mathbb{Z}^{2n}}\prod_{k=0}^{n-1}\exp\Big{\\{}-\frac{2g\pi\sin\pi
a}{n}[|\xi_{1}|^{2}\beta_{a}+\beta_{a}^{-1}|\xi_{2}|^{2}]\Big{\\}}.$ (4.19)

Given
$\xi_{i}=\mathrm{R}\sum_{r=1}^{n}e^{2\pi\mathrm{i}kr/n}(m_{r,1}^{(i)}+\mathrm{i}m_{r,2}^{(i)}),i=1,2$,
we have

$|\xi_{i}|^{2}=\mathrm{R}^{2}\sum_{r,s=1}^{n}\Big{[}\sum_{j=1}^{2}m^{(i)}_{r,j}m^{(i)}_{s,j}\Big{[}\cos\frac{2\pi(r-s)k}{n}\Big{]}+(m^{(i)}_{r,1}m^{(i)}_{s,2}-m^{(i)}_{s,1}m^{(i)}_{r,2})\sin\Big{[}\frac{2\pi(r-s)k}{n}\Big{]}\Big{]}.$
(4.20)

Then substitute eq. (4.20) into eq. (4.18) and after some simplification, one
can rewrite the result in terms of Siegel theta functions. Since the
computation is almost the same as [43], we just report the final result here

$Z^{\mathrm{cl}}_{n}(\mu)=[\Theta(\mathbf{0}|\eta\Omega)\Theta(\mathbf{0}|\eta\tilde{\Omega})]^{2},$
(4.21)

where $\Omega$ and $\tilde{\Omega}$ are $n\times n$ matrix with elements

$\Omega_{rs}=\frac{2\mathrm{i}}{n}\sum_{k=0}^{n-1}\sin(\pi
a)\beta_{a}\cos[\frac{2\pi k}{n}(r-s)],$ (4.22)
$\tilde{\Omega}_{rs}=\frac{2\mathrm{i}}{n}\sum_{k=0}^{n-1}\sin(\pi
a)\frac{1}{\beta_{a}}\cos[\frac{2\pi k}{n}(r-s)],$ (4.23)

and $\eta\equiv g\mathrm{R}^{2}$. The spectrum of the matrices $\Omega$ and
$\tilde{\Omega}$ are given by

$\\{\omega_{p}|p=1,2,\cdots,n-1\\}\cup\Big{\\{}\omega_{n}\equiv\mathrm{i}\sin\left(\frac{\mu}{2n}\right)\beta_{\frac{\mu}{2\pi
n}}\Big{\\}}$ (4.24)

and

$\\{\tilde{\omega}_{p}|p=1,2,\cdots,n-1\\}\cup\Big{\\{}\tilde{\omega}_{n}\equiv\mathrm{i}\sin\left(\frac{\mu}{2n}\right)\Big{/}\beta_{\frac{\mu}{2\pi
n}}\Big{\\}}$ (4.25)

respectively, where for $p=1,2,\cdots,n-1$, we have

$\omega_{p}=\mathrm{i}\sin\left[\pi\left(\frac{p}{n}-\frac{\mu}{2\pi
n}\right)\right]\beta_{\frac{p}{n}-\frac{\mu}{2\pi
n}}+\mathrm{i}\sin\left[\pi\left(\frac{p}{n}+\frac{\mu}{2\pi
n}\right)\right]\beta_{\frac{p}{n}+\frac{\mu}{2\pi n}},$ (4.26)
$\tilde{\omega}_{p}=\mathrm{i}\sin\left[\pi\left(\frac{p}{n}-\frac{\mu}{2\pi
n}\right)\right]\Big{/}\beta_{\frac{p}{n}-\frac{\mu}{2\pi
n}}+\mathrm{i}\sin\left[\pi\left(\frac{p}{n}+\frac{\mu}{2\pi
n}\right)\right]\Big{/}\beta_{\frac{p}{n}+\frac{\mu}{2\pi n}}.$ (4.27)

The matrices $\Omega$ and $\tilde{\Omega}$ have a common eigenbasis whose
normalized eigenvectors can be written as

$(e_{p})_{q}=\frac{1}{\sqrt{n}}e^{2\pi\mathrm{i}pq/n},\quad p,q=1,2,\cdots,n.$
(4.28)

Then $\Omega$ and $\tilde{\Omega}$ can be diagonalized simultaneously by
matrix $U$ whose elements are given by $U_{pq}=(e_{p})_{q}$.

We finally have

$\mathcal{F}_{n}(\mu,x)=s_{n,\mu}\frac{[\Theta(\mathbf{0}|\eta\Omega)\Theta(\mathbf{0}|\eta\tilde{\Omega})]^{2}}{\prod_{k=0}^{n-1}\beta_{a}[F_{a}(x)]^{2}}.$
(4.29)

Here $s_{n,\mu}$ can be fixed by requiring $\mathcal{F}_{n}(\mu,0)=1$ as (See
appendix B for details)

$s_{n,\mu}=2n\eta^{n}\cos^{n-1}\left(\frac{\mu}{2n}\right)\sin\left(\frac{\mu}{2n}\right).$
(4.30)

Using identities of the $\Theta$ function in the appendix B, we find another
expression of $\mathcal{F}_{n}(\mu,x)$

$\mathcal{F}_{n}(\mu,x)=s_{n,\mu}\frac{[\Theta(\mathbf{0}|\eta\Omega)\Theta(\mathbf{0}|\hat{\Omega}/\eta)]^{2}}{\prod_{p=1}^{n}(-\mathrm{i}\eta\tilde{\omega}_{p})\prod_{k=0}^{n-1}\beta_{a}[F_{a}(x)]^{2}},$
(4.31)

where

$\hat{\Omega}_{rs}=-\tilde{\Omega}^{-1}_{rs}=\frac{-1}{n}\sum_{p=1}^{n}\tilde{\omega}_{p}^{-1}\cos[\frac{2\pi
p}{n}(r-s)].$ (4.32)

###  4.2 Small $x$ regime

For $x\rightarrow 0$, the function $\beta_{a}(x)$ behave as

$\beta_{a}(x)=-\frac{\sin\pi a}{\pi}(\log x+f_{a})+\mathcal{O}(x),\quad
f_{a}\equiv 2\gamma_{E}+\psi(a)+\psi(1-a),$ (4.33)

where $\gamma_{E}$ is the Euler constant and $\psi(z)$ is the diagamma
function. Then

$\Theta(0|\eta\Omega)=1+\sum_{\mathbf{m}\in\mathbb{Z}^{n}/{\mathbf{0}}}x^{\frac{2\eta}{n}\sum_{k=0}^{n-1}\sin^{2}(\pi
a)\mathbf{m}^{\mathrm{t}}\cdot
C_{k/n}\cdot\mathbf{m}}e^{\frac{2\eta}{n}\sum_{k=0}^{n-1}\sin^{2}(\pi
a)f_{a}\mathbf{m}^{\mathrm{t}}\cdot C_{k/n}\cdot\mathbf{m}}+\cdots,$ (4.34)

where $(C_{k/n})_{rs}\equiv\cos[\frac{2\pi k}{n}(r-s)]$ and the following
relation holds

$\frac{2}{n}\sum_{k=0}^{n-1}\sin^{2}(\pi a)\mathbf{m}^{\mathrm{t}}\cdot
C_{k/n}\cdot\mathbf{m}=\sum_{j=1}^{n}m_{j}^{2}-\cos\left(\frac{\mu}{n\pi}\right)\sum_{j=1}^{n}m_{j}m_{j+1},$
(4.35)

where $m_{n+1}\equiv m_{1}$. This expression is obviously minimized when the
non-zero vector $\mathbf{m}$ has only one entry equals to $\pm 1$ and all
other elements are zero. There are $2n$ such vectors for which eq. (4.35)
equals to $1$. All these $2n$ vectors gives the same value for
$\mathbf{m}^{\mathrm{t}}\cdot C_{k/n}\cdot\mathbf{m}$, and the result is 1.
Using the following integral representation of diagamma function

$\psi(z)+\gamma_{E}=\int_{0}^{\infty}dt\frac{e^{-t}-e^{-zt}}{1-e^{-t}},$
(4.36)

we find the formula

$\begin{split}\frac{2}{n}\sum_{k=0}^{n-1}\sin^{2}(\pi
a)f_{a}=f_{\frac{\mu}{2\pi}}-2\log
n+{\textrm{Re}}~{}B_{\theta}\left(\frac{\mu}{2\pi},0\right)+{\textrm{Re}}~{}B_{\theta}\left(1-\frac{\mu}{2\pi},0\right),\end{split}$
(4.37)

where $B_{z}(a,b)$ is the incomplete Beta function

$B_{z}(a,b)=\int_{0}^{z}t^{a-1}(1-t)^{b-1}dt,$ (4.38)

and $\theta\equiv e^{2\pi\mathrm{i}/n}$.

Then the coefficient of $x^{\eta}$ is
$2n^{1-2\eta}\exp\\{\eta[f_{\mu/2\pi}+{\textrm{Re}}~{}B_{\theta}(\mu/2\pi,0)+{\textrm{Re}}~{}B_{\theta}(1-\mu/2\pi,0)]\\}$.

To find the small $x$ expansion of $\Theta(\mathbf{0}|\hat{\Omega}/\eta)$, we
first compute

$\tilde{\omega}_{p}^{-1}=\begin{cases}\frac{\mathrm{i}}{2\pi}\left(\log
x+\frac{1}{2}\left(f_{a^{+}}+f_{a^{-}}\right)\right)+\mathcal{O}\left(\frac{1}{\log
x}\right),&p=1,2,\cdots,n-1\\\ \frac{\mathrm{i}}{2\pi}\left(\log
x+f_{\frac{\mu}{2\pi n}}\right)+\mathcal{O}\left(\frac{1}{\log
x}\right),&p=n\end{cases}$ (4.39)

where $a^{\pm}\equiv\frac{k}{n}\pm\frac{\mu}{2\pi n}$. Then

$\Theta(0|\hat{\Omega}/\eta)=1+\sum_{\mathbf{m}\in\mathbb{Z}^{n}/{\mathbf{0}}}x^{\frac{1}{2n\eta}\sum_{k=1}^{n}\mathbf{m}^{\mathrm{t}}\cdot
C_{k/n}\cdot\mathbf{m}}e^{\frac{1}{4n\eta}\sum_{k=1}^{n-1}(f_{a^{+}}+f_{a^{-}})\mathbf{m}^{\mathrm{t}}\cdot
C_{k/n}\cdot\mathbf{m}}e^{\frac{1}{2n\eta}f_{\frac{\mu}{2\pi
n}}\mathbf{m}^{\mathrm{t}}\cdot C_{n/n}\cdot\mathbf{m}}+\cdots$ (4.40)

From the relation

$\frac{1}{2n}\sum_{k=1}^{n}\mathbf{m}^{\mathrm{t}}\cdot
C_{k/n}\cdot\mathbf{m}=\frac{1}{2}\sum_{j=1}^{n}m_{j}^{2},$ (4.41)

it’s easy to see that this expression is minimized when the non-zero vector
$\mathbf{m}$ has only one entry equals to $\pm 1$ and all other elements are
zero. There are $2n$ such vectors for which eq. (4.41) equals to
$\frac{1}{2}$. From the relation

$\frac{1}{2}\sum_{k=1}^{n-1}(f_{a^{+}}+f_{a^{-}})+f_{\frac{\mu}{2\pi
n}}=n(f_{\frac{\mu}{2\pi}}-2\log n),$ (4.42)

we see that the coefficient of $x^{\frac{1}{2\eta}}$ is
$2n\exp[\frac{1}{2\eta}(f_{\frac{\mu}{2\pi}}-2\log n)]$. Therefore we get the
small $x$ behavior of the scaling function

$\mathcal{F}_{n}(\mu,x)=1+\xi_{n,\mu}x^{\alpha}+\cdots$ (4.43)

where

$\xi_{n,\mu}=\begin{cases}4n\exp[\frac{1}{2\eta}(f_{\frac{\mu}{2\pi}}-2\log
n)],&\text{if }\frac{1}{2\eta}\leq\eta\\\
4n^{1-\eta}\exp\\{\eta[f_{\frac{\mu}{2\pi}}+{\textrm{Re}}~{}B_{\theta}\left(\frac{\mu}{2\pi},0\right)+{\textrm{Re}}~{}B_{\theta}\left(1-\frac{\mu}{2\pi},0\right)]\\},&\text{otherwise}\end{cases}$
(4.44)

and $\alpha=\min(\eta,\frac{1}{2\eta})$.

###  4.3 Decompactification regime

For fixed $x$, in the limit of large $\eta$, we have
$\Theta(\mathbf{0}|\eta\Omega)=1+\cdots$ and
$\Theta(\mathbf{0}|\eta\tilde{\Omega})=1+\cdots$, where $\cdots$ denotes terms
which vanish when taking the limit $\eta\rightarrow\infty$. Therefore, in the
large $\eta$ limit, we find

$\mathcal{F}_{n}(\mu,x)=\frac{s_{n,\mu}}{\prod_{k=0}^{n-1}F_{a}(1-x)F_{a}(x)},$
(4.45)

which recovering the correct quantum result with the proper $\eta$ dependent
normalisation.

##  5 Charged Rényi negativity

As explained in section 2, for the charged Rényi negativity defined in eq.
(3.15), we have the following relation

$\begin{split}&R_{n}(\mu)=\langle\mathcal{T}_{a}(u_{1})\tilde{\mathcal{T}}_{a}(v_{1})\tilde{\mathcal{T}}_{a}(u_{2})\mathcal{T}_{a}(v_{2})\rangle\\\
&=c_{n,\mu}^{2}\left(\frac{(u_{2}-u_{1})(v_{2}-v_{1})}{(v_{1}-u_{1})(v_{2}-u_{2})(v_{2}-u_{1})(u_{2}-v_{1})}\right)^{2\Delta_{n,\mu}}\mathcal{G}_{n}(\mu;y),\end{split}$
(5.1)

where we have introduced the ordered four point ratio

$y=\frac{(v_{1}-u_{1})(v_{2}-u_{2})}{(u_{2}-u_{1})(v_{2}-v_{1})}=\frac{x}{x-1}$
(5.2)

with $0<y<1$. The two scaling function $\mathcal{F}_{n}(\mu;x)$ and
$\mathcal{G}_{n}(\mu;y)$ are related as

$\mathcal{G}_{n}(\mu;x)=(1-y)^{4\Delta_{n,\mu}}\mathcal{F}_{n}\left(\mu;\frac{y}{y-1}\right).$
(5.3)

Since the scaling function $\mathcal{F}_{n}(\mu,x)$ given in eq. (4.29) is
only valid in the region $0<x<1$, while in our case $-\infty<\frac{y}{y-1}<0$.
So we must find the expression of the scaling function
$\mathcal{F}_{n}(\mu;x,\bar{x})$ for generic four-point ratio
$x,\bar{x}\in\mathbb{C}$. As explained in appendix A, for generic
$x\in\mathbb{C}$, $\mathcal{F}_{n}(\mu;x,\bar{x})$ reads

$\mathcal{F}_{n}(\mu;x,\bar{x})=\prod_{k=0}^{n-1}\frac{\mathrm{const}}{F_{a}(x)\bar{F}_{a}(1-\bar{x})+\bar{F}_{a}(\bar{x})F_{a}(1-x)}\sum_{\mathbf{m}\in\mathbb{Z}^{2n}}\prod_{k=0}^{n-1}e^{-S_{a}^{\mathrm{cl}}}.$
(5.4)

In this case, the classical action is given by eq. (A.27)

$S_{a}^{\mathrm{cl}}=\frac{2g\pi\sin(\pi
a)}{n}\left[\frac{|\tau_{a}|^{2}}{\beta_{a}}|\xi_{1}|^{2}+\frac{\alpha_{a}}{\beta_{a}}(\xi_{1}\bar{\xi}_{2}\bar{\gamma}+\bar{\xi}_{1}\xi_{2}\gamma)+\frac{|\xi_{2}|^{2}}{\beta_{a}}\right].$
(5.5)

For generic $x\in\mathbb{C}$, the real part $\alpha_{a}(x)$ of the modulus
$\tau_{a}(x)$ of a “fake torus” is nonzero

$\tau_{a}(x)\equiv\mathrm{i}\frac{F_{a}(1-x)}{F_{a}(x)}=\alpha_{a}(x)+\mathrm{i}\beta_{a}(x).$
(5.6)

Substitute the explicit expression of $\xi_{i}$ into eq. (5.5) and summing
over $k$ and $\mathbf{m}$, the classical part of the partition function can be
written as some Siegel theta function. This calculation is almost the same
with [12], and the final result can be obtained by tiny adjustment

$\mathcal{F}_{n}(\mu;x,\bar{x})=s_{n,\mu}\frac{\Theta(\mathbf{0}|\eta
G)^{2}}{\prod_{k=0}^{n-1}{\textrm{Re}}[F_{a}(x)\bar{F}_{a}(1-\bar{x})]}.$
(5.7)

The $2n\times 2n$ symmetric matrix $G$ introduced in the $\mathrm{r.h.s}$ of
eq. (5.7) is purely imaginary and can be written in terms of $n\times n$ block
matrices as

$G=2\mathrm{i}\begin{pmatrix}A&W\\\ W^{\mathrm{t}}&B\end{pmatrix},$ (5.8)

with

$\begin{split}&A_{rs}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{|\tau_{a}|^{2}}{\beta_{a}}\sin(\pi
a)\cos\Big{[}\frac{2\pi k}{n}(r-s)\Big{]},\\\
&B_{rs}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\beta_{a}}\sin(\pi
a)\cos\Big{[}\frac{2\pi k}{n}(r-s)\Big{]},\\\
&W_{rs}=-\frac{1}{n}\sum_{k=0}^{n-1}\frac{\alpha_{a}}{\beta_{a}}\sin(\pi
a)\sin\Big{[}\frac{2\pi k}{n}(r-s+\frac{1}{2})\Big{]}.\end{split}$ (5.9)

For $x\in(0,1)$, we have $\alpha_{a}(x)=0$ for every $k$ and therefore the off
diagonal blocks $W$ of $G$ are zero, then the Siegel theta function
$\Theta(\mathbf{0}|\eta G)$ factorizes into a product of two Siegel theta
functions $\Theta(\mathbf{0}|\eta A)\Theta(\mathbf{0}|\eta
B)=\Theta(\mathbf{0}|\eta\Omega)\Theta(\mathbf{0}|\eta\tilde{\Omega})$ and eq.
(5.7) gives back eq. (4.29).

Then we can write $\mathcal{G}(\mu;y)$ for $0<y<1$ as

$\mathcal{G}(\mu;y)=s_{n,\mu}(1-y)^{4\Delta_{n,\mu}}\frac{\Theta(\mathbf{0}|\eta
G(\frac{y}{y-1}))^{2}}{\prod_{k=0}^{n-1}{\textrm{Re}}[F_{a}(\frac{y}{y-1})\bar{F}_{a}(\frac{1}{1-y})]}.$
(5.10)

In the decompactification limit $\eta\rightarrow\infty$, we have

$\mathcal{G}(\mu;y)=\frac{s_{n,\mu}(1-y)^{4\Delta_{n,\mu}}}{\prod_{k=0}^{n-1}{\textrm{Re}}[F_{a}(\frac{y}{y-1})\bar{F}_{a}(\frac{1}{1-y})]}.$
(5.11)

##  6 Numerical results

In this section, we consider the charged moments and the charged Rényi
negativity for the complex harmonic chain with periodic boundary conditions.
We will use this lattice model to check the CFT formula obtained in section 4
and section 5 in the decompactification regime.

The Hamiltonian of the real harmonic chain made by $L$ sites reads

$H_{HC}=\sum_{j=0}^{L-1}\left(\frac{1}{2M}p_{j}^{2}+\frac{M\omega^{2}}{2}q_{j}^{2}+\frac{K}{2}(q_{j+1}-q_{j})^{2}\right),$
(6.1)

where periodic boundary conditions $q_{L}\equiv q_{0},p_{L}\equiv p_{0}$ are
imposed and variables $p_{j}$ and $q_{j}$ satisfy standard bosonic commutation
relations $[q_{i},q_{j}]=[p_{i},p_{j}]=0$ and
$[q_{i},p_{j}]=\mathrm{i}\delta_{ij}$. We can work with $M=K=1$ without loss
of generality. The lattice version of the complex non-compact bosonic field
theory is the sum of two of the above harmonic chain. In terms of the
variables $q^{(1)},p^{(1)}$ and $q^{(2)},p^{(2)}$, the Hamiltonian is

$H_{CHC}(p^{(1)}+\mathrm{i}p^{(2)},q^{(1)}+\mathrm{i}q^{(2)})=H_{HC}(p^{(1)},q^{(1)})+H_{HC}(p^{(2)},q^{(2)}).$
(6.2)

Since the bosonic field is not compactified and massless, we must compare the
continuum limit of eq. (6.2) with the regime $\eta\rightarrow\infty$ of the
CFT results computed in section 4 and section 5. The Hamiltonian eq. (6.2) can
be diagonalized by introducing the creation and annihilation operators
$a_{k},a_{k}^{\dagger}$ and $b_{k},b_{k}^{\dagger}$, satisfying
$[a_{k},a_{k^{\prime}}^{\dagger}]=\delta_{kk^{\prime}}$ and
$[b_{k},b_{k^{\prime}}^{\dagger}]=\delta_{kk^{\prime}}$. In terms of these
operators, the Hamiltonian eq. (6.2) is diagonal

$H_{CHC}=\sum_{k=0}^{L-1}\epsilon_{k}(a_{k}^{\dagger}a_{k}+b_{k}^{\dagger}b_{k}),\quad\epsilon_{k}=\sqrt{\mathbf{\omega}^{2}+\frac{4K}{M}\sin^{2}\left(\frac{\pi
k}{L}\right)}.$ (6.3)

While the $U(1)$ charge is

$Q=\sum_{k=0}^{L-1}\epsilon_{k}(a_{k}^{\dagger}a_{k}-b_{k}^{\dagger}b_{k}).$
(6.4)

The conserved charge is local and can also be written in the position space
and for a given subsystem $A$ reads

$Q_{A}=\sum_{j\in A}\epsilon_{k}(a_{j}^{\dagger}a_{j}-b_{j}^{\dagger}b_{j}).$
(6.5)

###  6.1 Charged moments

(a)

Figure 2: Numerical data of $Z_{1}^{\text{norm}}$ as a function of $\mu$ for
different $z=l_{0}/L$ in the complex harmonic chain. The full lines are the
CFT predictions. Here we consider $l_{0}=100,d=400,L=1000,\omega=10^{-7}$ and
$l_{0}=500,d=250,L=1500,\omega=10^{-7}$. The agreement is very well for $\mu$
near $\pi$, but it worsens as $\mu$ gets closer to $0$ and $2\pi$.

The charged moments factorise as

$Z_{n}(\mu)={\textrm{Tr}}\rho_{A}^{n}e^{\mathrm{i}\mu
Q_{A}}={\textrm{Tr}}[\rho_{A}^{n}e^{\mathrm{i}\mu
N_{A}^{a}}]\times{\textrm{Tr}}[\rho_{A}^{n}e^{-\mathrm{i}\mu N_{A}^{b}}].$
(6.6)

The correlation functions for real harmonic chain are

$\begin{split}\mathbf{Q}_{rs}\equiv\bra{0}q_{r}q_{s}\ket{0}=\frac{1}{2L}\sum_{k=0}^{L-1}\frac{1}{M\omega_{k}}\cos\left(\frac{2\pi
k(r-s)}{L}\right),\\\
\mathbf{P}_{rs}\equiv\bra{0}p_{r}p_{s}\ket{0}=\frac{1}{2L}\sum_{k=0}^{L-1}M\omega_{k}\cos\left(\frac{2\pi
k(r-s)}{L}\right).\end{split}$ (6.7)

when $r,s=0,1,\cdots,L-1$ run over the whole chain, $\mathbf{Q}$ and
$\mathbf{P}$ are $L\times L$ matrix, satisfy
$\mathbf{Q}\mathbf{P}=\frac{1}{4}\mathbf{I}$. The limit $\omega\rightarrow 0$
is ill defined since zero mode in $\mathbf{Q}_{rs}$ diverges. Therefore to
test our analytic result, we must keep $\omega>0$ and let $\omega L\ll 1$ in
order to stay in the conformal regime.

(a)

Figure 3: Numerical data of $Z_{n}^{\text{norm}}$ as a function of $z=l_{0}/L$
for different $n$ and $\mu$ in the complex harmonic chain. The full lines are
the CFT predictions (cf. eq. (6.13)). Here we consider $\omega=10^{-8},L=2000$
and $\mu=\frac{\pi}{4},\frac{\pi}{2},\pi$ for $n=1$, $\mu=\frac{\pi}{16},\pi$
for $n=2$ and $\mu=\frac{\pi}{3}$ for $n=3$. As shown in the figure, the
agreement is very well.

The charged Rényi entropies of any subsystem $A$ consist of $l$ lattice sites
can be computed from these correlators. Firstly, we have to consider the
correlation matrices $\mathbf{Q}_{A}$ and $\mathbf{P}_{A}$ obtained by
restricting the the indices of the correlation matrices $\mathbf{Q}$ and
$\mathbf{P}$ to the sites belonging to $A$. Then find the eigenvalues of the
$l\times l$ matrix $\mathbf{Q}_{A}\mathbf{P}_{A}$ which are denoted by
$\\{\sigma_{1}^{2},\cdots,\sigma_{l}^{2}\\}$. We also introduce the basis
$\ket{\mathbf{n}}\equiv\otimes_{k=1}^{l}\ket{n_{k}}$, defined by products of
Fock states of the number operator in the subsystem $A$, the reduced density
matrix of $A$ can be written as

$\rho_{A}=\sum_{\mathbf{n}}\prod_{k=1}^{l}\frac{1}{\sigma_{k}+1/2}\left(\frac{\sigma_{k}-1/2}{\sigma_{k}+1/2}\right)^{n_{k}}\ket{\mathbf{n}}\bra{\mathbf{n}}.$
(6.8)

Finally, the charged moments for real harmonic chain are given by

$\begin{split}{\textrm{Tr}}\rho_{A}^{n}e^{\mathrm{i}\mu
N_{A}}&=\sum_{\mathbf{n}}\prod_{k=1}^{l}\left[\frac{1}{\sigma_{k}+1/2}\left(\frac{\sigma_{k}-1/2}{\sigma_{k}+1/2}\right)^{n_{k}}\right]^{n}e^{\mathrm{i}n_{k}\mu}\\\
&=\prod_{k=1}^{l}\left[\left(\sigma_{k}+\frac{1}{2}\right)^{n}-e^{\mathrm{i}\mu}\left(\sigma_{k}-\frac{1}{2}\right)^{n}\right]^{-1}.\end{split}$
(6.9)

Therefore the charged moments of RDM of an arbitrary subsystem $A$ for a
complex harmonic chain are

$\begin{split}&Z_{n}(\mu)=|{\textrm{Tr}}\rho_{A}^{n}e^{\mathrm{i}\mu
N_{A}}|^{2}\\\
&=\prod_{k=1}^{l}\left[\left(\sigma_{k}+\frac{1}{2}\right)^{n}-e^{\mathrm{i}\mu}\left(\sigma_{k}-\frac{1}{2}\right)^{n}\right]^{-1}\left[\left(\sigma_{k}+\frac{1}{2}\right)^{n}-e^{-\mathrm{i}\mu}\left(\sigma_{k}-\frac{1}{2}\right)^{n}\right]^{-1}.\end{split}$
(6.10)

This method also hold when $A$ is the union of two disjoint intervals
$A_{1},A_{2}$, which is the situation we consider in the CFT approach.

Let us denote the number of sites in $A_{1}$ and $A_{2}$ by $l_{1}$ and
$l_{2}$ respectively, and denote the number of sites in the separation between
$A_{1}$ and $A_{2}$ by $d_{1}$ and $d_{2}$ in the periodic chain. Then
$l=l_{1}+l_{2}$ and $L=l_{1}+l_{2}+d_{1}+d_{2}$ must be imposed. For
simplicity, we will consider the configuration where all the intervals have
the same length and all the separations have the same size, namely

$l_{1}=l_{2}\equiv l_{0},\quad d_{1}=d_{2}\equiv d.$ (6.11)

In this situation, we have only one free parameter $l_{0}/L$ to character the
configuration since $d/L=1/2-l_{0}/L$.

To compare our numerical data with the CFT results obtained in the previous
sections, we must map the CFT formulas on the complex plane to the cylinder of
circumference $L$. This can be easily done by replace each length $y$ (e.g.
$l_{0},d,l_{0}+d$,etc.) with the corresponding chord length
$\frac{L}{\pi}\sin\frac{\pi y}{L}$. For the cross ratio

$x=\frac{\sin^{2}(\pi
l_{0}/L))}{\sin^{2}(\pi(l_{0}+d)/L)}=\sin^{2}\left(\frac{\pi
l_{0}}{L}\right).$ (6.12)

For the CFT on a cylinder of circumference $L$, we have

$\begin{split}Z_{n}(\mu,z)=&c_{n,\mu}^{2}s_{n,\mu}\left(\frac{\pi^{2}}{L^{2}\sin^{2}(\pi
z)\sin[\pi(z+1/2)]\sin[\pi(1/2-z)]}\right)^{4\Delta_{n,\mu}}\\\
&\times\prod_{k=0}^{n-1}\left[F_{\frac{k}{n}+\frac{\mu}{2\pi n}}(\sin^{2}(\pi
z))F_{\frac{k}{n}+\frac{\mu}{2\pi n}}(1-\sin^{2}(\pi
z))\right]^{-1},\end{split}$ (6.13)

where $z\equiv\frac{|u_{1}-v_{1}|}{L}=\frac{|u_{2}-v_{2}|}{L}$. In order to
eliminate the unknown parameter $c_{n,\mu}$, it is more convenient to
normalize the results through a fixed configuration. We choose the following
one

$\text{fixed configuration}:\quad
l_{1}=l_{2}=d_{1}=d_{2}=\Big{[}\frac{L}{4}\Big{]}$ (6.14)

and denote the corresponding charged moments by $Z_{n}^{\text{fixed}}(\mu)$.

In the remaining part, we assume that the length of chain $L$ multiples of 4
for simplicity. Then we have $Z_{n}^{\text{fixed}}(\mu)=Z_{n}(\mu,1/4)$. The
numerical data and the CFT results shown in the figures are compared in terms
of their normalised version

$Z_{n}^{\text{norm}}(\mu,z)=\frac{Z_{n}(\mu,z)}{Z_{n}^{\text{fixed}}(\mu)}=[\csc(2\pi
z)]^{4\Delta_{n,\mu}}\prod_{k=0}^{n-1}\frac{[F_{\frac{k}{n}+\frac{\mu}{2\pi
n}}(1/2)]^{2}}{F_{\frac{k}{n}+\frac{\mu}{2\pi n}}(\sin^{2}(\pi
z))F_{\frac{k}{n}+\frac{\mu}{2\pi n}}(1-\sin^{2}(\pi z))}.$ (6.15)

In the limit $L\rightarrow\infty$ with $\omega L$ kept fixed, our numerical
results should converge to the CFT computations for
$Z^{\mathrm{norm}}_{n}(\mu,x)$ in the decompactification regime.

The numerical results for the function $Z_{1}^{\mathrm{norm}}(\mu)$ are
reported in Fig. 2 for different $L$ and different sub subsystem sizes
$l_{0}$. As shown in the figure, the agreement between numerical data and CFT
prediction is excellent for $\mu$ near $\pi$, while it gets worse for $\mu$
close to $0$ and $2\pi$. We found that the convergence to the CFT results is
not uniform and it’s much more slower for $\mu$ near $0$ and $2\pi$. This can
be explained by the fact that the limit $\mu\rightarrow 0$ or $\mu\rightarrow
2\pi$ does not commute with the critical limit $\omega\rightarrow 0$ [26, 29].

In Fig. 3, we report the numerical data for the quantities
$Z_{n}^{\mathrm{norm}}(\mu,z)$ for various $n$ and $\mu$. As the figure shows,
the numerical results and the CFT predictions match very well.

(a)

Figure 4: Numerical data of $R_{1}^{\text{norm}}$ as a function of $\mu$ for
different $z=l_{0}/L$ in the complex harmonic chain. The full lines are the
CFT predictions (cf. eq. (6.19)). Here we consider
$l_{0}=100,L=2000,\omega=10^{-8}$ and $l_{0}=2000,L=5000,\omega=10^{-9}$.
Again, the agreement is very well for $\mu$ near $\pi$, but it worsens as
$\mu$ gets closer to $0$ and $2\pi$.

###  6.2 Charged Rényi negativity

In the Fock basis $\\{\ket{\mathbf{n}}\\}$, $Q_{2}^{T_{2}}=Q_{2}$ and the
operator $\mathcal{Q}_{A}=Q_{1}-Q^{T_{2}}_{2}=Q_{1}-Q_{2}$ becomes exactly the
charge imbalance operator. For the complex harmonic chain, the charged Rényi
negativity factorises as

$R_{n}(\mu)={\textrm{Tr}}[(\rho^{T_{2}}_{A})^{n}e^{\mathrm{i}\mu(Q_{1}-Q_{2})}]={\textrm{Tr}}[(\rho^{T_{2}}_{A})^{n}e^{\mathrm{i}\mu\mathcal{Q}_{A}^{a}}]\times{\textrm{Tr}}[(\rho^{T_{2}}_{A})^{n}e^{-\mathrm{i}\mu\mathcal{Q}_{A}^{b}}].$
(6.16)

In bosonic system, the net effect of partial transposition with respect to
$A_{2}$ is changing the sign of the momenta corresponding to $A_{2}$. Thus the
momenta correlators in the partial transposed density matrix can be obtained
from $\mathbf{P}_{A}$ by simply change the sign of the momenta that in
$A_{2}$, i.e.

$\mathbf{P}_{A}^{T_{2}}=\mathbf{R}_{A_{2}}\mathbf{P}_{A}\mathbf{R}_{A_{2}},$
(6.17)

where $\mathbf{R}_{A_{2}}$ is the $l_{0}\times l_{0}$ diagonal matrix with
elements $(\mathbf{R}_{A_{2}})_{rs}=(-1)^{\delta_{r\in A_{2}}}\delta_{rs}$. We
denote the eigenvalues of $\mathbf{Q}_{A}\mathbf{P}_{A}^{T_{2}}$ by
$\\{\tau_{1}^{2},\tau_{2}^{2},\cdots,\tau_{l}^{2}\\}$. Then we have

${\textrm{Tr}}[(\rho^{T_{2}}_{A})^{n}e^{\mathrm{i}\mu\mathcal{Q}_{A}^{a}}]=\prod_{j=1}^{l_{0}}\left[\left(\tau_{j}+\frac{1}{2}\right)^{n}-e^{\mathrm{i}\mu}\left(\tau_{j}-\frac{1}{2}\right)^{n}\right]^{-1}\prod_{k=l_{0}+1}^{l}\left[\left(\tau_{k}+\frac{1}{2}\right)^{n}-e^{-\mathrm{i}\mu}\left(\tau_{k}-\frac{1}{2}\right)^{n}\right]^{-1}.$
(6.18)

As before, in the large $L$ limit while with $\omega L$ kept fixed, our
numerical results should converge to the CFT computations for
$R^{\mathrm{norm}}_{n}(\mu,x)$ in the decompactification regime.

For the non-compact free boson CFT on a cylinder of circumference $L$, we have

$R_{n}^{\text{norm}}(\mu,z)=\frac{R_{n}(\mu,z)}{R_{n}^{\text{fixed}}(\mu)}=[\cot(\pi
z)]^{4\Delta_{n,\mu}}\prod_{k=0}^{n-1}\frac{{\textrm{Re}}[F_{\frac{k}{n}+\frac{\mu}{2\pi
n}}(-1)\bar{F}_{\frac{k}{n}+\frac{\mu}{2\pi
n}}(-2)]}{{\textrm{Re}}[F_{\frac{k}{n}+\frac{\mu}{2\pi n}}(\frac{\sin^{2}(\pi
z)}{\sin^{2}(\pi z)-1})\bar{F}_{\frac{k}{n}+\frac{\mu}{2\pi
n}}(\frac{1}{1-\sin^{2}(\pi z)})]}$ (6.19)

and

$\mathcal{G}_{n}^{\text{norm}}(\mu,z)=[\cos^{2}(\pi
z)]^{-4\Delta_{n,\mu}}R_{n}^{\text{norm}}(\mu,z).$ (6.20)

While for the harmonic chain, we have

$\mathcal{G}_{n,\text{lat}}^{\text{norm}}(\mu,l_{0}/L)=\left[4\sin\left[\pi\left(\frac{1}{2}-\frac{l_{0}}{L}\right)\right]\sin\left[\pi\left(\frac{1}{2}+\frac{l_{0}}{L}\right)\right]\sin^{2}\left(\frac{\pi
l_{0}}{L}\right)\right]^{4\Delta_{n,\mu}}{\textrm{Tr}}[(\rho^{T_{2}}_{A})^{n}e^{\mathrm{i}\mu\mathcal{Q}_{A}}].$
(6.21)

The numerical results for the function $R_{1}^{\mathrm{norm}}(\mu)$ are
reported in Fig. 4 for different $L$ and different sub subsystem sizes
$l_{0}$. As shown in the figure, the agreement between numerical data and CFT
prediction is pretty well for $\mu$ near $\pi$, while it gets worse for $\mu$
close to $0$ and $2\pi$ as before.

In Fig. 5, we report the numerical data for the quantities
$\mathcal{G}_{n}^{\mathrm{norm}}(\mu,x)$ for various $n$ and $\mu$. In this
case, as shown in the figure, the numerical results and the CFT predictions
matched excellently.

(a)

Figure 5: Numerical data of $\mathcal{G}_{n}^{\text{norm}}$ as a function of
$z=l_{0}/L$ for different $n$ and $\mu$ in the complex harmonic chain(cf. eq.
(6.21)). The full lines are the CFT predictions (cf. eq. (6.20)). Here we
consider $L=2000,\omega=10^{-9}$ for $n=2,3,4$ with $\mu=\pi$ and $\mu=\pi/2$
respectively. The agreement is very well for small $n$, but it worsens as $n$
goes larger.

##  7 Conclusion and discussion

In this paper, we consider the charged moments of reduced density matrices of
two disjoint regions in 1+1 dimensional compact free boson CFT. We discuss two
important regimes of the scaling function, one is the small $x$ regime, which
will be useful to study the conformal block of the fluxed twist fields. The
other region is the decompactification regime, in this regime, the bosonic
fields become non-compact, and will be convenient for numerical checks. We
also compute the charged Rényi negativity. We test our analytic results
against exact numerical computations in the complex harmonic chain, finding
perfect agreement.

We must mention we just obtained the charged moments and the charged Rényi
negativity, it would be very interesting to apply Fourier transformation and
analytic continue our results to obtain the corresponding symmetry resolved
entanglement entropy and charge imbalance resolved entanglement negativity.

It would be also interesting the further check our analytic formula by
considering the conformal block expansion of the four-point correlation
function of fluxed twist fields and applying the Zamolodchikov recursion
formula [49, 50] for each conformal block. In this method, each term in the
conformal block expansion can be analytically continued and this approach can
provide a good approximation for the symmetry resolved entanglement entropy
and charge imbalance resolved entanglement negativity.

In this paper, we focus on the charged moments and charged Rényi negativity of
complex boson with $U(1)$ symmetry. In 1+1 dimensional, for fermions with
$U(1)$ symmetry $\psi_{k}\rightarrow e^{\mathrm{i}\mu}\psi_{k}$, the
corresponding fluxed twist field admits a bosonisation formula. We can write
the complex fermionic field as $\psi_{k}\sim e^{\mathrm{i}\varphi_{k}}$, then
the fermion twist field is given by the vertex operator
$\mathcal{T}_{n,k,\mu}=e^{\mathrm{i}(\frac{k}{n}+\frac{\mu}{2\pi
n})\varphi_{k}}$, which have a different scaling dimension
$\Delta_{n,k,\mu}=\left(\frac{k}{n}+\frac{\mu}{2\pi n}\right)^{2}$. This can
be understood, since via bosonisation of $U(1)$ complex fermions, the
corresponding bosons transform by translation, and should instead satisfy
boundary condition $\varphi_{k}(e^{2\pi\mathrm{i}}z)=\varphi_{k}(z)+\mu$.

## Acknowledgments

This work was supported by the National Natural Science Foundation of China,
Grant No. 12005081.

##  Appendix A Correlation function of fluxed twist fields

In this section, we give some detail of the calculation of four point
correlation functions of the $U(1)$ twist fields $\mathcal{T}_{n,k,\mu}$. We
will call it generally $\mathcal{T}_{a}$, where $a=\frac{k}{n}+\frac{\mu}{2\pi
n}$ with $0<a<1$. This problem is already encountered in the theory of
orbifold CFT [48]. Here we can just borrow the results with slightly change.
However, for the reader’s convenience we briefly review this method here.

Let us consider a complex scalar field $X(z,\bar{z})$ defined on a Riemann
sphere. The action is given by

$\mathcal{S}[X,\bar{X}]=\frac{1}{4\pi}\int d^{2}z(\partial
X\bar{\partial}\bar{X}+\bar{\partial}X\partial\bar{X})$ (A.1)

If the field $\mathcal{T}_{n,\mu,k}$ inserted at the origin, then we have

$X(e^{2\pi\mathrm{i}}z,e^{-2\pi\mathrm{i}}\bar{z})=e^{2\pi\mathrm{i}a}X(z,\bar{z})+\lambda,$
(A.2)

where $\lambda\in\Lambda$. Let us now split the field $X$ into a classical
piece $X_{\mathrm{cl}}$ and a quantum fluctuation $X_{\mathrm{qu}}$ as
described in section 4. The classical field should behave the same as the full
field

$X_{\mathrm{cl}}(e^{2\pi\mathrm{i}}z,e^{-2\pi\mathrm{i}}\bar{z})=e^{2\pi\mathrm{i}a}X_{\mathrm{cl}}(z,\bar{z})+\lambda,$
(A.3)

which implies

$X_{\mathrm{qu}}(e^{2\pi\mathrm{i}}z,e^{-2\pi\mathrm{i}}\bar{z})=e^{2\pi\mathrm{i}a}X_{\mathrm{qu}}(z,\bar{z}),$
(A.4)

In order to determine the quantum Green functions in the presence of the twist
fields and classical solutions, one needs some global information that can be
obtained from the monodromy conditions for transporting $X_{\mathrm{qu}}$ and
$X_{\mathrm{cl}}$ around collections of twist fields that have net twist zero.
Such paths $\mathcal{C}_{i}$ are called closed loops. For any closed loop
$\mathcal{C}_{i}$, we must have

$\Delta_{\mathcal{C}_{i}}X_{\mathrm{qu}}=\oint_{\mathcal{C}_{i}}dz~{}\partial
X_{\mathrm{qu}}+\oint_{\mathcal{C}_{i}}d\bar{z}~{}\bar{\partial}X_{\mathrm{qu}}=0$
(A.5)

and

$\Delta_{\mathcal{C}_{i}}X_{\mathrm{cl}}=\oint_{\mathcal{C}_{i}}dz~{}\partial
X_{\mathrm{cl}}+\oint_{\mathcal{C}_{i}}d\bar{z}~{}\bar{\partial}X_{\mathrm{cl}}=\lambda,$
(A.6)

where now for our case $\lambda\in(1-e^{2\pi\mathrm{i}a})\Lambda$.

Now let us consider the following four-point function of the twist field
$\mathcal{T}_{a}$

$Z=\langle\mathcal{T}_{a}(z_{1},\bar{z}_{1})\tilde{\mathcal{T}}_{a}(z_{2},\bar{z}_{2})\mathcal{T}_{a}(z_{3},\bar{z}_{3})\tilde{\mathcal{T}}_{a}(z_{4},\bar{z}_{4})\rangle=\int[dX][d\bar{X}]e^{-\mathcal{S}[X,\bar{X}]}=Z_{\mathrm{qu}}\sum_{X_{\mathrm{cl}}}e^{-\mathcal{S}[X_{\mathrm{cl}},\bar{X}_{\mathrm{cl}}]}.$
(A.7)

We can first compute the Green function in the presence of four twist fields

$g(z,w;z_{i})=\frac{\langle-\frac{1}{2}\partial_{z}X\partial_{w}\bar{X}\mathcal{T}_{a}(z_{1})\tilde{\mathcal{T}}_{a}(z_{2})\mathcal{T}_{a}(z_{3})\tilde{\mathcal{T}}_{a}(z_{4})\rangle}{\langle\mathcal{T}_{a}(z_{1})\tilde{\mathcal{T}}_{a}(z_{2})\mathcal{T}_{a}(z_{3})\tilde{\mathcal{T}}_{a}(z_{4})\rangle}.$
(A.8)

This Green function should obey the asymptotic conditions:
$g(z,w;z_{i})\sim(z-w)^{-2}$ as $z\rightarrow w$ and for $z\rightarrow z_{i}$
we have $g(z,w;z_{i})\sim(z-z_{j})^{-a}$ when $j=1,3$ and
$g(z,w;z_{i})\sim(z-z_{j})^{-(1-a)}$ when $j=2,4$. The holomorphic fields for
the cut $z$-plane in this case are:

$\begin{split}\partial
X(z)\equiv\omega_{a}(z)=[(z-z_{1})(z-z_{3})]^{-a}[(z-z_{2})(z-z_{4})]^{-(1-a)},\\\
\partial\bar{X}(z)\equiv\omega_{1-a}(z)=[(z-z_{1})(z-z_{3})]^{-(1-a)}[(z-z_{2})(z-z_{4})]^{-a}.\end{split}$
(A.9)

Then the unique form of the Green function with the desired properties reads

$\begin{split}g(z,w;z_{i})=\omega_{a}(z)\omega_{1-a}&(w)\Big{[}a\frac{(z-z_{1})(z-z_{3})(w-z_{2})(w-z_{4})}{(z-w)^{2}}\\\
&+(1-a)\frac{(z-z_{2})(z-z_{4})(w-z_{1})(w-z_{3})}{(z-w)^{2}}+A(z_{i},\bar{z}_{i})\Big{]}.\end{split}$
(A.10)

We can find the expectation value of the stress tensor
$T=-\frac{1}{2}:\partial X\partial\bar{X}:$ in the presence of twist fields by
taking $w\rightarrow z$ of $g(z,w;z_{i})$ and subtracting the diverging
pieces. We have

$\begin{split}&\frac{\langle
T(z)\mathcal{T}_{a}(z_{1})\tilde{\mathcal{T}}_{a}(z_{2})\mathcal{T}_{a}(z_{3})\tilde{\mathcal{T}}_{a}(z_{4})\rangle}{\langle\mathcal{T}_{a}(z_{1})\tilde{\mathcal{T}}_{a}(z_{2})\mathcal{T}_{a}(z_{3})\tilde{\mathcal{T}}_{a}(z_{4})\rangle}=\lim_{w\rightarrow
z}(g(z,w;z_{i})-(z-w)^{-2})\\\
&=\frac{1}{2}a(1-a)\left(\frac{1}{z-z_{1}}+\frac{1}{z-z_{3}}-\frac{1}{z-z_{2}}-\frac{1}{z-z_{4}}\right)^{2}+\frac{A}{(z-z_{1})(z-z_{2})(z-z_{3})(z-z_{4})}.\end{split}$
(A.11)

From the operator product expansion

$T(z)\tilde{\mathcal{T}}_{a}(z_{2})\sim\frac{h_{a}\tilde{\mathcal{T}}_{a}(z_{2})}{(z-z_{2})^{2}}+\frac{\partial_{z_{2}}\tilde{\mathcal{T}}_{a}(z_{2})}{z-z_{2}},$
(A.12)

it’s easy to know that the conformal dimensions of the twist fields
$\tilde{\mathcal{T}}_{a}$ (and also of $\mathcal{T}_{a}$) are

$h_{a}=\bar{h}_{a}=\frac{1}{2}a(1-a)=\frac{1}{2}\left(\frac{k}{n}+\frac{\mu}{2\pi
n}\right)\left(1-\frac{k}{n}-\frac{\mu}{2\pi n}\right)$ (A.13)

and the scaling dimension is

$\Delta_{a}=h_{a}+\bar{h}_{a}=\left(\frac{k}{n}+\frac{\mu}{2\pi
n}\right)\left(1-\frac{k}{n}-\frac{\mu}{2\pi n}\right).$ (A.14)

The partition function satisfies the following differential equation
$(z_{ij}\equiv z_{i}-z_{j})$

$\partial_{z_{2}}\ln
Z_{\mathrm{qu}}(z_{i},\bar{z}_{i})=-2h_{a}\left(\frac{1}{z_{21}}+\frac{1}{z_{23}}-\frac{1}{z_{24}}\right)+\frac{A(z_{i},\bar{z}_{i})}{z_{21}z_{23}z_{24}}.$
(A.15)

It’s convenient to use $SL(2,\mathbb{C})$ symmetry
$z\rightarrow\frac{(z_{1}-z)(z_{3}-z_{4})}{(z_{1}-z_{3})(z-z_{4})}$ to fix the
locations of three of the four twist operators to
$z_{1}=0,z_{2}=x\equiv\frac{z_{12}z_{34}}{z_{13}z_{24}},z_{3}=1,z_{4}\rightarrow\infty$.

$\partial_{x}\ln
Z_{\mathrm{qu}}(x,\bar{x})=-2h_{a}\left(\frac{1}{x}-\frac{1}{1-x}\right)-\frac{A(x,\bar{x})}{x(1-x)}.$
(A.16)

We can use the global monodromy conditions A.5 to determine $A(x,\bar{x})$.
Before doing this, we must introduce the auxiliary correlation function

$h(\bar{z},w;z_{i})=\frac{\langle-\frac{1}{2}\partial_{\bar{z}}X\partial_{w}\bar{X}\mathcal{T}_{a}(z_{1})\tilde{\mathcal{T}}_{a}(z_{2})\mathcal{T}_{a}(z_{3})\tilde{\mathcal{T}}_{a}(z_{4})\rangle}{\langle\mathcal{T}_{a}(z_{1})\tilde{\mathcal{T}}_{a}(z_{2})\mathcal{T}_{a}(z_{3})\tilde{\mathcal{T}}_{a}(z_{4})\rangle}=B(z_{i},\bar{z}_{i})\bar{\omega}_{1-a}(\bar{z})\omega_{1-a}(w),$
(A.17)

which is determined up to the constant factor $B(z_{i},\bar{z}_{i})$ in the
same way as $g$ was. Then we have

$\oint_{\mathcal{C}_{i}}dz~{}g(z,w)+\oint_{\mathcal{C}_{i}}d\bar{z}~{}h(\bar{z},w)=0.$
(A.18)

Dividing by $\omega_{1-a}(w)$ and letting $w\rightarrow\infty$

$A\oint_{\mathcal{C}_{i}}dz~{}\omega_{a}+B\oint_{\mathcal{C}_{i}}d\bar{z}~{}\bar{\omega}_{1-a}=-(1-a)\oint_{\mathcal{C}_{i}}dz~{}(z-x)\omega_{a}.$
(A.19)

All the contour integrals appeared in the above equation can be expressed in
terms of the hypergeometric function $F(x)\equiv{{}_{2}F_{1}}(a,1-a,1,x)$ and
its derivative

$\begin{split}\oint_{\mathcal{C}_{1}}dz~{}\omega_{a}=2\pi\mathrm{i}e^{-\pi\mathrm{i}a}F(x),\qquad\oint_{\mathcal{C}_{2}}dz~{}\omega_{a}=2\pi\mathrm{i}F(x),\\\
\oint_{\mathcal{C}_{1}}d\bar{z}~{}\bar{\omega}_{1-a}=2\pi\mathrm{i}e^{-\pi\mathrm{i}a}\bar{F}(\bar{x}),\qquad\oint_{\mathcal{C}_{2}}d\bar{z}~{}\bar{\omega}_{1-a}=-2\pi\mathrm{i}\bar{F}(1-\bar{x}),\\\
-(1-a)\oint_{\mathcal{C}_{1}}dz~{}(z-x)\omega_{a}=2\pi\mathrm{i}e^{-\pi\mathrm{i}a}x(1-x)\frac{dF(x)}{dx},\\\
-(1-a)\oint_{\mathcal{C}_{2}}dz~{}(z-x)\omega_{a}=2\pi\mathrm{i}x(1-x)\frac{dF(x)}{dx}.\end{split}$
(A.20)

Solving eq. (A.19) for $A$

$A(x,\bar{x})=x(1-x)\partial_{x}\ln I(x,\bar{x}),\quad
I(x,\bar{x})=F(x)\bar{F}(1-\bar{x})+\bar{F}(\bar{x})F(1-x).$ (A.21)

Then

$Z_{\mathrm{qu}}(x,\bar{x})=\frac{\mathrm{const}}{|x(1-x)|^{2\Delta_{a}}}\frac{1}{I(x,\bar{x})}.$
(A.22)

Then to compute the classical action, we need to find properly normalized
classical solutions $X_{\mathrm{cl}}(z,\bar{z})$ and
$X_{\mathrm{cl}}(z,\bar{z})$. From the equations of motion, it’s easily sees
that $\partial_{z}X_{\mathrm{cl}}$ and $\partial_{z}\bar{X}_{\mathrm{cl}}$ are
holomorphic while $\partial_{\bar{z}}X_{\mathrm{cl}}$ and
$\partial_{\bar{z}}\bar{X}_{\mathrm{cl}}$ are antiholomorphic

$\begin{split}&\partial
X_{\mathrm{cl}}(z)=\mathfrak{a}~{}\omega_{a}(z),\quad\bar{\partial}X_{\mathrm{cl}}(\bar{z})=\mathfrak{b}~{}\bar{\omega}_{1-a}(\bar{z}),\\\
&\partial\bar{X}_{\mathrm{cl}}(z)=\tilde{\mathfrak{a}}~{}\omega_{a}(z),\quad\bar{\partial}\bar{X}_{\mathrm{cl}}(\bar{z})=\tilde{\mathfrak{b}}~{}\bar{\omega}_{a}(\bar{z}).\end{split}$
(A.23)

We can first construct two classical solutions which have simple global
monodromy:

$\Delta_{\mathcal{C}_{i}}X_{\mathrm{cl},j}=\Delta_{\mathcal{C}_{i}}\bar{X}_{\mathrm{cl},j}=2\pi\delta_{ij},\quad
i,j=1,2$ (A.24)

Let $\mathfrak{a}_{i},\mathfrak{b}_{i},\tilde{\mathfrak{a}}_{i}$ and
$\tilde{\mathfrak{b}}_{i}$ be the coefficients for $X_{\mathrm{cl},i}$ and
$\bar{X}_{\mathrm{cl},i}$ in eq. (A.24), then

$\begin{split}\mathfrak{a}_{1}=-e^{2\pi\mathrm{i}a}\tilde{\mathfrak{a}}_{1}=-\mathrm{i}e^{\pi\mathrm{i}a}\frac{\bar{F}(1-\bar{x})}{I(x,\bar{x})},\quad\mathfrak{a}_{2}=\tilde{\mathfrak{a}}_{2}=-\mathrm{i}\frac{\bar{F}(\bar{x})}{I(x,\bar{x})},\\\
\mathfrak{b}_{1}=-e^{2\pi\mathrm{i}a}\tilde{\mathfrak{b}}_{1}=-\mathrm{i}e^{\pi\mathrm{i}a}\frac{F(1-x)}{I(x,\bar{x})},\quad\mathfrak{b}_{2}=\tilde{\mathfrak{b}}_{2}=\mathrm{i}\frac{F(x)}{I(x,\bar{x})}.\end{split}$
(A.25)

Therefore the coefficients in eq. (A.23) reads

$\mathfrak{a}=\mathfrak{a}_{1}\lambda_{1}+\mathfrak{a}_{2}\lambda_{2},\quad\mathfrak{b}=\mathfrak{b}_{1}\lambda_{1}+\mathfrak{b}_{2}\lambda_{2},\quad\tilde{\mathfrak{a}}=\tilde{\mathfrak{a}}_{1}\bar{\lambda}_{1}+\tilde{\mathfrak{a}}_{2}\bar{\lambda}_{2},\quad\tilde{\mathfrak{b}}=\tilde{\mathfrak{b}}_{1}\bar{\lambda}_{1}+\tilde{\mathfrak{b}}_{2}\bar{\lambda}_{2},$
(A.26)

with $\lambda_{1,2}\in(1-e^{2\mathrm{i}\pi a})\Lambda$. The classical action
is then obtained straightforwardly

$S_{a}^{\mathrm{cl}}(\lambda_{1},\lambda_{2})=\pi\sin(\pi
a)\left[\frac{|\tau_{a}|^{2}}{\beta_{a}}|\xi_{1}|^{2}+\frac{\alpha_{a}}{\beta_{a}}(\xi_{1}\bar{\xi}_{2}\bar{\gamma}+\bar{\xi}_{1}\xi_{2}\gamma)+\frac{|\xi_{2}|^{2}}{\beta_{a}}\right].$
(A.27)

where $\gamma=-\mathrm{i}e^{-\mathrm{i}\pi a}$ and $\xi_{j}(j=1,2)$ are
generic vectors of the target space lattice $\Lambda$.

##  Appendix B Normalisation of $\mathcal{F}_{n}(\mu,x)$

The Siegel theta function $\Theta(\mathbf{0}|M)$ for any symmetric $n\times n$
matrix $M$ with positive imaginary part can be rewritten as

$\begin{split}\Theta(\mathbf{0}|M)=\sum_{\mathbf{m}\in\mathbb{Z}^{n}}e^{\mathrm{i}\pi\mathbf{m}^{\mathrm{t}}\cdot
M\cdot\mathbf{m}}=\int
d^{n}\mathbf{x}\sum_{\mathbf{m}\in\mathbb{Z}^{n}}\delta^{n}(\mathbf{x}-\mathbf{m})e^{\mathrm{i}\pi\mathbf{x}^{\mathrm{t}}\cdot
M\cdot\mathbf{x}}.\end{split}$ (B.1)

Using Poisson resummation formula

$\sum_{\mathbf{m}\in\mathbb{Z}^{n}}\delta^{n}(\mathbf{x}-\mathbf{m})=\sum_{\mathbf{p}\in\mathbb{Z}^{n}}e^{2\pi\mathrm{i}\mathbf{p}\cdot\mathbf{x}},$
(B.2)

we have

$\begin{split}\Theta(\mathbf{0}|M)=\sum_{\mathbf{p}\in\mathbb{Z}^{n}}\int
d^{n}\mathbf{x}~{}e^{\mathrm{i}\pi\mathbf{x}^{\mathrm{t}}\cdot
M\cdot\mathbf{x}+2\pi\mathrm{i}\mathbf{p}\cdot\mathbf{x}}=\sum_{\mathbf{p}\in\mathbb{Z}^{n}}\frac{1}{\sqrt{\det(-\mathrm{i}M)}}e^{-\mathrm{i}\pi\mathbf{p}^{\mathrm{t}}\cdot
M^{-1}\cdot\mathbf{p}}=\frac{\Theta(\mathbf{0}|-M^{-1})}{\sqrt{\det(-\mathrm{i}M)}}.\end{split}$
(B.3)

Now applying this formula to our problem. Firstly, we have

$\eta\tilde{\Omega}=-A(\Omega/\eta)^{-1}A,$ (B.4)

where $A=U^{\dagger}HU$ with
$H={\textrm{diag}}(\cdots,(-\omega_{p}\tilde{\omega}_{p})^{1/2},\cdots)$. Then
by setting $M=\eta\tilde{\Omega}$ in eq. (B.3), we find

$\Theta(\mathbf{0}|\eta\tilde{\Omega})=\frac{\sqrt{\det(\Omega/(\mathrm{i}\eta))}}{\det
A}\Theta(\mathbf{0}|A^{-1}(\Omega/\eta)A^{-1}).$ (B.5)

We also find

$\frac{\det(\Omega/(\mathrm{i}\eta))}{(\det
A)^{2}}=\prod_{k=1}^{n}\frac{-\mathrm{i}\omega_{k}}{(-\eta\omega_{k}\tilde{\omega}_{k})}=\prod_{k=1}^{n}(-\mathrm{i}\eta\tilde{\omega}_{k})^{-1}$
(B.6)

and

$[A(\Omega/\eta)^{-1}A]_{rs}=\frac{\eta}{n}\sum_{p=1}^{n}\tilde{\omega}_{p}^{-1}\cos[\frac{2\pi
p}{n}(r-s)]$ (B.7)

In the limit $x\rightarrow 0$, we have $F_{a}(x)\rightarrow 1$,
$\beta_{a}(x)\rightarrow\infty$, thus
$\Theta(\mathbf{0}|\eta\Omega)\rightarrow 1$. Then

$\begin{split}&\lim_{x\rightarrow
0}\frac{[\Theta(\mathbf{0}|\eta\Omega)\Theta(\mathbf{0}|\eta\tilde{\Omega})]^{2}}{\prod_{k=0}^{n-1}\beta_{a}[F_{a}(x)]^{2}}=\lim_{x\rightarrow
0}\frac{\det(\Omega/(\mathrm{i}\eta))}{(\det
A)^{2}\prod_{k=0}^{n-1}\beta_{\frac{k}{n}+\frac{\mu}{2\pi
n}}}=\eta^{-n}\lim_{x\rightarrow
0}\prod_{p=1}^{n}(-\mathrm{i}\tilde{\omega}_{p}\beta_{\frac{p-1}{n}+\frac{\mu}{2\pi
n}})^{-1}\\\
&=\eta^{-n}\Big{[}2\sin\left(\frac{\mu}{2n}\right)\Big{]}^{-1}\prod_{p=1}^{n-1}\left[\sin\Big{(}\frac{\pi
p}{n}+\frac{\mu}{2n}\Big{)}+\sin\Big{(}\frac{\pi
p}{n}-\frac{\mu}{2n}\Big{)}\right]^{-1}\\\
&=\frac{1}{2n\eta^{n}}\Big{[}\sin\left(\frac{\mu}{2n}\right)\Big{]}^{-1}\Big{[}\cos\Big{(}\frac{\mu}{2n}\Big{)}\Big{]}^{1-n},\end{split}$
(B.8)

where we have used the identity

$\prod_{p=1}^{n-1}\left[\sin\Big{(}\frac{\pi
p}{n}+\frac{\mu}{2n}\Big{)}+\sin\Big{(}\frac{\pi
p}{n}-\frac{\mu}{2n}\Big{)}\right]=n\Big{[}\cos\Big{(}\frac{\mu}{2n}\Big{)}\Big{]}^{n-1}.$
(B.9)

Thus, the normalisation constant $s_{n,\mu}$ of $\mathcal{F}_{n}(\mu,x)$ is
obtained.

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