aainstitutetext: Institute of Physics,
Sachivalaya Marg, Bhubaneshwar, India-751005bbinstitutetext: Homi Bhabha
National Institute,
Training School Complex, Anushakti Nagar, Mumbai, India-400085ccinstitutetext:
Chennai Mathematical Institute,
H1, SIPCOT IT Park, Siruseri 603103, India

# Linearized Einstein’s Equation around pure BTZ
from Entanglement Thermodynamics

Partha Paul c Pratik Roy<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

It is known that the linearized Einstein’s equation around the pure AdS can be
obtained from the constraint $\Delta S=\Delta\left<H\right>$, known as the
first law of entanglement, on the boundary CFT. The corresponding dual state
in the boundary CFT is the vacuum state around which the linear perturbation
is taken. In this paper we revisit this question, in the context of
$\textnormal{AdS}_{3}/\textnormal{CFT}_{2}$, with the state of the boundary
$\textnormal{CFT}_{2}$ as a thermal state. The corresponding dual geometry is
a planar BTZ black hole. By considering the linearized perturbation around
this black brane we show that Einstein’s equation follows from the first law
of entanglement.

##  1 Introduction

After the realization of a connection between gravity and thermodynamics
bekenstein ; bardeen ; hawking  various attempts have been made to understand
gravitational dynamics from horizon thermodynamics jacobson ; padmanabhan ;
verlinde . The discovery of AdS/CFT correspondence maldacena ; ooguri  led to
the new idea that the dynamics of spacetime can be understood from some sort
of entanglement between the degrees of freedom of the boundary CFT mav ;
faulkner ; myers ; speranza ; mosk ; czech ; swingle ; ted ; xian . See also
raamsdonk ; raamsdonk1 ; raamsdonk2 . In this note, following mav ; faulkner ;
myers , we explore this idea further.

Linear perturbations around a fixed reference state in the continuum field
theory satisfy the first law of entanglement, $\Delta S=\Delta\left<H\right>$
mav ; faulkner ; myers , where $S$ is the entanglement entropy of a spatial
region and $H$ is the modular Hamiltonian associated with that region. In the
AdS/CFT framework, each side of this first law can be computed using the dual
geometry. In mav ; myers , the vacuum of a holographic CFT, with corresponding
dual geometry pure AdS, was chosen as a fixed reference state. Considering
perturbations around pure AdS, mav ; myers  calculated $\Delta S$ and
$\Delta\left<H\right>$ hologrpahically and showed that to linear order in the
perturbation the first law of entanglement is satisfied, while inclusion of
higher order contributions gives the constraint
$\Delta\left<H\right>\geq\Delta S$ mav ; myers 111For related discussions one
can also see, e.g, jb ; numasawa ; nozaki ; allahbakhshi ; wong  . In mav ;
faulkner  linear perturbations were considered and it was shown that
Einstein’s equations linearized around pure AdS do follow from the first law
of entanglement, thus showing their equivalence at first order.

In this paper, we take the thermal state of a holographic CFT as the fixed
reference state and perturb it infinitesimally. The change in the entanglement
entropy and the modular Hamiltonian of a spatial region will satisfy the first
law of entanglement. Based on the holographic dictionary we then compute each
side of this relation using metric components of the dual geometry. We show
that for metric components of the dual geometry satisfying the linearized
Einstein’s equations, the first law of entanglement holds. Then we go the
other direction, i.e, we show that the first law of entanglement fixes the
metric uniquely if we demand that it holds in all frames of reference.

Entanglement entropy for a holographic field theory can be computed by
applying Ryu-Takayanagi formula rt  and its covariant generalization hrt . See
also colgain  for some useful discussions on holographic enetanglement entropy
in case of warped $\textnormal{AdS}_{3}$ geometries. Computing modular
Hamiltonian in the field theory side is not an easy task. There are only few
cases where it can be expressed as the integral of some local quantity, mainly
the stress tensor casini ; cardy . The modular Hamiltonian for a spatial
interval of a two dimensional CFT at finite temperature was calculated in
casini . Using the holographic dictionary one can obtain the boundary stress
tensor from the asymptotic behavior of the metric components myersbst ;
skenderis ; solodukhin , which can then be used to compute modular
Hamiltonian. See also erdmenger ; bala ; kostas1  for relevant discussions on
holographic stress tensor.

##  2 First Law of Entanglement $\Delta S=\Delta\left<H\right>$

In this section we briefly review the first law of entanglement. Relative
entropy quantifies distinguishability between two states of a quantum field
theory in the same Hilbert space ved 222 For relative entropy of excited
states in two dimensional CFT see ugajin ; jabbari .. Let us consider a
$d$-dimensional Minkowski spacetime $\mathbb{R}^{1,d-1}$. Also consider a
spatial region $A$ on a fixed time slice $t=t_{0}$ on this spacetime.

$A$$A^{C}$$\mathbb{R}^{1,d-1}$$x$$t$ Figure 1: Spatial region $A$ defined on a
constant time slice $t=t_{0}$ of $d-$dimensional Minkowski spacetime
$\mathbb{R}^{1,d-1}$. $A^{C},$ the compliment of $A$, denotes the rest of the
spacetime. $\sigma_{A}$ denotes the reduced density matrix defined for this
spatial region $A$, where $\sigma$ is a state (represented as a density
matrix) of a QFT defined on $\mathbb{R}^{1,d-1}$.

Let density matrices $\sigma$ and $\rho$ define two states of a QFT defined on
$\mathbb{R}^{1,d-1}$. Then the relative entropy of $\rho$ with respect to
$\sigma$ for the spatial region $A$ is defined as

$S_{A}(\rho_{A}|\sigma_{A})=-\text{Tr}(\rho_{A}\log\sigma_{A})-S_{A}(\rho_{A})=\text{Tr}(\rho_{A}\log\rho_{A})-\text{Tr}(\rho_{A}\log\sigma_{A}),$
(1)

where $\rho_{A}$ and $\sigma_{A}$ are the reduced density matrices associated
with the spatial region $A$.

Choose $\sigma$ as the reference state. With the definitions of entanglement
entropy of a region $A$, $S_{A}(\rho_{A})=-\text{Tr}(\rho_{A}\log\rho_{A})$,
and of the modular Hamiltonian associated with that region,
$H_{A}=-\log\sigma_{A}$, one can rewrite (1) as

$S_{A}(\rho_{A}|\sigma_{A})=\Delta\left<H_{A}\right>-\Delta S_{A},$ (2)

where

$\Delta S_{A}=S_{A}(\rho_{A})-S_{A}(\sigma_{A}),$ (3)

and

$\Delta\left<H_{A}\right>=\text{Tr}(\rho_{A}H_{A})-\text{Tr}(\sigma_{A}H_{A}).$
(4)

It has the nice positivity property that $S_{A}(\rho_{A}|\sigma_{A})\geq 0$
with the inequality saturated if and only if $\rho_{A}=\sigma_{A}$. Using this
property one can show easily that for $\rho_{A}$ very close to $\sigma_{A}$
the relative entropy vanishes at linear order of perturbation giving rise to
the constraint mav ; myers

$\Delta S=\Delta\left<H\right>.$ (5)

This is known as the first law of entanglement.

Below, we will be interested in the entanglement entropy of a single spatial
interval $A=(-R,+R)$ for a holographic state of a CFT in 1+1-dimensions. We
choose the initial holographic state to be a thermal state with temperature
$T=\frac{1}{2\pi}$. In this case the modular Hamiltonian is given by cardy ;
Hartman:2015apr 333We thank David Blanco for poiting out reference
Hartman:2015apr  to us.

$H_{A}=\frac{4\pi}{\sinh
R}\int_{-R}^{+R}dx\left[\sinh\left(\frac{R+x}{2}\right)\sinh\left(\frac{R-x}{2}\right)\right]T_{00}(x,t)$
(6)

where $T_{00}(x,t)$ is the time-time component of the field theory stress
tensor.

$x$$-R$$R$$t$$A$$A^{C}$$A^{C}$ Figure 2: Spatial region $A\equiv(-R,R)$ on a
fixed time slice $t=t_{0}$ of the boundary spacetime $\mathbb{R}^{1,1}$. Our
modular Hamiltonian (6) is defined for this region in a thermal state of a CFT
with temperature $T=\frac{1}{2\pi}$

##  3 Holographic Computation of $\Delta S$ and $\Delta\left<H\right>$

We consider a thermal state of 1+1-dimensional holographic CFT with
temperature $T=\frac{1}{2\pi}$ as a fixed reference state. We know that the
dual geometry of a thermal state of a holographic CFT is a black hole in AdS.
In 2+1 bulk dimensions, it is the BTZ black hole. We would consider the BTZ
black brane instead of black hole because the field theory is defined on
$\mathbb{R}^{1,1}$. The static BTZ black brane metric is444Our convention is
that we use indices $K,L$ for the three bulk coordinates $\\{z,t,x\\}$ and
indices $\mu,\,\nu$ for two boundary coordinates $\\{t,x\\}$. btz

$ds^{2}=\frac{1}{z^{2}}\left[\frac{dz^{2}}{1-z^{2}}-(1-z^{2})dt^{2}+dx^{2}\right]$
(7)

Here we have set the AdS radius of curvature, $L_{\textnormal{AdS}}=1$. With
this convention one can check that the inverse temperature of this black brane
is $\beta=2\pi$.

###  3.1 Holographic Computation of $\Delta S$

Let us consider a spatial interval $A=(-R,+R)$ at a fixed time $t=t_{0}$ in
the boundary spacetime $\mathbb{R}^{1,1}$. The entanglement entropy of this
interval can be computed using the Ryu-Takayanagi formula rt :

$S_{A}=\frac{\textnormal{Length}(\gamma_{A})}{4G_{N}},$ (8)

where $\gamma_{A}$ is a geodesic in the bulk homologous to $A$, and $G_{N}$ is
Newton’s constant. When the spacetime is not static, one needs to use its
covariant generalization hrt . We will deal with that in the case of boosted
black brane.

Let us now concentrate on our specific situation. The equation for the
geodesic, with prescribed boundary conditions $t=t_{0}=0,\ z=0,\ x=\pm R,$ is

$z^{2}\cosh^{2}R+\cosh^{2}x=\cosh^{2}R.$ (9)

Performing a coordinate transformation to Fefferman-Graham(FG) coordinates fg
,

$z^{2}=\left(1+\frac{\tilde{z}^{2}}{4}\right)^{-2}\tilde{z}^{2},$ (10)

we can write down the metric (7) as

$ds^{2}=\frac{1}{\tilde{z}^{2}}\left[d\tilde{z}^{2}-\left(1-\frac{\tilde{z}^{2}}{4}\right)^{2}dt^{2}+\left(1+\frac{\tilde{z}^{2}}{4}\right)^{2}dx^{2}\right].$
(11)

The advantage of writing down any asymptotically AdS spacetime in this
coordinate is to identify the boundary stress tensor easily using the
holographic prescription myersbst ; solodukhin . One can easily check that in
case of pure BTZ (11), the boundary stress tensor has the non-vanishing
components $T_{tt}=T_{xx}=\frac{1}{16\pi G_{N}}$.

$x$$-R$$R$$t$$A$$A^{C}$$A^{C}$$\gamma_{A}$ Figure 3: Pictorial representation
of the Ryu-Takayanagi proposal. Area of the minimal area curve, i.e., length
of the geodesic, $\gamma_{A}$ gives the holographic entanglement entropy for
the spatial region $A$ in the boundary.

In terms of FG coordinates, the extremal surface equation (9) transforms as

$\displaystyle\frac{\tilde{z}^{2}}{\left(1+\frac{\tilde{z}^{2}}{4}\right)^{2}}=$
$\displaystyle\ \frac{\cosh^{2}R-\cosh^{2}x}{\cosh^{2}R}$
$\displaystyle\Rightarrow\qquad\qquad\tilde{z}^{2}=$ $\displaystyle\
4\frac{\cosh R-\cosh x}{\cosh R+\cosh x}\ .$ (12)

We can treat $x$ as the intrinsic coordinate on the geodesic. The induced
metric on the geodesic before perturbation is

$g^{(0)}_{xx}=G^{(0)}_{KL}\frac{dx^{K}}{dx}\frac{dx^{L}}{dx},$ (13)

where $G^{(0)}_{KL}$ are the metric components of pure BTZ (11). The length
functional is

$A=\int_{-R}^{+R}dx\ \sqrt{g^{(0)}_{xx}}.$ (14)

Now we add some pure metric perturbation to the BTZ metric (11)555Linear
perturbations around BTZ black brane were also considered in jyotirmay  and
the first order correction to holographic entanglement entropy was calculated.
They have also discussed the dynamics of the shift of holographic entanglement
entropy.. Any such perturbation in the FG coordinates can be written as

$ds^{2}=\frac{1}{\tilde{z}^{2}}\left[d\tilde{z}^{2}-\left(1-\frac{\tilde{z}^{2}}{4}\right)^{2}dt^{2}+\left(1+\frac{\tilde{z}^{2}}{4}\right)^{2}dx^{2}+\tilde{z}^{2}H_{\mu\nu}(\tilde{z},x,t)dx^{\mu}dx^{\nu}\right].$
(15)

From now on (in the case of static BTZ black brane) everything will be done in
FG coordinates so we will drop the $\,\tilde{}\,$ sign over $\tilde{z}$ and
simply write $z$.

To linear order in perturbation $H_{\mu\nu}(z,x,t)$, the change in the length
functional (14) is

$\displaystyle\Delta A=$ $\displaystyle\ \int
dx\frac{1}{2}\sqrt{g^{(0)}}g^{(0)xx}\delta g_{xx},$ $\displaystyle=$
$\displaystyle\ \frac{1}{\sinh 2R}\int
dx[\sinh(R+x)\sinh(R-x)]H_{xx}(z(x),x,t).$ (16)

Hence to first order in the perturbation, the change in the entanglement
entropy is

$\Delta\hat{S}_{A}=\int dx[\sinh(R+x)\sinh(R-x)]H_{xx}(z(x),x,t).$ (17)

Here $\Delta\hat{S}_{A}=4G_{N}\sinh(2R)\Delta S_{A}$.

###  3.2 Holographic Computation of $\Delta\left<H\right>$

Equation (6) gives the modular Hamiltonian for a spatial interval $A=(-R,+R)$
of a thermal state with temperature $T=\frac{1}{2\pi}$. If we perturb this
state infinitesimally, the change in the modular Hamiltonian is

$\Delta\left<H_{A}\right>=\frac{4\pi}{\sinh
R}\int_{-R}^{+R}dx\left[\sinh\left(\frac{R+x}{2}\right)\sinh\left(\frac{R-x}{2}\right)\right]\Delta\left<T_{00}(x,t)\right>,$
(18)

where $\Delta\left<T_{00}(x,t)\right>$ is the change in the expectation value
of the time-time component of the field theory stress tensor due to
infinitesimal perturbations.

We have already mentioned that the boundary stress tensor can be found from
the asymptotic form of the asymptotically AdS bulk metric. From myersbst ;
skenderis ; solodukhin  we know that for a $d+1$-dimensional asymptotically
AdS bulk metric written in FG coordinates,

$ds^{2}=\frac{1}{z^{2}}\left[dz^{2}+\eta_{\mu\nu}dx^{\mu}dx^{\nu}+z^{d}g_{\mu\nu}(z,x)dx^{\mu}dx^{\nu}\right],$
(19)

the boundary stress tensor can be found from the following asymptotic
relation,

$\left<T_{\mu\nu}(x)\right>=\frac{d}{16\pi G_{N}}g_{\mu\nu}(z=0,x).$ (20)

Using the formula in our case, we have

$\left<T_{\mu\nu}(x,t)\right>=\frac{1}{8\pi
G_{N}}\left(\frac{1}{2}+H_{\mu\nu}(z=0,x,t)\right),$ (21)

where $\frac{1}{16\pi G_{N}}$ is the background boundary stress tensor. Hence
the change in the modular Hamiltonian is

$\Delta\braket{\hat{H}_{A}}=4\cosh
R\int_{-R}^{+R}dx\left[\sinh\left(\frac{R+x}{2}\right)\sinh\left(\frac{R-x}{2}\right)\right]H_{tt}(x,t).$
(22)

where $\Delta\braket{\hat{H}_{A}}=4G_{N}\sinh(2R)\Delta\left<H_{A}\right>$.

##  4 Proof That Einstein’s Equations Imply $\Delta S=\Delta\left<H\right>$

We now have the expressions for both $\Delta S$ and $\Delta\left<H\right>$ to
linear order in the bulk perturbation. In this section we will show that the
solutions of linearized Einstein’s equations satisfy the relation $\Delta
S=\Delta\left<H\right>$.

In $d+1$-dimensions with a cosmological constant
$\Lambda=-\frac{d(d-1)}{2L_{\textnormal{AdS}}^{2}}$, Einstein’s equations read
(recall that we have set $L_{\textnormal{AdS}}=1$)

$R_{AB}-\frac{1}{2}G_{AB}(R+d(d-1))=0.$ (23)

Using the metric (15) to linear order in $H_{\mu\nu},$ different components of
equations (23) (with $d=2$) read

$\displaystyle
32zH_{tt}(z,x,t)-(4-z^{2})\left[-(12+z^{2})\partial_{z}H_{tt}(z,x,t)-z(4-z^{2})\partial_{z}^{2}H_{tt}(z,x,t)\right]=$
$\displaystyle\ 0$ (24)
$\displaystyle-32zH_{xx}(z,x,t)+(4+z^{2})\left[(12-z^{2})\partial_{z}H_{xx}(z,x,t)+z(4+z^{2})\partial_{z}^{2}H_{xx}(z,x,t)\right]=$
$\displaystyle\ 0$ (25)
$\displaystyle(48+z^{4})\partial_{z}H_{tx}(z,x,t)+z(16-z^{4})\partial_{z}^{2}H_{tx}(z,x,t)=$
$\displaystyle\ 0$ (26) $\displaystyle-2$
$\displaystyle(4-z^{2})^{2}\partial_{t}H_{tx}(z,x,t)+2(16+z^{4})\partial_{x}H_{tt}$
$\displaystyle\ \qquad\qquad\qquad\
-z(16-z^{4})\,\partial_{z}\left(\partial_{t}H_{tx}(z,x,t)-\partial_{x}H_{tt}(z,x,t)\right)=\
0$ (27) $\displaystyle 2$
$\displaystyle(4+z^{2})^{2}\partial_{x}H_{tx}(z,x,t)-2(16+z^{4})\partial_{t}H_{xx}$
$\displaystyle\ \qquad\qquad\qquad\
-z(16-z^{4})\,\partial_{z}\left(\partial_{t}H_{xx}(z,x,t)-\partial_{x}H_{tx}(z,x,t)\right)=\
0$ (28)
$\displaystyle-2(4+z^{2})^{2}H_{tt}(z,x,t)+2(4-z^{2})^{2}H_{xx}(z,x,t)+z\left[16z\partial_{t}^{2}H_{xx}(z,x,t)\right.$
$\displaystyle-32z\partial_{t}\partial_{x}H_{tx}(z,x,t)+16z\partial_{x}^{2}H_{tt}(z,x,t)-(16-z^{4})\partial_{z}H_{tt}(z,x,t)$
$\displaystyle\left.+(16-z^{4})\partial_{z}H_{xx}(z,x,t)\right]=$
$\displaystyle\ 0$ (29)

Setting $z=0$ in (29) one can see that

$-H_{tt}(z=0,x,t)+H_{xx}(z=0,x,t)=0.$ (30)

This holds because the boundary theory is a conformal field theory and (30) is
the tracelessness condition of the boundary field theory stress tensor. Now
demanding the smoothness condition at $z=0$ and using (30) one arrives at the
following solutions for the perturbations:

$\displaystyle H_{tt}(z,x,t)=$ $\displaystyle\ (4-z^{2})H(x,t),$ (31)
$\displaystyle H_{xx}(z,x,t)=$ $\displaystyle\ (4+z^{2})H(x,t),$ (32)
$\displaystyle H_{tx}(z,x,t)=$ $\displaystyle\ h(t,x)$ (33)

with $h(x,t)$ and $H(x,t)$ restricted by the conditions,

$\displaystyle\partial_{t}h(x,t)=$ $\displaystyle\ 4\partial_{x}H(x,t),$ (34)
$\displaystyle\partial_{x}h(x,t)=$ $\displaystyle\ 4\partial_{t}H(x,t),$ (35)
$\displaystyle(\partial_{t}^{2}-\partial_{x}^{2})H(x,t)=$ $\displaystyle\ 0$
(36)

Now we consider the expressions for $\Delta\hat{S}_{A}$ and
$\Delta\braket{\hat{H}_{A}}$.

$\displaystyle\Delta\hat{S}_{A}=$ $\displaystyle\
\int_{-R}^{R}dx\sinh(R+x)\sinh(R-x)H_{xx}(z,x,t_{0})$ $\displaystyle=$
$\displaystyle\ \int_{-R}^{R}dx\sinh(R+x)\sinh(R-x)(4+z^{2})H(x,t_{0})$
$\displaystyle=$ $\displaystyle\ 16\cosh
R\int_{-R}^{+R}dx\sinh\left(\frac{R+x}{2}\right)\sinh\left(\frac{R-x}{2}\right)H(x,t_{0})$
(37)

and

$\displaystyle\Delta\braket{\hat{H}_{A}}=$ $\displaystyle\ 4\cosh
R\int_{-R}^{R}dx\sinh\left(\frac{R+x}{2}\right)\sinh\left(\frac{R-x}{2}\right)H_{tt}(0,x,t_{0})$
$\displaystyle=$ $\displaystyle\ 16\cosh
R\int_{-R}^{R}dx\sinh\left(\frac{R+x}{2}\right)\sinh\left(\frac{R-x}{2}\right)H(x,t_{0})$
(38)

Comparing (4) and (4) we see that indeed $\Delta
S_{A}=\Delta\left<H_{A}\right>$.

##  5 Proof That First Law of Entanglement Implies Einstein’s Equations

In this section we go the other direction. We will show that the constraint
(5) with the boundary condition (21) fixes the metric uniquely. We would
follow the strategy of mav  for the proof .

Let $H^{EE}_{\mu\nu}$ be the metric that solves Einstein’s equations
($\ref{EE1}-\ref{EE6}$) with the boundary conditions (21). We will show that
there is no metric other than $H^{EE}_{\mu\nu}$ with the boundary conditions
(21) which satisfies the same relation (5). Suppose we have another metric
$H_{\mu\nu}$ satisfying (5) with the same boundary condition (21). We will
show that

$\Delta_{\mu\nu}(z,x,t)\equiv H^{EE}_{\mu\nu}-H_{\mu\nu}=0,$ (39)

for all $z$. Just by demanding that both the metrics satisfy the same boundary
condition, we already have

$\Delta_{\mu\nu}(0,x,t)=0.$ (40)

Hence (22) tells us that

$\Delta(\Delta\left<H_{A}\right>)=0.$ (41)

Now the constraint (5) together with the expression (17) tells us that in a
fixed frame of reference (say, the $t=t_{0}$ frame) we have

$0=\int{dx\,\Delta_{xx}(z(x),x+x_{0},t_{0})[\sinh(R+x)\sinh(R-x)]}.$ (42)

In the integral (42) we have shifted the origin from $x=0$ to $x=x_{0}$. Now
we expand $\Delta_{\mu\nu}$ as

$\displaystyle\Delta_{\mu\nu}(z(x),x+x_{0},t_{0})=$ $\displaystyle\
\sum_{n=0}^{\infty}z^{n}\Delta^{(n)}_{\mu\nu}(x+x_{0},t_{0})$ $\displaystyle=$
$\displaystyle\
\sum_{n,m}z^{n}\frac{x^{2m}}{(2m)!}\partial_{x}^{2m}\Delta^{(n)}_{\mu\nu}(x_{0},t_{0}).$
(43)

Equation (42) thus becomes

$\sum_{n,m}\frac{1}{(2m)!}\partial_{x}^{2m}\Delta^{(n)}_{xx}(t_{0},x_{0})\int_{-R}^{+R}dx[z^{n}x^{2m}\sinh(R+x)\sinh(R-x)]=0.$
(44)

Substitung for $z$ the expression (12) and performing some simplifications, we
finally get

$\sum_{n,m}\frac{2^{n+1}}{(2m)!}\partial_{x}^{2m}\Delta^{(n)}_{xx}(t_{0},x_{0})I_{n,m}(R)=0,$
(45)

where we write

$I_{n,m}=\int_{0}^{R}dx\left[x^{2m}(\cosh R+\cosh x)^{-\frac{n}{2}+1}(\cosh
R-\cosh x)^{\frac{n}{2}+1}\right].$ (46)

We now need to expand (45) in powers of $R$ and set each coefficient to zero
to see what constraints do they impose on $\Delta_{xx}(t_{0},x_{0})$. Thus
expanding (45) around $R=0$, we get

$\displaystyle\sum_{n,m,j}\frac{2^{n+1}}{(2m)!}\partial_{x}^{2m}\Delta^{(n)}_{xx}(t_{0},x_{0})\frac{R^{j}}{j!}\left[\partial_{R}^{j}I_{n,m}(R)\right]_{R=0}=0.$
(47)

Vanishing of the $R^{J+3}-$th term requires

$\displaystyle\sum_{n,m}\frac{2^{n+1}}{(2m)!}\partial_{x}^{2m}\Delta^{(n)}_{xx}(t_{0},x_{0})\left[\partial_{R}^{J+3}I_{n,m}(R)\right]_{R=0}=0$
(48)

The LHS of (48) contains a summation over $n,\,m$ for a fixed $J$. In appendix
A we explicitly show that for both odd and even $J$ (for odd $J$ numerical
analysis is presented in appendix A), all the terms in the summation vanish
except when $n\leq J$. For $n=J$, we have $m=0$ and for $n<J$, we need $m\neq
0$ (Please see appendix A for details). The last non-vanishing term in the
summation (48) with a fixed $J$ is $n=J,\ m=0.$ This term can be expressed as
a linear combination of the lower order terms which establishes the result
that

$\Delta^{(J)}_{xx}(t_{0},x_{0})=0$ (49)

for $J=0,1,2,\cdots$

Having shown that the entanglement first law fixes the solution to linearized
Einstein’s equations in the rest frame, we now consider a boosted frame and
try to repeat the analysis above. Consider then the boosted BTZ black brane.
The coordinate transformations to go from the static BTZ black brane to the
boosted BTZ black brane are

$\displaystyle t=\gamma(t^{\prime}-\beta x^{\prime}),$ (50) $\displaystyle
x=\gamma(x^{\prime}-\beta t^{\prime}).$ (51)

Here $\beta,\gamma$ have their usual meanings from the special theory of
relativity. The metric of the boosted black brane is given by

$ds^{2}=\frac{1}{z^{2}}\left[\frac{dz^{2}}{1-z^{2}}+(-1+\gamma^{2}z^{2}){dt^{\prime}}^{2}+(1+\beta^{2}\gamma^{2}z^{2}){dx^{\prime}}^{2}-2\beta\gamma^{2}z^{2}dt^{\prime}dx^{\prime}\right]$
(52)

Notice that here, $z$ denotes the original radial coordinate and not the FG
coordinate $\tilde{z}$. Below, we will explicitly write $\tilde{z}$ for the FG
coordinate. We need to solve the spacelike geodesic equations in this geometry
with boundary conditions $z=0,\,t=0,\,x=\pm R$. Working in the geometry (52)
with these boundary conditions is equivalent to working in the geometry (7)
with the following boundary conditions

$\displaystyle z=0,\quad t=-\beta\gamma R,\quad x=\gamma R,$ $\displaystyle
z=0,\quad t=\beta\gamma R,\quad x=-\gamma R.$ (53)

Hence we will solve for the spacelike geodesics with metric (7) and boundary
conditions (5). Let $s$ be the proper length along the geodesic. Then we have
the following equations for the spacelike geodesics

$\displaystyle\frac{1}{z^{2}}\left[\frac{1}{1-z^{2}}\left(\frac{dz}{ds}\right)^{2}-(1-z^{2})\left(\frac{dt}{ds}\right)^{2}+\left(\frac{dx}{ds}\right)^{2}\right]=$
$\displaystyle\ 1,$ (54)
$\displaystyle\frac{1}{z^{2}}\left[\frac{dx}{ds}\right]=$ $\displaystyle\ p,$
(55)
$\displaystyle-\frac{1}{z^{2}}\left[(1-z^{2})\left(\frac{dt}{ds}\right)\right]=$
$\displaystyle\ e,$ (56)

where $p$ and $e$ are two constants of motion along the geodesic. We want to
write down the geodesic in a parametric form $z=z(x),~{}t=t(x)$. Hence we
write down the above three equations in the following form

$\displaystyle pz\frac{dz}{dx}=$ $\displaystyle\
\sqrt{1-(1-e^{2}+p^{2})z^{2}+p^{2}z^{4}}\,,$ (57)
$\displaystyle\frac{dt}{dx}=$ $\displaystyle\ -\frac{e}{p(1-z^{2})}\,.$ (58)

With the prescribed boundary conditions (5), the solutions are

$\tanh(\gamma R)\tanh t+\tanh(\beta\gamma R)\tanh x=0,$ (59)

and

$\cosh^{2}(\gamma R)z^{2}+\left[1-\frac{\sinh^{2}(\beta\gamma
R)}{\sinh^{2}(\gamma R)}\right]\cosh^{2}x=\left[1-\frac{\sinh^{2}(\beta\gamma
R)}{\sinh^{2}(\gamma R)}\right]\cosh^{2}(\gamma R).$ (60)

In terms of the FG coordinates, (60) becomes

$\tilde{z}^{2}=4\frac{\cosh(\gamma R)-\sqrt{(1-r)\cosh^{2}(\gamma
R)+r\cosh^{2}x}}{\cosh(\gamma R)+\sqrt{(1-r)\cosh^{2}(\gamma
R)+r\cosh^{2}x}},$ (61)

where

$r=\left[1-\frac{\sinh^{2}(\beta\gamma R)}{\sinh^{2}(\gamma R)}\right].$

The induced metric on the spacelike geodesic is

$ds^{2}_{ind}=\frac{\sinh^{2}(2\gamma R)}{[\cosh(2\gamma
R)-\cosh(2x)]^{2}}dx^{2}.$ (62)

Change in the geodesic length is

$\displaystyle\Delta A=\int_{-\gamma R}^{+\gamma R}dx\,\frac{\sinh(\gamma
R+x)\sinh(\gamma R-x)}{\sinh(2\gamma
R)}\left[\left(\frac{dt}{dx}\right)^{2}H_{tt}+2\left(\frac{dt}{dx}\right)H_{tx}+H_{xx}\right].$

An argument similar to that leading to (42) gives the following equation for
the boosted case

$\int_{-\gamma R}^{+\gamma R}dx\sinh(\gamma R+x)\sinh(\gamma
R-x)\left[\left(\frac{dt}{dx}\right)^{2}\Delta_{tt}+2\left(\frac{dt}{dx}\right)\Delta_{tx}+\Delta_{xx}\right]=0.$
(63)

Now assuming that $\Delta_{\mu\nu}(z,x,t)$ is an analytic function, we can
expand it in the following form

$\displaystyle\Delta_{\mu\nu}(\tilde{z},t+t_{0},x+x_{0})=$ $\displaystyle\
\sum_{n=0}^{\infty}\tilde{z}^{n}\Delta_{\mu\nu}^{(n)}(t+t_{0},x+x_{0}),$
$\displaystyle=$ $\displaystyle\
\sum_{n=0}^{\infty}\tilde{z}^{n}\left[\sum_{m_{t},m_{x}}\frac{\partial_{t}^{m_{t}}\partial_{x}^{m_{x}}\Delta_{\mu\nu}^{(n)}(t_{0},x_{0})}{m_{t}!m_{x}!}\right]t^{m_{t}}x^{m_{x}},$
$\displaystyle=$ $\displaystyle\
\sum_{n,m_{t},m_{x}}B^{n,m_{t},m_{x}}_{\mu\nu}\tilde{z}^{n}t^{m_{t}}x^{m_{x}}.$
(64)

(63) then becomes

$\sum_{n,m_{t},m_{x}}\left[B^{n,m_{t},m_{x}}_{tt}I^{n,m_{t},m_{x}}_{tt}(R)+B^{n,m_{t},m_{x}}_{tx}I^{n,m_{t},m_{x}}_{tx}(R)+B^{n,m_{t},m_{x}}_{xx}I^{n,m_{t},m_{x}}_{xx}(R)\right]=0,$
(65)

where

$\displaystyle I^{n,m_{t},m_{x}}_{tt}(\gamma R)=$ $\displaystyle\
\int_{-\gamma R}^{+\gamma R}dxx^{m_{x}}[\sinh(\gamma R+x)\sinh(\gamma
R-x)]\tilde{z}^{n}t^{m_{t}}\left(\frac{dt}{dx}\right)^{2},$ (66)
$\displaystyle I^{n,m_{t},m_{x}}_{tx}(\gamma R)=$ $\displaystyle\
2\int_{-\gamma R}^{+\gamma R}dxx^{m_{x}}[\sinh(\gamma R+x)\sinh(\gamma
R-x)]\tilde{z}^{n}t^{m_{t}}\left(\frac{dt}{dx}\right),$ (67) $\displaystyle
I^{n,m_{t},m_{x}}_{xx}(\gamma R)=$ $\displaystyle\ \int_{-\gamma R}^{+\gamma
R}dxx^{m_{x}}[\sinh(\gamma R+x)\sinh(\gamma R-x)]\tilde{z}^{n}t^{m_{t}}.$ (68)

We now use the same technique of expanding around $R=0$ as before, but the
calculation is difficult even for even integer $n$. One can examine term by
term the expansion series (65) and check which coefficients in a particular
term are non zero. In the appendix A we discuss the first few terms. Working
in this manner we finally arrive, apart from some constant factors, at the
following equations

$\displaystyle\Delta^{(0)}_{xx}-2\beta\Delta^{(0)}_{tx}+\beta^{2}\Delta^{(0)}_{tt}=$
$\displaystyle\ 0$
$\displaystyle\Delta^{(1)}_{xx}-2\beta\Delta^{(1)}_{tx}+\beta^{2}\Delta^{(1)}_{tt}=$
$\displaystyle\ 0$
$\displaystyle\Delta^{(2)}_{xx}-2\beta\Delta^{(2)}_{tx}+\beta^{2}\Delta^{(2)}_{tt}=$
$\displaystyle\ 0$ (69) $\displaystyle\vdots\qquad\qquad$

One can take any arbitrary value of $n$ and check that the pattern should be
the same, although for high values of $n$ computations become difficult. Now
(69) is a polynomial in $\beta$ and the coefficients of each $\beta^{k}$ must
vanish individually. From the coefficients of $\beta^{0},$ we get:

$\Delta^{(n)}_{xx}=0,$ (70)

from the coefficients of $\beta$:

$\Delta^{(n)}_{tx}=0,$ (71)

and from the coefficients of $\beta^{2}$:

$\Delta^{(n)}_{tt}=0,$ (72)

where $n=0,1,2,\ldots$

Thus we have shown that $(\ref{delmn})$ is true for all $z$. Note that we have
assumed that $\Delta_{\mu\nu}(z,x,t)$ is an analytic function.

##  6 Discussion

In this paper we have shown that linearized Einstein’s equations around the
BTZ black brane can be obtained from the first law of entanglement
thermodynamics, $\Delta S=\Delta\left<H\right>$, where the reference state was
taken to be a thermal state of the CFT which is dual to the black brane. It
would be interesting to check if non-linear Einstein’s equations can be
obtained from some constraints on the entanglenemnt entropy of the thermal
state of boundary CFT as well. In particular, in thomas  the vacuum state of
the boundary CFT was perturbed by some scalar primary or stress tensor
operators and it was shown that for such excited states, up to second order in
the perturbation, the entanglement entropy of all ball-shaped regions can be
obtained using the covariant prescription for holographic entanglement entropy
from the corresponding dual geometries. It was shown that the corresponding
dual spacetimes must satisfy Einstein’s equations up to second order in the
perturbation around AdS. It would be interesting to extend their work for the
thermal state of holographic CFTs.

One of the important points we would like to emphasize is that the Einstein
equations that we have derived hold outside the horizon. This is because the
spacelike geodesics used to compute the entanglement entropy do not see the
region behind the horizon. It will be very interesting if a similar method can
be used to prove Einstein’s equations behind the horizon. It seems that
generalization of this method to the two-sided eternal BTZ black hole may give
us some insights into the derivation of Einstein’s equations behind the
horizon.

It would also be interesting to check it for higher dimensional black holes
and also at non-linear order. We have only considered the metric perturbations
in the bulk such that only the boundary stress tensor has a non-trivial
expectation value. One can also consider other excitations as well, e.g.,
turning on scalar field in the bulk where the scalar operator will acquire
non-trivial expectation values in the boundary theory.

Note Added: While this article was under preparation, the work Blanco:2018riw
appeared in the arXiv.666We are grateful to David Blanco for bringing this
paper to our attention. Some of their results, although they take a different
approach, appear to agree with ours where they overlap.

## Acknowledgements

We are grateful to Shamik Banerjee for suggesting this problem and guiding us
throughout the project. We are also thankful to him for carefully reading the
draft and for correcting some parts of it. We are thankful to Amitabh Virmani
and Jyotirmay Bhattacharya for helpful discussions on related matters. We
would also like to acknowledge the hospitality at Chennai Mathematical
Institute during the workshop ‘Student Talks in Trending Topics in Theory
(ST4)’ where part of this work was done. PR is grateful to Debangshu Mukherjee
for his generous help and multiple discussions on this work. PR is partially
supported by a grant to CMI from the Infosys Foundation.

##  Appendix A  Analysis to show that $\Delta^{(n)}_{\mu\nu}(t_{0},x_{0})=0$
for all $n$

The $j$-th derivative of $I_{n,m}(R)$ (46) gives

$\displaystyle\partial_{R}^{j}I_{n,m}(R)=\sum_{i=0}^{j-1}\partial_{R}^{j-i-1}\left(\partial_{R}^{i}f_{n,m}(x,R)|_{x=R}\right),+\int_{0}^{R}\partial_{R}^{j}f_{n,m}(x,R)dx$
(73)

where

$f_{n,m}(x,R)=x^{2m}(\cosh R+\cosh x)^{-\frac{n}{2}+1}(\cosh R-\cosh
x)^{\frac{n}{2}+1}.$ (74)

Thus from (48) we have

$\displaystyle\sum_{n,m}\frac{2^{n+1}}{(2m)!}\partial_{x}^{2m}\Delta^{(n)}_{xx}(t_{0},x_{0})\left[\sum_{i=0}^{J+2}\partial_{R}^{J-i+2}\left(\partial_{R}^{i}f_{n,m}(x,R)|_{x=R}\right)\right]_{R=0}=0.$
(75)

First we will show that $\Delta_{xx}=0$ for even integers. For odd integers,
we check it numerically below. Writing $2l$ for the even integer $n$, and
simplifying a bit, we get

$\sum_{l=0}^{J+1}\sum_{m}\frac{(l+1)!\,2^{l}}{(2m)!}\partial_{x}^{2m}\Delta^{(2l)}_{xx}(t_{0},x_{0})C^{J,l,m}(R=0)=0,$
(76)

where

$C^{J,l,m}(R=0)=\left[\partial_{R}^{J-(l-1)}\left(R^{2m}\cosh^{1-l}R\,\sinh^{1+l}R\right)\right]_{R=0}.$
(77)

Notice that we have restricted $l$ to range from $0$ to $J+1.$ This is because
the terms with $l>J+1$ all vanish at $x=R$. Now, (76) can be written as

$\sum_{l}(l+1)!2^{l}\Delta^{(2l)}_{xx}(t_{0},x_{0})C^{J,l,0}(R=0)=-\sum_{l,m\neq
0}\frac{(l+1)!\,2^{l}}{(2m)!}\partial_{x}^{2m}\Delta^{(2l)}_{xx}(t_{0},x_{0})C^{J,l,m}(R=0).$
(78)

Setting $J=0,$ we get

$\Delta_{xx}^{(0)}(t_{0},x_{0})=0.$ (79)

$J=2$ gives

$\displaystyle\Delta_{xx}^{(2)}(t_{0},x_{0})=$ $\displaystyle\
-\frac{1}{2}\Delta_{xx}^{(0)}(t_{0},x_{0})+\frac{3}{8}\partial_{x}^{2}\Delta_{xx}^{(0)}(t_{0},x_{0}),$
$\displaystyle\Rightarrow\qquad\Delta_{xx}^{(2)}(t_{0},x_{0})=$
$\displaystyle\ 0.$ (80)

$J=4$ gives

$\displaystyle\Delta_{xx}^{(4)}(t_{0},x_{0})=$ $\displaystyle\
-\frac{2}{9}\Delta_{xx}^{(2)}(t_{0},x_{0})+\frac{1}{3}\partial_{x}^{2}\Delta_{xx}^{(2)}(t_{0},x_{0})-\frac{1}{9}\Delta_{xx}^{(0)}(t_{0},x_{0})$
$\displaystyle+\frac{5}{18}\partial_{x}^{2}\Delta_{xx}^{(0)}(t_{0},x_{0})+\frac{5}{144}\partial_{x}^{4}\Delta_{xx}^{(0)}(t_{0},x_{0})$
(81) $\displaystyle\Rightarrow\qquad\Delta_{xx}^{(4)}(t_{0},x_{0})=$
$\displaystyle\ 0$ (82)

Let’s check it for $J=2N$. It is easy to see that for all $l\geq N+1$ the
coefficients $C$ are zero for all $m$. For $l=N$

$C^{2N,N,0}(R=0)=[\partial_{R}^{N+1}(\cosh^{1-N}R\sinh^{1+N}R)]_{R=0}=(N+1)!\,.$
(83)

So the last non vanishing term in the $l$ series in (78) with $m=0$ is
$\Delta_{xx}^{(2N)}(t_{0},x_{0})$ and with $m\neq 0$ it will be a lower order
term. Now notice that this term is a linear combination of all the lower order
terms and their derivatives, which are zero. Hence
$\Delta_{xx}^{(2N)}(t_{0},x_{0})$ is also zero for $N=0,1,\ldots\ .$

### Numerical analysis for odd $n$

Consider the expansion (47):

$\sum_{n,m,j}\frac{2^{n+1}}{(2m)!}\partial_{x}^{2m}\Delta^{(n)}_{xx}(t,x_{0})R^{j}C^{j}_{n,m}(0)=0,$
(84)

where

$C^{j}_{n,m}(0)=\frac{1}{j!}\left[\partial^{j}_{R}I_{n,m}(R)\right]_{R=0}.$

We have already seen that the first two terms in this series do not give any
constraint and that the terms with odd powers of $R$ (i.e, odd $j$) impose
constraints on even $n$. Thus one should expect that terms with alternative
powers of $R$ (i.e, even $j$) should impose constraints on odd $n$. Let us
check the term of order $R^{4}.$ Using numerics one can see that the
coefficient $C^{4}_{n,m}(0)$ is zero for all $n,m$ except for $n=1,m=0$. The
relevant plots are shown in figure 5.

At order $R^{6}$, there are three non-zero coefficients, namely,
$C^{6}_{3,0},C^{6}_{1,0},C^{6}_{1,1}$. This implies that
$\Delta^{(3)}_{xx}(t_{0},x_{0})$ can be written in terms of
$\Delta^{(1)}_{xx}(t_{0},x_{0})$ and
$\partial_{x}^{2}\Delta^{(1)}_{xx}(t_{0},x_{0})$, and hence
$\Delta^{(3)}_{xx}(t_{0},x_{0})$ is also 0. Figure 5 shows the behavior of
coefficients at $R=0$.

At order $R^{8}$, there are several non-zero coefficients which are shown in
figure 6. From the plot, one can deduce that $\Delta^{(5)}(t_{0},x_{0})$ can
be written as a linear combination of the lower order terms and their
derivatives, and hence is 0.

From this pattern, we conclude that in general the higher order coefficients
can be written in terms of lower order coefficients. This means that

$\Delta^{(n)}_{xx}(t_{0},x_{0})=0$

for all $n$.

Figure 4: $C^{4}_{n,m}$ vs $R$ for different values of $n$ and $m$.

Figure 5: $C^{6}_{n,m}$ vs $R$ for different values of $n$ and $m$.

Figure 6: $C^{8}_{n,m}$ vs $R$ for different values of $n$ and $m$.

These figures show for a particular $j$ what coefficients $C^{j}_{n,m}$ are
non-zero as $R\to 0$.

#### Some numerical computations for boosted black brane:

Here we have plotted different coefficients of $R^{4}$-th and $R^{5}$-th terms
in the expansion series (65) around $R=0$. They are denoted by
$C^{4}_{\mu\nu}$ and $C^{5}_{\mu\nu}$ respectively.

Figure 7: $C^{4}_{xx}$ vs $R$ for different values of $n,\ m_{t}$ and $m_{x}$.

Figure 8: $C^{4}_{tx}$ vs $R$ for different values of $n,\ m_{t}$ and $m_{x}$.

Figure 9: $C^{4}_{tt}$ vs $R$ for different values of $n,\ m_{t}$ and $m_{x}$.

Figure 10: $C^{5}_{xx}$ vs $R$ for different values of $n,\ m_{t}$ and
$m_{x}$.

Figure 11: $C^{5}_{tx}$ vs $R$ for different values of $n,\ m_{t}$ and
$m_{x}$.

Figure 12: $C^{5}_{tt}$ vs $R$ for different values of $n,\ m_{t}$ and
$m_{x}$.

Figures 8, 8 and 10 show that all the coefficients of $R^{4}$-th term are zero
as $R\to 0$ except when $n=1$ and $m_{t}=m_{x}=0$ which give the first
equation in (69) with $\beta=0.5$. Similarly for $R^{5}$-th term the only non-
zero coefficient with maximum value of $n$ as $R\to 0$ is $n=2$ and
$m_{t}=m_{x}=0$. This can be expressed as a linear combination of the terms
with lower values of $n$ which are already zero. These give the second
equation in (69).

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[a]W. G. Parrott

# The search for new physics in $B\to K\ell^{+}\ell^{-}$ and $B\to
K\nu\bar{\nu}$ using precise lattice QCD form factors

C. Bouchard  C. T. H. Davies

###### Abstract

We present HPQCD’s improved scalar, vector and tensor form factors for $B\to
K$ semileptonic decays, using the heavy-HISQ formalism for more accurate
normalisation of the weak currents. Working with masses close to the physical
$b$ on the finest ensemble and including three ensembles with physical light
quarks, we cover the full physical $q^{2}$ range with good precision. Our
uncertainties at $q^{2}=0$ are a factor of three better than earlier work.

We compare Standard Model observables using our form factors to experimental
measurements for the rare flavour changing neutral current processes $B\to
K\ell^{+}\ell^{-}$ and $B\to K\nu\bar{\nu}$ and discuss the significance of
the tensions that arise.

##  1 Introduction

$\overline{b}$$\overline{s}$$u$$u$$\ell^{-}$$\ell^{+}$$(c1)!0.5!(c3)-(1cm,0)$$\bar{t}$$\bar{t}$$W^{+}$ Figure 1: Feynman diagram for a $B^{+}\to K^{+}\ell^{+}\ell^{-}$ decay. Table 1: Gluon field ensembles used in this work, numbered in column 1. Column 2 gives the approximate value of $a$ for each set. Column 3 gives the spatial ($N_{x}$) and temporal ($N_{t}$) dimensions of each lattice in lattice units and column 4, the number of configurations and time sources used in each case. Columns 5-7 give the masses of the valence and sea quarks in lattice units, noting that $m_{u}=m_{d}=m_{l}$ and the valence and sea masses are the same in the case of $m_{l}$. Column 8 gives the heavy masses on each ensemble. In each case, the lowest mass is tuned to that of the charm. The sea charm mass is not included here, but is close to the valence value. For more details, see [1] Set |  $a$(fm) | $N_{x}^{3}\times N_{t}$ | $n_{\mathrm{cfg}}\times n_{\mathrm{src}}$ | $am_{l}^{\mathrm{sea/val}}$ | $am_{s}^{\mathrm{sea}}$ | $am_{s}^{\mathrm{val}}$ | $am_{h}$
---|---|---|---|---|---|---|---
1 | 0.15 | $32^{3}\times 48$ | $998\times 16$ | 0.00235 | 0.0647 | 0.0678 | 0.8605
2 | 0.12 | $48^{3}\times 64$ | $985\times 16$ | 0.00184 | 0.0507 | 0.0527 | 0.643
3 | 0.088 | $64^{3}\times 96$ | $620\times 8$ | 0.00120 | 0.0363 | 0.036 | 0.433, 0.683, 0.8
4 | 0.15 | $16^{3}\times 48$ | $1020\times 16$ | 0.013 | 0.065 | 0.0705 | 0.888
5 | 0.12 | $24^{3}\times 64$ | $1053\times 16$ | 0.0102 | 0.0509 | 0.0545 | 0.664, 0.8, 0.9
6 | 0.09 | $32^{3}\times 96$ | $499\times 16$ | 0.0074 | 0.037 | 0.0376 | 0.449, 0.566, 0.683, 0.8
7 | 0.059 | $48^{3}\times 144$ | $413\times 8$ | 0.0048 | 0.024 | 0.0234 | 0.274, 0.45, 0.6, 0.8
8 | 0.044 | $64^{3}\times 192$ | $375\times 4$ | 0.00316 | 0.0158 | 0.0165 | 0.194, 0.45, 0.6, 0.8

$B\to K\ell\bar{\ell}$ decays involving $b\to s$ flavour changing neutral
currents are highly suppressed loop processes in the standard model (SM) (see
Figure 1) and are good places to look for new physics. The increasing quantity
of experimental data being collected demands better theoretical bounds on SM
quantities to reveal new physics.

The uncertainties in current SM predictions are dominated by the hadronic form
factors, which previously used the non-relativistic (NRQCD) formalism for the
$b$ quark [2], or the Fermilab ([3]) interpretation [4]. Both of these methods
rely on $\mathcal{O}(\alpha_{s})$ perturbation theory to match the lattice
weak current operators to their continuum counterparts.

In this proceedings, we present the first fully relativistic calculation,
using the HISQ formalism for all valence quarks and working on MILC
$N_{f}=2+1+1$ gluon field ensembles that include HISQ quarks in the sea [5,
6]. This eliminates the aforementioned perturbative matching, and associated
errors.

Here, we summarise the content of [1], which covers the lattice calculation of
the $B\to K$ form factors, and [7], which contains the SM predictions arising
from these new form factor determinations. See these references for more
detail.

##  2 Lattice calculation details

###  2.1 Heavy HISQ

The calculation detailed in [1] uses the heavy-HISQ method. Because working at
the physical $b$ mass requires extremely fine (and so costly) lattice
spacings, $a$, to achieve small discretisation effects, we instead work with a
set of generic heavy quarks of mass $m_{h}$, which result in heavy mesons of
mass $M_{H}$, covering a range of values. In Table 1, we detail each of the 8
MILC ensembles that we use in this work. We use a variety of heavy masses on
each ensemble, generally in the range $am_{c}\leq am_{h}\leq 0.8$. We must
then perform an extrapolation to the physical $b$ mass $m_{h}\to m_{b}$
($M_{H}\to M_{B}$). This method ultimately allows us to calculate the form
factors across the full physical $q^{2}$ range, and evaluate our results at
any mass between $M_{D}$ and $M_{B}$.

###  2.2 Form factor calculation

We wish to calculate the scalar, vector and tensor form factors,
$f_{0}(q^{2})$, $f_{+}(q^{2})$ and $f_{T}(q^{2})$ as functions of the
4-momentum transfer $q^{2}=(M_{H}-E_{K})^{2}-(\vec{p}_{K})^{2}$, where we work
in the rest frame of the parent $H$ meson. We can construct form factors using
the following expressions,

$\begin{split}Z_{V}\bra{K}V^{\mu}_{\mathrm{latt}}\ket{\hat{H}}&=f_{+}(q^{2})\Big{(}p_{H}^{\mu}+p_{K}^{\mu}-\frac{M_{H}^{2}-M_{K}^{2}}{q^{2}}q^{\mu}\Big{)}+f_{0}(q^{2})\frac{M_{H}^{2}-M_{K}^{2}}{q^{2}}q^{\mu},\\\
\bra{K}S_{\mathrm{latt}}\ket{H}&=\frac{M_{H}^{2}-M_{K}^{2}}{m_{h}-m_{s}}f_{0}(q^{2}),\\\
Z_{T}(\mu)\bra{\hat{K}}T^{k0}_{\mathrm{latt}}\ket{\hat{H}}&=\frac{2iM_{H}p_{K}^{k}}{M_{H}+M_{K}}f_{T}(q^{2},\mu),\end{split}$
(1)

where the scalar, vector and tensor matrix elements are extracted from multi-
exponential Bayesian fits to three-point correlation functions calculated on
the lattice (see [1] for details). The vector current is normalised via
$Z_{V}$ using the PCVC relation [8, 9]. The tensor current renormalisation
factor $Z_{T}$ is calculated in [10], using RI-SMOM.

Applying twisted boundary conditions to the strange daughter quark for a range
of different momenta gives us good coverage of the $q^{2}$ range on each
ensemble.

####  2.2.1 Extrapolating to the continuum

Taking our form factor data points at different $am_{h}$, and $q^{2}$ values,
we perform a modified $z$ expansion using the Bourreley-Caprini-Lellouch (BCL)
parameterisation [11],

$\begin{split}f_{0}(q^{2})&=\frac{\mathcal{L}}{1-\frac{q^{2}}{M^{2}_{H_{s0}^{*}}}}\sum_{n=0}^{N-1}a_{n}^{0}z^{n}\\\
f_{+,T}(q^{2})&=\frac{\mathcal{L}}{1-\frac{q^{2}}{M^{2}_{H_{s}^{*}}}}\sum_{n=0}^{N-1}a_{n}^{+,T}\Big{(}z^{n}-\frac{n}{N}(-1)^{n-N}z^{N}\Big{)},\end{split}$
(2)

where
$z(q^{2})=\frac{\sqrt{t_{+}-q^{2}}-\sqrt{t_{+}}}{\sqrt{t_{+}-q^{2}}+\sqrt{t_{+}}}$
and $\mathcal{L}$ is a hard pion chiral logarithm term [12]. The coefficients,

$\begin{split}a_{n}^{0,+,T}=&\Big{(}\frac{M_{D}}{M_{H}}\Big{)}^{\zeta_{n}}\Big{(}1+\rho_{n}^{0,+,T}\log\Big{(}\frac{M_{H}}{M_{D}}\Big{)}\Big{)}\times(1+\mathcal{N}^{0,+,T}_{n})\times\\\
&\sum^{N_{ijkl}-1}_{i,j,k,l=0}d_{ijkln}^{0,+,T}\Big{(}\frac{\Lambda_{\text{QCD}}}{M_{H}}\Big{)}^{i}\Big{(}\frac{am_{h}^{\text{val}}}{\pi}\Big{)}^{2j}\Big{(}\frac{a\Lambda_{\text{QCD}}}{\pi}\Big{)}^{2k}(x_{\pi}-x_{\pi}^{\mathrm{phys}})^{l}\end{split}$
(3)

contain terms allowing for discretisation effects ($am_{h}^{\text{val}}/\pi$
and $a\Lambda_{\text{QCD}}/\pi$) and quark mistunings
($\mathcal{N}^{0,+,T}_{n}$ and $x_{\pi}=M^{2}_{\pi}/(4\pi f_{\pi})^{2}$) (see
[1]), as well as a Heavy Quark Effective Theory inspired fit to the heavy
meson mass $M_{H}$. The coefficients $\rho_{n}^{0,+,T}$, $d_{ijkln}^{0,+,T}$
and $\zeta_{0}$ are tunable fit coefficients, with $\zeta_{n\neq 0}=0$. This
combined $q^{2}$ and $M_{H}$ fit allows us to extrapolate our results to the
physical point; $a=0$ and $M_{H}=M_{B}$.

Figure 2: The scalar, vector and tensor form factors, with poles removed, in
$z$ space. Data points on each ensemble are included as well as a solid band
indicating the continuum result for $B\to K$.

###  2.3 Results

Figure 2 shows the continuum form factors alongside the data points on each
ensemble, in $z$ space, with the pole removed. We see that $z$ space
polynomials are very nearly linear, and that the fit is well behaved. Data on
set 8, our finest ensemble, at $am_{h}=0.8$ (labelled $m0.8$), our highest
mass, closely approaches the physical point, as we would expect, given that
this is very close to the physical $B$ mass. The data points labelled as sets
9 and 10 are correlated $B_{s}\to\eta_{s}$ data on sets 6 and 7, taken from
[13] (see [1]).

Figure 3: The scalar, vector and tensor form factors, at $q^{2}=0$ and
$q^{2}_{\mathrm{max}}$, plotted against $M_{H}$. Other results [14, 15, 2, 4,
16] are included for comparison. For a discussion of the scale, $\mu$, see
[1].

Because our heavy-HISQ approach requires a fit in the heavy mass, with
$am_{c}\leq am_{h}<am_{b}$ we are able to evaluate the form factors at any
mass between $M_{D}$ and $M_{B}$. Figure 3 shows the scalar, vector and tensor
form factors, at extremal $q^{2}$ values (note $f_{+}(0)=f_{0}(0)$), plotted
against $M_{H}$. We can see that we agree well with previous work [14, 15, 2,
4, 16], across the $M_{H}$ and $q^{2}$ range, with the exception of
$f_{T}(q^{2}_{\mathrm{max}},2~{}\mathrm{GeV})$ and
$f_{+}(q^{2}_{\mathrm{max}})$ from ETMC [14, 15]. Our uncertainties at
$q^{2}=0$ are improved by a factor of three over previous work.

##  3 Phenomenology

###  3.1 $B\to K\ell^{+}\ell^{-}$

We can write the SM differential decay rate for $B\to K\ell^{+}\ell^{-}$ in
terms of $f_{0}$, $f_{+}$ and $f_{T}$, as well as the Wilson coefficients
$W_{i}$. This lengthy expression is discussed in detail in [7]. Schematically,

$\frac{d\Gamma^{B\to
K\ell^{+}\ell^{-}}}{dq^{2}}=\mathcal{F}_{1}|F_{P}(f_{0},f_{+},W_{i})|^{2}+\mathcal{F}_{2}f_{+}^{2}+\mathcal{F}_{3}|F_{V}(f_{+},f_{T},W_{i})|^{2}+\mathcal{F}_{4}|f_{+}F_{P}^{*}(f_{0},f_{+},W_{i})|,$
(4)

where $\mathcal{F}_{i}$ are known functions of kinematic factors and $W_{i}$.
This expression does not account for production of $c\bar{c}$ resonances, so
is not valid in all regions of $q^{2}$. We can compare this differential decay
rate (or equivalently the differential branching fraction $\mathcal{B}$,
noting that $\mathcal{B}=\Gamma\tau_{B}$, for $B$ lifetime $\tau_{B}$),
directly with experiment in different $q^{2}$ regions, to look for evidence of
new physics.

Figure 4: Left: our result for $d\mathcal{B}(B^{+}\to
K^{+}e(\mu)^{+}e(\mu)^{-})/dq^{2}$ (blue band), compared with experiment [17,
18, 19, 20, 21, 22]. We linearly interpolate across both vetoed regions in
grey, and the region between them. Right: the same result, but integrated
across the full $q^{2}$ range, compared with experiment [23, 24, 19, 18, 17,
25, 21]. Our value is carried down the page in grey. The dotted lines indicate
additional QED uncertainty (see [7]). Figure 5: The ratio of experimental
branching fractions [19, 21, 22] to our result, over some theoretically clean
$q^{2}$ bins. (1(.1)-6, and 15-22$\mathrm{GeV}^{2}$ respectively). Caps on
error bars indicate 1, 3 and 5 $\sigma$.

The left of Figure 4 shows our result for $d\mathcal{B}(B^{+}\to
K^{+}e(\mu)^{+}e(\mu)^{-})/dq^{2}$, across the full $q^{2}$ range, as well as
data from experiment. Grey bands indicated vetoed regions where $c\bar{c}$
resonances dominate and mean that Eq. (4) is no longer valid. We see that our
result lies considerably above experiment, particularly at low $q^{2}$. On the
right of the same figure, we show the result when integrated across the full
$q^{2}$ range, interpolating across the vetoed regions, as is done in
experiment.

Focusing specifically on the most recent experimental results in $q^{2}$ bins
away from the $c\bar{c}$ resonances, we plot experimental branching fractions,
divided by our result, in Figure 5. Error bars have caps at 1, 3 and 5
$\sigma$. We see that we are in $3-5\sigma$ tension with LHCb [19] results in
these theoretically clean regions of $q^{2}$.

###  3.2 $B\to K\nu\bar{\nu}$

Figure 6: The branching fraction $B^{+}\to K^{+}\nu\bar{\nu}$ compared with
other theoretical work [26, 27, 28, 29, 30, 31, 32] (left) and experimental
bounds (right) [33, 34]. Experimental bounds in blue are current 90%
confidence limits. The red points indicate the expected future uncertainty
from Belle II at 5 and 50 $\mathrm{ab}^{-1}$ [35], centred on our result.

A much cleaner channel, free from resonances, that we can also probe, is $B\to
K\nu\bar{\nu}$, although experimental measurements are much less mature here.
We can write the short distance (SD) differential branching fraction,

$\frac{d\mathcal{B}(B\to
K\nu\bar{\nu})_{\text{SD}}}{dq^{2}}=\frac{(\eta_{\mathrm{EW}}G_{F})^{2}\alpha_{\mathrm{EW}}^{2}X_{t}^{2}}{32\pi^{5}\sin^{4}\theta_{W}}\tau_{B}|V_{tb}V^{*}_{ts}|^{2}|\vec{p}_{K}|^{3}f^{2}_{+}(q^{2}),$
(5)

where the values used are given in [7]. A well determined $~{}10\%$ correction
is added to account for long distance effects in the case $B^{+}\to K^{+}$
(see [7]).

In Figure 6, we present our result for the branching fraction, as well as
other theoretical and experimental work. Our number is slightly larger and
more precise than previous theoretical work, and well within the current
experimental confidence bounds. We have similar uncertainty to that expected
from Belle II with $50\mathrm{ab}^{-1}$ of data [35].

##  4 Conclusions

We present the first fully relativistic determination of the scalar, vector
and tensor form factors for $B\to K$ decays. We have good agreement with
previous determinations, and improved uncertainty at low $q^{2}$.

Using these form factors, we can calculate the SM branching fraction for $B\to
Ke(\mu)^{+}e(\mu)^{-}$ in $q^{2}$ bins which are free from $c\bar{c}$
resonances and measured experimentally. We find tensions above $3\sigma$ in
the ratio of our result with LHCb [19], and above $5\sigma$ in one instance
for the low $q^{2}$ bin.

We also determine the theoretically clean branching fraction $B\to
K\nu\bar{\nu}$, finding a value which is more precise than previous
experimental work. Our result has an uncertainty below 10%, which is
commensurate with the uncertainty expected from Belle II in future, with
$50\mathrm{ab}^{-1}$ of data [35].

##  5 Acknowledgements

We are grateful to the MILC collaboration for the use of their configurations
and their code, which we use to generate quark propagators and construct
correlators. We would also like to thank L. Cooper, J. Harrison, D. Hatton and
G. P. Lepage for useful discussions and B. Chakraborty, J. Koponen and A. T.
Lytle for generating propagators/correlators in previous projects that we
could make use of here. Computing was done on the Cambridge Service for Data
Driven Discovery (CSD3) supercomputer, part of which is operated by the
University of Cambridge Research Computing Service on behalf of the UK Science
and Technology Facilities Council (STFC) DiRAC HPC Facility. The DiRAC
component of CSD3 was funded by BEIS via STFC capital grants and is operated
by STFC operations grants. We are grateful to the CSD3 support staff for
assistance. Funding for this work came from STFC.

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