# Null Infinity and Unitary Representation of The Poincare Group

Shamik Banerjee  Institute of Physics,
Sachivalaya Marg, Bhubaneshwar, India-751005
and
Homi Bhabha National Institute, Anushakti Nagar, Mumbai, India-400085
<EMAIL_ADDRESS>
###### Abstract

Following Pasterski-Shao-Strominger we construct a new basis of states in the
single-particle Hilbert space of massless particles as a linear combination of
standard Wigner states. Under Lorentz transformation the new basis states
transform in the Unitary Principal Continuous Series representation. These
states are obtained if we consider the little group of a null momentum
direction rather than a null momentum. The definition of the states in terms
of the Wigner states makes it easier to study the action of space-time
translation in this basis. We show by taking into account the effect of space-
time translation that the dynamics of massless particles described by these
states takes place completely on the null-infinity of the Minkowski space. We
then second quantize the theory in this basis and obtain a manifestly Poincare
invariant (field) theory of free massless particles living on null-infinity.
This theory has unitary time evolution. The null-infinity arises in this case
purely group-theoretically without any reference to bulk space-time. Action of
BMS is particularly natural in this picture.

As a by-product we generalize the conformal primary wave-functions for
massless particles in a way which makes the action of space-time translation
simple. Using these wave-functions we write down a modified Mellin(-Fourier)
transformation of the S-matrix elements. The resulting amplitude is Poincare
covariant. Under Poincare transformation it transforms like products of
primaries of inhomogeneous $SL(2,\mathbb{C})$ ($ISL(2,\mathbb{C})$) inserted
at various points of null-infinity. $ISL(2,\mathbb{C})$ primaries are defined
in the paper.

###### Contents

  1. I Introduction And Summary Of Main Results
  2. II Lorentz Group In (2+1) Dimensions
    1. II.1 Null Momenta
    2. II.2 Standard Rotation
    3. II.3 Little Group of A Null Momentum Direction
    4. II.4 Hilbert Space Representation
      1. II.4.1 Change of Basis
      2. II.4.2 Action of $U(\Lambda)$
  3. III Representation on Wave Functions (Packets)
    1. III.1 Differential Operators and Casimir
      1. III.1.1 Translation
      2. III.1.2 Special Conformal Transformation (SCT)
      3. III.1.3 Scale Transformation (ST)
      4. III.1.4 Value of the Casimir for $\Delta$-Representation
      5. III.1.5 Comments on The Representation of The Conformal Algebra
  4. IV  (3+1) Dimensions
    1. IV.1 Little Group of a Null Momentum Direction
    2. IV.2 Construction Of New Basis
    3. IV.3 Action of The Lorentz Group
    4. IV.4 Reality Condition on The Generators
  5. V Action of The Space-time Translation Operators
    1. V.1 Derivation
      1. V.1.1 A Geometric Argument
  6. VI Creation And Annihilation Fields At Null-Infinity
  7. VII Primary Of $ISL(2,\mathbb{C})$
  8. VIII Constraints From Translational Invariance
    1. VIII.0.1 4-point function
    2. VIII.0.2 3-point function
    3. VIII.0.3 2-point function
  9. IX A Hint of Supertranslation
  10. X Conformal Primary Wave-Functions With Translational Invariance
  11. XI Acknowledgement

##  I Introduction And Summary Of Main Results

In $(3+1)$ dimensions the Lorentz group $SL(2,\mathbb{C})$ acts as the group
of global conformal transformations on the celestial sphere ($S^{2}$) at null-
infinity. A series of very interesting recent papers Pasterski:2016qvg ;
Pasterski:2017kqt ; Pasterski:2017ylz  have constructed external state wave
functions for massive and massless particles which transform like 2-D
conformal-primary operators under $SL(2,\mathbb{C})$ (Lorentz) transformation.
In particular, the detailed analysis of Pasterski:2017kqt  shows that the
"Unitary principal continuous series representation" of the Lorentz group
plays a central role in this whole construction. One of the interesting points
about this basis of "conformal primary wave-functions" is that the 4-D
scattering amplitudes, expressed in this basis, transform like the correlation
function of products of $SL(2,\mathbb{C})$ primaries inserted at various
points of the celestial sphere at null-infinity. This suggests a relation
between 4-D scattering amplitudes and a 2-D Euclidean CFT defined on the
celestial sphere. This 4D-2D correspondence is a hint of flat-space holography
Pasterski:2016qvg ; Pasterski:2017kqt ; Pasterski:2017ylz ; Cheung:2016iub ;
deBoer:2003vf ; Bagchi:2016bcd ; Bagchi:2016geg ; Cardona:2017keg ;
Banerjee:2017jeg ; Schreiber:2017jsr .

In this paper we further explore this construction by looking at the single-
particle Hilbert space of massless particles. Following Pasterski:2016qvg ;
Pasterski:2017kqt ; Pasterski:2017ylz  we define a new basis in the single
particle Hilbert space which is related to the standard Wigner-states by
Mellin transformation. These new states transform in the "Unitary principal
continuous series" representation of $SL(2,\mathbb{C})$ (Lorentz group).
Geometrically speaking these new states are obtained if, instead of a null
momentum vector, we consider the little-group of a null momentum direction. A
null momentum direction is defined as the set of all null momentum vectors
which point in the same direction and so can be represented by a single point
of the celestial sphere in the momentum space. These states are naturally
written as $\ket{h,\bar{h},z,\bar{z}}$ where $(z,\bar{z})$ are the
stereographic coordinates of the celestial sphere in the momentum space.
$(h,\bar{h})$ are little-group indices defined as,

$h=\frac{1+i\lambda-\sigma}{2},\ \bar{h}=\frac{1+i\lambda+\sigma}{2}$

where $\lambda\in\mathbb{R}$ and $\sigma$ is the helicity of the massless
particle. It is shown in the paper that under a unitary $SL(2,\mathbb{C})$
(Lorentz) transformation $U(\Lambda)$ the state transforms as,

$U(\Lambda)\ket{h,\bar{h},z,\bar{z}}=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\ket{h\
,\bar{h}\ ,\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}},\
\Lambda=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in{SL(2,\mathbb{C})},$

These states are somewhat analogous to position eigenkets in non-relativistic
quantum mechanics. In the state $\ket{h,\bar{h},z,\bar{z}}$ the massless
particle can be thought of as sitting at the point $(z,\bar{z})$ of the
celestial sphere in the momentum space and $(h,\bar{h})$ are internal quantum
numbers. These states form a complete set.

Now the definition of the states $\ket{h,\bar{h},z,\bar{z}}$ in terms of the
Wigner-states makes it easier to study the action of space-time translation on
them. It turns out that the simplest way to represent the action of the space-
time translation on these states is to first consider the "position eigenket
in Heisenberg-picture" defined as,

$\ket{h,\bar{h},u,z,\bar{z}}=e^{iHu}\ket{h,\bar{h},z,\bar{z}}$

where $H$ is the Hamiltonian. These states are analogous to the states
$\ket{t,\vec{x}}$ in non-relativistic quantum mechanics. The action of the
Inhomogeneous $SL(2,\mathbb{C})$ ($ISL(2,\mathbb{C})$) or the Poincare group
on the Heisenberg-picture states are shown to be,

$U(\Lambda)\ket{h,\bar{h},u,z,\bar{z}}=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\ket{h\
,\bar{h}\ ,\frac{u\ (1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}}$
$e^{-il.P}\ket{h,\bar{h},u,z,\bar{z}}=\ket{h,\bar{h},u+f(z,\bar{z},l),z,\bar{z}}$

where $l^{\mu}$ is the space-time translation vector and

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$

So a massless particle described by the quantum state
$\ket{h,\bar{h},z,\bar{z}}$ can be thought of as living in a space(-time) with
three coordinates given by $(u,z,\bar{z})$. The action of the Poincare group
on the $(u,z,\bar{z})$ space(-time) is given by,

$\Lambda(u,z,\bar{z})=\bigg{(}\frac{u\ (1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}}\
,\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$
$T(l)(u,z,\bar{z})=(u+f(z,\bar{z},l),z,\bar{z})$

Here the coordinate $u$ is time-like because we get it from Hamiltonian
evolution of the state.

Now, as far as the action of the Poincare group is concerned, we can identify
the three dimensional space-time as the null-infinity in Minkowski space with
$(u,z,\bar{z})$ identified as the Bondi coordinates at infinity. So from now
on we will refer to this space as null-infinity but there is of course no
"bulk space-time" metric here. The null-infinity arises purely group-
theoretically and the dynamics of a free massless particle takes place in this
space.

We then second-quantize the theory using the basis states
$\ket{h,\bar{h},z,\bar{z}}$ and write down Heisenberg-Picture creation and
annihilation fields which live on the null-infinity. So a manifestly Poincare-
invariant and unitary many-body theory of free massless particles can be
formulated completely on ($(u,z,\bar{z})$ space-time) null-infinity without
any reference to bulk space-time.

The transition from the momentum space to null-infinity is direct.
$(z,\bar{z})$ start as coordinates in the momentum-space but once we take into
account the dynamics, $(z,\bar{z})$ together with the time coordinate $u$
transmute into space-time coordinates, albeit of the boundary. This is also
consistent with the current understanding of the relation between infrared
structure of gauge and gravity theories and asymptotic symmetries of flat
space-time Strominger:2013jfa ; Strominger:2013lka ; He ; Kapec:2014opa ;
Strominger:2014pwa ; Kapec:2016jld ; Campiglia:2014yka ; Campiglia:2015yka ;
Campiglia:2015kxa ; Kapec:2017gsg ; Strominger:2017zoo ; Weinberg:1965nx ;
Cachazo:2014fwa ; Sen:2017xjn ; Sen:2017nim ; Laddha:2017ygw ;
Chakrabarti:2017ltl ; Chakrabarti:2017zmh ; Laddha:2017vfh ; Hawking:2016msc .

This free theory has some interesting properties. For example, the transition
amplitude given by,

$\displaystyle\bra{h,\bar{h},z,\bar{z}}e^{-iH(u-u^{\prime})}\ket{h^{\prime},\bar{h}^{\prime},z^{\prime},\bar{z}^{\prime}}$
$\displaystyle=\bra{h,\bar{h},u,z,\bar{z}}\ket{h^{\prime},\bar{h}^{\prime},u^{\prime},z^{\prime},\bar{z}^{\prime}}$
$\displaystyle=\frac{\delta_{\sigma\sigma^{\prime}}}{2\pi}\frac{\Gamma\big{(}i(\lambda^{\prime}-\lambda)\big{)}}{(1+z\bar{z})^{i(\lambda^{\prime}-\lambda)}}\frac{\delta^{2}(z^{\prime}-z)}{\big{(}-i(u^{\prime}-u+i0+)\big{)}^{i(\lambda^{\prime}-\lambda)}}$

is $SL(2,\mathbb{C})$ (Lorentz) covariant and is manifestly invariant under
space-time translation because of the Dirac delta function. In fact the
transition amplitude retains its form under more general "BMS
supertranslation" given by
$(u,z,\bar{z})\rightarrow(u+g(z,\bar{z}),z,\bar{z})$ where $g(z,\bar{z})$ is
an arbitrary smooth function on the sphere. So we get a hint of
supertranslation in a purely non-gravitational context. We discuss these
things in detail in the paper.

This paper consists of two parts. In the first part we do the $(2+1)$ case as
a warm-up and also to check the correctness the procedure. In the first part
we do not discuss the action of space-time translation. The second part
consists of the $(3+1)$ dimensional case which is our main interest. The two
parts are more or less independent.

Throughout this paper we will use the mostly positive metric signature and use
the conventions of Weinberg Weinberg:1995mt .

##  II Lorentz Group In (2+1) Dimensions

In $(2+1)$ dimensions Lorentz group is $SO(2,1)$. In this paper we shall
consider $SL(2,\mathbb{R})$ which is the double cover of $SO(2,1)$.
$SL(2,\mathbb{R})$ is the group of real two by two matrices with determinant
$1$. The group elements will be denoted by,

$\Lambda=\begin{pmatrix}a&b\\\ c&d\end{pmatrix},\ ad-bc=1,\
(a,b,c,d)\in\mathbb{R}$ (1)

Let us now consider a momentum vector $P=(P^{0},P^{1},P^{2})$ and associate a
real symmetric two by two matrix, denoted by the same letter $P$, defined as

$P=\begin{pmatrix}P^{0}-P^{2}&P^{1}\\\ P^{1}&P^{0}+P^{2}\end{pmatrix},\ \det
P=(P^{0})^{2}-(P^{1})^{2}-(P^{2})^{2}=-P^{2}$ (2)

So the determinant gives the norm of the three vector $P$.

The $SL(2,\mathbb{R})$ matrix $\Lambda$ acts on $P$ according to the following
rule,

$P^{\prime}=\Lambda P\Lambda^{T}$ (3)

Under this transformation $P^{\prime}$ is again a real symmetric matrix and,
$\det P=\det P^{\prime}$ . Therefore $\\{\pm\Lambda\\}$ induces a Lorentz
transformation of the three vector $P$.

###  II.1 Null Momenta

Let us now specialize to the case of null momenta. When $P$ is null we can
define a real variable $z$ as,

$z=\frac{P^{1}}{P^{0}+P^{2}}$ (4)

The variable $z$ has the property that when $P$ transforms under Lorentz
transformation as $P\rightarrow P^{\prime}=\Lambda P\Lambda^{T}$ , $z$
transforms as,

$z\rightarrow z^{\prime}=\Lambda
z=\frac{P^{1^{\prime}}}{P^{0^{\prime}}+P^{2^{\prime}}}=\frac{az+b}{cz+d}$ (5)

The geometrical interpretation of the variable $z$ is that it parametrizes the
space of null momentum directions rather than a specific null vector. To see
this let us consider the set of all null vectors which point in the positive
$2$-direction. These vectors have the form $k^{\mu}=(E,0,E)$ where $E>0$. It
is easy to check that the set of all such null vectors get mapped to the point
$z=0$.

Now let us make a rotation in the $1-2$ plane through an angle $\theta$. Under
this transformation $k^{\mu}\rightarrow
k^{\prime\mu}=(E,E\sin\theta,E\cos\theta)$. This is the most general form of a
null vector which makes an angle $\theta$ with the positive $2$-axis. As a
result of this rotation,

$z=0\rightarrow z^{\prime}=\tan\frac{\theta}{2}$ (6)

So we can see that the parameter $z$ depends only on the direction of the null
ray and the set of all null vectors that point in the same direction
correspond to a single value of $z$. In fact, once we choose the set
$\\{k^{\mu}=(E,0,E),E>0\\}$ as the reference null direction with
$(z=0,\theta=0)$ then there is an invertible correspondence between the space
of null directions and values of $z$ given by Eq-(6).

Geometrically the space of null momentum directions in (2+1) dimensions is a
circle parametrized by the angle $\theta$ and $z$ is the stereographic
coordinate on the circle. The Lorentz group $SL(2,\mathbb{R})$ acts on the $z$
coordinate according to Eq-5.

###  II.2 Standard Rotation

A standard rotation, $R(z)$, is a rotation in the $1-2$ plane which takes the
reference null direction, $\bigg{\\{}k^{\mu}=(E,0,E),E>0\big{|}z=0\bigg{\\}}$
to the null direction
$\bigg{\\{}k^{\mu}=(E,E\sin\theta,E\cos\theta),E>0\big{|}z=\tan\frac{\theta}{2}\bigg{\\}}$.
In matrix form,

$R(z)=\pm\begin{pmatrix}\cos\frac{\theta}{2}&\sin\frac{\theta}{2}\\\ \\\
-\sin\frac{\theta}{2}&\cos\frac{\theta}{2}\end{pmatrix},\
z=\tan\frac{\theta}{2}$ (7)

###  II.3 Little Group of A Null Momentum Direction

In the standard Wigner construction little group of a null vector plays a
central role. We want to consider instead the little group of a null
direction. This is the subgroup of the Lorentz group which does not change the
direction of a null vector. As a result, the set of all null vectors which
point in the same direction, is closed under the action of this subgroup. Let
us now explicitly write down the subgroup in (2+1) dimensions.

The generators of $SL(2,\mathbb{R})$ can be written as $(J_{0},K_{1},K_{2})$,
where $J_{0}$ is the generator of rotation in the $1-2$ plane, $K_{1}$ is the
generator of boost in the $1$-direction and $K_{2}$ is the generator of boost
in the $2$-direction. Algebra satisfied by these generators is,

$[J_{0},K_{1}]=iK_{2},\ [J_{0},K_{2}]=-iK_{1},\ [K_{1},K_{2}]=-iJ_{0}$ (8)

Let us now consider the little group of the reference null direction given by
the set $\bigg{\\{}k^{\mu}=(E,0,E),E>0\big{|}z=0\bigg{\\}}$. The little group
of the null vector, $(1,0,1)$, is generated by the element, $A=-J_{0}-K_{1}$.
It is well known that the one-parameter subgroup generated by $A$ is non-
compact and isomorphic to the group of translation. Now in order to get the
little group of the reference null direction we just have to add the boost
generator $K_{2}$ to this. Therefore, the little group, $D(z=0)$, of the
reference null direction, $\bigg{\\{}k^{\mu}=(E,0,E),E>0\big{|}z=0\bigg{\\}}$,
is generated by two elements, $A=-J_{0}-K_{1}$ and $K_{2}$. One can check
that, $[K_{2},A]=iA$. Similarly the little group of the null direction,
$\bigg{\\{}k^{\mu}=(E,E\sin\theta,E\cos\theta),E>0\big{|}z=\tan\frac{\theta}{2}\bigg{\\}}$,
is given by conjugation, $D(z)=R(z)D(z=0)R(z)^{-1}$.

###  II.4 Hilbert Space Representation

Let us consider massless single particle quantum states in $(2+1)$ dimensions.
The single particle states are denoted by $\ket{p}$ with $p^{2}=0$. The
Lorentz-invariant inner product between the states is given by,
$\bra{p_{1}}\ket{p_{2}}=(2\pi)^{2}\ 2|\vec{p}_{1}|\
\delta^{2}(\vec{p}_{1}-\vec{p}_{2})$.

Let $\Lambda$ be an arbitrary Lorentz transformation and $U(\Lambda)$ its
unitary representation in the Hilbert space which acts on $\ket{p}$ as,

$U(\Lambda)\ket{p}=\ket{\Lambda p}$ (9)

If we take $\Lambda$ to be the boost in the $2$-direction with velocity $v$
then

$U(\Lambda)\ket{E,0,E}=\ket{e^{-\eta}E,0,e^{-\eta}E}$ (10)

where $\eta$ is the rapidity defined as, $\tanh\eta=v$.

Now the $SL(2,\mathbb{R})$ matrix which generates boost in the 2-direction is
given by ,

$B_{2}(\eta)=\pm\begin{pmatrix}e^{\frac{\eta}{2}}&0\\\ \\\
0&e^{-\frac{\eta}{2}}\end{pmatrix}$ (11)

The transformation of the $z$ variable under boost in the $2$-direction is,
$z\rightarrow e^{\eta}z$. So it is a scale transformation for which $z=0$ is a
fixed point. This is another way of seeing that $K_{2}$ is a generator of the
little group of the reference null direction
$\bigg{\\{}k^{\mu}=(E,0,E),E>0\big{|}z=0\bigg{\\}}$.

Let us mention one more important point. Here we are considering single
particle states $\ket{p}$ annihilated by the single non-compact generator
$A=-J_{0}-K_{1}$ of the little group of a null vector, i.e,

$A\ket{E,0,E}=0,E>0$ (12)

####  II.4.1 Change of Basis

The works of Pasterski:2017kqt ; Pasterski:2016qvg ; Cheung:2016iub ;
deBoer:2003vf ; Pasterski:2017ylz  suggest the introduction of the following
states parametrized by a complex number $\Delta$,

$\ket{\Delta,z=0}\ =\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{0}^{\infty}dE\
E^{\Delta-1}\ket{E,0,E}$ (13)

Now if we apply a boost in the $2$-direction with rapidity $\eta$ we get,

$U\big{(}B_{2}(\eta)\big{)}\ket{\Delta,z=0}=\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{0}^{\infty}dE\
E^{\Delta-1}\ket{e^{-\eta}E,0,e^{-\eta}E}\ =e^{\eta\Delta}\ket{\Delta,z=0}$
(14)

Let us now define states $\ket{\Delta,z}$ which are related to
$\ket{\Delta,z=0}$ by standard rotation $R(z)$ in the $1-2$ plane, i.e,

$\boxed{\ket{\Delta,z}:=\frac{1}{(1+z^{2})^{\Delta}}\
U\big{(}R(z)\big{)}\ket{\Delta,z=0}=N_{\Delta}(z)U(R(z))\ket{\Delta,z=0}}$
(15)

We can write down the states explicitly as,

$\ket{\Delta,z}=\frac{1}{(2\pi)^{\frac{3}{2}}}\frac{1}{(1+z^{2})^{\Delta}}\int_{0}^{\infty}dEE^{\Delta-1}\ket{E,E\sin\theta,E\cos\theta}$
(16)

The inner product between two such states can be written as,

$\displaystyle\bra{\Delta_{2},z_{2}}\ket{\Delta_{1},z_{1}}$ (17)
$\displaystyle=\frac{1}{(2\pi)^{3}}\frac{1}{(1+z_{2}^{2})^{\Delta_{2}^{*}}(1+z_{1}^{2})^{\Delta_{1}}}\times$
$\displaystyle\int_{0}^{\infty}\int_{0}^{\infty}dE_{2}dE_{1}{E_{2}}^{\Delta_{2}^{*}-1}{E_{1}}^{\Delta_{1}-1}\bra{E_{2},E_{2}\sin\theta_{2},E_{2}\cos\theta_{2}}\ket{E_{1},E_{1}\sin\theta_{1},E_{1}\cos\theta_{1}}$

where $*$ denotes complex conjugation. This integral is convergent if we
choose $\Delta$ to be of the form $(\frac{1}{2}+i\lambda,\
\lambda\in\mathbb{R})$ Pasterski:2017kqt . With this choice of $\Delta$ the
states that we have defined are delta function normalizable, i.e,

$\bra{\Delta_{2},z_{2}}\ket{\Delta_{1},z_{1}}=\delta(\lambda_{2}-\lambda_{1})\delta(z_{2}-z_{1})$
(18)

where $\Delta_{i}=\frac{1}{2}+i\lambda_{i}$.

####  II.4.2 Action of $U(\Lambda)$

Let us now calculate $U(\Lambda)\ket{\Delta,z}$ where $\Lambda$ is an
arbitrary Lorentz transformation. If we use the definition of the state given
in Eq-(15) we get,

$\displaystyle U(\Lambda)\ket{\Delta,z}$
$\displaystyle=N_{\Delta}(z)U(\Lambda)\ U(R(z))\ket{\Delta,z=0}$ (19)
$\displaystyle=N_{\Delta}(z)U(R(\Lambda z))\ U^{-1}(R(\Lambda z))\ U(\Lambda)\
U(R(z))\ket{\Delta,z=0}$ $\displaystyle=N_{\Delta}(z)U(R(\Lambda z))\
W(\Lambda,z)\ket{\Delta,z=0}$

where $W(\Lambda,z)$ is the Lorentz transformation given by,

$W(\Lambda,z)=U^{-1}(R(\Lambda z))\ U(\Lambda)\ U(R(z))=U(R^{-1}(\Lambda
z)\Lambda R(z))$ (20)

Now using the definition of the state $\ket{\Delta,z=0}$ we get,

$W(\Lambda,z)\ket{\Delta,z=0}=\frac{1}{(2\pi)^{\frac{3}{2}}}\int_{0}^{\infty}dE\
E^{\Delta-1}W(\Lambda,z)\ket{E,0,E}$ (21)

So we have to find out the effect of the transformation $R^{-1}(\Lambda
z)\Lambda R(z)$ on the set of null vectors of the form $(E,0,E)$, pointing in
the positive $2$-direction which is our reference direction. Since Lorentz
transformation is linear it is sufficient to consider its effect on the vector
$k^{\mu}=(1,0,1)$. The transformations act on $k^{\mu}$ in the following way :

1) $R(z)$ sends the vector $k^{\mu}=(1,0,1)$ to the vector
$k^{\prime\mu}=(1,\sin\theta,\cos\theta)$. The corresponding $z$ transforms
from $z=0$ to $z=\tan\frac{\theta}{2}$.

2) Now we make an arbitrary Lorentz transformation $\Lambda$ given by the
$SL(2,\mathbb{R})$ matrix,

$\Lambda=\begin{pmatrix}a&b\\\ c&d\end{pmatrix},\ ad-bc=1$ (22)

This transformation sends $k^{\prime}$ to $\Lambda k^{\prime}$ and
$z=\tan\frac{\theta}{2}$ to

$\Lambda
z=\frac{a\tan\frac{\theta}{2}+b}{c\tan\frac{\theta}{2}+d}=\tan\frac{\Lambda\theta}{2}$
(23)

where $\Lambda\theta$ is the angle made by $\Lambda k^{\prime}$ with the
positive $2$-direction.

The components of $\Lambda k^{\prime}$ are given by,

$(\Lambda k^{\prime})^{0}=\frac{(cz+d)^{2}+(az+b)^{2}}{1+z^{2}},\
z=\tan\frac{\theta}{2}$ (24)

$(\Lambda k^{\prime})^{1}=\frac{2(az+b)(cz+d)}{1+z^{2}},\
z=\tan\frac{\theta}{2}$ (25)

$(\Lambda k^{\prime})^{2}=\frac{(cz+d)^{2}-(az+b)^{2}}{1+z^{2}},\
z=\tan\frac{\theta}{2}$ (26)

We also have the relation,

$(\Lambda k^{\prime})^{2}=(\Lambda k^{\prime})^{0}\cos\Lambda\theta,\ (\Lambda
k^{\prime})^{1}=(\Lambda k^{\prime})^{0}\sin\Lambda\theta$ (27)

3) Now the final rotation $R^{-1}(\Lambda z)$ rotates the vector $\Lambda
k^{\prime}$ in 1-2 plane through an angle $-\Lambda\theta$ and the resulting
vector points again in the positive 2-direction. Similarly the rotation brings
$\Lambda z=\tan\frac{\Lambda\theta}{2}$ back to $z=0$ which is the reference
null direction $\bigg{\\{}(E,0,E),E>0\bigg{|}z=0\bigg{\\}}$. The form of the
matrix $R^{-1}(\Lambda z)$ is ,

$R^{-1}(\Lambda
z)=\pm\begin{pmatrix}\cos\frac{\Lambda\theta}{2}&-\sin\frac{\Lambda\theta}{2}\\\
\\\ \sin\frac{\Lambda\theta}{2}&\cos\frac{\Lambda\theta}{2}\end{pmatrix},\
\Lambda z=\tan\frac{\Lambda\theta}{2}$ (28)

Therefore the resulting vector can be written as,

$\bigg{(}R^{-1}(\Lambda z)\Lambda
R(z)\bigg{)}(1,0,1)=\bigg{(}\frac{(cz+d)^{2}+(az+b)^{2}}{1+z^{2}},\ 0,\
\frac{(cz+d)^{2}+(az+b)^{2}}{1+z^{2}}\bigg{)}$ (29)

Here we have made use of the facts that the $0$-th component of a vector does
not change under rotation in the $1-2$ plane and the transformed vector is a
null vector.

For a general vector $k^{\mu}$ of the form $(E,0,E)$ we can write,

$\bigg{(}R^{-1}(\Lambda z)\Lambda
R(z)\bigg{)}(E,0,E)=B_{2}(\eta)(E,0,E)=(e^{-\eta}E,0,e^{-\eta}E),\ E>0$ (30)

where we have defined the boost factor $e^{-\eta}$ as,

$\boxed{e^{-\eta}=\frac{(cz+d)^{2}+(az+b)^{2}}{1+z^{2}}=(cz+d)^{2}\
\frac{1+(\Lambda z)^{2}}{1+z^{2}},\ \Lambda z=\frac{az+b}{cz+d}}$ (31)

With this information we can now derive the transformation law as,

$\displaystyle U(\Lambda)\ket{\Delta,z}$
$\displaystyle=N_{\Delta}(z)U(R(\Lambda z))\ \underbrace{U^{-1}(R(\Lambda z))\
U(\Lambda)\ U(R(z))}\ket{\Delta,z=0}$ (32)
$\displaystyle=N_{\Delta}(z)U(R(\Lambda z))\ W(\Lambda,z)\ket{\Delta,z=0}$
$\displaystyle=N_{\Delta}(z)U(R(\Lambda z))\ e^{\eta\Delta}\ket{\Delta,z=0},\
\text{We have used Eq-$\ref{LG}$, Eq-$\ref{LT}$, Eq-$\ref{beigenvalue}$}$
$\displaystyle=e^{\eta\Delta}\frac{N_{\Delta}(z)}{N_{\Delta}(\Lambda z)}\
\underbrace{N_{\Delta}(\Lambda z)U(R(\Lambda z))\ket{\Delta,z=0}}$
$\displaystyle=e^{\eta\Delta}\frac{N_{\Delta}(z)}{N_{\Delta}(\Lambda
z)}\ket{\Delta,\Lambda z},\ \text{We have used Eq-$\ref{define}$}$
$\displaystyle=\frac{1}{(cz+d)^{2\Delta}}\
\frac{(1+z^{2})^{\Delta}}{(1+(\Lambda z)^{2})^{\Delta}}\ \frac{(1+(\Lambda
z)^{2})^{\Delta}}{(1+z^{2})^{\Delta}}\ \ket{\Delta,\Lambda z}$
$\displaystyle=\frac{1}{(cz+d)^{2\Delta}}\ket{\Delta,\Lambda z}$

So a general Lorentz transformation $\Lambda$ acts on the states
$\ket{\Delta,z}$ like,

$\boxed{U(\Lambda)\ket{\Delta,z}=\frac{1}{(cz+d)^{2\Delta}}\ket{\Delta,\Lambda
z},\ \Lambda=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in SL(2,\mathbb{R})}$ (33)

Now using this it is easy to see that the states
$\big{\\{}\ket{\Delta,\Lambda}\big{\\}}$, for a fixed $\Delta$, form a
representation of the Lorentz group. In order to see that let us consider two
Lorentz transformations $\Lambda_{1}$ and $\Lambda_{2}$.

$\displaystyle U(\Lambda_{2})U(\Lambda_{1})\ket{\Delta,z}$
$\displaystyle=U(\Lambda_{2})\bigg{(}\frac{1}{(c_{1}z+d_{1})^{2\Delta}}\ket{\Delta,\Lambda_{1}z}\bigg{)}$
(34) $\displaystyle=\frac{1}{(c_{1}z+d_{1})^{2\Delta}}\
U(\Lambda_{2})\ket{\Delta,\Lambda_{1}z}$
$\displaystyle=\frac{1}{(c_{1}z+d_{1})^{2\Delta}}\
\frac{1}{(c_{2}(\Lambda_{1}z)+d_{2})^{2\Delta}}\ket{\Delta,\Lambda_{2}\Lambda_{1}z}$
$\displaystyle=\frac{1}{(c_{\Lambda_{2}\Lambda_{1}}z+d_{\Lambda_{2}\Lambda_{1}})^{2\Delta}}\ket{\Delta,\Lambda_{2}\Lambda_{1}z}$
$\displaystyle=U(\Lambda_{2}\Lambda_{1})\ket{\Delta,z}$

Where,

$\Lambda_{2}\Lambda_{1}=\begin{pmatrix}a_{\Lambda_{2}\Lambda_{1}}&b_{\Lambda_{2}\Lambda_{1}}\\\
c_{\Lambda_{2}\Lambda_{1}}&d_{\Lambda_{2}\Lambda_{1}}\end{pmatrix}=\begin{pmatrix}a_{2}&b_{2}\\\
c_{2}&d_{2}\end{pmatrix}\ \begin{pmatrix}a_{1}&b_{1}\\\
c_{1}&d_{1}\end{pmatrix}$ (35)

Therefore,

$U(\Lambda_{2})U(\Lambda_{1})=U(\Lambda_{2}\Lambda_{1})$ (36)

and so $U(\Lambda)$ form a representation of the Lorentz group when acting on
the states $\big{\\{}\ket{\Delta,z}\big{\\}}$ according to Eq-33. So we have
representations of the Lorentz group parametrized by
{$\Delta=\frac{1}{2}+i\lambda,\lambda\in\mathbb{R}\big{\\}}$.

This is also a unitary representation as one can easily check that,

$\big{(}U(\Lambda)\ket{\Delta^{\prime},z^{\prime}},U(\Lambda)\ket{\Delta,z}\big{)}=\big{(}\ket{\Delta^{\prime},z^{\prime}},\ket{\Delta,z}\big{)}$
(37)

The representation labelled by a fixed $\Delta$ is known as the "Unitary
principal continuous series" representation of the Lorentz group bargman
which has also appeared in several recent investigations of CFT Gadde:2017sjg
; Hogervorst:2017sfd ; Simmons-Duffin:2017nub .

##  III Representation on Wave Functions (Packets)

The inner product between the states is given by,

$\bra{\Delta^{\prime},z^{\prime}}\ket{\Delta,z}=\delta(\lambda^{\prime}-\lambda)\delta(z^{\prime}-z),\
\Delta=\frac{1}{2}+i\lambda,\ \lambda\in\mathbb{R}$ (38)

So we can write the completeness relation as,

$\ket{\Psi}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}d\lambda\ dz\
\ket{\Delta,z}\bra{\Delta,z}\ket{\Psi}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}d\lambda\
dz\ \ket{\Delta,z}\Psi(\Delta,z)$ (39)

where $\ket{\Psi}$ is an arbitrary massless one-particle state in the QFT.
Using this the inner product between two states can be written as,

$\bra{\Phi}\ket{\Psi}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}d\lambda\
dz\ \Phi^{*}(\Delta,z)\ \Psi(\Delta,z)$ (40)

where $*$ denotes complex conjugation. We want to find out the effect of an
arbitrary Lorentz transformation $\Lambda$ on the wave function
$\Psi(\Delta,z)$. In order to do that we can write,

$\displaystyle\big{(}U(\Lambda^{-1})\Psi\big{)}(\Delta,z)$
$\displaystyle=\big{(}\ket{\Delta,z},U(\Lambda^{-1})\ket{\Psi}\big{)}$ (41)
$\displaystyle=\big{(}U(\Lambda^{-1})^{\dagger}\ket{\Delta,z},\ket{\Psi}\big{)}=\big{(}U(\Lambda)\ket{\Delta,z},\ket{\Psi}\big{)}$
$\displaystyle=\big{(}U(\Lambda)\ket{\Delta,z},\ket{\Psi}\big{)}$
$\displaystyle=\bigg{(}\frac{1}{(cz+d)^{2\Delta}}\ket{\Delta,\Lambda
z},\ket{\Psi}\bigg{)}$ $\displaystyle=\frac{1}{(cz+d)^{2\Delta^{*}}}\
\Psi(\Delta,\Lambda z),\ \Delta^{*}=\frac{1}{2}-i\lambda$

Therefore the action of the Lorentz group on the wave-functions is,

$\boxed{\big{(}U(\Lambda^{-1})\Psi\big{)}\big{(}\Delta,z\big{)}=\frac{1}{(cz+d)^{2\Delta^{*}}}\
\Psi\bigg{(}\Delta,\frac{az+b}{cz+d}\bigg{)},\
\Delta^{*}=\frac{1}{2}-i\lambda}$ (42)

###  III.1 Differential Operators and Casimir

The Lorentz group $SL(2,\mathbb{R})$ is the group of conformal transformations
of the line (or circle) parametrized by $z$. So let us consider dilatation,
special conformal transformation and translation separately.

####  III.1.1 Translation

Let us consider translation given by,

$T(\alpha)z=z^{\prime}=z+\alpha,\ \Lambda=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}=\begin{pmatrix}1&\alpha\\\ 0&1\end{pmatrix}$ (43)

Now after a straightforward calculation, which we describe in the Appendix, we
get,

$U\big{(}T(\alpha)\big{)}=e^{i\alpha\big{(}J_{0}-K_{1}\big{)}}=e^{\alpha
L_{-1}}$ (44)

So,

$(L_{-1})^{\dagger}=-L_{-1}$ (45)

Now using Eq-42, we get,

$L_{-1}=i(J_{0}-K_{1})=-\frac{d}{dz}$ (46)

####  III.1.2 Special Conformal Transformation (SCT)

Let us consider the SCT given by,

$S(\alpha)z=z^{\prime}=\frac{z}{1-\alpha z}\ ,\ \Lambda=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}=\begin{pmatrix}1&0\\\ -\alpha&1\end{pmatrix}$ (47)

After a starightforward calculation we get,

$U\big{(}S(\alpha)\big{)}=e^{i\alpha(J_{0}+K_{1})}=e^{\alpha L_{1}}$ (48)

So,

$(L_{1})^{\dagger}=-L_{1}$ (49)

In differential operator form we get,

$L_{1}=i(J_{0}+K_{1})=-z^{2}\frac{d}{dz}-2\Delta^{*}z$ (50)

####  III.1.3 Scale Transformation (ST)

Consider the scale transformation given by,

$D(\alpha)z=z^{\prime}=e^{\alpha}z\ ,\ \Lambda=\begin{pmatrix}a&b\\\
c&d\end{pmatrix}=\begin{pmatrix}e^{\frac{\alpha}{2}}&0\\\
0&e^{-\frac{\alpha}{2}}\end{pmatrix}$ (51)

It is easy to check that,

$U\big{(}D(\alpha)\big{)}=e^{i\alpha K_{2}}=e^{\alpha L_{0}}$ (52)

So,

${L_{0}}^{\dagger}=-L_{0}$ (53)

So in differential operator form,

$L_{0}=iK_{2}=-z\frac{d}{dz}-\Delta^{*}$ (54)

####  III.1.4 Value of the Casimir for $\Delta$-Representation

The Casimir of the Lorentz group is given by,

$C=-(J_{0})^{2}+(K_{1})^{2}+(K_{2})^{2}$ (55)

Let us now consider a representation with a specific value of $\Delta$.
According to our previous calculation, in such a representation,

$J_{0}=\frac{i}{2}(z^{2}+1)\frac{d}{dz}+i\Delta^{*}z$ (56)

$K_{1}=\frac{i}{2}(z^{2}-1)\frac{d}{dz}+i\Delta^{*}z$ (57)

$K_{2}=iz\frac{d}{dz}+i\Delta^{*}$ (58)

Now one can easily check that,

$C\Psi(\Delta,z)=\bigg{(}-(J_{0})^{2}+(K_{1})^{2}+(K_{2})^{2}\bigg{)}\Psi(\Delta,z)=\Delta\Delta^{*}\
\Psi(\Delta,z)=\big{(}\lambda^{2}+\frac{1}{4}\big{)}\Psi(\Delta,z)$ (59)

So,

$\boxed{C=\Delta^{*}\Delta=\frac{1}{4}+\lambda^{2}}\ ,\
\Delta=\frac{1}{2}+i\lambda,\ \lambda\in\mathbb{R}$ (60)

This is the value of the Casimir obtained in bargman .

####  III.1.5 Comments on The Representation of The Conformal Algebra

The algebra of Lorentz generators in $(2+1)$ dimensions can be written as,

$\big{[}L_{0},L_{1}\big{]}=-L_{1},\ \big{[}L_{0},L_{-1}\big{]}=L_{-1},\
\big{[}L_{1},L_{-1}\big{]}=2L_{0}$ (61)

This is also the algebra of global conformal group $SL(2,\mathbb{R})$ in one
dimension. But as we have seen the reality properties of the generators are
different if we think of the $SL(2,\mathbb{R})$ as acting unitarily on the
Hilbert space of massless single particle states of a $(2+1)$ dimensional QFT.
For example in the standard highest weight representation of the conformal
group,

$L_{0}^{\dagger}=L_{0},\ L_{1}^{\dagger}=L_{-1},\ L_{-1}^{\dagger}=L_{1}$ (62)

whereas in our case unitarity requires that,

$L_{0}^{\dagger}=-L_{0},\ L_{1}^{\dagger}=-L_{1},\ L_{-1}^{\dagger}=-L_{-1}$
(63)

##  IV  (3+1) Dimensions

In four dimensions the same argument goes through unchanged except that the
states now acquire an extra helicity index. For the sake of convenience we
will collect the necessary formulas.

In four dimensions we can associate a hermitian matrix with a four momentum
$P^{\mu}=(P^{0},P^{1},P^{2},P^{3})$ as,

$P=\begin{pmatrix}P^{0}-P^{3}&P^{1}+iP^{2}\\\
P^{1}-iP^{2}&P^{0}+P^{3}\end{pmatrix},\ \det P=-P^{2}$ (64)

In $(3+1)$ dimensions $SL(2,\mathbb{C})$ is the double cover of the Lorentz
group $SO(3,1)$ and an $SL(2,\mathbb{C})$ matrix $\Lambda$ acts on $P$ as,

$P\rightarrow P^{\prime}=\Lambda P\Lambda^{\dagger},\
\Lambda=\begin{pmatrix}a&b\\\ c&d\end{pmatrix},\ ad-bc=1$ (65)

In four dimensions the space of null directions is a two-sphere (a space-like
cross section of the future light-cone in the momentum space). The
stereographic coordinate of the two-sphere can be defined as,

$z=\frac{P^{1}+iP^{2}}{P^{0}+P^{3}}$ (66)

Here the projection is from the south-pole of the sphere so that the north-
pole has coordinate $z=0$, which corresponds to the family of null vectors,
$\big{\\{}(E,0,0,E)\big{|}E>0\big{\\}}$, pointing in the positive
$3$-direction. One can also check that under Lorentz transformation $\Lambda$,
$z$ transforms as,

$z\rightarrow
z^{\prime}=\frac{{P^{\prime}}^{1}+i{P^{\prime}}^{2}}{{P^{\prime}}^{0}+{P^{\prime}}^{3}}=\frac{az+b}{cz+d}$
(67)

Now if we introduce spherical polar coordinates in momentum space then we can
write a null vector $P^{\mu}$ as,

$P^{\mu}=(E,E\sin\theta\cos\phi,E\sin\theta\sin\phi,E\cos\theta)$ (68)

In this prametrization $(\theta,\phi)$ become coordinates on the two-sphere.
Its relation to the stereographic coordinates $z$ is given by,

$z=\tan\frac{\theta}{2}e^{i\phi}$ (69)

###  IV.1 Little Group of a Null Momentum Direction

In four dimensions the little group of the standard null direction
$\bigg{\\{}(E,0,0,E),E>0\big{|}z=0\bigg{\\}}$ is generated by the elements
$\big{(}J_{3},K_{3},A,B\big{)}$ where $A=J_{2}-K_{1}$ and $B=-J_{1}-K_{2}$.
The commutators are given by,

$[A,B]=0\ ,[J_{3},A]=iB,\ [J_{3},B]=-iA,\ [J_{3},K_{3}]=0,\ [K_{3},A]=iA,\
[K_{3},B]=iB$ (70)

###  IV.2 Construction Of New Basis

Let us define the following state,

$\ket{\lambda,\sigma,z=0,\bar{z}=0}=\frac{1}{\sqrt{8\pi^{4}}}\int_{0}^{\infty}dE\
E^{i\lambda}\ket{E,0,0,E\ ;\sigma},\ \lambda\in\mathbb{R}$ (71)

where $\sigma$ is the helicity. The momentum states have the standard
normalisation given by,

$\bra{p_{1},\sigma_{1}}\ket{p_{2},\sigma_{2}}=(2\pi)^{3}2|\vec{p}_{1}|\delta^{3}(\vec{p}_{1}-\vec{p}_{2})\delta_{\sigma_{1}\sigma_{2}}$
(72)

The action of the little group on the state
$\ket{\lambda,\sigma,z=0,\bar{z}=0}$ is given by,

$A\ket{\lambda,\sigma,z=0,\bar{z}=0}=B\ket{\lambda,\sigma,z=0,\bar{z}=0}=0$
(73)

$U\big{(}R_{3}(\phi)\big{)}\ket{\lambda,\sigma,z=0,\bar{z}=0}=e^{i\sigma\phi}\ket{\lambda,\sigma,z=0,\bar{z}=0}$
(74)

$U\big{(}B_{3}(\eta)\big{)}\ket{\lambda,\sigma,z=0,\bar{z}=0}=e^{\eta\Delta}\ket{\lambda,\sigma,z=0,\bar{z}=0},\
\Delta=1+i\lambda$ (75)

where $\phi$ is the angle of rotation around the 3-axis and $\eta$ is the
rapidity of the boost in the 3-direction.

Now let us define the states $\ket{\lambda,\sigma,z,\bar{z}}$ as,

$\boxed{\ket{\lambda,\sigma,z,\bar{z}}=\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}\
U\big{(}R(z,\bar{z})\big{)}\ket{\lambda,\sigma,z=0,\bar{z}=0}=N(z,\bar{z})\
U\big{(}R(z,\bar{z})\big{)}\ket{\lambda,\sigma,z=0,\bar{z}=0}}$ (76)

where $U\big{(}R(z,\bar{z})\big{)}$ is a unitary rotation operator defined as,

$U\big{(}R(z,\bar{z})\big{)}=e^{-i\phi J_{3}}\ e^{-i\theta J_{2}}\ e^{i\phi
J_{3}},\ z=\tan\frac{\theta}{2}e^{i\phi}$ (77)

The rotation operator we have defined takes the standard null direction
$\big{\\{}(E,0,0,E)\big{|}z=0\big{\\}}$ to the direction given by
$\bigg{\\{}(E,E\sin\theta\cos\phi,E\sin\theta\sin\phi,E\cos\theta)\bigg{|}z=\tan\frac{\theta}{2}e^{i\phi}\bigg{\\}}$.

Now with our choice of normalization the inner product between the states is
given by,

$\bra{\lambda_{1},\sigma_{1},z_{1},\bar{z}_{1}}\ket{\lambda_{2},\sigma_{2},z_{2},\bar{z}_{2}}=\delta(\lambda_{1}-\lambda_{2})\delta^{2}(z_{1}-z_{2})\delta_{\sigma_{1}\sigma_{2}}$
(78)

where $\delta^{2}(z_{1}-z_{2})=\delta(\real z_{1}-\real
z_{2})\delta(\imaginary z_{1}-\imaginary z_{2})$.

###  IV.3 Action of The Lorentz Group

Now we want to compute $U(\Lambda)\ket{\lambda,\sigma,z,\bar{z}}$. The
computation is straightforward and we will summarize only the essential
elements with obvious notation. Just as before,

$\displaystyle U(\Lambda)\ket{\lambda,\sigma,z,\bar{z}}$
$\displaystyle=N(z,\bar{z})U(\Lambda)U(R(z,\bar{z}))\ket{\lambda,\sigma,z=0,\bar{z}=0}$
(79) $\displaystyle=N(z,\bar{z})U(\Lambda
R(z,\bar{z}))\ket{\lambda,\sigma,z=0,\bar{z}=0}$
$\displaystyle=N(z,\bar{z})U(R(\Lambda z,\Lambda\bar{z}))U(R^{-1}(\Lambda
z,\Lambda\bar{z})\Lambda R(z,\bar{z}))\ket{\lambda,\sigma,z=0,\bar{z}=0}$
$\displaystyle=N(z,\bar{z})U(R(\Lambda
z,\Lambda\bar{z}))U(W(\Lambda,z,\bar{z}))\ket{\lambda,\sigma,z=0,\bar{z}=0}$

Now the Lorentz transformation $W(\Lambda,z,\bar{z})$ belongs to the little
group of the standard null direction $\big{\\{}(E,0,0,E)\big{|}z=0\big{\\}}$
and we need to calculate this.

We will evaluate this by multiplying the corresponding $SL(2,\mathbb{C})$
matrices. Please see the appendix for our convention.

The matrices are given by,

$R(z,\bar{z})=\begin{pmatrix}\cos\frac{\theta}{2}&e^{i\phi}\sin\frac{\theta}{2}\\\
-e^{-i\phi}\sin\frac{\theta}{2}&\cos\frac{\theta}{2}\end{pmatrix}=\frac{1}{\sqrt{(1+|z|^{2})}}\begin{pmatrix}1&z\\\
-\bar{z}&1\end{pmatrix},\ z=\tan\frac{\theta}{2}e^{i\phi}$ (80)

$\Lambda=\begin{pmatrix}a&b\\\ c&d\end{pmatrix}\in SL(2,\mathbb{C})$ (81)

$R^{-1}(\Lambda z,\Lambda\bar{z})=\frac{1}{\sqrt{(1+|\Lambda
z|^{2})}}\begin{pmatrix}1&-\Lambda z\\\ \Lambda\bar{z}&1\end{pmatrix},\ $ (82)

where,

$\Lambda z=\frac{az+b}{cz+d},\
\Lambda\bar{z}=\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}$ (83)

If we multiply these matrices the little group element $W(\Lambda,z,\bar{z})$
can be written as,

$W(\Lambda,z,\bar{z})=\begin{pmatrix}e^{\frac{\alpha}{2}}&0\\\
0&e^{-\frac{\alpha}{2}}\end{pmatrix}\begin{pmatrix}1&0\\\
-\beta&1\end{pmatrix}=D(\alpha)S(\beta)$ (84)

where

$e^{\alpha}=\frac{1+|z|^{2}}{1+|\Lambda z|^{2}}\frac{1}{(cz+d)^{2}}$ (85)

We do not need the expression of $\beta$. So we get,

$U(W)=U(D(\alpha))U(S(\beta))$ (86)

Now it is easy to check that ,

$U(D(\alpha))=e^{\alpha L_{0}+\bar{\alpha}\bar{L}_{0}},\
D(\alpha)z=e^{\alpha}z$ (87)

where

$L_{0}=\frac{iK_{3}-J_{3}}{2},\ \bar{L}_{0}=-{L_{0}}^{\dagger}$ (88)

Similarly,

$U(S(\beta))=e^{\beta L_{1}+\bar{\beta}\bar{L}_{1}},\
S(\beta)z=\frac{z}{1-\beta z}$ (89)

where,

$L_{1}=\frac{iK_{1}-J_{1}}{2}+i\frac{iK_{2}-J_{2}}{2},\
\bar{L}_{1}=-{L_{1}}^{\dagger}$ (90)

For later use let us also write down the generator of translation in $z$. It
is given by,

$U(T(\gamma))=e^{\gamma L_{-1}+\bar{\gamma}\bar{L}_{-1}},\
T(\gamma)z=z+\gamma$ (91)

where

$L_{-1}=\frac{J_{1}-iK_{1}}{2}-i\frac{J_{2}-iK_{2}}{2},\
\bar{L}_{-1}=-{L_{-1}}^{\dagger}$ (92)

Now let us take note of the fact that the generator of the special conformal
transformations on $z$ can be written as,

$L_{1}=\frac{1}{2}(B-iA),\ \bar{L}_{1}=-\frac{1}{2}(B+iA)$ (93)

where $A$ and $B$ are the elements of the little group of the null vector
$(E,0,0,E)$. These are the generators which are set to zero on the states
$\ket{E,0,0,E;\sigma}$ to satisfy the requirement of a finite number of
polarization states of a massless particle. Using these facts we get,

$\displaystyle U(\Lambda)\ket{\lambda,\sigma,z,\bar{z}}$
$\displaystyle=N(z,\bar{z})U(R(\Lambda
z,\Lambda\bar{z}))U(W(\Lambda,z,\bar{z}))\ket{\lambda,\sigma,0,0}$ (94)
$\displaystyle=N(z,\bar{z})U(R(\Lambda
z,\Lambda\bar{z}))U(D(\alpha))U(S(\beta))\ket{\lambda,\sigma,0,0}$
$\displaystyle=N(z,\bar{z})U(R(\Lambda
z,\Lambda\bar{z}))U(D(\alpha))\ket{\lambda,\sigma,0,0}$
$\displaystyle=N(z,\bar{z})U(R(\Lambda
z,\Lambda\bar{z}))(e^{\alpha})^{L_{0}}(e^{\bar{\alpha}})^{\bar{L}_{0}}\ket{\lambda,\sigma,0,0}$
$\displaystyle=N(z,\bar{z})U(R(\Lambda
z,\Lambda\bar{z}))\bigg{[}\frac{1+|z|^{2}}{1+|\Lambda
z|^{2}}\frac{1}{(cz+d)^{2}}\bigg{]}^{\frac{\Delta-\sigma}{2}}\bigg{[}\frac{1+|z|^{2}}{1+|\Lambda
z|^{2}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2}}\bigg{]}^{\frac{\Delta+\sigma}{2}}\ket{\lambda,\sigma,0,0}$
$\displaystyle=\frac{1}{(1+|z|^{2})^{\Delta}}\frac{(1+|z|^{2})^{\Delta}}{(1+|\Lambda
z|^{2})^{\Delta}}\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}U(R(\Lambda
z,\Lambda\bar{z}))\ket{\lambda,\sigma,0,0}$
$\displaystyle=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\ket{\lambda,\sigma,\Lambda
z,\Lambda\bar{z}}$

where we have used the definition of the states
$\ket{\lambda,\sigma,z,\bar{z}}$ and have defined,

$\boxed{h=\frac{\Delta-\sigma}{2}=\frac{1+i\lambda-\sigma}{2},\
\bar{h}=\frac{\Delta+\sigma}{2}=\frac{1+i\lambda+\sigma}{2}}$ (95)

Here $\bar{h}$ is not the complex conjugate of $h$.

Therefore, if we rename the state $\ket{\lambda,\sigma,z,\bar{z}}$ as
$\ket{h,\bar{h},z,\bar{z}}$, we can write,

$\boxed{U(\Lambda)\ket{h,\bar{h},z,\bar{z}}=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\ket{h,\bar{h},\Lambda
z,\Lambda\bar{z}}=\bigg{(}\frac{d\Lambda
z}{dz}\bigg{)}^{h}\bigg{(}\frac{d\Lambda\bar{z}}{d\bar{z}}\bigg{)}^{\bar{h}}\ket{h,\bar{h},\Lambda
z,\Lambda\bar{z}},}$ (96)

where $\Lambda\in SL(2,\mathbb{C})$. Using this transformation law one can
easily check that,

$\bigg{(}U(\Lambda)\ket{h^{\prime},\bar{h}^{\prime},z^{\prime},\bar{z}^{\prime}},U(\Lambda)\ket{h,\bar{h},z,\bar{z}}\bigg{)}=\bigg{(}\ket{h^{\prime},\bar{h}^{\prime},z^{\prime},\bar{z}^{\prime}},\ket{h,\bar{h},z,\bar{z}}\bigg{)}$
(97)

So the basic kets form a unitary representation of the Lorentz group and
transform like (Lorentz) $SL(2,\mathbb{C})$ primary operators of weight
$(h,\bar{h})$ on the plane.

###  IV.4 Reality Condition on The Generators

The Lorentz group acts on the two-sphere as the group of global conformal
transformation. In the last section we decomposed the six generators as
$\big{(}L_{-1},L_{0},L_{1}\big{)}$ and
$\big{(}\bar{L}_{-1},\bar{L}_{0},\bar{L}_{1}\big{)}$ following the standard
convention of 2-D CFT. Then we saw that the unitarity imposes the reality
condition,

${L_{n}}^{\dagger}=-\bar{L}_{n},\ {\bar{L}_{n}}^{\dagger}=-L_{n},\ n=-1,0,1$
(98)

The commutator of these generators are given by,

$[L_{m},L_{n}]=(m-n)L_{m+n},\ [\bar{L}_{m},\bar{L}_{n}]=(m-n)\bar{L}_{m+n},\
[L_{m},\bar{L}_{n}]=0$ (99)

So in this case unitarity imposes a reality condition on the conformal
generators which is different from the reality condition in standard highest-
weight representation.

##  V Action of The Space-time Translation Operators

Let us now discuss the action of the space-time translations on the states
$\ket{h,\bar{h},z,\bar{z}}$. This will also give us the physical/geometrical
interpretation of these states.

There are four space-time translation operators given by,
$P^{\mu}=\big{(}P^{0}=H,P^{1},P^{2},P^{3}\big{)}$. Here $H$ is the Hamiltonian
- the generator of time translation. The generators transform as four-vector
under Lorentz transformation, i.e,

$U(\Lambda)^{-1}P^{\mu}U(\Lambda)={\Lambda^{\mu}}_{\nu}P^{\nu}$ (100)

Let us consider the following family of time dependent (Heisenberg-Picture)
states defined as,

$\ket{h,\bar{h},u,z,\bar{z}}=e^{iHu}\ket{h,\bar{h},z,\bar{z}}=\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}\
e^{iHu}\ U(R(z,\bar{z}))\ket{h,\bar{h},0,0}$ (101)

where $u$ is a real number. These states are somewhat analogous to the
Heisenberg picture states $\ket{t,\vec{x}}$ in the non-relativistic quantum
mechanics. But now we also have the "internal degrees of freedom" parametrized
by $(h,\bar{h})$ or $(\lambda,\sigma)$. The completeness relation can be
rewritten as,

$\int_{-\infty}^{\infty}d\lambda\int
d^{2}z\ket{h,\bar{h},u,z,\bar{z}}\bra{h,\bar{h},u,z,\bar{z}}=1$ (102)

This is analogous to the completeness relation $\int
d^{n}\vec{x}\ket{t,\vec{x}}\bra{t,\vec{x}}=1$ in non-relativistic quantum
mechanics. We can calculate the matrix elements of the unitary translation
operators in the basis $\big{\\{}\ket{h,\bar{h},z,\bar{z}}\big{\\}}$ using its
definition given in terms of the standard Wigner states. But for physical
interpretation it will be more useful to have the transformation law of the
time dependent states $\\{\ket{h,\bar{h},u,z,\bar{z}}\\}$ under Poincare
transformations. In other words we first construct the three parameter family
of states and then see how they look like from a different frame of reference.
Let us first state the results.

Under a Lorentz transformation $\Lambda$,

$\boxed{U(\Lambda)\ket{h,\bar{h},u,z,\bar{z}}=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\ket{h\
,\bar{h}\ ,\frac{u\ (1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}}}$ (103)

and under space-time translation through a vector $l^{\mu}$, we get

$\boxed{e^{-il.P}\ket{h,\bar{h},u,z,\bar{z}}=\ket{h,\bar{h},u+f(z,\bar{z},l),z,\bar{z}}}$
(104)

where

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$
(105)

From Eq-103 and Eq-104 we can see that on the Heisenberg-Picture states
$\ket{h,\bar{h},u,z,\bar{z}}$ Poincare action is realized completely
geometrically modulo the little group factors. In fact we can think of a space
parametrized by three coordinates $(u,z,\bar{z})$ on which the Poincare group
action is given by,

$\Lambda(u,z,\bar{z})=\bigg{(}\frac{u\ (1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}}\
,\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (106)
$T(l)(u,z,\bar{z})=(u+f(z,\bar{z},l),z,\bar{z})$ (107)

where $f$ is given above in Eq-105. This action of the Poincare group on the
$(u,z,\bar{z})$ space is the same as the action of the Poincare group at null-
infinity in Minkowski space if we identify $(u,z,\bar{z})$ with the Bondi
coordinates. We would like to emphasize that the coordinates $(u,z,\bar{z})$
arise purely group-theoretically and the coordinate $u$ has a time-like
character because we get it by unitary Hamiltonian evolution. Also there is no
Poincare invariant metric in the $(u,z,\bar{z})$ space.

Therefore the Heisenberg picture states $\ket{h,\bar{h},u,z,\bar{z}}$
transform like the primary of the (Asymptotic) Poincare or inhomogeneous
$SL(2,\mathbb{C})$ denoted by $ISL(2,\mathbb{C})$.

From a physical point of view, in the basis $\ket{h,\bar{h},z,\bar{z}}$ for
single-particle quantum states, massless particles can be thought of as living
at null-infinity. More importantly, the Poincare invariant dynamics takes
place at infinity. This has the flavour of "holography". We will come back to
this later.

###  V.1 Derivation

We will not give the details of the algebra - leading to Eq-103 and Eq-104 \-
which is identical to whatever we have done before in the case Lorentz
transformations. Let us just mention that the state
$\ket{h,\bar{h},z=0,\bar{z}=0}$ has the following crucial property which is
required for this thing to work. It satisfies,

$e^{ia(H-P^{3})}\ket{h,\bar{h},z=0,\bar{z}=0}=e^{iaP^{1}}\ket{h,\bar{h},z=0,\bar{z}=0}=e^{iaP^{2}}\ket{h,\bar{h},z=0,\bar{z}=0}=\ket{h,\bar{h},z=0,\bar{z}=0}$
(108)

where $a$ is a real number and ,

$\ket{h,\bar{h},z=0,\bar{z}=0}=\frac{1}{\sqrt{8\pi^{4}}}\int_{0}^{\infty}dE\
E^{i\lambda}\ket{E,0,0,E\ ;\sigma}$ (109)

This is the same standard state given in Eq-71 defined at the beginning except
that we have replaced $(\lambda,\sigma)$ with $(h,\bar{h})$.

####  V.1.1 A Geometric Argument

Let us now give a geometric argument which makes it clear that why we obtained
this result, i.e, why we get an action of the Poincare group on a three
dimensional space instead of the bulk four dimensional Minkowski space-time.
This is related to the nature of the basis states and the geometry of the
(conformally compactified) Minkowski space-time.

So let us consider the state $\ket{h,\bar{h},z=0,\bar{z}=0}$ but now from the
point of view of the Poincare group. We know that the little group of a null
direction in the momentum space is generated by four elements,
$\big{\\{}J_{3},K_{3},A=J_{2}-K_{1},B=-J_{1}-K_{2}\big{\\}}$ and as has
already been discussed the state $\ket{h,\bar{h},z=0,\bar{z}=0}$ can be
associated with the null direction given by,
$\big{\\{}(E,0,0,E),E>0\big{|}z=0,\bar{z}=0\big{\\}}$. Now when space-time
translations are included one can check from the definition that the three
translations $\big{\\{}e^{ia(H-P^{3})},e^{iaP^{1}},e^{iaP^{2}}\big{\\}}$ keep
the state $\ket{h,\bar{h},z=0,\bar{z}=0}$ invariant. So let us adjoin the
three generators to the little (algebra) group of a null direction. As a
result we get a subgroup of the Poincare group generated by seven elements -
$\big{\\{}J_{3},K_{3},A=J_{2}-K_{1},B=-J_{1}-K_{2},H-P^{3},P^{1},P^{2}\big{\\}}$.
The seven generators are closed under commutation. Let us denote the subgroup
generated by them by $NH$.

The geometric significance of $NH$ is that it is isomorphic to the little
group of a null-hyperplane in Minkowski space. It is easy to check this.

For example let us consider the null-hyperplane passing through the origin of
the Minkowski space. The equation of this null-hyperplane is,

$X.P=0,\ P^{2}=0$ (110)

Here $P$ is a representative from the equivalence class $\big{\\{}P\sim\alpha
P\big{|}\alpha>0\big{\\}}$ specifying the direction of the null normal. The
null geodesic generators of the null hyperplane are also parallel to $P$.

Now let us consider a Lorentz transformation $\Lambda$, in the active sense,
which maps the point $X$ lying on the null hyperplane to $\Lambda X$. Now if
$\Lambda X$ also belongs to the null hyperplane then, $(\Lambda
X).P=0=X.(\Lambda^{-1}P)$. Therefore we must have,

$\Lambda^{-1}P\sim P\rightarrow\Lambda^{-1}P=\alpha\ P$ (111)

for some $\alpha>0$. If we impose $\alpha=1$ and choose $P$ to be $(1,0,0,1)$
then we get the standard little group of massless particles generated by
$\big{\\{}J_{3},A,B\big{\\}}$. But since any other positive value of $\alpha$
is allowed, boost in the 3-direction is also an element of the little group.
In this way we get four Lorentz generators,
$\big{\\{}J_{3},K_{3},A,B\big{\\}}$ as elements of the little group of a null-
hypersurface.

Now for translations given by, $X\rightarrow X^{\prime}=X+a$, the condition
that $X^{\prime}$ belongs to the null-hyperplane gives

$a.P=0$ (112)

This has three solutions corresponding to translations in the direction of $P$
itself and in the two space-like directions orthogonal to $P$. These are
generated by $\big{\\{}H-P^{3},P^{1},P^{2}\big{\\}}$ if we take $P$ to be
$(1,0,0,1)$.

So altogether we get seven generators
$\big{\\{}J_{3},K_{3},A,B,H-P^{3},P^{1},P^{2}\big{\\}}$ which map the null-
hyperplane to itself and we can associate the state
$\ket{h,\bar{h},z=0,\bar{z}=0}$ with the null-hypersurface $X.P=0$ with
$P\sim(1,0,0,1)$.

The remaining three generators of the Poincare group act non-trivially on the
space of null-hyperplanes. Now a null-hyperplane can be thought of as the past
light-cone of a point of the future null-infinity (or the future light-cone of
a point of the past null-infinity)111See for example Penrose:1967wn .. So the
space of null-hyperplanes in Minkowski space can either be thought of as the
future null-infinity or the past null-infinity. The number of parameters also
match. There are three-parameter family of null hyperplanes corresponding to
the three Poincare generators which act non-trivially on the space of null
hyperplanes and the null infinity is also three dimensional. This shows that
we can think of the massless particle, described by a quantum state associated
with a null-hyperplane, as sitting at one point at null infinity.

Let us now analyse the state
$\ket{h,\bar{h},u,z,\bar{z}}=e^{iHu}U(R(z,\bar{z}))\ket{h,\bar{h},z=0,\bar{z}=0}$.
As we have already described, we can associate the state
$\ket{h,\bar{h},z=0,\bar{z}=0}$ with the null-hyperplane $X.P=0$ with
$P\sim(1,0,0,1)$ which passes through the origin. Now $U(R(z,\bar{z}))$
rotates the state and the rotated state corresponds to the null-hyperplane
$X.P^{\prime}=0$ where $P^{\prime}=R(z,\bar{z})P$. In this way we generate the
2-parameter family of null-hyperplanes all of which pass through the origin.
They represent the family of states
$\big{\\{}\ket{h,\bar{h},z,\bar{z}}\big{\\}}$ with varying $(z,\bar{z})$. Now
the rest of the null hyperplanes are generated by time translating this two
parameter family of null hyperplanes. This is essentially the action of
$e^{iHu}$ on the states $\ket{h,\bar{h},z,\bar{z}}$. This explains that why
the states $\ket{h,\bar{h},u,z,\bar{z}}$ transform under the asymptotic
Poinacre group.

In passing we would like to point out that the generators
$\big{\\{}J_{3},K_{3},A=J_{2}-K_{1},B=-J_{1}-K_{2},H-P^{3},P^{1},P^{2}\big{\\}}$
of the Poincare group are also the kinematic generators in the Light-front
quantization. It will be interesting to see if the states in the light-front
quantization are related to the asymptotic states
$\ket{h,\bar{h},u,z,\bar{z}}$ by some non-local transformation.

##  VI Creation And Annihilation Fields At Null-Infinity

In this section we change our notation a little bit and define,

$\ket{\lambda,\sigma,z,\bar{z}}=\ket{h,\bar{h},z,\bar{z}},\
h=\frac{1+i\lambda-\sigma}{2},\ \bar{h}=\frac{1+i\lambda+\sigma}{2}$ (113)

Let us now introduce Heisenberg-Picture creation operator
$A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})$ corresponding to the states
$\ket{\lambda,\sigma,u,z,\bar{z}}$ such that,

$U(\Lambda)\ A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})\ U(\Lambda)^{-1}\\\
=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\
A^{\dagger}_{\lambda,\sigma}\bigg{(}\frac{u\
(1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (114)

and

$e^{-il.P}A^{\dagger}_{\lambda,\sigma}e^{il.P}=A^{\dagger}_{\lambda,\sigma}(u+f(z,\bar{z},l),z,\bar{z})$
(115)

where

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$
(116)

Similarly the transformation property of the corresponding annihilation
operator $A_{\lambda,\sigma}(u,z,\bar{z})$ is given by,

$U(\Lambda)\ A_{\lambda,\sigma}(u,z,\bar{z})\ U(\Lambda)^{-1}\\\
=\frac{1}{(cz+d)^{2\bar{h}^{*}}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2h^{*}}}\
A_{\lambda,\sigma}\bigg{(}\frac{u\
(1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (117)

and

$e^{-il.P}A_{\lambda,\sigma}e^{il.P}=A_{\lambda,\sigma}(u+f(z,\bar{z},l),z,\bar{z})$
(118)

where $*$ denotes complex conjugation.

Now the question is how are these creation/annihilation operators related to
the standard creation/annihilation operators in the momentum eigenstate basis
? The simplest anwer is given by,

$A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}\int_{0}^{\infty}dE\
E^{i\lambda}e^{iEu}\ a^{\dagger}(p,\sigma)$ (119)
$A_{\lambda,\sigma}(u,z,\bar{z})=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1-i\lambda}\int_{0}^{\infty}dE\
E^{-i\lambda}e^{-iEu}\ a(p,\sigma)$ (120)

where

$p=E\bigg{(}1,\frac{z+\bar{z}}{1+z\bar{z}},\frac{-i(z-\bar{z})}{1+z\bar{z}},\frac{1-z\bar{z}}{1+z\bar{z}}\bigg{)},\
z=\tan\frac{\theta}{2}e^{i\phi}$ (121)

and $a^{\dagger}(p,\sigma)$ is the creation operator in the momentum-helicity
basis $\ket{p,\sigma}$.

This is essentially a rewriting of the relation between the basis states

$\ket{\lambda,\sigma,u,z,\bar{z}}=\frac{1}{\sqrt{8\pi^{4}}}\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{1+i\lambda}U\big{(}R(z,\bar{z})\big{)}\int_{0}^{\infty}dE\
E^{i\lambda}e^{iEu}\ket{E,0,0,E,\sigma}$ (122)

in second-quantized notation.

The (anti) commutator between the creation and annihilation operators is given
by,

$\displaystyle{[A_{\lambda,\sigma}(u,z,\bar{z}),A^{\dagger}_{\lambda^{\prime},\sigma^{\prime}}(u^{\prime},z^{\prime},\bar{z}^{\prime})]_{\pm}}$
$\displaystyle=\bra{\lambda,\sigma,u,z,\bar{z}}\ket{\lambda^{\prime},\sigma^{\prime},u^{\prime},z^{\prime},\bar{z}^{\prime}}$
(123)
$\displaystyle=\frac{\delta_{\sigma\sigma^{\prime}}}{2\pi}\frac{\Gamma\big{(}i(\lambda^{\prime}-\lambda)\big{)}}{(1+z\bar{z})^{i(\lambda^{\prime}-\lambda)}}\frac{\delta^{2}(z^{\prime}-z)}{\big{(}-i(u^{\prime}-u+i0+)\big{)}^{i(\lambda^{\prime}-\lambda)}}$

Eq-($\ref{amc1},\ref{amc2}$) and the transformation properties under Poincare
group show that we should interpret $A_{\lambda,\sigma}(u,z,\bar{z})$ and
$A^{\dagger}_{\lambda,\sigma}(u,z,\bar{z})$ as the positive and negative
frequency annihilation and creation fields living on null infinity in the
Minkowski space. So we have gone directly from the momentum space to the
boundary of the Minkowski space-time. In other words, $(z,\bar{z})$ start as
coordinates in the momentum-space but once we take into account the dynamics,
$(z,\bar{z})$ together with the time coordinate $u$ transmute into coordinates
of the boundary of the Minkowski space-time.

It will be interesting to relate this to the notion of asymptotic quantization
Ashtekar:1978zz . One difference is that here we do not arrive at the quantum
theory on the null-infinity by quantizing a classical theory living on null-
infinity. It will be interesting to clarify these things.

##  VII Primary Of $ISL(2,\mathbb{C})$

Let us consider a Poincare transformation $(l,\Lambda)$ which acts on
Minkowski coordinates as
$X^{\mu}\rightarrow{\Lambda^{\mu}}_{\nu}X^{\nu}+l^{\mu}$.

Now for the sake of convenience we define a primary operator of (Poincare)
$ISL(2,\mathbb{C})$ as any Heisenberg picture operator
$\phi_{h,\bar{h}}(u,z,\bar{z})$ transforming as,

$U(\Lambda)\ \phi_{h,\bar{h}}(u,z,\bar{z})\ U(\Lambda)^{-1}\\\
=\frac{1}{(cz+d)^{2h}}\frac{1}{(\bar{c}\bar{z}+\bar{d})^{2\bar{h}}}\
\phi_{h,\bar{h}}\bigg{(}\frac{u\
(1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d}\
,\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}$ (124)
$e^{-il.P}\phi_{h,\bar{h}}(u,z,\bar{z})e^{il.P}=\phi_{h,\bar{h}}(u+f(z,\bar{z},l),z,\bar{z})$
(125)

where

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$

The $(h,\bar{h})$ are defined to be,

$h=\frac{1+i\lambda-\sigma}{2}$ $\bar{h}=\frac{1+i\lambda+\sigma}{2}$

We emphasize that the "primary" here does not refer to any highest-weight
representation. The basic creation and annihilation fields at null-infinity
that we have defined in the last section are examples of such primary
operators.

##  VIII Constraints From Translational Invariance

We will now work out some kinematical constraints on correlation functions of
$ISL(2,\mathbb{C})$ primaries in Minkowski vacuum which follow from "bulk"
translational invariance. Under space-time translation by an arbitrary four
vector $l^{\mu}$, $u$ shifts by,

$u\rightarrow
u+f(z,\bar{z},l)=u+\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$
(126)

whereas $z$ does not change.

The correlation functions are invariant under (Poincare) $ISL(2,\mathbb{C})$
transformation, i.e,

$\bra{\Omega}\prod_{i=1}^{n}U(l,\Lambda)\ \phi_{i}(P_{i})\
U(l,\Lambda)^{-1}\ket{\Omega}=\bra{\Omega}\prod_{i=1}^{n}\phi_{i}(P_{i})\ket{\Omega}$
(127)

where $\phi_{i}(P_{i})$ is some $ISL(2,\mathbb{C})$ primary operator inserted
at the point $P_{i}=(u_{i},z_{i},\bar{z}_{i})$ and $U(l,\Lambda)$ is a
(Poincare) $ISL(2,\mathbb{C})$ transformation. $\ket{\Omega}$ is the Poincare
invariant vacuum.

Let us start from the 4-point function.

####  VIII.0.1 4-point function

We denote the 4-point function by $A(P_{1},P_{2},P_{3},P_{4})$. We first make
a Lorentz transformation, $\Lambda$, to map the points
$(z_{1},z_{2},z_{3},z_{4})\rightarrow(z,1,0,\infty)$ where $z$ is the cross
ratio of the four points given by,

$z=\frac{(z_{1}-z_{3})(z_{2}-z_{4})}{(z_{1}-z_{4})(z_{2}-z_{3})}$ (128)

Under this Lorentz transformation the $u$’s also transform to some other
values say $u^{\prime}$. So we can write,

$A(P_{1},P_{2},P_{3},P_{4})=G(\\{\Lambda,z_{i},\bar{z}_{i},h_{i},\bar{h}_{i})\\})\
A(P_{1}^{\prime},P_{2}^{\prime},P_{3}^{\prime},P_{4}^{\prime})$ (129)

where

$P_{1}^{\prime}=(u_{1}^{\prime},z,\bar{z}),\
P_{2}^{\prime}=(u_{2}^{\prime},1,1),\ P_{3}^{\prime}=(u_{3}^{\prime},0,0),\
P_{4}^{\prime}=(u_{4}^{\prime},\infty,\infty)$ (130)

and $G$ has no $u$ dependence. Therefore we need to consider the 4-point
function at these special values of $z_{i}$’s. Now translational invariance
requires that,

$A(P_{1}^{\prime},P_{2}^{\prime},P_{3}^{\prime},P_{4}^{\prime})=A(P_{1}^{\prime\prime},P_{2}^{\prime\prime},P_{3}^{\prime\prime},P_{4}^{\prime\prime})$
(131)

where

$\displaystyle
P_{1}^{\prime\prime}=\big{(}u_{1}^{\prime}+f(z,\bar{z},l\big{)},z,\bar{z}),P_{2}^{\prime\prime}=\big{(}u_{2}^{\prime}+f(1,1,l),1,1\big{)},P_{3}^{\prime\prime}=\big{(}u_{3}^{\prime}+f(0,0,l),0,0\big{)},$
(132) $\displaystyle
P_{4}^{\prime\prime}=\big{(}u_{4}^{\prime}+f(\infty,\infty,l),\infty,\infty\big{)}$

Here we have used the fact that under translation $z_{i}$ does not change. For
$z=0,1,\infty$ we have,

$f(0,0,l)=l^{0}-l^{3},\ f(1,1,l)=l^{0}-l^{1},\ f(\infty,\infty,l)=l^{0}+l^{3}$
(133)

Now if we take the translations to be infinitesimal then we get four
differential equations given by,

$A_{1}+A_{2}+A_{3}+A_{4}=0$ (134)
$A_{1}-A_{3}+\frac{1-z\bar{z}}{1+z\bar{z}}A_{4}=0$ (135)
$A_{2}+\frac{z+\bar{z}}{1+z\bar{z}}A_{4}=0$ (136)
$\frac{z-\bar{z}}{1+z\bar{z}}A_{4}=0$ (137)

where $A_{i}=\frac{\partial A}{\partial u_{i}^{\prime}}$.

Now if $z\neq\bar{z}$ then we get, $A_{1}=A_{2}=A_{3}=A_{4}=0$. Therefore the
most general non-trivil solution for the 4-point function can be written as,

$A(P_{1},P_{2},P_{3},P_{4})=\delta(\imaginary z)\
\tilde{A}(P_{1},P_{2},P_{3},P_{4})$ (138)

where $z$ is the the cross ratio of the four points
$(z_{1},z_{2},z_{3},z_{4})$. So the four point function will be zero unless
the cross ratio of the four insertion points are real. This is the same
constraint obtained in Pasterski:2017ylz  from the study of Gluon scattering
amplitude in the basis of conformal primary wave functions. Recently this has
also been studied in detail in Lam:2017ofc .

####  VIII.0.2 3-point function

In the case of the 3-point function one can repeat the same procedure except
that now there is no cross ratio. We get 3 differential equations given by,

$A_{1}+A_{2}+A_{3}=0$ (139) $A_{1}-A_{3}=0$ (140) $A_{2}=0$ (141)

This has the trivial solution $A_{1}=A_{2}=A_{3}=0$. Therefore the most
general non-trivial solution of the 3-point function is,

$A(P_{1},P_{2},P_{3})=0$ (142)

This corresponds to the fact that in $(-+++)$ signature 3-point scattering
amplitude of massless particles vanish for generic null momenta.

####  VIII.0.3 2-point function

The general solution of the 2-point function can be written as,

$A(P_{1},P_{2})=N(\lambda_{1},\lambda_{2})\
\frac{\delta_{\sigma_{1}+\sigma_{2},0}}{(1+z_{1}\bar{z}_{1})^{i(\lambda_{1}+\lambda_{2})}}\
\frac{\delta^{2}(z_{1}-z_{2})}{\big{(}u_{1}-u_{2})^{i(\lambda_{1}+\lambda_{2})}}$
(143)

where $N(\lambda_{1},\lambda_{2})$ is a prefactor which cannot be determined
solely from the Poincare invariance. There should also be an $i\epsilon$
prescription to take care of the singularity at $u_{1}=u_{2}$ but this cannot
be determined just from symmetry consideration.

##  IX A Hint of Supertranslation

Let us first consider the transition amplitude for a free massless particle in
the $\ket{\lambda,\sigma,z,\bar{z}}$ basis, given by

$\displaystyle\bra{\lambda,\sigma,z,\bar{z}}e^{-iH(u-u^{\prime})}\ket{\lambda^{\prime},\sigma^{\prime},z^{\prime},\bar{z}^{\prime}}$
$\displaystyle=\bra{\lambda,\sigma,u,z,\bar{z}}\ket{\lambda^{\prime},\sigma^{\prime},u^{\prime},z^{\prime},\bar{z}^{\prime}}$
(144)
$\displaystyle=\frac{\delta_{\sigma\sigma^{\prime}}}{2\pi}\frac{\Gamma\big{(}i(\lambda^{\prime}-\lambda)\big{)}}{(1+z\bar{z})^{i(\lambda^{\prime}-\lambda)}}\frac{\delta^{2}(z^{\prime}-z)}{\big{(}-i(u^{\prime}-u+i0+)\big{)}^{i(\lambda^{\prime}-\lambda)}}$

The Dirac delta function arises because a free particle does not change its
direction 222In this picture we can only talk about the direction of motion or
the direction of the momentum vector because the magnitude has been integrated
out. of motion. Due to the presence of $\delta^{2}(z-z^{\prime})$ the
amplitude is manifestly invariant under space-time translations under which
$u\rightarrow u+f(z,\bar{z},l)$ and $z$ remains unchanged. Now it is easy to
see that for the same reason the amplitude is in fact invariant under more
general transformation $u\rightarrow u+g(z,\bar{z})$ where $g(z,\bar{z})$ is
an arbitrary smooth function on the sphere. These can be identified with BMS
supertranslations Bondi:1962px . At this stage supertranslation invariance is
just an accidental symmetry which is manifest in this basis.

The dependence of a general two point function on the coordinates of the null-
infinity is completely fixed by the Poincare invariance. It is clear that the
2-point function given in Eq-143 is also invariant under BMS
supertranslations. This is also an accidental symmetry of the correlation
function.

The supertranslation invariance of the transition amplitude of a free particle
or the 2-point function is trivial from physical point of view
Strominger:2013jfa . In fact this cannot be the case for higher-point
functions (in some interacting theory) because BMS is spontaneously broken in
the Minkowski vacuum Strominger:2013jfa . But, in some sense, this shows that
perhaps it is "natural" to consider extension of the Poincare group to the BMS
group once the field theory is formulated on null-infinity. This is also
suggested by Strominger’s conjecture about BMS invariance of the gravitational
S-matrix Strominger:2013jfa  and recent advances in understanding the relation
between the infrared structure of gravity and asymptotic symmetry
Strominger:2013lka ; He ; Kapec:2014opa ; Strominger:2014pwa ; Kapec:2016jld ;
Campiglia:2014yka ; Campiglia:2015yka ; Campiglia:2015kxa ; Kapec:2017gsg ;
Strominger:2017zoo ; Weinberg:1965nx ; Cachazo:2014fwa ; Schwab:2014xua ;
Bern:2014oka ; Broedel:2014fsa ; Avery:2015gxa ; Sen:2017xjn ; Sen:2017nim ;
Laddha:2017ygw ; Chakrabarti:2017ltl ; Chakrabarti:2017zmh ; Laddha:2017vfh ;
Campoleoni:2017mbt ; Hawking:2016msc  of flat space-time.

In fact the conjecture of Barnich:2009se ; Banks:2003vp  and the recent works
He ; Kapec:2014opa ; Strominger:2014pwa ; Kapec:2016jld ; Campiglia:2014yka ;
Campiglia:2015yka ; Campiglia:2015kxa ; Kapec:2017gsg ; Bagchi:2016bcd ;
Bagchi:2016geg  suggest that that the superrotation symmetry plays a central
role in a holographic reformulation of flat space physics. So the full
symmetry group should perhaps be the (extended) BMS including superrotation.
The representation of the BMS group in three space-time dimensions has been
studied in Barnich:2014kra ; Campoleoni:2016vsh .

We would like to emphasize that this BMS should not be thought of as an
asymptotic symmetry group Bondi:1962px ; Banks:2003vp ; Barnich:2009se . There
is no dynamical gravity in this picture and in fact we did not require the
presence of any "bulk space-time". The transition from momentum space to null-
infinity was direct. Here the Supertranslation is some geometric
transformation of the background space-time parametrized by $(u,z,\bar{z})$
under which $2$-point function or the commutator (Eq-123) turns out to be
symmetric. But so far the symmetry group which acts on states in the Hilbert
space is still the four dimensional Poincare group or $ISL(2,\mathbb{C})$.

The fact that we get a hint of supertranslation in a purely non-gravitational
setting is perhaps an indication of holography in asymptotically flat space-
time. This is somewhat similar to the situation in AdS3-CFT2 correspondence.
The infinite dimensional Virasoro symmetry of a two-dimensional conformal
field theory can be understood in a purely field theory setting without
dynamical gravity. But in asymptotically AdS3 spaces Virasoro can also be
understood as the asymptotic symmetry group of the bulk gravity theory. These
two facts nicely fit together in AdS3-holography. Similarly the goal here is
to understand BMS in a purely non-gravitational setting. If such an
understanding can be reached in a consistent manner then that will perhaps be
an indication of flat-space holography.

##  X Conformal Primary Wave-Functions With Translational Invariance

For simplicity we consider the massless scalar conformal primary wave
functions Pasterski:2017kqt ; Cheung:2016iub ; deBoer:2003vf . Scalar
conformal primary wave functions in (3+1) dimensions are given by,

$\Phi_{\Delta}^{\pm}(X^{\mu}|z,\bar{z})=\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{\Delta}\int_{0}^{\infty}dEE^{\Delta-1}e^{\pm
iEQ\cdot X}e^{-\epsilon
E}=\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{\Delta}\frac{(\mp
i)^{\Delta}\Gamma(\Delta)}{(-Q\cdot X\mp i\epsilon)^{\Delta}}\ ,\
\Delta=1+i\lambda$ (145)

where $X^{\mu}$ is an arbitrary point in the Minkowski space and
$Q^{\mu}(z,\bar{z})$ is a standard null vector given by

$Q^{\mu}(z,\bar{z})=\bigg{(}1,\frac{z+\bar{z}}{1+z\bar{z}},\frac{-i(z-\bar{z})}{1+z\bar{z}},\frac{1-z\bar{z}}{1+z\bar{z}}\bigg{)}$
(146)

which under a $SL(2,\mathbb{C})$ transformation transforms as ,

$Q^{\mu}(\Lambda z,\Lambda\bar{z})=\frac{1+z\bar{z}}{|az+b|^{2}+|cz+d|^{2}}\
{\Lambda^{\mu}}_{\nu}Q^{\nu}(z,\bar{z}),\ \Lambda z=\frac{az+b}{cz+d},\
\Lambda\bar{z}=\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}$ (147)

where ${\Lambda^{\mu}}_{\nu}$ is the corresponding Lorentz transformation
matrix. It was shown in Pasterski:2017kqt  that the wave-functions given in
Eq-145 form a complete set of delta function normalizable solutions of the
massless Klein-Gordan equation if $\Delta=1+i\lambda$.

Under Lorentz transformation the scalar conformal primary wave functions
transform as,

$\Phi_{\Delta}^{\pm}\bigg{(}{\Lambda^{\mu}}_{\nu}X^{\nu}\bigg{|}\frac{az+b}{cz+d},\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}=(cz+d)^{\Delta}(\bar{c}\bar{z}+\bar{d})^{\Delta}\
\Phi_{\Delta}^{\pm}(X^{\mu}|z,\bar{z})$ (148)

To see the action of the space-time translation in a simple manner let us
introduce an extra parameter $u$ and new wave functions labelled by
$\Delta(=1+i\lambda)$ and three parameters $(u,z,\bar{z})$. The new wave
functions are given by,

$\boxed{\Phi_{\Delta}^{\pm}(X^{\mu}|u,z,\bar{z})=\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{\Delta}\int_{0}^{\infty}dEE^{\Delta-1}e^{\pm
iE(Q\cdot X+u)}e^{-\epsilon
E}=\bigg{(}\frac{1}{1+z\bar{z}}\bigg{)}^{\Delta}\frac{(\mp
i)^{\Delta}\Gamma(\Delta)}{(-Q\cdot X-u\mp i\epsilon)^{\Delta}}}\ $ (149)

This still solves the massless Kelin-Gordon equation and for a fixed value of
$u$ the family of wave functions parametrized by $(\Delta,u,z,\bar{z})$ form a
complete set just like the wave functions given in Eq-145 . Now one can easily
check that under Lorentz transformation,

$\Phi_{\Delta}^{\pm}\bigg{(}{\Lambda^{\mu}}_{\nu}X^{\nu}\bigg{|}\frac{u(1+z\bar{z})}{|az+b|^{2}+|cz+d|^{2}},\frac{az+b}{cz+d},\frac{\bar{a}\bar{z}+\bar{b}}{\bar{c}\bar{z}+\bar{d}}\bigg{)}=(cz+d)^{\Delta}(\bar{c}\bar{z}+\bar{d})^{\Delta}\
\Phi_{\Delta}^{\pm}(X^{\mu}|u,z,\bar{z})$ (150)

and under spacetime translation by four vector $l^{\mu}$ ,

$\Phi_{\Delta}^{\pm}(X^{\mu}+l^{\mu}\big{|}u+f(z,\bar{z},l),z,\bar{z})=\Phi_{\Delta}^{\pm}(X^{\mu}|u,z,\bar{z})$
(151)

where

$f(z,\bar{z},l)=\frac{(l^{0}-l^{3})-(l^{1}-il^{2})z-(l^{1}+il^{2})\bar{z}+(l^{0}+l^{3})z\bar{z}}{1+z\bar{z}}$
(152)

So the wave functions $\Phi_{\Delta}^{\pm}(X^{\mu}|u,z,\bar{z})$ transform in
a simple manner under the Poincare group and it is natural to identify the
parameters $(u,z,\bar{z})$ as the Bondi coordinates of the null infinity in
Minkowski space.

The above procedure has a simple geometric interpretation. As we have
described in section-(V.1.1) the states $\ket{h,\bar{h},u,z,\bar{z}}$ are
associated with null-hyperplanes in the Minkowski space. The original
2-parameter family of conformal primary wave functions, given by
$\Phi_{\Delta}^{\pm}(X^{\mu}|z,\bar{z})$ (Eq-145), are singular along the null
hyperplanes $X\cdot Q(z,\bar{z})=0$ passing through the origin. They
correspond to the states $\ket{h,\bar{h},u=0,z,\bar{z}}$. So to get the rest
of the states with nonzero $u$ we have translate the family $X\cdot
Q(z,\bar{z})=0$ along the time axis. This gives rise to the wave functions
$\Phi_{\Delta}^{\pm}(X^{\mu}|u,z,\bar{z})$ (Eq-149) singular along the null-
hyperplanes $X\cdot Q(z,\bar{z})+u=0$. This 3-parameter family exhausts all
the null-hyperplanes in the Minkowski space and correspond to the family of
states $\ket{h,\bar{h},u,z,\bar{z}}$ located on null-infinity. This makes it
clear that why the Poincare group has simple action on the wave-functions
$\Phi_{\Delta}^{\pm}(X^{\mu}|u,z,\bar{z})$.

Now suppose instead of plane-waves we use the wave functions
$\big{\\{}\Phi_{\Delta_{i}}^{\pm}(X^{\mu}|u_{i},z_{i},\bar{z}_{i})\big{\\}}$
as the wave functions of the external particles then we can write down a
modified Mellin-transform as,

$\boxed{\tilde{A}(\\{h_{i},\bar{h}_{i},u_{i},z_{i},\bar{z}_{i}\\})=\prod_{i=1}^{n}\bigg{(}\frac{1}{1+z_{i}\bar{z}_{i}}\bigg{)}^{\Delta_{i}}\int_{0}^{\infty}dE_{i}\
E_{i}^{\Delta_{i}-1}e^{-i\epsilon_{i}E_{i}u_{i}}A(\\{E_{i},z_{i},\bar{z}_{i}\\})}$
(153)

where $h_{i}=\bar{h}_{i}=\frac{\Delta_{i}}{2}$ and we have also defined
$\epsilon_{i}=+1$ for an outgoing particle and $\epsilon_{i}=-1$ for an
incoming particle. This is a slightly modified form of the Mellin amplitude
defined in Pasterski:2016qvg ; Pasterski:2017ylz .

$\tilde{A}$ amplitude behaves like correlation function of $ISL(2,\mathbb{C})$
primaries inserted at points at null infinity and transform under the
Asymptotic Poincare Group as,

$\tilde{A}\big{(}\\{h_{i},\bar{h}_{i},\Lambda u_{i},\Lambda
z_{i},\Lambda\bar{z}_{i}\\}\big{)}=\prod_{i=1}^{n}(cz_{i}+d)^{2h_{i}}(\bar{c}\bar{z}_{i}+\bar{d})^{2\bar{h}_{i}}\
\tilde{A}\big{(}\\{h_{i},\bar{h}_{i},u_{i},z_{i},\bar{z}_{i}\\}\big{)}$ (154)
$\tilde{A}\big{(}\\{h_{i},\bar{h}_{i},u_{i}+f(z_{i},\bar{z}_{i},l),z_{i},\bar{z}_{i}\\}\big{)}=\tilde{A}\big{(}\\{h_{i},\bar{h}_{i},u_{i},z_{i},\bar{z}_{i}\\}\big{)}$
(155)

where

${}\Lambda
u_{i}=\frac{(1+z_{i}\bar{z}_{i})u_{i}}{|az_{i}+b|^{2}+|cz_{i}+d|^{2}}$

and $f(z,\bar{z},l)$ is given by Eq-152.

It will be interesting to compute the $\tilde{A}$ amplitude for some field
theories along the lines of Pasterski:2017ylz ; Cardona:2017keg ;
Banerjee:2017jeg ; Schreiber:2017jsr .

The $\tilde{A}$ amplitudes can be thought of as the "scattering amplitude with
asymptotic states" given by $\ket{h_{i},\bar{h}_{i},u_{i},z_{i},\bar{z}_{i}}$.
It is very likely that a relation of the type,

$\tilde{A}(\\{h_{i},\bar{h}_{i},u_{i},z_{i},\bar{z}_{i}\\})\sim\bra{\Omega}\prod_{i=1}^{n}\phi_{(h_{i},\bar{h}_{i})}(u_{i},z_{i},\bar{z}_{i})\ket{\Omega}$
(156)

exists where $\phi_{(h_{i},\bar{h}_{i})}$ are $ISL(2,\mathbb{C})$ primaries at
null-infinity and the correlator in the R.H.S is computed in a theory defined
at null-infinity. The correlation function should perhaps be u-ordered
because, as we have seen, the coordinate $u$ has time-like character. It will
be interesting to make these things more precise.

##  XI Acknowledgement

It is a pleasure to thank Arjun Bagchi, Nabamita Banerjee, Sayantani
Bhattacharya, Bidisha Chakrabarty, Abhijit Gadde, Sudip Ghosh, Sachin Jain,
Dileep Jatkar, Arnab Kundu, Alok Laddha, R.Loganayagam, Gautam Mandal, Sunil
Mukhi, Amin Nizami, Partha Paul, Ashoke Sen, Yogesh Srivastava and Amitabh
Virmani for useful discussions and Alok Laddha for useful comments on the
draft. I would also like to thank the string theory groups in HRI Allahabad,
ICTS Bengaluru, IISER Pune and TIFR Mumbai for hospitality at various stages
of the work. I would like to thank the organizers and the participants of the
National Strings Meeting 2017 held in NISER Bhubaneswar where part of this
work was presented.

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