compat=1.0.0 $\kappa$-deformed scalar field Jerzy Kowalski-Glikman
Wojciech Wiślicki Doctor of Philosophy Physical Sciences National Centre for
Nuclear Research Department BP2 Factulty of Physics

Doctoral Thesis

_Author:_

_Supervisors:_

A thesis submitted in fulfillment of the requirements
for the degree of
in the

###### Abstract

Abstract In the following work we will introduce and discuss in detail a
particular model of complex $\kappa$-deformed scalar field, whose behaviour
under $C$, $P$, $T$ transformation is particularly transparent from both a
formal and phenomenological point of view. We will begin by introducing the
key mathematical structure at the basis of our investigation, namely the
$\kappa$-Poincaré (Hopf) algebra and the $\kappa$-Minkowski spacetime. We will
then investigate the behaviour of general two-particle states under deformed
boost. After this we will introduce the action of our $\kappa$-deformed
complex scalar field. From it, we will derive the equations of motion, as well
as the Noether charges due to the continuous symmetries. The peculiar features
of $\kappa$-deformation in general, and of our model in particular, allow for
very non-trivial interaction between discrete and continuous symmetries, of
which we will investigate the phenomenological consequences (particularly in
terms of difference of lifetime of decaying particles). To conclude, we will
obtain the $\kappa$-deformed propagator of the $\kappa$-deformed complex
scalar field, and the imaginary part of the 1-loop contribution to it, ending
with additional phenomenological consequences. The third chapter is new,
unpublished work.

I, , declare that this thesis titled, ?? and the work presented in it are my
own. I confirm that:

  * •

This work was done wholly or mainly while in candidature for a research degree
at the National Centre for Nuclear Research.

  * •

Where any part of this thesis has previously been submitted for a degree or
any other qualification at the National Centre for Nuclear Research or any
other institution, this has been clearly stated.

  * •

Where I have consulted the published work of others, this is always clearly
attributed.

  * •

Where I have quoted from the work of others, the source is always given. With
the exception of such quotations, this thesis is entirely my own work.

  * •

I have acknowledged all main sources of help.

  * •

Where the thesis is based on work done by myself jointly with others, I have
made clear exactly what was done by others and what I have contributed myself.

Signed:

Date:

W niniejszej pracy wprowadzimy i przedyskutujemy w szczegółach model
zespolonego, $\kappa$-zdeformowanego pola skalarnego, którego zachowanie pod
wpływem transformacji C, P i T można jasno przedstawić, zarówno z formalnego,
jak z fenomenologicznego punktu widzenia. Zaczniemy od wprowadzenia struktur
matematycznych, leżących u podstaw naszych badań, a mianowicie algebry
$\kappa$-Poincarego (Hopfa) oraz czasoprzestrzeni $\kappa$-Minkowskiego.
Następnie zbadamy zachowanie ogólnych stanów dwucząstkowych pod wpływem
zdeformowanego pchnięcia Lorentza. Potem wprowadzimy działanie
$\kappa$-zdeformowanego, zespolonego pola skalarnego, po czym wyprowadzimy
równania ruchu i ładunki Noether odpowiadające symetriom ciągłym. Pewne
szczególne własności $\kappa$-deformacji, zarówno te ogólne, jak i dotyczące
tylko naszego modelu, umożliwiają na istnienie nietrywialnch związków między
symetriami dyskretnymi i ciągłymi, z których wyprowadzimy konsekwencje
fenomenologiczne i poddamy je badaniom (szczególnie konsekwencje dotyczące
różnic czasów życia niestabilnych cząstek i antycząstek). Obliczymy także
$\kappa$-zdeformowany propagator $\kappa$-zdeformowanego, zespolonego pola
skalarnego, jak również część urojoną jednopętlowej poprawki do propagatora, a
następnie zakończymy dyskusją konsekwencji fenomenologicznych. Trzeci rozdział
to nowa, nieopublikowana praca.

###### Acknowledgements.

I would like to thank my supervisors Jurek and Wojtek. Their deep knowledge
and understanding of physics has been the source of many interesting
discussions, and their guidance has made my Ph.D. studies a wonderful
experience. For the same reasons, I would also like to thank Giulia, Michele,
Giacomo, and Josua. I would like to thank Prof. Stanisław Mrówczyński for the
interesting discussions, and for giving me the opportunity to assist him as a
tutor in his Statistical Mechanics lectures. I am also deeply grateful to
Przemek for his help in making these four years possible in the first place. I
would also like to thank Monika for all the incredible help, and for the very
interesting discussions during my frequent trips to her office, as well as for
the enjoyable Polish lectures. Additionally, I would like to thank Ubaldo and
Anatolii for the wonderful time spent together. A special thanks goes also to
my family for their help, I wouldn’t have been able to study for so many years
without their support. Finally, my partner Alice has accompanied me throughout
all my studies, in a voyage which has been made even more beautiful (and much
more chaotic) by the birth of our son Levi. To both of them goes all my
gratitude.
Part of this work has been supported by funds provided by the Polish National
Science Center, the project number 2019/33/B/ST2/00050, and by the NAWA
scholarship financed by the STER program. My participation to COST Action
CA18108 activities has been financed by COST Action. Part of this work’s
bibliography has been collected using [1].

###### Contents

  1. 0 $\kappa$-Minkowski and $\kappa$-Poincaré
    1. 1 Introduction to $\kappa$-deformation
    2. 2 Formal definition of Hopf algebra
    3. 3 From $\kappa$-Poincaré to $\kappa$-Minkowski
    4. 4 From $\kappa$-Minkowski to $\kappa$-Poincaré
    5. 5 Finite boosts for two-particle states
      1. 1 Fifth column:
      2. 2 Fourth column:
      3. 3 Third column:
      4. 4 Second column:
      5. 5 First column:
      6. 6 Expansion of the final matrix up to $1/\kappa$
      7. 7 Check of the consistency of the first order expansion of $L^{\prime}_{g}$
    6. 6 Two-particle kinematics
  2. 1 Complex scalar field on $\kappa$-Minkowski spacetime
    1. 1 Introduction
    2. 2 Weyl maps
    3. 3 Action for the complex scalar field
    4. 4 Action of derivatives on $\star$ products
      1. 1 $A=0$
      2. 2 $A=i$
      3. 3 $A=4$
      4. 4 $A=+$
      5. 5 $\Delta_{+}^{\dagger}=\kappa^{2}\Delta_{+}^{-1}$
    5. 5 Integration-by-parts relation for $\partial_{i}$ going $\leftarrow$
    6. 6 Integration-by-parts relation for $\partial_{0}$ going $\leftarrow$
    7. 7 Integration-by-parts relation for $\partial_{i}$ going $\rightarrow$
    8. 8 Integration-by-parts relation for $\partial_{0}$ going $\rightarrow$
    9. 9 Mass terms integration by parts
    10. 10 EoM and surface terms $\Pi^{0}$, $\Pi^{i}$, and $\Pi^{4}$ for both orderings of the action
    11. 11 Complex scalar field and its properties
      1. 1 $C,P,T$ in the non-deformed context: a short review
      2. 2 $C,P,T$ in the $\kappa$-deformed context
    12. 12 Off-shell action in momentum space
      1. 1 First ordering of the action
      2. 2 Second ordering of the action
      3. 3 Sum of the two actions
    13. 13 Symplectic form for the two actions
      1. 1 Example in the non-deformed case
      2. 2 The $\kappa$-deformed case
    14. 14 Conserved charges: geometric approach
      1. 1 Non-deformed case
        1. Translation charges
        2. Boost and rotation charges
          1. Crucial step
      2. 2 $\kappa$-deformed case
        1. Translation charges
          1. Non-deformed contraction:
          2. Deformed contraction:
        2. Notes on the above (overall) procedure
        3. Boosts
          1. Crucial step
          2. The (very) peculiar features of boosts
          3. The role of $\zeta(p)$
        4. Rotations
      3. 3 Is it $\delta A\wedge\delta B=-\delta B\wedge\delta A$, or $\delta A\wedge\delta B=S(\delta B\wedge\delta A)$?
      4. 4 And what if we add contractions with vector fields?
    15. 15 The algebra of the $\kappa$-deformed charges
      1. $[\mathscr{N}_{i},\mathscr{P}_{j}]=-i\eta_{ij}\mathscr{P}_{0}$
      2. $[\mathscr{N}_{i},\mathscr{N}_{j}]=-i\epsilon_{ijk}\mathscr{M}_{k}$
    16. 16 $C$ and its commutator with the boosts
    17. 17 $\cancel{CPT}$ $\Leftrightarrow$ ~~Lorentz~~?
      1. 1 Greenberg’s argument
      2. 2 Jost theorem
      3. 3 Comments on Jost and Greenberg theorems
        1. $CPT$ properties of $\langle 0|\phi_{1}^{\dagger}(x_{1})\star\phi_{2}(x_{2})|0\rangle$
        2. Lorentz properties of $\langle 0|\phi_{1}^{\dagger}(x_{1})\star\phi_{2}(x_{2})|0\rangle$
    18. 18 Phenomenological consequences of deformed $CPT$ transformations
      1. 1 Estimation of deformed decay width
  3. 2 $\kappa$-deformed propagator, and 1-loop correction to it
    1. 1 Introduction
    2. 2 Feynman propagator
      1. 1 From the generating functional
      2. 2 As inverse of the field operator
    3. 3 Example of loop correction to the propagator: non-deformed case
      1. 1 Using renormalization
    4. 4 Example of loop correction to the propagator: $\kappa$\- deformed case
      1. 1 Dimensional regularization and cut-off in the $\kappa$-deformed context
        1. Summary of the individual diagrams and total 1-loop amplitude
    5. 5 Mass distribution width
    6. 6 Summary

To A.B. and L.B-B.

##  Chapter 0 $\kappa$-Minkowski and $\kappa$-Poincaré

###  1 Introduction to $\kappa$-deformation

Immediately after both general relativity and quantum mechanics were
formalized, physicists were already trying to understand what a quantum theory
of gravity would look like [2]. Several approaches were tried, each with their
pros and cons (an extensive discussion of each approach would be out of place
here, but one can consult [3] for a review). From a theoretical point of view,
the wealth of approaches was due to the vastly different nature of general
relativity and quantum mechanics, both conceptually and formally. Moreover,
from the more pragmatic point of view, effects of quantum gravitational origin
were thought to appear when particles with Plank energy $E_{p}$ scatter with
impact parameter given by the Plank length $l_{p}$. This means that very early
on, there was significant doubt that any quantum gravitational effect could be
phenomenologically relevant, even in the distant future [4]. Moreover, the
measurement of phenomena happening at distances comparable or smaller than
$l_{p}$ would require apparatuses so large that they inevitably collapse to a
black hole [5], which means there may be some limitation from first principles
in the conventional definition of spacetime as a continuous structure (this
was already noted in [4], [6], [7]). Despite these advances, however, the lack
of phenomenological feedback seemed to be destined to be a long lasting issue
in quantum gravity (QG) research. It was therefore unexpected when it was
suggested that, switching the focus to cumulative effects of QG, a
phenomenology was not only possible, but within reach [8]. Indeed, instead of
focussing on high energy scattering processes with very small impact
parameter, one could try to study the cumulative effects of QG, which can for
example manifest themselves with a slightly different propagation speed for
photons of different energies [9], [10], [53], [12], [13], [35], and may also
affect neutrino properties [15], [16], [17], [18], [19], [20]. One of the most
effective ways to go about studying potential QG effects is therefore not to
study any model of QG directly, but a potential effective theory of QG.

One of the most studied model in this sense is based on a deformation of the
Poincaré algebra, called $\kappa$-Poincaré algebra, where $\kappa$ is a
deformation parameter with dimensions of energy. Its origin can be traced back
to the 1990’s, particularly to [21], [22], [23], [24], [25], [26]. The idea is
to deform the canonical Poincaré algebra to investigate how symmetries might
change in some well defined physical limit of QG.

At the same time, new models were being proposed where in addition to a
constant speed $c$, there is also a length scale $l=\frac{\hbar}{\kappa}$ (or
$l=\frac{1}{\kappa}$ if $\hbar=1$) which is independent of the observer [27],
[28], [29], [30]. This length scale also implies a momentum scale in momentum
space; the idea is therefore that such observer-independent scale can be used
to define some non-linear dispersion relation while keeping the theory Lorentz
invariant (where ‘Lorentz invariance’ in this context is defined in the
appropriate way). Intuitively speaking, such an additional scale would allow
for a non-trivial geometry of momentum space, which is usually flat in non-
deformed theories which lack such a scale. To give just a small example, one
can start from the action of a classical particle with non-deformed dispersion
relation moving in a curved spacetime [75], [76]

$\displaystyle S=\int
d\tau\,\,\dot{x}^{\mu}\tensor{e}{{}_{\mu}^{a}}(x)p_{a}-N(\eta^{ab}p_{a}p_{b}+m^{2})$

From this, one can by analogy write down the action of a particle moving in
flat spacetime, but curved momentum space

$\displaystyle S_{\kappa}=-\int
d\tau\,\,x^{a}\tensor{E}{{}_{a}^{\mu}}(k,\kappa)\dot{k}_{\mu}-N(\mathscr{C}_{\kappa}(k)+m^{2}).$
(1)

where $\tensor{E}{{}_{a}^{\mu}}$ is the tetrad which is related to the
momentum-space metric by

$\displaystyle
G^{\mu\nu}(k,\kappa)=\eta^{ab}\tensor{E}{{}_{a}^{\mu}}(k,\kappa)\tensor{E}{{}_{b}^{\nu}}(k,\kappa),$
(2)

and $\mathscr{C}_{\kappa}(k)$ is the deformed dispersion relation (once again,
here we are using a deformation parameter with dimensions of energy). For
example, one could choose

$\displaystyle\mathscr{C}_{\kappa}(k)=-4\kappa^{2}\sinh^{2}\left(\frac{k_{0}}{2\kappa}\right)+\mathbf{k}^{2}e^{\frac{k_{0}}{\kappa}}.$
(3)

One can show that

$\displaystyle S_{\kappa}=\int
d\tau\,\,\dot{x}^{0}k_{0}-e^{\frac{k_{0}}{\kappa}}\mathbf{x}\cdot\dot{\mathbf{k}}+N(\mathscr{C}_{\kappa}(k)+m^{2})$
(4)

which is invariant under a deformed set of transformations which satisfy the
so called $\kappa$-Poincaré algebra (which will be discussed in detail in
later sections). The symplectic form is then given by

$\displaystyle\Omega_{\kappa}=\int d\tau\,\,\delta k_{0}\wedge\delta
x^{0}+e^{\frac{k_{0}}{\kappa}}\delta\mathbf{k}_{i}\wedge\delta\mathbf{x}^{i}-\frac{\mathbf{x}^{i}}{\kappa}e^{\frac{k_{0}}{\kappa}}\delta
k_{0}\wedge\delta\mathbf{k}_{i}$ (5)

which implies

$\displaystyle\\{x^{0},p_{0}\\}=1\qquad\\{\mathbf{x}^{i},\mathbf{k}_{j}\\}=e^{-\frac{k_{0}}{\kappa}}\delta^{i}_{j}\qquad\\{x^{0},\mathbf{x}^{i}\\}=-\frac{1}{\kappa}\mathbf{x}^{i}.$
(6)

We see that, upon quantization, spacetime coordinates do not commute. These
type of non-commutativity defines the so called $\kappa$-Minkowski spacetime.
This kind of models showcases interesting phenomena, such as relative locality
[33], [34], [35], and are related to $\kappa$-Poincaré algebras [36], [65],
[66], [67].

Finally, in $2+1$ dimensions, one can show that $\kappa$-deformation naturally
emerges when considering point-like particles coupled with gravity (see for
example [38],[37], [75] and references therein). This is due to the fact that
the Newton constant $G_{N}$ is dimensionful in $2+1$ dimensions, and it has
dimensions of energy. The procedure is fairly straightforward and can be
summarized in the following steps [76],[75]:

  * 1)

From the total action of gravity and point-like particles coupled to it, one
can get the explicit solutions of the (finitely many) gravitational degrees of
freedom;

  * 2)

Substitute these solutions back into the action, obtaining a new action which
describes the motion of point-like particles ‘dressed’ in their own
gravitational field;

  * 3)

This new action can then be shown to correspond to a $\kappa$-deformed model,
much like the one described by eq. (4) (although in the case of $2+1$
particles with gravity we have a different kind of deformations).

Furthermore, one can argue that the deformed symmetry group $\mathscr{P}^{3}$
acting on the Hilbert space $\mathscr{H}^{3}$ of $2+1$ gravity with particles
is expected to be a subgroup of the full group $\mathscr{P}^{4}$ acting on the
Hilbert space $\mathscr{H}^{4}$ of particles coupled with gravity in $3+1$
dimensions [38]. All this motivates the study of fields in the context of
$\kappa$-Poincaré and $\kappa$-Minkowski.

$\kappa$-deformation models have also proven to be interesting from the
phenomenological point of view, ranging from symmetry violation, to string
theory, astronomy, particle physics, and black holes [51], [52], [53], [54],
[55], [56], [57], [58].

In this work, we will investigate the kinematical properties of particles in
$\kappa$-Minkowski spacetime (chapter •), and we will then introduce the
$\kappa$-deformed complex scalar field, studying its properties and computing
the Noether charges (chapter 1). Finally, we will compute the propagator and
the imaginary part of the 1-loop correction to it (chapter 2). Each of the
chapters will contain a final section investigating possible experimental
signature of the theoretical results obtained earlier in the chapter. There
have been already several approaches building fields on top of
$\kappa$-Minkowski spacetime, or using $\kappa$-Minkowski algebra in general,
see for example [39], [86], [40], [41], [42], [43], [44], [45], [46], [47],
[48], [80], [49], [50]. In contrast to these models, the construction which we
will present in this work allows for a much simpler, canonical definition of
deformed symmetries in the $\kappa$-deformed context. The particularly simple
form of the discrete symmetries in turn allows for an interesting (and easy to
interpret) phenomenology.

Of course, key ingredients in all our constructions are the $\kappa$-Minkowski
spacetime, and the $\kappa$-Poincaré algebra, so we will dedicate the first
two section of the current chapter to a brief exposition of their properties.
It turns out, however, that it is not just the deformed $\kappa$-Poincaré
algebra to be important, but the Hopf algebra structure of which the
$\kappa$-Poincaré algebra is but a part [61]. The mathematical aspects of
$\kappa$-deformations and the importance of the relevant Hopf algebra have
been investigated in detail in [62], [63], [64], [65], [68], [69], and the
important physical ultra-relativistic and non-relativistic limits in [71],
[70], [72]. The constructions of $\kappa$-Minkowski spacetime and the
$\kappa$-Poincaré Hopf algebra, as well as the relations between them, have
been extensively studied for decades. Here we will highlight how one can
obtain $\kappa$-Minkowski spacetime starting from the $\kappa$-Poincaré Hopf
algebra following [59], [37] and references therein as a guide. We will then
show the opposite direction, namely how one can obtain the $\kappa$-Poincaré
Hopf algebra structure starting from $\kappa$-Minkowski spacetime, following
the approach presented in [60]. Both the approaches can be useful when working
in the context of $\kappa$-deformation, and each of them sheds light on some
feature of $\kappa$-deformation in general. Since these notions are well known
in the literature, we will not delve into too much details, referring the
reader to the provided references for a deeper treatment. Nevertheless, we
will be explicit in out computations, and show each important step.

###  2 Formal definition of Hopf algebra

Before beginning our discussion, we give a mathematical definition of a Hopf
algebra. A Hopf algebra is technically defined as a bi-algebra with an
antipode, satisfying several compatibility conditions. To understand what is a
bi-algebra with antipode, we will need to introduce the notion of algebra, co-
algebra, and antipode separately. Using the same notation as [73], we can
define a unital algebra as a vector field $\mathscr{A}$ over a field
$\mathbb{K}$ together with two maps, the product map
$\mu:\mathscr{A}\otimes\mathscr{A}\rightarrow\mathscr{A}$, and the unit map
$\eta:\mathbb{K}\rightarrow\mathscr{A}$, which need to satisfy the
associativity and unity axiom defined respectively by the following
commutative diagrams [73]

A co-algebra can basically be defined by reversing all the arrows. In
particular, a co-algebra is a vector field $\mathscr{C}$ over a field
$\mathbb{K}$ together with two maps, the co-product map
$\Delta:\mathscr{C}\rightarrow\mathscr{C}\otimes\mathscr{C}$ and the co-unit
map $\epsilon:\mathscr{C}\rightarrow\mathbb{K}$, which need to satisfy the co-
associativity and co-unit axiom, which are defined respectively by the
following commutative diagrams [73]

A bi-algebra $B$ is therefore both an algebra and a co-algebra, with
additional compatibility conditions. In fact, we now have a structure with
product $\mu$, unity $\eta$, co-product $\Delta$, and co-unity $\epsilon$, and
they must satisfy [73]

$\displaystyle\Delta(\mu(a,b))=\mu(\Delta(a),\Delta(b)),\qquad\epsilon(\mu(a,b))=\epsilon(a)\epsilon(b)$
(7)

for all $a,b\in B$, and where

$\displaystyle\Delta(\text{id})=\text{id}\otimes\text{id}\qquad\epsilon(\text{id})=1$
(8)

where $1\in\mathbb{K}$. Finally, a Hopf algebra $\mathscr{H}$ is a bi-algebra
with an additional map $S:\mathscr{H}\rightarrow\mathscr{H}$ such that

$\displaystyle S(\mu(a,b))=\mu(S(a),S(b))\qquad
S(\mathbb{1})=\mathbb{1}\qquad(S\otimes S)\circ\Delta=\tau\circ\Delta\circ S$
(9)

where $a,b\in\mathscr{H}$ and
$\tau:\mathscr{H}\otimes\mathscr{H}\rightarrow\mathscr{H}\otimes\mathscr{H}$
is the flip map defined by

$\displaystyle\tau(a\otimes b)=b\otimes a$ (10)

for all $a\in\mathscr{H}$.

In what follows we will concentrate on the $\kappa$-Poincaré Hopf algebra. All
the above equalities and commutative diagrams are satisfied by the product,
co-product, unity, co-unity, and antipode, but we will not check them
explicitly here. Our aim with the discussion in this section and the next is
to highlight how the $\kappa$-Poincaré Hopf algebra, and more in particular
its co-product sector, is related to the $\kappa$-Minkowski spacetime.

###  3 From $\kappa$-Poincaré to $\kappa$-Minkowski

The algebra sector of the $\kappa$-Poincaré Hopf algebra in the so called
bicrossproduct basis is given by

$\displaystyle[M_{i},k_{j}]=i\epsilon_{ijl}k_{l},\qquad[M_{i},k_{0}]=0,$ (11)
$\displaystyle[N_{i},k_{j}]=i\delta_{ij}\left(\frac{\kappa}{2}(1-e^{-2k_{0}/\kappa})+\frac{\mathbf{k}^{2}}{2\kappa}\right)-\frac{i}{\kappa}k_{i}k_{j},\qquad[N_{i},k_{0}]=ik_{i},$
(12)
$\displaystyle[M_{i},M_{j}]=i\epsilon_{ijk}M_{k},\qquad[M_{i},N_{j}]=i\epsilon_{ijk}N_{k}\qquad[N_{i},N_{j}]=-i\epsilon_{ijk}M_{k}.$
(13)

The co-algebra sector is given by

$\displaystyle\Delta k_{i}=k_{i}\otimes\mathbb{1}+e^{-k_{0}/\kappa}\otimes
k_{i},\qquad\Delta k_{0}=k_{0}\otimes\mathbb{1}+\mathbb{1}\otimes k_{0},$ (14)
$\displaystyle\Delta N_{i}=N_{i}\otimes\mathbb{1}+e^{-k_{0}/\kappa}\otimes
N_{i}+\frac{1}{\kappa}\epsilon_{ijk}k_{j}\otimes M_{k},$ (15)
$\displaystyle\Delta M_{i}=M_{i}\otimes\mathbb{1}+\mathbb{1}\otimes M_{i}$
(16)

and finally the antipode sector is given by

$\displaystyle S(k_{0})=-k_{0},\qquad S(k_{i})=-k_{i}e^{k_{0}/\kappa},$ (17)
$\displaystyle S(M_{i})=-M_{i},\qquad
S(N_{i})=-e^{k_{0}/\kappa}\left(N_{i}-\frac{1}{\kappa}\epsilon_{ijk}k_{j}M_{k}\right).$
(18)

The co-units $\epsilon(k)$, $\epsilon(N_{i})$, $\epsilon(M_{j})$ are all zero.
Starting from this Hopf algebra, and in particular from the co-algebra sector,
we can build the spacetime associated to the algebra. The construction is
quite intuitive, because one first introduces the objects $x^{\mu}$
(interpreted as ‘coordinates’, the dual objects of the momenta $k_{\mu}$), and
then one defines the action of the algebra generators $k_{\mu}$, $N_{i}$,
$M_{j}$ on them. Having done this, the co-algebra sector is used to define the
action of $k_{\mu}$, $N_{i}$, $M_{j}$ on products $x^{\mu}x^{\nu}$, and one
immediately gets the coordinates commutation relation defining
$\kappa$-Minkowski spacetime from the action of $k_{\alpha}$ on
$x^{\mu}x^{\nu}$ (or more in general on polynomials of $x^{\mu}$).
Furthermore, one can also check that such commutation relations are invariant
under the action of $N_{i}$, $M_{j}$, showing that the Poincaré algebra indeed
describes the symmetries of $\kappa$-Minkowski spacetime. More in detail,
using the symbol $\triangleright$ to indicate the action of one of the
generators of the $\kappa$-Poincaré algebra on spacetime variables, we have
the definitions [37]

$\displaystyle k_{\mu}\triangleright x^{\nu}=-i\delta_{\mu}^{\nu},\qquad
N_{i}\triangleright\hat{x}^{j}=i\hat{x}^{0}\delta_{i}^{j},\qquad
N_{i}\triangleright\hat{x}^{0}=ix_{i},\qquad k_{\mu}\triangleright
1=\epsilon(k_{\mu})=0$ (19)

We then use the co-products to define the actions on polynomials of $x^{\mu}$
using the so called Sweedler notation. Writing everything explicitly, we have

$\displaystyle\Delta f\triangleright g_{1}g_{2}=(\Delta f)(g_{1},g_{2})$
$\displaystyle=\sum_{\alpha}(f_{\alpha}^{(1)}\triangleright
g_{1})(f_{\alpha}^{(2)}\triangleright g_{2})$ (20)

where $g_{1},g_{2}$ are some coordinates. The objects with (1) and (2)
represent respectively the ‘first’ and ‘second’ terms in each of the tensor
products in the definition of co-product in eq. (14), (15), (16). To make
things more concrete, take for example $k_{i}$ whose co-product rule is given
in eq. (14). We have

$\displaystyle k_{i}\triangleright\hat{x}^{0}\hat{x}^{j}$
$\displaystyle=\sum_{\alpha}((k_{i})_{(1)}^{\alpha}\triangleright\hat{x}^{0})((k_{i})_{(2)}^{\alpha}\triangleright\hat{x}^{j})$
(21)
$\displaystyle=((k_{i})_{(1)}^{1}\triangleright\hat{x}^{0})((k_{i})_{(2)}^{1}\triangleright\hat{x}^{j})+((k_{i})_{(1)}^{2}\triangleright\hat{x}^{0})((k_{i})_{(2)}^{2}\triangleright\hat{x}^{j})$
(22)
$\displaystyle=(k_{i}\triangleright\hat{x}^{0})(\mathbb{1}\triangleright\hat{x}^{j})+(e^{-k_{0}/\kappa}\triangleright\hat{x}^{0})(k_{i}\triangleright\hat{x}^{j})$
(23)
$\displaystyle=-i\delta_{i}^{0}\hat{x}^{j}+\left[\left(1-\frac{k_{0}}{\kappa}+\frac{k_{0}^{2}}{2\kappa^{2}}-\dots\right)\triangleright\hat{x}^{0}\right](-i\delta_{i}^{j})$
(24)
$\displaystyle=0-ix_{0}\delta_{i}^{j}+\frac{(k_{0}\triangleright\hat{x}^{0})}{\kappa}i\delta_{i}^{j}-\frac{k_{0}\triangleright(k_{0}\triangleright\hat{x}^{0})}{2\kappa^{2}}i\delta_{i}^{j}+\dots$
(25)
$\displaystyle=-i\hat{x}^{0}\delta_{i}^{j}+\frac{1}{\kappa}\delta_{i}^{j}+0-\dots$
(26) $\displaystyle=-ix_{0}\delta_{i}^{j}+\frac{1}{\kappa}\delta_{i}^{j}$ (27)

Notice that we also used the relations $k^{2}\triangleright
x=k\triangleright(k\triangleright x)$, and recall that $\epsilon(k)=0$ in the
$\kappa$-Poincaré Hopf algebra. In the same way, we also have

$\displaystyle k_{i}\triangleright\hat{x}^{j}\hat{x}^{0}$
$\displaystyle=\sum_{\alpha}((k_{i})_{(1)}^{\alpha}\triangleright\hat{x}^{j})((k_{i})_{(2)}^{\alpha}\triangleright\hat{x}^{0})$
(28)
$\displaystyle=((k_{i})_{(1)}^{1}\triangleright\hat{x}^{j})((k_{i})_{(2)}^{1}\triangleright\hat{x}^{0})+((k_{i})_{(1)}^{2}\triangleright\hat{x}^{j})((k_{i})_{(2)}^{2}\triangleright\hat{x}^{0})$
(29)
$\displaystyle=(k_{i}\triangleright\hat{x}^{j})(\mathbb{1}\triangleright\hat{x}^{0})+(e^{-k_{0}/\kappa}\triangleright\hat{x}^{j})(k_{i}\triangleright\hat{x}^{0})$
(30) $\displaystyle=-i\hat{x}^{0}\delta_{i}^{j}.$ (31) $\displaystyle
k_{0}\triangleright\hat{x}^{0}\hat{x}^{j}$
$\displaystyle=\sum_{\alpha}((k_{0})_{(1)}^{\alpha}\triangleright\hat{x}^{0})((k_{0})_{(2)}^{\alpha}\triangleright\hat{x}^{j})$
(32)
$\displaystyle=((k_{0})_{(1)}^{1}\triangleright\hat{x}^{0})((k_{0})_{(2)}^{1}\triangleright\hat{x}^{j})+((k_{0})_{(1)}^{2}\triangleright\hat{x}^{0})((k_{0})_{(2)}^{2}\triangleright\hat{x}^{j})$
(33)
$\displaystyle=(k_{0}\triangleright\hat{x}^{0})(\mathbb{1}\triangleright\hat{x}^{j})+(\mathbb{1}\triangleright\hat{x}^{0})(k_{0}\triangleright\hat{x}^{j})$
(34) $\displaystyle=-i\hat{x}^{j}$ (35) $\displaystyle
k_{0}\triangleright\hat{x}^{j}\hat{x}^{0}$
$\displaystyle=\sum_{\alpha}((k_{0})_{(1)}^{\alpha}\triangleright\hat{x}^{j})((k_{0})_{(2)}^{\alpha}\triangleright\hat{x}^{0})$
(36)
$\displaystyle=((k_{0})_{(1)}^{1}\triangleright\hat{x}^{j})((k_{0})_{(2)}^{1}\triangleright\hat{x}^{0})+((k_{0})_{(1)}^{2}\triangleright\hat{x}^{j})((k_{0})_{(2)}^{2}\triangleright\hat{x}^{0})$
(37)
$\displaystyle=(k_{0}\triangleright\hat{x}^{j})(\mathbb{1}\triangleright\hat{x}^{0})+(\mathbb{1}\triangleright\hat{x}^{j})(k_{0}\triangleright\hat{x}^{0})$
(38) $\displaystyle=-i\hat{x}^{j}$ (39)

Furthermore, one can also obtain the relation

$\displaystyle k_{i}\triangleright\hat{x}^{j}\hat{x}^{k}$
$\displaystyle=\sum_{\alpha}((k_{i})_{(1)}^{\alpha}\triangleright\hat{x}^{j})((k_{i})_{(2)}^{\alpha}\triangleright\hat{x}^{k})$
(40)
$\displaystyle=((k_{i})_{(1)}^{1}\triangleright\hat{x}^{j})((k_{i})_{(2)}^{1}\triangleright\hat{x}^{k})+((k_{i})_{(1)}^{2}\triangleright\hat{x}^{j})((k_{i})_{(2)}^{2}\triangleright\hat{x}^{k})$
(41)
$\displaystyle=(k_{i}\triangleright\hat{x}^{j})(\mathbb{1}\triangleright\hat{x}^{k})+(e^{-k_{0}/\kappa}\triangleright\hat{x}^{j})(k_{i}\triangleright\hat{x}^{k})$
(42)
$\displaystyle=-i\delta_{i}^{j}\hat{x}^{k}+\left[\left(1-\frac{k_{0}}{\kappa}+\frac{k_{0}^{2}}{2\kappa^{2}}-\dots\right)\triangleright\hat{x}^{j}\right](-i\delta_{i}^{k})$
(43) $\displaystyle=-i\delta_{i}^{j}\hat{x}^{k}-i\delta_{i}^{k}\hat{x}^{j}$
(44)

One can therefore deduce

$\displaystyle
k_{i}\triangleright\hat{x}^{0}\hat{x}^{j}-k_{i}\triangleright\hat{x}^{j}\hat{x}^{0}=\frac{\delta_{i}^{j}}{\kappa}\qquad
k_{0}\triangleright\hat{x}^{0}\hat{x}^{j}-k_{0}\triangleright\hat{x}^{j}\hat{x}^{0}=0$
(45) $\displaystyle
k_{i}\triangleright\hat{x}^{j}\hat{x}^{k}-k_{i}\triangleright\hat{x}^{k}\hat{x}^{j}=0$
(46)

and therefore

$\displaystyle[\hat{x}^{0},\hat{x}^{j}]=\frac{i}{\kappa}\hat{x}^{j}\qquad[\hat{x}^{i},\hat{x}^{j}]=0$
(47)

which are the commutation relations for the coordinates defining
$\kappa$-Minkowski spacetime. For the second commutator, notice that we used
$k_{0}\triangleright\hat{x}^{i}=0$. This algebra is called $\mathfrak{an}(3)$
algebra.

One can then show that the commutators (47) is invariant under boosts and
rotations. The invariance under rotations is trivial since, as can be seen
from eq. (11), (13), (16), (18) the rotation sector is completely not deformed
in the $\kappa$-Poincaré Hopf algebra, and so it acts as usual. For the
boosts, using eq. (19), as well as the previous computations one gets [61]

$\displaystyle N_{i}\triangleright\hat{x}^{0}\hat{x}^{j}$
$\displaystyle=(N_{i}\triangleright\hat{x}^{0})(\mathbb{1}\triangleright\hat{x}^{j})+(e^{-k_{0}/\kappa}\triangleright\hat{x}^{0})(N_{i}\triangleright\hat{x}^{j})+\frac{1}{\kappa}\epsilon_{ijk}(k_{j}\triangleright\hat{x}^{0})(M_{k}\triangleright\hat{x}^{j})$
$\displaystyle=ix_{i}\hat{x}^{j}+\left[\left(1-\frac{k_{0}}{\kappa}+\frac{k_{0}^{2}}{2\kappa^{2}}-\dots\right)\triangleright\hat{x}^{0}\right](i\hat{x}^{0}\delta_{i}^{j})+0$
(48)
$\displaystyle=ix_{i}\hat{x}^{j}+i(\hat{x}^{0})^{2}\delta_{i}^{j}-\frac{\hat{x}^{0}\delta_{i}^{j}}{\kappa}$
(49) $\displaystyle N_{i}\triangleright\hat{x}^{j}\hat{x}^{0}$
$\displaystyle=(N_{i}\triangleright\hat{x}^{j})(\mathbb{1}\triangleright\hat{x}^{0})+(e^{-k_{0}/\kappa}\triangleright\hat{x}^{j})(N_{i}\triangleright\hat{x}^{0})+\frac{1}{\kappa}\epsilon_{ijk}(k_{j}\triangleright\hat{x}^{j})(M_{k}\triangleright\hat{x}^{0})$
$\displaystyle=i\hat{x}^{0}\delta_{i}^{j}\hat{x}^{0}+i\hat{x}^{j}x_{i}+0$ (50)
$\displaystyle=i(\hat{x}^{0})^{2}\delta_{i}^{j}+i\hat{x}^{j}x_{i}$ (51)

and therefore

$\displaystyle
N_{i}\triangleright[\hat{x}^{0},\hat{x}^{j}]=N_{i}\triangleright\frac{i\hat{x}^{j}}{\kappa}+i[x_{i},\hat{x}^{j}]=N_{i}\triangleright\frac{i\hat{x}^{j}}{\kappa}$
(52)

which shows that indeed the commutation relation (47) are invariant under
$\kappa$-deformed boosts. Notice that we used the fact that
$M_{k}\triangleright\hat{x}^{0}=0$ (spatial rotations do not act on the time
direction) and eq. (47).

Although we used the bicrossproduct basis for the computations up to now, this
is by no means the only choice. In particular, another choice of coordinates
for momentum space is the so called classical basis, which is related to the
bicrossproduct one by the following coordinate change

$\displaystyle
P_{0}(k_{0},\mathbf{k})=\kappa\sinh\frac{k_{0}}{\kappa}+\frac{\mathbf{k}^{2}}{2\kappa}e^{\frac{k_{0}}{\kappa}},\qquad
P_{i}(k_{0},\mathbf{k})=k_{i}e^{\frac{k_{0}}{\kappa}}$ (53) $\displaystyle
P_{4}(k_{0},\mathbf{k})=\kappa\cosh\frac{k_{0}}{\kappa}-\frac{\mathbf{k}^{2}}{2\kappa}e^{\frac{k_{0}}{\kappa}}.$
(54)

One can check by direct computations [74], [75] that

$\displaystyle-P_{0}^{2}+\mathbf{P}^{2}+P_{4}^{2}=\kappa^{2}\qquad
P_{+}:=P_{0}+P_{4}>0\qquad P_{4}>0.$ (55)

In this basis, the algebra sector of the $\kappa$-Poincaré Hopf algebra reads

$\displaystyle[M_{i},P_{j}]=i\epsilon_{ijk}P_{k},\quad[M_{i},P_{0}]=0,\quad[N_{i},P_{j}]=i\delta_{ij}P_{0},\quad[N_{i},P_{0}]=iP_{i},$
(56)
$\displaystyle[M_{i},M_{j}]=i\epsilon_{ijk}M_{k},\quad[M_{i},N_{j}]=i\epsilon_{ijk}N_{k},\quad[N_{i},N_{j}]=i\epsilon_{ijk}M_{k}.$
(57)

The co-algebra sector is given by

$\displaystyle\Delta P_{i}=\frac{1}{\kappa}P_{i}\otimes
P_{+}+\mathbb{1}\otimes P_{i},$ (58) $\displaystyle\Delta
P_{0}=\frac{1}{\kappa}P_{0}\otimes P_{+}+\sum_{k}\frac{P_{k}}{P_{+}}\otimes
P_{k}+\frac{\kappa}{P_{+}}\otimes P_{0},$ (59) $\displaystyle\Delta
P_{4}=\frac{1}{\kappa}P_{4}\otimes P_{+}-\sum_{k}\frac{P_{k}}{P_{+}}\otimes
P_{k}-\frac{\kappa}{P_{+}}\otimes P_{0},$ (60) $\displaystyle\Delta
N_{i}=N_{i}\otimes\mathbb{1}+\frac{\kappa}{P_{+}}\otimes
N_{i}+\epsilon_{ijk}\frac{P_{j}}{P_{+}}\otimes M_{k},$ (61)
$\displaystyle\Delta M_{i}=M_{i}\otimes\mathbb{1}+\mathbb{1}\otimes M_{i},$
(62)

and the antipodes by

$\displaystyle S(P_{i})=-\frac{\kappa P_{i}}{P_{+}},\qquad
S(P_{0})=-P_{0}+\frac{\mathbf{P}^{2}}{P_{+}},\qquad S(P_{4})=P_{4},$ (63)
$\displaystyle S(M_{i})=-M_{i},\qquad
S(N_{i})=-N_{i}\frac{P_{+}}{\kappa}+\frac{1}{\kappa}\epsilon_{ijk}P_{j}M_{k}.$
(64)

Once again, the co-units $\epsilon(k)$, $\epsilon(N_{i})$, $\epsilon(M_{j})$
are all zero. Furthermore, notice also that rotations are non-deformed in both
basis. A very interesting property of the classical basis is the fact that the
algebra coincides with the non-deformed Poincaré algebra, so the totality of
the effect of $\kappa$-deformation is found in the co-product and antipode
sector. This also shows that $\kappa$-deformation is important at the Hopf
algebra level, and not merely at the algebra level. Because of the
construction of $\kappa$-Minkowski showed above, the co-products cover a
particularly important role since they allow the definition of
$\kappa$-Minkowski spacetime. More details can be found in [75], [59].

###  4 From $\kappa$-Minkowski to $\kappa$-Poincaré

The first step in going from $\kappa$-Minkowski to $\kappa$-Poincaré is to
find an explicit representation of the $\mathfrak{an}(3)$ algebra in eq. (47).
We choose (with a slight abuse of notation, we call the explicit matrices
representing the algebra with the same names as the elements of the algebra
itself)

$\hat{x}^{0}=-\frac{i}{\kappa}\,\left(\begin{array}[]{ccc}0&\mathbf{0}&1\\\
\mathbf{0}^{T}&\tilde{\mathbf{0}}&\mathbf{0}^{T}\\\
1&\mathbf{0}&0\end{array}\right)\quad\hat{x}^{i}=\frac{i}{\kappa}\,\left(\begin{array}[]{ccc}0&{(\epsilon^{i})\,{}^{T}}&0\\\
\epsilon^{i}&\tilde{\mathbf{0}}&\epsilon^{i}\\\
0&-(\epsilon^{i})\,{}^{T}&0\end{array}\right).$ (71)

The central $\tilde{\mathbf{0}}$ is a $3\times 3$ null matrix, while
$\mathbf{0}=(0,0,0)$ and $\epsilon^{i}$ is a unit column vector, such that
$(\epsilon^{1})^{T}=(1,0,0)$, $(\epsilon^{2})^{T}=(0,1,0)$, and
$(\epsilon^{1})^{T}=(0,0,1)$. We now compute the group elements of $AN(3)$

$\hat{e}_{k}=e^{ik_{i}\hat{x}^{i}}e^{ik_{0}\hat{x}^{0}}\,.$ (72)

Notice that because of the non-trivial algebra we need to carefully define the
exponentiation, i.e. on how we decide to define the group element. Here we
choose the so called time-to-the-right convention, in which
$e^{ik_{j}\hat{x}^{j}}$ is to the left of $e^{ik_{0}\hat{x}^{0}}$. Because of
this choice, the objects $k_{0},\mathbf{k}$ in the exponent (which are
dimensionally momenta) coincide with the bicrossproduct basis introduced in
the previous section. With a different choice of group element, for example

$\displaystyle\hat{e}_{q}=e^{iq_{\mu}\hat{x}^{\mu}}$ (73)

one would get different coordinates for momentum space (in this particular
case, we would be introducing the so called normal basis [75], but we will not
address this further). Furthermore, notice that the representation matrices
are $5\times 5$.

We have

$\displaystyle\begin{pmatrix}0&\bm{0}&1\\\ \bm{0}&\bm{0}&\bm{0}\\\
1&\bm{0}&0\end{pmatrix}^{2n}=\begin{pmatrix}1&\bm{0}&0\\\
\bm{0}&\bm{0}&\bm{0}\\\
0&\bm{0}&1\end{pmatrix}\qquad\begin{pmatrix}0&\bm{0}&1\\\
\bm{0}&\bm{0}&\bm{0}\\\
1&\bm{0}&0\end{pmatrix}^{2n+1}=\begin{pmatrix}0&\bm{0}&1\\\
\bm{0}&\bm{0}&\bm{0}\\\ 1&\bm{0}&0\end{pmatrix}$ (74)

and therefore

$\displaystyle e^{ik_{0}\hat{x}^{0}}$
$\displaystyle=\sum_{n}\frac{(ik_{0}x^{0})^{n}}{n!}=\sum_{n}\frac{1}{n!}\left(\frac{k_{0}}{\kappa}\right)^{n}\begin{pmatrix}0&\bm{0}&1\\\
\bm{0}&\bm{0}&\bm{0}\\\ 1&\bm{0}&0\end{pmatrix}^{n}$ (75)
$\displaystyle=\left(\begin{array}[]{ccc}\cosh\frac{k_{0}}{\kappa}&\mathbf{0}&\sinh\frac{k_{0}}{\kappa}\\\
&&\\\ \mathbf{0}&\mathbf{1}&\mathbf{0}\\\ &&\\\
\sinh\frac{k_{0}}{\kappa}&\mathbf{0}&\cosh\frac{k_{0}}{\kappa}\end{array}\right)$
(81)

Similarly one can check that

$\displaystyle\begin{pmatrix}0&\epsilon^{T}&0\\\ \epsilon&\bm{0}&\epsilon\\\
0&-\epsilon^{T}&0\end{pmatrix}^{2}=\begin{pmatrix}1&\bm{0}&1\\\
\bm{0}&\bm{0}&\bm{0}\\\
-1&\bm{0}&-1\end{pmatrix}\qquad\begin{pmatrix}0&\epsilon^{T}&0\\\
\epsilon&\bm{0}&\epsilon\\\ 0&-\epsilon^{T}&0\end{pmatrix}^{3}=0$ (82)

This behaviour is the reason for the ‘$N$’ in the name ’$AN(3)$’, which stands
for ‘nilpotent’. Therefore

$\displaystyle
e^{ik_{i}\hat{x}^{i}}=1+ik_{i}\hat{x}^{i}+\frac{1}{2}(ik_{i}\hat{x}^{i})^{2}=\begin{pmatrix}1+\frac{\mathbf{k}^{2}}{2\kappa^{2}}&\frac{\mathbf{k}^{T}}{\kappa}&\frac{\mathbf{k}^{2}}{2\kappa^{2}}\\\
\frac{\mathbf{k}}{\kappa}&\bm{1}&\frac{\mathbf{k}}{\kappa}\\\
-\frac{\mathbf{k}^{2}}{2\kappa^{2}}&-\frac{\mathbf{k}^{T}}{\kappa}&1-\frac{\mathbf{k}^{2}}{2\kappa^{2}}.\end{pmatrix}$
(83)

Multiplying these two matrices we get

$\displaystyle\hat{e}_{k}=\begin{pmatrix}\cosh\frac{k_{0}}{\kappa}+\frac{\mathbf{k}^{2}}{2\kappa^{2}}e^{\frac{k_{0}}{\kappa}}&\frac{\mathbf{k}^{T}}{\kappa}&\sinh\frac{k_{0}}{\kappa}+\frac{\mathbf{k}^{2}}{2\kappa^{2}}e^{\frac{k_{0}}{\kappa}}\\\
\frac{\mathbf{k}}{\kappa}e^{\frac{k_{0}}{\kappa}}&\bm{1}&\frac{\mathbf{k}}{\kappa}e^{\frac{k_{0}}{\kappa}}\\\
\sinh\frac{k_{0}}{\kappa}-\frac{\mathbf{k}^{2}}{2\kappa^{2}}e^{\frac{k_{0}}{\kappa}}&-\frac{\mathbf{k}^{T}}{\kappa}&\cosh\frac{k_{0}}{\kappa}-\frac{\mathbf{k}^{2}}{2\kappa^{2}}e^{\frac{k_{0}}{\kappa}}\end{pmatrix}$
(84)

where we used the relation $\cosh x+\sinh x=e^{x}$. Notice now that we can
make the following substitutions

$\displaystyle{P_{0}}(k_{0},\mathbf{k})$ $\displaystyle=$
$\displaystyle\kappa\sinh\frac{k_{0}}{\kappa}+\frac{\mathbf{k}^{2}}{2\kappa}\,e^{{k_{0}}/\kappa},$
(85) $\displaystyle P_{i}(k_{0},\mathbf{k})$ $\displaystyle=$ $\displaystyle
k_{i}\,e^{{k_{0}}/\kappa},$ (86) $\displaystyle{P_{4}}(k_{0},\mathbf{k})$
$\displaystyle=$
$\displaystyle\kappa\cosh\frac{k_{0}}{\kappa}-\frac{\mathbf{k}^{2}}{2\kappa}\,e^{{k_{0}}/\kappa}$
(87)

to bring the above group element $\hat{e}_{k}$ into the following form

$\displaystyle\hat{e}_{P(k)}=\frac{1}{\kappa}\begin{pmatrix}\tilde{P}_{4}&\kappa\mathbf{P}/P_{+}&P_{0}\\\
\mathbf{P}&\kappa\times\mathbb{1}_{3\times 3}&\mathbf{P}\\\
\tilde{P}_{0}&-\kappa\mathbf{P}/P_{+}&P_{4}\end{pmatrix}$ (88)

where

$\displaystyle\tilde{P}_{0}=P_{0}-\frac{\mathbf{P}^{2}}{P_{+}}=-S(P_{0})\qquad\tilde{P}_{4}=P_{4}+\frac{\mathbf{P}^{2}}{P_{+}}$
(89)

Incidentally, one can use the group property
$\hat{e}_{P(k)}\hat{e}_{Q(l)}=\hat{e}_{P(k)\oplus Q(l)}$ and
$\hat{e}^{-1}_{P}=\hat{e}_{S(P)}$ to obtain the deformed antipode and sum in
the classical basis.

$\displaystyle
S(P_{0})=-P_{0}+\frac{\mathbf{P}^{2}}{P_{0}+P_{4}}=\frac{\kappa^{2}}{P_{0}+P_{4}}-P_{4}\,,$
(90) $\displaystyle
S(\mathbf{P})=-\frac{\kappa\mathbf{P}}{P_{0}+P_{4}}\,,\quad S(P_{4})=P_{4}.$
(91) $\displaystyle(P\oplus Q)_{0}$
$\displaystyle=\frac{1}{\kappa}\,P_{0}(Q_{0}+Q_{4})+\frac{\mathbf{P}\mathbf{Q}}{P_{0}+P_{4}}+\frac{\kappa}{P_{0}+P_{4}}\,Q_{0}$
(92) $\displaystyle(P\oplus Q)_{i}$
$\displaystyle=\frac{1}{\kappa}\,P_{i}(Q_{0}+Q_{4})+Q_{i}$ (93)
$\displaystyle(P\oplus Q)_{4}$
$\displaystyle=\frac{1}{\kappa}\,P_{4}(Q_{0}+Q_{4})-\frac{\mathbf{P}\mathbf{Q}}{P_{0}+P_{4}}-\frac{\kappa}{P_{0}+P_{4}}\,Q_{0}$
(94)

One can of course do the same thing in the bicrossproduct basis. Notice the
relationship between these equations and the co-products in eq. (58), (59),
(60).

Notice that the momenta defined in eq. (85) are indeed again the classical
basis defined in eq. (53), (54). By construction, therefore, they satisfy the
constraints in eq. (55). Furthermore, notice that the group elements
$\hat{e}_{P(k)}$ are uniquely identified once one knows the quantities
$P_{0},P_{i},P_{4}$, which are the components on the last column of the matrix
(88). Indeed, calling $\mathscr{O}=(0,0,0,0,\kappa)^{T}$, we have

$\displaystyle\hat{e}_{P(k)}\mathscr{O}=(P_{0},\mathbf{P},P_{4})^{T}$ (95)

and the action of the group $AN(3)$ on $\mathscr{O}$ is transitive, meaning
that acting with all $g\in AN(3)$ on $\mathscr{O}$ reaches all the points of
the manifold associated to the group. We already know that relations (55)
hold, which means that the 4-dimesional manifold associated to $AN(3)$ is half
of de Sitter space, namely the half where $P_{+}>0$. Notice that the curvature
of such a space is constant and equal to $1/\kappa^{2}$. Furthermore, since
each point is uniquely described by the coordinates $(P_{0},\mathbf{P})$
(after fixing both $P_{0}$ and $\mathbf{P}$, $P_{4}$ is not a free parameter,
and can be obtained through the constraint), the group manifold $AN(3)$ is
therefore our momentum space manifold. This gives a clear picture of how the
non-commutative nature of spacetime coordinates is linked to a constant
curvature of momentum space. Notice that to obtain coordinates spanning the
other half of de Sitter space defined by the first constraint in eq. (55), one
applies the same group elements to the new origin
$\mathscr{O}^{*}=(0,0,0,0,-\kappa)^{T}$. Alternatively, one can define a
matrix

$\displaystyle\mathfrak{z}=e^{\pi\kappa\hat{x}^{0}}$ (96)

which allows us to define

$\displaystyle\hat{e}^{*}_{k}=\hat{e}_{k}\mathfrak{z}$ (97)

and then apply these new objects $\hat{e}^{*}$ to the same origin
$\mathscr{O}=(0,0,0,0,\kappa)$. In both cases, we obtain the coordinates

$\displaystyle{P^{*}_{0}}(k_{0},\mathbf{k})$ $\displaystyle=$
$\displaystyle-\kappa\sinh\frac{k_{0}}{\kappa}-\frac{\mathbf{k}^{2}}{2\kappa}\,e^{{k_{0}}/\kappa},$
(98) $\displaystyle P^{*}_{i}(k_{0},\mathbf{k})$ $\displaystyle=$
$\displaystyle-k_{i}\,e^{{k_{0}}/\kappa},$ (99)
$\displaystyle{P^{*}_{4}}(k_{0},\mathbf{k})$ $\displaystyle=$
$\displaystyle-\kappa\cosh\frac{k_{0}}{\kappa}+\frac{\mathbf{k}^{2}}{2\kappa}\,e^{{k_{0}}/\kappa}.$
(100)

These starred coordinates satisfy $P_{+}^{*}=P_{0}^{*}+P_{4}^{*}<0$, and one
can show that

$\displaystyle S(P^{*})=-S(P)$ (101)

and that the deformed sum is not modified if one or both of the momenta is
starred. In other words, we have

$\displaystyle(P^{(*)}\oplus Q^{(*)})_{0}$
$\displaystyle=\frac{1}{\kappa}\,P^{(*)}_{0}(Q^{(*)}_{0}+Q^{(*)}_{4})+\frac{\mathbf{P}^{(*)}\mathbf{Q}^{(*)}}{P^{(*)}_{0}+P^{(*)}_{4}}+\frac{\kappa}{P^{(*)}_{0}+P^{(*)}_{4}}\,Q^{(*)}_{0}$
(102) $\displaystyle(P^{(*)}\oplus Q^{(*)})_{i}$
$\displaystyle=\frac{1}{\kappa}\,P^{(*)}_{i}(Q^{(*)}_{0}+Q^{(*)}_{4})+Q^{(*)}_{i}$
(103) $\displaystyle(P^{(*)}\oplus Q^{(*)})_{4}$
$\displaystyle=\frac{1}{\kappa}\,P^{(*)}_{4}(Q^{(*)}_{0}+Q^{(*)}_{4})-\frac{\mathbf{P}^{(*)}\mathbf{Q}^{(*)}}{P^{(*)}_{0}+P^{(*)}_{4}}-\frac{\kappa}{P^{(*)}_{0}+P^{(*)}_{4}}\,Q^{(*)}_{0},$
(104)

where the (∗) indicates that either $P$ or $Q$ or both or neither can be
starred momenta, and the deformed sum formula works in the same way regardless
[78].

We now address the issue of how one can get the $\kappa$-Poincaré structure
out of the $AN(3)$ group, and in particular the co-product and antipode
sectors. We will not treat in detail all the coproducts and antipodes for all
generators, and we will use either the classical or the bicrossproduct basis
depending on which is easier to work with. The interested reader is referred
back to [60] for more details.

We start with the coproduct and antipode of momenta. As was already discussed
in the previous section, coproducts describe the way the $\kappa$-Poincaré
algebra act on products, be it products of coordinates like before, or of
states like we will consider now. In particular, we have

$\displaystyle\Delta k_{\mu}(|p\rangle\otimes|q\rangle)=(p\oplus
q)_{\mu}|p\rangle\otimes|q\rangle$ (105)

where $p\oplus q$ is define through the group multiplication

$\displaystyle\hat{e}_{p}\hat{e}_{q}=\hat{e}_{p\oplus q}.$ (106)

In the same way, the antipode $S(k)_{\mu}$ is related to inverse group
elements by the definition

$\displaystyle\hat{e}^{-1}_{k}=\hat{e}_{S(k)}.$ (107)

One can therefore multiply two copies of eq. (84) and deduce the sum of
momenta, and then take the inverse of eq. (84) directly to obtain the
antipode. Alternatively, in a more illuminating and less computation intensive
way, one can work directly with the definition in eq. (72). Indeed, using the
Baker-Campbell-Hausdorff formula one can show that

$\displaystyle[X,Y]=sY\qquad\implies\qquad
e^{X}e^{Y}=\exp\left(X+\frac{s}{1-e^{-s}}Y\right)$ (108)

which means that since because of eq. (47) we have

$\displaystyle[ik_{0}x^{0},i\mathbf{k}\mathbf{x}]=-k_{0}\mathbf{k}[x^{0},\mathbf{x}]=-k_{0}\mathbf{k}\frac{i}{\kappa}\mathbf{x}=-\frac{k_{0}}{\kappa}(i\mathbf{k}\mathbf{x}),$
(109)

then

$\displaystyle e^{ik_{0}x^{0}}e^{i\mathbf{k}\mathbf{x}}$
$\displaystyle=\exp\left(ik_{0}x^{0}+\frac{-\frac{k_{0}}{\kappa}}{1-e^{\frac{k_{0}}{\kappa}}}i\mathbf{k}\mathbf{x}\right)$
(110) $\displaystyle e^{i\mathbf{q}\mathbf{x}}e^{iq_{0}x^{0}}$
$\displaystyle=\exp\left(iq_{0}x^{0}+\frac{\frac{k_{0}}{\kappa}}{1-e^{-\frac{k_{0}}{\kappa}}}i\mathbf{q}\mathbf{x}\right)=\exp\left(iq_{0}x^{0}+\frac{-\frac{k_{0}}{\kappa}}{1-e^{\frac{k_{0}}{\kappa}}}ie^{\frac{k_{0}}{\kappa}}\mathbf{q}\mathbf{x}\right)$
(111)

and therefore sending
$\mathbf{q}\rightarrow\mathbf{k}e^{-\frac{k_{0}}{\kappa}}$ and
$q_{0}\rightarrow k_{0}$ we prove

$\displaystyle
e^{ik_{0}x^{0}}e^{i\mathbf{k}\mathbf{x}}=e^{ie^{-\frac{k_{0}}{\kappa}}\mathbf{k}\mathbf{x}}e^{ik_{0}x^{0}}.$
(112)

Using this relation, we get

$\displaystyle\hat{e}_{k}\hat{e}_{q}$
$\displaystyle=e^{i\mathbf{k}\mathbf{x}}e^{ik_{0}x^{0}}e^{i\mathbf{q}\mathbf{x}}e^{iq_{0}x^{0}}=e^{i\left(\mathbf{k}+e^{-\frac{k_{0}}{\kappa}}\mathbf{q}\right)\mathbf{x}}e^{i(k_{0}+q_{0})x^{0}}$
(113)

Furthermore, we can also recover the inverse group element $\hat{e}_{k}^{-1}$
using a similar procedure. In fact we can use the same steps as before, but at
the end we must impose that both the exponent go to zero, so that we get

$\displaystyle\mathbf{q}=-e^{\frac{k_{0}}{\kappa}}\mathbf{k}\qquad
q_{0}=-k_{0}$ (114)

and therefore

$\displaystyle\hat{e}_{k}^{-1}=e^{-ie^{\frac{k_{0}}{\kappa}}\mathbf{k}}e^{-ik_{0}x^{0}}$
(115)

We now consider the bicrossproduct coordinate functions on $AN(3)$ defined as
[75]

$\displaystyle f^{B}_{\mu}(g\in
AN(3))=f^{B}_{\mu}\left(e^{i\mathbf{k}\mathbf{x}}e^{ik_{0}x^{0}}\right):=k_{\mu}=(k_{0},\mathbf{k})$
(116)

These are important because functions on a group have a natural structure of
Hopf algebra [75]. In fact, using the same notation as [75], the algebra
$Fun(G)$ of functions on $G$ with multiplication and identity defined as usual

$\displaystyle(f_{1},f_{2})(g)=f_{1}(g)f_{2}(g)\qquad I(g)=1$ (117)

This algebra $Fun(G)$ is of course commutative and associative (because
pointwise multiplication of functions with codomain in $\mathbb{C}$ is of
course commutative and associative). From this we can naturally define all of
the additional structure needed for a Hopf algebra, namely coproduct, co-unit,
and antipode

$\displaystyle\Delta$ $\displaystyle:Fun(G)\rightarrow Fun(G\times
G)\qquad\epsilon:Fun(G)\rightarrow\mathbb{C}\qquad S:Fun(G)\rightarrow Fun(G)$
(118)

with the definitions

$\displaystyle(\Delta f)(g_{1},g_{2})$
$\displaystyle=\sum_{\alpha}f_{\alpha}^{(1)}\otimes
f_{\alpha}^{(2)}(g_{1},g_{2})=f(g_{1}g_{2})$ (119)
$\displaystyle\epsilon(g)=f(1)\qquad(S(f))(g)=f(g^{-1})$ (120)

Therefore, using these definitions and the bicrossproduct coordinate
functions, we immediately find

$\displaystyle\Delta\left(f^{B}_{0}\right)(\hat{e}_{k},\hat{e}_{q})$
$\displaystyle=f^{B}_{0}(\hat{e}_{k}\hat{e}_{q})$ (121)
$\displaystyle=f^{B}_{0}\left(e^{i\left(\mathbf{k}+e^{-\frac{k_{0}}{\kappa}}\mathbf{q}\right)\mathbf{x}}e^{i(k_{0}+q_{0})x^{0}}\right)$
(122) $\displaystyle=k_{0}+q_{0}$ (123) $\displaystyle=\left(f^{B}_{0}\otimes
1+1\otimes f^{B}_{0}\right)(\hat{e}_{k},\hat{e}_{q})$ (124)
$\displaystyle\Delta\left(\mathbf{f}^{B}_{i}\right)(\hat{e}_{k},\hat{e}_{q})$
$\displaystyle=\mathbf{f}^{B}_{i}(\hat{e}_{k}\hat{e}_{q})$ (125)
$\displaystyle=\mathbf{f}^{B}_{i}\left(e^{i\left(\mathbf{k}+e^{-\frac{k_{0}}{\kappa}}\mathbf{q}\right)\mathbf{x}}e^{i(k_{0}+q_{0})x^{0}}\right)$
(126) $\displaystyle=\mathbf{k}+e^{-\frac{k_{0}}{\kappa}}\mathbf{q}$ (127)
$\displaystyle=\left(\mathbf{f}^{B}_{i}\otimes
1+e^{-\frac{k_{0}}{\kappa}}\otimes\mathbf{f}^{B}_{i}\right)(\hat{e}_{k},\hat{e}_{q})$
(128)

and similarly for the co-unit

$\displaystyle\epsilon(f^{B}_{0})=f^{B}_{0}(1)=f^{B}_{0}\left(e^{i0\mathbf{x}}e^{i0x^{0}}\right)=0$
(129)
$\displaystyle\epsilon(\mathbf{f}^{B}_{i})=\mathbf{f}^{B}_{i}(1)=\mathbf{f}^{B}_{i}\left(e^{i0\mathbf{x}}e^{i0x^{0}}\right)=0$
(130)

and the antipode

$\displaystyle
S(f^{B}_{0})(\hat{e}_{k})=f^{B}_{0}(\hat{e}_{k}^{-1})=f^{B}_{0}\left(e^{-ie^{\frac{k_{0}}{\kappa}}\mathbf{k}}e^{-ik_{0}x^{0}}\right)=-k_{0}$
(131) $\displaystyle
S(\mathbf{f}^{B}_{i})(\hat{e}_{k})=f_{i}(\hat{e}_{k}^{-1})=f_{i}\left(e^{-ie^{\frac{k_{0}}{\kappa}}\mathbf{k}}e^{-ik_{0}x^{0}}\right)=-e^{\frac{k_{0}}{\kappa}}\mathbf{k}$
(132)

With a slight abuse of notation, these relations are traditionally written as

$\displaystyle\Delta k_{0}=k_{0}\otimes\mathbb{1}+\mathbb{1}\otimes
k_{0}\qquad\Delta\mathbf{k}_{i}=\mathbf{k}_{i}\otimes\mathbb{1}+e^{-\frac{k_{0}}{\kappa}}\otimes\mathbf{k}_{i}$
(133) $\displaystyle S(k)_{0}=-k_{0}\qquad
S(\mathbf{k})_{i}=-e^{\frac{k_{0}}{\kappa}}\mathbf{k}_{i}$ (134)

so that we recover eq. (14) and (17). In the same way, using the classical
basis, we recover eq. (58), (59), (63).

To get the coproduct of boosts and rotations, we follow a slightly different
route. Indeed, knowing that $AN(3)$ is our momentum manifold, we want to find
some action of $SO(1,3)$ on it. The trick to find it is to notice that both
$AN(3)$ and $SO(1,3)$ are subgroups of $SO(1,4)$. We can then use something
called Iwasawa decomposition

$\displaystyle\mathfrak{so}(1,4)=\mathfrak{so}(1,3)\oplus\mathfrak{a}\oplus\mathfrak{n}$

where $\mathfrak{a}$ is generated by $\hat{x}^{0}$, $\mathfrak{n}$ is
generated by $\hat{x}^{i}$. In other words, given $g\in SO(1,4)$, then given
$K_{g}\in SO(1,3)$ and $g\in AN(3)$, then there exist two unique elements
$K^{\prime}_{g}\in SO(1,3)$ and $g^{\prime}\in AN(3)$ such that

$\displaystyle K_{g}g=g^{\prime}K^{\prime}_{g}\qquad\Leftrightarrow\qquad
K_{g}g(K^{\prime}_{g})^{-1}=g^{\prime}.$

In this way, we obtain the action of $SO(1,3)$ on a single group element. To
get the co-product, we then need to define the action on the product $gh$,
where $g,h\in AN(3)$. We do this by the following relation.

$\displaystyle(gh)^{\prime}=K_{g}ghK^{\prime}_{gh}=(K_{g}g{K^{\prime}}_{g}^{-1})(K^{\prime}_{g}hK^{\prime}_{gh})$
(135)

We will see shortly that $K^{\prime}_{gh}$ is irrelevant for the definition of
the coproducts of $SO(1,3)$ generators, so that we only need $K_{g}$ and
$K^{\prime}_{g}$. We are only interested in infinitesimal transformations, so
we assume

$\displaystyle K_{g}\approx 1+i\xi^{a}\mathfrak{k}_{a}\qquad
K^{\prime}_{g}\approx 1+i\xi^{a}h^{b}_{a}(g)\mathfrak{k}_{b}.$ (136)

Notice that $h^{b}_{a}(g)$ is in general momentum dependent, a fact which is
responsible for the non-trivial boost co-product. Concentrating on boosts, and
in particular choosing a boost in the direction 1, and using the classical
basis for simplicity, explicitly we have

$\displaystyle\begin{pmatrix}1&\xi&0&0&0\\\ \xi&1&0&0&0\\\ 0&0&1&0&0\\\
0&0&0&1&0\\\
0&0&0&0&1\end{pmatrix}\begin{pmatrix}\bar{P}_{4}&\frac{\kappa\mathbf{P}_{i}}{P_{+}}&P_{0}\\\
\mathbf{P}_{i}&\kappa&\mathbf{P}_{i}\\\
\bar{P}_{0}&-\frac{\kappa\mathbf{P}_{i}}{P_{+}}&P_{4}\end{pmatrix}$ (137)
$\displaystyle=$
$\displaystyle\begin{pmatrix}\bar{P}^{\prime}_{4}&\frac{\kappa\mathbf{P}^{\prime}_{i}}{P^{\prime}_{+}}&P^{\prime}_{0}\\\
\mathbf{P}^{\prime}_{i}&\kappa&\mathbf{P}^{\prime}_{i}\\\
\bar{P}^{\prime}_{0}&-\frac{\kappa\mathbf{P}^{\prime}_{i}}{P^{\prime}_{+}}&P^{\prime}_{4}\end{pmatrix}\begin{pmatrix}1&\bar{\xi}^{1}&\bar{\xi}^{2}&\bar{\xi}^{3}&0\\\
\bar{\xi}^{1}&1&\bar{\rho}^{3}&-\bar{\rho}^{2}&0\\\
\bar{\xi}^{2}&-\bar{\rho}^{3}&1&\bar{\rho}^{1}&0\\\
\bar{\xi}^{3}&\bar{\rho}^{2}&-\bar{\rho}^{1}&1&0\\\ 0&0&0&0&1\end{pmatrix}$
(138)

where the parameters $\bar{\xi}^{i}$ and $\bar{\rho}^{j}$ can be uniquely
determined by dong the product and comparing both sides of the equivalence
[75]. After doing the computations, one obtains

$\displaystyle
K_{g}^{\prime}=\begin{pmatrix}1&\frac{\kappa}{P_{+}}\xi&0&0&0\\\
\frac{\kappa}{P_{+}}\xi&1&\frac{P_{2}}{P_{+}}\xi&\frac{P_{3}}{P_{+}}\xi&0\\\
0&-\frac{P_{2}}{P_{+}}\xi&1&0&0\\\ 0&-\frac{P_{3}}{P_{+}}\xi&0&1&0\\\
0&0&0&0&1\end{pmatrix}=\mathbb{1}+\frac{\kappa}{P_{+}}\xi
N_{1}+\epsilon_{ijk}\frac{P_{j}}{P_{+}}M_{k}.$ (139)

In other words, we have up to first order

$\displaystyle(gh)^{\prime}=(\mathbb{1}+\xi
N_{1})g\left(\mathbb{1}+\frac{\kappa}{P_{+}}\xi
N_{1}+\epsilon_{ijk}\frac{P_{j}}{P_{+}}M_{k}\right)hK^{\prime}_{gh}$ (140)

Now, notice that a multiplication from the right by $K^{\prime}_{gh}$ cannot
influence the momenta in a group element $h$. This is due to the fact that
$K^{\prime}_{gh}$ will have all zero entries in the last column and row,
except $(K^{\prime}_{gh})_{44}=1$, and a multiplication from the right of such
a matrix cannot influence the last row of $h$, which contains the momenta.
Therefore, we have up to first order

$\displaystyle(gh)^{\prime}=gh+\xi\left((N_{1}g)(h)+\left(\frac{\kappa}{P_{+}}g\right)(N_{1}h)\right)+\epsilon_{1jk}\left(\frac{P_{j}}{P_{+}}g\right)(M_{k}h).$
(141)

and generalizing this relation to a boost in a generic direction we get
immediately the co-product in eq. (61). Finally, for the antipode, one can
just switch $g\mapsto g^{-1}$ and $h\mapsto g$ into eq. (141), and impose that
the result gives $\mathbb{1}$, which means that we get the antipode in eq.
(64). For the rotation, the whole process is repeated, and one gets that the
co-product and the antipode are not deformed, in accordance with eq. (62) and
(64).

###  5 Finite boosts for two-particle states

What we computed above is the co-product of generators of the algebra on
tensor product states. To get a finite boost for two particles, one needs to
exponentiate this infinitesimal generators to get a finite transformation.
However, due to the non-trivial nature of co-products both in the classical
and bicrossproduct basis, this is often not practical. Luckily for us,
however, the Iwasawa decomposition that was used in section 4 holds in
general, and we only used a first order expansion for the involved matrices
because we only needed the generators co-products [77]. Here, we employ once
again the same decomposition, but this time we will use the full
transformation matrices to obtain a finite boost for a two-particle state.
Contrary to the previous sections, the computations and discussions for the
rest of the chapter are original results. Because of the importance of the
final result for the kinematics of a two-particle state in the
$\kappa$-deformed context, we will explicitly perform all the computations in
full.

We start therefore from

$\displaystyle Lg=g^{\prime}L^{\prime}_{g}$ (142)

where the boost on the RHS can in general be momentum dependent.

We need to recover what is $L^{\prime}_{g}$. Written explicitly in terms of
matrices, and considering without loss of generality $L$ as a boost in the
$x^{1}$ direction, we have

$\displaystyle L=\begin{pmatrix}\gamma&-\beta\gamma&0&0&0\\\
-\beta\gamma&\gamma&0&0&0\\\ 0&0&1&0&0\\\ 0&0&0&1&0\\\ 0&0&0&0&1\end{pmatrix}$
(143) $\displaystyle
g=\begin{pmatrix}\tilde{P}_{4}&\kappa\mathbf{P}_{1}/P_{+}&\kappa\mathbf{P}_{2}/P_{+}&\kappa\mathbf{P}_{3}/P_{+}&P_{0}\\\
\mathbf{P}_{1}&\kappa&0&0&\mathbf{P}_{1}\\\
\mathbf{P}_{2}&0&\kappa&0&\mathbf{P}_{2}\\\
\mathbf{P}_{3}&0&0&\kappa&\mathbf{P}_{3}\\\
\tilde{P}_{0}&-\kappa\mathbf{P}_{1}/P_{+}&-\kappa\mathbf{P}_{2}/P_{+}&-\kappa\mathbf{P}_{3}/P_{+}&P_{4}\end{pmatrix}$
(144) $\displaystyle
g^{\prime}=\begin{pmatrix}\tilde{P}^{\prime}_{4}&\kappa\mathbf{P}^{\prime}_{1}/P^{\prime}_{+}&\kappa\mathbf{P}^{\prime}_{2}/P^{\prime}_{+}&\kappa\mathbf{P}^{\prime}_{3}/P^{\prime}_{+}&P^{\prime}_{0}\\\
\mathbf{P}^{\prime}_{1}&\kappa&0&0&\mathbf{P}^{\prime}_{1}\\\
\mathbf{P}^{\prime}_{2}&0&\kappa&0&\mathbf{P}^{\prime}_{2}\\\
\mathbf{P}^{\prime}_{3}&0&0&\kappa&\mathbf{P}^{\prime}_{3}\\\
\tilde{P}^{\prime}_{0}&-\kappa\mathbf{P}^{\prime}_{1}/P^{\prime}_{+}&-\kappa\mathbf{P}^{\prime}_{2}/P^{\prime}_{+}&-\kappa\mathbf{P}^{\prime}_{3}/P^{\prime}_{+}&P^{\prime}_{4}\end{pmatrix}$
(145)

where $\tilde{P}_{4}$ and $\tilde{P}_{0}$ are defined in eq. (89). For what
concerns the matrix $L^{\prime}_{g}$, we will use the generic matrix

$\displaystyle L^{\prime}_{g}=\begin{pmatrix}L_{00}&L_{01}&L_{02}&L_{03}&0\\\
L_{10}&L_{11}&L_{12}&L_{13}&0\\\ L_{20}&L_{21}&L_{22}&L_{23}&0\\\
L_{30}&L_{31}&L_{32}&L_{33}&0\\\ 0&0&0&0&1\end{pmatrix}$ (146)

####  1 Fifth column:

$\displaystyle\begin{pmatrix}\gamma P_{0}-\beta\gamma\mathbf{P}_{1}\\\
\gamma\mathbf{P}_{1}-\beta\gamma P_{0}\\\ \mathbf{P}_{2}\\\ \mathbf{P}_{3}\\\
P_{4}\end{pmatrix}\overset{!}{=}\begin{pmatrix}P^{\prime}_{0}\\\
\mathbf{P}^{\prime}_{1}\\\ \mathbf{P}^{\prime}_{2}\\\
\mathbf{P}^{\prime}_{3}\\\ P^{\prime}_{4}\end{pmatrix}$ (147)

which translates into the equalities

$\displaystyle P^{\prime}_{0}$ $\displaystyle=\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}$ (148) $\displaystyle\mathbf{P}^{\prime}_{1}$
$\displaystyle=\gamma\mathbf{P}_{1}-\beta\gamma P_{0}$ (149)
$\displaystyle\mathbf{P}^{\prime}_{2}$ $\displaystyle=\mathbf{P}_{2}$ (150)
$\displaystyle\mathbf{P}^{\prime}_{3}$ $\displaystyle=\mathbf{P}_{3}$ (151)
$\displaystyle P^{\prime}_{4}$ $\displaystyle=P_{4}$ (152)

####  2 Fourth column:

$\displaystyle\begin{pmatrix}\gamma\kappa\mathbf{P}_{3}/P_{+}\\\
-\beta\gamma\kappa\mathbf{P}_{3}/P_{+}\\\ 0\\\ \kappa\\\
-\kappa\mathbf{P}_{3}/P_{+}\end{pmatrix}\overset{!}{=}\begin{pmatrix}L_{03}\tilde{P}^{\prime}_{4}+L_{13}\kappa\mathbf{P}^{\prime}_{1}/P_{+}+L_{23}\kappa\mathbf{P}^{\prime}_{2}/P_{+}+L_{33}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
L_{03}\mathbf{P}^{\prime}_{1}+L_{13}\kappa\\\
L_{03}\mathbf{P}^{\prime}_{2}+L_{23}\kappa\\\
L_{03}\mathbf{P}^{\prime}_{3}+L_{33}\kappa\\\
L_{03}\tilde{P}^{\prime}_{0}-L_{13}\kappa\mathbf{P}^{\prime}_{1}/P_{+}-L_{23}\kappa\mathbf{P}^{\prime}_{2}/P_{+}-L_{33}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
\end{pmatrix}$ (153)

Summing the equations coming from the first and fifth row we get

$\displaystyle
L_{03}(\tilde{P}^{\prime}_{4}+\tilde{P}^{\prime}_{0})=(\gamma-1)\kappa\mathbf{P}_{3}/P_{+}$
(154)

and using eq. (89) and (148) we get

$\displaystyle
L_{03}=(\gamma-1)\kappa\frac{\mathbf{P}_{3}}{P_{+}P_{+}^{\prime}}$ (155)

and therefore

$\displaystyle\boxed{L_{03}=(\gamma-1)\kappa\frac{\mathbf{P}_{3}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}}$ (156)

From the third row we get

$\displaystyle L_{23}=-\frac{\mathbf{P}_{2}}{\kappa}L_{03}$ (157)

and therefore

$\displaystyle\boxed{L_{23}=(1-\gamma)\frac{\mathbf{P}_{2}\mathbf{P}_{3}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}}$ (158)

From the fourth row we get

$\displaystyle L_{33}=-\frac{\mathbf{P}_{3}}{\kappa}L_{03}+1$ (159)

and therefore

$\displaystyle\boxed{L_{33}=(1-\gamma)\frac{\mathbf{P}_{3}^{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}+1}$ (160)

Finally, from the second row we get

$\displaystyle
L_{13}=-\beta\gamma\frac{\mathbf{P}_{3}}{P_{+}}-\frac{\mathbf{P}^{\prime}_{1}}{\kappa}L_{03}$
(161)

and therefore

$\displaystyle\boxed{L_{13}=-\beta\gamma\frac{\mathbf{P}_{3}}{P_{+}}+(1-\gamma)\frac{(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})\mathbf{P}_{3}}{P_{+}(\gamma P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}}$
(162)

####  3 Third column:

$\displaystyle\begin{pmatrix}\gamma\kappa\mathbf{P}_{2}/P_{+}\\\
-\beta\gamma\kappa\mathbf{P}_{2}/P_{+}\\\ \kappa\\\ 0\\\
-\kappa\mathbf{P}_{2}/P_{+}\end{pmatrix}\overset{!}{=}\begin{pmatrix}L_{02}\tilde{P}^{\prime}_{4}+L_{12}\kappa\mathbf{P}^{\prime}_{1}/P_{+}+L_{22}\kappa\mathbf{P}^{\prime}_{2}/P_{+}+L_{32}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
L_{02}\mathbf{P}^{\prime}_{1}+L_{12}\kappa\\\
L_{02}\mathbf{P}^{\prime}_{2}+L_{22}\kappa\\\
L_{02}\mathbf{P}^{\prime}_{3}+L_{32}\kappa\\\
L_{02}\tilde{P}^{\prime}_{0}-L_{12}\kappa\mathbf{P}^{\prime}_{1}/P_{+}-L_{22}\kappa\mathbf{P}^{\prime}_{2}/P_{+}-L_{32}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
\end{pmatrix}$ (163)

Summing the equations coming from the first and fifth row we get

$\displaystyle
L_{02}(\tilde{P}^{\prime}_{4}+\tilde{P}^{\prime}_{0})=(\gamma-1)\kappa\mathbf{P}_{2}/P_{+}$
(164)

and using eq. (89) and (148) we get

$\displaystyle
L_{02}=(\gamma-1)\kappa\frac{\mathbf{P}_{2}}{P_{+}P_{+}^{\prime}}$ (165)

and therefore

$\displaystyle\boxed{L_{02}=(\gamma-1)\kappa\frac{\mathbf{P}_{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}}$ (166)

From the fourth row we get

$\displaystyle L_{32}=-\frac{\mathbf{P}_{3}}{\kappa}L_{02}$ (167)

and therefore

$\displaystyle\boxed{L_{32}=(1-\gamma)\frac{\mathbf{P}_{2}\mathbf{P}_{3}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}}$ (168)

From the third row we get

$\displaystyle L_{22}=-\frac{\mathbf{P}_{2}}{\kappa}L_{02}+1$ (169)

and therefore

$\displaystyle\boxed{L_{22}=(1-\gamma)\frac{\mathbf{P}_{2}^{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}+1}$ (170)

Finally, from the second row we get

$\displaystyle
L_{12}=-\beta\gamma\frac{\mathbf{P}_{2}}{P_{+}}-\frac{\mathbf{P}^{\prime}_{1}}{\kappa}L_{02}$
(171)

and therefore

$\displaystyle\boxed{L_{12}=-\beta\gamma\frac{\mathbf{P}_{2}}{P_{+}}+(1-\gamma)\frac{(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})\mathbf{P}_{2}}{P_{+}(\gamma P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}}$
(172)

####  4 Second column:

$\displaystyle\begin{pmatrix}\gamma\kappa\mathbf{P}_{1}/P_{+}-\beta\gamma\kappa\\\
-\beta\gamma\kappa\mathbf{P}_{1}/P_{+}+\gamma\kappa\\\ 0\\\ 0\\\
-\kappa\mathbf{P}_{1}/P_{+}\end{pmatrix}\overset{!}{=}\begin{pmatrix}L_{01}\tilde{P}^{\prime}_{4}+L_{11}\kappa\mathbf{P}^{\prime}_{1}/P_{+}+L_{21}\kappa\mathbf{P}^{\prime}_{2}/P_{+}+L_{31}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
L_{01}\mathbf{P}^{\prime}_{1}+L_{11}\kappa\\\
L_{01}\mathbf{P}^{\prime}_{2}+L_{21}\kappa\\\
L_{01}\mathbf{P}^{\prime}_{3}+L_{31}\kappa\\\
L_{01}\tilde{P}^{\prime}_{0}-L_{11}\kappa\mathbf{P}^{\prime}_{1}/P_{+}-L_{21}\kappa\mathbf{P}^{\prime}_{2}/P_{+}-L_{31}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
\end{pmatrix}$ (173)

Summing the equations coming from the first and fifth row we get

$\displaystyle
L_{01}(\tilde{P}^{\prime}_{4}+\tilde{P}^{\prime}_{0})=(\gamma-1)\kappa\mathbf{P}_{1}/P_{+}-\beta\gamma\kappa$
(174)

and using eq. (89) and (148) we get

$\displaystyle
L_{01}=(\gamma-1)\kappa\frac{\mathbf{P}_{1}}{P_{+}P_{+}^{\prime}}-\frac{\beta\gamma\kappa}{P^{\prime}_{+}}$
(175)

and therefore

$\displaystyle\boxed{L_{01}=(\gamma-1)\kappa\frac{\mathbf{P}_{1}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}-\frac{\beta\gamma\kappa}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}}$ (176)

From the fourth row we get

$\displaystyle L_{31}=-\frac{\mathbf{P}_{3}}{\kappa}L_{01}$ (177)

and therefore

$\displaystyle\boxed{L_{31}=(1-\gamma)\frac{\mathbf{P}_{1}\mathbf{P}_{3}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}+\frac{\beta\gamma\mathbf{P}_{3}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}}$ (178)

From the third row we get

$\displaystyle L_{21}=-\frac{\mathbf{P}_{2}}{\kappa}L_{01}$ (179)

and therefore

$\displaystyle\boxed{L_{21}=(1-\gamma)\frac{\mathbf{P}_{1}\mathbf{P}_{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}+\frac{\beta\gamma\mathbf{P}_{2}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}}$ (180)

Finally, from the second row we get

$\displaystyle
L_{11}=-\beta\gamma\frac{\mathbf{P}_{1}}{P_{+}}-\frac{\mathbf{P}^{\prime}_{1}}{\kappa}L_{01}+\gamma$
(181)

and therefore

$\displaystyle\boxed{L_{11}=-\beta\gamma\frac{\mathbf{P}_{1}}{P_{+}}+(1-\gamma)\frac{(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})\mathbf{P}_{1}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}+\frac{\beta\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})}{\gamma P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}+\gamma}$ (182)

####  5 First column:

$\displaystyle\begin{pmatrix}\gamma\tilde{P}_{4}-\beta\gamma\mathbf{P}_{1}\\\
\gamma\mathbf{P}_{1}-\beta\gamma\tilde{P}_{4}\\\ \mathbf{P}_{2}\\\
\mathbf{P}_{3}\\\
\tilde{P}_{0}\end{pmatrix}\overset{!}{=}\begin{pmatrix}L_{00}\tilde{P}^{\prime}_{4}+L_{10}\kappa\mathbf{P}^{\prime}_{1}/P_{+}+L_{20}\kappa\mathbf{P}^{\prime}_{2}/P_{+}+L_{30}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
L_{00}\mathbf{P}^{\prime}_{1}+L_{10}\kappa\\\
L_{00}\mathbf{P}^{\prime}_{2}+L_{20}\kappa\\\
L_{00}\mathbf{P}^{\prime}_{3}+L_{30}\kappa\\\
L_{00}\tilde{P}^{\prime}_{0}-L_{10}\kappa\mathbf{P}^{\prime}_{1}/P_{+}-L_{20}\kappa\mathbf{P}^{\prime}_{2}/P_{+}-L_{30}\kappa\mathbf{P}^{\prime}_{3}/P_{+}\\\
\end{pmatrix}$ (183)

Summing the equations coming from the first and fifth row we get

$\displaystyle
L_{00}=\gamma\frac{P_{4}}{P^{\prime}_{+}}+\frac{P_{0}}{P^{\prime}_{+}}+(\gamma-1)\frac{\mathbf{P}^{2}}{P_{+}P^{\prime}_{+}}-\beta\gamma\frac{\mathbf{P}_{1}}{P^{\prime}_{+}}$
(184)

and therefore

$\displaystyle L_{00}$ $\displaystyle=\frac{\gamma P_{4}+P_{0}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}+(\gamma-1)\frac{\mathbf{P}^{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}$
$\displaystyle-\beta\gamma\frac{\mathbf{P}_{1}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}$ (185a)

From the third row we have

$\displaystyle L_{20}=\frac{\mathbf{P}_{2}}{\kappa}(1-L_{00})$ (186)

and therefore

$\displaystyle L_{20}$
$\displaystyle=\frac{\mathbf{P}_{2}}{\kappa}\Big{[}1-\Big{(}\frac{\gamma
P_{4}+P_{0}}{\gamma P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}$
$\displaystyle+(\gamma-1)\frac{\mathbf{P}^{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}-\beta\gamma\frac{\mathbf{P}_{1}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}\Big{)}\Big{]}$ (187a)

From the fourth row we have

$\displaystyle L_{30}=\frac{\mathbf{P}_{3}}{\kappa}(1-L_{00})$ (188)

and therefore

$\displaystyle L_{30}$
$\displaystyle=\frac{\mathbf{P}_{3}}{\kappa}\Big{[}1-\Big{(}\frac{\gamma
P_{4}+P_{0}}{\gamma P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}$
$\displaystyle+(\gamma-1)\frac{\mathbf{P}^{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}-\beta\gamma\frac{\mathbf{P}_{1}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}\Big{)}\Big{]}$ (189a)

Finally, from the second row we have

$\displaystyle L_{10}=\frac{1}{\kappa}\left(\gamma\mathbf{P}_{1}-\beta\gamma
P_{4}-\beta\gamma\frac{\mathbf{P}^{2}}{P_{+}}-(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})L_{00}\right)$ (190)

and therefore

$\displaystyle L_{10}$
$\displaystyle=\frac{1}{\kappa}\Big{[}\gamma\mathbf{P}_{1}-\beta\gamma
P_{4}-\beta\gamma\frac{\mathbf{P}^{2}}{P_{+}}-(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})\Big{(}\frac{\gamma P_{4}+P_{0}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}$
$\displaystyle+(\gamma-1)\frac{\mathbf{P}^{2}}{P_{+}(\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4})}-\beta\gamma\frac{\mathbf{P}_{1}}{\gamma
P_{0}-\beta\gamma\mathbf{P}_{1}+P_{4}}\Big{)}\Big{]}$ (191a)

####  6 Expansion of the final matrix up to $1/\kappa$

In most phenomenological applications, an expansion of the full results up to
first order in $\frac{1}{\kappa}$ is in general useful for numerical
estimations of the effect of $\kappa$-deformation on canonical, non-deformed
quantities. Because of this, we here expand up to first order the matrix
elements computed in the previous section. We make use of the following
relations.

$\displaystyle\frac{a}{b+cP_{4}}\approx\frac{a}{c\kappa}.$ (192)

More in general, we also have

$\displaystyle\frac{a+bP_{4}}{c+dP_{4}}$
$\displaystyle\approx\frac{b}{d}+\frac{b}{d\kappa}\left(\frac{ad}{b}-\frac{c}{d}\right)$
(193)

We also have

$\displaystyle\frac{a\kappa}{P_{+}(b+P_{4})}\approx\frac{a}{\kappa}$ (194)
$\displaystyle\frac{a\kappa}{b+P_{4}}\approx a-\frac{ab}{\kappa}$ (195)

where $a,b,c,d$ do not depend on $\kappa$. We have

$\displaystyle\eqref{L00}\mapsto\boxed{L_{00}\approx\gamma+\frac{(1-\gamma)}{\kappa}\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]}$
(196)
$\displaystyle\eqref{L10}\mapsto\boxed{L_{10}\approx-\beta\gamma+\frac{\gamma}{\kappa}\mathbf{P}_{1}-\frac{\gamma}{\kappa}(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})}$ (197)
$\displaystyle\eqref{L20}\mapsto\boxed{L_{20}\approx\frac{(1-\gamma)}{\kappa}\mathbf{P}_{2}}$
(198)
$\displaystyle\eqref{L30}\mapsto\boxed{L_{30}\approx\frac{(1-\gamma)}{\kappa}\mathbf{P}_{3}}$
(199)
$\displaystyle\eqref{L01}\mapsto\boxed{L_{01}\approx-\beta\gamma+\frac{\gamma}{\kappa}\mathbf{P}_{1}-\frac{\gamma}{\kappa}(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})=L_{10}}$ (200)
$\displaystyle\eqref{L11}\mapsto\boxed{L_{11}\approx\gamma+\beta\gamma\frac{(\gamma-1)}{\kappa}\mathbf{P}_{1}-\frac{\beta^{2}\gamma^{2}}{\kappa}P_{0}}$
(201)
$\displaystyle\eqref{L21}\mapsto\boxed{L_{21}\approx\frac{\beta\gamma\mathbf{P}_{2}}{\kappa}}$
(202)
$\displaystyle\eqref{L31}\mapsto\boxed{L_{31}\approx\frac{\beta\gamma\mathbf{P}_{3}}{\kappa}}$
(203)
$\displaystyle\eqref{L02}\mapsto\boxed{L_{02}\approx(\gamma-1)\frac{\mathbf{P}_{2}}{\kappa}=-L_{20}}$
(204)
$\displaystyle\eqref{L12}\mapsto\boxed{L_{12}\approx-\beta\gamma\frac{\mathbf{P}_{2}}{\kappa}}$
(205) $\displaystyle\eqref{L22}\mapsto\boxed{L_{22}\approx 1}$ (206)
$\displaystyle\eqref{L32}\mapsto\boxed{L_{32}\approx 0}$ (207)
$\displaystyle\eqref{L03}\mapsto\boxed{L_{03}\approx(\gamma-1)\frac{\mathbf{P}_{3}}{\kappa}}$
(208)
$\displaystyle\eqref{L13}\mapsto\boxed{L_{13}\approx-\beta\gamma\frac{\mathbf{P}_{3}}{\kappa}}$
(209) $\displaystyle\eqref{L23}\mapsto\boxed{L_{23}\approx 0}$ (210)
$\displaystyle\eqref{L33}\mapsto\boxed{L_{33}\approx 1}$ (211)

We can now finally write the matrix in extended form. We have

$\displaystyle{\tiny
L^{\prime}_{g}=\begin{pmatrix}\gamma+\frac{(1-\gamma)}{\kappa}\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]&-\beta\gamma+\frac{\gamma}{\kappa}\mathbf{P}_{1}-\frac{\gamma}{\kappa}(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})&(\gamma-1)\kappa\frac{\mathbf{P}_{2}}{\kappa}&(\gamma-1)\frac{\mathbf{P}_{3}}{\kappa}&0\\\
-\beta\gamma+\frac{\gamma}{\kappa}\mathbf{P}_{1}-\frac{\gamma}{\kappa}(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})&\gamma+\beta\gamma\frac{(\gamma-1)}{\kappa}\mathbf{P}_{1}-\frac{\beta^{2}\gamma^{2}}{\kappa}P_{0}&-\beta\gamma\frac{\mathbf{P}_{2}}{\kappa}&-\beta\gamma\frac{\mathbf{P}_{3}}{\kappa}&0\\\
\frac{(1-\gamma)}{\kappa}\mathbf{P}_{2}&\frac{\beta\gamma\mathbf{P}_{2}}{\kappa}&1&0&0\\\
\frac{(1-\gamma)}{\kappa}\mathbf{P}_{3}&\frac{\beta\gamma\mathbf{P}_{3}}{\kappa}&0&1&0\\\
0&0&0&0&1\end{pmatrix}}$ (212)

This can be rewritten more simply as a contribution with
$\kappa\rightarrow\infty$ plus the first order in $\frac{1}{\kappa}$ obtaining

$\displaystyle
L^{\prime}_{g}=L^{\prime}_{g}(\kappa=\infty)+\frac{1}{\kappa}\tilde{L}^{\prime}_{g}$
(213)

where

$\displaystyle
L^{\prime}_{g}(\kappa=\infty)=\begin{pmatrix}\gamma&-\beta\gamma&0&0&0\\\
-\beta\gamma&\gamma&0&0&0\\\ 0&0&1&0&0\\\ 0&0&0&1&0\\\ 0&0&0&0&1\end{pmatrix}$
(214)
$\displaystyle{\tiny\frac{1}{\kappa}\tilde{L}^{\prime}_{g}=\frac{1}{\kappa}\begin{pmatrix}(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]&\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})&(\gamma-1)\mathbf{P}_{2}&(\gamma-1)\mathbf{P}_{3}&0\\\
\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})&\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}&-\beta\gamma\mathbf{P}_{2}&-\beta\gamma\mathbf{P}_{3}&0\\\
(1-\gamma)\mathbf{P}_{2}&\beta\gamma\mathbf{P}_{2}&0&0&0\\\
(1-\gamma)\mathbf{P}_{3}&\beta\gamma\mathbf{P}_{3}&0&0&0\\\
0&0&0&0&0\end{pmatrix}}$ (215)

####  7 Check of the consistency of the first order expansion of
$L^{\prime}_{g}$

Notice that if $L^{\prime}_{g}$ is a legitimate Lorentz transformation, then
by definition one must have

$\displaystyle L^{\prime}_{g}\eta{L^{\prime}}^{T}_{g}=\eta$ (216)

where $\eta=\text{diag}(+---\text{const})$. We know that the zeroth order in
perturbation satisfies the relation

$\displaystyle
L^{\prime}_{g}(\kappa=\infty)\eta{L^{\prime}}^{T}_{g}(\kappa=\infty)=\eta$
(217)

by definition, which means that we must have (ignoring terms of order
$1/\kappa^{2}$)

$\displaystyle
L^{\prime}_{g}(\kappa=\infty)\eta(\tilde{L}^{\prime})^{T}_{g}+\tilde{L}^{\prime}_{g}\eta[L^{\prime}_{g}(\kappa=\infty)]^{T}=0$
(218)

We have

$\displaystyle
L^{\prime}_{g}(\kappa=\infty)\eta(\tilde{L}^{\prime})^{T}_{g}=\begin{pmatrix}A&B&(1-\gamma)\mathbf{P}_{2}&(1-\gamma)\mathbf{P}_{3}&0\\\
C&D&\beta\gamma\mathbf{P}_{2}&\beta\gamma\mathbf{P}_{3}&0\\\
(1-\gamma)\mathbf{P}_{2}&\beta\gamma\mathbf{P}_{2}&0&0&0\\\
(1-\gamma)\mathbf{P}_{3}&\beta\gamma\mathbf{P}_{3}&0&0&0\\\
0&0&0&0&0\end{pmatrix}$ (219)

where

$\displaystyle A$
$\displaystyle=\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]+\beta\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]$ (220) $\displaystyle C$
$\displaystyle=-\beta\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]-\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]$ (221) $\displaystyle B$
$\displaystyle=\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]+\beta\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]$
(222) $\displaystyle D$
$\displaystyle=-\beta\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]-\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]$
(223)

Furthermore we have

$\displaystyle\tilde{L}^{\prime}_{g}\eta(L^{\prime}_{g}(\kappa=\infty))^{T}=\begin{pmatrix}\tilde{A}&\tilde{B}&(\gamma-1)\mathbf{P}_{2}&(\gamma-1)\mathbf{P}_{3}&0\\\
\tilde{C}&\tilde{D}&-\beta\gamma\mathbf{P}_{2}&-\beta\gamma\mathbf{P}_{3}&0\\\
(\gamma-1)\mathbf{P}_{2}&-\beta\gamma\mathbf{P}_{2}&0&0&0\\\
(\gamma-1)\mathbf{P}_{3}&-\beta\gamma\mathbf{P}_{3}&0&0&0\\\
0&0&0&0&0\end{pmatrix}$ (224)

where

$\displaystyle\tilde{A}$
$\displaystyle=\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]+\beta\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]$ (225) $\displaystyle\tilde{B}$
$\displaystyle=-\beta\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]-\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]$ (226) $\displaystyle\tilde{C}$
$\displaystyle=\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]+\beta\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]$
(227) $\displaystyle\tilde{D}$
$\displaystyle=-\beta\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]-\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]$
(228)

The sum of the two matrices is therefore

$\displaystyle\begin{pmatrix}A+\tilde{A}&B+\tilde{B}&0&0&0\\\
C+\tilde{C}&D+\tilde{D}&0&0&0\\\ 0&0&0&0&0\\\ 0&0&0&0&0\\\
0&0&0&0&0\end{pmatrix}$ (229)

so that we only need to compute the remaining quantities. We do it one by one.
First notice that $A=\tilde{A}$, so we need to show that $A=0$. Indeed, we
have

$\displaystyle A$
$\displaystyle=\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]+\beta\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]$ (230)
$\displaystyle=\mathbf{P}_{1}[-\beta\gamma^{2}(1-\gamma)+\beta\gamma^{2}-\beta\gamma^{3}]$
(231) $\displaystyle+P_{0}[\gamma(1-\gamma^{2})+\beta^{2}\gamma^{3}]$ (232)
$\displaystyle=0$ (233)

where we used the fact that
$1-\gamma^{2}=(1-\beta^{2}-1)/(1-\beta^{2})=-\beta^{2}\gamma^{2}$. Secondly,
notice that also $D=\tilde{D}$, and therefore we must have $D=0$. Indeed we
have

$\displaystyle D$
$\displaystyle=-\beta\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]-\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]$
(234)
$\displaystyle=\mathbf{P}_{1}[-\beta\gamma^{2}+\beta\gamma^{3}-\beta\gamma^{3}+\beta\gamma^{2}]$
(235) $\displaystyle+P_{0}[-\beta^{2}\gamma^{3}+\beta^{2}\gamma^{3}]$ (236)
$\displaystyle=0$ (237)

Then we need to check $B+\tilde{B}$ and $C+\tilde{C}$. Starting with the first
one we have

$\displaystyle B+\tilde{B}$
$\displaystyle=\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]+\beta\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]$
(238)
$\displaystyle+\left\\{-\beta\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]-\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]\right\\}$ (239)
$\displaystyle=\beta\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]-\beta\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]$
(240)
$\displaystyle=\mathbf{P}_{1}[\beta^{2}\gamma^{3}-\beta^{2}\gamma^{2}+\beta^{2}\gamma^{2}-\beta^{2}\gamma^{3}]+P_{0}[-\beta^{3}\gamma^{3}-\beta\gamma(1-\gamma^{2})]$
(241) $\displaystyle=0$ (242)

where we used the fact that
$1-\gamma^{2}=(1-\beta^{2}-1)/(1-\beta^{2})=-\beta^{2}\gamma^{2}$. Finally, we
have

$\displaystyle C+\tilde{C}$
$\displaystyle=-\beta\gamma(1-\gamma)\left[(1+\gamma)P_{0}-\beta\gamma\mathbf{P}_{1}\right]-\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]$
$\displaystyle+\gamma[\gamma\mathbf{P}_{1}-\gamma(\gamma\mathbf{P}_{1}-\beta\gamma
P_{0})]+\beta\gamma[\beta\gamma(\gamma-1)\mathbf{P}_{1}-\beta^{2}\gamma^{2}P_{0}]$
(243) $\displaystyle=B+\tilde{B}=0$ (244)

which concludes the check. Therefore, the matrix

$\displaystyle
L^{\prime}_{g}=L^{\prime}_{g}(\kappa=\infty)+\frac{1}{\kappa}\tilde{L}^{\prime}_{g}$
(245)

satisfies

$\displaystyle L^{\prime}_{g}\eta(L^{\prime}_{g})^{T}=\eta$ (246)

and it is therefore a valid Lorentz transformation.

###  6 Two-particle kinematics

We now have the finite boost for a two-particle state, and we can therefore
turn to understanding the phenomenological consequences of our computations.

As we already discussed in the context of the boost co-product in section 4,
multiplication of the matrices $Lg,L^{\prime}_{g}h$ from the right by an
arbitrary Lorentz matrix does not change the entries of their last columns so
that $|(Lg(L^{\prime}_{g})^{-1})(P)\rangle=|(Lg)(P)\rangle$ and in writing the
states we can neglect the Lorentz group element on the right of $g$. Therefore
we define the action of the finite Lorentz transformation $\mathscr{L}$ on a
two particle state as [77]

$\displaystyle\Delta\mathscr{L}\triangleright|g(P)\rangle\otimes|h(Q)\rangle=|(Lg)(P)\rangle|(L^{\prime}_{g}h)(Q)\rangle$
(247)

where $(Lg)(P)$ denotes (in the case of the boost along the first axis) the
four top components of the last column of the product of the matrices $L$
(which is a canonical boost) and $g$ (eq. (144)), in components explicitly

$\displaystyle Lg(P)=(\gamma P_{0}-\beta\gamma P_{1},-\beta\gamma P_{0}+\gamma
P_{1},P_{2},P_{3})$ (248)

The components of $L^{\prime}_{g}h$ can be similarly computed using eq. (213),
(214), and (215), as we will discuss in more details shortly.

We now consider the decay of a particle of mass $M$, originally at rest, into
two particles of mass $m$. Starting from the first order expansion in
$1/\kappa$ of the four-momentum deformed composition rule $(P\oplus Q)_{0}$
and $(P\oplus Q)_{i}$ we get

$\displaystyle(P\oplus Q)_{0}\approx
P_{0}+Q_{0}+\frac{\mathbf{P}\mathbf{Q}}{\kappa}\qquad(P\oplus Q)_{i}\approx
P_{i}+Q_{i}+\frac{P_{i}Q_{0}}{\kappa}$ (249)

and we need to impose

$\displaystyle P_{0}+Q_{0}+\frac{\mathbf{P}\mathbf{Q}}{\kappa}=M\qquad
P_{i}+Q_{i}+\frac{P_{i}Q_{0}}{\kappa}=0.$ (250)

We need to find the spatial momenta such that these two relations are
satisfied. The conservation of spatial momenta tells us that the momenta
$\mathbf{P}$ and $\mathbf{Q}$ are parallel, which means that we can align them
with one of our axis, and we can write $P_{\mu}=(P_{0},P,0,0)$ and
$Q_{\mu}=(Q_{0},Q,0,0)$ without loss of generality. We have therefore

$\displaystyle P_{0}+Q_{0}+\frac{PQ}{\kappa}=M\qquad
P+Q+\frac{PQ_{0}}{\kappa}=0.$ (251)

From the conservation of momenta, one can get the relation

$\displaystyle P\approx-Q+\frac{QQ_{0}}{\kappa}$ (252)

which substituted back into the sum of energies gives (at first order in
$1/\kappa$)

$\displaystyle
Q\approx\sqrt{M^{2}-4m^{2}}\left(\frac{1}{2}+\frac{3M}{8\kappa}\right)$ (253)

Substituting back into (252), we also get

$\displaystyle
P\approx-\sqrt{M^{2}-4m^{2}}\left(\frac{1}{2}+\frac{M}{8\kappa}\right).$ (254)

Notice that the modulus $P$ differs from that of $Q$ by
$\sqrt{M^{2}-4m^{2}}\,M/4\kappa$. Notice also that another solution would have
been to switch the RHS of eq. (253) and (254), which however amounts to the
same solution. Notice that in the formal limit $\kappa\rightarrow\infty$ the
spatial momenta are equal and opposite. The deformation effects is here
considered only up to first order in $1/\kappa$, but can be understood without
needing to approximate. Indeed, since we are imposing $P\oplus Q=0$, it is
clear that one must either have $P=P^{\prime}$ and $Q=S(P^{\prime})$ or vice
versa. What we obtain in eq. (253) and (254) is just a first order expansion
of this all-orders result. We will see in chapter 1 that this is not only a
general kinematical property due to the presence of $\kappa$-deformation, but
it is also comes out naturally when dealing with particles and antiparticles
at the same time.

We can now boost the two particles. Since the boost is in the $x^{1}$
direction but the two particles can go in arbitrary direction in the center of
mass frame, we consider spherical coordinates with the polar axis oriented
along the $x$ axis, so that the coordinates of the momenta $P_{\mu}$ and
$Q_{\mu}$ are given by

$\displaystyle P_{\mu}$
$\displaystyle=(P_{0},P\cos\theta,P\sin\theta\sin\phi,P\sin\theta\cos\phi)=:(P_{0},P_{1},P_{2},P_{3})$
(255) $\displaystyle Q_{\mu}$
$\displaystyle=(Q_{0},Q\cos\theta,Q\sin\theta\sin\phi,Q\sin\theta\cos\phi)=:(Q_{0},Q_{1},Q_{2},Q_{3}).$
(256)

When computing the boost, we have two possibilities, depending on which
particle is the first and which is second in (247), namely $L\triangleright
P_{\mu}$ and $L^{\prime}_{g}\triangleright Q_{\mu}$, or $L\triangleright
Q_{\mu}$ and $L^{\prime}_{g}\triangleright P_{\mu}$. Since the non-deformed
boost acts on momenta as usual, we will only concentrate in the action of
$L^{\prime}_{g}$. Furthermore, since the second case can be obtained from the
first one by simply switching $P\longleftrightarrow Q$, we will only consider
the first one. Using equations (143), (213), (214), (215), (253), (254), we
can write down the first order expansion in powers of $1/\kappa$ of the
momenta boosted using $L^{\prime}_{g}$. We have

$\displaystyle(L^{\prime}_{g}\triangleright Q)_{0}$
$\displaystyle=\gamma(Q_{0}-\beta{Q}_{1})+\frac{1}{\kappa}\Big{[}(\gamma-1)({P}_{2}{Q}_{2}+{P}_{3}{Q}_{3}-\gamma{P}_{1}{Q}_{1}+\beta\gamma
Q_{0}{P}_{1})$
$\displaystyle+P_{0}(Q_{0}-Q_{0}\gamma^{2}+\beta\gamma^{2}{Q}_{1})\Big{]}$
(257) $\displaystyle(L^{\prime}_{g}\triangleright Q)_{1}$
$\displaystyle=\gamma({Q}_{1}-\beta
Q_{0})+\frac{1}{\kappa}\Big{[}-\beta\gamma({P}_{2}{Q}_{2})+{P}_{3}{Q}_{3}-{P}_{1}(Q_{0}-\beta{Q}_{1})(\gamma-1)\gamma$
$\displaystyle+P_{0}({Q}_{1}-{Q}_{1}\gamma^{2}+Q_{0}\beta\gamma^{2})\Big{]}$
(258) $\displaystyle(L^{\prime}_{g}\triangleright Q)_{2}$
$\displaystyle={Q}_{2}+\frac{1}{\kappa}[{P}_{2}(Q_{0}-Q_{0}\gamma+{Q_{1}}\beta\gamma)]$
(259) $\displaystyle(L^{\prime}_{g}\triangleright Q)_{3}$
$\displaystyle={Q}_{3}+\frac{1}{\kappa}[{P}_{3}(Q_{0}-Q_{0}\gamma+{Q_{1}}\beta\gamma)]$
(260)

One can also explicitly write the expressions of $L\triangleright P$ and
$L^{\prime}_{g}\triangleright Q$ obtained by substituting eq. (253), (254) in
the formulae, which are the following.

$\displaystyle(L^{\prime}_{g}\triangleright Q)_{0}$
$\displaystyle=\frac{1}{2}\gamma\Big{(}\beta\cos(\theta)\sqrt{M^{2}-4m^{2}}+\sqrt{5M^{2}-4m^{2}}\Big{)}$
$\displaystyle+\frac{1}{40\kappa}\Big{[}5\beta\gamma\cos(\theta)\sqrt{M^{2}-4m^{2}}\Big{(}3M-2\sqrt{5M^{2}-4m^{2}}\Big{)}$
$\displaystyle+4m^{2}\Big{(}5\gamma^{2}-\frac{12\gamma
M}{\sqrt{5M^{2}-4m^{2}}}-5\Big{)}+3M\Big{(}\gamma\sqrt{5M^{2}-4m^{2}}-15\Big{(}\gamma^{2}-1\Big{)}M\Big{)}$
$\displaystyle-5(\gamma-1)^{2}\cos(2\theta)\Big{(}4m^{2}-M^{2}\Big{)}\Big{]}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(261) $\displaystyle(L^{\prime}_{g}\triangleright Q)_{1}$
$\displaystyle=-\frac{1}{2}\gamma\Big{(}\beta\sqrt{5M^{2}-4m^{2}}+\cos(\theta)\sqrt{M^{2}-4m^{2}}\Big{)}$
$\displaystyle+\frac{1}{8\kappa\sqrt{5M^{2}-4m^{2}}}\Big{[}\beta\gamma\Big{(}(\gamma-2)\cos(2\theta)\Big{(}4m^{2}-M^{2}\Big{)}\sqrt{5M^{2}-4m^{2}}$
$\displaystyle+4m^{2}\Big{(}3M-\gamma\sqrt{5M^{2}-4m^{2}}\Big{)}-3M^{2}\Big{(}M-3\gamma\sqrt{5M^{2}-4m^{2}}\Big{)}\Big{)}$
$\displaystyle+\cos(\theta)\sqrt{M^{2}-4m^{2}}\Big{(}M\Big{(}10(\gamma-1)M-3\gamma\sqrt{5M^{2}-4m^{2}}\Big{)}$
$\displaystyle-8(\gamma-1)m^{2}\Big{)}\Big{]}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(262) $\displaystyle(L^{\prime}_{g}\triangleright Q)_{2}$
$\displaystyle=\frac{1}{2}\sin(\theta)\sqrt{M^{2}-4m^{2}}\cos(\phi)$
$\displaystyle-\frac{1}{8\kappa}\Big{[}\sin(\theta)\sqrt{M^{2}-4m^{2}}\cos(\phi)\Big{(}2\beta\gamma\cos(\theta)\sqrt{M^{2}-4m^{2}}$
$\displaystyle+2(\gamma-1)\sqrt{5M^{2}-4m^{2}}-3M\Big{)}\Big{]}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(263) $\displaystyle(L^{\prime}_{g}\triangleright Q)_{3}$
$\displaystyle=\frac{1}{2}\sin(\theta)\sqrt{M^{2}-4m^{2}}\sin(\phi)$
$\displaystyle-\frac{1}{8\kappa}\Big{[}\sin(\theta)\sqrt{M^{2}-4m^{2}}\sin(\phi)\Big{(}2\beta\gamma\cos(\theta)\sqrt{M^{2}-4m^{2}}$
$\displaystyle+2(\gamma-1)\sqrt{5M^{2}-4m^{2}}-3M\Big{)}\Big{]}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(264) $\displaystyle(L\triangleright P)_{0}$
$\displaystyle=\frac{1}{2}\gamma\left(\beta\cos(\theta)\sqrt{M^{2}-4m^{2}}+\sqrt{5M^{2}-4m^{2}}\right)$
$\displaystyle+\frac{\gamma
M\left(\beta\cos(\theta)\sqrt{M^{2}-4m^{2}}+\frac{M^{2}-4m^{2}}{\sqrt{5M^{2}-4m^{2}}}\right)}{8\kappa}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(265) $\displaystyle(L\triangleright P)_{1}$
$\displaystyle=-\frac{1}{2}\gamma\left(\beta\sqrt{5M^{2}-4m^{2}}+\cos(\theta)\sqrt{M^{2}-4m^{2}}\right)$
$\displaystyle+\frac{\gamma
M\left(\frac{\beta\left(4m^{2}-M^{2}\right)}{\sqrt{5M^{2}-4m^{2}}}-\cos(\theta)\sqrt{M^{2}-4m^{2}}\right)}{8\kappa}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(266) $\displaystyle(L\triangleright P)_{2}$
$\displaystyle=-\frac{1}{2}\sin(\theta)\sqrt{M^{2}-4m^{2}}\cos(\phi)$
$\displaystyle-\frac{M\sin(\theta)\sqrt{M^{2}-4m^{2}}\cos(\phi)}{8\kappa}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(267) $\displaystyle(L\triangleright P)_{3}$
$\displaystyle=-\frac{1}{2}\sin(\theta)\sqrt{M^{2}-4m^{2}}\sin(\phi)$
$\displaystyle-\frac{M\sin(\theta)\sqrt{M^{2}-4m^{2}}\sin(\phi)}{8\kappa}+O\Big{(}\Big{(}\frac{1}{\kappa}\Big{)}^{2}\Big{)}$
(268)

Notice that, in absence of deformation, the components of momenta along
$x^{2}$ and $x^{3}$ (i.e. those components perpendicular to the boost
direction) of $L^{\prime}_{g}\triangleright Q$ and $L\triangleright P$ are
equal and opposite. However, deformation introduces additional terms which
make even these components not equal and opposite after boost, which in turn
gives a deformed angular distribution. To show this, we plot the angular
dependence of the modulus of the deformed boosted momenta to compare it to the
non-deformed case. In figures below all the quantities on the axes are
expressed in GeV.

In the center of mass (COM) frame, the two momenta are distributed on a
sphere. Because of deformation, however, the spatial momenta $P$ and $Q$ are
different in modulus (see eq. (253), (254)). In Figure 1 we show the
distribution of momenta in the COM frame, together with the distribution in
the non-deformed case. To highlight the qualitative difference, for the next
plots we chose hypothetically $M=10\;\mbox{GeV}$, $m=0.1\;\mbox{GeV}$,
$\kappa=10^{2}\;\mbox{GeV}$. Furthermore, we chose the domains
$\phi\in\left[-\frac{\pi}{10},\frac{3\pi}{2}+\frac{\pi}{10}\right]$ and
$\phi\in\left[0,\frac{3\pi}{2}\right]$ for the distributions of the deformed
$P$ and the deformed $Q$ respectively, in order to clearly show how they are
related to the non-deformed one. Notice that the non-deformed case contains
only a single surface because both spatial momenta have the same modulus.

Figure 1: Angular distribution of deformed and non-deformed momenta in the COM
frame, with parameters $M=10\;\mbox{GeV}$, $m=0.1\;\mbox{GeV}$,
$\kappa=10^{2}\;\mbox{GeV}$. Notice the onion-like structure, with the non-
deformed momentum being innermost, followed by the deformed $P$, and finally
the deformed $q$. Here, $P$ and $Q$ are antipodal also in the
$\kappa$-deformed context (see eq. (250)). To give a more concrete numerical
estimate, the modulus of the non-deformed momentum is $5.00$ GeV, the modulus
of $P$ is $5.12$ GeV, while the modulus of $Q$ is $5.37$ GeV.

We then boost our momenta. In particular, $P_{\mu}$ will be canonically
boosted while $Q_{\mu}$ will be boosted in a deformed way (recall however that
both $P_{\mu}$ and $Q_{\mu}$ still have some effects of deformation in their
modulus). In this case we chose $\gamma=5$ while the other parameters remain
the same. We also restricted the domain in $\phi$ of all distributions to
better show the features of the surfaces, and we obtain Figure 2. Notice that
the deformed boost not only modifies the amplitude of the boosted momentum,
but also its angular distribution. Finally, notice that switching
$P\leftrightarrow Q$ in the above formula makes manifest that the
$\kappa$-deformed boost contribution is much greater than the center of mass
deformation of the moduli.

Figure 2: Angular distribution of deformed and non-deformed momenta in the
boosted frame, with parameters $\gamma=5$, $M=10\;\mbox{GeV}$,
$m=0.1\;\mbox{GeV}$, $\kappa=10^{2}\;\mbox{GeV}$. Recall that both $P$ and $Q$
feel the effects of deformation already in the modulus (see eq. (254) and
(253)). When boosting, $P$ gets boosted with a canonical non-deformed boost,
while $Q$ is instead boosted in a deformed way.

From a phenomenological point of view, eq. (6) and (260) show that the
difference between the two boosted momenta mostly depend on the
$\frac{1}{\kappa}$ contribution coming from the deformed boost (large values
of boost parameter $\gamma$ highlight these contributions). This in turn is
important for the study of two-particle correlations and interference
patterns, since these are often expressed as a function of the difference of
momenta (or lifetime) of the involved particles. Of particular relevance, the
study of favoured mesons from the decays $\Phi^{0}(1020)\rightarrow
K^{0}\bar{K}^{0}$, $\Psi(3770)\rightarrow D^{0}\bar{D}^{0}$,
$\Upsilon(10580)\rightarrow B^{0}\bar{B}^{0}$ and $\Upsilon(10860)\rightarrow
B^{0}_{s}\bar{B}^{0}_{s}$.

##  Chapter 1 Complex scalar field on $\kappa$-Minkowski spacetime

###  1 Introduction

Given $\kappa$-Minkowski spacetime, one can begin building physical quantities
on top of it. In particular, one can introduce fields over $\kappa$-Minkowski
spacetime. The whole process is however not trivial, and because of the many
computations involved, it will be easy to lose track of the progress or the
direction of our investigations. As such, in the remainder of this section we
will provide the reader with a roadmap of this chapter. This chapter will be
divided in the following main sections:

  * $i)$

In order to introduce a field in $\kappa$-Minkowski spacetime, one could
directly work in the bicrossproduct basis using the coordinate representation
given in eq. (71). It is however much simpler to work with canonical,
commuting coordinates, with the caveat that one needs to switch to the
classical basis. One can do this through the so called Weyl map (the canonical
product of functions is also deformed to a non-commutative $\star$ product
under Weyl map). Therefore, the first step will be to introduce the Weyl map
and its properties, after which the rest of the discussion will be more
transparent. This will be done in section 2;

  * $ii)$

We are now in the position to introduce the action of a complex scalar field,
and we will do so immediately after having introduced the Weyl map. Because of
the properties of the $\star$ product, our total action $S$ will be defined in
terms of two possible partial actions $S_{1}$ and $S_{2}$. This will be done
in section 3;

  * $iii)$

The first thing to understand about the action is how to compute physically
meaningful quantities starting from it, like for example the equations of
motion or the Noether charges. However, because of the non-trivial co-product,
the derivatives do not satisfy the Leibniz rule. We will therefore first
introduce the action of derivatives on $\star$ products from in section 4.
After this we can immediately introduce the integration-by-parts relations,
from section 5 to 9;

  * $iv)$

We now have both an action, and the integration-by-parts relation which means
that we can vary the action introduced in section 3 to obtain both the
equations of motion (EoM) and the surface terms. We will do this in section
10;

  * $v)$

We can now build the on-shell field since we know the EoM. Once we have it, we
can also discuss its properties under the discrete transformations $C$, $P$,
and $T$. We will do this in section 11;

  * $vi)$

Going a bit off-topic, it is useful to introduce at this point the off-shell
momentum-space action, in section 12. This is not needed for the rest of this
chapter, but will be needed in chapter 2 when we will obtain the propagator;

  * $vii)$

Since we now have the EoM and the on-shell field, the other important
quantities to obtain from the action defined in section 3 are the charges. We
will proceed along the following steps:

    * $a)$

To compute the charges, we introduce the symplectic form in section 13. The
role of this object is twofold. On one hand it allows us to compute the
creation/annihilation operators algebra. On the other hand, we can use it to
compute the charges using the geometric approach, which we describe in section
14. This will lead us to introduce deformed rules for the contraction of
vector fields with forms. Using them, we will obtain both the translation
(section 14), boost (section 2) and the rotation charges (section 2);

    * $b)$

We will conclude the computation of the charges with some mathematical
considerations on the properties of the Poisson brackets of
creation/annihilation operators, and the role of the antipode in the
interaction between forms and vector fields (section 3 and 4).

  * $viii)$

We now finally have all the charges, as well as the creation/annihilation
operators algebra. The natural next step is to check what is the algebra
satisfied by the charges. We check it in section 15;

  * $ix)$

We now have all the charges coming from continuous symmetries, and as we
discussed in point $v)$ we also have discrete symmetries. We are therefore
ready to discuss the interaction between them. The first surprising result is
the non-commutativity of the $C$ operator with the boost charge
$\mathscr{N}_{i}$. We show the detailed computations of this fact in section
16;

  * $x)$

Another important and non-trivial result is related to the Greenberg’s
theorem, which (in canonical axiomatic QFT) shows how invariance $CPT$
transformation is equivalent to Lorentz invariance. Because our result of
point $ix)$, we already know that in our model $CPT$ invariance is subtly
violated. We investigate the details of Greenberg’s theorem and its
application to the deformed context in section 17;

  * $xi)$

We will conclude with some phenomenological considerations concerning possible
signatures of the fact that $[C,\mathscr{N}_{i}]\neq 0$ in section 18.

The Weyl map introduced in point $i)$ of the above list is a quantity which
(at least in canonical quantum mechanics) has been known for decades. Here we
just review some of the material in the literature which we need to proceed
forward, but it does not constitute original work. Similarly, the deformation
of the Lorenz rule for derivatives and the integration-by-parts obtained in
point $iii)$ have been already discussed in [86] in the context of the
bicrossproduct basis, and only their translation to the classical basis
discussed in this work is original. All other points are original results.

###  2 Weyl maps

We begin with point $i)$ in the roadmap. In our context, a Weyl map is a
function which allows for every object built on top of the group manifold
$AN(3)$ to be translated into commutative spacetime with $\star$ product. This
includes functions of $x$, integrals, derivatives. Since this is a standard
construction, we only pinpoint the most important features and definitions,
referring the reader to [86], [80], [85], [78] for a more complete discussion.
Conceptually speaking, a Weyl map is not strictly needed for any of the
computations in the present and following chapters. Indeed, since $AN(3)$ is a
Lie group, there are well defined concepts of differentiation and integration,
which suffice for our purpose. Nevertheless, being able to switch to a more
conventional spacetime, where coordinates are not matrices any more, greatly
simplifies the computations, and allows for a more clear interpretation of the
results. For clarity, since we will need to talk about the parity
transformation $P$, from here on we will indicate the classical basis with $p$
and not $P$.

By definition, we have

$\displaystyle\mathscr{W}(\hat{e}_{k}(\hat{x}))=e_{p(k)}(x),\quad\mathscr{W}^{-1}(e_{p(k)}(x))=\hat{e}_{k}(\hat{x}),$
(1)

where

$\displaystyle e_{p(k)(x)}=e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})},$
(2)

and where $k$ are momenta in the bicrossproduct basis, and $p$ are the momenta
in the classical basis as a function of the $k$ momenta, see eq. (85), (98)
(recall also the definition of $\hat{e}^{*}$ in eq. (97)). Of course, we need
to take care also of the sum of momenta and antipodes, and we have the
definitions (we will only write down the transformation in one direction,
since the inverse is then obvious, and we also avoid writing the dependence on
both $x$ and $\hat{x}$, since they are also obvious)

$\displaystyle\mathscr{W}(({\hat{e}}_{k})^{-1})=e_{S(p(k))},\qquad\mathscr{W}(\hat{e}_{k}\hat{e}_{l})=e_{p(k)\oplus
q(l)}=:e_{p}\star e_{q}$ (3)

where the second equation defines the $\star$ product. Notice that, since the
product in $AN(3)$ is not commutative, then neither is the $\star$ product.
Furthermore, notice that since by definition we have

$\displaystyle e_{p}^{\dagger}\star e_{p}=e_{p}\star e_{p}^{\dagger}=1$ (4)

then we have

$\displaystyle e_{p}^{\dagger}=e_{S(p)}.$ (5)

Furthermore, calling $\hat{\partial}$ the derivatives in the context of the
$AN(3)$ group defined by

$\displaystyle\hat{\partial}_{\mu}\,\hat{e}_{k}$
$\displaystyle=ip_{\mu}(k)\,\hat{e}_{k}\,,\quad\hat{\partial}_{4}\,\hat{e}_{k}=i(\kappa-
p_{4}(k))\,\hat{e}_{k}$ $\displaystyle\hat{\partial}_{\mu}\,\hat{e}^{*}_{k}$
$\displaystyle=ip^{*}_{\mu}(k)\,\hat{e}^{*}_{k}\,,\quad\hat{\partial}_{4}\,\hat{e}^{*}_{k}=i(\kappa-p^{*}_{4}(k))\,\hat{e}^{*}_{k}$
(6)

we define the action of the Weyl map action by

$\displaystyle\mathscr{W}(\hat{\partial}_{\mu}\hat{e}_{k})=\partial_{\mu}e_{p}\qquad\mathscr{W}(\hat{\partial}_{4}\hat{e}_{k})=\partial_{4}e_{p},\quad\partial_{4}e_{p}=i(\kappa-
p_{4})e_{p}.$ (7)

Finally, we have the definition

$\displaystyle\widehat{\int}\hat{e}_{k}=\int_{\mathbb{R}^{4}}d^{4}x\,\mathscr{W}(\hat{e}_{k}(\hat{x}))$
(8)

for plane waves. Notice that the integral on the LHS is not an integral over
$AN(3)$. This in turn allows us to define fields using the Fourier transform
[86], [80], [85], [78], [81], [82]

$\displaystyle\phi(x)=\int_{AN(3)}d\mu(p)\tilde{\phi}(p)e_{p}(x)\qquad
d\mu(p)=\frac{d^{4}p}{p_{4}/\kappa}$ (9)

restricted by the constraints in eq. (55). Of course, one can define the same
thing at the group level and not in canonical spacetime by

$\displaystyle\hat{\phi}(\hat{x})=\int_{AN(3)}d\mu(p)\tilde{\phi}(p)\hat{e}_{k}(\hat{x})$
(10)

with the definitions

$\displaystyle\mathscr{W}(\hat{\phi}(\hat{x}))=\phi(x)\qquad\mathscr{W}(\hat{\phi}(\hat{x})\hat{\psi}(\hat{x}))=\phi(x)\star\psi(x)$
(11)

and therefore

$\displaystyle\widehat{\int}\hat{\phi}(\hat{x})\hat{\psi}(\hat{x})=\int_{\mathbb{R}^{4}}d^{4}x\,\phi(x)\star\psi(x).$
(12)

In other words, we can safely translate quantities defined purely in terms of
(derivatives and integrals of) group elements in $AN(3)$ into more familiar
terms, with canonical Riemannian integrals defined in $\mathbb{R}^{4}$ and
derivatives acting on canonical plane waves, paying attention to the fact that
products of different objects need to be defined as $\star$ product. Of
course, such a set of definitions is not unique, for example here we chose to
relate canonical plane waves with group elements defined according to the
time-to-the-right convention (recall eq. (72)). Instead, we could have chosen
a Weyl map with the definition

$\displaystyle\mathscr{W}^{-1}(e_{p})=e^{ip_{\mu}\hat{x}^{\mu}}$ (13)

which, as we saw in eq. (73), corresponds to a different choice of basis in
momentum space. Different Weyl maps can offer different advantages and
disadvantages, depending on the problem at hand, for a more in depth
discussion see [86], [80], [85], [78].

In what follows, we will use the Weyl map defined as above, and we will assume
that it has already been applied.

###  3 Action for the complex scalar field

We can now go to point $ii)$ in the roadmap in section 1. Because of what has
been said in section 2, and recalling that we need to get back the non-
deformed action in the formal $\kappa\rightarrow\infty$ limit, we have two
possible choices for the action, namely

$\displaystyle S_{1}$
$\displaystyle=\int_{\mathbb{R}^{4}}d^{4}x\,\,(\partial^{\mu}\phi)^{\dagger}\star(\partial_{\mu}\phi)-m^{2}\phi^{\dagger}\star\phi$
(14) $\displaystyle S_{2}$ $\displaystyle=\int_{\mathbb{R}^{4}}d^{4}x\,\
(\partial_{\mu}\phi)\star(\partial^{\mu}\phi)^{\dagger}-m^{2}\phi\star\phi^{\dagger}.$
(15)

Because of this, and for reasons that will become apparent later, we choose
the following action.

$\displaystyle\boxed{S=\frac{1}{2}\int_{\mathbb{R}^{4}}d^{4}x\,\,[(\partial^{\mu}\phi)^{\dagger}\star(\partial_{\mu}\phi)+(\partial^{\mu}\phi)\star(\partial_{\mu}\phi)^{\dagger}-m^{2}(\phi^{\dagger}\star\phi+\phi\star\phi^{\dagger})]}.$
(16)

We now need to extract the equations of motion (EoM) and the conserved Noether
charges associated with continuous symmetries out of this action. Notice that
the quantities $\partial^{\dagger}$ are non-trivial objects which will be
defined (and whose properties will be investigated) in the next section.

###  4 Action of derivatives on $\star$ products

We now have a star product in a commutative spacetime, and we want to know how
derivatives act on the star product of two functions. We are at point $iii)$
of our roadmap in section 1. We know that the action cannot be Leibnizian,
since otherwise one would get the following contradiction.

$\displaystyle i(p\oplus q)_{\mu}e_{p\oplus q}$
$\displaystyle=\partial_{\mu}(e_{p}\star e_{q})=(\partial_{\mu}e_{p})\star
e_{q}+e_{p}\star\partial_{\mu}e_{q}=i(p+q)_{\mu}e_{p\oplus q}$ (17)

Indeed, from the first equality in eq. (17) we already see that the action
must be defined in terms of the co-product. Since we are using plane waves
whose momenta are in the classical basis, we will need to model the action of
derivatives on products over eq. (58), (59), (60). We define therefore the
following $\kappa$-deformed Leibniz rules, in analogy with those defined
already111In [86] the co-product rules in the bicrossproduct basis are used to
define the action of derivatives on products, instead of the classical basis
that we use here. Furthermore, the model considered in [86] is completely
different from the present one, since they consider only one of the two
orderings of the action (the equivalent of our eq. (14)) and they have a
completely different field decomposition. The action of derivatives on
products, however, is only a manifestation of the co-product, and it is
therefore model-independent. in [86].

$\displaystyle\boxed{\partial_{0}(\phi\star\psi)=\frac{1}{\kappa}(\partial_{0}\phi)\star(\Delta_{+}\psi)+\kappa(\Delta_{+}^{-1}\phi)\star(\partial_{0}\psi)+i(\Delta_{+}^{-1}\partial_{i}\phi)\star(\partial_{i}\psi)}$
(18)
$\displaystyle\qquad\qquad\qquad\boxed{\partial_{i}(\phi\star\psi)=\frac{1}{\kappa}(\partial_{i}\phi)\star(\Delta_{+}\psi)+\phi\star(\partial_{i}\psi)}$
(19)
$\displaystyle\qquad\qquad\qquad\qquad\boxed{\Delta_{+}(\phi\star\psi)=\frac{1}{\kappa}(\Delta_{+}\phi)\star(\Delta_{+}\psi)}.$
(20)

Furthermore, we introduce the objects $\partial^{\dagger}$ which are related
to antipodes. More in detail, are defined in such a way that

$\displaystyle\partial e_{p}\propto
pe_{p}\qquad\Leftrightarrow\qquad\partial^{\dagger}e_{p}\propto S(p)e_{p}.$
(21)

As such, we expect them to be related to eq. (63). We define

$\displaystyle\boxed{\partial_{i}^{\dagger}=\kappa\Delta_{+}^{-1}\partial_{i}},\qquad\boxed{\partial_{0}^{\dagger}=\partial_{0}-i\Delta_{+}^{-1}\partial^{2}},\qquad\boxed{\partial_{4}^{\dagger}=-\partial_{4}},\qquad\boxed{\Delta_{+}^{\dagger}=\kappa^{2}\Delta_{+}^{-1}}$
(22) $\displaystyle\boxed{\Delta_{+}=i\partial_{0}+\kappa+i\partial_{4}}.$
(23)

Notice that for the definition of $\Delta_{+}$ we used the conventions
highlighted in eq. (6). If we define the linear combinations

$\Xi(x)=a_{\mathbf{p}}\,e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}+b^{\dagger}_{\mathbf{p^{*}}}\,e^{i(S(\omega^{*}_{p})t-S(\mathbf{p}^{*})\mathbf{x})}\equiv\Xi_{(+)}(x)+\Xi_{(-)}(x)$
(24)
$\Xi^{\dagger}(x)=a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}+b_{\mathbf{p^{*}}}\,e^{i(\omega^{*}_{p}t-\mathbf{p}^{*}\mathbf{x})}\equiv\Xi^{\dagger}_{(+)}(x)+\Xi^{\dagger}_{(-)}(x)$
(25)

then we have

$\displaystyle(\partial_{A}\Xi)^{\dagger}=\partial_{A}^{\dagger}\Xi^{\dagger}.$
(26)

The reason for this particular choice of linear combination $\Xi$ will become
clear after we introduce the complex fields in subsequent sections (at this
point, $a,a^{\dagger},b,b^{\dagger}$ are just some complex coefficients). We
check the validity of eq. (26) for $A=0,i,4$.

####  1 $A=0$

In this case

$\displaystyle\partial_{0}\Xi=-i\omega_{p}a_{\mathbf{p}}\,e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}+iS(\omega_{p}^{*})b^{\dagger}_{\mathbf{p^{*}}}\,e^{i(S(\omega^{*}_{p})t-S(\mathbf{p}^{*})\mathbf{x})}$
(27)

which means that

$\displaystyle(\partial_{0}\Xi)^{\dagger}=i\omega_{p}a_{\mathbf{p}}^{\dagger}\,e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}-iS(\omega_{p}^{*})b_{\mathbf{p^{*}}}\,e^{i(\omega^{*}_{p}t-\mathbf{p}^{*}\mathbf{x})}.$
(28)

On the other hand

$\displaystyle\partial_{0}^{\dagger}\Xi^{\dagger}$
$\displaystyle=(\partial_{0}-i\Delta_{+}^{-1}\partial^{2})\Xi^{\dagger}$ (29)
$\displaystyle=\underbrace{\left(-iS(\omega_{p})+\frac{i}{S(\omega_{p})+p_{4}}S(\mathbf{p})^{2}\right)}_{=iS(S(\omega_{p}))}a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}$
$\displaystyle+\underbrace{\left(i\omega_{p}^{*}+\frac{i}{-\omega_{p}^{*}-p_{4}^{*}}(\mathbf{p}^{*})^{2}\right)}_{=-iS(\omega_{p}^{*})}b_{\mathbf{p^{*}}}\,e^{i(\omega^{*}_{p}t-\mathbf{p}^{*}\mathbf{x})}$
(30)

where we also used $S(p_{4})=p_{4}$ which is true both off-shell and on-shell,
so that we only need to verify that $S(S(\omega_{p}))=\omega_{p}$. We have

$\displaystyle S(S(\omega_{p}))$
$\displaystyle=-S(\omega_{p})+\frac{1}{S(\omega_{p})+p_{4}}S(\mathbf{p})^{2}$
(31)
$\displaystyle=\omega_{p}-\frac{\mathbf{p}^{2}}{\omega_{p}+p_{4}}+\frac{\omega_{p}+p_{4}}{\kappa^{2}}\frac{\kappa^{2}\mathbf{p}^{2}}{(\omega_{p}+p_{4})^{2}}$
(32) $\displaystyle=\omega_{p}$ (33)

where we used

$\displaystyle
S(p_{0})+p_{4}=-p_{0}+\frac{\mathbf{p}^{2}}{p_{0}+p_{4}}+p_{4}=\frac{\kappa^{2}}{p_{0}+p_{4}}.$
(34)

Notice the important detail that the definition
$\hat{\partial}_{4}\,\hat{e}^{*}_{k}=i(\kappa-p^{*}_{4}(k))\,\hat{e}^{*}_{k}$
in equation (6) translated through the Weyl map states that

$\displaystyle\partial_{4}\,e^{-i(\omega_{p}^{*}t-\mathbf{p}^{*}\mathbf{x})}=i(\kappa-p^{*}_{4}(k))\,e^{-i(\omega_{p}^{*}t-\mathbf{p}^{*}\mathbf{x})}$
(35)

while in eq. (29), when we use $\Delta_{+}^{-1}$ in the second term in the
RHS, we are applying $\partial_{4}$ to
$e^{i(\omega_{p}^{*}t-\mathbf{p}^{*}\mathbf{x})}=e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}$,
so that in the second term in the RHS of eq. (29) we get the quantity

$\displaystyle\frac{1}{\omega_{p}+p_{4}}=\frac{1}{-\omega_{p}^{*}-p_{4}^{*}}.$
(36)

####  2 $A=i$

In this case

$\displaystyle\partial_{i}\Xi=i\mathbf{p}_{i}a_{\mathbf{p}}\,e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}-iS(\mathbf{p}^{*})_{i}b^{\dagger}_{\mathbf{p^{*}}}\,e^{i(S(\omega^{*}_{p})t-S(\mathbf{p}^{*})\mathbf{x})}$
(37)

and therefore

$\displaystyle(\partial_{i}\Xi)^{\dagger}=-i\mathbf{p}_{i}a_{\mathbf{p}}^{\dagger}\,e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}+iS(\mathbf{p}^{*})_{i}b_{\mathbf{p^{*}}}\,e^{i(\omega^{*}_{p}t-\mathbf{p}^{*}\mathbf{x})}.$
(38)

On the other hand we have

$\displaystyle\partial_{i}^{\dagger}\Xi^{\dagger}$
$\displaystyle=(\kappa\Delta_{+}^{-1}\partial_{i})\Xi^{\dagger}$ (39)
$\displaystyle=\underbrace{\left(\kappa\frac{i}{S(\omega_{p})+p_{4}}S(\mathbf{p})_{i}\right)}_{=-iS(S(\mathbf{p}))_{i}}a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}$
$\displaystyle+\underbrace{\left(-\kappa\frac{i}{-\omega_{p}^{*}-p_{4}^{*}}\mathbf{p}^{*}_{i}\right)}_{=iS(\mathbf{p})_{i}}b_{\mathbf{p^{*}}}\,e^{i(\omega^{*}_{p}t-\mathbf{p}^{*}\mathbf{x})}$
(40)

and also

$\displaystyle\kappa\frac{i}{S(\omega_{p})+p_{4}}S(\mathbf{p})_{i}=-i\kappa\frac{\omega_{p}+p_{4}}{\kappa^{2}}\frac{\kappa\mathbf{p}_{i}}{\omega_{p}+p_{4}}=-i\mathbf{p}_{i}$
(41)

which concludes the verification.

####  3 $A=4$

We have

$\displaystyle\partial_{4}\Xi=i(\kappa-
p_{4})a_{\mathbf{p}}\,e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}+i\underbrace{(\kappa-S(p_{4}))}_{=\kappa-
p_{4}}b^{\dagger}_{\mathbf{p^{*}}}\,e^{i(S(\omega^{*}_{p})t-S(\mathbf{p}^{*})\mathbf{x})}$
(42)

where in the first term in the RHS we just used eq. (6) and in the second term
we used the fact that

$\displaystyle
e^{i(S(\omega^{*}_{p})t-S(\mathbf{p}^{*})\mathbf{x})}=e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}$
(43)

and then we used again the definition (6). Therefore we have

$\displaystyle(\partial_{4}\phi)^{\dagger}=-i(\kappa-
p_{4})a_{\mathbf{p}}^{\dagger}\,e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}-i(\kappa-
p_{4})b_{\mathbf{p^{*}}}\,e^{i(\omega^{*}_{p}t-\mathbf{p}^{*}\mathbf{x})}$
(44)

On the other hand we have

$\displaystyle\partial_{4}^{\dagger}\phi^{\dagger}$
$\displaystyle=-\partial_{4}\phi^{\dagger}$ (45)
$\displaystyle=-i(\kappa-S(p_{4}))a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}-i(\kappa-
p_{4})b_{\mathbf{p^{*}}}\,e^{i(\omega^{*}_{p}t-\mathbf{p}^{*}\mathbf{x})}$
(46)

which concludes the check.

####  4 $A=+$

From subsection (1) and (3) together with the definition (23) we have that
$(\Delta_{+}\phi)^{\dagger}=\Delta_{+}^{\dagger}\phi^{\dagger}$. In fact, we
know that $(\partial_{0}\phi)^{\dagger}=\partial_{0}^{\dagger}\phi^{\dagger}$
then we also know that
$(\alpha\partial_{0}\phi)^{\dagger}=\alpha^{\dagger}(\partial_{0}\phi)^{\dagger}=\alpha^{\dagger}\partial_{0}^{\dagger}\phi^{\dagger}$
for any constant $\alpha$, and the same goes for $\partial_{4}$.

####  5 $\Delta_{+}^{\dagger}=\kappa^{2}\Delta_{+}^{-1}$

We have

$\displaystyle\Delta_{+}^{\dagger}$
$\displaystyle=-i\partial_{0}^{\dagger}+\kappa-i\partial_{4}^{\dagger}$ (47)
$\displaystyle=-i\left(\partial_{0}-i\Delta_{+}^{-1}\partial^{2}\right)+\kappa+i\partial_{4}$
(48)
$\displaystyle=-i\partial_{0}-\frac{\partial^{2}}{i\partial_{0}+\kappa+i\partial_{4}}+\kappa+i\partial_{4}$
(49)

Furthermore, recall that by definition we have

$\displaystyle(\kappa+i\partial_{4})^{2}=\kappa^{2}-\partial_{0}^{2}+\partial^{2}$
(51)

and therefore we end up with

$\displaystyle\Delta_{+}^{\dagger}=\frac{\partial_{0}^{2}-\partial^{2}+\kappa^{2}-\partial_{0}^{2}+\partial^{2}}{\Delta_{+}}=\frac{\kappa^{2}}{\Delta_{+}}$
(52)

which proves the formula.

###  5 Integration-by-parts relation for $\partial_{i}$ going $\leftarrow$

We now need explicitly what are the integration by parts relations that one
can get using eq. (18), (19), (20), (22), (23). This is a preparatory step,
and we will need it after the introduction of the action of the complex scalar
field in order to compute the continuous charges.

For the spatial integration by parts we have

$\displaystyle\boxed{(\partial_{i}\phi)^{\dagger}\star(\partial_{i}\psi)=\partial_{i}\left[(\partial_{i}\phi)^{\dagger}\star\psi\right]-\frac{\Delta_{+}}{\kappa}\left[(\partial^{2}\phi)^{\dagger}\star\psi\right]}.$
(53)

which can be obtained using the following steps

$\displaystyle(\partial_{i}\phi)^{\dagger}\star(\partial_{i}\psi)$
$\displaystyle\overset{\eqref{4.14bis}}{=}\partial_{i}\left[(\partial_{i}\phi)^{\dagger}\star\psi\right]-\frac{1}{\kappa}[\partial_{i}(\partial_{i}\phi)^{\dagger}]\star(\Delta_{+}\psi)$
(54)
$\displaystyle\overset{\eqref{4.15bis}}{=}\partial_{i}\left[(\partial_{i}\phi)^{\dagger}\star\psi\right]-\Delta_{+}\left[(\Delta_{+}^{-1}\partial_{i}(\partial_{i}\phi)^{\dagger})\star\psi\right]$
(55)
$\displaystyle=\partial_{i}\left[(\partial_{i}\phi)^{\dagger}\star\psi\right]-\frac{\Delta_{+}}{\kappa}\left[(\kappa\Delta_{+}^{-1}\partial_{i}(\partial_{i}\phi)^{\dagger})\star\psi\right]$
(56)
$\displaystyle\overset{\eqref{4.16}}{=}\partial_{i}\left[(\partial_{i}\phi)^{\dagger}\star\psi\right]-\frac{\Delta_{+}}{\kappa}\left[(\partial_{i}^{\dagger}(\partial_{i}\phi)^{\dagger})\star\psi\right]$
(57)
$\displaystyle=\partial_{i}\left[(\partial_{i}\phi)^{\dagger}\star\psi\right]-\frac{\Delta_{+}}{\kappa}\left[(\partial^{2}\phi)^{\dagger}\star\psi\right].$
(58)

###  6 Integration-by-parts relation for $\partial_{0}$ going $\leftarrow$

We do the same thing for the temporal part. We have

$\displaystyle\boxed{(\partial_{0}\phi)^{\dagger}\star(\partial_{0}\psi)=\frac{\partial_{0}}{\kappa}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-i\partial_{i}[(\Delta_{+}^{-1}\partial_{i}{\partial_{0}\phi})^{\dagger}\star\psi]-\frac{\Delta_{+}}{\kappa}\left[(\partial_{0}^{2}\phi)^{\dagger}\star\psi\right]}$
(59)

which can be proven using the following steps.

$\displaystyle\kappa\,(\partial_{0}\phi)^{\dagger}\star(\partial_{0}\psi)$
(60)
$\displaystyle\overset{\eqref{4.13bis}}{=}\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-\frac{1}{\kappa}(\Delta_{+}\partial_{0}(\partial_{0}\phi)^{\dagger})\star(\Delta_{+}\psi)-i(\partial_{i}(\partial_{0}\phi)^{\dagger})\star(\partial_{i}\psi)$
(61)
$\displaystyle\overset{\eqref{4.15bis}}{=}\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-\Delta_{+}[(\partial_{0}(\partial_{0}\phi)^{\dagger})\star\psi]-i(\partial_{i}(\partial_{0}\phi)^{\dagger})\star(\partial_{i}\psi)$
(62)
$\displaystyle\overset{\eqref{4.14bis}}{=}\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-\Delta_{+}[(\partial_{0}(\partial_{0}\phi)^{\dagger})\star\psi]-i\partial_{i}[(\partial_{i}({\partial_{0}\phi})^{\dagger})\star\psi]$
$\displaystyle+\frac{i}{\kappa}(\partial^{2}({\partial_{0}\phi})^{\dagger})\star(\Delta_{+}\psi)$
(63)
$\displaystyle\overset{\eqref{4.15bis}}{=}\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-\Delta_{+}[(\partial_{0}(\partial_{0}\phi)^{\dagger})\star\psi]-i\partial_{i}[(\partial_{i}({\partial_{0}\phi})^{\dagger})\star\psi]$
$\displaystyle+i\Delta_{+}[(\Delta_{+}^{-1}\partial^{2}({\partial_{0}\phi})^{\dagger})\star\psi]$
(64)
$\displaystyle=\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-i\partial_{i}[(\partial_{i}({\partial_{0}\phi})^{\dagger})\star\psi]$
$\displaystyle-\Delta_{+}\left\\{[(\partial_{0}-i\Delta_{+}^{-1}\partial^{2})(\partial_{0}\phi)^{\dagger}]\star\psi\right\\}$
(65)
$\displaystyle\overset{\eqref{4.16}}{=}\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-i\partial_{i}[\Delta_{+}\kappa^{-1}(\partial_{i}{\partial_{0}\phi})^{\dagger}\star\psi]$
$\displaystyle-\Delta_{+}\left\\{[(\partial_{0}-i\Delta_{+}^{-1}\partial^{2})(\partial_{0}\phi)^{\dagger}]\star\psi\right\\}$
(66)
$\displaystyle\overset{\eqref{4.16}}{=}\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-i\kappa\partial_{i}[(\Delta_{+}^{-1}\partial_{i}{\partial_{0}\phi})^{\dagger}\star\psi]$
$\displaystyle-\Delta_{+}\left\\{[(\partial_{0}-i\Delta_{+}^{-1}\partial^{2})(\partial_{0}\phi)^{\dagger}]\star\psi\right\\}$
(67)
$\displaystyle\overset{\eqref{4.16}}{=}\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-i\kappa\partial_{i}[(\Delta_{+}^{-1}\partial_{i}{\partial_{0}\phi})^{\dagger}\star\psi]-\Delta_{+}\left\\{[\partial_{0}^{\dagger}(\partial_{0}\phi)^{\dagger}]\star\psi\right\\}$
(68)
$\displaystyle=\partial_{0}\left[(\Delta_{+}(\partial_{0}\phi)^{\dagger})\star\psi\right]-i\kappa\partial_{i}[(\Delta_{+}^{-1}\partial_{i}{\partial_{0}\phi})^{\dagger}\star\psi]-\Delta_{+}\left[(\partial_{0}^{2}\phi)^{\dagger}\star\psi\right].$
(69)

###  7 Integration-by-parts relation for $\partial_{i}$ going $\rightarrow$

Working in the same way, we obtain

$\displaystyle\boxed{(\partial_{i}\psi)\star(\partial_{i}\phi)^{\dagger}=\kappa\partial_{i}(\psi\star[\Delta_{+}^{-1}(\partial_{i}\phi)^{\dagger}])-\psi\star(\partial^{2}\phi)^{\dagger}}.$
(70)

Indeed, we have

$\displaystyle\frac{1}{\kappa}(\partial_{i}\psi)\star(\partial_{i}\phi)^{\dagger}\overset{\eqref{4.14bis}}{=}\partial_{i}(\psi\star[\Delta_{+}^{-1}(\partial_{i}\phi)^{\dagger}])-\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{i}\phi)^{\dagger}]$
(71)

and therefore

$\displaystyle(\partial_{i}\psi)\star(\partial_{i}\phi)^{\dagger}$
$\displaystyle\overset{\eqref{4.14bis}}{=}\kappa\partial_{i}(\psi\star[\Delta_{+}^{-1}(\partial_{i}\phi)^{\dagger}])-\psi\star[\kappa\Delta_{+}^{-1}\partial_{i}(\partial_{i}\phi)^{\dagger}]$
(72)
$\displaystyle\overset{\eqref{4.16}}{=}\kappa\partial_{i}(\psi\star[\Delta_{+}^{-1}(\partial_{i}\phi)^{\dagger}])-\psi\star(\partial^{2}\phi)^{\dagger}.$
(73)

###  8 Integration-by-parts relation for $\partial_{0}$ going $\rightarrow$

One can show that

$\displaystyle(\partial_{0}\psi)\star(\partial_{0}\phi)^{\dagger}$
$\displaystyle=\partial_{0}(\psi\star[\kappa\Delta_{+}^{-1}(\partial_{0}\phi)^{\dagger}])-i\partial_{i}(\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger}])$
$\displaystyle-[\psi\star(\partial_{0}^{2}\phi)^{\dagger}]+\left(\frac{i}{\kappa}\partial_{0}+i\frac{\partial_{4}}{\kappa}\right)[\psi\star(\partial_{0}^{2}\phi)^{\dagger}]$
(74a)

Indeed we have

$\displaystyle\frac{1}{\kappa}(\partial_{0}\psi)\star(\partial_{0}\phi)^{\dagger}$
$\displaystyle\overset{\eqref{4.13bis}}{=}\partial_{0}(\psi\star[\Delta_{+}^{-1}(\partial_{0}\phi)^{\dagger}])-\kappa(\Delta_{+}^{-1}\psi)\star(\Delta_{+}^{-1}\partial_{0}(\partial_{0}\phi)^{\dagger})$
$\displaystyle-i(\Delta_{+}^{-1}\partial_{i}\psi)\star(\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger})$
(75)
$\displaystyle\overset{\eqref{4.15bis}}{=}\partial_{0}(\psi\star[\Delta_{+}^{-1}(\partial_{0}\phi)^{\dagger}])-\Delta_{+}^{-1}[\psi\star(\partial_{0}(\partial_{0}\phi)^{\dagger})]$
$\displaystyle-i\kappa^{-1}\Delta_{+}^{-1}[(\partial_{i}\psi)\star(\partial_{i}(\partial_{0}\phi)^{\dagger})]$
(76)
$\displaystyle\overset{\eqref{4.14bis}}{=}\partial_{0}(\psi\star[\Delta_{+}^{-1}(\partial_{0}\phi)^{\dagger}])-\Delta_{+}^{-1}[\psi\star(\partial_{0}(\partial_{0}\phi)^{\dagger})]$
$\displaystyle-i\Delta_{+}^{-1}\Big{[}\partial_{i}(\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger}])-\psi\star[\Delta_{+}^{-1}\partial^{2}(\partial_{0}\phi)^{\dagger}]\Big{]}$
(77)
$\displaystyle=\partial_{0}(\psi\star[\Delta_{+}^{-1}(\partial_{0}\phi)^{\dagger}])-i\Delta_{+}^{-1}\partial_{i}(\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger}])$
$\displaystyle-\Delta_{+}^{-1}[\psi\star(\partial_{0}-i\Delta_{+}^{-1}\partial^{2})(\partial_{0}\phi)^{\dagger}]$
(78)
$\displaystyle\overset{\eqref{4.16}}{=}\partial_{0}(\psi\star[\Delta_{+}^{-1}(\partial_{0}\phi)^{\dagger}])-i\Delta_{+}^{-1}\partial_{i}(\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger}])$
$\displaystyle-\Delta_{+}^{-1}[\psi\star(\partial_{0}^{2}\phi)^{\dagger}]$
(79)

We now expand $\Delta_{+}^{-1}$ (recall that
$\Delta_{+}=i\partial_{0}+\kappa+i\partial_{4}$). By expanding we mean writing
the differential operator $\kappa\Delta_{+}^{-1}$ in terms of a sum of
derivatives in the numerator. In other words, after applying
$\kappa\Delta_{+}^{-1}$ to a star product, we get some function of momenta at
the denominator, and we expand such a function around a point in momentum
space which is convenient for us. At this point, we convert this function back
to action of derivatives on the original star product. The net effective
result is that we have expanded out original differential operator as if it
was already a function of momenta. We have

$\displaystyle\kappa\Delta_{+}^{-1}=\frac{1}{i\frac{\partial_{0}}{\kappa}+1+i\frac{\partial_{4}}{\kappa}}$
(80)

Notice that we cannot expand around $(\partial_{0},\partial_{i})=(0,0)$ (here
we abuse notation a bit by addressing derivatives as if they are already
momenta, but one should always keep in mind our previous discussion) because
in this case $m=0$ on-shell, and we are assuming $m\neq 0$ instead.
Furthermore, notice that if we expand around $\partial_{0}=0$, then the
expansion up to first order would be

$\displaystyle\kappa\Delta_{+}^{-1}=\kappa\Delta_{+}^{-1}(\partial_{0}=0)+\frac{\delta\kappa\Delta_{+}^{-1}}{\delta\partial_{0}}\Bigg{|}_{\partial_{0}=0}\partial_{0}+O(\partial_{0}^{2})$
(81)

where we used eq. (6), and in particular the zeroth order would become

$\displaystyle\kappa\Delta_{+}^{-1}(\partial_{0}=0)=\frac{1}{1-\sqrt{1+\frac{m^{2}}{\kappa^{2}}}}\neq
1$ (82)

which is however problematic, because one would not recover the correct
equation of motion (since the temporal component of the kinetic term will have
a different rescaling than the spatial part). We need to find an expansion
point $p$ such that $\kappa\Delta_{+}^{-1}(p)=1$. Therefore, we expand around
$\partial_{0}=-\partial_{4}=\partial_{4}^{\dagger}$. We have

$\displaystyle\kappa\Delta_{+}^{-1}$
$\displaystyle=\kappa\Delta_{+}^{-1}(\partial_{0}=-\partial_{4})+\frac{\delta\kappa\Delta_{+}^{-1}}{\delta\partial_{0}}\Bigg{|}_{\partial_{0}=-\partial_{4}}(\partial_{0}-(-\partial_{4}))+O((\partial_{0}+\partial_{4})^{2})$
(83)
$\displaystyle=1-\frac{\frac{i}{\kappa}}{\left(i\frac{\partial_{0}}{\kappa}+1+i\frac{\partial_{4}}{\kappa}\right)^{2}}\Bigg{|}_{\partial_{0}=-\partial_{4}}(\partial_{0}+\partial_{4})=1-\frac{i}{\kappa}\partial_{0}-i\frac{\partial_{4}}{\kappa}$
(84)

Coming back to our integration-by-parts relation, without loss of generality
we can write

$\displaystyle(\partial_{0}\psi)\star(\partial_{0}\phi)^{\dagger}$
$\displaystyle=\partial_{0}(\psi\star[\kappa\Delta_{+}^{-1}(\partial_{0}\phi)^{\dagger}])-i\partial_{i}(\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger}])-[\psi\star(\partial_{0}^{2}\phi)^{\dagger}]$
$\displaystyle+i\partial_{i}\left(\frac{i}{\kappa}\partial_{0}+i\frac{\partial_{4}}{\kappa}\right)(\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger}])$
$\displaystyle+\left(\frac{i}{\kappa}\partial_{0}+i\frac{\partial_{4}}{\kappa}\right)[\psi\star(\partial_{0}^{2}\phi)^{\dagger}]$
(85)

However, the term
$i\partial_{i}\left(\frac{i}{\kappa}\partial_{0}+i\frac{\partial_{4}}{\kappa}\right)(\psi\star[\Delta_{+}^{-1}\partial_{i}(\partial_{0}\phi)^{\dagger}])$
can be ignored because it contain a double total derivative (we are
considering $\partial_{4}$ as an independent derivative with respect to the
others, because indeed $p_{4}$ can have a value independent from the value of
$p_{0}$ or $p_{i}$). This proves eq. (74).

Notice that, since we do not have any prefactor in front of the kinetic part
of the on-shell operator, we also don’t need to modify the mass term to obtain
the correct on-shell relation, like it was needed in the previous ordering.
Furthermore, notice also that we only expanded the $\Delta_{+}^{-1}$ which
acted globally on each star product, while there would be no meaning in
expanding the factors $\Delta_{+}^{-1}$ inside the star product. Furthermore,
this expansion does not mean that our results are approximate. Indeed, higher
order term in the expansion of $\kappa\Delta_{+}^{-1}$ would contribute
factors proportional to $\partial^{2}$ or higher, and all these contributions
vanish in any case because of the integral.

###  9 Mass terms integration by parts

Recalling that $\Delta_{+}=i\partial_{0}+\kappa+i\partial_{4}$ for the first
ordering we have

$\displaystyle\boxed{m^{2}\phi^{\dagger}\star\psi=-\frac{i\partial_{0}}{\kappa}(m^{2}\phi^{\dagger}\star\psi)-\frac{i\partial_{4}}{\kappa}(m^{2}\phi^{\dagger}\star\psi)+\frac{\Delta_{+}}{\kappa}(m^{2}\phi^{\dagger}\star\psi)}$
(86)

which can be proven trivially from

$\displaystyle m^{2}\phi^{\dagger}\star\psi$
$\displaystyle=-\left(\frac{\Delta_{+}}{\kappa}-1\right)(m^{2}\phi^{\dagger}\star\psi)+\frac{\Delta_{+}}{\kappa}(m^{2}\phi^{\dagger}\star\psi).$
(87)

For the second ordering we don’t have a contribution coming from this
reasoning.

###  10 EoM and surface terms $\Pi^{0}$, $\Pi^{i}$, and $\Pi^{4}$ for both
orderings of the action

We now have all the necessary tools, i.e. eq. (53), (59), (70), (74), (86), to
integrate by parts both orderings of the action, i.e. eq. (14) and (15), which
we rewrite here for simplicity

$\displaystyle\eqref{S1}\rightarrow\qquad S_{1}$
$\displaystyle=\int_{\mathbb{R}^{4}}d^{4}x\,\,(\partial^{\mu}\phi)^{\dagger}\star(\partial_{\mu}\phi)-m^{2}\phi^{\dagger}\star\phi$
(88) $\displaystyle\eqref{S2}\rightarrow\qquad S_{2}$
$\displaystyle=\int_{\mathbb{R}^{4}}d^{4}x\,\
(\partial_{\mu}\phi)\star(\partial^{\mu}\phi)^{\dagger}-m^{2}\phi\star\phi^{\dagger}.$
(89)

We are at point $iv)$ of our roadmap in section 1. We can now compute the
variation of the action in eq. (16)

$\displaystyle\delta
S_{1}=\frac{1}{2}\int_{\mathbb{R}^{4}}d^{4}x\,(\partial_{\mu}\delta\phi)^{\dagger}\star\partial^{\mu}\phi+(\partial_{\mu}\phi)^{\dagger}\star\partial^{\mu}\delta\phi-m^{2}\delta\phi^{\dagger}\star\phi-m^{2}\phi^{\dagger}\star\delta\phi$
(90)

which can be immediately rewritten using integration by parts (53), (59), (86)