††institutetext:  aCentre for High Energy Physics,
Indian Institute of Science,
C. V. Raman Avenue, Bangalore 560012, India.
bDepartment of Theoretical Physics,
Tata Institute of Fundamental Research,
Homi Bhabha Road, Mumbai 400005, India.

# Precision tests of bulk entanglement entropy

Barsha G. Chowdhurya, Justin R. Davida, Semanti Duttaa, Jyotirmoy Mukherjeeb
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We consider linear superpositions of single particle excitations in a scalar
field theory on $AdS_{3}$ and evaluate their contribution to the bulk
entanglement entropy across the Ryu-Takayanagi surface. We compare the
entanglement entropy of these excitations obtained using the Faulkner-
Lewkowycz-Maldacena formula to the entanglement entropy of linear
superposition of global descendants of a conformal primary in a large $c$ CFT
obtained using the replica trick. We show that the closed from expressions for
the entanglement entropy in the small interval expansion both in gravity and
the CFT precisely agree. The agreement serves as a non-trivial check of the
FLM formula for the quantum corrections to holographic entropy which also
involves a contribution from the back reacted minimal area. Our checks
includes an example in which the state is time dependent and spatially in-
homogenous as well another example involving a coherent state with a Bañados
geometry as its holographic dual.

##  1 Introduction

Entanglement has played a key role in recent developments in black hole
physics, emergence of space time in holography and quantum gravity. This has
been possible due to the discovery of the Ryu-Takayanagi formula for
entanglement entropy of a CFT which admits a holographic dual. The formula
expresses the entanglement entropy across an entangling surface in the CFT in
terms of the area of the minimal surface in the bulk Ryu:2006bv . The Ryu-
Takayanagi formula is classical and valid in the leading of the gravitational
coupling $G_{N}$. The one loop quantum corrections to this formula have been
proposed by Faulkner, Lewkowycz and Maldacena, it states Faulkner:2013ana

$\displaystyle S_{\rm EE}^{\rm CFT}(A)=\frac{{\rm
Area}(\gamma_{A})}{4G_{N}}+S_{\rm bulk}^{\rm EE}(\Sigma_{A}).$ (1)

Here $A$ is the subregion of interest in the boundary CFT, $\gamma_{A}$ is the
minimal Ryu-Takayanagi surface and $\Sigma_{A}$ is the region which extends
between $\gamma_{A}$ and $A$. $S_{\rm bulk}^{\rm EE}$ is the entanglement of
all fields present in the bulk effective field theory. The geometry for a 2d
CFT is shown in figure 5. The FLM proposal and its generalizations
Engelhardt:2014gca ; Dong:2017xht ; Jafferis:2015del ; Dong:2016eik ;
Faulkner:2017vdd  have played a fundamental role in our recent understanding
of quantum gravity. The proposal is very similar to the generalised entropy
for a black hole which is defined by the same equation as in (1), but with $A$
replaced by the horizon of the black hole and $S_{\rm bulk}$ is replaced by
the Von-Neumann entropy of all the fields outside the black hole horizon
Hawking:1971bv ; Bekenstein:1973ur ; Hawking:1975vcx . This similarity and the
fact that for a general time dependent situation we need to extremize over all
possible surfaces Hubeny:2007xt  led to the notion of the quantum extremal
surface and extended the definition of the generalized entropy to any sub-
region in quantum gravity Engelhardt:2014gca .

Inspite of the impact of the FLM formula, it has been rarely tested on states
which break symmetries. One route to obtain one loop corrections to the Ryu-
Takayanagi formula which several works take is to evaluate the quantum
corrections by a path integral approach rather than a direct computation of
the bulk entanglement entropies Barrella:2013wja ; Chen:2013kpa ;
Datta:2013hba ; Headrick:2015gba . An early check of the FLM formula involved
using it to evaluate the shift in the central charge of 2 CFT’s related by a
renormalization group flow trigged by a double trace deformation
Miyagawa:2015sql ; Sugishita:2016iel . A more direct check of the FLM formula
involves evaluating the single interval entanglement entropy of a single
particle excitation of a minimally coupled massive scalar in $AdS_{3}$
Belin:2018juv . Here the state considered was the lowest lying state which is
dual to a primary in $CFT_{2}$. The entanglement entropies both on the LHS and
the RHS of the equation (1) were evaluated in the short distance approximation
and were shown to agree to the leading and sub-leading orders. A follow up of
this test in which the single particle excitation was boosted to create a time
dependent state was done in Agon:2020fqs .

In this paper we generalise the check initiated in Belin:2018juv  to other
single particle excitations of the massive scalar in global $AdS_{3}$. In fact
all single particle excitations of the minimally coupled scalar are dual to
global or $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ descendants of the primary
in $CFT_{2}$. Schematically, let us write this map as

$\displaystyle|\psi_{m,n}\rangle_{\rm bulk}=a^{\dagger}_{m,n}|0\rangle_{\rm
bulk}$
$\displaystyle\qquad\longleftrightarrow\qquad(L_{-1})^{n_{1}}(\bar{L}_{-1})^{n_{2}}|h,h\rangle_{CFT},$
(2) $\displaystyle{\rm with}$ $\displaystyle 2n+|m|=n_{1}+n_{2},\qquad
m=n_{1}-n_{2},$

where $a_{m,n}^{\dagger}$ refers to the creation operator of the modes of a
minimally coupled bulk scalar of mass

$\displaystyle M^{2}=4h(h-1).$ (3)

$L_{-1},\bar{L}_{-1}$ are the raising operators of the left and right
$SL(2,\mathbb{R})$’s of the CFT.

In 2d CFT a detailed study of the entanglement properties of descendants was
initiated in Chowdhury:2021qja  111See Palmai:2014jqa ; Caputa:2015tua ;
Taddia:2016dbm ; Brehm:2020zri  for earlier work on the 2nd Rényi entropy of
descendants. . Though the descendants are related to the primary by the
$SL(2,\mathbb{R})$ symmetry of the theory, their single interval entanglement
differs from the primary, it depends on the weight of the primary, the level
of the descendant, and the 3-point functions in the CFT. For CFT’s with large
central charge $c$ and for primaries with weight $h<<c$, one result obtained
in Chowdhury:2021qja  is the following. Consider descendants of the
holomorphic primary of weight $(h,0)$ defined by $L_{-1}^{l}|h,0\rangle$, the
short distance expansion of the single interval entanglement entropy is given
by

$\displaystyle S\big{(}\rho_{L_{-1}^{l}|h,0\rangle}\big{)}=2(h+l)(1-\pi
x\cot\pi
x)-\frac{\Gamma(\frac{3}{2})\Gamma(2h+1)}{\Gamma(2h+\frac{3}{2})}\times\left(\frac{\Gamma(2h+l)}{\Gamma(2h)l!}\right)^{2}(\pi
x)^{4h}+\cdots,$

where $2\pi x$ is the size of the interval on the circumference of the spatial
cylinder of the CFT. These results were obtained using the replica trick in
the CFT. In the above expansion, we have slightly abused the notion of the
short distance expansion. For the leading term which admits an analytical
expansion in $x$ we have kept all the orders, while we have kept only the
leading non-analytical term $(\pi x)^{4h}$. Note that the information of the
descendant appears in the change of weight of the leading order term
$h\rightarrow h+l$ while the sub-leading, non-analytical term acquires a
factor depending on the level of the descendant.

In this paper, with the aim of evaluating both sides of the equation in (1)
for arbitrary single particle excitations we first generalize the CFT analysis
so that we have the short distance expansion of the single interval
entanglement of an arbitrary linear combination of descendants. An example of
the result is the following, consider the state

$\displaystyle|\hat{\Psi}\rangle=\sum_{l=0}^{\infty}c_{l}L_{-1}^{l}|h,h\rangle.$
(5)

The short distance entanglement is given by

$\displaystyle S(\rho_{|\hat{\Psi}\rangle})$ $\displaystyle=$
$\displaystyle\sum_{l,l^{\prime}=0}^{\infty}\frac{c_{l}c_{l^{\prime}}^{*}\hat{g}_{ll^{\prime}}(x)}{\langle\hat{\Psi}|\hat{\Psi}\rangle}+2h(1-\pi
x\cot\pi x)$
$\displaystyle\qquad-\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}\frac{(\pi
x)^{8h}}{\langle\hat{\Psi}|\hat{\Psi}\rangle^{2}}\times\left(\sum_{l,l^{\prime}=0}^{\infty}c_{l}c_{l^{\prime}}^{*}D_{ll^{\prime}}(h,2h)\right)^{2}+\cdots,$

where the coefficients $g_{ll^{\prime}}(x)$ which depend on the interval can
be read out from (29) and

$\displaystyle
D_{ll^{\prime}}(h,2h)=\frac{\Gamma(2h+l)\Gamma(2h+l^{\prime})}{(\Gamma(2h))^{2}}.$
(7)

and $\langle\hat{\Psi}|\hat{\Psi}\rangle$ is the norm of the state defined in
(5). Observe that for an arbitrary linear combination, the simple dependence
of the entanglement entropy of the excited state $(1-\pi x\cot\pi x)$ is
modified non-trivially to include the coefficients $\hat{g}_{ll^{\prime}}(x)$.
The non-analytical dependence $(\pi x)^{8h}$ acquires a pre-factor depending
on the linear combination and the weight of the primary.

With these large $c$ CFT results at hand we are in a position to perform a
precision test of the FLM formula (1). To evaluate the RHS, we generalize the
methods of Belin:2018juv  and construct single particle wave functions of a
few low lying descendants. We then evaluate their stress tensor and construct
the back-reacted metric. This turns out to be a non-trivial exercise for the
linear combination of excitations because these break rotational symmetry or
spatial homogeneity in the boundary CFT as well as time translational symmetry
of global $AdS_{3}$. Using the back reacted metric we can obtain the
corrections to the Ryu-Takayanagii minimal area contribution in (1), which
also requires care for states which break spatial homogeneity. We then proceed
to evaluate the bulk entanglement entropy for these excitations. This is done
by mapping the reduced density matrix of the single excitation in global $AdS$
to a state in the Rindler BTZ. To complete the evaluation of the entanglement
entropy, we need the Bogoliubov coefficients relating the states
$|\psi_{m,n}\rangle$ to states in Rindler BTZ. We notice that the Bogoliubov
coefficients simplify in the short interval limit and obey a nice scaling
property as shown in table 2. On summing up the corrected minimal area term
and the contribution to $S_{\rm bulk}$ we obtain the RHS of (1).

In all cases studied in this paper we demonstrate precise agreement with the
CFT results. This agreement depends crucially on both the corrected minimal
area and the scaling property of the Bogoliubov coefficients. There is also a
non-trivial cancellation between the two terms on the RHS of (1) which occur
at lower order in the short distance expansion. This cancellation was also
observed for the primary in Belin:2018juv , here however since some of the
states break rotational symmetry the cancellation is a non-trivial check of
the gravitational Gauss law due to Wald Wald:1993nt ; Iyer:1994ys  on states
which break the isometries of $AdS$.

If the coefficients in the linear combination are chosen as follows

$\displaystyle c_{l}=(1-z\bar{z})^{h}\frac{z^{l}}{l!},$ (8)

where $z$ is a complex number, then the state reduces to one of the coherent
states constructed in Caputa:2022zsr . These coherent states were argued to be
dual to the Bañados geometries constructed in Banados:1998gg . These states
are semi-classical, that is $h\sim O(c)$ and $c>>1$ but with $h/c$ finite and
lesser than unity. We compare the single interval entanglement entropy for the
coherent state obtained using our CFT methods and show that it precisely
agrees with that obtained by the Ryu-Takayanagi formula in the Bañados
geometry. This agreement not only serves as a check on the identification of
geometry dual to the coherent state but also a check on our CFT results for
the single interval entanglement entropy for all descendants. In particular,
this agreement is a check on the coefficients $\hat{g}_{ll^{\prime}}(x)$ in
(1) as well as the coefficients $D_{ll^{\prime}}$.

The organization of this paper is as follows. Section 2 evaluates the single
interval entanglement entropy of $SL(2,\mathbb{R})$ descendants as well as
their linear combination in the short distance expansion using the replica
trick in 2d CFT. The results are then applied to a coherent state. Section 3
first sets up the dictionary relating a primary and its descendants of a CFT
to single particle excitations of a minimally coupled massive scalar in
$AdS_{3}$. Then the section proceeds to evaluate each of the 2 terms in the
FLM formula for $2$ low lying states in detail and compares the results with
that of the CFT. One of the states break both time translation symmetry and
rotational symmetry of $AdS_{3}$. Lastly, this section compares the
entanglement entropy of the Bañados state against the coherent state
constructed in the CFT. Section 4 contains our conclusions. The appendices
contain the details required for both the CFT and gravity calculation.
Appendix C contains the details of the calculation of the Bogoliubov
coefficients that relate excitations in global $AdS_{3}$ to the Rindler BTZ
for 4 low lying states. Finally appendix D contains the check of the FLM
formula for the quantum corrections to single interval entanglement entropy
for 4 more low lying single particle excitations against that obtained from
CFT.

##  2 Entanglement of excited states in CFT

In this section we briefly review the set up to evaluate the single interval
entanglement entropy for an excited state in 2d CFT. Consider the theory on a
cylinder with coordinates $(t,\varphi)$ with $\varphi\sim\varphi+2\pi$. We
define the holomorphic and the anti-holomorphic co-ordinates on the cylinder
to be $y=\varphi+it,\bar{y}=\varphi-it$. Let us first restrict our discussions
to excitations in the holomorphic sector of the CFT. In the path integral
language an excited state $|{\cal O}\rangle$ is obtained by placing the
operator ${\cal O}$ which need not be a primary at $t=-\infty$ and performing
the path integral on the cylinder till $t=0$. We wish to study the single
interval entanglement entropy of this state, let the interval be $[0,2\pi x]$
at $t=0$. The entanglement entropy is given by evaluating the Von-Neumann
entropy of the reduced density matrix

$\rho_{\cal O}={\rm Tr}_{[0,2\pi x]}\Big{(}(|{\cal O}\rangle\langle{\cal
O}|\Big{)}.$ (9)

In the path integral, this reduced density matrix is obtained by evaluating
the path integral on the cylinder as shown in figure with the operators ${\cal
O}$ placed at $t\rightarrow-\infty$ and ${\cal O}^{*}$ placed at
$t\rightarrow\infty$. Note that the partial trace over the complement of the
interval $[0,2\pi x]$ leaves an open cut at the interval on the cylinder as
shown in figure 1 222Figures 1, 2, 3, 4, are taken from Chowdhury:2021qja .

Figure 1:  This figure shows the cut cylinder which represents the path
integral for the density matrix $\rho_{{\it\mathcal{O}}}$.

The entanglement entropy is obtained by the analytical continuation of the
Rényi entropies using the expression

$\hat{S}_{n}(\rho_{\cal O})=\frac{1}{1-n}\log\big{(}{\rm Tr}\rho_{\cal
O}^{n}\big{)},\qquad\hat{S}(\rho)=\lim_{n\rightarrow 1}\hat{S}_{n}(\rho_{\cal
O}).$ (10)

The above formula includes the universal entanglement entropy of the vacuum.
It is convenient to subtract this contribution, for this, it is sufficient to
evaluate the ratio of traces of the density matrices corresponding of the
operator ${\cal O}$ and that of the vacuum.

$S_{n}(\rho_{\cal O})=\frac{1}{1-n}\log\Big{(}\frac{{\rm Tr}\rho_{\cal
O}^{n}}{{\rm Tr}\rho_{(0)}^{n}}\Big{)},\qquad S(\rho)=\lim_{n\rightarrow
1}S_{n}(\rho_{\cal O}).$ (11)

Here $\rho_{(0)}$ refers to the density matrix without any operator
insertions. Using the path integral and conformal transformations the ratio of
the traces of the density matrices can be written as the following $2n$-point
function on the plane

$\frac{{\rm Tr}\rho_{\cal O}^{n}}{{\rm
Tr}\rho_{(0)}^{n}}=\frac{1}{\big{(}\langle{\cal O}|{\cal
O}\rangle\big{)}^{n}}\Big{\langle}\prod_{k=0}^{n-1}w\circ{\cal
O}(w_{k})\prod_{k^{\prime}=0}^{n-1}\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k})\Big{\rangle}.$ (12)

Here $w\circ{\cal O}(z)$ refers to the action of the conformal transformation
$w(z)$ on the operator ${\cal O}$ 333If ${\cal O}$ is a primary of weight $h$,
then $w\circ{\cal O}(z)=(\frac{\partial w}{\partial z})^{h}O(w(z))$ where

$w(z)=\left(\frac{z-u}{z-v}\right)^{\frac{1}{n}},$ (13)

and

$u=e^{2\pi ix},\qquad v=1.$ (14)

This is the map that takes the $n$-branched cylinder to the plane and the
operator ${\cal O}$ placed at $t\rightarrow-\infty$ on each cylinder to the
following point on the plane.

$w_{k}=e^{\frac{2\pi i(k+x)}{n}}=\lim_{z\rightarrow
0_{k}}\left(\frac{z-u}{z-v}\right)^{\frac{1}{n}},$ (15)

where $0_{k}$ refers to the position of the operator ${\cal O}$ on the $k$-th
cylinder. Each cylinder is mapped to a wedge on the complex plane. Similarly
$\hat{w}\circ{\cal O}(\hat{z})$ refers to the conformal transformation

$\hat{w}(\hat{z})=\left(\frac{\frac{1}{\hat{z}}-u}{\frac{1}{\hat{z}}-v}\right)^{\frac{1}{n}}.$
(16)

Thus $\hat{w}(\hat{z})$ is the map that takes the $n$-branched cylinder to the
plane and the operator ${\cal O}^{*}$ placed at $t\rightarrow\infty$ on each
cylinder to the following point on the plane.

$\hat{w}_{k}=\lim_{\hat{z}\rightarrow\hat{0}_{k}}\left(\frac{\frac{1}{\hat{z}}-u}{\frac{1}{\hat{z}}-v}\right)^{\frac{1}{n}}=e^{\frac{2\pi
ik}{n}}.$ (17)

Th operator positions in the $2n$-point function in (12) are shown in figure
2.

Figure 2:  The figure shows the uniformized plane for the $n=8$ replica
surface. Each cylinder is mapped to a wedge on the uniformed plane. There are
$2n$ operators with a pair of operators on each wedge. The operators are
located on the unit circe separated by an arc length of $2\pi x$.

Note that the operators on a given wedge are located on the unit circle and
separated by an arc distance of $2\pi x$. The norm $\langle{\cal O}|{\cal
O}\rangle$ can be evaluated by using the maps $w(z),\hat{w}(z)$ given in (13),
(16) with $n=1$.

###  2.1 Short distance expansion

In general the $2n$-point function cannot be evaluated exactly but we can
resort to the short distance expansion developed in Sarosi:2016oks ;
Chowdhury:2021qja . Let us define

${\cal C}_{(2n)}=\Big{\langle}\prod_{k=0}^{n-1}w\circ{\cal
O}(w_{k})\prod_{k^{\prime}=0}^{n-1}\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k^{\prime}})\Big{\rangle}.$ (18)

In the short distance expansion since the distance between the operators
${\cal O}$ and ${\cal O}^{*}$ in a given wedge is small, the leading
contribution to the $2n$-point function arises when one factorizes the
correlator into $n$ 2-point functions as shown in figure 3.

Figure 3:  A $n=8$ uniformized plane showing the contraction structure of the
$2n$-point function for the leading term in the single interval entanglement
entropy. The $2n$-point function is factored into $n$ 2-point function with
pairs of operators on the same wedge.

The sub-leading contribution is obtained by factorizing the correlator as
demonstrated in the figure 4.

Figure 4: The figure shows a $n=8$ uniformized plane with the factorisation of
the $2n$-point function into $(n-2)$, 2-point functions which are contracted
on the same wedge and a $4$-point functions involving operators on a pair of
wedges which are in blue. The contributions from all such pairs are summed
over to obtain the sub-leading contribution to the single interval
entanglement entropy.

We write both these contributions as

$\displaystyle{\cal C}_{2n}$ $\displaystyle=$ $\displaystyle{\cal
C}_{2n}^{(0)}+{\cal C}_{2n}^{(1)}+\cdots,$ (19) $\displaystyle{\cal
C}_{2n}^{(0)}$ $\displaystyle=$ $\displaystyle\prod_{k=0}^{n-1}\langle
w\circ{\cal O}(w_{k})\hat{w}\circ{\cal O}^{*}(\hat{w}_{k})\rangle,$ (20)
$\displaystyle{\cal C}_{2n}^{(1)}$ $\displaystyle=$
$\displaystyle\sum_{i,j=0,i\neq j}^{n-1}\left(\prod_{k=0,k\neq
i,j}^{n-1}\langle w\circ{\cal O}(w_{k})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k})\rangle\right)\Big{\langle}w\circ{\cal O}(w_{i})w\circ{\cal
O}^{*}(\hat{w}_{i})w\circ{\cal O}(w_{j})w\circ{\cal
O}^{*}(\hat{w}_{j})\Big{\rangle}_{c}.$

where the subscript ‘$c$’ refers to the connected correlator. Thus the first
sub-leading correction arises from the $4$-point function of the operators on
the replica geometry in which a pair of operators are placed on one of the
wedges and another pair in another wedge. The rest of the operators are
contracted as pairs on the same wedge. The complete contribution involves a
sum over all possible pairs of wedges involved in the $4$-point function. Once
the $2n$-point function is evaluated we can substitute it in the expression
for the entanglement entropy and obtain

$S(\rho_{\cal O})=\lim_{n\rightarrow 1}\frac{1}{1-n}\log\Big{(}\frac{{\cal
C}_{2n}}{(\langle{\cal O}|{\cal O}\rangle)^{n}}\Big{)}.$ (22)

#### Leading contribution: 2-point function

In the short distance approximation the leading term (20), is completely
determined by the dimension of the operator if it is a primary, if it is a
descendant, the result is dependent on the dimension, the level and nature of
the descendant and the central charge of the theory. We do not need to know
the details of the theory to evaluate the leading term. In this paper we will
restrict ourselves to descendants created by the action of the global part of
the Virasoro group. If ${\cal O}$ is a primary of weight $(h,0)$ 444For
clarity in presentation, we choose to discuss only the holomorphic sector
first, we will generalise the expressions for operators with weight $(h,h)$
subsequently. then the two point function on any of the wedges is given by

$\frac{1}{\langle{\cal O}|{\cal O}\rangle}\big{\langle}w\circ{\cal
O}(w_{k})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k})\big{\rangle}=\left(\frac{\sin\pi x}{n\sin\frac{\pi
x}{n}}\right)^{2h},\quad k=0,\cdots n-1.$ (23)

Global descendants of the primary ${\cal O}$ are given by

$\partial^{l}{\cal O}|0\rangle=(L_{-1})^{l}|h\;0\rangle.$ (24)

The two point function between descendants of arbitrary levels on a given
wedge in the replica geometry is given by

$\displaystyle\langle w\circ\partial^{l}{\cal
O}(w_{k})\hat{w}\circ\partial^{l^{\prime}}{\cal O}(\hat{w}_{k})\rangle$
$\displaystyle=$
$\displaystyle\partial_{z}^{l}\partial_{\hat{z}}^{l^{\prime}}G(z,\hat{z},n)\Big{|}_{(z,\hat{z})=(0_{k},\hat{0}_{k})},$
(25) $\displaystyle G(z,\hat{z},n)$ $\displaystyle=$
$\displaystyle\big{(}\partial_{z}w(z)\partial_{\hat{z}}\hat{w}(\hat{z})\big{)}^{h}\left(\frac{1}{w(z)-\hat{w}(\hat{z})}\right)^{2h}.$

This expression for the two point function between global descendants can be
easily derived from the two point function of the primary and the fact that
one needs to take appropriate number of derivatives to obtain the two point
function of global descendants. More details of this can be found in
Chowdhury:2021qja . Since we are interested in entanglement entropy, it is
sufficient to expand the function $G(z,\hat{z},n)$ to the leading order around
$n\rightarrow 1$ 555We are ignoring an overall phase $(-1)^{h}$ which arises
in $G(z,\hat{z},n)$ due to the fact our norm is defined by the map
$\hat{I}(z)=\frac{1}{z}$ rather than the $SL(2,R)$ map $I(z)=-\frac{1}{z}$
used in Chowdhury:2021qja . This phase always cancels on dividing by the norm
of the operator which also contains the same phase.

$\displaystyle
G(z,\hat{z},n)=\frac{1}{(1-z\hat{z})^{2h}}-(n-1)\times\frac{h}{(1-z\hat{z})^{2h}}\left\\{2+\log\Big{(}\frac{z-u}{z-v}\Big{)}+\log\Big{(}\frac{1-u\hat{z}}{1-v\hat{z}}\Big{)}\right.$
$\displaystyle\qquad\qquad\left.+\frac{2}{(u-v)(1-z\hat{z})}\Big{[}(z-u)(1-v\hat{z})\log\Big{(}\frac{z-u}{z-v}\Big{)}+(v-z)(1-u\hat{z})\log\Big{(}\frac{1-u\hat{z}}{1-v\hat{z}}\Big{)}\Big{]}\right\\}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad+O((n-1)^{2}).$ (26)

For subsequent purpose, it is useful to label the coefficients of the Taylor
series expansion of the function $G(z,\hat{z},n)$. Let us define

$g_{ll^{\prime}}(x,n)=\left.\partial_{z}^{l}\partial_{\hat{z}}^{l^{\prime}}G(z,\hat{z},n)\right|_{(z,\hat{z})=(0_{k},\hat{0}_{k})}.$
(27)

From the expansion in (2.1), we see that the expansion of these coefficients
around $n=1$ is of the form

$g_{ll^{\prime}}(x,n)=\frac{\Gamma(2h+l)l!}{\Gamma(2h)}\delta_{ll^{\prime}}+(n-1)\hat{g}_{ll^{\prime}}(x)+O((n-1)^{2}),$
(28)

where

$\displaystyle\hat{g}_{ll^{\prime}}(x)$ $\displaystyle=$
$\displaystyle\left.\partial_{z}^{l}\partial_{\hat{z}}^{l^{\prime}}\hat{G}(z,\hat{z})\right|_{(z,\hat{z})=(0_{k},\hat{0}_{k})}$
(29) $\displaystyle\hat{G}(z,\hat{z})$ $\displaystyle=$
$\displaystyle-\frac{h}{(1-z\hat{z})^{2h}}\left\\{2+\log\Big{(}\frac{z-u}{z-v}\Big{)}+\log\Big{(}\frac{1-u\hat{z}}{1-v\hat{z}}\Big{)}\right.$
$\displaystyle\qquad\qquad\left.+\frac{2}{(u-v)(1-z\hat{z})}\Big{[}(z-u)(1-v\hat{z})\log\Big{(}\frac{z-u}{z-v}\Big{)}+(v-z)(1-u\hat{z})\log\Big{(}\frac{1-u\hat{z}}{1-v\hat{z}}\Big{)}\Big{]}\right\\}.$

We can think of $G(z,\hat{z},n)$ as the generating function from which we can
derive the two point function using (25). Note that the norm of the descendant
can be read out from (28) from the coefficient of $\delta_{ll^{\prime}}$.

We can now use these results to evaluate the leading contribution to the
$2n$-point function for a linear combination of a primary and its descendants

$|\Psi\rangle=\sum_{l=0}^{\infty}\alpha_{l}(L_{-1})^{l}|h\;0\rangle.$ (30)

with the norm

$\langle\Psi|\Psi\rangle=\sum_{l=0}^{\infty}|\alpha_{l}|^{2}\frac{\Gamma(2h+l)l!}{\Gamma(2h)}.$
(31)

The leading contribution ${\cal C}_{2n}^{(0)}$ is given by

$\frac{{\cal
C}_{2n}^{(0)}}{(\langle\Psi|\Psi\rangle)^{n}}=1+(n-1)\sum_{l,l^{\prime}=0}^{\infty}\frac{\alpha_{l}\alpha_{l^{\prime}}^{*}\hat{g}_{ll^{\prime}}(x)}{\langle\Psi|\Psi\rangle}+O((n-1)^{2}).$
(32)

#### Sub-leading term: 4 point function

Let us now examine the 2nd term (2.1) in the short distance expansion of the
$2n$ point function ${\cal C}_{2n}$. This term involves the four point
function which in general depends on the theory considered. Let us first
examine the $4$ point function of conformal primaries places on wedge labelled
as $j$ and $k$. Using the conformal block decomposition Perlmutter:2015iya ,
we can expand the four point function as

$\displaystyle\langle w\circ{\cal O}(w_{j})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{j})w\circ{\cal O}(w_{k})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k})\rangle_{c}=(B_{j}\hat{B}_{j}B_{k}\hat{B}_{k})^{h}\times$
(33)
$\displaystyle\frac{1}{(w_{j}-\hat{w}_{j})^{2h}(w_{k}-\hat{w}_{k})^{2h}}\left(\sum_{q=1}^{\infty}\chi_{\rm
vac,q}w^{2}{}_{2}F_{1}(q,q,2q,w)+\sum_{p}C_{{\cal O}{\cal O}{\cal
O}_{p}}C^{{\cal O}_{p}}_{\;\;{\cal O}{\cal O}}w^{h_{p}}{\cal
F}(c,h,h_{p},w)\right).$

The cross ratio $w$ is defined as

$w=\frac{(w_{j}-\hat{w}_{j})(w_{k}-\hat{w}_{k})}{(w_{j}-w_{k})(\hat{w}_{j}-\hat{w}_{k})}=\left(\frac{\sin\frac{\pi
x}{n}}{\sin\frac{\pi}{n}(j-k)}\right)^{2}.$ (34)

The first term in the round bracket in (33) is the expansion of the Virasoro
block corresponding to the stress tensor exchange in terms of the global
$SL(2,C)$ blocks represented by the hypergeometric function
${}_{2}F_{1}(q,q,2q,w)$. The first two coefficients which is sufficient for
our purpose are

$\chi_{1}=0,\qquad\qquad\qquad\chi_{2}=\frac{2h^{2}}{c},$ (35)

where $c$ is the central charge of the CFT. The second term in the round
bracket represent the contribution of the Virasoro blocks of the primaries of
dimension $h_{p}$ of the theory. $C_{{\cal O}{\cal O}{\cal O}_{p}}C^{{\cal
O}_{P}}_{\;\;{\cal O}{\cal O}}$ is the product of the corresponding structure
constants, the Virasoro block admits the following expansion

${\cal F}(c,h,h_{p};w)=1+O(w).$ (36)

The $B$’s are derivatives of the conformal transformations to the uniformised
plane, they are given by

$\displaystyle B_{k}$ $\displaystyle=$ $\displaystyle\lim_{z\rightarrow
0_{j}}\partial_{z}w(z)=\frac{1}{n}e^{\frac{2\pi i(x+k)}{n}}(1-e^{-2\pi ix}),$
(37) $\displaystyle\hat{B}_{k}$ $\displaystyle=$
$\displaystyle\lim_{\hat{z}\rightarrow\hat{0}_{j}}\partial_{\hat{z}}\hat{w}(\hat{z})=\frac{1}{n}e^{\frac{2\pi
ik}{n}}(1-e^{2\pi ix}).$

For holographic CFT’s we can be more specific about the $4$-point function.
For a generalised free field $h<<c$, therefore the stress tensor block does
not contribute at the leading order. The leading operator exchanged is the
composite $:{\cal O}^{2}:$ with $h_{p}=2h$ and the product of structure
constants $C_{{\cal O}{\cal O}{\cal O}_{p}}C^{{\cal O}_{p}}_{\;\;{\cal O}{\cal
O}}=2$ Sarosi:2016oks ; Belin:2018juv ; Chowdhury:2021qja . Therefore in
generalised free field theory, the leading contribution to the $4$-point
function of primaries of weight $h$ is given by

$\displaystyle\langle w\circ{\cal O}(w_{j})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{j})w\circ{\cal O}(w_{k})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k})\rangle_{c}=(B_{j}\hat{B}_{j}B_{k}\hat{B}_{k})^{h}\times$
(38)
$\displaystyle\frac{1}{(w_{j}-\hat{w}_{j})^{2h}(w_{k}-\hat{w}_{k})^{2h}}\left[2\left(\frac{\sin\frac{\pi
x}{n}}{\sin\frac{\pi}{n}(j-k)}\right)^{4h}\right]+\cdots.$

The second class of operators in holographic theories are operators with
conformal dimensions $h\sim c$, then the leading contribution to the $4$-point
function is given by the stress tensor exchange and determined by $\chi_{2}$

$\displaystyle\langle w\circ{\cal O}(w_{j})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{j})w\circ{\cal O}(w_{k})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k})\rangle_{c}=(B_{j}\hat{B}_{j}B_{k}\hat{B}_{k})^{h}\times$
(39)
$\displaystyle\frac{1}{(w_{j}-\hat{w}_{j})^{2h}(w_{k}-\hat{w}_{k})^{2h}}\left[\frac{2h^{2}}{c}\left(\frac{\sin\frac{\pi
x}{n}}{\sin\frac{\pi}{n}(j-k)}\right)^{4}\right]+\cdots.$

From expressions (37) we see that in the leading short distance expansion, we
can approximate the prefactors occuring in (38) and (39) using

$\displaystyle\lim_{x\rightarrow
0}\frac{B_{j}}{w_{j}-\hat{w}_{j}}=\frac{1}{n},\qquad\qquad\lim_{x\rightarrow
0}\frac{\hat{B}_{j}}{w_{j}-\hat{w}_{j}}=-\frac{1}{n}.$ (40)

Note that once this limit is taken, these factors become independent of $j$,
the 2-point functions on the $n-2$ wedges in (33) are also independent of the
location of the wedge. This allows to perform the sum over $j,k$ using the
following Calabrese:2010he

$\displaystyle\sum_{j,k,j\neq
k}^{n-1}\left(\frac{1}{\sin\frac{\pi}{n}(j-k)}\right)^{2h_{p}}$
$\displaystyle=$ $\displaystyle\sum_{l=1}^{n-1}\frac{n-l}{\big{(}\sin\frac{\pi
l}{n}\big{)}^{2h_{p}}},$ $\displaystyle=$
$\displaystyle(n-1)\frac{\Gamma(\frac{3}{2})\Gamma(h_{p}+1)}{2\Gamma(h_{p}+\frac{3}{2})}+O((n-1)^{2}).$

The important point to note is that this sum is proportional to $n-1$ and
since we are interested in entanglement entropy, there all the rest of the
factors present in the expression for ${\cal C}_{2n}^{(1)}$ can be evaluated
in the limit $n\rightarrow 1$. Therefore the $n-2$, 2-point functions present
in ${\cal C}_{2n}^{(1)}$ reduce to their norm. Proceeding, we obtain the
following result for the sub-leading corrections when ${\cal O}$ is a primary

$\displaystyle\frac{{\cal C}_{2n}^{(1)}}{\big{(}\langle{\cal O}|{\cal
O}\rangle\big{)}^{n}}=(n-1)\frac{\Gamma(\frac{3}{2})\Gamma(2h_{p}+1)}{2\Gamma(2h_{p}+\frac{3}{2})}C_{{\cal
O}{\cal O}{\cal O}_{p}}C^{{\cal O}_{p}}_{\;\;{\cal O}{\cal
O}}\big{(}\sin\frac{\pi x}{n}\big{)}^{2h_{p}}+O((n-1)^{2}).$ (42)

Here for generalised free field theory we have

$C_{{\cal O}{\cal O}{\cal O}_{p}}C^{{\cal O}_{p}}_{\;\;{\cal O}{\cal
O}}=2,\qquad\qquad h_{p}=2h,$ (43)

and for operators of large conformal dimensions, that is $\frac{h}{c}\sim
O(1)$ with large $c$, the stress tensor exchange is dominant and we can
replace

$C_{{\cal O}{\cal O}{\cal O}_{p}}C^{{\cal O}_{P}}_{\;\;{\cal O}{\cal
O}}\rightarrow\frac{2h^{2}}{c},\qquad\qquad h_{p}=2.$ (44)

We can now consider the 4-point functions of arbitrary descendants and proceed
similarly. In the leading short distance expansion we obtain

$\displaystyle\lim_{x\rightarrow 0}\langle w\circ\partial^{l}{\cal
O}(w_{j})\hat{w}\circ\partial^{l^{\prime}}{\cal
O}^{*}(\hat{w}_{j})w\circ\partial^{m}{\cal
O}(w_{k})\hat{w}\circ\partial^{m^{\prime}}{\cal
O}^{*}(\hat{w}_{k})\rangle_{c}=$ (45)
$\displaystyle\frac{(B_{j}^{h+l}\hat{B}_{j}^{h+l^{\prime}}B_{k}^{h+m}\hat{B}_{k}^{h+m^{\prime}})}{(w_{j}-\hat{w}_{j})^{2h+l+l^{\prime}}(w_{k}-\hat{w}_{k})^{2h+m+m^{\prime}}}\times\left(\frac{\sin\frac{\pi
x}{n}}{\sin\frac{\pi}{n}(j-k)}\right)^{2h_{p}}D_{ll^{\prime}}(h,h_{p},n)D_{mm^{\prime}}(h,h_{p},n)$

Thus the four point function changes by a numerical factor
$D_{ll^{\prime}}(h,h_{p},n)$, called the dressing factor in Chowdhury:2021qja
. The details of this derivation and various examples worked out examples can
be found in Chowdhury:2021qja . Again examining the leading short distance
limit for the factors $B_{k},\hat{B}_{k}$ given in in (40) we can perform the
sum over $j,k$ using (2.1). Thus we need the dressing factor only in the limit
$n\rightarrow 1$, which can be obtained by the deformed norm as follows
Chowdhury:2021qja . Let us define the maps

$s(z)=\frac{z}{z-1},\qquad\hat{s}(\hat{z})=\frac{1}{1-\hat{z}},$ (46)

Then the deformed norm is given by

$\displaystyle D_{ll^{\prime}}(h,h_{p})$ $\displaystyle=$
$\displaystyle\lim_{n\rightarrow 1}D_{ll^{\prime}}(h,h_{p},n),$ (47)
$\displaystyle D_{ll^{\prime}}(h,h_{p})$ $\displaystyle=$
$\displaystyle\left.\partial_{z}^{l}\partial_{\hat{z}}^{l^{\prime}}H(z,\hat{z})\right|_{(z,\hat{z})=(0,0)},$
$\displaystyle H(z,\hat{z})$ $\displaystyle=$
$\displaystyle(\partial_{z}s(z)\partial_{\hat{z}}s(\hat{z}))^{h}\left(\frac{1}{s(z)-\hat{s}(\hat{z})}\right)^{2h-h_{p}},$
$\displaystyle=$
$\displaystyle\frac{1}{(1-z\hat{z})^{2h-h_{p}}(1-z)^{h_{p}}(1-\hat{z})^{h_{p}}}.$

Observe that when $h_{p}=0$, the deformed norm reduces to the norm between
descendants

$D_{ll^{\prime}}(h,0)=\frac{\Gamma(2h+l)l!}{\Gamma(2h)}\delta_{ll^{\prime}}.$
(48)

Using all these inputs, let us evaluate the sub leading contribution (39) to
the $2n$-point function corresponding to the state $|\Psi\rangle$ given in
(30).

$\displaystyle\frac{{\cal C}_{2n}^{(1)}}{(\langle\Psi|\Psi\rangle)^{n}}$
$\displaystyle=$
$\displaystyle(n-1)\frac{\Gamma(\frac{3}{2})\Gamma(h_{p}+1)}{2\Gamma(h_{p}+\frac{3}{2})}C_{{\cal
O}{\cal O}{\cal O}_{p}}C^{{\cal O}_{p}}_{\;\;{\cal O}{\cal O}}\frac{(\sin\pi
x)^{2h_{p}}}{\langle\Psi|\Psi\rangle^{2}}$
$\displaystyle\qquad\qquad\times\left(\sum_{l,l^{\prime}=0}^{\infty}\alpha_{l}\alpha_{l^{\prime}}^{*}D_{ll^{\prime}}(h,h_{p})\right)^{2}+O((n-1)^{2}).$

###  2.2 Primaries

We can apply the result for the leading and sub-leading contributions to the
$2n$-point function and obtain the leading approximations to the short
distance expansions of the entanglement entropy. Consider the case of the
primary ${\cal O}$ with weight $(h,0)$, let the state be normalised to unity
modulo the phase $(-1)^{2h}$ which occurs since we are using the
transformation $\hat{I}(z)=\frac{1}{z}$ to define the norm. Then we can
evaluate the coefficient which determined the $2$ point functions in (29)

$\hat{g}_{00}=-2h(1-\pi x\cot\pi x),$ (50)

and then using (28) we find that

$g_{00}=1-(n-1)2h(1-\pi x\cot\pi x).$ (51)

Substituting in the expression for the leading correction to the $2n$ point
function, we obtain

$\frac{{C}_{2n}^{(0)}}{\big{(}\langle{\cal O}|{\cal
O}\rangle\big{)}^{n}}=1-(n-1)2h(1-\pi x\cot\pi x)+O((n-1)^{2}).$ (52)

Let us now proceed to evaluate the sub-leading correction for a generalised
free field, then the leading correction is obtained by setting $h_{p}=2h$ and
$C_{{\cal O}{\cal O}{\cal O}_{p}}C^{{\cal O}{\cal O}{\cal O}_{p}}=2$. The
subleading correction to the $2n$ point function can be read out from (42),

$\frac{{C}_{2n}^{(1)}}{\big{(}\langle{\cal O}|{\cal
O}\rangle\big{)}^{n}}=(n-1)\frac{\Gamma(\frac{3}{2})\Gamma(2h+1)}{\Gamma(2h+\frac{3}{2})}(\sin\pi
x)^{4h}+O((n-1)^{2}).$ (53)

Substituting (52) and (53) into the expression for the entanglement entropy
(22), we obtain

$S\big{(}\rho_{|h,0\rangle}\big{)}=2h(1-\pi x\cot\pi
x)-\frac{\Gamma(\frac{3}{2})\Gamma(2h+1)}{\Gamma(2h+\frac{3}{2})}(\pi
x)^{4h}+\cdots.$ (54)

This is the leading short distance expansion of the entanglement entropy for
generalised free fields for which the leading contribution to the $4$-point
function arises from the composite operator $:{\cal O}^{2}:$.

We now proceed with the case when the operator ${\cal O}$ corresponds to the
primary with equal holomorphic and ant-holomorphic weights, so the state we
consider is $|h,h\rangle$. Going through the same steps and the fact that
2-point functions factorize into holomorphic and anti-holomorphic parts, we
obtain for the leading correction to the $2n$-point function

$\displaystyle\frac{{C}_{2n}^{(0)}}{\big{(}\langle{\cal O}|{\cal
O}\rangle\big{)}^{n}}=1-(n-1)4h(1-\pi x\cot\pi x)+O((n-1)^{2}).$ (55)

The simplest way to see this is to realize that the 2 point function of these
primaries are given by taking $h\rightarrow 2h$ in the equation (23) leading
to

$\frac{1}{\langle{\cal O}|{\cal O}\rangle}\big{\langle}w\circ{\cal
O}(w_{k},\bar{w}_{k})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k},\bar{\hat{w}}_{k}))\big{\rangle}=\left(\frac{\sin\pi
x}{n\sin\frac{\pi x}{n}}\right)^{4h},\quad k=0,\cdots n-1.$ (56)

Taking the complex conjugates of the maps in (13), (16) gives the action of
the maps on the anti-holomorphic coordinates. Taking $n$ products of the above
two point function and then performing the $n\rightarrow 1$ limit results in
(55). The sub-leading contribution to the $2n$-point function can be evaluated
using the $4$-point function of these primaries in generalized free field
theory. This is given by++

$\displaystyle\langle w\circ{\cal O}(w_{j},\bar{w}_{j})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{j},\bar{\hat{w}}_{j})w\circ{\cal
O}(w_{k},\bar{w}_{k})\hat{w}\circ{\cal
O}^{*}(\hat{w}_{k},\bar{\hat{w}}_{k})\rangle_{c}=|B_{j}\hat{B}_{j}B_{k}\hat{B}_{k}|^{2h}\times$
$\displaystyle\frac{1}{|w_{j}-\hat{w}_{j}|^{4h}|w_{k}-\hat{w}_{k}|^{4h}}\left[2\left(\frac{\sin\frac{\pi
x}{n}}{\sin\frac{\pi}{n}(j-k)}\right]^{8h}\right]+\cdots.$ (57)

Using the same steps followed for operators with holomorphic weights, we
obtain the following correction to the $2n$ point function

$\displaystyle\frac{{\cal C}_{2n}^{(1)}}{\big{(}\langle{\cal O}|{\cal
O}\rangle\big{)}^{n}}=(n-1)\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}(\sin\pi
x)^{8h}+O((n-1)^{2}).$ (58)

Putting (55) and (58) together, we obtain the following expression for the
leading corrections to the single interval entanglement entropy of primaries
of weight $(h,h)$.

$\displaystyle S\big{(}\rho_{|h,h\rangle}\big{)}=4h(1-\pi x\cot\pi
x)-\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}(\pi
x)^{8h}+\cdots.$ (59)

###  2.3 Descendants

Let us first consider the descendants of holomorphic primaries, that is
operators of weight $(h,0)$ which are given by

$|\Psi^{(l)}\rangle=(\partial_{z})^{l}{\cal
O}|0\rangle=(L_{-1})^{l}|h,0\rangle.$ (60)

The norm of this state is given by

$\langle\Psi^{(l)}|\Psi^{(l)}\rangle=\langle
h|L_{1}^{l}L_{-1}^{l}|h\rangle=\frac{\Gamma(2h+l)l!}{\Gamma(2h)}.$ (61)

The contribution to the $2n$-point function of these descendants can be read
out from (32) using the fact that $\alpha_{l}=1$ for the given value of $l$
and zero for the rest.

$\displaystyle\frac{{\cal
C}_{2n}^{(0)}}{\big{[}\langle\Psi^{(l)}|\Psi^{(l)}\rangle\big{]}^{n}}=1-(n-1)\frac{\hat{g}_{ll}(x)}{\langle\Psi^{(l)}|\Psi^{(l)}\rangle}+O((n-1)^{2}),$
(62)

where $\hat{g}_{ll}$ can be found from (29). Since there are equal number of
derivatives with respect to $z$ and $\hat{z}$, we can look at the terms in the
function $\hat{G}(z,\hat{z})$ which are functions of the product $z\hat{z}$,
these are given by

$\displaystyle\hat{G}(z,\hat{z})|_{z\hat{z}}$ $\displaystyle=$
$\displaystyle-\frac{h}{(1-z\hat{z})^{2h}}\left\\{2+\log\Big{(}\frac{u}{v}\Big{)}\right.$
$\displaystyle+\left.\frac{2}{(u-v)(1-z\hat{z})}\Big{[}2z\hat{z}(u-v)-(u+z\hat{z}v)\log\Big{(}\frac{u}{v}\Big{)}\Big{]}\right\\}.$

Then expanding in powers of $z\hat{z}$, we obtain

$\displaystyle\hat{g}_{ll}(x)=-\frac{\Gamma(2h+l)l!}{\Gamma(2h)}2(h+l)(1-\pi
x\cot\pi x).$ (64)

Using (62) and (64) we obtain

$\frac{{\cal
C}_{2n}^{(0)}}{\big{[}\langle\Psi^{(l)}|\Psi^{(l)}\rangle\big{]}^{n}}=1-(n-1)2(h+l)(1-\pi
x\cot\pi x)+O((n-1)^{2}).$ (65)

The sub-leading term is evaluated using (2.1), for which we need the
coefficient $D_{ll}(h,h_{p})$ defined in (47). From its definition we see that

$D_{ll}(h,2h)=\left(\frac{\Gamma(2h+l)}{\Gamma(2h)}\right)^{2}.$ (66)

Substituting in (22), we obtain the leading contributions to the entanglement
entropy of descendants of a holomorphic primary of weight $(h,0)$

$\displaystyle S\big{(}\rho_{L_{-1}^{l}|h,0\rangle}\big{)}=2(h+l)(1-\pi
x\cot\pi
x)-\frac{\Gamma(\frac{3}{2})\Gamma(2h+1)}{\Gamma(2h+\frac{3}{2})}\times\left(\frac{\Gamma(2h+l)}{\Gamma(2h)l!}\right)^{2}(\pi
x)^{4h}+\cdots.$

Let us consider the holomorphic descendants of primaries with weight $(h,h)$.
These states are defined by

$|\hat{\Psi}^{(l,0)}\rangle=\big{(}L_{-1})^{l}|h,h\rangle.$ (68)

Since the $2$-point function factorises into holomorphic and anti-holomorphic
factors, going through the same analysis we obtain the following result for
the leading correction to the $2n$-point function on the uniformized plane

$\displaystyle\frac{{\cal
C}_{2n}^{(0)}}{\big{[}\langle\hat{\Psi}^{(l,0)}|\hat{\Psi}^{(l,0)}\rangle\big{]}^{n}}=1-(n-1)2(2h+l)(1-\pi
x\cot\pi x).$ (69)

Again going through the same steps for the sub-leading correction we obtain

$\frac{{\cal
C}_{2n}^{(1)}}{\big{[}\langle\hat{\Psi}^{(l,0)}|\hat{\Psi}^{(l,0)}\rangle\big{]}^{n}}=(n-1)\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}\left(\frac{\Gamma(2h+l)}{\Gamma(2h)l!}\right)^{2}(\pi
x)^{8h}+\cdots.$ (70)

Note that the dressing factor occurs only for the holomorphic sector, while in
the remaining terms $h$ is replaced by $h\rightarrow 2h$. Substituting (69)
and (70) into the expression for the entanglement entropy we obtain

$\displaystyle S\big{(}\rho_{L_{-1}^{l}|h,h\rangle}\big{)}=2(2h+l)(1-\pi
x\cot\pi
x)-\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}\times\left(\frac{\Gamma(2h+l)}{\Gamma(2h)l!}\right)^{2}(\pi
x)^{8h}+\cdots.$

By repeating the analysis for states of form
$L_{-1}^{l}\bar{L}_{-1}^{l}|h,h\rangle$, we obtain

$\displaystyle
S\big{(}\rho_{L_{-1}^{l}\bar{L}_{-1}^{l}|h,h\rangle}\big{)}=4(h+l)(1-\pi
x\cot\pi
x)-\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}\times\left(\frac{\Gamma(2h+l)}{\Gamma(2h)l!}\right)^{4}(\pi
x)^{8h}+\cdots.$

###  2.4 Linear combinations of descendants

Since entanglement entropy of excited states involves a $2n$ point function,
it is certainly does not obey the linear superposition rule when one considers
the entanglement of linear combinations of excited states. It is useful to
study arbitrary linear combinations of excited states involving primaries and
descendants. This is because as we will see subsequently such linear
combinations span the complete Hilbert space of single particle states of a
minimally coupled scalar in $AdS_{3}$. We will also note that linear
combinations of excited states need not to be homogenous in the spatial
directions and therefore this offers a means to study entanglement entropy of
spatially non-homogenous states.

First let us consider the case of linear combination of holomorphic global
descendants of a primary with weight $(h,0)$

$\displaystyle|\Psi\rangle=\sum_{l=0}^{\infty}c_{l}L_{-1}^{l}|h,0\rangle.$
(73)

Then the norm of this state is given in (31), using (32) and (2.1) in the
expression for entanglement entropy we obtain

$\displaystyle S(\rho_{|\Psi\rangle})$ $\displaystyle=$
$\displaystyle\sum_{l,l^{\prime}=0}^{\infty}\frac{c_{l}c_{l^{\prime}}^{*}\hat{g}_{ll^{\prime}}(x)}{\langle\Psi|\Psi\rangle}$
$\displaystyle\qquad-\frac{\Gamma(\frac{3}{2})\Gamma(2h+1)}{\Gamma(2h+\frac{3}{2})}\frac{(\pi
x)^{4h}}{\langle\Psi|\Psi\rangle^{2}}\times\left(\sum_{l,l^{\prime}=0}^{\infty}c_{l}c_{l^{\prime}}^{*}D_{ll^{\prime}}(h,2h)\right)^{2}+\cdots.$

Here $\hat{g}_{ll^{\prime}}(x)$ is evaluated using (29).
$D_{ll^{\prime}}(h,2h)$ is obtained from its definition in (47), which is
given by

$D_{ll^{\prime}}(h,2h)=\frac{\Gamma(2h+l)\Gamma(2h+l^{\prime})}{\big{(}\Gamma(2h)\big{)}^{2}}.$
(75)

Again it is important to mention that the result in (2.4) are the leading
contributions to the short distance expansion of the single interval
entanglement entropy. We can easily generalise the expression for the linear
combination of holomorphic global descendants of a primary with weight
$(h,h)$.

$|\hat{\Psi}\rangle=\sum_{l=0}^{\infty}c_{l}L_{-1}^{l}|h,h\rangle.$ (76)

Going through the same analysis, we obtain

$\displaystyle S(\rho_{|\hat{\Psi}\rangle})$ $\displaystyle=$
$\displaystyle\sum_{l,l^{\prime}=0}^{\infty}\frac{c_{l}c_{l^{\prime}}^{*}\hat{g}_{ll^{\prime}}(x)}{\langle\hat{\Psi}|\hat{\Psi}\rangle}+2h(1-\pi
x\cot\pi x)$
$\displaystyle\qquad-\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}\frac{(\pi
x)^{8h}}{\langle\hat{\Psi}|\hat{\Psi}\rangle^{2}}\times\left(\sum_{l,l^{\prime}=0}^{\infty}c_{l}c_{l^{\prime}}^{*}D_{ll^{\prime}}(h,2h)\right)^{2}+\cdots.$

Again this expression is easy to obtain when one observes that the 2-point
function on each wedge factorizes into products of holomorphic and anti-
holmorphic parts. This leads to the additional term in the first line of
(2.4), then the change of $h\rightarrow 2h$ to some of the terms in the second
line is due to the same reason which results in (2.3). As a simple check of
(2.4), note that it reduces to (2.3) on choosing $c_{l}=1$ for one particular
$l$ and vanishing for the rest.

###  2.5 Primary and level one descendant

For the purposes of section 3, it is useful to write down the explicit formula
of the entanglement entropy of the following linear combination

$|\Phi\rangle=c_{0}|h\;h\rangle+c_{1}L_{-1}|h\;h\rangle.$ (78)

The norm of this state is given by

$\langle\Phi|\Phi\rangle=|c_{0}|^{2}+2h|c_{1}|^{2}.$ (79)

To evaluate the leading corrections we use (2.4), for which we need the
coefficients

$\displaystyle\hat{g}_{00}$ $\displaystyle=$ $\displaystyle-2h(1-\pi x\cot\pi
x),$ (80) $\displaystyle\hat{g}_{01}$ $\displaystyle=$ $\displaystyle
h(i+\cot\pi x)(-2\pi x+\sin 2\pi x),$ $\displaystyle\hat{g}_{10}$
$\displaystyle=$ $\displaystyle h(-i+\cot\pi x)(-2\pi x+\sin 2\pi x),$
$\displaystyle\hat{g}_{11}$ $\displaystyle=$ $\displaystyle-4h(h+1)(1-\pi
x\cot\pi x).$

We have evaluated these coefficients by using (29). Let us write the
entanglement entropy as

$S(\rho_{|\Phi\rangle})=S^{(0)}(\rho_{|\Phi\rangle})+S^{(1)}(\rho_{|\Phi\rangle}),$
(81)

to represent the leading and subleading contributions. Then using (2.4), the
leading short distance contribution to the entanglement entropy of the linear
combination in (78) is given by

$\displaystyle S^{(0)}(\rho_{|\Phi\rangle})$ $\displaystyle=$
$\displaystyle-\frac{|c_{0}|^{2}\hat{g}_{00}+|c_{1}|^{2}\hat{g}_{11}+c_{0}c_{1}^{*}\hat{g}_{01}+c_{1}c_{0}^{*}\hat{g}_{10}}{|c_{0}|^{2}+2h|c_{1}|^{2}}+2h(1-\pi
x\cot\pi x),$ $\displaystyle=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\times\Bigg{(}\Big{[}4h|c_{0}|^{2}+4h(2h+1)|c_{1}|^{2}\Big{]}(1-\pi
x\cot\pi x)$ $\displaystyle+h\Big{[}(c_{0}c_{1}^{*}+c_{1}c_{0}^{*})\cot\pi
x+i(c_{0}c_{1}^{*}-c_{1}c_{0}^{*})\Big{]}(2\pi x-\sin 2\pi x)\Bigg{)}.$

Proceeding with the evaluation of the sub-leading correction using the second
term in (2.4) we obtain

$\displaystyle S^{(1)}(\rho_{|\Phi\rangle})$ $\displaystyle=$
$\displaystyle-\frac{\Gamma(\frac{3}{2})\Gamma(4h+1)}{\Gamma(4h+\frac{3}{2})}\frac{(\sin\pi
x)^{8h}}{(c_{0}|^{2}+2h|c_{1}|^{2})^{2}}\times\Big{(}|c_{0}|^{2}+2h(c_{0}c_{1}^{*}+c_{0}^{*}c_{1})+(2h)^{2}|c_{1}|^{2}\Big{)}^{2},$

One of our goals in section 3 is to reproduce both (2.5) and (2.5) from
holography. The expressions shows that leading contribution to the
entanglement entropy of the state depends on the spatial coordinate $x$ in a
non-trivial way, not just proportional to the function $(1-\pi x\cot\pi x)$
which is the case of the primaries or any of the descendants as in (54), (59),
(2.3, (2.3).

#### Comparison with modular Hamiltonian

It is known that that the leading contributions to the entanglement of the
excited state can be obtained by evaluating the expectation value of the
modular Hamiltonian arising from the vacuum $K^{(0)}$ in the excited state
Belin:2018juv .

$S^{(0)}(\rho_{|\Phi\rangle})=\frac{\langle\Phi|K^{(0)}|\Phi\rangle}{\langle\Phi|\Phi\rangle}.$
(84)

where the modular Hamiltonian of an interval in the vacuum on the cylinder is
given by

$\displaystyle K^{0}=\int_{0}^{2\pi
x}d\varphi\left(\frac{\cos(\varphi-\frac{\theta}{2})}{\sin\frac{\theta}{2}}-\cot\frac{\theta}{2}\right)\left(T_{yy}(\varphi)+\bar{T}_{\bar{y}\bar{y}}(\varphi)\right).$
(85)

Here we have taken the length of the cylinder to be $2\pi$ and the length of
the interval along the circumference at $t=0$ to be $2\pi x$. A short review
detailing the derivation for the above expression for the modular Hamiltonian
is provided in the appendix A. $T_{yy}$ and $\bar{T}_{\bar{y}\bar{y}}$ are the
holomorphic and anti-holomorphic components of the stress tensor on the
cylinder. At the $t=0$ slice on the cylinder they are given by

$\displaystyle
T_{yy}(\varphi)=\sum_{n=-\infty}^{\infty}L_{n}\exp(in\varphi),\qquad\qquad\bar{T}_{\bar{y}\bar{y}}(\varphi)=\sum_{n=-\infty}^{\infty}\bar{L}_{n}\exp(-in\varphi).$
(86)

Note that evaluating the expectation value of the modular Hamiltonian on the
primary $|h,h\rangle$, yields

$\langle h,h|K^{(0)}|h,h\rangle=4h(1-\pi x\cot\pi x).$ (87)

which is the entanglement entropy of the excited state. Here it is only the
$L_{0}$ and $\bar{L}_{0}$ term in the expansion of the stress tensor in (86)
which contributes.

Since the leading contribution to the entanglement entropy of the linear
combination in (78) is a non-trivial function it is interesting to see how the
entanglement Hamiltonian reproduces the expression (2.5). For this let us
evaluate the following expectation values

$\displaystyle\langle h\;h|K^{(0)}L_{-1}|h\;h\rangle$ $\displaystyle=$
$\displaystyle 2h\int_{0}^{2\pi
x}d\varphi\left(\frac{\cos(\varphi-\frac{\theta}{2})}{\sin\frac{\theta}{2}}-\cot\frac{\theta}{2}\right)e^{i\varphi},$
$\displaystyle=$ $\displaystyle h(i+\cot\pi x)(2\pi x-\sin 2\pi x).$

Similarly

$\displaystyle\langle h\;h|L_{1}K^{(0)}|h\;h\rangle$ $\displaystyle=$
$\displaystyle 2h\int_{0}^{2\pi
x}d\varphi\left(\frac{\cos(\varphi-\frac{\theta}{2})}{\sin\frac{\theta}{2}}-\cot\frac{\theta}{2}\right)e^{-i\varphi},$
$\displaystyle=$ $\displaystyle h(-i+\cot\pi x)(2\pi x-\sin 2\pi x).$

Finally we also have

$\displaystyle\langle h\;h|L_{1}K^{(0)}L_{-1}|h\;h\rangle$ $\displaystyle=$
$\displaystyle(h+1)4h(1-\pi x\cot\pi x)+4h^{2}(1-\pi x\cot\pi x),$
$\displaystyle=$ $\displaystyle 4h(2h+1)(1-\pi x\cot\pi x).$

Here the second term comes from the anti-holomorphic part of the stress tensor
in (86). Now using (87), (2.5), (2.5), (2.5) we can evaluate

$\displaystyle\frac{\langle\Phi|K^{(0)}|\Phi\rangle}{\langle\Phi|\Phi\rangle}$
$\displaystyle=$
$\displaystyle\frac{1}{|a_{0}|^{2}+2h|a_{1}|^{2}}\times\Bigg{(}\Big{[}4h|c_{0}|^{2}+4h(2h+1)|c_{1}|^{2}\Big{]}(1-\pi
x\cot\pi x)$ $\displaystyle+h\Big{[}(c_{0}c_{1}^{*}+c_{1}c_{0}^{*})\cot\pi
x+i(c_{0}c_{1}^{*}-c_{1}c_{0}^{*})\Big{]}(2\pi x-\sin 2\pi x)\Bigg{)}.$

where the state $|\Phi\rangle$ is given in (78). Observe that the above
equation precisely agrees with (2.5), which provides a cross check for the
leading contribution to the entanglement entropy evaluated using the replica
method.

###  2.6 Coherent states

In Caputa:2022zsr  a class of coherent state was studied both in the CFT and
the bulk. These states involve specific linear combinations of the primary and
the descendants including Virasoro descendants. The primaries had conformal
dimensions $\frac{h}{c}\sim O(1)$ and $c>>1$, therefore it is possible to
construct geometries dual to these states. One such state considered by
Caputa:2022zsr  is the following linear combination 666The primary considered
in Caputa:2022zsr  had equal holomorphic and anti-holomorphic weights. Here we
have taken the primary to have only holomorphic weight for simplicity, though
the calculation can be easily generalized.

$|\Psi_{k}(z,\bar{z})\rangle=(1-z_{k}\bar{z}_{k})^{h_{k}}\sum_{l=0}^{\infty}\frac{z_{k}^{l}}{k^{n}l!}L_{-k}^{l}|h\;0\rangle,$
(92)

where

$\displaystyle h_{k}=\frac{1}{k}\Big{[}h+\frac{c}{24}(k^{2}-1)\Big{]}.$ (93)

Based on the expectation value of the stress tensor in this state it was
proposed that this coherent state is dual to Ba$\tilde{\rm{n}}$ados geometries
Banados:1998gg  and an expression for its entanglement entropy for arbitrary
$k$ was derived using the Ryu-Takanayagi formula. For the special case of
$k=1$, $|\Psi_{1}(z,\bar{z})\rangle$ when the state is a coherent state of
global descendants, the Heavy-Heavy-Light-Light correlator was used to
demonstrate agreement with the expression for entanglement entropy derived
from the Ba$\tilde{\rm{n}}$ados geometry.

Here we will use the general expression for the leading and sub-leading
corrections to the entanglement entropy for a linear combination of global
descendants to evaluate the single interval entanglement entropy. The state we
are interested is $k=1$ of (92), which is given by

$|\Psi_{1}(z,\bar{z})\rangle=(1-z\bar{z})^{h}\sum_{l=0}^{\infty}\frac{z^{l}}{l!}L_{-1}^{l}|h\;0\rangle$
(94)

where we have dropped the subscript in $z_{1},\bar{z}_{1}$. Observe that from
(61), this state is of unit norm, the coefficients for the linear combination
of global descendants are given by

$c_{l}=(1-z\hat{z})^{h}\frac{z^{l}}{l!},\qquad
c_{l}^{*}=(1-z\bar{z})^{h}\frac{\bar{z}^{l}}{l!}$ (95)

With these coefficients and using the definition (29) we have

$\sum_{l,l^{\prime}}c_{l}c_{l^{\prime}}^{*}g_{ll^{\prime}}(x,n)=(1+z\bar{z})^{2h}\hat{G}(z,\bar{z})$
(96)

Therefore from (32) and (22) the leading contribution to the entanglement
entropy is given by

$\displaystyle S^{(0)}(\rho_{|\Psi_{1}\rangle})$ $\displaystyle=$
$\displaystyle
h\left\\{2+\log\Big{(}\frac{z-u}{z-v}\Big{)}+\log\Big{(}\frac{1-u\bar{z}}{1-v\bar{z}}\Big{)}\right.$
$\displaystyle\qquad\left.+\frac{2}{(u-v)(1-z\bar{z})}\Big{[}(z-u)(1-v\hat{z})\log\Big{(}\frac{z-u}{z-v}\Big{)}+(v-z)(1-u\bar{z})\log\Big{(}\frac{1-u\bar{z}}{1-v\bar{z}}\Big{)}\Big{]}\right\\}$
$\displaystyle=$ $\displaystyle
h\Bigg{\\{}2+\frac{\Big{(}2z+2\bar{z}uv-(1+z\bar{z})(u+v)\Big{)}}{(u-v)(1-z\bar{z})}\log\Big{[}\frac{(z-u)(1-v\bar{z})}{(z-v)(1-u\bar{z})}\Big{]}\Bigg{\\}}.$

It is important to note that here $(z,\bar{z})$ are just variables
parameterising the coherent state in (94) and $(u,v)$ carry the information of
the location of the interval. This result also gives us a physical
interpretation for the generating function $\hat{G}(z,\bar{z})$. It is the
leading contribution to the entanglement entropy of the coherent state in
(94).

We can proceed to evaluate the sub-leading contribution. The state is based on
the primary with dimensions $\frac{h}{c}\sim O(1)$. Therefore, the leading
contribution will be due to the stress tensor exchange with $h_{p}=2$ and

$C_{{\cal O}{\cal O}{\cal O}_{p}}C^{{\cal O}_{p}}_{\;{\cal O}{\cal
O}}\rightarrow\frac{2h^{2}}{c},$ (98)

as discussed around (35) and (44) From (2.1) we see that we need the dressing
factor, which is given by

$\sum_{l,l^{\prime}=0}^{\infty}c_{l}c_{l^{\prime}}^{*}D_{ll^{\prime}}(h,2)=\frac{(1-z\bar{z})^{2}}{(1-z)^{h}(1-\bar{z})^{h}},$
(99)

here we have used the definition (47) and $c_{l}$ from (95). Finally using
(2.1) and the definition (22), we obtain the following contribution at the
sub-leading level

$\displaystyle S^{(1)}(\rho_{|\Psi_{1}\rangle})$ $\displaystyle=$
$\displaystyle-\frac{8h^{2}}{15c}\frac{(1-z\bar{z})^{4}}{(1-z)^{2h}(1-\bar{z})^{2h}}(\pi
x)^{4}+\cdots.$ (100)

In section 3.5, we will show the entanglement entropy evaluated using the Ryu-
Takayanagi formula on the Ba$\tilde{\rm{n}}$ados geometry corresponding to the
coherent state (94) precisely agree at both with the leading (2.6) and the
sub-leading (100) terms in the short distance expansion.

##  3 Entanglement of single particle states in gravity

In this section we evaluate the single interval entanglement entropy of
excited states using the formula proposed by Faulkner-Lewkowycz-Maldacena.
Before we proceed we review the proposal. We restrict our attention to
excitations which do not change the asymptotic geometry. Consider the interval
on the boundary of the asymptotically $AdS_{3}$ geometry labelled as $A$ and
its complement $\bar{A}$ in figure 5. The bulk geometry at the $t=0$ slice
corresponds to the interior of the circle in the same figure. The FLM proposal
states that the result for the entanglement entropy for the interval $A$ is
given by the

$S_{\rho_{A}}=\frac{A(\gamma_{A})}{4G_{N}}+S_{\rm bulk}(\Sigma_{A}).$ (101)

Here $\gamma_{A}$ is the minimal geodesic in the bulk joining the end points
of the interval $A$, and $A(\gamma_{A})$ is its length. $\Sigma_{A}$ is a bulk
spatial region contained between $\gamma_{A}$ and the boundary as shown in
figure 5. $S_{\rm bulk}(\Sigma_{A})$ is the bulk entanglement entropy
evaluated by viewing the bulk as an effective quantum field theory.

$r_{m}$III$\gamma_{A}$$\Sigma_{A}$$\Sigma_{\bar{A}}$$\bar{A}$$A$$\varphi=2\pi
x$$\varphi=0$ Figure 5: The $t=0$ slice of $AdS_{3}$. We consider the
entanglement between system $A$ and $\bar{A}$ in the boundary. The minimal
surface $\gamma_{A}$ is the geodesic in the bulk connecting the end points of
$A$. $\gamma_{A}$ splits the bulk into right(left) wedge denoted by
$\Sigma_{A}(\Sigma_{\bar{A}})$. $\gamma_{A}$ consists of: Branch I, where
$\varphi^{\prime}(r)<0$ and Branch II with $\varphi^{\prime}(r)>0$. When
excitations breaks the isometry in $\varphi$, we need to evaluate the minimal
area for the above branches separately.

By the standard rules of AdS/CFT, a primary operator ${\cal O}$ with
dimensions $(h,h)$ in the CFT is dual to a minimally coupled scalar $\phi$
propagating in the bulk whose mass is given by

$M^{2}=\Delta(\Delta-2)=4h(h-1),\qquad\quad\Delta=2h,$ (102)

here we have assumed the radius of $AdS_{3}$ is unity. In fact all global
descendants of the operator ${\cal O}$ are dual to the single particle
excitations of the scalar $\phi$. This then allows us to perform a precision
check of the FLM formula. We consider various single particle excitations or
their linear combinations of the bulk geometry. The bulk geometry is deformed
by these excitations and it is necessary to compute the back reacted geometry.
The shift in minimal area due to the deformed geometry together with the
$S_{\rm bulk}(\Sigma_{A})$ in the FLM formula can be evaluated in the short
interval expansion using the methods developed by Belin:2018juv . Combining
both the terms in (101) we can evaluate the entanglement entropy of the single
particle states and compare it to the results obtained from CFT section 2.
This check extends that done in Belin:2018juv  for the lowest energy state.

This section is organised as follows, in sub-section 3.1 we review the duality
which relates descendants to single particle excitations of the scalar
propagating in $AdS$. We focus on $6$ low lying states which are dual to the
following descendants in the CFT

$\displaystyle|\hat{\Psi}^{(1,0)}\rangle=L_{-1}|h\;h\rangle,$
$\displaystyle|\hat{\Psi}^{(2,0)}\rangle=L_{-1}^{2}|h\;h\rangle,$ (103)
$\displaystyle|\hat{\Psi}^{(1,1)}\rangle=L_{-1}\bar{L}_{-1}|h\;h\rangle,$
$\displaystyle|\hat{\Psi}^{(2,2)}\rangle=L_{-2}\bar{L}_{-2}|h\;h\rangle,$
$\displaystyle|\Phi\rangle=c_{0}|h\;h\rangle+c_{1}L_{-1}|h\;h\rangle,$
$\displaystyle|\Upsilon\rangle=c_{0}|h\;h\rangle+c_{1}L_{-1}\bar{L}_{-1}|h\;h\rangle.$

In section 3.2, we obtain the back reacted geometry for these states, then in
section 3.3, we obtain the shift in minimal area and finally in section 3.4,
we evaluate the bulk entanglement entropy. This requires the knowledge of the
reduced density matrix of the excited states in the bulk which is obtained
using the Bogoliubov transformations which decompose the creation operators of
these states into operators interior and exterior of the entangling region
$\Sigma_{A}$.

###  3.1 Construction of single particle states

In this section we construct the single particle states in the bulk which are
dual to the conformal primary of dimensions $(h,h)$ and its global
descendants. The AdS/CFT dictionary instructs us to examine the minimally
coupled scalar of mass (102). The action of this scalar along with that of the
metric is given by

$S=\int d^{3}x\sqrt{-g}\left(\frac{1}{16\pi
G_{N}}\left(R+2\right)-\frac{1}{2}\left(\nabla\phi\right)^{2}-\frac{1}{2}M^{2}\phi^{2}\right).$
(104)

We start with the global $AdS$ solution, which corresponds to the vacuum of
the CFT

$ds^{2}=-(r^{2}+1)dt^{2}+\frac{dr^{2}}{r^{2}+1}+r^{2}d\varphi^{2}\qquad\varphi\sim\varphi+2\pi.$
(105)

The equations of motion of the scalar in this background is given by

$(\nabla^{2}-M^{2})\phi(x)=0\,.$ (106)

We expand the solutions in terms of modes

$\phi(t,r,\varphi)=\sum_{m,n}\left(a_{m,n}e^{-i\Omega_{m,n}t}e^{im\varphi}f_{m,n}(r)+a^{\dagger}_{m,n}e^{i\Omega_{m,n}t}e^{-im\phi}f^{*}_{m,n}(r)\right)\,.$
(107)

Here the sum over $m$ runs over the set of integers due to the periodic
boundary conditions on $\varphi$. $n$ labels the radial wave function. The
solutions which are regular at $r=0$ are given by

$\displaystyle f_{m,n}(r)$ $\displaystyle=$ $\displaystyle
C_{m,n}r^{m}(1+r^{2})^{\frac{\Omega_{m,n}}{2}}{}_{2}F_{1}\Big{(}\frac{1}{2}(m-2h+2+\Omega_{m,n}),\frac{1}{2}(m+2h+\Omega_{m,n}),1+m;-r^{2}\Big{)},$
$\displaystyle{\rm for}\;\;m>0,$ $\displaystyle f_{m,n}(r)$ $\displaystyle=$
$\displaystyle
C_{m,n}r^{-m}(1+r^{2})^{\frac{\Omega_{m,n}}{2}}{}_{2}F_{1}\Big{(}\frac{1}{2}(-m-2h+2+\Omega_{m,n}),\frac{1}{2}(-m+2h+\Omega_{m,n}),1-m;-r^{2}\Big{)},$
$\displaystyle{\rm for}\;\;m<0.$

Demanding that these functions are normalizable, or bounded at
$r\rightarrow\infty$ results in the quantization condition

$\Omega_{m,n}=2h+|m|+2n,\qquad n=0,1,2,\cdots,{\rm with}\;m\in\mathbb{Z}.$
(109)

The constants $C_{m,n}$ are fixed using the standard Klein-Gordan inner
product which is given by

$2\Omega_{m,n}\int
drd\varphi\sqrt{-g}g^{tt}(r)f_{m,n}(r,\varphi)f_{m^{\prime},n^{\prime}}^{*}(r,\varphi)=\delta_{n,n^{\prime}}\delta_{m,m^{\prime}}\,.$
(110)

The explicit forms of the wave functions for some low lying states are given
in table 1

$m$ | $n$ | $f_{m,n}(r)$ | $L_{0}+\bar{L}_{0}$ | $L_{0}-\bar{L}_{0}$
---|---|---|---|---
$0$ | $0$ | $\frac{1}{\sqrt{2\pi}\left(r^{2}+1\right)^{h}}$ | $2h$ | 0
$0$ | $1$ | $\frac{1}{\sqrt{2\pi}}\frac{2hr^{2}-1}{\left(r^{2}+1\right)^{h+1}}$ | $2h+2$ | $0$
$0$ | $2$ | $\frac{1}{\sqrt{2\pi}}\frac{h(2h+1)r^{4}-2(2h+1)r^{2}+1}{\left(r^{2}+1\right)^{h+2}}$ | $2h+4$ | $0$
$1$ | $0$ | $\frac{\sqrt{h}r}{\sqrt{\pi}\left(r^{2}+1\right)^{h+\frac{1}{2}}}$ | $2h+1$ | $1$
$2$ | $0$ | $\frac{\sqrt{h(2h+1)}r^{2}}{\sqrt{2\pi}\left(r^{2}+1\right)^{h+1}}$ | $2h+2$ | $2$
Table 1:  This table lists the explicit wave functions of the single particle
states for low values of $m$ and $n$. The last 2 columns list out the quantum
numbers of $L_{0},\bar{L}_{0}$ of the corresponding dual state in the CFT.

From the canonical commutation relations of $\phi$ and $\dot{\phi}$ we obtain
that the commutation relations

$[a_{m,n},a^{\dagger}_{m,n}]=\delta_{n,n^{\prime}}\delta_{m,m^{\prime}}$ (111)

Therefore single particle states on the global $AdS_{3}$ vacuum are given by

$|\psi_{m,n}\rangle=a_{m,n}^{\dagger}|0\rangle.$ (112)

It is clear from the construction, that $m$ and $\Omega_{m,n}$ are the eigen
values of the operators corresponding to translations in $\varphi$ and the
global time $t$ respectively. Let us use these quantum numbers to relate these
states to the global descendants of a primary in the dual CFT. Consider the
descendant

$|\hat{\Psi}^{(n_{1},n_{2})}\rangle=\big{(}L_{-1})^{n_{1}}\big{(}\bar{L}_{-1}\big{)}^{n_{2}}|h\;h\rangle.$
(113)

Then from the global $SL(2,\mathbb{R})$ algebra we have the following
properties for these states

$\displaystyle(L_{0}+\bar{L}_{0})|\hat{\Psi}^{(n_{1},n_{2})}\rangle$
$\displaystyle=$
$\displaystyle(2h+n_{1}+n_{2})|\hat{\Psi}^{(n_{1},n_{2})}\rangle,$ (114)
$\displaystyle(L_{0}-\bar{L}_{0})|\hat{\Psi}^{(n_{1},n_{2})}\rangle$
$\displaystyle=$
$\displaystyle(n_{1}-n_{2})|\hat{\Psi}^{(n_{1},n_{2})}\rangle.$

From the matching of symmetries we know that the operator $L_{0}-\bar{L}_{0}$
corresponds to the generator of translations along the angular direction
$\varphi$ in the bulk. The identification of the global time in $AdS_{3}$ with
the time of the CFT, we see that $L_{0}+\bar{L}_{0}$ corresponds to the
generator of translations along the global time $t$ in the bulk. This implies
that we have the relations

$\displaystyle 2h+n_{1}+n_{2}$ $\displaystyle=$
$\displaystyle\Omega_{m,n}=2h+|m|+2n,$ (115) $\displaystyle n_{1}-n_{2}$
$\displaystyle=$ $\displaystyle m,$

which gives

$n_{1}=\frac{|m|+m}{2}+n,\qquad n_{2}=\frac{|m|-m}{2}+n$ (116)

These relations allow us to identify the single particle excitations in the
bulk given in (112) with the global descendants (113) of the primary. The last
2 columns in table 1 are obtained using (116). To summarize, we identify bulk
single particle excitations of the minimally coupled scalar with the global
descendants of a primary as

$|\psi_{m,n}\rangle\rightarrow|\hat{\Psi}^{(n_{1},n_{2})}\rangle.$ (117)

where $(n_{1},n_{2})$ are related to $(m,n)$ using the relations (116).

There is a simple check we can perform to verify the identification in (117).
The $SL(2,\mathbb{R})$ generators of the global part of the Virasoro algebra
of the CFT can be identified with the isometries of $AdS_{3}$ Maldacena:1998bw
. The vector fields corresponding to the left moving $SL(2,\mathbb{R})$ are
given by

$\displaystyle L_{0}=i\partial_{u},$ $\displaystyle
L_{-1}=ie^{-iu}\left[\frac{\cosh 2\rho}{\sinh
2\rho}\partial_{u}-\frac{1}{\sinh
2\rho}\partial_{v}+\frac{i}{2}\partial_{\rho}\right],$ $\displaystyle
L_{1}=ie^{iu}\left[\frac{\cosh 2\rho}{\sinh 2\rho}\partial_{u}-\frac{1}{\sinh
2\rho}\partial_{v}-\frac{i}{2}\partial_{\rho}\right].$ (118)

Here we have used the co-ordinates

$u=t+\varphi,\qquad v=t-\varphi,\qquad\sinh\rho=r.$ (119)

The right moving isometries are given by the interchange $u\leftrightarrow v$.
Since we have identified the bulk isometries with that of global part of the
Virasoro algebra, the states $|\psi_{m,n}\rangle$ must be related to
$|\psi_{0,0}\rangle$ by the action of the differential operator
$(L_{-1})^{n_{1}}(L_{-1})^{n_{2}}$. Indeed, it is easy to verify that the
state $m=1,n=0$ can be obtained by acting $L_{-1}$ on the state $m=0,n=0$

$\displaystyle
L_{-1}\Big{[}f_{0,0}(r)e^{-i2ht}\Big{]}=\sqrt{2h}f_{1,0}(r)e^{-i(2h+1)t-i\varphi}.$
(120)

Similarly we have the relation

$\displaystyle
L_{-1}^{2}\Big{[}f_{0,0}(r)e^{-i2ht}\Big{]}=\sqrt{(4h)(2h+1)}f_{2,0}e^{-i(2h+2)t-2i\varphi}.$
(121)

Note that the one particle states in the bulk are of unit norm and the
prefactors in (120) and (121) ensure that the normalization agree with that in
the CFT.

Now that we have the mapping between single particle states in the bulk to the
global descendants of the CFT, we can consider linear combinations of these
states and we obtain the dictionary

$\displaystyle|\Phi\rangle$ $\displaystyle\rightarrow$ $\displaystyle
c_{0}|\psi_{0,0}\rangle+\sqrt{2h}c_{1}|\psi_{1,0}\rangle,$ (122)
$\displaystyle|\Upsilon\rangle$ $\displaystyle\rightarrow$ $\displaystyle
c_{0}|\psi_{0,0}\rangle+2hc_{1}|\psi_{0,1}\rangle.$

Here the states $|\Phi\rangle$ and $|\Upsilon\rangle$ are states in the CFT
defined in (103). The relative factors which depend on $h$ are necessary so
that the normalizations agree with that in the CFT. The states
$|\psi_{m,n}\rangle$ are unit normalized.

###  3.2 Backreacted geometry

Once we excite global $AdS_{3}$ by any of the states we discussed in the
previous section, the energy density induced by the excited state back reacts
when $G_{N}$ is non-vanishing and deforms the geometry. The leading
corrections to the entanglement entropy would then be obtained by evaluating
the length of minimal surface in the deformed geometry. At the leading oder in
$G_{N}$ we can evaluate the back reacted geometry by solving the Einsteins
equations with the stress tensor as source. Following the approach in
Belin:2018juv , we evaluate the expectation value of the stress tensor of the
scalar on the excited states in (103). The stress tensor is given by

$T_{\mu\nu}=:\partial_{\mu}\phi\partial_{\nu}\phi-\frac{1}{2}g_{\mu\nu}\left((\nabla\phi)^{2}+m^{2}\phi^{2}\right):.$
(123)

The stress tensor is normal ordered, this ensures that its expectation value
on the vacuum vanishes and therefore it implies that we are not sensitive to
the UV-divergent zero point energy. These UV effects are state independent and
do not affect the calculations. They cancel on considering the entanglement
entropy between excited states and the vacuum.

To evaluate the expectation value $\langle\psi|T_{\mu\nu}|\psi\rangle$ we
substitute the expansion of $\phi$ given in (107) and use the creation
annihilation algebra in (111). This was done for the primary state
$|\psi_{0,0}\rangle$ in Belin:2018juv , here we extend this analysis for the
states listed in (103). We provide the details for the descendant
$|\psi_{0,1}\rangle\rightarrow|\hat{\Psi}^{(1,1)}\rangle$. This state does not
have angular momentum so it makes the analysis very similar to that done for
the primary in Belin:2018juv . The second state we consider in detail is the
linear combination of the primary and the descendant referred to
$c_{0}|\psi_{0,0}\rangle+\sqrt{2h}c_{1}|\psi_{1,0}\rangle\rightarrow|\Phi\rangle$.
This back reacted geometry corresponding to this state is time dependent and
also has angular momentum. The construction of the back reacted geometry is
more involved in this case. The analysis for the rest of the states in the
list (103) is provided in appendix D.

####  First excited state with zero angular momentum: $|\psi_{0,1}\rangle$

The non-zero components of the expectation value of the stress tensor on this
state are given by

$\displaystyle\langle\psi_{0,1}|T_{tt}|\psi_{0,1}\rangle=\frac{2h\left(4(2h-1)hr^{2}\left(hr^{2}-1\right)+2h+1\right)+2}{\pi\left(r^{2}+1\right)^{2h+1}},$
(124)
$\displaystyle\langle\psi_{0,1}|T_{rr}|\psi_{0,1}\rangle=\frac{2h\left(4hr^{2}\left(hr^{2}-1\right)+3\right)+2}{\pi(1+r^{2})^{2h+3}},$
$\displaystyle\langle\psi_{0,1}|T_{\varphi\varphi}|\psi_{0,1}\rangle$
$\displaystyle=\frac{2r^{2}\left(h\left(r^{2}\left(4hr^{2}\left(h(1-2h)r^{2}+7h+1\right)-22h-9\right)+3\right)-r^{2}+1\right)}{\pi\left(r^{2}+1\right)^{2h+3}}.$

As a cross check, we have verified that these components satisfy the
conservation law

$\nabla^{\mu}\langle\psi_{0,1}|T_{\mu\nu}|\psi_{0,1}\rangle=0.$ (125)

The plot of the expectation value of the energy density of this state,
together with the state $|\psi_{0,0}\rangle$ and $|\psi_{0,2}\rangle$ all
carrying zero angular momentum is given in figure 6. Note as the level of the
descendant increases, the energy density is more de-localized and has
increasing number of extrema.

(a)

(b)

(c)

Figure 6:  Expectation value of the energy of the first 3 states with zero
angular momentum for a few values of the weight $h$. (a) Energy density in the
state $|\psi_{0,0}\rangle$. (b) Energy density in the state
$|\psi_{0,1}\rangle$. (c) Energy density in the state $|\psi_{0,1}\rangle$.
Note energy density is more de-localized with increasing number of extrema in
descendants.

The components of the stress tensor in (124) have only radial dependence and
the ones that are non-zero are only the diagonal ones. This is the same
property of the expectation value of the stress tensor in the primary
$|\psi_{0,0}\rangle$, so we can use the same metric ansatz as in Belin:2018juv
which is given by

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle-(r^{2}+G_{1}(r)^{2})dt^{2}+\frac{dr^{2}}{r^{2}+G_{2}(r)^{2}}+r^{2}d\phi^{2},$
(126) $\displaystyle G_{1}(r)$ $\displaystyle=$ $\displaystyle
1+a(r)G_{N},\qquad\qquad G_{2}(r)=1+b(r)G_{N}.$

Plugging this ansatz in the Einstein’s equation

$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R-g_{\mu\nu}=8\pi
G_{N}\langle\psi|T_{\mu\nu}|\psi\rangle\,,$ (127)

we obtain the following differential equations for the functions $a(r),b(r)$
at $O(G_{N})$

$\displaystyle
b^{\prime}(r)=-\frac{8r}{(1+r^{2})^{2h+2}}\Big{[}2+2h(1+2h)+8h^{2}(2h-1)r^{2}(hr^{2}-1)\Big{]},$
(128)
$\displaystyle(1+r^{2})a^{\prime}(r)-2ra(r)+2rb(r)=\frac{8r}{(1+r^{2})^{2h+1}}\Big{[}2+6h+8h^{2}r^{2}(hr^{2}-1)\Big{]}.$

These equations result from considering the $tt$ and the $rr$ components of
the Einstein’s equation respectively. All the other Einstein’s equations are
trivial or trivially satisfied once these equations hold. The solutions for
these equations are given by

$\displaystyle b(r)$
$\displaystyle=A+8\left(4h^{3}r^{4}+2hr^{2}+h+1\right)\left(r^{2}+1\right)^{-2h-1},$
$\displaystyle a(r)$
$\displaystyle=\left(16h\frac{r^{2}}{\left(r^{2}+1\right)^{2h+1}}\right)+A+B(1+r^{2}),$
(129)

where $A$ and $B$ are the constants of integration. It is clear that we need
to set $B=0$ so that the metric asymptotes to $AdS_{3}$ at
$r\rightarrow\infty$. We can fix the constant $A$ by demanding that the stress
tensor evaluated from the bulk using the Fefferman-Graham coordinates agrees
with the expectation value of the stress tensor of the CFT in the state
$|\hat{\Psi}^{(1,1)}\rangle$. The details of the construction of Fefferman-
Graham expansion for the metric in (126) is given in appendix B. From (275),
we obtain the following relation

$\langle\psi_{0,1}|T_{tt}(t,\varphi)|\psi_{0,1}\rangle\Big{|}_{\rm
FG}=\frac{1}{4G_{N}}\left(-\frac{1}{2}-G_{N}A\right).$ (130)

The stress tensor of the CFT on the cylinder is given by

$T_{tt}(t,\varphi)=\sum_{n=-\infty}^{\infty}(L_{n}e^{in(t+\varphi)}+\bar{L}_{n}e^{in(t-\varphi)})-\frac{c+\bar{c}}{24}.$
(131)

The expectation value of the CFT stress tensor

$\displaystyle\frac{\langle\hat{\Psi}^{(1,1)}|T_{tt}(t,\varphi)|\hat{\Psi}^{(1,1)}\rangle}{\langle\hat{\Psi}^{(1,1)}|\hat{\Psi}^{(1,1)}\rangle}$
$\displaystyle=$ $\displaystyle-\frac{c+\bar{c}}{24}+2h+2,$ $\displaystyle=$
$\displaystyle-\frac{c}{12}+2h+2,$

we have used $c=\bar{c}$ to obtain the last line of the above equation. The
Brown-Henneaux formula relates the central charge to the Newton’s constant

$\frac{1}{G_{N}}=\frac{2c}{3}.$ (133)

Then requiring (130 ) and (3.2) to agree and using (133), we obtain

$A=-8(h+1).$ (134)

In Belin:2018juv  a similar constant of integration which occurs for the back
reacted metric corresponding to the state $\psi_{0,0}\rangle$ was fixed by
using the fact that the conical defect of the metric measures the energy of
the particle. We find that the above method is more suitable to generalise to
the situation when the geometry depends on time and is not isometric in the
angular direction. Using (134) and $B=0$, we can write the back reacted metric
for the state $|\psi_{0,1}\rangle$ given in (126)

$\displaystyle G_{1}(r)$ $\displaystyle=$ $\displaystyle
1-8G_{N}(h+1)+\left(16hG_{N}\frac{r^{2}}{\left(r^{2}+1\right)^{2h+1}}\right),$
$\displaystyle G_{2}(r)$ $\displaystyle=$ $\displaystyle
1-8G_{N}(h+1)+8G_{N}\frac{\left(4h^{3}r^{4}+2hr^{2}+h+1\right)}{\left(r^{2}+1\right)^{2h+1}}.$
(135)

####  Time dependent state with non-zero angular momentum:
$c_{0}|\psi_{0,0}\rangle+c_{1}|\psi_{1,0}\rangle$

On evaluating the expectation value of the stress tensor for the state

$|\hat{\phi}\rangle=c_{0}|\psi_{0,0}\rangle+\sqrt{2h}c_{1}|\psi_{1,0}\rangle,$
(136)

we note that all the components of the stress tensor are non-vanishing and
have angular as well as time dependence. They are given by

$\displaystyle\frac{\langle\hat{\phi}|T_{tt}|\hat{\phi}\rangle}{\langle\hat{\phi}|\hat{\phi}\rangle}=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\frac{1}{\pi}\bigg{\\{}2h(2h-1)\left(r^{2}+1\right)^{1-2h}|c_{0}|^{2}+4h^{2}\left(r^{2}+1\right)^{-2h}\left(4h^{2}r^{2}-2hr^{2}+1\right)|c_{1}|^{2}$
$\displaystyle+4h^{2}(2h-1)r\left(r^{2}+1\right)^{\frac{1}{2}-2h}\cos(t+\varphi)\left(c_{1}c_{0}^{*}+c_{0}c_{1}^{*}\right)$
$\displaystyle-4ih^{2}(2h-1)r\left(r^{2}+1\right)^{\frac{1}{2}-2h}\sin(t+\varphi)\left(c_{1}c_{0}^{*}-c_{0}c_{1}^{*}\right)\bigg{\\}},$
$\displaystyle\frac{\langle\hat{\phi}|T_{rr}|\hat{\phi}\rangle}{\langle\hat{\phi}|\hat{\phi}\rangle}=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\frac{1}{\pi}\bigg{\\{}2h\left(r^{2}+1\right)^{-2h-1}|c_{0}|^{2}+8h^{3}r^{2}\left(r^{2}+1\right)^{-2(h+1)}|c_{1}|^{2}$
$\displaystyle+4h^{2}r\left(r^{2}+1\right)^{-2h-\frac{3}{2}}\cos(t+\varphi)\left(c_{1}c_{0}^{*}+c_{0}c_{1}^{*}\right)-4ih^{2}r\left(r^{2}+1\right)^{-2h-\frac{3}{2}}\sin(t+\varphi)\left(c_{1}c_{0}^{*}-c_{0}c_{1}^{*}\right)\bigg{\\}},$
$\displaystyle\frac{\langle\hat{\phi}|T_{\varphi\varphi}|\hat{\phi}\rangle}{\langle\hat{\phi}|\hat{\phi}\rangle}=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\frac{1}{\pi}\bigg{\\{}-\frac{2hr^{2}\left(r^{2}+1\right)^{-2h-1}\left((2h-1)r^{2}-1\right)}{\pi}|c_{0}|^{2}$
$\displaystyle-4h^{2}r^{4}\left(r^{2}+1\right)^{-2(h+1)}\left(4h^{2}r^{2}-2h\left(r^{2}+3\right)-1\right)|c_{1}|^{2}$
$\displaystyle-4h^{2}r^{3}\left(r^{2}+1\right)^{-2h-\frac{3}{2}}\left((2h-1)r^{2}-2\right)\cos(t+\varphi)\left(c_{1}c_{0}^{*}+c_{0}c_{1}^{*}\right)$
$\displaystyle+4ih^{2}r^{3}\left(r^{2}+1\right)^{-2h-\frac{3}{2}}\left((2h-1)r^{2}-2\right)\sin(t+\varphi)\left(c_{1}c_{0}^{*}-c_{0}c_{1}^{*}\right)\bigg{\\}},$
$\displaystyle\frac{\langle\hat{\phi}|T_{t\varphi}|\hat{\phi}\rangle}{\langle\hat{\phi}|\hat{\phi}\rangle}=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\frac{1}{\pi}\bigg{\\{}4h^{2}(2h+1)r^{2}\left(r^{2}+1\right)^{-2h-1}|c_{1}|^{2}+2h^{2}r\left(r^{2}+1\right)^{-2h-\frac{1}{2}}\cos(t+\varphi)\left(c_{1}c_{0}^{*}+c_{0}c_{1}^{*}\right)$
$\displaystyle-2ih^{2}r\left(r^{2}+1\right)^{-2h-\frac{1}{2}}\sin(t+\varphi)\left(c_{1}c_{0}^{*}-c_{0}c_{1}^{*}\right)\bigg{\\}},$
$\displaystyle\frac{\langle\hat{\phi}|T_{rt}|\hat{\phi}\rangle}{\langle\hat{\phi}|\hat{\phi}\rangle}=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\frac{1}{\pi}\bigg{\\{}2h^{2}r^{2}\left(r^{2}+1\right)^{-2h-\frac{3}{2}}\sin(t+\varphi)\left(c_{1}c_{0}^{*}+c_{0}c_{1}^{*}\right)$
$\displaystyle+2ih^{2}r^{2}\left(r^{2}+1\right)^{-2h-\frac{3}{2}}\cos(t+\varphi)\left(c_{1}c_{0}^{*}-c_{0}c_{1}^{*}\right)\bigg{\\}},$
$\displaystyle\frac{\langle\hat{\phi}|T_{r\varphi}|\hat{\phi}\rangle}{\langle\hat{\phi}|\hat{\phi}\rangle}=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\frac{1}{\pi}\bigg{\\{}2h^{2}\left(r^{2}+1\right)^{-2h-\frac{1}{2}}\sin(t+\varphi)\left(c_{1}c_{0}^{*}+c_{0}c_{1}^{*}\right)$
$\displaystyle+2ih^{2}\left(r^{2}+1\right)^{-2h-\frac{1}{2}}\cos(t+\varphi)\left(c_{1}c_{0}^{*}-c_{0}c_{1}^{*}\right)\bigg{\\}}.$
(137)

Just to recall these are evaluated by substituting the mode expansion (107)
into the expression for the stress tensor in (123) and using the algebra of
creation and annihilation operators. We have verified that these components
satisfy the conservation law. Another simple check is that on setting
$c_{1}=0$, observe that stress tensor reduces to that of the primary state
evaluated in Belin:2018juv . Also setting $c_{0}=0$, we see that it reduces to
the stress tensor evaluated in the appendix for the state
$|\psi_{1,0}\rangle$.

It is clear from the expectation values in (3.2), the stress energy is both
time dependent as well spatially in-homogenous. It is interesting to plot the
expectation value of the energy density as a function of the radial distance
for different angles at $t=0$. These plots are given in figure 7.

(a)

(b)

(c)

(d)

(e)

Figure 7: Expectation value of the energy for the state
$|\hat{\phi}\rangle=c_{0}|\psi_{0,0}\rangle+\sqrt{2h}c_{1}|\psi_{1,0}\rangle$
with $c_{0}=c_{1}=1$ and $t=0$. (a) Energy density at $\varphi=0$. (b) Energy
density at $\varphi=\frac{\pi}{2}$. (c) Energy density at $\varphi=\pi$.(d)
Energy density at $\varphi=\frac{3\pi}{2}$. (e) Energy density at
$\varphi=2\pi$.

To solve Einstein’s equation we substitute the stress tensor(3.2) as the
source in (127) and take the following ansatz

$\displaystyle ds^{2}$ $\displaystyle=$
$\displaystyle\Big{[}1+r^{2}+J_{1}(t,r,\varphi)\Big{]}dt^{2}+2J_{2}(t,r,\varphi)drdt+2J_{3}(r)dtd\varphi$
(138)
$\displaystyle\qquad\qquad+\frac{dr^{2}}{1+r^{2}+J_{4}(t,r,\varphi)}+r^{2}d\varphi^{2}.$

Here we have introduced arbitrary functions for the $4$ components. We make a
further ansatz for the dependence of these $4$ components which is given
below.

$\displaystyle
J_{1}(t,r,\varphi)\text{=}\frac{G_{N}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\bigg{\\{}2|c_{0}|^{2}a_{0,0}(r)+2(2h)|c_{1}|^{2}a_{1,0}(r)\bigg{\\}},$
(139) $\displaystyle
J_{2}(t,r,\varphi)\text{=}\frac{G_{N}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\sqrt{2h}\bigg{\\{}[R_{2}(r)(c_{1}c_{0}^{*}+c_{0}c_{1}^{*})\sin(t+\varphi)$
$\displaystyle\qquad\qquad\qquad+i\tilde{R}_{2}(r)(c_{1}c_{0}^{*}-c_{0}c_{1}^{*})\cos(t+\varphi)\bigg{\\}},$
$\displaystyle
J_{3}(t,r,\varphi)\text{=}\frac{G_{N}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\bigg{\\{}|c_{0}|^{2}b_{0,0}(r)+(2h)|c_{1}|^{2}b_{1,0}(r)$
$\displaystyle+\sqrt{2h}[R_{3}(r)(c_{1}c_{0}^{*}+c_{0}c_{1}^{*})\cos(t+\varphi)-i\tilde{R}_{3}(r)(c_{1}c_{0}^{*}-c_{0}c_{1}^{*})\sin(t+\varphi)]\bigg{\\}},$
$\displaystyle
J_{4}(t,r,\varphi)\text{=}\frac{G_{N}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\bigg{\\{}2|c_{0}|^{2}d_{0,0}(r)+2(2h)|c_{1}|^{2}d_{1,0}(r)$
$\displaystyle+2\sqrt{2h}[R_{4}(r)(c_{1}c_{0}^{*}+c_{0}c_{1}^{*})\cos(t+\varphi)-i\tilde{R}_{4}(r)(c_{1}c_{0}^{*}-c_{0}c_{1}^{*})\sin(t+\varphi)]\bigg{\\}}.$

One can arrive at this ansatz by requiring that when either $c_{0}$ or $c_{1}$
is set to zero the metric reduces to the back reacted solution for the primary
$|\psi_{0,0}\rangle$ or the state $|\psi_{1,1}\rangle$ respectively. Therefore
from the results of Belin:2018juv  which is also reviewed in appendix D, we
get

$\displaystyle a_{0,0}(r)$ $\displaystyle=$ $\displaystyle-8h,\qquad
b_{0,0}(r)=0,\qquad d_{0,0}(r)=-8h+\frac{8h}{\left(r^{2}+1\right)^{2h-1}},$
(140)

and (346) we get

$\displaystyle a_{1,0}(r)$ $\displaystyle=$ $\displaystyle
4\left(r^{2}+1\right)^{-2h}-4(2h+1),$ (141) $\displaystyle b_{1,0}(r)$
$\displaystyle=$ $\displaystyle 4\left(r^{2}+1\right)^{-2h},$ $\displaystyle
d_{1,0}(r)$ $\displaystyle=$ $\displaystyle
4\left(4h^{2}r^{2}+2h+1\right)\left(r^{2}+1\right)^{-2h}-4(2h+1).$

The rest of the terms in the ansatz are either proportional to ${\rm
Re}(c_{0}c_{1}^{*})$ or ${\rm Im}(c_{0}c_{1}^{*})$. From the stress tensor in
(3.2), we see that these terms come together with either $\sin(t+\varphi)$ or
$\cos(t+\varphi)$. There is a simplifying case of setting $c_{0}=c_{1}=1$
which switches off the terms proportional to ${\rm Im}(c_{0}c_{1}^{*})$. This
also helps in arriving at the ansatz in (138), (139).

We substitute the metric in the Einstein equation with the stress tensor given
in (3.2) as the source and solve for
$R_{2},\tilde{R}_{2},R_{3},\tilde{R}_{3},R_{4},\tilde{R}_{4}$. Though the
Einstein’s equations seem to over constrain these functions, there is a
consistent solution. The $tt$ component of the Einstein’s equations results in
2 equations, one which arises as the coefficient of ${\rm
Re}\;(c_{0}c_{1}^{*})$ which is given by

$\displaystyle
r(r^{2}+1)R_{4}^{\prime}(r)+R_{4}(r)=-\frac{16(2h-1)h\sqrt{2h}\;r^{3}}{(1+r^{2})^{2h-\frac{1}{2}}}\qquad:tt$
(142)

The equation which arises as the coefficient for ${\rm Im}\;(c_{0}c_{1}^{*})$
results in an identical equation for $\tilde{R}_{4}(r)$,

$\displaystyle
r(r^{2}+1)\tilde{R}_{4}^{\prime}(r)+\tilde{R}_{4}(r)=-\frac{16(2h-1)h\sqrt{2h}\;r^{3}}{(1+r^{2})^{2h-\frac{1}{2}}}\qquad:tt$
(143)

This feature is seen for all the components of the Einstein equations. That
is, the equations which arise as the coefficient of ${\rm
Re}\;(c_{0}c_{1}^{*})$ are equations that determine the functions
$R_{2},R_{3},R_{4}$, while the equations which arise as the coefficient of
${\rm Im}\;(c_{0}c_{1}^{*})$ determine the functions
$\tilde{R}_{2},\tilde{R}_{3},\tilde{R}_{4}$ are identical. Therefore we write
the remaining equations that result from the Einstein equations just from the
${\rm Re}\;(c_{0}c_{1}^{*})$.

$\displaystyle
r(1+r^{2})R_{2}(r)+2r^{2}R_{4}(r)-R_{3}(r)=\frac{16h\sqrt{2h}\;r^{3}}{(1+r^{2})^{2h-\frac{1}{2}}},\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad:rr$
$\displaystyle(1+r^{2})^{2}R_{2}^{\prime}(r)+r(1+r^{2})R_{2}(r)+(1+r^{2})\Big{[}rR_{4}^{\prime}(r)+R_{4}(r)\Big{]}=-\frac{16h\sqrt{2h}\;r[(2h-1)r^{2}-1]}{(1+r^{2})^{2h-\frac{1}{2}}},\quad:\varphi\varphi$
$\displaystyle(1+r^{2})^{2}(rR_{2}^{\prime}(r)-rR_{3}^{\prime\prime}(r)+R_{3}^{\prime}(r))+(1-r^{4})R_{2}(r)-2rR_{4}(r)=\frac{16h\sqrt{2h}\;r^{2}}{(1+r^{2})^{2h-\frac{1}{2}}},\qquad\qquad\quad\quad:t\varphi$
$\displaystyle
r(1+r^{2})R_{3}^{\prime}(r)-2(1+r^{2})R_{3}(r)+r(1+r^{2})R_{2}(r)+2r^{2}R_{4}(r)=\frac{16h\sqrt{2h}\;r^{3}}{(1+r^{2})^{2h-\frac{1}{2}}},\qquad\qquad\quad\quad:r\varphi$
$\displaystyle-(1+r^{2})R_{3}^{\prime}(r)+2r(R_{3}(r)+R_{4}(r))+(1+r^{2})R_{2}(r)=\frac{16h\sqrt{2h}\;r^{2}}{(1+r^{2})^{2h-\frac{1}{2}}}\qquad\qquad\qquad\qquad\qquad:rt$

To emphasize again, the equations for $\tilde{R}_{2},\tilde{R}_{3}$ and
$\tilde{R}_{4}$ are identical to the above equations with the
$R_{2}\rightarrow\tilde{R}_{2},R_{3}\rightarrow\tilde{R}_{3}$ and
$R_{4}\rightarrow\tilde{R}_{4}$. The solutions to these equations are given by

$\displaystyle R_{4}(r)$ $\displaystyle=$
$\displaystyle\frac{A\sqrt{r^{2}+1}}{r}+\frac{4\sqrt{2h}\left(2hr^{2}+1\right)}{r\left(r^{2}+1\right)^{2h-\frac{1}{2}}},$
$\displaystyle R_{2}(r)$ $\displaystyle=$
$\displaystyle\frac{B}{\sqrt{r^{2}+1}}-\frac{8\sqrt{2h}}{\left(r^{2}+1\right)^{2h+\frac{1}{2}}},$
$\displaystyle R_{3}(r)$ $\displaystyle=$
$\displaystyle(2A+B)r\sqrt{1+r^{2}},$ (145)

and

$\displaystyle\tilde{R}_{4}(r)$ $\displaystyle=$
$\displaystyle\frac{\tilde{A}\sqrt{r^{2}+1}}{r}+\frac{4\sqrt{2h}\left(2hr^{2}+1\right)}{r\left(r^{2}+1\right)^{2h-\frac{1}{2}}},$
$\displaystyle\tilde{R}_{2}(r)$ $\displaystyle=$
$\displaystyle\frac{\tilde{B}}{\sqrt{r^{2}+1}}-\frac{8\sqrt{2h}}{\left(r^{2}+1\right)^{2h+\frac{1}{2}}},$
$\displaystyle\tilde{R}_{3}(r)$ $\displaystyle=$
$\displaystyle(2\tilde{A}+\tilde{B})r\sqrt{1+r^{2}}.$ (146)

Observe that the solutions $R$ and $\tilde{R}$ just differ in the constants of
integration. Further more demanding that the metric in (138) asymptotes to
$AdS_{3}$ at $r\rightarrow\infty$, we must have

$2A+B=0,\qquad\qquad 2\tilde{A}+\tilde{B}=0.$ (147)

We just need to fix two constants $A$ and $\tilde{A}$. For this we can appeal
to the same method followed in the case of the state $|\psi_{0,1}\rangle$.
From the construction of the Fefferman-Graham expansion of the metric in (138)
given in the appendix B we see that the expectation value of the boundary
stress tensor in the state (136) is given by (B)

$\displaystyle\langle\hat{\phi}|T^{FG}_{tt}(t,\varphi)|\hat{\phi}\rangle=\frac{1}{4G_{N}}\left[-\frac{1}{2}+\frac{4G_{N}\Big{[}2h|c_{0}|^{2}+2h(2h+1)|c_{1}|^{2}\Big{]}}{|c_{0}|^{2}+2h|c_{1}|^{2}}\right.$
(148)
$\displaystyle\left.-\frac{AG_{N}\sqrt{2h}\Big{[}\big{(}c_{0}c_{1}^{*}+c_{1}c_{0}^{*}\big{)}\cos(t+\varphi)+i\big{(}c_{0}^{*}c_{1}-c_{1}^{*}c_{0}\big{)}\sin(t+\varphi)\Big{]}}{|c_{0}|^{2}+2h|c_{1}|^{2}}\right].$

The expectation value of the stress tensor of the corresponding dual state
$|\Phi\rangle$ in the CFT using (131) is given by

$\displaystyle\frac{\langle\Phi|T_{tt}(t,\varphi)|\Phi\rangle}{\langle\Phi|\Phi\rangle}$
$\displaystyle=$
$\displaystyle-\frac{c}{12}+\frac{2h|c_{0}|^{2}+2h(2h+1)|c_{1}|^{2}}{|c_{0}|^{2}+2h|c_{1}|^{2}}$
$\displaystyle+\frac{2h}{|c_{0}|^{2}+2h|c_{1}|^{2}}\Big{[}(c_{0}^{*}c_{1}+c_{0}c_{1}^{*})\cos(t+\varphi)+i(c_{0}^{*}c_{1}-c_{1}^{*}c_{0})\sin(t+\varphi)\Big{]}.$

Then comparing (148) and (3.2) to agree we find that

$A=-\tilde{A}=-4\sqrt{2h},$ (150)

and then using (147) we get

$B=-\tilde{B}=8\sqrt{2h}.$ (151)

To summarize, the back reacted geometry corresponding to the state (136) is
given by the metric (138) with the functions (139) along with (140), (141) and
with (145) and (146).

The back reacted geometry for the rest of the states in the list (103) are
constructed in the appendix D using the same methods.

###  3.3 Perturbed minimal area

Now that the back reacted geometry is constructed we can use it to evaluate
the entanglement entropy from the bulk using the Ryu-Takayanagi formula. This
is the first term in the FLM formula in (101), the length of the minimal
geodesic between the two end points of the interval on the boundary CFT. This
is shown as the curve $\gamma_{A}$ in figure 5. The curve connects these ends
points at the time $t=0$. Since we are interested in the difference in
entanglement entropy between the ground state and the single particle excited
state, it is sufficient to consider the correction to the minimal length. We
have found the back reacted metric to the leading order in $G_{N}$, therefore
to the leading order the change in minimal area is given by

$\delta A=A[g_{0}+\delta g]-A[g_{0}]$ (152)

Where $A[g]$ refers to the length of the geodesic $\gamma_{A}$ evaluated with
the metric $g$ and $g_{0}$ is the metric of global $AdS_{3}$ and $g_{0}+\delta
g$ is the back reacted metric.

The minimal length in the global $AdS_{3}$ metric $g_{0}$, is given by
minimizing the functional

$A[g_{0}]=2\int_{r_{\min}}^{\infty}dr\sqrt{\frac{1}{r^{2}+1}+r^{2}(\varphi^{\prime}(r))^{2}}\,,$
(153)

where $r_{m}$ is the turning point on the geodesic. The geodesic is determined
by the differential equation

$\displaystyle\varphi^{\prime}(r)=-\frac{r_{m}}{r\sqrt{\left(r^{2}+1\right)\left(r^{2}-r_{m}^{2}\right)}}\qquad:{\rm
Branch\;I},$ (154)
$\displaystyle\varphi^{\prime}(r)=\frac{r_{m}}{r\sqrt{\left(r^{2}+1\right)\left(r^{2}-r_{m}^{2}\right)}}\qquad:{\rm
Branch\;II}$

The two signs for the slopes are allowed due to the fact that there is a
turning point ar $r_{m}$. In Branch I, the angle increases as $r$ decreases
towards $r_{m}$ from infinity and in Branch II, the angle continues to
increase as $r$ increases from $r_{m}$ to infinty. The integration constants
in Branch I part of the geodesic is fixed by demanding that the angle
$\varphi$ is the initial location of the geodesic say $\varphi=0$ at
$r\rightarrow\infty$. In Branch II, the integration constant is fixed by
demanding that the angle agrees with that of Branch I at the turning point, so
that the geodesic is continuous at $r=r_{m}$. We will choose the initial angle
$\varphi=0$ since this is one of the end points of interval we have chosen as
we see from (14). This results in the following equation for $\varphi$

$\displaystyle\varphi$ $\displaystyle=$
$\displaystyle\frac{\pi}{2}-\arctan\left(\frac{r^{2}+\sqrt{(1+r^{2})(r^{2}-r_{m}^{2})}}{r_{m}}\right),\qquad{\rm
Branch\;I}$ $\displaystyle=$
$\displaystyle\frac{\pi}{2}-\arctan\left(\frac{r^{2}-\sqrt{(1+r^{2})(r^{2}-r_{m}^{2})}}{r_{m}}\right),\qquad{\rm
Branch\;II}.$

Let us define

$\pi x=\arctan\frac{1}{r_{m}}={\rm arccot}\;r_{m}\equiv\frac{\theta}{2}.$
(156)

In Branch I, the angle $\varphi$ increases from $0$ to $\pi x$ as $r$ takes
values from $\infty$ to $r_{m}$ and in Branch II, the angle $\varphi$
increases from $\pi x$ to $2\pi x$. Therefore the total size of the interval
is $2\pi x=\theta$. We need to be careful about obtaining the curves precisely
and locating the end points of the geodesic at $\varphi=0$ and
$\varphi=\theta$ because we will need to deal with the situation when the
perturbed metric breaks the isometry in the $\varphi$ direction and therefore
the answer will dependent on the location of the end points and not just the
opening angle. We have chosen the end points of the geodesic so as to agree
with the choice we made in (14) for the end points of the interval in the CFT.

Since we are interested in the correction to the leading order in $G_{N}$ as
given in (152), it is easy to see that the geodesic is unchanged at this order
since it is the minimal curve and change in the area at order $G_{N}$ is
obtained by substituting the shifted metric and integrating it along the
geodesic. Let the coefficient of $g_{rr}$ be given as in (3.3)

$g_{rr}=\frac{1}{1+r^{2}+J_{4}(t,r,\varphi)}.$ (157)

This form of this metric coefficient is general and it also takes care of the
situation when the change in metric is independent of the angle or time. Note
that the the coefficient $g_{\varphi\varphi}$ and $g_{r\varphi}$ remains
unaltered in the back reacted metrics (126), (138). Therefore the shift in the
minimal area is given by

$\delta
A=-\frac{1}{2}\int_{\gamma_{A}}drJ_{4}(t=0,r,\varphi)\frac{(r^{2}-r_{\rm
min}^{2})^{\frac{1}{2}}}{r(1+r^{2})^{\frac{3}{2}}},$ (158)

where the integral is taken along the geodesic curve given in (3.3). We set
$t=0$ since we have chosen this time slice to evaluate the entanglement
entropy.

#### Area shift for the state $|\psi_{0,1}\rangle$

Let us now apply this to the back reacted geometry corresponding to the states
$|\psi_{0,1}\rangle$. Reading out the perturbed metric from (126) with (3.2),
(134) and applying in the expression (158), we are led to evaluate the
integral

$\displaystyle\delta A_{|\psi_{0,1}\rangle}=2G_{N}\int_{r_{\rm
min}}^{\infty}dr\left[8(h+1)-8\frac{(4h^{3}r^{4}+2hr^{2}+h+1)}{(r^{2}+1)^{2h+1}}\right]\frac{\sqrt{r^{2}-r_{m}}}{r(1+r^{2})^{\frac{3}{2}}}.$
(159)

Note that here the perturbed metric is independent of the angle $\varphi$.
Hence to obatin the total minimal area we have just doubled the contribution
from Branch II.

$\displaystyle\frac{\delta A_{|\psi_{0,1}\rangle}}{4G_{N}}$ $\displaystyle=$
$\displaystyle 4(h+1)(1-r_{m}{\rm arccot}\;r_{m})$
$\displaystyle-8h^{3}r_{m}^{-4h}\frac{\Gamma(2h)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2}\big{)}}{}_{2}F_{1}\Big{(}2h+\frac{5}{2},2h,2h+\frac{3}{2},-\frac{1}{r_{m}^{2}}\Big{)}$
$\displaystyle-4hr_{m}^{-4h-2}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{5}{2})}{}_{2}F_{1}\Big{(}2h+\frac{5}{2},2h+1,2h+\frac{5}{2},-\frac{1}{r_{m}^{2}}\Big{)}$
$\displaystyle-2(h+1)r_{m}^{-4h-4}\frac{\Gamma(2h+2)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{7}{2})}{}_{2}F_{1}\Big{(}2h+\frac{5}{2},2h+2,2h+\frac{7}{2},-\frac{1}{r_{m}^{2}}\Big{)}.$

Substituting for $r_{m}$ in terms of the interval length $\pi x$ from (156) ,
we see from the above expression, that there are non-analytic terms in the
expansion in terms of $\pi x$, keeping the leading term we obtain

$\displaystyle\frac{\delta A_{|\psi_{0,1}\rangle}}{4G_{N}}=4(h+1)(1-\pi
x\cot\pi
x)-(2h)^{2}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}(\pi
x)^{4h}+\cdots.$ (161)

On comparing this expression with the short distance expansion for the
entanglement entropy evaluated for the corresponding dual state in CFT given
in (2.3) with $l=1$, we see that the leading term agrees. However in the CFT
the leading non-analytical term starts at $(\pi x)^{8h}$ as opposed to $(\pi
x)^{4h}$ in the minimal area. We will see in the next section, that the $(\pi
x)^{4h}$ term is cancelled by the leading contribution form the bulk
entanglement entropy $S_{\rm bulk}(\Sigma_{A})$ and the next order term will
precisely match with the CFT result. This phenomenon was also observed for the
primary state $|\psi_{0,0}\rangle$ in the analysis of Belin:2018juv .

#### Area shift for the state $|\hat{\phi}\rangle$, with angular momentum

The perturbed metric corresponding to the state $|\hat{\phi}\rangle$ defined
in (136) is given in (138) with correction to the metric coefficient $g_{rr}$
given in (139). Substituting this correction in the expression for the shifted
area (158), we obtain

$\displaystyle\frac{\delta{A_{|\hat{\phi}\rangle}}}{G_{N}}=$
$\displaystyle\frac{1}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\bigg{\\{}-2|c_{0}|^{2}\int_{r_{m}}^{\infty}dr\frac{d_{0,0}(r)}{(1+r^{2})^{3/2}}\sqrt{1-\frac{r_{m}^{2}}{r^{2}}}$
$\displaystyle-$ $\displaystyle
4h|c_{1}|^{2}\int_{r_{m}}^{\infty}dr\frac{d_{1,0}(r)}{(1+r^{2})^{3/2}}\sqrt{1-\frac{r_{m}^{2}}{r^{2}}}$
$\displaystyle-\sqrt{2h}(c_{1}c_{0}^{*}+c_{0}c_{1}^{*})\int_{\gamma_{A}}r\frac{R_{4}(r)}{(1+r^{2})^{3/2}}\sqrt{1-\frac{r_{m}^{2}}{r^{2}}}\cos\varphi$
$\displaystyle+i\sqrt{2h}(c_{1}c_{0}^{*}-c_{0}c_{1}^{*})\int_{\gamma_{A}}dr\frac{\tilde{R}_{4}(r)}{(1+r^{2})^{3/2}}\sqrt{1-\frac{r_{m}^{2}}{r^{2}}}\sin\varphi\bigg{\\}},$
(162)
$\displaystyle\equiv\frac{\delta{A}_{1}}{G_{N}}+\frac{\delta{A}_{2}}{G_{N}}+\frac{\delta{A}_{3}}{G_{N}}+\frac{\delta{A}_{4}}{G_{N}}.$

Note that for the first 2 terms we have doubled the contribution from the
branche II since the integrand does not depend on the angle. For the last 2
terms we have to substitute the value of the angle $\varphi$ given in (3.3)
along the geodesic and perform the integral.

Let us proceed with the integrals for each of the 4 terms. Substituting the
expression for $d_{0,0}(r)$ from (140) for $\delta A_{1}$, we obtain

$\displaystyle\frac{\delta A_{1}}{4G_{N}}$ $\displaystyle=$
$\displaystyle\frac{|c_{0}|^{2}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\int_{r_{m}}^{\infty}dr\Big{(}4h-\frac{4h}{(1+r^{2})^{2h-1}}\Big{)}\frac{\sqrt{r^{2}-r_{m}^{2}}}{r(r^{2}+1)^{\frac{3}{2}}},$
$\displaystyle=$
$\displaystyle\frac{|c_{0}|^{2}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\left[4h(1-r_{m}{\rm
arccot}\;{r_{m}})-\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}{}_{2}F_{1}\Big{(}2h+\frac{1}{2},2h,2h+\frac{3}{2};-\frac{1}{r_{m}^{2}}\Big{)}\right].$

On keeping the leading term in the non-analytical power of the interval $\pi
x$, we obtain

$\displaystyle\frac{\delta
A_{1}}{4G_{N}}=\frac{|c_{0}|^{2}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\left[4h(1-\pi
x\cot\pi x)-\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}(\pi
x)^{4h}\right]+\cdots.$ (164)

Using $d_{1,0}(r)$ in (141), we get

$\displaystyle\frac{\delta A_{2}}{4G_{N}}$ $\displaystyle=$
$\displaystyle\frac{h|c_{1}|^{2}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\int_{r_{m}}^{\infty}dr\Big{(}4(2h+1)-4\frac{4h^{2}r^{2}+2h+1}{(r^{2}+1)^{2h}}\Big{)}\frac{\sqrt{r^{2}-r_{m}^{2}}}{r(r^{2}+1)^{\frac{3}{2}}},$
$\displaystyle=$
$\displaystyle\frac{|c_{1}|^{2}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\left[4h(2h+1)(1-r_{m}\arctan(\frac{1}{r_{m}}))\right.$
$\displaystyle-(2h)^{2}r_{m}^{-4h}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}{}_{2}F_{1}\Big{(}2h+\frac{3}{2},2h,2h+\frac{3}{2};-\frac{1}{r_{m}^{2}}\Big{)}$
$\displaystyle\left.\qquad-2h(2h+1)r_{m}^{-4h-2}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{5}{2})}{}_{2}F_{1}\Big{(}2h+\frac{3}{2},2h+1,2h+\frac{5}{2};-\frac{1}{r_{m}^{2}}\Big{)}\right].$

We can again keep only the leading term in the non-analytic power of the
interval length which results in

$\displaystyle\frac{\delta A_{2}}{4G_{N}}$ $\displaystyle=$
$\displaystyle\frac{|c_{1}|^{2}}{(|c_{0}|^{2}+2h|c_{1}|^{2})}\left[4h(2h+1)(1-\pi
x\cot\pi
x)-(2h)^{2}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}(\pi
x)^{4h}\right]+\cdots.$

Let us now proceed to evaluate the term $\delta A_{3}$ in (3.3) which depends
on the angular position along the geodesic. For this we need to solve for
$\cos\varphi$ in terms of the radial position $r$ along the 2 branches using
(3.3). After some elementary manipulations we obtain

$\displaystyle\cos\varphi=\frac{r_{m}^{2}\sqrt{1+r^{2}}+\sqrt{r^{2}-r_{m}^{2}}}{r(1+r_{m}^{2})},\qquad{\rm
Branch\;I},$ (167)
$\displaystyle\cos\varphi=\frac{r_{m}^{2}\sqrt{1+r^{2}}-\sqrt{r^{2}-r_{m}^{2}}}{r(1+r_{m}^{2})},\qquad{\rm
Branch\;II}.$

Substituting this along with the expression for $R_{4}(r)$ from (145), we
obtain

$\displaystyle\frac{\delta A_{3}}{4G_{N}}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{h}(c_{1}c_{0}^{*}+c_{1}^{*}c_{0})}{\sqrt{2}(|c_{0}|^{2}+2h|c_{1}|^{2})}\int_{r_{m}}^{\infty}dr\frac{r_{m}^{2}\sqrt{(1+r^{2}}}{r(1+r_{m}^{2})}\left(\frac{4\sqrt{2h(r^{2}+1)}}{r}-\frac{4\sqrt{2h}(2hr^{2}+1)}{r(1+r^{2})^{2h-\frac{1}{2}}}\right)\frac{\sqrt{r^{2}-r_{m}^{2})}}{r(1+r^{2})^{\frac{3}{2}}}.$

Here we have used the value of $A$ from (150). We have also taken care of the
orientation of the Branch I with respect to Branch II to add the contribution
from the 2 branches and write it as one integral from $r_{m}$ to $\infty$.
Performing the integral we obtain

$\displaystyle\frac{\delta A_{3}}{4G_{N}}$ $\displaystyle=$
$\displaystyle-\frac{\sqrt{h}(c_{1}c_{0}^{*}+c_{1}^{*}c_{0})}{\sqrt{2}(|c_{0}|^{2}+2h|c_{1}|^{2})}\left[2\sqrt{2h}r_{m}(\frac{r_{m}}{1+r_{m}^{2}}+{\rm
arccot}\;r_{m}))\right.$
$\displaystyle-2\sqrt{2h}\frac{r_{m}^{2}}{(1+r_{m}^{2})}r_{m}^{-4h}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}{}_{2}F_{1}(2h+\frac{1}{2},2h,2h+\frac{3}{2};-\frac{1}{r_{m}^{2}})$
$\displaystyle-\left.2\sqrt{2h}\frac{r_{m}^{2}}{(1+r_{m}^{2})}r_{m}^{-4h-2}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{5}{2})}{}_{2}F_{1}(2h+\frac{1}{2},2h+1,2h+\frac{5}{2};-\frac{1}{r_{m}^{2}})\right].$

Keeping the leading non-analytical power in the short interval expansion
results in

$\displaystyle\frac{\delta
A_{3}}{4G_{N}}=\frac{c_{1}c_{0}^{*}+c_{1}^{*}c_{0}}{|c_{0}|^{2}+2h|c_{1}|^{2}}\left(h\cot\pi
x(2\pi x-\sin 2\pi
x)-2h\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}(\pi
x)^{4h}\right)+\cdots.$

Finally we evaluate the term $\delta A_{4}$ in (3.3). For this we need the
value of $\sin\varphi$ in both the branches of the minimal curve. Substituting
the expression for $\varphi\phi$ in the branches from (3.3) and after standard
manipulations we obtain

$\displaystyle\sin\varphi=\frac{r_{m}(\sqrt{1+r^{2}}-\sqrt{r^{2}-r_{m}^{2}})}{r(1+r_{m}^{2})},\qquad{\rm
Branch\;I},$ (171)
$\displaystyle\sin\varphi=\frac{r_{m}(\sqrt{1+r^{2}}+\sqrt{r^{2}-r_{m}^{2}})}{r(1+r_{m}^{2})},\qquad{\rm
Branch\;II}.$

Then from (146), (150) and the definition of $\delta_{4}A$ we get

$\displaystyle\frac{\delta A_{4}}{4G_{N}}$ $\displaystyle=$ $\displaystyle
i\frac{\sqrt{h}(c_{1}c_{0}^{*}-c_{1}^{*}c_{0})}{\sqrt{2}(|c_{0}|^{2}+2h|c_{1}|^{2})}\int_{r_{m}}^{\infty}dr\frac{r_{m}\sqrt{(1+r^{2}}}{r(1+r_{m}^{2})}\left(\frac{4\sqrt{2h(r^{2}+1)}}{r}+\frac{4\sqrt{2h}(2hr^{2}+1)}{r(1+r^{2})^{2h-\frac{1}{2}}}\right)\frac{\sqrt{r^{2}-r_{m}^{2})}}{r(1+r^{2})^{\frac{3}{2}}}.$

Here again we have added the contribution along the 2 branches of the geodesic
taking into account its orientation. Performing the integral we obtain

$\displaystyle\frac{\delta A_{4}}{4G_{N}}$ $\displaystyle=$ $\displaystyle
i\frac{\sqrt{h}(c_{1}c_{0}^{*}-c_{1}^{*}c_{0})}{\sqrt{2}(|c_{0}|^{2}+2h|c_{1}|^{2})}\left[2\sqrt{2h}(-\frac{r_{m}}{1+r_{m}^{2}}+{\rm
arccot}\;r_{m}))\right.$
$\displaystyle+2\sqrt{2h}\frac{r_{m}}{(1+r_{m}^{2})}r_{m}^{-4h}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}{}_{2}F_{1}(2h+\frac{1}{2},2h,2h+\frac{3}{2};-\frac{1}{r_{m}^{2}})$
$\displaystyle+\left.2\sqrt{2h}\frac{r_{m}}{(1+r_{m}^{2})}r_{m}^{-4h-2}\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{5}{2})}{}_{2}F_{1}(2h+\frac{1}{2},2h+1,2h+\frac{5}{2};-\frac{1}{r_{m}^{2}})\right].$

We retain the leading contribution of the non-analytical terms in the short
interval expansion to get

$\displaystyle\frac{\delta
A_{4}}{4G_{N}}=i\frac{c_{1}^{*}c_{0}-c_{1}c_{0}^{*}}{|c_{0}|^{2}+2h|c_{1}|^{2}}\left(h(2\pi
x-\sin 2\pi
x)-2h\frac{\Gamma(2h+1)\Gamma(\frac{3}{2})}{\Gamma(2h+\frac{3}{2})}(\pi
x)^{4h+1}\right)+\cdots.$

Summing up the contributions of all the minimal area corrections from (164),
(3.3), (3.3) and (3.3) we get

$\displaystyle\frac{\delta A}{4G_{N}}$ $\displaystyle=$
$\displaystyle\frac{\delta A_{1}}{4G_{N}}+\frac{\delta
A_{2}}{4G_{N}}+\frac{\delta A_{3}}{4G_{N}}+\frac{\delta A_{4}}{4G_{N}},$ (175)
$\displaystyle=$
$\displaystyle\frac{1}{|c_{0}|^{2}+2h|c_{1}|^{2}}\times\Bigg{(}\Big{[}4h|c_{0}|^{2}+4h(2h+1)|c_{1}|^{2}\Big{]}(1-\pi
x\cot\pi x)$ $\displaystyle+h\Big{[}(c_{0}c_{1}^{*}+c_{1}c_{0}^{*})\cot\pi
x+i(c_{0}c_{1}^{*}-c_{1}c_{0}^{*})\Big{]}(2\pi x-\sin 2\pi x)\Bigg{)}$
$\displaystyle-\frac{\Gamma(\frac{3}{2})\Gamma(2h+1)}{\Gamma(2h+\frac{3}{2})}\frac{(\pi
x)^{4h}}{(c_{0}|^{2}+2h|c_{1}|^{2})}\times\Big{(}|c_{0}|^{2}+2h(c_{0}c_{1}^{*}+c_{0}^{*}c_{1})+(2h)^{2}|c_{1}|^{2}\Big{)}$
$\displaystyle+\cdots$

On comparing the leading correction, which consists of analytical terms in the
short distance expansion to the expression obtained from the CFT analysis in
(2.5) to the first 2 lines of the above equation we see that they precisely
agree. We would like to emphasize that the single particle state
$|\hat{\phi}\rangle$ in (136) breaks the isometry both in time and angular
directions of global $AdS_{3}$. The evaluation of the corrections to minimal
area involved integrating corrections to the $g_{rr}$ component of the metric
which break the angular isometry and the agreement of the leading term is a
non-trivial check of the back reacted geometry. In the next section we will
show that the sub-leading non-analytical term in the interval length, these
are the terms in the last line of (175) cancel with a contribution form the
bulk entanglement entropy and the next leading terms are proportional to $(\pi
x)^{8h}$ which will agree with the CFT analysis in (2.5).

###  3.4 Bulk entanglement entropy

In this section we will follow the method introduced in Belin:2018juv  to
evaluate the bulk entanglement entropy $S_{\rm bulk}(\Sigma_{A})$ of the
region $\Sigma_{A}$. This is the contribution to the holographic entanglement
entropy thinking of the bulk as an effective field theory. For our purposes we
can restrict ourselves to the scalar field. Then if $|\psi\rangle$ is the
single particle state of interest, then the reduced density matrix is given by

$\rho_{\rm bulk}={\rm
Tr}_{\bar{\Sigma}_{A}}\Big{(}|\psi\rangle\langle\psi|\Big{)}.$ (176)

where $\bar{\Sigma}_{A}$ is the region outside the entanglement wedge as
indicated in figure 5. Then from this reduced density matrix, one can evaluate
the entanglement entropy using the Von-Neumann’s formula. Since this involves
a trace over a region in curved space, it is non-trivial to perform. It is
convenient to use the co-ordinate system in which the minimal surface
$\gamma_{A}$ in global $AdS_{3}$ is mapped to a horizon in the new geometry
Casini:2011kv . The map arises by parametrising the $AdS_{3}$ hyperboloid in 2
different ways as follows

$\displaystyle X_{0}$ $\displaystyle=$ $\displaystyle\sqrt{r^{2}+1}\sin t,$
$\displaystyle=$ $\displaystyle\sqrt{\rho^{2}-1}\sinh\tau,$ $\displaystyle
X_{1}$ $\displaystyle=$ $\displaystyle\sqrt{r^{2}+1}\cos t,$ $\displaystyle=$
$\displaystyle\rho\cosh\eta\cosh x+\sqrt{\rho^{2}-1}\sinh\eta\cosh\tau,$
$\displaystyle X_{2}$ $\displaystyle=$ $\displaystyle
r\sin(\varphi-\frac{\theta}{2}),$ $\displaystyle=$ $\displaystyle\rho\sinh x,$
$\displaystyle X_{3}$ $\displaystyle=$ $\displaystyle
r\cos(\varphi-\frac{\theta}{2}),$ $\displaystyle=$
$\displaystyle\cosh\eta\sqrt{\rho^{2}-1}\cosh\tau+\rho\sinh\eta\cosh x,$

Here $\theta$ is defined in (156) and the $AdS_{3}$ hyperboloid is defined by

$-X_{0}^{2}-X_{1}^{2}+X_{2}^{2}+X_{3}^{3}=-1.$ (178)

It is easy to see that the metric in the $(\rho,\tau,x)$ co-ordinates reduces
to

$ds^{2}=-(\rho^{2}-1)d\tau^{2}+\frac{d\rho^{2}}{\rho^{2}-1}+\rho^{2}dx^{2},\qquad
x\in\mathbb{R}.$ (179)

From the parametrization in (3.4), we see that $\tau$ is periodic in the
imaginary directions $\tau\sim\tau+2\pi i$ . The equations in (3.4) also
define the transformation from the global $AdS_{3}$ co-orodinates
$(r,t,\varphi)$ to the Rindler BTZ coordinates $(\rho,\tau,x)$ which is given
by

$\displaystyle r$ $\displaystyle=$
$\displaystyle\sqrt{\rho^{2}\sinh^{2}x+\left(\sqrt{\rho^{2}-1}\cosh\eta\cosh\tau+\rho\cosh
x\sinh\eta\right)^{2}}\,,$ $\displaystyle t$ $\displaystyle=$
$\displaystyle\arctan\left(\frac{\sinh\tau\sqrt{\rho^{2}-1}}{\rho\cosh
x\cosh\eta+\sqrt{\rho^{2}-1}\cosh\tau\sinh\eta}\right)\,,$
$\displaystyle\varphi-\frac{\theta}{2}$ $\displaystyle=$
$\displaystyle\arctan\left(\frac{\rho\sinh
x}{\sqrt{\rho^{2}-1}\cosh\tau\cosh\eta+\rho\cosh x\sinh\eta}\right)\,.$ (180)

The usefulness of the Rindler co-ordinates is that the horizon at $t=0,\rho=1$
is the image of the Ryu-Takayanagi minimal surface in global $AdS_{3}$ once we
make the identification

$\cosh\eta=\frac{1}{\sin\frac{\theta}{2}}=\sqrt{r_{m}^{2}+1}.$ (181)

From (3.4) we see that the horizon is a surface in the co-ordinates
$r,\varphi$ given by

$r^{2}=\sinh^{2}x+\cosh^{2}x\sinh^{2}\eta,\qquad\tan(\varphi-\frac{\theta}{2})=\frac{\sinh
x}{\cosh x\sinh\eta}.$ (182)

Eliminating $x$ from the above equations and solving $\varphi$ in terms of
$r$, we obtain 2 branches

$\displaystyle\tan(\varphi-\frac{\theta}{2})=-\frac{\sqrt{r^{2}-r_{m}^{2}}}{r_{m}(\sqrt{1+r^{2}})},\qquad{\rm
Branch\;I},$ (183)
$\displaystyle\tan(\varphi-\frac{\theta}{2})=\frac{\sqrt{r^{2}-r_{m}^{2}}}{r_{m}(\sqrt{1+r^{2}})},\qquad{\rm
Branch\;II}.$

Using standard manipulations, we can write these equations as

$\displaystyle\tan\varphi=\frac{r_{m}(\sqrt{1+r^{2}}-\sqrt{r^{2}-r_{m}^{2}})}{r_{m}^{2}\sqrt{1+r^{2}}+\sqrt{r^{2}-r_{m}^{2}}},\qquad{\rm
Branch\;I},$ (184)
$\displaystyle\tan\varphi=\frac{r_{m}(\sqrt{1+r^{2}}+\sqrt{r^{2}-r_{m}^{2}})}{r_{m}^{2}\sqrt{1+r^{2}}-\sqrt{r^{2}-r_{m}^{2}}},\qquad{\rm
Branch\;II}.$

We see that the above equations precisely agree with the parametrization of
the minimal surface in (167) and (171), This concludes our proof that the
horizon of the Rindler BTZ metric in (179) coincides with the Ryu-Takanayagi
surface using the co-ordinate transformation (3.4).

The Rindler BTZ coordinates (3.4) does not cover the entire global $AdS_{3}$,
this can be seen easily by the fact the range of $\varphi$ is restricted once
one chooses a definite branch of $\arctan$. The modes of the single particle
states have support in both the branches, but one of the branches is to the
left of the horizon in the Penrose diagram just as in the Rindler patch
discription of flat Minkowski space. Therefore the left part of the Rindler
BTZ corresponds to the region $\bar{\Sigma}_{A}$ in pure $AdS_{3}$, while the
right Rindler wedge corresponds to the region $\Sigma_{A}$. Tracing over the
region in the left wedge corresponds to tracing over the region
$\bar{\Sigma}_{A}$. Just as the transformation of the global Minkowski vacuum
to the Rindler space, the global $AdS_{3}$ vacuum can be written as a
thermofield double state in the Rindler left and right vacua

$\displaystyle|0\rangle=\sum_{n}e^{-\frac{2\pi
E_{n}}{2}}|n^{*}\rangle_{L}|n\rangle_{R}.$ (185)

Here $L,R$ refers to the left and right Rindler Hilbert spaces, the
temperature of the thermofield double is decided from the temperature of the
Rindler BTZ geometry (179). From the geometry we see that the inverse
temperature $\beta=2\pi$. The label for the states $n$ is continuous, which
implies the sum over $n$ is an integral. The state $|n^{*}\rangle_{L}$ is the
CPT conjugate of the corresponding state on the right Rindler patch.

To obtain the states corresponding to the single particle excitations in the
Rindler BTZ, we expand the scalar field in these coordinates as

$\displaystyle\phi(\rho,\tau,x)=\sum_{I\in L,R}\int_{\omega>0}\frac{d\omega
dk}{4\pi^{2}}\Big{(}e^{-i\omega\tau}b_{\omega,k,I}(\rho,x)g_{\omega,k,I}(\rho,x)+e^{i\omega\tau}b^{\dagger}_{\omega,k,I}(\rho,x)g^{*}_{\omega,k,I}(\rho,x)\Big{)}.$

Here $g_{\omega,k,I}(\rho,x)$ are the left and right eigen modes of the
Laplacian in Rindler space which are given by Hamilton:2006az

$\displaystyle g_{\omega,k,I}$ $\displaystyle=$ $\displaystyle
N_{\omega,k}e^{ikx}\rho^{-2h}\Big{(}1-\frac{1}{\rho^{2}}\Big{)}^{-\frac{i\omega}{2}}{}_{2}F_{1}\Big{[}h-\frac{i(\omega+k)}{2},h-\frac{i(\omega-k)}{2},2h;\frac{1}{\rho^{2}}\Big{]}$
$\displaystyle{\rm where}\quad I\in\\{L,R\\}.$

The normalization constant is given by

$N_{\omega,k}=\left[\frac{\Gamma\Big{(}h+\frac{i}{2}(\omega+k)\Big{)}\Gamma\Big{(}h-\frac{i}{2}(\omega+k)\Big{)}\Gamma\Big{(}h+\frac{i}{2}(\omega-k)\Big{)}\Gamma\Big{(}h-\frac{i}{2}(\omega-k)\Big{)}}{2\omega\Gamma(2h)^{2}\Gamma(i\omega)\Gamma(-i\omega)}\right]^{\frac{1}{2}}.$
(188)

This normalization is fixed by demanding that the creation and annihilation
operators obey the commutation relations Almheiri:2014lwa  777One way to fix
the normalization is examine the equal time canonical commutation relation
$[\phi(\rho,\tau,x),\partial_{0}(\rho^{\prime},\tau,x^{\prime})]$ close to the
horizon $\rho,\rho^{\prime}\sim 1$.

$[b_{\omega,k,I},b^{\dagger}_{\omega^{\prime},k^{\prime},I^{\prime}}]=4\pi^{2}\delta(\omega-\omega^{\prime})\delta(k-k^{\prime})\delta_{II^{\prime}}.$
(189)

Let us now recall some known properties of the global $AdS_{3}$ vacuum and the
relations between the creation annihilation operators in the two co-ordinate
systems. Since the $AdS_{3}$ vacuum is written in terms of thermofield double
as given in (185), the trace over $\bar{\Sigma}_{A}$ or the left Rindler wedge
leads to the thermal state

$\rho_{0}={\rm Tr}_{\bar{\Sigma}_{A}}\Big{(}|0\rangle\langle
0|\Big{)}=e^{-2\pi H_{R}}.$ (190)

here $\hat{H}_{R}$ is the Hamiltonian of the single particle excitations in
the right wedge given by

$H_{R}=\sum_{\omega,k}w~{}b_{\omega,k,R}^{\dagger}b_{\omega,k,R}.$ (191)

Furthermore the thermofield double satisfies the usual properties

$\displaystyle
b_{\omega,k,L}|0\rangle=e^{-\pi\omega}b^{\dagger}_{\omega,-k,R}|0\rangle,\qquad
b^{\dagger}_{\omega,k,L}|0\rangle=e^{\pi\omega}b_{\omega,-k,R}|0\rangle,$
(192)

From the fact that the field $\phi$ can be expanded in modes in either the
Rindler coordinates or in the global $AdS_{3}$ coordinates, we must have the
relation

$\displaystyle
a_{m,n}=\sum_{I,\omega,k}(\alpha_{m,n;\omega,k,I}b_{\omega,k,I}+\beta_{m,n;\alpha,k,l}b^{\dagger}_{\omega,k,I}),$
(193)

where $\alpha_{m,n;\omega,k,I}$, $\alpha_{m,n;\omega,k,I}$ are the Bogoliubov
coefficients relating the creation and annihilation operators in the two
coordinates. Here the we use the notation

$\sum_{\omega}=\int_{0}^{\infty}\frac{d\omega}{2\pi},\qquad\sum_{k}=\int_{-\infty}^{\infty}\frac{dk}{2\pi}.$
(194)

Since the global vacuum is annihilated by the operators $a_{m,n}$, we see that
Bogoliubov coefficients must satisfy the relations

$\alpha_{\omega,k,L}=-e^{\pi\omega}\beta^{*}_{\omega,-k,R},\qquad\beta^{*}_{\omega,k,L}=-e^{-\pi\omega}\alpha_{\omega,-k,R}.$
(195)

The canonical commutation relations (111) together with (193) results in the
following property of the Bogoliubov coefficients

$\sum_{I,\omega,k}\big{(}\alpha_{m,n;\omega,k,I}\alpha^{*}_{m^{\prime},n^{\prime};\omega,k,I}-\beta^{*}_{m,n;\omega,k,I}\beta_{m^{\prime},n^{\prime};\omega,k,I}\big{)}=\delta_{m,m^{\prime}}\delta_{n,n^{\prime}}.$
(196)

This constraint on the Bogoliubov coefficients can be re-written in terms of
only the coefficients in the right patch using (195)

$\sum_{\omega,k}\big{[}\alpha_{m,n;\omega,k,R}\alpha^{*}_{m^{\prime},n^{\prime};\omega,k,R}(1-e^{-2\pi\omega})-\beta_{m,n;\omega,k,R}\beta^{*}_{m^{\prime},n^{\prime};\omega,k,R}(1-e^{2\pi\omega})\big{]}=\delta_{m,m^{\prime}}\delta_{n,n^{\prime}}.$
(197)

Finally, using (193) and (195) we can write the single particle state in
global $AdS_{3}$ as an excitation involving only the right sector

$|\psi_{m,n}\rangle=\sum_{\omega,k}\left[(1-e^{-2\pi\omega})\alpha_{m,n;\omega,k,R}^{*}b^{\dagger}_{\omega,k,R}+(1-e^{2\pi\omega})\beta_{m,n;\omega,k,R}b_{\omega,k,R}\right]|0\rangle.$
(198)

This is the key relation which allows to perform the trace over the left
sector or $\bar{\Sigma}_{A}$ easily. The trace over the left sector results in
a thermal vacuum over which we have single particle excitation.

We can now provide the general formula for the bulk entanglement entropy of
excited states in terms of a density matrix written in BTZ Rindler modes. Let
us consider the following single particle state

$\displaystyle|\psi\rangle$ $\displaystyle=$
$\displaystyle\sum_{m,n}B_{m,n}a^{\dagger}_{mn}|0\rangle,$ $\displaystyle=$
$\displaystyle\sum_{m,n,\omega,k}B_{m,n}\left[(1-e^{-2\pi\omega})\alpha_{m,n;\omega,k,R}^{*}b^{\dagger}_{\omega,k,R}+(1-e^{2\pi\omega})\beta_{m,n;\omega,k,R}b_{\omega,k,R}\right]|0\rangle.$

Here the coefficients $B_{m,n}$ are such that the state is normalised to unity

$\displaystyle\sum_{mn}|B_{m,n}|^{2}=1.$ (200)

All the operators in (3.4) belong to the right sector, from now on we ignore
the subscript $R$, writing the density matrix of the state $|\psi\rangle$, we
have

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle\sum_{m,n,\omega,k}B_{m,n}\left[(1-e^{-2\pi\omega})\alpha_{m,n;\omega,k}^{*}b^{\dagger}_{\omega,k}+(1-e^{2\pi\omega})\beta_{m,n;\omega,k}b_{\omega,k}\right]|0\rangle\langle
0|$
$\displaystyle\times\sum_{m^{\prime},n^{\prime},\omega^{\prime},k^{\prime}}B_{m^{\prime},n^{\prime}}^{*}\left[(1-e^{-2\pi\omega^{\prime}})\alpha_{m^{\prime},n^{\prime};\omega^{\prime},k^{\prime}}b_{\omega^{\prime},k^{\prime}}+(1-e^{2\pi\omega^{\prime}})\beta_{m^{\prime},n^{\prime};\omega^{\prime},k^{\prime}}^{*}b_{\omega^{\prime},k^{\prime}}^{\dagger}\right].$

Now tracing out the left sector we obtain

$\displaystyle\rho_{\rm bulk}\equiv{\rm Tr}_{\cal H_{L}}\rho.$ (202)

For the excited state in (3.4), we obtain the following reduced density matrix

$\displaystyle\rho_{\rm bulk}$ $\displaystyle=$
$\displaystyle\sum_{m,n,\omega,k}B_{m,n}\left[(1-e^{-2\pi\omega})\alpha_{m,n;\omega,k}^{*}b^{\dagger}_{\omega,k}+(1-e^{2\pi\omega})\beta_{m,n;\omega,k}b_{\omega,k}\right]\rho_{0}$
$\displaystyle\times\sum_{m^{\prime},n^{\prime},\omega^{\prime},k^{\prime}}B_{m^{\prime},n^{\prime}}^{*}\left[(1-e^{-2\pi\omega^{\prime}})\alpha_{m^{\prime},n^{\prime};\omega^{\prime},k^{\prime}}b_{\omega^{\prime},k^{\prime}}+(1-e^{2\pi\omega^{\prime}})\beta_{m^{\prime},n^{\prime};\omega^{\prime},k^{\prime}}^{*}b_{\omega^{\prime},k^{\prime}}^{\dagger}\right].$

and $\rho_{0}$ is defined in (190).

We wish to evaluate the difference in single interval entanglement between the
single particle excitations and the ground state in the CFT. When we consider
the bulk enatnglement entropy this translates as,

$\displaystyle S_{\rm bulk}(\Sigma_{A})=\lim_{n\rightarrow 1}S_{n\;:{\rm
bulk}}(\Sigma_{A}),\qquad S_{n\;:{\rm
bulk}}(\Sigma_{A})=\frac{1}{1-n}\log\frac{\rm Tr(\rho_{\rm bulk})^{n}}{{\rm
Tr\rho_{0}^{n}}}.$ (204)

Here the division by the denominator ensures that the contribution to the
single interval bulk entanglement from the vacuum is subtracted just as in the
case of the CFT calculation in (11). In principle it is possible to evaluate
the trace ${\rm Tr}(\rho_{\rm bulk})^{n}$ using the creation and annihilation
operator algebra and the following two point function

$\displaystyle{\rm
Tr}(\rho_{0}b^{\dagger}_{\omega,k}b_{\omega^{\prime},k^{\prime}})=\frac{4\pi^{2}\delta(\omega-\omega^{\prime})\delta(k-k^{\prime})}{e^{2\pi\omega}-1}.$
(205)

For this we would need the Bogoliubov coefficients and also perform integrals
over $\omega$ and $k$ and then analytically continue in the replica index $n$
to obtain the entanglement entropy. In the CFT analysis in section 2, we saw
that it was possible to analytically continue in $n$ while performing the
short interval expansion.

In the next sub-sections 3.4.1, 3.4.2, we follow Belin:2018juv  and obtain the
leading and sub-leading corrections to the short interval expansion of the
bulk entanglement entropy. The analysis in Belin:2018juv  was done for the
state $|\psi_{0,0}\rangle$, here we generalise the analysis to a linear
combination of single particle states. Then we apply our result to to evaluate
the bulk entanglement entropies for the states $|\psi_{0,1}\rangle$ and the
state $|\hat{\phi}\rangle$ using the knowledge of the Bogoliubov coefficients
for these states evaluated in appendix C. One of the important results of
evaluating the Bogoliubov coefficients for higher level excitations is that in
the short distance limit of interest the Bogoliubov coefficients of all higher
excitations are proportional to the Bogoliubov coefficients for the lowest
energy excitation $|\psi_{0,0}\rangle$. The proportionality constant depends
on the weight of the primary and is proportional to the norm of the
corresponding state in the CFT. For the first few levels, this behaviour can
be seen in results summarised in the table 2. This behaviour of the Bogoliubov
coefficients will enable us to finally obtain the bulk entanglement entropy
and compare with the CFT analysis.

$m$ | $n$ | $\lim_{r\rightarrow\infty}|\psi_{m,n}\rangle$ | $\lim_{\eta\rightarrow\infty}\frac{\alpha_{m,n;\omega,k}}{\alpha_{0,0;\omega,k}}=\lim_{\eta\rightarrow\infty}\frac{\beta_{m,n;\omega,k}}{\beta_{0,0;\omega,k}}$
---|---|---|---
$0$ | $0$ | $\frac{1}{\sqrt{2\pi}}\frac{1}{r^{2h}}$ | $1$
$0$ | $1$ | $\frac{1}{\sqrt{2\pi}}\frac{2h}{r^{2h}}e^{-it}$ | $2h$
$0$ | $2$ | $\frac{1}{\sqrt{2\pi}}\frac{h(2h+1)}{r^{2h}}e^{-2it}$ | $h(2h+1)$
$1$ | $0$ | $\frac{1}{\sqrt{2\pi}}\frac{\sqrt{2h}}{r^{2h}}e^{-it}e^{i\varphi}$ | $\sqrt{2h}$
$2$ | $0$ | $\frac{1}{\sqrt{2\pi}}\frac{\sqrt{h(2h+1)}}{r^{2h}}e^{-2it}e^{2i\varphi}$ | $\sqrt{h(2h+1)}$
Table 2:  This table demonstrates the scaling Bogoliubov coefficients for
different modes with respect to the lowest mode $|\psi_{0,0}\rangle$ in short
interval limit $\eta\rightarrow\infty$. The third column shows the behaviour
of the normalized wave function of the corresponding state at asymptotic
infinity. The fourth column evaluates the ratio of the Bogoliubov coefficients
in the short distance limit. Note that the scaling coefficient is proportional
to the norm of the corresponding state in the CFT.

####  3.4.1 Short interval expansion of bulk entanglement: first order

When the entangling intervals are small, the density matrix $\rho_{\rm bulk}$
should admit a short interval expansion from $\rho_{0}$. This can roughly be
seen by examining the explicit form of the Bogoliubov coeffiicents evaluated
in appendix C which are proportional to $(\cosh\eta)^{-2h}$ and the fact the
the density matrix in (LABEL:rhobulk) is quadratic in the Bogoliubov
coeffiicents. From the relation (181), we see that

${\rm for}\;\theta\rightarrow 0,\qquad\eta\rightarrow\infty.$ (206)

This then implies that the density matrix admits a short interval expansion.
Therefore we define

$\delta\rho=\rho_{\rm bulk}-\rho_{0}.$ (207)

Consider the expansion

$\displaystyle{\rm Tr}(\rho_{\rm bulk})^{n}$ $\displaystyle=$
$\displaystyle{\rm Tr}(\rho_{0}+\delta\rho)^{n}={\rm Tr}(\rho_{0}^{n})+n{\rm
Tr}(\rho_{0}^{n-1}\delta\rho)+O(\delta\rho^{2}),$ $\displaystyle=$
$\displaystyle(1-n){\rm Tr}(\rho_{0}^{n})+n{\rm Tr}(\rho_{0}^{n-1}\rho_{\rm
bulk}).$

where we have retained only the first order corrections to the density matrix.
Evaluating the bulk entanglement using (204) we obtain

$\displaystyle S_{\rm bulk}^{(1)}(\Sigma_{A})$
$\displaystyle=\lim_{n\rightarrow 1}S_{n\;:{\rm bulk}}(\Sigma_{A})$
$\displaystyle=-\partial_{n}S_{n\;:{\rm bulk}}(\Sigma_{A})\big{|}_{n=1}$
$\displaystyle=-{\rm Tr}(\rho_{\rm bulk}\log\rho_{0})+{\rm
Tr}(\rho_{0}\log\rho_{0}).$ (209)

Here we have used the fact that the density matrices are normalised i.e. ${\rm
Tr}\rho_{0}={\rm Tr}\rho_{\rm bulk}=1$. Substituting $\rho_{bulk}$ from
(LABEL:rhobulk), we obtain

$\displaystyle S_{\rm bulk}^{(1)}(\Sigma_{A})=$ (210)
$\displaystyle\sum_{\stackrel{{\scriptstyle
m_{1},n_{1},m_{3},n_{3}}}{{\omega_{1},\omega_{2},\omega_{3},k_{1},k_{2},k_{3}}}}\Bigg{(}{\rm
Tr}\Big{\\{}\rho_{0}B_{m_{1},n_{1}}^{*}\left[(1-e^{-2\pi\omega_{1}})\alpha_{m_{1},n_{1};\omega_{1},k_{1}}b_{\omega_{1},k_{1}}+(1-e^{2\pi\omega_{1}})\beta_{m_{1},n_{1};\omega_{1},k_{1}}^{*}b_{\omega_{1},k_{1}}^{\dagger}\right]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\times
2\pi\omega_{2}b^{\dagger}_{\omega_{2},k_{2}}b_{\omega_{2},k_{2}}$
$\displaystyle\qquad\qquad\times
B_{m_{3},n_{3}}\left[(1-e^{-2\pi\omega_{3}})\alpha_{m_{3},n_{3};\omega_{3},k_{3}}^{*}b^{\dagger}_{\omega_{3},k_{3}}+(1-e^{2\pi\omega_{3}})\beta_{m_{3},n_{3};\omega_{3},k_{3}}b_{\omega_{3},k_{3}}\right]\Big{\\}}\Bigg{)}$
$\displaystyle\qquad\qquad\qquad+{\rm Tr}(\rho_{0}\log\rho_{0}).$

After some thought, it is easy to see that the term on the last line cancels
the self contraction terms involving $b^{\dagger}_{\omega_{2},k_{2}}$ and
$b_{\omega_{2},k_{2}}$. To demonstrate this we must use the Wick rules (205),
the constraint among the Bogoliubov coefficients (197) together with the
normalization (200). Evaluating the remaining contractions we obtain

$\displaystyle S_{\rm
bulk}^{(1)}(\Sigma_{A})=2\pi\sum_{\stackrel{{\scriptstyle
m_{1},n_{1},m_{2},n_{2}}}{{\omega,k}}}\omega\Big{[}B_{m_{1},n_{1}}^{*}B_{m_{2},n_{2}}(\alpha_{m_{1},n_{1};w,k}\alpha^{*}_{m_{2},n_{2};\omega,k}+\beta_{m_{1},n_{1};w,k}^{*}\beta_{m_{2},n_{2};\omega,k})\Big{]}.$

To declutter the subsequent expressions it is convenient to introduce the
notation

$B\cdot\alpha_{i}^{*}=\sum_{m,n}B_{m,n}\alpha^{*}_{m,n;\omega_{i},k_{i}},\qquad
B\cdot\beta_{i}=\sum_{m,n}B_{m,n}\beta_{m,n;\omega_{i},k_{i}}.$ (212)

Using this, the first order correction to the bulk entanglement entropy is
written as

$\displaystyle S_{\rm
bulk}^{(1)}(\Sigma_{A})=2\pi\sum_{\omega_{1},k_{1}}\omega\Big{(}|B\cdot\alpha_{1}^{*}|^{2}+|B\cdot\beta_{1}|^{2}\Big{)}.$
(213)

Let us proceed to apply this expression to evaluate the leading order bulk
entanglement for the two states $|\psi_{0,1}\rangle$ and the linear
combination $|\hat{\phi}\rangle$ defined in (136).

#### First order bulk entanglement for the state $|\psi_{0,1}\rangle$

To use the expression (213) for the first order correction to the bulk
entanglement entropy corresponding to the state $|\psi_{0,1}\rangle$ we need
to set $B_{0,1}=1$ while the remaining coefficients vanish. Then we obtain

$\displaystyle S_{\rm
bulk}^{(1)}(\Sigma_{A})|_{|\psi_{0,1}\rangle}=2\pi\sum_{\omega,k}\omega(|\alpha_{0,1;\omega,k}|^{2}+|\beta_{0,1;\omega,k}|^{2}\Big{)}.$
(214)

The relevant Bogoliubov coefficients have been evaluated in appendix C. They
are given in equations (C) and (C). Since the $\alpha$ and $\beta$
coefficients come as sum, rather than the difference as in the
constraint(197), we can take the short distance or the $\eta\rightarrow\infty$
limit point wise in the integral over the frequencies and momenta. We see from
the results in the appendix that the Bogoliubov coefficients behave as

$\displaystyle\lim_{\eta\rightarrow\infty}\alpha_{0,1;\omega,k}=\frac{2h}{(\cosh\eta)^{2h}}F(\omega,k),$
(215)
$\displaystyle\lim_{\eta\rightarrow\infty}\beta_{0,1;\omega,k}=-\frac{2h}{(\cosh\eta)^{2h}}F(\omega,k).$

where

$\displaystyle
F(\omega,k)=\frac{2^{2h}\sqrt{\omega}}{\Gamma(2h)\sqrt{4\pi}}\left|\Gamma(iw)\Gamma\Big{(}h+i\frac{\omega-k}{2}\Big{)}\Gamma\Big{(}h-i\frac{\omega-k}{2}\Big{)}\right|.$
(216)

It is relevant to point out an important property of the Bogoliubov
coefficients of the excited state compared to the ground state. The Bogoliubov
coefficient of the ground state $|\psi_{0,0}\rangle$ which was evaluated in
Belin:2018juv , they are given by

$\displaystyle\alpha_{0,0;\omega,k}=\frac{1}{(\cosh\eta)^{2h}}\left[\frac{e^{\eta}-i}{e^{\eta}+i}\right]^{i\omega}F(\omega,k),$
(217)
$\displaystyle\beta_{0,0;\omega,k}=-\frac{1}{(\cosh\eta)^{2h}}\left[\frac{e^{\eta}+i}{e^{\eta}-i}\right]^{i\omega}F(\omega,k).$

From (215) and (217), in the short interval limit, we obtain the scaling
property

$\displaystyle\lim_{\eta\rightarrow\infty}\alpha_{0,1;\omega,k}=2h\lim_{\eta\rightarrow\infty}\alpha_{0,0;\omega,k},\qquad\qquad\lim_{\eta\rightarrow\infty}\beta_{0,1;\omega,k}=2h\lim_{\eta\rightarrow\infty}\beta_{0,0;\omega,k}.$
(218)

Therefore in the short interval limit, the leading behaviour of the Bogoliubov
coefficients for the excited state is the same as that of the ground state but
for a pre-factor. In the appendix C, we observe this property for all the $4$
excited states we study in this paper. The results are summarized in table 2.
The scaling property of Bogoliubov coefficients in (218) makes it easy to
evaluate the correction to bulk entanglement entropy in (214). Substituting,
we obtain the bulk entanglement entropy

$\displaystyle S_{\rm