as222One can integrate by parts one of the two terms, and then obtain the
other by taking the Hermitian conjugate. Of course, the same relation is
obtained if one starts with the co-product rules for the derivatives
$\partial^{\dagger}$, and then proceeds from there.

$\displaystyle\delta
S_{1}=\frac{1}{2}\int_{\mathbb{R}^{4}}d^{4}x\,\Bigg{\\{}-$
$\displaystyle\frac{\Delta_{+}}{\kappa}\left[(\partial_{\mu}^{\dagger}(\partial^{\mu})^{\dagger}-m^{2})\phi^{\dagger}\star\delta\phi\right]+\partial_{A}\left(\Pi_{1}^{A}\star\delta\phi\right)$
$\displaystyle-$
$\displaystyle\frac{\kappa}{\Delta_{+}}\left[\delta\phi^{\dagger}\star(\partial_{\mu}\partial^{\mu}-m^{2})\phi\right]+\partial_{A}^{\dagger}\left(\delta\phi^{\dagger}\star\left(\Pi_{1}^{A}\right)^{\dagger}\right)\Bigg{\\}}$
(91)

where

$\displaystyle\Pi^{0}_{1}=(\Pi_{0})_{1}$
$\displaystyle=\frac{1}{\kappa}(\Delta_{+}\partial_{0}^{\dagger}+im^{2})\phi^{\dagger}$
(92) $\displaystyle\Pi^{i}_{1}=-(\Pi_{i})_{1}$
$\displaystyle=(-\partial_{i}(1+i\Delta_{+}^{-1}\partial_{0}))\phi^{\dagger}$
(93) $\displaystyle\Pi^{4}_{1}=(\Pi_{4})_{1}$
$\displaystyle=-i\frac{m^{2}\phi^{\dagger}}{\kappa}.$ (94)

For the other ordering we proceed in a similar manner, with the only
difference that we use the integration-by-parts relations (70), (74). We have

$\displaystyle\delta
S_{2}=\frac{1}{2}\int_{\mathbb{R}^{4}}d^{4}x\,\partial^{\mu}\phi\star(\partial_{\mu}\delta\phi)^{\dagger}+\partial^{\mu}\delta\phi\star(\partial_{\mu}\phi)^{\dagger}-m^{2}\phi\star\delta\phi^{\dagger}-m^{2}\delta\phi\star\phi^{\dagger}$
(95)

which can be rewritten as

$\displaystyle\delta
S_{2}=\frac{1}{2}\int_{\mathbb{R}^{4}}d^{4}x\,\Bigg{\\{}-$
$\displaystyle\left[\delta\phi\star(\partial_{\mu}^{\dagger}(\partial^{\mu})^{\dagger}-m^{2})\phi^{\dagger}\right]+\partial_{A}\left(\delta\phi\star\Pi^{A}\right)$
$\displaystyle-$
$\displaystyle\left[(\partial_{\mu}\partial^{\mu}-m^{2})\phi\star\delta\phi^{\dagger}\right]+\partial_{A}^{\dagger}\left(\left(\Pi^{A}\right)^{\dagger}\star\delta\phi^{\dagger}\right)\Bigg{\\}}$
(96)

where

$\displaystyle\Pi^{0}_{2}=(\Pi_{0})_{2}$
$\displaystyle=\left(\frac{\kappa}{\Delta_{+}}\partial_{0}^{\dagger}+\frac{i}{\kappa}(\partial_{0}^{\dagger})^{2}\right)\phi^{\dagger}$
(97) $\displaystyle\Pi^{i}_{2}=-(\Pi_{i})_{2}$
$\displaystyle=-\frac{\kappa}{\Delta_{+}}(\partial_{i}^{\dagger}+i\partial_{i}\partial_{0}^{\dagger})\phi^{\dagger}$
(98) $\displaystyle\Pi^{4}_{2}=(\Pi_{4})_{2}$
$\displaystyle=+i\frac{(\partial_{0}^{\dagger})^{2}}{\kappa}\phi^{\dagger}.$
(99)

We see that the equations of motion are the canonical Klein-gordon equation

$\displaystyle(\partial_{\mu}\partial^{\mu}-m^{2})\phi=0\qquad(\partial_{\mu}^{\dagger}(\partial^{\mu})^{\dagger}-m^{2})\phi^{\dagger}=0.$
(100)

Notice that in momentum space,
$\partial_{\mu}^{\dagger}(\partial^{\mu})^{\dagger}\leftrightarrow
S(p)_{\mu}S(p)^{\mu}=p_{\mu}p^{\mu}\leftrightarrow\partial_{\mu}\partial^{\mu}$),
where one can check explicitly that

$\displaystyle S(p)_{\mu}S(p)^{\mu}$
$\displaystyle=S(\omega_{p})^{2}-S(\mathbf{p})^{2}$ (101)
$\displaystyle=\left(-\omega_{p}+\frac{\mathbf{p}^{2}}{\omega_{p}+p_{4}}\right)^{2}-\left(-\frac{\kappa\mathbf{p}}{\omega_{p}+p_{4}}\right)^{2}$
(102)
$\displaystyle=\omega_{p}^{2}-2\frac{\omega_{p}\mathbf{p}^{2}}{\omega_{p}+p_{4}}+\frac{\mathbf{p}^{4}}{(\omega_{p}+p_{4})^{2}}-\frac{\kappa^{2}\mathbf{p}^{2}}{(\omega_{p}+p_{4})^{2}}$
(103)
$\displaystyle=\omega_{p}^{2}+\mathbf{p}^{2}\frac{-(\omega_{p}+p_{4})^{2}}{(\omega_{p}+p_{4})^{2}}$
(104) $\displaystyle=p_{\mu}p^{\mu}$ (105)

and therefore

$\displaystyle S(p)_{\mu}S(p)^{\mu}=p_{\mu}p^{\mu}$ (106)

The fields satisfying such equations are usually written as linear
combinations of plane waves and their Hermitian conjugate. Here, because of
the curved nature of momentum space, it is not so automatic. Nevertheless, we
will see in the next section that fields satisfying these equations, in the
context of $\kappa$-deformation, can be described in a particularly clear way.

###  11 Complex scalar field and its properties

We are now in point $v)$ of our roadmap. As it is clear from eq. (55), the
$AN(3)$ group manifold is a 4-dimensional submanifold of a 5-dimensional
space. Furthermore, we have plane waves, i.e. group elements $g\in AN(3)$. To
define a field on it which satisfies the EoM in eq. (100) we can use both
these facts and define [78]

$\displaystyle\phi(x)=\int
d^{5}p\,\,2\kappa\delta(p_{A}p^{A}+\kappa^{2})\theta(p_{+})\theta(p_{4})\delta(p_{\mu}p^{\mu}-m^{2})\tilde{\phi}(p)e^{-ipx}$
(107)

where $\phi(x)$ is a function in commutative spacetime. Notice that the
quantity

$\displaystyle d^{5}p\,\,2\kappa\delta(p_{A}p^{A}+\kappa^{2})$ (108)

is the metric in $\mathbb{R}^{5}$ restricted to the group manifold with the
constraint in eq. (55) (the $\kappa$ is only present for dimensional reasons).
In the same way, the Heaviside $\theta$ functions restrict ourselves to the
correct part of the manifold, namely where $P_{+}>0$ and $P_{4}>0$. Finally,
the remaining Dirac delta imposes the on-shell condition. In order to write
this field in a more familiar way, one can first split the constraints as
follows.

$\displaystyle\delta(p_{\mu}p^{\mu}-m^{2})=\delta(p_{\mu}p^{\mu}-m^{2})\theta(p_{0}-m)+\delta(p_{\mu}p^{\mu}-m^{2})\theta(-p_{0}-m)$
(109)

We use the notation

$\displaystyle\phi_{+}(x)=\int
d^{5}p\,\,2\kappa\delta(p_{A}p^{A}+\kappa^{2})\theta(p_{+})\theta(p_{4})\delta(p_{\mu}p^{\mu}-m^{2})\theta(p_{0}-m)\tilde{\phi}(p)e^{-ipx}$
(110) $\displaystyle\phi_{-}(x)=\int
d^{5}p\,\,2\kappa\delta(p_{A}p^{A}+\kappa^{2})\theta(p_{+})\theta(p_{4})\delta(p_{\mu}p^{\mu}-m^{2})\theta(-p_{0}-m)\tilde{\phi}(p)e^{-ipx}$
(111)

for simplicity. We can then change variable in $\phi_{-}$ by sending $p\mapsto
S(p)$. Because of eq. (106) and (63), we know that the arguments of the Dirac
delta are invariant. One can also show starting from eq. (63) that

$\displaystyle
S(p_{+})=-p_{0}+\frac{\mathbf{p}^{2}}{p_{+}}+S(p)_{4}=\frac{\kappa^{2}}{p_{+}}$
(112)

so that also the Heaiside $\theta(p_{+})$ and $\theta(p_{4})$ are left
invariant. One can then change variable $p\mapsto-p=p^{*}$ and pick up some
additional factors coming from the determinant of the Jacobian obtained in the
coordinate change $p\mapsto S(p)$. Furthermore, one can apply the Dirac
deltas, reducing the integration from $d^{5}p$ to $d^{3}p$, with other factors
added to the integrand (more details can be found in [78]). The result is that
the field can be written as

$\displaystyle\phi(x)$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\,a_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+\int\frac{d^{3}p^{*}}{\sqrt{2|\omega_{\mathbf{p}^{*}}|}}\,\,\zeta(p)b^{\dagger}_{\mathbf{p}^{*}}\,e^{i(S(\omega_{\mathbf{p}^{*}})t-S(\mathbf{p}^{*})\mathbf{x})}$
(113) $\displaystyle\phi^{\dagger}(x)$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\,a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}+\int\frac{d^{3}p^{*}}{\sqrt{2|\omega_{\mathbf{p}^{*}}|}}\,\,\zeta(p)b_{\mathbf{p}^{*}}\,e^{i(\omega_{\mathbf{p}^{*}}t-\mathbf{p}^{*})\mathbf{x}}$
(114)

Notice that if $p>0$, then $S(p^{*})>0$, so that both $\phi(x)$ and
$\phi^{\dagger}(x)$ are a linear combination of positive and negative energy
states. Notice that there is an additional function $\zeta(p)$ which will be
important when talking about the action. Furthermore, notice also that the ∗
can be eliminated by all the expressions by simply change variable
$p^{*}\mapsto-p$. Therefore, it may be convenient to use instead the fields

$\displaystyle\phi(x)$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\,a_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)b^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}$
(115) $\displaystyle\phi^{\dagger}(x)$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\,a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}+\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)b_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p})\mathbf{x}}$
(116)

for simplicity. The reason for using the variables $p^{*}$ instead of $-p$ is
to make clear that the minus sign in $-p$ needs to also be present in front of
$p_{4}$, and not only $p_{0},\mathbf{p}$. This is important because we need
the relation defined in (101) to hold. Having clarified this point, we can
safely switch from $p^{*}$ to $-p$ without issues.

####  1 $C,P,T$ in the non-deformed context: a short review

We provide a short review of the definition of $C,P,T$ transformations in the
non-deformed case, which will be the basis for our treatment of the deformed
case. The non-deformed fields are the same as eq. (115), (116) but in the
limit $\kappa\rightarrow\infty$, so we have

$\displaystyle\phi_{U}(x)$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,a_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}|}}\,\,b^{\dagger}_{\mathbf{p}}\,e^{i(\omega_{\mathbf{p}}t-\mathbf{p})\mathbf{x})}$
(117) $\displaystyle\phi_{U}^{\dagger}(x)$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,a^{\dagger}_{\mathbf{p}}\,e^{i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,b_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p})\mathbf{x}}.$
(118)

  * $(P)$

By definition, a $P$ transformation is defined by
$(t,\mathbf{x})\mapsto(t,-\mathbf{x})$ and therefore
$\mathscr{P}\phi(t,\mathbf{x})\mathscr{P}^{-1}=\phi(t,-\mathbf{x})$.

$\displaystyle\mathscr{P}\phi_{U}(t,\mathbf{x})\mathscr{P}^{-1}$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}\mathscr{P}a_{\mathbf{p}}\mathscr{P}^{-1}e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}+\mathscr{P}b_{\mathbf{p}}^{\dagger}\mathscr{P}^{-1}e^{i(\omega_{p}t-\mathbf{p}\mathbf{x})}$
(119)
$\displaystyle\overset{!}{=}\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}a_{\mathbf{p}}e^{-i(\omega_{p}t+\mathbf{p}\mathbf{x})}+b_{\mathbf{p}}^{\dagger}e^{i(\omega_{p}t+\mathbf{p}\mathbf{x})}$
(120)

which means that

$\displaystyle\mathscr{P}a_{\mathbf{p}}\mathscr{P}^{-1}=a_{-\mathbf{p}}\qquad\mathscr{P}b_{\mathbf{p}}\mathscr{P}^{-1}=b_{-\mathbf{p}}$
(121)

and the same for $a^{\dagger},b^{\dagger}$.

  * $(T)$

By definition, a $T$ transformation is defined by
$(t,\mathbf{x})\mapsto(-t,\mathbf{x})$ and therefore
$\mathscr{T}\phi(t,\mathbf{x})\mathscr{T}^{-1}=\phi(-t,\mathbf{x})$. Recall
that this is an anti-unitary operator. One can show easily this because
$\mathscr{T}x\mathscr{T}^{-1}=x$,
$\mathscr{T}\mathbf{p}\mathscr{T}^{-1}=-\mathbf{p}$ imply

$\displaystyle[x,p]=i\xrightarrow{\mathscr{T}}-[x,p]=\mathscr{T}i\mathscr{T}^{-1}\overset{!}{=}-i.$
(122)

We have

$\displaystyle\mathscr{T}\phi(t,\mathbf{x})\mathscr{T}^{-1}$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}\mathscr{T}a_{\mathbf{p}}\mathscr{T}^{-1}e^{+i(\omega_{p}t-\mathbf{p}\mathbf{x})}+\mathscr{T}b_{\mathbf{p}}^{\dagger}\mathscr{T}^{-1}e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}$
(123)
$\displaystyle\overset{!}{=}\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}a_{\mathbf{p}}e^{-i(-\omega_{p}t-\mathbf{p}\mathbf{x})}+b_{\mathbf{p}}^{\dagger}e^{i(-\omega_{p}t+\mathbf{p}\mathbf{x})}$
(124)

which means that

$\displaystyle\mathscr{T}a_{\mathbf{p}}\mathscr{T}^{-1}=a_{-\mathbf{p}}\qquad\mathscr{T}b_{\mathbf{p}}\mathscr{T}^{-1}=b_{-\mathbf{p}}$
(125)

and the same for $a^{\dagger},b^{\dagger}$.

  * $(C)$

Contrary to the previous two transformation, this is not related to spacetime
coordinates. Its action is defined by the relation
$\mathscr{C}\phi\mathscr{C}^{-1}=\phi^{\dagger}$. We have

$\displaystyle\mathscr{C}\phi(t,\mathbf{x})\mathscr{C}^{-1}$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}\mathscr{C}a_{\mathbf{p}}\mathscr{C}^{-1}e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}+\mathscr{C}b_{\mathbf{p}}^{\dagger}\mathscr{C}^{-1}e^{+i(\omega_{p}t-\mathbf{p}\mathbf{x})}$
(126)
$\displaystyle\overset{!}{=}\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}a_{\mathbf{p}}^{\dagger}e^{+i(\omega_{p}t-\mathbf{p}\mathbf{x})}+b_{\mathbf{p}}e^{-i(\omega_{p}t+\mathbf{p}\mathbf{x})}$
(127)

which means that

$\displaystyle\mathscr{C}a_{\mathbf{p}}\mathscr{C}^{-1}=b_{\mathbf{p}}\qquad\mathscr{C}b_{\mathbf{p}}\mathscr{C}^{-1}=a_{\mathbf{p}}$
(128)

and the same for $a^{\dagger},b^{\dagger}$.

####  2 $C,P,T$ in the $\kappa$-deformed context

The $P$ and $T$ transformations can consistently be defined as acting like in
the non-deformed case (they leave
$[\hat{x}^{0},\hat{x}^{i}]=\frac{i}{\kappa}\hat{x}^{i}$ invariant). Therefore,
they behave as in the non-deformed case analysed before

$\displaystyle\mathscr{T}\phi(t,\mathbf{x})\mathscr{T}^{-1}$
$\displaystyle=\phi(-t,\mathbf{x})\quad\implies\quad\mathscr{T}a_{\mathbf{p}}\mathscr{T}^{-1}=a_{-\mathbf{p}}$
(129) $\displaystyle\mathscr{P}\phi(t,\mathbf{x})\mathscr{P}^{-1}$
$\displaystyle=\phi(t,-\mathbf{x})\quad\implies\quad\mathscr{P}a_{\mathbf{p}}\mathscr{P}^{-1}=a_{-\mathbf{p}}$
(130)

The only non-trivial transformation is therefore the $C$ transformation.
However, because of the presence of the antipode $S(\cdot)$ in the fields in
eq. (115), (116), also $C$ can be shown to behave like in the non-deformed
case (in its action on $a,a^{\dagger},b,b^{\dagger}$) when acting on fields.

Indeed, take the $\phi_{+}(x)$ component first

$\displaystyle\mathscr{C}\phi_{+}(x)\mathscr{C}^{-1}=\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}\zeta(p)\,\mathscr{C}a_{\mathbf{p}}\mathscr{C}^{-1}e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}.$
(131)

On the other hand we have

$\displaystyle\phi^{\dagger}_{-}(x)=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)b_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p})\mathbf{x}}$
(132)

which means that

$\displaystyle\mathscr{C}a_{\mathbf{p}}\mathscr{C}^{-1}=b_{\mathbf{p}}.$ (133)

At the same time, we also have

$\displaystyle\mathscr{C}\phi_{-}(x)\mathscr{C}^{-1}=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\mathscr{C}b^{\dagger}_{\mathbf{p}}\mathscr{C}^{-1}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}$
(134)

and

$\displaystyle\phi^{\dagger}_{+}(x)=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\,a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}$
(135)

which means that

$\displaystyle\mathscr{C}b^{\dagger}_{\mathbf{p}}\mathscr{C}^{-1}=a^{\dagger}_{\mathbf{p}}.$
(136)

Notice that the definition of the $C$ transformation is trivial also in the
deformed case, and we stress once again that this is due to the presence of
the antipode in the field definition. This is in contrast with other
approaches to charge conjugation, which do not give rise to such simple
transformation rules.

###  12 Off-shell action in momentum space

It is instructive (and we will need it in chapter 2) to compute the off-shell
action in momentum space (we are in point $vi)$ of our roadmap in section 1).
This gives a unique insight on the on-shell relation in momentum space, as
well as a clear picture of the invariance of the action under both discrete
$C,P,T$ transformation, and continuous $\kappa$-Poincaré transformations.

To do so we will need the off-shell field. We cannot use the same construction
as before because it includes the Dirac delta $\delta(p_{\mu}p^{\mu}-m^{2})$.
Therefore, we define

$\displaystyle\phi_{off}(x)$
$\displaystyle=\int_{\mathscr{I}^{+}}\frac{d^{4}p}{p_{4}/\kappa}\,\,\zeta(p)\,\left\\{a_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+b^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\right\\}$
(137) $\displaystyle\phi_{off}^{\dagger}(x)$
$\displaystyle=\int_{\mathscr{I}^{+}}\frac{d^{4}p}{p_{4}/\kappa}\,\,\zeta(p)\,\left\\{a^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}+b_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}\right\\}.$
(138)

Since we have two ordering of the action, namely eq. (14) and (15), we will
compute them separately.

####  1 First ordering of the action

Using the fact that
$(\partial_{A}\phi)^{\dagger}\equiv\partial_{A}^{\dagger}\phi^{\dagger}$ and
substituting eq. (137), (138) into (14) we get

$\displaystyle S=\int d^{4}x$
$\displaystyle\int\frac{d^{4}p}{p_{4}/\kappa}\frac{d^{4}q}{q_{4}/\kappa}\Bigg{\\{}(p_{\mu}q^{\mu}-m^{2})a_{\mathbf{p}}^{\dagger}a_{\mathbf{q}}e^{-i(S(\omega_{p})\oplus\omega_{q})t}e^{i(S(\mathbf{p})\oplus\mathbf{q})\mathbf{x}}$
(139)
$\displaystyle-(p_{\mu}S(q)^{\mu}-m^{2})a_{\mathbf{p}}^{\dagger}b_{\mathbf{q}}^{\dagger}e^{-i(S(\omega_{p})\oplus
S(\omega_{q}))t}e^{i(S(\mathbf{p})\oplus S(\mathbf{q}))\mathbf{x}}$ (140)
$\displaystyle-(S(p)_{\mu}q^{\mu}-m^{2})b_{\mathbf{p}}a_{\mathbf{q}}e^{-i(\omega_{p}\oplus\omega_{q})t}e^{i(\mathbf{p}\oplus\mathbf{q})\mathbf{x}}$
(141)
$\displaystyle+(S(p)_{\mu}S(q)^{\mu}-m^{2})b_{\mathbf{p}}b_{\mathbf{q}}^{\dagger}e^{-i(\omega_{p}\oplus
S(\omega_{q}))t}e^{-i(\mathbf{p}\oplus
S(\mathbf{q}))\mathbf{x}}\Bigg{\\}}\zeta(p)\zeta(q)$ (142)

For the moment we will ignore the terms (140) and (141). Their treatment is
quite non-trivial, and we will defer their treatment to future studies.
Therefore, integrating in $d^{4}x$,

$\displaystyle S=$
$\displaystyle\int\frac{d^{4}p}{p_{4}/\kappa}\frac{d^{4}q}{q_{4}/\kappa}\Bigg{\\{}(p_{\mu}q^{\mu}-m^{2})a_{\mathbf{p}}^{\dagger}a_{\mathbf{q}}\delta(S(\omega_{p})\oplus\omega_{q})\delta(S(\mathbf{p})\oplus\mathbf{q})$
(143)
$\displaystyle+(S(p)_{\mu}S(q)^{\mu}-m^{2})b_{\mathbf{p}}b_{\mathbf{q}}^{\dagger}\delta(\omega_{p}\oplus
S(\omega_{q}))\delta(\mathbf{p}\oplus
S(\mathbf{q}))\Bigg{\\}}\zeta(p)\zeta(q)$ (144) $\displaystyle=$
$\displaystyle\int\frac{d^{4}p}{p_{4}/\kappa}\frac{d^{4}q}{q_{4}/\kappa}\Bigg{\\{}(p_{\mu}q^{\mu}-m^{2})a_{\mathbf{p}}^{\dagger}a_{\mathbf{q}}\delta(S(p)\oplus
q)$ (145)
$\displaystyle+(S(p)_{\mu}S(q)^{\mu}-m^{2})b_{\mathbf{p}}b_{\mathbf{q}}^{\dagger}\delta(p\oplus
S(q))\Bigg{\\}}\zeta(p)\zeta(q)$ (146)

where we used the notation

$\displaystyle\delta(S(\omega_{p})\oplus\omega_{q})\delta(S(\mathbf{p})\oplus\mathbf{q})$
$\displaystyle=\delta(S(p)\oplus q)$ (147)
$\displaystyle\qquad\delta(\omega_{p}\oplus
S(\omega_{q}))\delta(\mathbf{p}\oplus S(\mathbf{q}))$
$\displaystyle=\delta(p\oplus S(q)).$ (148)

We now use a known result in the literature [85] which states that

$\displaystyle\delta(p\oplus
S(q))=\frac{|p_{+}|^{3}}{\kappa^{3}}\delta(S(p)\oplus
q)\qquad\delta(S(p)\oplus q)=\frac{p_{4}}{\kappa}\delta(p-q)$ (149)

The additional $\frac{p_{4}}{\kappa}$ factor eliminates one of the factors in
one of the two metrics, so that the final expression for the firs off-shell
action in momentum space is

$\displaystyle
S=\int\frac{d^{4}p}{p_{4}/\kappa}\zeta^{2}(p)\left[(p_{\mu}p^{\mu}-m^{2})a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+(S(p)_{\mu}S(p)^{\mu}-m^{2})b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}\frac{|p_{+}|^{3}}{\kappa^{3}}\right]$
(150)

where the red term is the factor coming from the delta.

####  2 Second ordering of the action

Using the switched convention, i.e. substituting eq. (137) and (138) into eq.
(15), we obtain

$\displaystyle S=\int d^{4}x$
$\displaystyle\int\frac{d^{4}p}{p_{4}/\kappa}\frac{d^{4}q}{q_{4}/\kappa}\Bigg{\\{}(p_{\mu}q^{\mu}-m^{2})a_{\mathbf{q}}a_{\mathbf{p}}^{\dagger}e^{-i(\omega_{q}t-\mathbf{q}\mathbf{x})}\star
e^{-i(S(\omega_{p})t-S(\mathbf{p})\mathbf{x})}$ (151)
$\displaystyle+(S(p)_{\mu}S(q)^{\mu}-m^{2})b_{\mathbf{q}}^{\dagger}b_{\mathbf{p}}e^{i(S(\omega_{q})t-S(\mathbf{q})\mathbf{x})}\star
e^{i(\omega_{p}t-\mathbf{p}\mathbf{x})}\Bigg{\\}}\zeta(p)\zeta(q)$ (152)
$\displaystyle=\int d^{4}x$
$\displaystyle\int\frac{d^{4}p}{p_{4}/\kappa}\frac{d^{4}q}{q_{4}/\kappa}\Bigg{\\{}(p_{\mu}q^{\mu}-m^{2})a_{\mathbf{q}}a_{\mathbf{p}}^{\dagger}e^{-i(\omega_{q}\oplus
S(\omega_{p}))t}e^{i(\mathbf{q}\oplus S(\mathbf{p}))\mathbf{x}}$ (153)
$\displaystyle+(S(p)_{\mu}S(q)^{\mu}-m^{2})b_{\mathbf{q}}^{\dagger}b_{\mathbf{p}}e^{-i(S(\omega_{q})\oplus\omega_{p})t}e^{-i(S(\mathbf{q})\oplus\mathbf{p})\mathbf{x}}\Bigg{\\}}\zeta(p)\zeta(q)$
(154)

The two missing terms have been eliminated for the same reason as before, and
we will study them in detail in forthcoming publications.

These are exactly the same terms as in the previous calculations with only two
exceptions:

  * 1)

In the previous action we had the products
$a_{\mathbf{p}}^{\dagger}a_{\mathbf{q}}$ and
$b_{\mathbf{p}}b_{\mathbf{q}}^{\dagger}$, while with this switched convention
we have the other order, i.e. $a_{\mathbf{q}}a_{\mathbf{p}}^{\dagger}$ and
$b_{\mathbf{q}}^{\dagger}b_{\mathbf{p}}$;

  * 2)

The Dirac deltas are switched with respect to the computations with the
previous ordering. More explicitly, the exponential multiplying the term
$a_{\mathbf{q}}a_{\mathbf{p}}^{\dagger}$ here is the same that was multiplying
the term $b_{\mathbf{p}}b_{\mathbf{q}}^{\dagger}$, and vice versa.

Therefore, all the computations proceed in the same way but the extra
coefficient is now found with the $a$, $a^{\dagger}$ operators, i.e. we have

$\displaystyle
S=\int\frac{d^{4}p}{p_{4}/\kappa}\zeta^{2}(p)\left[\frac{|p_{+}|^{3}}{\kappa^{3}}(p_{\mu}p^{\mu}-m^{2})a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+(S(p)_{\mu}S(p)^{\mu}-m^{2})b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}\right]$
(155)

Notice that the on-shell conditions are not swapped, because their formulation
only depends on the property
$(\partial_{A}\phi)^{\dagger}=\partial_{A}^{\dagger}\phi^{\dagger}$, and not
on the ordering of the star products.

####  3 Sum of the two actions

Using the definition of the total action given by eq. (16), and substituting
eq. (150) and (155), we obtain (recall that we are still in the classical
case, but upon quantization one should remember that normal-ordering should be
applied)

$\displaystyle
S=\int\frac{d^{4}p}{p_{4}/\kappa}\left[1+\frac{|p_{+}|^{3}}{\kappa^{3}}\right]\zeta^{2}(p)\left\\{(p_{\mu}p^{\mu}-m^{2})a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+(S(p)_{\mu}S(p)^{\mu}-m^{2})b_{\mathbf{p}}^{\dagger}b_{\mathbf{p}}\right\\}.$
(156)

If we decided to go on-shell, the computations would proceed as usual, but
with an additional Dirac delta imposing the shell condition, which means that
the action would be exactly zero.

###  13 Symplectic form for the two actions

We are now at point $vii)a)$ of our roadmap in section 1.

####  1 Example in the non-deformed case

Let us consider a complex scalar field with the Lagrangian

$L=-\left(\partial^{\mu}\phi^{\dagger}\partial_{\mu}\phi+m^{2}\phi^{\dagger}\phi\right)$
(157)

The variation of this Lagrangian is

$\delta
L=-\left(\partial^{\mu}\delta\phi^{\dagger}\partial_{\mu}\phi+\partial^{\mu}\phi^{\dagger}\partial_{\mu}\delta\phi+m^{2}\delta\phi^{\dagger}\phi+m^{2}\phi^{\dagger}\delta\phi\right)$
(158)

This variation contains terms proportional to the field equations and the
total derivative term

$\delta L={\rm
EoM}-\partial^{\mu}\left(\delta\phi^{\dagger}\partial_{\mu}\phi\right)-\partial_{\mu}\left(\partial^{\mu}\phi^{\dagger}\delta\phi\right)$
(159)

Therefore the variation of the action is

$\delta S=\int_{M}{\rm EoM}-\int_{\partial
M}d\sigma_{\mu}\left(\delta\phi^{\dagger}\partial^{\mu}\phi+\partial^{\mu}\phi^{\dagger}\delta\phi\right)$
(160)

The second, boundary term in this expression is the so-called presymplectic
structure or Liouville form $\theta$, if we choose the boundary $\partial M$
to be a Cauchy surface, for example the surface of constant time $t=0$. Then
we can find the symplectic form

$\omega=\delta\theta=\frac{1}{2}\int_{t=0}d^{3}x\,\left(\delta\phi^{\dagger}\wedge\frac{\partial}{\partial
t}\delta\phi-\frac{\partial}{\partial
t}\delta\phi^{\dagger}\wedge\delta\phi\right)_{t=0}$ (161)

Now we take the field decomposition

$\phi(t,\mathbf{x})=\int\frac{d^{3}p}{\sqrt{2E}}\,a_{\mathbf{p}}\,e^{-i(Et-{\mathbf{p}}{\mathbf{x}})}+b^{\dagger}_{\mathbf{p}}\,e^{i(Et-{\mathbf{p}}{\mathbf{x}})}$
(162)

so that

$\delta\phi(t,\mathbf{x})=\int\frac{d^{3}p}{\sqrt{2E}}\,\delta
a_{\mathbf{p}}\,e^{-i(Et-{\mathbf{p}}{\mathbf{x}})}+\delta
b^{\dagger}_{\mathbf{p}}\,e^{i(Et-{\mathbf{p}}{\mathbf{x}})}$ (163)

and plug it to (161) to obtain the symplectic form, whose inverse defines for
us the Poisson brackets, and (upon quantization) the commutators.

$\displaystyle\omega$ $\displaystyle=\frac{i}{2}\int d^{3}x\int
d^{3}p\,d^{3}q\left(\delta
a^{\dagger}_{\mathbf{p}}\,e^{-i{\mathbf{p}}{\mathbf{x}}}+\delta
b_{\mathbf{p}}\,e^{i{\mathbf{p}}{\mathbf{x}}}\right)\wedge\left(-\delta
a_{\mathbf{q}}\,e^{i{\mathbf{q}}{\mathbf{x}}}+\delta
b^{\dagger}_{\mathbf{q}}\,e^{-i{\mathbf{q}}{\mathbf{x}}}\right)$
$\displaystyle-\frac{i}{2}\int d^{3}x\int d^{3}q\,d^{3}p\left(\delta
a^{\dagger}_{\mathbf{p}}\,e^{-i{\mathbf{p}}{\mathbf{x}}}-\delta
b_{\mathbf{p}}\,e^{i{\mathbf{p}}{\mathbf{x}}}\right)\wedge\left(\delta
a_{\mathbf{q}}\,e^{i{\mathbf{q}}{\mathbf{x}}}+\delta
b^{\dagger}_{\mathbf{q}}\,e^{-i{\mathbf{q}}{\mathbf{x}}}\right)$

Defining

$\Omega=\int{d^{3}p}d^{3}q\,\,\omega(p,q)$ (164)

we find

$\Omega(p,q)\sim-\delta a^{\dagger}_{\mathbf{p}}\wedge\delta
a_{\mathbf{q}}-\delta b^{\dagger}_{\mathbf{p}}\wedge\delta b_{\mathbf{q}}$
(165)

from which the commutational relations

$[a_{\mathbf{p}},a^{\dagger}_{\mathbf{q}}]\sim[b_{\mathbf{p}},b^{\dagger}_{\mathbf{q}}]\sim\delta(p-q)$
(166)

####  2 The $\kappa$-deformed case

The extension of this reasoning to the case of fields in the context of
$\kappa$-Poincaré algebra is straightforward. In particular we have already
computed the surface term in previous sections, and we are only interested in
those terms whose boundary is spacelike, i.e. we only need to consider the
following presymplectic forms.

$\displaystyle\qquad\theta_{1}=-\int_{\mathbb{R}^{3}}d^{3}x\left(\Pi^{0}_{1}\star\delta\phi+\delta\phi^{\dagger}\star\left(\Pi^{0}_{1}\right)^{\dagger}\right)$
$\displaystyle=-\int_{\mathbb{R}^{3}}d^{3}x$
$\displaystyle\left\\{\frac{1}{\kappa}\left(\Delta_{+}\partial_{0}^{\dagger}+im^{2}\right)\phi^{\dagger}\star\delta\phi+\delta\phi^{\dagger}\star\left[\frac{1}{\kappa}\left(\Delta_{+}\partial_{0}^{\dagger}+im^{2}\right)\phi^{\dagger}\right]^{\dagger}\right\\}.$
(167)
$\displaystyle\qquad\qquad\theta_{2}=-\int_{\mathbb{R}^{3}}d^{3}x\left(\delta\phi\star\Pi^{0}_{2}+\left(\Pi^{0}_{2}\right)^{\dagger}\star\delta\phi^{\dagger}\right)$
$\displaystyle=-\int_{\mathbb{R}^{3}}d^{3}x$
$\displaystyle\left\\{\delta\phi\star\left(\frac{\kappa}{\Delta_{+}}\partial_{0}^{\dagger}+\frac{i}{\kappa}(\partial_{0}^{\dagger})^{2}\right)\phi^{\dagger}+\left[\left(\frac{\kappa}{\Delta_{+}}\partial_{0}^{\dagger}+\frac{i}{\kappa}(\partial_{0}^{\dagger})^{2}\right)\phi^{\dagger}\right]^{\dagger}\star\delta\phi^{\dagger}\right\\}.$
(168)

Notice that the sign in the RHS of these two presymplectic forms is such that
we get the correct symplectic form in the canonical limit
$\kappa\rightarrow\infty$ (and therefore the correct canonical commutator
between $a$ and $a^{\dagger}$,a nd same for $b$ and $b^{\dagger}$).
Furthermore, we are not writing down explicitly the wedge product for the
moment because at the moment it is not present, it will be when we will
consider $\delta\theta_{1}$ and $\delta\theta_{2}$. To simplify the
computations, notice that

$\displaystyle\Pi_{1}^{0}$
$\displaystyle=\left(\frac{\Delta_{+}}{\kappa}\partial_{0}^{\dagger}+\frac{im^{2}}{\kappa}\right)\phi^{\dagger}$
(169)
$\displaystyle=\left(\Delta_{+}(\partial_{0}-i\Delta_{+}^{-1}\partial^{2})+im^{2}\right)\frac{\phi^{\dagger}}{\kappa}$
(170)
$\displaystyle=\left(\frac{\Delta_{+}}{\kappa}\partial_{0}+\frac{i}{\kappa}(m^{2}-\partial^{2})\right)\phi^{\dagger}$
(171)
$\displaystyle=\left(\frac{\Delta_{+}}{\kappa}\partial_{0}-\frac{i\partial_{0}^{2}}{\kappa}\right)\phi^{\dagger}$
(172)
$\displaystyle=\left(\frac{i\partial_{0}+\kappa+i\partial_{4}}{\kappa}\partial_{0}-\frac{i\partial_{0}^{2}}{\kappa}\right)\phi^{\dagger}$
(173)
$\displaystyle=\left[\frac{\kappa+i\partial_{4}}{\kappa}\partial_{0}\right]\phi^{\dagger}.$
(174)

and in the same way one gets

$\displaystyle\left[\frac{\kappa+i\partial_{4}}{\kappa}\partial_{0}\right]=\left(\frac{\Delta_{+}}{\kappa}\partial_{0}^{\dagger}+\frac{im^{2}}{\kappa}\right)=\left(\frac{\kappa}{\Delta_{+}}\partial_{0}^{\dagger}+\frac{i(\partial_{0}^{\dagger})^{2}}{\kappa}\right)^{\dagger}$
(175)

and therefore

$\displaystyle\left(\frac{\kappa}{\Delta_{+}}\partial_{0}^{\dagger}+\frac{i(\partial_{0}^{\dagger})^{2}}{\kappa}\right)=\left[\frac{\kappa+i\partial_{4}}{\kappa}\partial_{0}\right]^{\dagger}.$
(176)

Notice that the quantity $\partial_{4}$ acting on any plane wave brings down
$i\kappa-ip_{4}$, and since $p_{4}=S(p_{4})$, then
$\kappa+i\partial_{4}\rightarrow\kappa+i(i\kappa-ip_{4})=p_{4}$ (recall that
$S(p_{4})=p_{4}$). The presymplectic forms can therefore be simplified to

$\displaystyle\theta_{1}=-\int_{\mathbb{R}^{3}}d^{3}x\frac{p_{4}}{\kappa}\left(\partial_{0}\phi^{\dagger}\star\delta\phi+\delta\phi^{\dagger}\star\partial_{0}^{\dagger}\phi\right)$
(177)
$\displaystyle\theta_{2}=-\int_{\mathbb{R}^{3}}d^{3}x\frac{p_{4}}{\kappa}\left(\delta\phi\star\partial_{0}^{\dagger}\phi^{\dagger}+\partial_{0}\phi\star\delta\phi^{\dagger}\right).$
(178)

Taking the delta, we get

$\displaystyle\delta\theta_{1}=-\int_{\mathbb{R}^{3}}d^{3}x\frac{p_{4}}{\kappa}\left(\delta\partial_{0}\phi^{\dagger}\overset{\wedge}{\star}\delta\phi-\delta\phi^{\dagger}\overset{\wedge}{\star}\delta\partial_{0}^{\dagger}\phi\right)$
(179)
$\displaystyle\delta\theta_{2}=-\int_{\mathbb{R}^{3}}d^{3}x\frac{p_{4}}{\kappa}\left(-\delta\phi\overset{\wedge}{\star}\partial_{0}^{\dagger}\delta\phi^{\dagger}+\partial_{0}\delta\phi\overset{\wedge}{\star}\delta\phi^{\dagger}\right).$
(180)

The notation $\overset{\wedge}{\star}$ means that we are still taking the
$\star$ product among plane waves, and also the wedge product between $\delta
a,\delta a^{\dagger},\delta b,\delta b^{\dagger}.$ We can now explicitly
compute both symplectic forms using the fields in eq. (115), (116). Starting
from the first, we have

$\displaystyle\delta\theta_{1}$
$\displaystyle=-\int_{\mathbb{R}^{3}}\frac{p_{4}}{\kappa}\left\\{\delta\partial_{0}\phi^{\dagger}\overset{\wedge}{\star}\delta\phi-\delta\phi^{\dagger}\overset{\wedge}{\star}\delta\partial_{0}^{\dagger}\phi\right\\}$
(181)
$\displaystyle=-i\int_{\mathbb{R}^{3}}\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\frac{d^{3}q}{\sqrt{2\omega_{\mathbf{q}}}}\,\zeta(q)\,\zeta(p)\
\frac{p_{4}}{\kappa}\times$
$\displaystyle\Big{\\{}(-S(\omega_{\mathbf{q}})-S(\omega_{\mathbf{p}}))a^{\dagger}_{\mathbf{q}}\wedge
a_{\mathbf{p}}e^{-i(S(\omega_{\mathbf{q}})\oplus\omega_{\mathbf{p}})t}e^{i(S(\mathbf{q})\oplus\mathbf{p})\mathbf{x}}$
(182)
$\displaystyle+(-\omega_{\mathbf{q}}-\omega_{\mathbf{p}})b_{\mathbf{q}}\wedge
b^{\dagger}_{\mathbf{p}}e^{-i(\omega_{\mathbf{q}}\oplus
S(\omega_{\mathbf{p}}))t}e^{i(\mathbf{q}\oplus S(\mathbf{p}))\mathbf{x}}$
(183)
$\displaystyle-(S(\omega_{\mathbf{q}})+\omega_{\mathbf{p}})a^{\dagger}_{\mathbf{q}}\wedge
b^{\dagger}_{\mathbf{p}}e^{-i(S(\omega_{\mathbf{q}})\oplus
S(\omega_{\mathbf{p}}))t}e^{i(S(\mathbf{q})\oplus S(\mathbf{p}))\mathbf{x}}$
(184)
$\displaystyle-(\omega_{\mathbf{q}}+S(\omega_{\mathbf{p}}))a_{\mathbf{q}}\wedge
b_{\mathbf{p}}e^{-i(\omega_{\mathbf{q}}\oplus\omega_{\mathbf{p}})t}e^{i(\mathbf{q}\oplus\mathbf{p})\mathbf{x}}\Big{\\}}$
(185)

We now use

$\displaystyle\int_{\mathbb{R}^{3}}e^{-i(S(\omega_{q})t-S(\mathbf{q})\mathbf{x})}\star
e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}=\delta\left(\mathbf{p}-\mathbf{q}\right)$
(186)
$\displaystyle\int_{\mathbb{R}^{3}}e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}\star
e^{-i(S(\omega_{q})t-S(\mathbf{q})\mathbf{x})}=\frac{|q_{+}|^{3}}{\kappa^{3}}\delta\left(\mathbf{p}-\mathbf{q}\right)$
(187)

so that the first two terms give

$\displaystyle\boxed{\delta\theta_{1}=i\int\frac{d^{3}p}{2\omega_{\mathbf{p}}}\zeta^{2}(p)\frac{p_{4}}{\kappa}\left(2S(\omega_{\mathbf{p}})a_{\mathbf{p}}^{\dagger}\wedge
a_{\mathbf{p}}+\frac{p_{+}^{3}}{\kappa^{3}}2\omega_{\mathbf{p}}b_{\mathbf{p}}\wedge
b_{\mathbf{p}}^{\dagger}\right)}$ (188)

and the other two go away because the Dirac delta impose on-shell that
$\omega_{\mathbf{q}}=-S(\omega_{\mathbf{p}})$. Indeed, one can check for
example that

$\displaystyle\int_{\mathbb{R}^{3}}e^{-i\omega_{p}t+i\mathbf{p}\mathbf{x}}\star
e^{-i\omega_{q}t+i\mathbf{q}\mathbf{x}}=e^{-i(\omega_{p}\oplus\omega_{q})t}\delta(\mathbf{p}\oplus\mathbf{q})=e^{-i(\omega_{p}\oplus\omega_{q})t}\delta\left(\frac{q_{+}}{\kappa}\mathbf{p}+\mathbf{q}\right)$
(189)

If we first integrate in $d^{3}p$, the quantity $|q_{+}|^{3}/\kappa^{3}$ can
be brought outside the delta obtaining
$\frac{\kappa^{3}}{|q_{+}|^{3}}\delta(\mathbf{p}+\mathbf{q}\frac{\kappa}{q_{+}})$.
Therefore we have

$\displaystyle
e^{-i(\omega_{p}\oplus\omega_{q})t}\delta\left(\frac{q_{+}}{\kappa}\mathbf{p}+\mathbf{q}\right)=\frac{\kappa^{3}}{|q_{+}|^{3}}e^{-i(\omega_{p}\oplus\omega_{q})t}\delta\left(\mathbf{p}+\frac{\kappa}{q_{+}}\mathbf{q}\right)$
(190)

The Dirac deltas imply also a modification of the energy $\omega_{p}$. In
fact, we have

$\displaystyle\omega_{q}^{2}-\mathbf{q}^{2}=m^{2}=\omega_{p}^{2}-\mathbf{p}^{2}=\omega_{p}^{2}-\frac{\kappa^{2}}{q_{+}^{2}}\mathbf{q}^{2}$
(191)

and therefore

$\displaystyle\omega_{p}^{2}=m^{2}+\frac{\kappa^{2}}{q_{+}^{2}}\mathbf{q}^{2}=S(\omega_{q})^{2}.$
(192)

In other words, the Dirac delta imposes

$\displaystyle\omega_{p}=-S(\omega_{q}).$ (193)

Notice that the LHS contains a minus sign because on-shell we have
$\omega_{p}>0$ and $S(\omega_{q})<0$, so taking the square root we need to
take this into account. The other exponential is treated in the same way,
since one can change variables $p\mapsto S(p)$, obtaining the same result, so
that the time-dependent parts go away.

For the second one, we proceed in a similar manner, and we have

$\displaystyle\delta\theta_{2}$
$\displaystyle=\int_{\mathbb{R}^{3}}\frac{p_{4}}{\kappa}\left\\{\delta\phi\overset{\wedge}{\star}\partial_{0}^{\dagger}\delta\phi^{\dagger}-\delta\partial_{0}\phi\overset{\wedge}{\star}\delta\phi^{\dagger}\right\\}$
(194)
$\displaystyle=i\int_{\mathbb{R}^{3}}\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\frac{d^{3}q}{\sqrt{2\omega_{\mathbf{q}}}}\,\zeta(q)\,\zeta(p)\
\frac{p_{4}}{\kappa}\times$
$\displaystyle\Big{\\{}(\omega_{\mathbf{q}}+\omega_{\mathbf{p}})\delta
a_{\mathbf{p}}\wedge\delta
a^{\dagger}_{\mathbf{q}}e^{-i(\omega_{\mathbf{p}}\oplus
S(\omega_{\mathbf{q}}))t}e^{i(\mathbf{p}\oplus S(\mathbf{q}))\mathbf{x}}$
(195) $\displaystyle+(S(\omega_{\mathbf{p}})+S(\omega_{\mathbf{q}}))\delta
b^{\dagger}_{\mathbf{p}}\delta\delta
b_{\mathbf{q}}e^{-i(S(\omega_{\mathbf{p}})\oplus\omega_{\mathbf{q}})t}e^{i(S(\mathbf{p})\oplus\mathbf{q})\mathbf{x}}$
(196) $\displaystyle+(S(\omega_{\mathbf{p}})+\omega_{\mathbf{q}})\delta
b^{\dagger}_{\mathbf{p}}\delta\delta
a^{\dagger}_{\mathbf{q}}e^{-i(S(\omega_{\mathbf{p}})\oplus
S(\omega_{\mathbf{q}}))t}e^{i(S(\mathbf{p})\oplus S(\mathbf{q}))\mathbf{x}}$
(197) $\displaystyle+(\omega_{\mathbf{p}}+S(\omega_{\mathbf{q}}))\delta
b_{\mathbf{p}}\delta\delta
a_{\mathbf{q}}e^{-i(\omega_{\mathbf{p}}\oplus\omega_{\mathbf{q}})t}e^{i(\mathbf{p}\oplus\mathbf{q})\mathbf{x}}\Big{\\}}$
(198)

so that, proceeding in the same way as before, we have

$\displaystyle\boxed{\delta\theta_{2}=\int\frac{d^{3}p}{2\omega_{\mathbf{p}}}\zeta^{2}(p)\frac{p_{4}}{\kappa}\left(\frac{p_{+}^{3}}{\kappa^{3}}2\omega_{\mathbf{p}}\delta
a_{\mathbf{p}}\wedge\delta
a_{\mathbf{p}}^{\dagger}+2S(\omega_{\mathbf{p}})\delta
b_{\mathbf{p}}^{\dagger}\wedge\delta b_{\mathbf{p}}\right)}$ (199)

Because of the definition of the action, the total symplectic form will be

$\displaystyle\delta\theta_{T}=\frac{\delta\theta_{1}+\delta\theta_{2}}{2}$
(200)

and therefore we have

$\displaystyle\boxed{\delta\theta_{T}=i\int\frac{d^{3}p}{2\omega_{p}}\zeta(p)^{2}\left[\frac{|p_{+}|^{3}}{\kappa^{3}}\omega_{p}-S(\omega_{p})\right]\frac{p_{4}}{\kappa}\left\\{a_{\mathbf{p}}\wedge
a_{\mathbf{p}}^{\dagger}\,-b_{\mathbf{p}}^{\dagger}\wedge
b_{\mathbf{p}}\,\right\\}}.$ (201)

This result can be simplified using the definition

$\displaystyle\boxed{\alpha(p)=\frac{\zeta(p)^{2}}{2\omega_{p}}\left[\frac{|p_{+}|^{3}}{\kappa^{3}}\omega_{p}-S(\omega_{p})\right]\frac{p_{4}}{\kappa}}$
(202)

and we get

$\displaystyle\boxed{\delta\theta_{T}=i\int
d^{3}p\,\,\alpha(p)\left\\{a_{\mathbf{p}}\wedge
a_{\mathbf{p}}^{\dagger}\,-b_{\mathbf{p}}^{\dagger}\wedge
b_{\mathbf{p}}\,\right\\}}.$ (203)

This result would imply the Poisson brackets

$\displaystyle\left\\{a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\right\\}$
$\displaystyle=-i\frac{1}{\xi(p)^{2}}\frac{\kappa}{p_{4}}\frac{2\omega_{p}}{\frac{|p_{+}|^{3}}{\kappa^{3}}\omega_{p}-S(\omega_{p})}\delta(\mathbf{p}-\mathbf{q})$
(204) $\displaystyle\left\\{b_{\mathbf{p}},b_{\mathbf{q}}^{\dagger}\right\\}$
$\displaystyle=-i\frac{1}{\xi(p)^{2}}\frac{\kappa}{p_{4}}\frac{2\omega_{p}}{\frac{|p_{+}|^{3}}{\kappa^{3}}\omega_{p}-S(\omega_{p})}\delta(\mathbf{p}-\mathbf{q}).$
(205)

which can also be simplified using eq. (202) as follows

$\displaystyle\left\\{a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\right\\}$
$\displaystyle=-\frac{i}{\alpha(p)}\delta(\mathbf{p}-\mathbf{q})$ (206)
$\displaystyle\left\\{b_{\mathbf{p}},b_{\mathbf{q}}^{\dagger}\right\\}$
$\displaystyle=-\frac{i}{\alpha(p)}\delta(\mathbf{p}-\mathbf{q}).$ (207)

Notice that, if one uses the canonical quantization map
$[A,B]=i\hbar\\{A,B\\}$ and assume $\hbar=1$, then we would get the
commutators

$\displaystyle\left[a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\right]$
$\displaystyle=\frac{1}{\alpha(p)}\delta(\mathbf{p}-\mathbf{q})$ (208)
$\displaystyle\left[b_{\mathbf{p}},b_{\mathbf{q}}^{\dagger}\right]$
$\displaystyle=\frac{1}{\alpha(p)}\delta(\mathbf{p}-\mathbf{q}).$ (209)

where the LHS is positive and the coefficient goes to 1 for
$\kappa\rightarrow\infty$, as it should. We will comment on how to technically
get these commutations relations in section 4 below.

###  14 Conserved charges: geometric approach

The starting point is the symplectic form in eq. (203). For simplicity, we
will call
$\Omega=\delta\theta_{T}=\frac{\delta\theta_{1}+\delta\theta_{2}}{2}$. In
order to get the charges from the symplectic form we need to do the following.

Assuming that the charges come from a symmetry described by some continuous
vector field $\xi$ in spacetime, then we compute [87] (this approach is also
used quite often in GR, for example see [89], [90])

$\displaystyle-\delta_{\xi}\lrcorner\,\Omega\overset{!}{=}\delta Q_{\xi}$
(210)

where $\delta$ is the exterior derivative in phase space (i.e. the set of
solutions of the EoM) and $Q_{\xi}$ is the charge associated to the vector
$\xi$. The object $\delta_{\xi}$ is a vector field in phase space generated by
the killing vector field $\xi$. In other words, $\delta_{\xi}A$ measures the
infinitesimal variation of the object $A$ in phase space due to the symmetry
due to the action along $\xi$ in spacetime. The symbol $\lrcorner$ is used to
indicete contraction of the vector field with forms. Notice that each charge
obtained in this way is a symmetry of the symplectic form. Indeed one can
easily show that

$\displaystyle\pounds_{\delta_{\xi}}\Omega=\delta_{\xi}\lrcorner\,\delta\Omega+\delta(\delta_{\xi}\lrcorner\,\Omega)=0+\delta^{2}Q_{\xi}=0$
(211)

where we used the fact that $\delta^{2}=0$ since $\delta$ is the exterior
derivative in phase space, and also $\delta\Omega=0$ because $\Omega$ is an
exact 2-form. The charges built in this way are independent of the dynamics
because the symplectic form $\Omega$ and the vector field $\delta_{\xi}$ are
both time independent and non-dynamical by construction, and they are defined
on-shell.

To see how the procedure of getting the charges using this formalism, we first
tackle the problem in the canonical non-deformed QFT context.

####  1 Non-deformed case

In this case, the symplectic form is given by the $\kappa\rightarrow\infty$
limit of eq. (203) above, and it reads

$\displaystyle\Omega^{\kappa\rightarrow\infty}$ $\displaystyle=i\int
d^{3}p\,(\delta a_{\mathbf{p}}\wedge\delta a_{\mathbf{p}}^{\dagger}\,-\delta
b_{\mathbf{p}}^{\dagger}\wedge\delta b_{\mathbf{p}}).$ (212)

##### Translation charges

In this case, the translation can be described by the vectors $\partial_{0}$
or $\partial_{i}$, depending on whether we are talking about time translation
or spatial translations.

Starting with the time translations, we first need to understand what
$\delta_{\partial_{0}}a_{\mathbf{p}}$ is (it will be analogous for $b$). We
know that after a translation we have

$\displaystyle a_{\mathbf{p}}\mapsto
e^{i\epsilon\omega_{p}}a_{\mathbf{p}}=a_{\mathbf{p}}+i\epsilon\omega_{p}a_{\mathbf{p}}$
(213)

and therefore

$\displaystyle\delta_{\partial_{0}}a_{\mathbf{p}}=i\epsilon\omega_{p}a_{\mathbf{p}}\quad\Leftrightarrow\quad\delta_{\partial_{0}}a_{\mathbf{p}}^{\dagger}=-i\epsilon\omega_{p}a_{\mathbf{p}}^{\dagger}$
(214)

where the second expression can be obtained from the first one by simply
taking the hermitian conjugate. Therefore, we have (the wedge product goes
away after the contraction because the creation/annihilation operators are not
themselves forms)

$\displaystyle-\delta_{\partial_{0}}\lrcorner\,\Omega^{\kappa\rightarrow\infty}$
$\displaystyle=-i\int d^{3}p\,(\delta_{\partial_{0}}a_{\mathbf{p}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}\delta_{\partial_{0}}a_{\mathbf{p}}^{\dagger}-\delta_{\partial_{0}}b_{\mathbf{p}}^{\dagger}\delta
b_{\mathbf{p}}+\delta
b_{\mathbf{p}}^{\dagger}\delta_{\partial_{0}}b_{\mathbf{p}})$ (215)
$\displaystyle=-i\int d^{3}p\,[i\epsilon\omega_{p}a_{\mathbf{p}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}(-i\epsilon\omega_{p}a_{\mathbf{p}}^{\dagger})$
$\displaystyle-(-i\epsilon\omega_{p}b_{\mathbf{p}}^{\dagger})\delta
b_{\mathbf{p}}+\delta
b_{\mathbf{p}}^{\dagger}i\epsilon\omega_{p}b_{\mathbf{p}}]$ (216)
$\displaystyle=\epsilon\int d^{3}p\,[\omega_{p}a_{\mathbf{p}}\delta
a_{\mathbf{p}}^{\dagger}\,+\delta
a_{\mathbf{p}}\omega_{p}a_{\mathbf{p}}^{\dagger}+\omega_{p}b_{\mathbf{p}}^{\dagger}\delta
b_{\mathbf{p}}+\delta b_{\mathbf{p}}^{\dagger}\omega_{p}b_{\mathbf{p}}]$ (217)
$\displaystyle=\epsilon\delta\int
d^{3}p\,\omega_{p}(a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\,+b_{\mathbf{p}}^{\dagger}b_{\mathbf{p}})$
(218)

and therefore we get the canonical charge apart from the irrelevant prefactor
$\epsilon$.

$\displaystyle\mathscr{P}^{\kappa\rightarrow\infty}_{0}=\int
d^{3}p\,\,\omega_{p}\left\\{a_{\mathbf{p}}^{\dagger}\,a_{\mathbf{p}}\,+b_{\mathbf{p}}\,b_{\mathbf{p}}^{\dagger}\,\right\\}$
(219)

We can also check that eq. (213) gives the correct field transformation.

$\displaystyle\delta_{\partial_{\mu}}\phi$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\left[i\epsilon^{\mu}p_{\mu}\,a_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}-i\epsilon^{\mu}p_{\mu}b^{\dagger}_{\mathbf{p}}\,e^{i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}\right]$
$\displaystyle=\epsilon^{\mu}\partial_{\mu}\,\phi(x).$ (220)

Of course, once we quantize these charges, the operators will need to be put
in normal order. The same exact reasoning will work for the spatial
translation charges, it is sufficient to substitute $\omega_{p}$ with
$\mathbf{p}_{i}$ everywhere.

##### Boost and rotation charges

In this case, the procedure is again the same, but this time we need to use as
a vector field $\xi$ the generator of boost transformation, and then we need
to understand how creation/annihilation operator change under this
transformation, i.e. we need to understand what is the object
$\delta_{\Lambda}a_{\mathbf{p}}$.

In the case of boosts the field Fourier components transform as follows

$\displaystyle\delta^{B}a_{\mathbf{p}}=i\omega_{\mathbf{p}}\,\lambda^{i}\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\,+i\,a_{\mathbf{p}}\lambda^{i}\frac{\mathbf{p}_{i}}{2\omega_{\mathbf{p}}},\quad$
$\displaystyle\delta^{B}a^{\dagger}_{\mathbf{p}}=i\omega_{\mathbf{p}}\,\lambda^{i}\frac{\partial
a^{\dagger}_{\mathbf{p}}}{\partial\mathbf{p}^{i}}+i\,a_{\mathbf{p}}^{\dagger}\lambda^{i}\frac{\mathbf{p}_{i}}{2\omega_{\mathbf{p}}}$
(221)

and analogous expressions for $b_{\mathbf{p}}$, $b^{\dagger}_{\mathbf{p}}$. We
have therefore

$\displaystyle-\delta^{B}\lrcorner\,\Omega^{\kappa\rightarrow\infty}$
$\displaystyle=i\int d^{3}p\,(\delta^{B}a^{\dagger}_{\mathbf{p}}\delta
a_{\mathbf{p}}\,-\delta
a_{\mathbf{p}}^{\dagger}\delta^{B}a_{\mathbf{p}}-\delta^{B}b_{\mathbf{p}}\delta
b^{\dagger}_{\mathbf{p}}+\delta
b_{\mathbf{p}}\delta^{B}b^{\dagger}_{\mathbf{p}})$
$\displaystyle={\lambda}^{i}\int
d^{3}p\,\omega_{\mathbf{p}}\left(\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}-\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}\delta b_{\mathbf{p}}+\delta
b_{\mathbf{p}}^{\dagger}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\right)$
$\displaystyle+{\lambda}^{i}\int
d^{3}p\,\frac{\mathbf{p}_{i}}{2\omega_{\mathbf{p}}}\left(a_{\mathbf{p}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}+\delta
b_{\mathbf{p}}^{\dagger}\,b_{\mathbf{p}}-b_{\mathbf{p}}^{\dagger}\delta
b_{\mathbf{p}}\right)$ $\displaystyle=\frac{1}{2}{\lambda}^{i}\delta\int
d^{3}p\,\omega_{\mathbf{p}}\left(\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}a_{\mathbf{p}}^{\dagger}\,-a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}-\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}+b_{\mathbf{p}}^{\dagger}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\right)$ (222)

which gives the non-deformed boost charge

$\displaystyle{\cal N}_{i}^{\kappa\rightarrow\infty}=\frac{1}{2}\int
d^{3}p\,\omega_{\mathbf{p}}\left(\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}a_{\mathbf{p}}^{\dagger}\,-a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}-\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}+b_{\mathbf{p}}^{\dagger}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\right).$ (223)

The case of rotations $R$ is analogous, and starting from the transformations

$\displaystyle\delta^{R}a_{\mathbf{p}}$
$\displaystyle=i\epsilon^{ijk}\rho_{i}\mathbf{p}_{j}\frac{\partial}{\partial\mathbf{p}^{k}}a_{\mathbf{p}}\qquad\delta^{R}a_{\mathbf{p}}^{\dagger}=i\epsilon^{ijk}\rho_{i}\mathbf{p}_{j}\frac{\partial}{\partial\mathbf{p}^{k}}a_{\mathbf{p}}^{\dagger}$
(224)

and analogously for $b_{\mathbf{p}},b^{\dagger}_{\mathbf{p}}$, one gets the
charge

$\displaystyle{\cal M}_{i}=\frac{1}{8}\int
d^{3}p\,\epsilon_{ijk}\mathbf{p}^{j}\left(\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}_{k}}a_{\mathbf{p}}^{\dagger}\,-a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}_{k}}-\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}_{k}}b_{\mathbf{p}}+b_{\mathbf{p}}^{\dagger}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}_{k}}\right).$ (225)

Once again, we can obtain the field transformations from eq. (1) and (224),
and one respectively gets

$\displaystyle\delta^{B}\phi(x)$
$\displaystyle=i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\left[\omega_{\mathbf{p}}\,\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\,+a_{\mathbf{p}}\frac{\mathbf{p}_{i}}{2\omega_{\mathbf{p}}}\right]e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+\left[\omega_{\mathbf{p}}\,\frac{\partial
b^{\dagger}_{\mathbf{p}}}{\partial\mathbf{p}^{i}}+b_{\mathbf{p}}^{\dagger}\frac{\mathbf{p}_{i}}{2\omega_{\mathbf{p}}}\right]e^{i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle=-i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\left[a_{\mathbf{p}}\,\omega_{\mathbf{p}}\frac{\partial}{\partial\mathbf{p}_{i}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+b^{\dagger}_{\mathbf{p}}\,\omega_{\mathbf{p}}\frac{\partial}{\partial\mathbf{p}_{i}}\,e^{i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}\right]$
$\displaystyle=i\lambda_{i}\,x^{i}\frac{\partial}{\partial t}\,\phi(x)$ (226)

and

$\displaystyle\delta^{R}\phi(x)$
$\displaystyle=i\rho^{i}\,\epsilon_{ik}{}^{j}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\left[a_{\mathbf{p}}\,\mathbf{p}_{k}\frac{\partial}{\partial\mathbf{p}_{j}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+b^{\dagger}_{\mathbf{p}}\,\mathbf{p}_{k}\frac{\partial}{\partial\mathbf{p}_{j}}\,e^{i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}\right]$
$\displaystyle=i\rho^{i}\,\epsilon_{ik}{}^{j}\,x^{j}\frac{\partial}{\partial
x^{k}}\,\phi(x)$ (227)

which are the standard spacetime boost and rotation transformations.

Notice that for the boosts we actually have antisymmetrization when writing
them in spacetime. In fact, we only wrote down the contribution from
$\omega_{\mathbf{p}}\frac{\partial}{\partial\mathbf{p}_{i}}$, but we should
also include the contribution of
$\mathbf{p}_{i}\frac{\partial}{\partial\omega_{\mathbf{p}}}$. However, since
we are only integrating in $d^{3}p$, we first need to transform the derivative
in $\omega_{\mathbf{p}}$ into a derivative in $\mathbf{p}_{i}$ (which will
allow us to integrate by parts). Moreover, since we are on-shell (which means
that $\omega_{\mathbf{p}}=\sqrt{\mathbf{p}^{2}+m^{2}}$), one can show by chain
rule that

$\displaystyle\frac{\partial}{\partial\mathbf{p}_{i}}=\frac{\partial\omega_{\mathbf{p}}}{\partial\mathbf{p}_{i}}\frac{\partial}{\partial\omega_{\mathbf{p}}}=\frac{\mathbf{p}^{i}}{\omega_{\mathbf{p}}}\frac{\partial}{\partial\omega_{\mathbf{p}}}=-\frac{\mathbf{p}_{i}}{\omega_{\mathbf{p}}}\frac{\partial}{\partial\omega_{\mathbf{p}}}$
(228)

and we get the same contribution as before.

Given the importance of the boost for the following discussion, as a double
check of the correctness of eq. (223), we also sketch the computation of the
boost charge from scratch using Noether theorem in non-deformed QFT. The
action is

$\displaystyle\mathscr{L}=(\partial_{\mu}\phi)^{\dagger}(\partial^{\mu}\phi)-m^{2}\phi^{\dagger}\phi$
(229)

and the on-shell fields for such an action are given by

$\displaystyle\phi=\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}a_{\mathbf{p}}e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}+b_{\mathbf{p}}^{\dagger}e^{i(\omega_{p}t-\mathbf{p}\mathbf{x})}$
(230)
$\displaystyle\phi^{\dagger}=\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}a_{\mathbf{p}}^{\dagger}e^{i(\omega_{p}t-\mathbf{p}\mathbf{x})}+b_{\mathbf{p}}e^{-i(\omega_{p}t-\mathbf{p}\mathbf{x})}.$
(231)

The energy-momentum tensor is given by

$\displaystyle T_{\mu}^{\nu}$
$\displaystyle=\frac{\partial\mathscr{L}}{\partial(\partial_{\nu}\phi)}\partial_{\mu}\phi+\frac{\partial\mathscr{L}}{\partial(\partial_{\nu}\phi^{\dagger})}\partial_{\mu}\phi^{\dagger}-\delta^{\nu}_{\mu}\mathscr{L}$
(232)
$\displaystyle=\frac{1}{2}(\partial_{\mu}\phi)^{\dagger}(\partial^{\nu}\phi)+\frac{1}{2}(\partial_{\mu}\phi)(\partial^{\nu}\phi)^{\dagger}-\frac{1}{2}(\partial_{\alpha}\phi)^{\dagger}(\partial^{\alpha}\phi)+\frac{1}{2}m^{2}\phi^{\dagger}\phi.$
(233)

By Noether theorem, the charge is given by

$\displaystyle\mathscr{L}_{0i}=i\int_{\mathbb{R}^{3}}(x_{0}T^{0}_{i}-x_{i}T^{0}_{0})$
(234)

(the additional $i$ factor is added by hand, and it does not modify the
discussion) and using eq. (232) and eq. (230), (231) one can do the explicit
computations, which we will not do explicitly here. Notice however the
following subtlety. In the term $x_{i}T^{0}_{0}$, the $x_{i}$ factor can be
described as a derivative in momentum space acting on the spatial exponent,
but since this exponent will contain $\mathbf{p}-\mathbf{q}$ (the two momenta
are necessary because $T_{\mu\nu}$ contains product of fields, each of which
is an integral in momentum space, so we need two integration variables) we
have two ways of going so. This will result in two possible ways to compute
the coefficient $x_{i}T^{0}_{0}$, namely

$\displaystyle-\int_{\mathbb{R}^{3}}x_{i}T^{0}_{0}$
$\displaystyle=-\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}\frac{d^{3}q}{\sqrt{2\omega_{q}}}\Bigg{\\{}(\omega_{p}\omega_{q}+\mathbf{p}\mathbf{q}+m^{2})\times$
$\displaystyle\times\left(-ia_{\mathbf{p}}a_{\mathbf{q}}^{\dagger}e^{-i(\omega_{p}-\omega_{q})t}+ib_{\mathbf{p}}^{\dagger}b_{\mathbf{q}}e^{i(\omega_{p}-\omega_{q})t}\right)\frac{\partial}{\partial\mathbf{p}^{i}}\delta(\mathbf{p}-\mathbf{q})\Bigg{\\}}$
(235) $\displaystyle-\int_{\mathbb{R}^{3}}x_{i}T^{0}_{0}$
$\displaystyle=-\int\frac{d^{3}p}{\sqrt{2\omega_{p}}}\frac{d^{3}q}{\sqrt{2\omega_{q}}}\Bigg{\\{}(\omega_{p}\omega_{q}+\mathbf{p}\mathbf{q}+m^{2})\times$
$\displaystyle\times\left(ia_{\mathbf{p}}a_{\mathbf{q}}^{\dagger}e^{-i(\omega_{p}-\omega_{q})t}-ib_{\mathbf{p}}^{\dagger}b_{\mathbf{q}}e^{i(\omega_{p}-\omega_{q})t}\right)\frac{\partial}{\partial\mathbf{q}^{i}}\delta(\mathbf{p}-\mathbf{q})\Bigg{\\}}$
(236)

After integration by parts, in both cases one term will eliminate the
coefficient $x_{0}T^{0}_{i}$, and the final charges will be

$\displaystyle N_{i}^{(p)}=-\int
d^{3}p\,\Bigg{\\{}\omega_{p}\left(-\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}a_{\mathbf{p}}^{\dagger}+\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}\right)\Bigg{\\}}.$
(237) $\displaystyle N_{i}^{(q)}=\int
d^{3}p\,\Bigg{\\{}\omega_{p}\left(-a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}+b_{\mathbf{p}}^{\dagger}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\right)\Bigg{\\}}.$ (238)

However, this means that we can write without loss of generality that the
boost charge $N_{i}$ is given by

$\displaystyle N_{i}=\frac{1}{2}(N_{i}^{(p)}+N_{i}^{(q)})$ (239)

which is exactly eq. (223).

###### Crucial step

Notice that in the last passage of Eq. (1) we used the fact that (we consider
only the $a,a^{\dagger}$ case since the same reasoning applies to
$b,b^{\dagger}$)

$\displaystyle\frac{\partial a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}=\frac{1}{2}\delta\left(\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}^{i}}a_{\mathbf{p}}^{\dagger}\,-a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}\right)+\frac{1}{2}\frac{\partial}{\partial\mathbf{p}^{i}}\left(a_{\mathbf{p}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\right).$ (240)

The second term on the RHS of the last equation is still not a total
derivative because of the presence of $\omega_{\mathbf{p}}$ inside the
integrand. After integration by parts, this term contributes a factor

$\displaystyle-\frac{\mathbf{p}_{i}}{2\omega_{\mathbf{p}}}\left(a_{\mathbf{p}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\right)$ (241)

to the integrand, which cancels the second term in the second passage of (1).

####  2 $\kappa$-deformed case

Here we work with the full symplectic form in eq. (203). In obtaining this
equation we used the following property of the wedge product of two $1$-forms
$e$ and $f$.

$\displaystyle e\wedge f=-f\wedge e.$ (242)

In particular, this property was used when summing the contribution to the
symplectic form coming from the two different orderings of the action. This is
a non-trivial assumption in the presence of deformation, and we will discuss
it more in section 3.

##### Translation charges

The idea is basically the same, with the only exception that now we need to
consider the presence of antipodes. Because of this, the first thing to do is
to understand how the creation/annihilation operator transform under
translation.

Judging by the definition of the fields in eq. (115) and (116), we assume the
following behaviour

$\displaystyle a_{\mathbf{p}}\mapsto
e^{i\epsilon\omega_{p}}a_{\mathbf{p}}\qquad a_{\mathbf{p}}^{\dagger}\mapsto
e^{i\epsilon S(\omega_{p})}a_{\mathbf{p}}^{\dagger}\qquad
b_{\mathbf{p}}^{\dagger}\mapsto e^{i\epsilon
S(\omega_{p})}b_{\mathbf{p}}^{\dagger}\qquad b_{\mathbf{p}}\mapsto
e^{i\epsilon\omega_{p}}b_{\mathbf{p}}$ (243)

and the same for spatial momenta, or in other words

$\displaystyle\delta_{\partial_{0}}a_{\mathbf{p}}$
$\displaystyle=i\epsilon\omega_{p}a_{\mathbf{p}}\qquad\delta_{\partial_{0}}a_{\mathbf{p}}^{\dagger}=i\epsilon
S(\omega_{p})a_{\mathbf{p}}^{\dagger}$
$\displaystyle\delta_{\partial_{0}}b_{\mathbf{p}}^{\dagger}$
$\displaystyle=i\epsilon
S(\omega_{p})b_{\mathbf{p}}^{\dagger}\qquad\delta_{\partial_{0}}b_{\mathbf{p}}=i\epsilon\omega_{p}b_{\mathbf{p}}$
(244)

These transformations are correspond to the field transformation

$\displaystyle\delta_{\partial_{\mu}}\phi$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\zeta(p)\left[i\epsilon^{\mu}p_{\mu}\,a_{\mathbf{p}}\,e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}+i\epsilon^{\mu}S(p)_{\mu}b^{\dagger}_{\mathbf{p}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\right]$
$\displaystyle=-\epsilon^{\mu}\partial_{\mu}\,\phi(x)$ (245)

for the index $\mu=0$. However, for these transformations to be well defined,
then in order to ensure the invariance of the products $aa^{\dagger}$ and
$bb^{\dagger}$ we need to interpret them as a star products. This star product
is defined in the phase space and not in canonical spacetime, otherwise the
steps leading to the computation of the charges are faulted by the fact that
we cannot write $ae^{(\cdot)}\star be^{[\cdot]}=abe^{(\cdot)}\star
e^{[\cdot]}$. However, things are easily solved by defining

$\displaystyle ae^{(\cdot)}\star be^{[\cdot]}=(a\star_{PS}b)\,e^{(\cdot)}\star
e^{[\cdot]}.$ (246)

In fact, we have

$\displaystyle
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}:=a_{\mathbf{p}}\star_{PS}a_{\mathbf{p}}^{\dagger}\overset{def.}{\mapsto}(a_{\mathbf{p}}\star_{PS}a_{\mathbf{p}}^{\dagger})\cdot\left(e^{i\epsilon\omega_{p}}\star
e^{i\epsilon
S(\omega_{p})}\right)=a_{\mathbf{p}}\star_{PS}a_{\mathbf{p}}^{\dagger}$ (247)

This definition of $aa^{\dagger}$ as $a\star_{PS}a^{\dagger}$ does not
influence the creation/annihilation operators algebra or the definition of
normal ordering, so everything remains well defined.

However, there is now the issue of how to treat the contraction of a vector
field $\delta_{\xi}$ with the form $\delta a_{\mathbf{p}}\wedge\delta
a_{\mathbf{p}}^{\dagger}$. In fact, the objects inside the wedge product are
now to be treated in the deformed case.

###### Non-deformed contraction:

While we are in the context of forms, the product between the forms $\delta a$
and $\delta a^{\dagger}$ is treated using the canonical wedge product with its
properties. However, when we contract with $\delta_{\xi}$, one has canonically
that

$\displaystyle\delta_{\xi}\lrcorner\,(\delta a_{\mathbf{p}}\wedge\delta
a_{\mathbf{p}}^{\dagger})=\delta_{\xi}a_{\mathbf{p}}\delta
a_{\mathbf{p}}^{\dagger}-\delta
a_{\mathbf{p}}\delta_{\xi}a_{\mathbf{p}}^{\dagger}$ (248)

and the wedge product is not present. Notice that the change of sign in the
second term can be understood in terms of the following steps.

  * •

The vector field always acts on the first objects it encounters in the wedge
product. Because of this, to compute its action on the second term of the
wedge product, we first need to use the formula

$\displaystyle v\wedge w=(-)^{p\cdot q}w\wedge v$ (249)

where $v$ and $w$ are respectively a $p$-form and a $q$-form. In our case, we
have the wedge product of two $1$-forms, and therefore we have

$\displaystyle\delta a_{\mathbf{p}}\wedge\delta
a_{\mathbf{p}}^{\dagger}=-\delta a_{\mathbf{p}}^{\dagger}\wedge\delta
a_{\mathbf{p}}.$ (250)

  * •

Now the vector field acts on the first component, giving

$\displaystyle-(\delta_{\xi}a_{\mathbf{p}}^{\dagger})\delta a_{\mathbf{p}}.$
(251)

  * •

Now the object $\delta_{\xi}a_{\mathbf{p}}^{\dagger}$ is just some function,
i.e. a $0$-form, and therefore it commutes with $\delta a$, so its placement
is arbitrary.

###### Deformed contraction:

In this case, we define:333We write them down for $a$, but the same goes for
$b$

$\displaystyle\delta_{\xi}\lrcorner\,(\delta a_{\mathbf{p}}\wedge\delta
a_{\mathbf{p}}^{\dagger})=(\delta_{\xi}a_{\mathbf{p}})\delta
a_{\mathbf{p}}^{\dagger}+\delta
a_{\mathbf{p}}[S(\delta_{\xi})a_{\mathbf{p}}^{\dagger}].$ (252)

The main difference between this axiom and the one in eq. (248) is the
presence of the antipode instead of the normal minus sign in the second term.
Notice that, since $S(AB)=S(A)S(B)$, and since
$S(A)S(B)\xrightarrow{\kappa\rightarrow\infty}(-A)(-B)=AB$, the object
$S(\delta_{\xi})$ in eq. (252) must contain an additional minus sign when
necessary, for consistency with the non-deformed axiom. For example, if
$\delta_{\xi}=\epsilon\mathbf{p}_{i}\frac{\partial}{\partial\mathbf{p}_{i}}$
where $\epsilon$ is just some constant, then $S(\delta_{\xi})=-\epsilon
S(\mathbf{p})_{i}\frac{\partial}{\partial S(\mathbf{p})_{i}}$, which has the
correct $\kappa\rightarrow\infty$ limit.

Coming back to the charges, we first notice that we can rewrite

$\displaystyle\Omega$ $\displaystyle=i\int d^{3}p\,\alpha\,(\delta
a_{\mathbf{p}}\wedge\delta a_{\mathbf{p}}^{\dagger}\,-\delta
b_{\mathbf{p}}^{\dagger}\wedge\delta b_{\mathbf{p}})$ (253)
$\displaystyle=-i\int d^{3}p\,\alpha\,(\delta
a_{\mathbf{p}}^{\dagger}\wedge\delta a_{\mathbf{p}}\,-\delta
b_{\mathbf{p}}\wedge\delta b_{\mathbf{p}}^{\dagger}).$ (254)

Then, contracting with the vector field $\delta_{\partial_{\mu}}$, using the
assumption in eq. (252), and using eq. (2) we finally get

$\displaystyle-\delta_{\partial_{0}}\lrcorner\,\Omega$ $\displaystyle=i\int
d^{3}p\,\alpha\,(\delta_{\partial_{0}}a_{\mathbf{p}}^{\dagger}\delta
a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}^{\dagger}S(\delta_{\partial_{0}})a_{\mathbf{p}}-\delta_{\partial_{0}}b_{\mathbf{p}}\delta
b_{\mathbf{p}}^{\dagger}-\delta
b_{\mathbf{p}}S(\delta_{\partial_{0}})b_{\mathbf{p}}^{\dagger})$ (255)
$\displaystyle=i\int d^{3}p\,\alpha\,[i\epsilon
S(\omega_{p})a_{\mathbf{p}}^{\dagger}\delta a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}^{\dagger}(i\epsilon S(\omega_{p})a_{\mathbf{p}})$
$\displaystyle-(i\epsilon\omega_{p}b_{\mathbf{p}}^{\dagger})\delta
b_{\mathbf{p}}-\delta
b_{\mathbf{p}}^{\dagger}i\epsilon\omega_{p}b_{\mathbf{p}}]$ (256)
$\displaystyle=-\epsilon\int
d^{3}p\,\alpha\,[S(\omega_{p})a_{\mathbf{p}}^{\dagger}\delta
a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}^{\dagger}S(\omega_{p})a_{\mathbf{p}}-\omega_{p}b_{\mathbf{p}}^{\dagger}\delta
b_{\mathbf{p}}-\delta b_{\mathbf{p}}^{\dagger}\omega_{p}b_{\mathbf{p}}]$ (257)
$\displaystyle=\epsilon\delta\int
d^{3}p\,\alpha\,[-S(\omega_{p})a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\,+\omega_{p}b_{\mathbf{p}}^{\dagger}b_{\mathbf{p}}]$
(258)

which, apart from the irrelevant $\epsilon$ factor, gives the correct time
translation charge (of course, in order to compare the final expression for
the charges, normal ordering is always assumed. As stated before, the normal
ordering is not affected by whether we consider the product of
creation/annihilation operators as $aa^{\dagger}$ or $a\star a^{\dagger}$.).
Notice also that we have used the fact that $\delta(AB)=(\delta A)B+A\delta
B$. Of course, the same reasoning gives the correct space translation charges,
since the procedure works in exactly the same way. We get therefore the
following translation charges

$\displaystyle\boxed{\mathscr{P}_{0}=\int
d^{3}p\,\,\alpha(p)\left\\{-S(\omega_{p})a_{\mathbf{p}}^{\dagger}\,a_{\mathbf{p}}\,+\omega_{p}b_{\mathbf{p}}\,b_{\mathbf{p}}^{\dagger}\,\right\\}}$
(259) $\displaystyle\boxed{\mathscr{P}_{i}=\int
d^{3}p\,\,\alpha(p)\left\\{-S(\mathbf{p})_{i}a_{\mathbf{p}}^{\dagger}\,a_{\mathbf{p}}\,+\mathbf{p}_{i}b_{\mathbf{p}}\,b_{\mathbf{p}}^{\dagger}\right\\}}$
(260)

Notice that by construction we have the formal limit ($\zeta$ goes to $1$ for
$\kappa\rightarrow\infty$ by definition)

$\displaystyle\lim_{\kappa\rightarrow\infty}\alpha(p)=1.$ (261)

Furthermore, since $\omega_{p}>0$ and $-S(\omega_{p})>0$, the energy has the
correct sign, and in the limit $\kappa\rightarrow\infty$ we get back the
correct translation charges of non-deformed complex scalar field. As a final
comment, in [78] the same charges were calculated using a direct approach
based on the Noether theorem. However, it became clear while writing this
thesis that the direct approach presented in [78] only reproduces some of the
terms of the full charges computed here (the charges in [78] are however still
correct, in light of the geometric approach presented here). The source of
this incongruence is an interesting topic, deserving of further investigation,
and it will be tackled in future publications.

##### Notes on the above (overall) procedure

There are several things to notice about the above procedure.

  * 1)

The most obvious thing to address is that there seems to be no reason as to
why we chose to write

$\displaystyle\Omega$ $\displaystyle=-i\int d^{3}p\,\alpha\,(\delta
a_{\mathbf{p}}^{\dagger}\wedge\delta a_{\mathbf{p}}\,-\delta
b_{\mathbf{p}}\wedge\delta b_{\mathbf{p}}^{\dagger})$ (262)

and not for example

$\displaystyle\Omega=-i\int d^{3}p\,\alpha\,(\delta
a_{\mathbf{p}}^{\dagger}\wedge\delta a_{\mathbf{p}}\,+\delta
b_{\mathbf{p}}^{\dagger}\wedge\delta b_{\mathbf{p}})$ (263)

which seems at least as good. The only sensible answer to this is to look at
the single symplectic forms $\delta\theta_{1}$ and $\delta\theta_{2}$, we have

$\displaystyle\delta\theta_{1}\propto\delta\phi^{\dagger}\wedge\delta\phi\qquad\delta\theta_{2}\propto\delta\phi\wedge\delta\phi^{\dagger}$
(264)

In both cases, the off diagonal terms go away and we are only left with the
wedge products between $a,a^{\dagger}$ and between $b,b^{\dagger}$. However,
both in $\delta\theta_{1}$ and $\delta\theta_{2}$, since $\phi$ contains $a$
and $b^{\dagger}$ (and of course $\phi^{\dagger}$ contains $a^{\dagger}$ and
$b$), each time that we have $\delta a\wedge\delta a^{\dagger}$ we also have
$\delta b^{\dagger}\wedge\delta b$. Therefore, the correct final symplectic
charge must be of the kind in eq. (262) and not the one in eq. (263).

Of course, in non-deformed QFT, either (262) or (263) are equally fine, but in
that case there is no antipode in the formula for the contraction of the
symplectic form and an external vector field $\delta_{\xi}$, while there is
one in the deformed case in (252).

Therefore, because of the presence of the antipode in (252), it seems that the
structure of the symplectic form in the deformed case is more rigid than in
the non-deformed case, because although we are free to chose the overall sign,
we cannot arbitrarily change the order of only a part of the symplectic form
before contracting with $\delta_{\xi}$.

  * 2)

Having tackled the first important point, then there is the issue of why we
chose the overall sign as in eq. (254) and not the one in (253). The answer is
that one can go from (253) to (254) by means of the following exchange

$\displaystyle a\leftrightarrow b\qquad a^{\dagger}\leftrightarrow
b^{\dagger}$ (265)

so that, performing the same exact computations, instead of the charge

$\displaystyle Q_{1}=\int
d^{3}p\,\alpha\,[-S(\omega_{p})a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\,+\omega_{p}b_{\mathbf{p}}^{\dagger}b_{\mathbf{p}}]$
(266)

we would end up with the charge

$\displaystyle Q_{2}=\int
d^{3}p\,\alpha\,[-S(\omega_{p})b_{\mathbf{p}}^{\dagger}b_{\mathbf{p}}+\omega_{p}a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\,].$
(267)

However, for all practical purposes, the charge $Q_{2}$ is the same as $Q_{1}$
because both $a,a^{\dagger}$ and $b,b^{\dagger}$ live on the same manifold.
The important thing is that one of them has energy $\omega_{p}$ and the other
$-S(\omega_{p})$ while living on the same manifold, the name of the operators
are not important in this context.

  * 3)

One last thing to notice is that the assumption (252) is not in contradiction
with the properties of the differential operator
$d_{F}=\epsilon^{A}\partial_{A}+\omega_{\mu\nu}L_{\mu\nu}$ that we used in the
computation of the charges (at leat at first sight). In fact, while $d_{F}$ is
the exterior derivative in spacetime, we are dealing with (the properties of)
the exterior derivative $\delta$ in phase space, which is however the space of
solutions to the EoM. Therefore, the behaviour of $\delta$ can be different
than the one of $d_{F}$.

##### Boosts

Analogously to what has been done before, we can now concentrate on the boost
charges. Once again, we need to understand what is the action of
$\delta_{\Lambda}$ on $a,a^{\dagger},b,b^{\dagger}$.

One approach would be to start from the assumed Lorentz transformations

$\displaystyle\delta^{B}\phi(x)$
$\displaystyle=i\lambda_{i}\,x^{i}\frac{\partial}{\partial t}\,\phi(x)$
$\displaystyle=-i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\,a_{\mathbf{p}}\,\omega_{\mathbf{p}}\frac{\partial}{\partial\mathbf{p}_{i}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle-i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\zeta(p)\,b^{\dagger}_{\mathbf{p}}\,S(\omega_{\mathbf{p}})\frac{\partial}{\partial
S(\mathbf{p})_{i}}\,e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}.$
(268)

and integrating by parts obtaining

$\displaystyle\delta^{B}\phi(x)=i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\omega_{\mathbf{p}}\,\zeta(p)\,\left[\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}_{i}}+a_{\mathbf{p}}\frac{\sqrt{2\omega_{\mathbf{p}}}}{\zeta(p)\omega_{p}}\frac{\partial}{\partial\mathbf{p}_{i}}\left(\frac{\omega_{\mathbf{p}}}{\sqrt{2\omega_{\mathbf{p}}}}\zeta(p)\right)\right]e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle+i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,S(\omega_{\mathbf{p}})\,\zeta(p)\,\left[\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})_{i}}+b_{\mathbf{p}}^{\dagger}\frac{\sqrt{2\omega_{\mathbf{p}}}}{\zeta(p)S(\omega_{p})}\frac{\partial}{\partial
S(\mathbf{p})_{i}}\left(\frac{S(\omega_{\mathbf{p}})}{\sqrt{2\omega_{\mathbf{p}}}}\zeta(p)\right)\right]e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}.$
$\displaystyle\delta^{B}\phi(x)=i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\,\zeta(p)\,\Bigg{\\{}\omega_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}}{\partial\mathbf{p}_{i}}+a_{\mathbf{p}}\frac{\sqrt{2\omega_{\mathbf{p}}}}{\zeta(p)}\frac{\partial}{\partial\mathbf{p}_{i}}\left(\frac{\omega_{\mathbf{p}}}{\sqrt{2\omega_{\mathbf{p}}}}\zeta(p)\right)\Bigg{\\}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle+i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\,\zeta(p)\,\Bigg{\\{}S(\omega_{\mathbf{p}})\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})_{i}}+b_{\mathbf{p}}^{\dagger}\frac{\sqrt{2\omega_{\mathbf{p}}}}{\zeta(p)}\frac{\partial}{\partial
S(\mathbf{p})_{i}}\left(\frac{S(\omega_{\mathbf{p}})}{\sqrt{2\omega_{\mathbf{p}}}}\zeta(p)\right)\Bigg{\\}}e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}.$

Therefore, limiting ourselves to $\phi$ since $\phi^{\dagger}$ is analogous,
we would have the following relations

$\displaystyle\delta^{B}a_{\mathbf{p}}$
$\displaystyle=-i\lambda^{i}\,\omega_{p}\left[\frac{\partial}{\partial\mathbf{p}^{i}}+\frac{1}{\zeta(p)\sqrt{\omega_{\mathbf{p}}}}\frac{\partial}{\partial\mathbf{p}^{i}}(\zeta(p)\sqrt{\omega_{\mathbf{p}}})\right]a_{\mathbf{p}}$
(269) $\displaystyle\delta^{B}a_{\mathbf{p}}^{\dagger}$
$\displaystyle=-i\lambda^{i}\,S(\omega_{p})\left[\frac{\partial}{\partial
S(\mathbf{p})^{i}}+\frac{\sqrt{\omega_{\mathbf{p}}}}{\zeta(p)S(\omega_{p})}\frac{\partial}{\partial
S(\mathbf{p})^{i}}\left(\frac{S(\omega_{\mathbf{p}})}{\sqrt{\omega_{\mathbf{p}}}}\zeta(p)\right)\right]a_{\mathbf{p}}^{\dagger}$
(270) $\displaystyle\delta^{B}b_{\mathbf{p}}$
$\displaystyle=-i\lambda^{i}\,\omega_{p}\left[\frac{\partial}{\partial\mathbf{p}^{i}}+\frac{1}{\zeta(p)\sqrt{\omega_{\mathbf{p}}}}\frac{\partial}{\partial\mathbf{p}^{i}}(\zeta(p)\sqrt{\omega_{\mathbf{p}}})\right]b_{\mathbf{p}}$
(271) $\displaystyle\delta^{B}b_{\mathbf{p}}^{\dagger}$
$\displaystyle=-i\lambda^{i}\,S(\omega_{p})\left[\frac{\partial}{\partial
S(\mathbf{p})^{i}}+\frac{\sqrt{\omega_{\mathbf{p}}}}{\zeta(p)S(\omega_{p})}\frac{\partial}{\partial
S(\mathbf{p})^{i}}\left(\frac{S(\omega_{\mathbf{p}})}{\sqrt{\omega_{\mathbf{p}}}}\zeta(p)\right)\right]b_{\mathbf{p}}^{\dagger}$
(272)

We will however now see that the above transformations need to be modified
into

$\displaystyle\delta^{B}a_{\mathbf{p}}$
$\displaystyle=-i\lambda^{i}\,\omega_{\mathbf{p}}\left[\frac{\partial}{\partial\mathbf{p}^{i}}+\frac{1}{2}\frac{1}{\omega_{\mathbf{p}}}\frac{\partial[\omega_{\mathbf{p}}S(\alpha(p))]}{\partial\mathbf{p}^{i}}\right]a_{\mathbf{p}}$
(273) $\displaystyle\delta^{B}a_{\mathbf{p}}^{\dagger}$
$\displaystyle=-i\lambda^{i}\,S(\omega_{\mathbf{p}})\left[\frac{\partial}{\partial
S(\mathbf{p})^{i}}+\frac{1}{2}\frac{1}{S(\omega_{\mathbf{p}})}\frac{\partial[S(\omega_{\mathbf{p}})\alpha(p)]}{\partial
S(\mathbf{p})^{i}}\right]a_{\mathbf{p}}^{\dagger}$ (274)
$\displaystyle\delta^{B}b_{\mathbf{p}}$
$\displaystyle=-i\lambda^{i}\,\omega_{\mathbf{p}}\left[\frac{\partial}{\partial\mathbf{p}^{i}}+\frac{1}{2}\frac{1}{\omega_{\mathbf{p}}}\frac{\partial[\omega_{\mathbf{p}}\alpha(p)]}{\partial\mathbf{p}^{i}}\right]b_{\mathbf{p}}$
(275) $\displaystyle\delta^{B}b_{\mathbf{p}}^{\dagger}$
$\displaystyle=-i\lambda^{i}\,S(\omega_{\mathbf{p}})\left[\frac{\partial}{\partial
S(\mathbf{p})^{i}}+\frac{1}{2}\frac{1}{S(\omega_{\mathbf{p}})}\frac{\partial[S(\omega_{\mathbf{p}})S(\alpha(p))]}{\partial
S(\mathbf{p})^{i}}\right]b_{\mathbf{p}}^{\dagger}$ (276)

in order to obtain a conserved charge. The reasoning here is as follows. We
know that the action is invariant under $\kappa$-Poincaré transformations,
which means that by Noether theorem there are conserved charges. More in
detail, there is a one-to-one correspondence between symmetries and conserved
charges. One usually goes in one direction, obtaining the conserved charges as
a result of exploiting the symmetries of the action. Here we go in the other
direction, obtaining the charges, and then defining the symmetry of the system
by showing that charges are conserved.

Notice that in this whole process, there is no way to know a priori the
transformation laws of the fields under a symmetry of the action. A symmetry
of the action does not necessarily translate to a symmetry of the field inside
the action; quite the opposite, the fields change under some transformation,
and the action is such combination of fields and their derivatives which is
invariant. Eq. (2) represents a canonical field transformation under non-
deformed boost; it is therefore not surprising that it does not give rise to a
charge since it is not a symmetry of our system.

We now first show that eq. (273), (274), (275), (276) give rise to a conserved
charge, and then we will show to what field transformation they correspond to.
We have

$\displaystyle-$ $\displaystyle\delta^{B}\lrcorner\,\Omega=-i\int
d^{3}p\,\alpha(p)\left(\delta^{B}a_{\mathbf{p}}^{\dagger}\delta
a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}^{\dagger}S(\delta^{B})a_{\mathbf{p}}-\delta^{B}b_{\mathbf{p}}\delta
b_{\mathbf{p}}^{\dagger}-\delta
b_{\mathbf{p}}S(\delta^{B})b_{\mathbf{p}}^{\dagger}\right)$
$\displaystyle=-\lambda^{i}\,\int
d^{3}p\,\alpha(p)\left(S(\omega_{p})\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}\delta
a_{\mathbf{p}}\,-\delta a_{\mathbf{p}}^{\dagger}\,S(\omega_{p})\frac{\partial
a_{\mathbf{p}}}{\partial S(\mathbf{p})^{i}}-\omega_{p}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\delta b_{\mathbf{p}}^{\dagger}+\delta
b_{\mathbf{p}}\omega_{p}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}\right)$
$\displaystyle+\lambda^{i}\,\int
d^{3}p\frac{1}{2}\frac{\partial[S(\omega_{\mathbf{p}})\alpha(p)]}{\partial
S(\mathbf{p})^{i}}\left(a_{\mathbf{p}}\delta a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\right)-\frac{1}{2}\frac{\partial[\omega_{\mathbf{p}}\alpha(p)]}{\partial\mathbf{p}^{i}}\left(b_{\mathbf{p}}\delta
b_{\mathbf{p}}^{\dagger}-\delta b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}\right)$
(277) $\displaystyle=-\frac{1}{2}{\lambda}^{i}\delta\int
d^{3}p\,\alpha(p)\left\\{S(\omega_{p})\left[\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}\,-a_{\mathbf{p}}^{\dagger}\,\frac{\partial
a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}}\right]+\omega_{p}\left[b_{\mathbf{p}}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}-\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}^{\dagger}\right]\right\\}$
(278)

which gives the boost charge

$\displaystyle\boxed{{\cal N}_{i}=-\frac{1}{2}\int
d^{3}p\,\alpha(p)\left\\{S(\omega_{p})\left[\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}\,-a_{\mathbf{p}}^{\dagger}\,\frac{\partial
a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}}\right]+\omega_{p}\left[b_{\mathbf{p}}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}-\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}^{\dagger}\right]\right\\}}.$
(279)

###### Crucial step

Once again, in order to be able to take the exterior derivative outside, we
used the fact that

$\displaystyle\frac{\partial a_{\mathbf{p}}}{\partial S(\mathbf{p})^{i}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}$
$\displaystyle=\delta\left(\frac{\partial a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}^{\dagger}\,-a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}\right)$
$\displaystyle-\frac{\partial\delta a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}^{\dagger}\,+a_{\mathbf{p}}\frac{\partial\delta
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}$ (280)

and

$\displaystyle\frac{\partial a_{\mathbf{p}}}{\partial S(\mathbf{p})^{i}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}$
$\displaystyle=\frac{\partial}{\partial
S(\mathbf{p})^{i}}\left(a_{\mathbf{p}}\delta a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\right)$
$\displaystyle+\frac{\partial\delta a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}^{\dagger}\,-a_{\mathbf{p}}\frac{\partial\delta
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}$ (281)

and therefore, adding (2) and (2) we finally get

$\displaystyle\frac{\partial a_{\mathbf{p}}}{\partial S(\mathbf{p})^{i}}\delta
a_{\mathbf{p}}^{\dagger}\,-\delta a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}$
$\displaystyle=\frac{1}{2}\delta\left(\frac{\partial a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}^{\dagger}\,-a_{\mathbf{p}}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{i}}\right)$
$\displaystyle+\frac{1}{2}\frac{\partial}{\partial
S(\mathbf{p})^{i}}\left(a_{\mathbf{p}}\delta a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\right).$ (282)

Of course, the same exact reasoning works for the $b,b^{\dagger}$ operators.
Here, notice that after integration by parts the second term produces the
following contribution to $-\delta_{\partial_{\Lambda}}\,\lrcorner\,\Omega$

$\displaystyle-\frac{1}{2}\frac{\partial[S(\omega_{\mathbf{p}})\alpha(p)]}{\partial
S(\mathbf{p})^{i}}\left(a_{\mathbf{p}}\delta a_{\mathbf{p}}^{\dagger}\,-\delta
a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}\right)+\frac{1}{2}\frac{\partial[\omega_{\mathbf{p}}\alpha(p)]}{\partial\mathbf{p}^{i}}\left(b_{\mathbf{p}}\delta
b_{\mathbf{p}}^{\dagger}-\delta
b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}\right),$ (283)

and once again it eliminates the relevant term in the computation of the boost
charge.

We can now turn to the transformation of the field due to eq. (273), (274),
(275), (276). We have

$\displaystyle\delta^{B}\phi(x)=i\lambda_{i}\,x^{i}\frac{\partial}{\partial
t}\,\phi(x)$
$\displaystyle-i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\,\zeta\,\Bigg{\\{}\omega_{\mathbf{p}}\left[\frac{1}{2}\frac{1}{\omega_{\mathbf{p}}}\frac{\partial[\omega_{\mathbf{p}}S(\alpha)]}{\partial\mathbf{p}^{i}}-\frac{\sqrt{\omega_{\mathbf{p}}}}{\zeta\omega_{p}}\frac{\partial}{\partial\mathbf{p}^{i}}\left(\frac{\omega_{\mathbf{p}}}{\sqrt{\omega_{\mathbf{p}}}}\zeta\right)\right]a_{\mathbf{p}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle-S(\omega_{\mathbf{p}})\left[\frac{1}{2}\frac{1}{S(\omega_{\mathbf{p}})}\frac{\partial[S(\omega_{\mathbf{p}})S(\alpha)]}{\partial
S(\mathbf{p})^{i}}-\frac{\sqrt{\omega_{\mathbf{p}}}}{\zeta
S(\omega_{p})}\frac{\partial}{\partial
S(\mathbf{p})^{i}}\left(\frac{S(\omega_{\mathbf{p}})}{\sqrt{\omega_{\mathbf{p}}}}\zeta\right)\right]b_{\mathbf{p}}^{\dagger}e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\Bigg{\\}}$
(284)
$\displaystyle\delta^{B}\phi^{\dagger}(x)=-i\lambda_{i}\,x^{i}\frac{\partial}{\partial
t}\,\phi^{\dagger}(x)$
$\displaystyle-i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\,\zeta\,\Bigg{\\{}\omega_{\mathbf{p}}\left[\frac{1}{2}\frac{1}{\omega_{\mathbf{p}}}\frac{\partial[\omega_{\mathbf{p}}\alpha]}{\partial\mathbf{p}^{i}}-\frac{\sqrt{\omega_{\mathbf{p}}}}{\zeta\omega_{p}}\frac{\partial}{\partial\mathbf{p}^{i}}\left(\frac{\omega_{\mathbf{p}}}{\sqrt{\omega_{\mathbf{p}}}}\zeta\right)\right]b_{\mathbf{p}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle-S(\omega_{\mathbf{p}})\left[\frac{1}{2}\frac{1}{S(\omega_{\mathbf{p}})}\frac{\partial[S(\omega_{\mathbf{p}})\alpha]}{\partial
S(\mathbf{p})^{i}}-\frac{\sqrt{\omega_{\mathbf{p}}}}{\zeta
S(\omega_{p})}\frac{\partial}{\partial
S(\mathbf{p})^{i}}\left(\frac{S(\omega_{\mathbf{p}})}{\sqrt{\omega_{\mathbf{p}}}}\zeta\right)\right]a_{\mathbf{p}}^{\dagger}e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\Bigg{\\}}$
(285)

To understand these additional terms, let us use eq. (202), and let us assume
that we want to impose

$\displaystyle\xi(p)\,\,:\,\,\alpha(p)=1.$ (286)

This is a natural choice, since it eliminates this factor from the charges. In
this case, we have

$\displaystyle\zeta(p)^{-2}=\frac{1}{2\omega_{\mathbf{p}}}\,\frac{p_{4}}{\kappa}\,\left[\frac{|p_{+}|^{3}}{\kappa^{3}}\omega_{\mathbf{p}}-S(\omega_{\mathbf{p}})\right]$
(287)

and one can show that, to the leading order in $\frac{1}{\kappa}$, the boost
transformation becomes

$\displaystyle\delta^{B}\phi(x)$
$\displaystyle=i\lambda_{i}\,x^{i}\frac{\partial}{\partial
t}\,\phi(x)+i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\Bigg{\\{}\frac{\mathbf{p}_{i}}{\kappa}\left(\frac{m^{2}}{\omega_{\mathbf{p}}^{2}}-2\right)a_{\mathbf{p}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle+\frac{\mathbf{p}_{i}}{\kappa}\left(\frac{5}{2}-\frac{m^{2}}{2\omega_{\mathbf{p}}^{2}}\right)b_{\mathbf{p}}^{\dagger}e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\Bigg{\\}}$
(288) $\displaystyle\delta^{B}\phi^{\dagger}(x)$
$\displaystyle=-i\lambda_{i}\,x^{i}\frac{\partial}{\partial
t}\,\phi(x)^{\dagger}+i\lambda^{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\Bigg{\\{}\frac{\mathbf{p}_{i}}{\kappa}\left(\frac{m^{2}}{\omega_{\mathbf{p}}^{2}}-2\right)b_{\mathbf{p}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle+\frac{\mathbf{p}_{i}}{\kappa}\left(\frac{5}{2}-\frac{m^{2}}{2\omega_{\mathbf{p}}^{2}}\right)a_{\mathbf{p}}^{\dagger}e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\Bigg{\\}}$
(289)

We see that particles and antiparticles, upon boost, get an additional
translation. Furthermore, notice also that using the convention in eq. (287),
this additional translation is the same for both $a^{\dagger}$ and
$b^{\dagger}$. This is already apparent looking at eq. (274), (276), since if
$\alpha=1$ the quantities $\alpha$ and $S(\alpha)$ disappear from the
formulae444The antipode does not act on objects independent of momenta, so
that $S(1)=1$.. If we used instead the convention

$\displaystyle\zeta(p)\,\,:\,\,\zeta^{2}(p)\left[1+\frac{|p_{+}|^{3}}{\kappa}\right]\frac{\kappa}{p_{4}}=1$
(290)

then one gets different translations for $a^{\dagger}$ and $b^{\dagger}$, as
is again apparent from eq. (274), (276). In this case, in fact, we would have
$S(\alpha)\neq\alpha$.

Repeating the same calculations as before, in this case one gets the following
boost transformation for the fields.

$\displaystyle\delta^{B}\phi(x)$
$\displaystyle=i\lambda_{i}\,x^{i}\frac{\partial}{\partial
t}\,\phi(x)+i\lambda_{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\Bigg{\\{}\frac{\mathbf{p}_{i}}{\kappa}\left(-\frac{19}{16\sqrt{2}}\right)a_{\mathbf{p}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle+\frac{\mathbf{p}_{i}}{\kappa}\left[\frac{3}{16\sqrt{2}}\left(9+\frac{4m^{2}}{\omega_{\mathbf{p}}^{2}}\right)\right]b_{\mathbf{p}}^{\dagger}e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\Bigg{\\}}$
(291) $\displaystyle\delta^{B}\phi^{\dagger}(x)$
$\displaystyle=-i\lambda_{i}\,x^{i}\frac{\partial}{\partial
t}\,\phi(x)^{\dagger}+i\lambda^{i}\,\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\,\Bigg{\\{}\frac{\mathbf{p}_{i}}{\kappa}\left(-\frac{11}{16\sqrt{2}}\right)b_{\mathbf{p}}e^{-i(\omega_{\mathbf{p}}t-\mathbf{p}\mathbf{x})}$
$\displaystyle+\frac{\mathbf{p}_{i}}{\kappa}\left(\frac{19}{16\sqrt{2}}+\frac{3m^{2}}{4\sqrt{2}\omega_{\mathbf{p}}^{2}}\right)a_{\mathbf{p}}^{\dagger}e^{-i(S(\omega_{\mathbf{p}})t-S(\mathbf{p})\mathbf{x})}\Bigg{\\}}.$
(292)

In this case, the additional translations for $a^{\dagger}$ and $b^{\dagger}$
are different.

###### The (very) peculiar features of boosts

We already saw in eq. (143) that a single particle state gets boosted
canonically. This is not just an assumption, since it can be proven by using
the algebra of the charges. Anticipating a little the results of section 15,
we will show that all the charges that we compute in this section satisfy the
canonical, non-deformed Poincaré algebra. This means in particular that
$[\mathscr{N}_{i},\mathscr{P}_{j}]=-i\eta_{ij}\mathscr{P}_{0}$. We now define
single-particle states as usual, namely given the vacuum state $|0\rangle$ we
have

$\displaystyle a^{\dagger}_{\mathbf{p}}|0\rangle=|\mathbf{p}\rangle_{a}\qquad
b^{\dagger}_{\mathbf{p}}|0\rangle=|\mathbf{p}\rangle_{b}.$ (293)

Since we already computed the translation charges (see eq. (259), (260)), we
can use them to obtain what is the eigenvalue of the above momentum
eigenstates. For example, recalling eq. (208), (209), we have

$\displaystyle\mathscr{P}_{0}|\mathbf{q}\rangle_{a}$
$\displaystyle=\int\,d^{3}p\,\alpha[-S(\omega_{\mathbf{p}})a^{\dagger}_{\mathbf{p}}a_{\mathbf{p}}]a^{\dagger}_{\mathbf{q}}|0\rangle$
$\displaystyle=\int\,d^{3}p\,\alpha[-S(\omega_{\mathbf{p}})\frac{1}{\alpha}\delta(\mathbf{p}-\mathbf{q})]a^{\dagger}_{\mathbf{p}}|0\rangle$
$\displaystyle=-S(\omega_{\mathbf{p}})|\mathbf{p}\rangle_{a}$ (294)

and in the same way one obtains

$\displaystyle\mathscr{P}_{i}|\mathbf{p}\rangle_{a}=-S(\mathbf{p})_{i}|\mathbf{p}\rangle_{a}\qquad\mathscr{P}_{i}|\mathbf{p}\rangle_{b}=\mathbf{p}_{i}|\mathbf{p}\rangle_{b}$
(295)
$\displaystyle\mathscr{P}_{0}|\mathbf{p}\rangle_{a}=-S(\omega_{\mathbf{p}})|\mathbf{p}\rangle_{a}\qquad\mathscr{P}_{0}|\mathbf{p}\rangle_{b}=\omega_{\mathbf{p}}|\mathbf{p}\rangle_{b}.$
(296)

We can now use the commutation relation
$[\mathscr{N}_{i},\mathscr{P}_{j}]=-i\eta_{ij}\mathscr{P}_{0}$ to show that

$\displaystyle-i\lambda^{j}\mathscr{N}_{j}\,\mathscr{P}_{i}|\mathbf{p}\rangle_{a}$
$\displaystyle=iS(p)_{i}\lambda^{j}\mathscr{N}_{j}|\mathbf{p}\rangle_{a}=-i\mathscr{P}_{i}\,\lambda^{j}\mathscr{N}_{j}|\mathbf{p}\rangle_{a}+\lambda_{i}S(\omega_{\mathbf{p}})|\mathbf{p}\rangle_{a}$
(297)
$\displaystyle-i\lambda^{j}\mathscr{N}_{j}\,\mathscr{P}_{0}|\mathbf{p}\rangle_{a}$
$\displaystyle=iS(\omega_{\mathbf{p}})\lambda^{j}\mathscr{N}_{j}|\mathbf{p}\rangle_{a}=-i\mathscr{P}_{0}\,\lambda^{j}\mathscr{N}_{j}|\mathbf{p}\rangle_{a}+\lambda^{i}S(p)_{i}|\mathbf{p}\rangle_{a}$
(298)

which immediately imply

$-i\lambda^{j}\mathscr{N}_{j}|\mathbf{p}\rangle_{a}=|\mathbf{p}+\bm{\lambda}\omega_{\mathbf{p}}\rangle_{a}$
(299)

which is the canonical action of an infinitesimal boosts. One can then obtain
the finite boost, which is therefore the canonical one.

If we now define a wave packet using the fields in eq. (115), (116) (we can
concentrate on particles created by $a^{\dagger}$, since the same
considerations will be valid about those created by $b^{\dagger}$), we have

$\displaystyle\phi^{\dagger}|0\rangle$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\zeta(p)\,a^{\dagger}_{\mathbf{p}}e^{-i(S(\omega_{\mathbf{p}})-S(\mathbf{p})\mathbf{x})}|0\rangle$
$\displaystyle=\int\frac{d^{3}p}{\sqrt{2\omega_{\mathbf{p}}}}\,\zeta(p)e^{-i(S(\omega_{\mathbf{p}})-S(\mathbf{p})\mathbf{x})}|\mathbf{p}\rangle$
$\displaystyle=|\varphi(x)\rangle$ (300)

and acting now with a boost, since single particle states behave canonically,
we will get a canonically boosted single particle wave packet. The $\varphi$
in this equation describes the packet obtained by an appropriate definition of
$\zeta(p)$. On the other hand, if we first boost the field (obtaining eq. (2)
or (2), or any other expression based on different choices of $\zeta(p)$), and
only then apply the boosted field to the vacuum state, we get a wave packet
with a different distribution due to the presence of the additional factors in
eq. (2), (2).

This is a puzzling property of boosts which needs to be more deeply
understood. One of the possible solutions would be to understand whether there
are differences between active and passive boosts. Indeed, one can consider
the boost acting on the state $|\varphi(x)\rangle$ as a passive boost (the
wave packet has already been created, and only later a boost is applied, which
changes the coordinates describing the distribution). On the other hand, eq.
(2), (2) may be considered as active boosts (the field is boosted before any
particle is created). In a passive boost, the particle does not need to
exchange energy with the environment, since we (the observers) are moving at a
different speed, describing what we see from a different point of view. On the
other hand, to physically create an active boost on a field one needs to
interact with it in some way, for example an electron can be actively boosted
by switching on an electric field in order to accelerate it, and then
switching it off, obtaining an electron in uniform rectilinear motion. In
canonical, non-deformed QFT, the net result is the same and indeed active and
passive transformations are completely equivalent. It is however possible that
in the deformed context there may be different contributions in the two cases,
leading to the difference just described. However in this context we are only
considering free particles, so that such considerations would be premature. We
will leave the investigation of this puzzle to forthcoming publications.

###### The role of $\zeta(p)$

Notice that in order to get a better idea for the boost of a field in the
deformed context we needed to assume a specific value for $\zeta(p)$. In the
above discussions, we used two natural definitions, namely eq. (287) and
(290). At first sight, therefore, it would seem that the physical results of
our model depend on our choice of the arbitrary factor $\zeta(p)$, which is of
course absurd.

This is however not the case. Notice, in fact, that the factor $\zeta(p)$ can
be found inside $\alpha$ in eq. (202), which is in turn present in the charges
(up to now we only have the translation charges in eq. (259), (260), and the
boost charge in eq. (279), but we will see that it also appears in the
rotation charge), symplectic form (see eq. (203)), and creation/annihilation
operators algebra (eq. (208), (209)). However, it also appears in the off-
shell action in momentum space in eq. (156). This means that if we sue the
convention for $\zeta(p)$ in eq. (287), then the generators of the algebra are
much simpler, but of course the off-shell action now has a global prefactor in
the integrand given by

$\displaystyle\left(1+\frac{|p_{+}|^{3}}{\kappa^{3}}\right)\left\\{\frac{1}{2\omega_{p}}\left[\frac{|p_{+}|^{3}}{\kappa^{3}}\omega_{p}-S(\omega_{p})\right]\frac{p_{4}}{\kappa}\right\\}^{-1}.$
(301)

On the other hand, with convention (290), the global prefactor in the
momentum-space off-shell action in eq. (156) is 1, but the charges now contain
a global prefactor in the integrand

$\displaystyle\left(1+\frac{|p_{+}|^{3}}{\kappa^{3}}\right)^{-1}\frac{1}{2\omega_{p}}\left[\frac{|p_{+}|^{3}}{\kappa^{3}}\omega_{p}-S(\omega_{p})\right]\frac{p_{4}}{\kappa}.$
(302)

Therefore, there is a do ut des scenario between the algebra generators and
the off-shell momentum-space action, which is important because, as we will
see in chapter 2, it is used in the definition of the propagator by using the
path integral in the $\kappa$-deformed context.

In other words, if we are in the situation described by eq. (287), then after
a boost particles and antiparticles receive the same additional translation,
but their propagator is modified by some non-trivial prefactor. On the other
hand, we can choose to work with convention given by eq. (290), in which case
after a boost particles and antiparticles receive a different additional
translation, but the propagator will be the traditional one. One may very well
choose another convention, in which case the behaviour of particles and
antiparticles under boosts and their propagator have mixed features, in
between the two extremal case discussed above.

##### Rotations

For rotations, we can proceed in the same manner as for boost, and we only
need to understand how to write $\delta_{R}a$ and $\delta_{R}a^{\dagger}$ (the
case of $b$ and $b^{\dagger}$ will be analogous). We start from

$\displaystyle\delta^{R}a_{\mathbf{p}}$
$\displaystyle=i\epsilon^{ijk}\rho_{i}\mathbf{p}_{j}\frac{\partial}{\partial\mathbf{p}^{k}}a_{\mathbf{p}}\qquad\delta^{R}a_{\mathbf{p}}^{\dagger}=i\epsilon^{ijk}\rho_{i}S(\mathbf{p})_{j}\frac{\partial}{\partial
S(\mathbf{p})^{k}}a_{\mathbf{p}}^{\dagger}$
$\displaystyle\delta^{R}b_{\mathbf{p}}$
$\displaystyle=i\epsilon^{ijk}\rho_{i}\mathbf{p}_{j}\frac{\partial}{\partial\mathbf{p}^{k}}b_{\mathbf{p}}\qquad\delta^{R}b_{\mathbf{p}}^{\dagger}=i\epsilon^{ijk}\rho_{i}S(\mathbf{p})_{j}\frac{\partial}{\partial
S(\mathbf{p})^{k}}b_{\mathbf{p}}^{\dagger}$ (303)

and we have

$\displaystyle-\delta_{R}\lrcorner\,\Omega$ $\displaystyle=i\int
d^{3}p\,\alpha\,(\delta_{\partial_{R}}a_{\mathbf{p}}^{\dagger}\delta
a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}^{\dagger}S(\delta_{\partial_{R}})a_{\mathbf{p}}$
$\displaystyle-\delta_{\partial_{R}}b_{\mathbf{p}}\delta
b_{\mathbf{p}}^{\dagger}-\delta
b_{\mathbf{p}}S(\delta_{\partial_{R}})b_{\mathbf{p}}^{\dagger})$ (304)
$\displaystyle=i\epsilon^{ij}\int
d^{3}p\,\alpha\,\Big{(}S(\mathbf{p})_{[i}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial S(\mathbf{p})^{j]}}\delta
a_{\mathbf{p}}\,-\delta
a_{\mathbf{p}}^{\dagger}\,S(\mathbf{p})_{[i}\frac{\partial
a_{\mathbf{p}}}{\partial S(\mathbf{p})^{j]}}$
$\displaystyle-\mathbf{p}_{[i}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{j]}}\delta b_{\mathbf{p}}^{\dagger}+\delta
b_{\mathbf{p}}\mathbf{p}_{[i}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{j]}}\Big{)}$ (305)
$\displaystyle=i\epsilon^{ij}\delta\int
d^{3}p\,\alpha\,\Big{(}S(\mathbf{p})_{[i}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{j]}}a_{\mathbf{p}}\,-a_{\mathbf{p}}^{\dagger}\,S(\mathbf{p})_{[i}\frac{\partial
a_{\mathbf{p}}}{\partial S(\mathbf{p})^{j]}}$
$\displaystyle-\mathbf{p}_{[i}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{j]}}b_{\mathbf{p}}^{\dagger}+b_{\mathbf{p}}\mathbf{p}_{[i}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{j]}}\Big{)}$ (306)

which give the rotation charge

$\displaystyle{\cal M}_{i}=-\epsilon_{i}{}^{jk}\frac{1}{8}\int
d^{3}q\,\left(S(\mathbf{q})_{j}\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{k}}a_{\mathbf{q}}-a_{\mathbf{q}}^{\dagger}S(\mathbf{q})_{j}\frac{\partial
a_{\mathbf{q}}}{\partial
S(\mathbf{q})^{k}}+b_{\mathbf{q}}\mathbf{q}_{j}\frac{\partial
b_{\mathbf{q}}^{\dagger}}{\partial\mathbf{q}^{k}}-\mathbf{q}_{j}\frac{\partial
b_{\mathbf{q}}}{\partial\mathbf{q}^{k}}b_{\mathbf{q}}^{\dagger}\right).$ (307)

The considerations in this case are the same as for the boost charges.

####  3 Is it $\delta A\wedge\delta B=-\delta B\wedge\delta A$, or $\delta
A\wedge\delta B=S(\delta B\wedge\delta A)$?

We are at point $vii)b)$ of our roadmap in section 1. We have introduced the
antipode in the contraction rule in eq. (252), but so far we have not
discussed whether this inclusion is consistent with other rules concerning
forms, vector fields, and their interaction.

The first thing to tackle is the one reported in the title of this subsection.
Indeed, in obtaining the symplectic form in eq. (203), we used the fact that
$\delta A\wedge\delta B=-\delta B\wedge\delta A$ (here $A,B$ represent the
quantities $a,a^{\dagger},b,b^{\dagger}$). Furthermore, we used it in two
ways. The first one was when we were taking the exterior derivative of an
object like $(\delta B)A$. Indeed canonically we have

$\displaystyle\delta[(\delta B)A]=-\delta B\wedge\delta A$ (308)

but we did not address whether this should instead be substituted by

$\displaystyle\delta[(\delta B)A]=S(\delta B\wedge\delta A).$ (309)

The second time was when we added the two symplectic forms $\delta\theta_{1}$
and $\delta\theta_{2}$ in eq. (188) and (199) coming from the two action
orderings, since one of the two had a global minus sign to compensate a global
ordering change in the wedge products. In other words, to go from eq. (188)
and (199) to eq. (203) we used the fact that

$\displaystyle\delta A\wedge\delta B=-\delta B\wedge\delta A.$ (310)

Notice however that these two relations are not uncorrelated. Indeed eq. (308)
can be considered as a consequence of eq. (310) because the exterior
derivative only acts on the first object, and then we switch places using eq.
(310). In other words, we have

$\displaystyle\delta(\delta BA)=\delta(A\delta B)=\delta A\wedge\delta
B=-\delta B\wedge\delta A.$ (311)

In what follows, we will consider what happens when we try to obtain a
symplectic form using the assumption (309). We will show that the only way to
obtain an explicitly time-independent symplectic form is by imposing

$\displaystyle S(\delta A\wedge\delta B)=-\delta A\wedge\delta B.$ (312)

In other words, the antipode acting on wedge products (without any
contraction) behaves like a canonical minus sign. We stress that the time
independence of the symplectic form is a necessary requirement if one wants
time-independent commutators between creation/annihilation operators, as well
as time independent charges.

Indeed, using eq. (309), and keeping in mind eq. (177) and (178), one can
easily see that the time dependent coefficients in both $\delta\theta_{1}$ and
$\delta\theta_{2}$ are of the type555Here we discuss about only one of these
time-dependent terms, but for all the others the same considerations hold.

$\displaystyle\left[\omega_{\mathbf{q}}\delta b_{\mathbf{q}}\wedge\delta
a_{\mathbf{p}}-S(\omega_{\mathbf{p}})S(\delta b_{\mathbf{q}}\wedge\delta
a_{\mathbf{p}})\right]e^{i(\mathbf{q}\oplus\mathbf{p})\mathbf{x}}e^{-i(\omega_{\mathbf{q}}\oplus\omega_{\mathbf{p}})t}.$
(313)

We know from eq. (193) that the Dirac delta which comes out of the exponential
imposes $\omega_{\mathbf{q}}=-S(\omega_{\mathbf{p}})$, so that the time-
dependent term becomes

$\displaystyle-S(\omega_{\mathbf{p}})\left[\delta
b_{S(\mathbf{p})}\wedge\delta a_{\mathbf{p}}+S(\delta
b_{S(\mathbf{p})}\wedge\delta
a_{\mathbf{p}})\right]e^{-i([-S(\omega_{\mathbf{p}})]\oplus\omega_{\mathbf{p}})t}.$
(314)

Notice that, without the assumption in eq. (309), this term would be
identically zero. If instead we use the assumption in eq. (309), then this
time-dependent term666Crucially
$[-S(\omega_{\mathbf{p}})]\oplus\omega_{\mathbf{p}}\neq 0$, this can be easily
realized by noticing that the $\kappa\rightarrow\infty$ of this sum is
$\omega_{\mathbf{p}}+\omega_{\mathbf{p}}\neq 0$. would remain, unless

$\displaystyle\boxed{S(\delta a_{\mathbf{q}}\wedge\delta
b_{S(\mathbf{q})}\,)=-\delta a_{\mathbf{q}}\wedge\delta b_{S(\mathbf{q})}}$
(315)

i.e. if the antipode acting on a wedge product is exactly the same as the
canonical minus acting on it, which is what we wanted to show.

####  4 And what if we add contractions with vector fields?

In the previous subsections we have seen that

$\displaystyle\delta B\wedge\delta A=S(\delta A\wedge\delta B)=-\delta
A\wedge\delta B.$ (316)

Eq. (316) however, only deals with wedge products in the absence of
contraction with vector fields. In case we need to contract with some vector
fields, because of the antipode in eq. (252), one needs to take care of
whether to use the antipode or the canonical minus sign when changing the
order of the wedge product involved in a contraction.

Given any two functions $f,g$ in phase space, their Poisson bracket is defined
as [91]

$\displaystyle\\{f,g\\}=\omega(X_{f},X_{g})$ (317)

where $X_{f}$ and $X_{g}$ are the vector fields generated by $f$ and $g$
respectively, i.e.

$\displaystyle X_{f}\,\lrcorner\,\omega=-\delta f\qquad
X_{g}\,\lrcorner\,\omega=-\delta g$ (318)

and where $\omega$ is the symplectic 2-form (compare the above equations with
eq. (210)). In our case, we will use the symplectic form (203). In this case
the functions on momentum space would be the creation/annihilation operators
themselves.

For simplicity, we first concentrate on the $a,a^{\dagger}$ part. In this
case, recalling the definition (252) it is easy to see that, assuming the
symplectic form in eq. (203), we have

$\displaystyle
X_{a_{\mathbf{q}}}=i\frac{1}{S(\alpha(q))}\frac{\partial}{\partial
a_{\mathbf{q}}^{\dagger}}\qquad
X_{a^{\dagger}_{\mathbf{q}}}=i\frac{1}{\alpha(q)}\frac{\partial}{\partial
a_{\mathbf{q}}}$ (319)

In fact we have

$\displaystyle X_{a_{\mathbf{q}}}\,\lrcorner\,\Omega$ $\displaystyle=i\int
d^{3}p\,\alpha(p)\,\left[i\frac{1}{S(\alpha(q))}\frac{\partial}{\partial
a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,(\delta a_{\mathbf{p}}\wedge\delta
a_{\mathbf{p}}^{\dagger})\right]$ (320) $\displaystyle=i\int
d^{3}p\,\alpha(p)\,\left[0+\delta
a_{\mathbf{p}}S\left(i\frac{1}{S(\alpha(q))}\frac{\partial}{\partial
a_{\mathbf{q}}^{\dagger}}\right)a_{\mathbf{p}}^{\dagger}\right]$ (321)
$\displaystyle=-\int d^{3}p\,\alpha(p)\,\delta
a_{\mathbf{p}}\frac{1}{\alpha(q)}\underbrace{\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
a_{\mathbf{q}}^{\dagger}}}_{=\delta(\mathbf{p}-\mathbf{q})}$ (322)
$\displaystyle=-\delta a_{\mathbf{q}}$ (323) $\displaystyle
X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,\Omega$ $\displaystyle=i\int
d^{3}p\,\alpha(p)\,\left[i\frac{1}{\alpha(q)}\frac{\partial}{\partial
a_{\mathbf{q}}}\,\lrcorner\,(\delta a_{\mathbf{p}}\wedge\delta
a_{\mathbf{p}}^{\dagger})\right]$ (324) $\displaystyle=i\int
d^{3}p\,\alpha(p)\,\left[i\frac{1}{\alpha(q)}\underbrace{\frac{\partial
a_{\mathbf{p}}}{\partial
a_{\mathbf{q}}}}_{=\delta(\mathbf{p}-\mathbf{q})}\delta
a_{\mathbf{p}}^{\dagger}+0\right]$ (325) $\displaystyle=-\delta
a_{\mathbf{q}}^{\dagger}$ (326)

Notice that here we used the assumption $S(a)=a$ and
$S(a^{\dagger})=a^{\dagger}$, so that the antipode does not act on the single
creation/annihilation operator but only on their wedge product. There are some
subtleties that need to be sorted out.

First of all, notice that the vector field $X_{a_{\mathbf{q}}}$ is defined in
such a way that it requires the application of the antipode to ‘work as
intended’. For this reason, in order to get the correct value for the
commutator $\\{a,a^{\dagger}\\}$ we must first contract $\Omega$ with
$X_{a_{\mathbf{q}}}$ and only after this we can contract with
$X_{a_{\mathbf{q}}^{\dagger}}$, and not vice versa. In this way, we
immediately get

$\displaystyle\\{a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\\}=X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,(X_{a_{\mathbf{p}}}\,\lrcorner\,\Omega)=-\frac{i}{\alpha}\delta(\mathbf{p}-\mathbf{q})$
(327)

which is in accordance with the canonical result (see for example eq. (206)).
It is important to notice that at first sight one has

$\displaystyle
X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,(X_{a_{\mathbf{p}}}\,\lrcorner\,\Omega)\neq-
X_{a_{\mathbf{p}}}\,\lrcorner\,(X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,\Omega)$
(328)

since, as can be seen from eq. (324), we have

$\displaystyle
X_{a_{\mathbf{p}}}\,\lrcorner\,(X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,\Omega)=-\frac{i}{S(\alpha)}\delta(\mathbf{p}-\mathbf{q}).$
(329)

However, as explained above, intuitively speaking one can say that in this
case the vector field $X_{a_{\mathbf{q}}}$ is not ‘working in the way it was
built to work’, since it requires the application of an antipode. More
technically, keeping in mind eq. (252), notice that when we are contracting
with vector fields, we are not free in general to switch the order of the
wedge products unless we use the antipode. An example of this fact is given by
the computation of the charges using the symplectic form. Take for example the
time translation charge (the computations of the charges are done in section
14, here we just use an example for the argument’s sake). The symplectic form
is given by (203), and the creation/annihilation operator transformations are
given by (2), and at the end we get the charge in eq. (259).

However, if one wants to change the order of the wedge products involved in a
contraction with an external vector field (this point is crucial), then one
gets exactly the same result if one uses eq. (316) (with the antipode).
Indeed, using the symplectic form (consider just the $a,a^{\dagger}$ case and
ignoring the factor $\alpha$ for simplicity)

$\displaystyle\Omega=-i\int d^{3}p\,\alpha\,S(\delta
a_{\mathbf{p}}\wedge\delta a_{\mathbf{p}}^{\dagger})$ (330)

then the charge becomes

$\displaystyle-\delta_{\partial_{0}}\lrcorner\,\Omega$ $\displaystyle=i\int
d^{3}p\,\,\,\alpha\,S(\delta_{\partial_{0}})a_{\mathbf{p}}\delta
a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}S(S(\delta_{\partial_{0}}))a_{\mathbf{p}}^{\dagger}$ (331)
$\displaystyle=i\int
d^{3}p\,\,\,\alpha\,S(\delta_{\partial_{0}})a_{\mathbf{p}}\delta
a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}\delta_{\partial_{0}}a_{\mathbf{p}}^{\dagger}$ (332)
$\displaystyle=-\epsilon\int
d^{3}p\,\,\,\alpha\,S(\omega_{p})a_{\mathbf{p}}\delta a_{\mathbf{p}}\,+\delta
a_{\mathbf{p}}S(\omega_{p})a_{\mathbf{p}}^{\dagger}$ (333)
$\displaystyle=\epsilon\delta\int
d^{3}p\,\,\,\alpha\,[-S(\omega_{p})a_{\mathbf{p}}a_{\mathbf{p}}^{\dagger}]$
(334)

which gives exactly the same result. Notice that, in the presence of a
contraction with a vector field, the prescription in eq. (316) needs means
that the antipode is applied last, meaning that

$\displaystyle X\,\lrcorner\,(\delta A\wedge\delta B)=S(X\,\lrcorner\,(\delta
B\wedge\delta A)).$ (335)

In other words, given the contraction

$\displaystyle X\,\lrcorner\,(\delta A\wedge\delta B)=(XA)\delta B+\delta
A(S(X)B)$ (336)

we define

$\displaystyle\boxed{S(X\,\lrcorner\,(\delta B\wedge\delta A))=(S(X)A)\delta
B+\delta A(XB)}.$ (337)

Notice that, under the assumption that $S$ does not act on $A,\delta
A,B,\delta B$ individually (in this section they represent
$a,b,a^{\dagger},b^{\dagger}$), eq. (335) explains the above computations
regarding charges. Alternatively, one could drop the assumption that
$S(\cdot)$ does not act on $A,\delta A,B,\delta B$ individually, and simply
define

$\displaystyle\boxed{X\,\lrcorner\,(\delta A\wedge\delta
B)=S(X)\,\lrcorner\,(\delta B\wedge\delta A)}.$ (338)

In light of these considerations, and also keeping in mind that eq. (316)
implies that

$\displaystyle\delta A\wedge\delta B=-S(\delta A\wedge\delta B)$ (339)

and, if we apply the same point of view that antipodes should be applied last
as before, we see that there are several ways to define the contraction of a
2-form with two vector fields. In particular, since $X_{a_{\mathbf{q}}}$ has
been designed to account for the presence of an antipode, and since as shown
in eq. (335) the antipode is applied last, notice that we have

$\displaystyle-S(X_{a_{\mathbf{p}}}\,\lrcorner\,(X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,\Omega))=\frac{i}{S(S(\alpha))}\delta(\mathbf{p}-\mathbf{q})=\frac{i}{\alpha}\delta(\mathbf{p}-\mathbf{q})$
(340)

which gives the correct value for the Poisson bracket defined as

$\displaystyle\\{a_{\mathbf{q}}^{\dagger},a_{\mathbf{p}}\\}=-S(X_{a_{\mathbf{p}}}\,\lrcorner\,(X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,\Omega))=\frac{i}{\alpha}\delta(\mathbf{p}-\mathbf{q}).$
(341)

Therefore, once a prescription for the contraction of two vector fields with a
2-form is given, when switching the order of contractions one needs to impose
an additional global $-S$ to the computations. Notice that we indeed recover
the property

$\displaystyle\\{a_{\mathbf{q}}^{\dagger},a_{\mathbf{p}}\\}=-\\{a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\\}$
(342)

Of course, if instead of starting from the symplectic form in eq. (203) we
started from the one in eq. (253), then the vector fields
$X_{a_{\mathbf{q}}},X_{a_{\mathbf{q}}^{\dagger}}$ would now be given by

$\displaystyle
X_{a_{\mathbf{q}}}=-i\frac{1}{\alpha(q)}\frac{\partial}{\partial
a_{\mathbf{q}}^{\dagger}}\qquad
X_{a^{\dagger}_{\mathbf{q}}}=-i\frac{1}{S(\alpha(q))}\frac{\partial}{\partial
a_{\mathbf{q}}}$ (343)

In fact we would have

$\displaystyle X_{a_{\mathbf{q}}}\,\lrcorner\,\Omega$ $\displaystyle=-i\int
d^{3}p\,\alpha(p)\,\left[-i\frac{1}{\alpha(q)}\frac{\partial}{\partial
a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,(\delta
a_{\mathbf{p}}^{\dagger}\wedge\delta a_{\mathbf{p}})\right]$ (344)
$\displaystyle=-\int d^{3}p\,\alpha(p)\,\delta
a_{\mathbf{p}}\frac{1}{\alpha(q)}\underbrace{\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
a_{\mathbf{q}}^{\dagger}}}_{=\delta(\mathbf{p}-\mathbf{q})}$ (345)
$\displaystyle=-\delta a_{\mathbf{q}}$ (346) $\displaystyle
X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,\Omega$ $\displaystyle=-i\int
d^{3}p\,\alpha(p)\,\left[-i\frac{1}{S(\alpha(q))}\frac{\partial}{\partial
a_{\mathbf{q}}}\,\lrcorner\,(\delta a_{\mathbf{p}}^{\dagger}\wedge\delta
a_{\mathbf{p}})\right]$ (347) $\displaystyle=-i\int
d^{3}p\,\alpha(p)\,\left[0+\delta
a_{\mathbf{p}}^{\dagger}S\left(-i\frac{1}{S(\alpha(q))}\frac{\partial}{\partial
a_{\mathbf{q}}}\right)\delta a_{\mathbf{p}}\right]$ (348)
$\displaystyle=-i\int
d^{3}p\,\alpha(p)\,\left[0-i\frac{1}{\alpha(q)}\underbrace{\frac{\partial
a_{\mathbf{p}}}{\partial
a_{\mathbf{q}}}}_{=\delta(\mathbf{p}-\mathbf{q})}\delta
a_{\mathbf{p}}^{\dagger}\right]$ (349) $\displaystyle=-\delta
a_{\mathbf{q}}^{\dagger}$ (350)

Once again, following the fact that this time $X_{a_{\mathbf{q}}^{\dagger}}$
has been defined to accommodate for the antipode, it must be contracted first,
so that we end up with

$\displaystyle\\{a^{\dagger}_{\mathbf{p}},a_{\mathbf{q}}\\}=X_{a_{\mathbf{q}}}\,\lrcorner\,(X_{a_{\mathbf{p}}^{\dagger}}\,\lrcorner\,\Omega)=\frac{i}{\alpha}\delta(\mathbf{p}-\mathbf{q})$
(351)

which is in perfect agreement with eq. (341). Once again, if we switch the
order of contraction, we need to add a global $-S$, and indeed one has

$\displaystyle\\{a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\\}=-S(X_{a_{\mathbf{q}}^{\dagger}}\,\lrcorner\,(X_{a_{\mathbf{p}}}\,\lrcorner\,\Omega))=-\frac{i}{\alpha}\delta(\mathbf{p}-\mathbf{q})$
(352)

which is in accordance with eq. (327). Of course, the same reasoning gives the
correct propagators for $b,b^{\dagger}$, and all the cross-commutators
involving mixing of $a,a^{\dagger}$ with $b,b^{\dagger}$ are zero.
Furthermore, one also gets $\\{a,a\\}=\\{b,b\\}=0$.

To clarify and summarize the above discussion, we write the main steps below:

  * 1)

We start from some symplectic form $\omega$ and from two quantities $f,g$ of
which we want to compute the Poisson brackets. For the moment, we restrict our
attention to the case when $f,g$ are some creation/annihilation operators
which are part of the phase space of the theory (and therefore $\delta
f,\delta g$ appear somewhere in the definition of $\omega$);

  * 2)

Use the definitions $X_{f}\,\lrcorner\,\omega=-\delta f$ and
$X_{g}\,\lrcorner\,\omega=-\delta g$ to obtain the vector fields
$X_{g},X_{f}$;

  * 3)

The definition of the vector fields themselves will imply that there is one
correct order of contraction of $X_{g},X_{f}$ with $\omega$. Using this
contraction, one defines the relevant Poisson bracket, say for example
$\\{f,g\\}=X_{g}\,\lrcorner\,(X_{f}\,\lrcorner\,\Omega)$;

  * 4)

We now want to relate the brackets $\\{f,g\\}$ and $\\{g,f\\}$. Notice that,
because of the previous point, only $\\{f,g\\}$ contains the appropriate
contractions of $X_{f},X_{g}$ with $\omega$. In order to correctly define
$\\{g,f\\}$, therefore, we make use of a global $-S$ to be added to the
reversed contraction, defining
$\\{g,f\\}=-S(X_{f}\,\lrcorner\,(X_{g}\,\lrcorner\,\Omega))$, i.e.

$\displaystyle\boxed{\\{f,g\\}=X_{g}\,\lrcorner\,(X_{f}\,\lrcorner\,\Omega)\qquad\Leftrightarrow\qquad\\{g,f\\}=-S(X_{f}\,\lrcorner\,(X_{g}\,\lrcorner\,\Omega))}.$
(353)

Notice of course that this gives the correct $\kappa\rightarrow\infty$ limit.

  * 5)

When $f,g$ are creation/annihilation operators, the above procedure ensures
that one gets the crucial property $\\{f,g\\}=-\\{g,f\\}$, as well as the
correct value for the Poisson brackets.

###  15 The algebra of the $\kappa$-deformed charges

We are now at point $viii)$ of our roadmap in section 1. We now verify that
the translation, rotation, and boost charges, which are respectively eq.
(259), (260), (307), (279), satisfy the canonical, non-deformed Poincaré
algebra.

$\displaystyle[\mathscr{M}_{i},\mathscr{P}_{j}]=i\epsilon_{ijk}\mathscr{P}_{k}\qquad[\mathscr{M}_{i},\mathscr{P}_{0}]=0\qquad[\mathscr{N}_{i},\mathscr{P}_{j}]=-i\eta_{ij}\mathscr{P}_{0}\qquad[\mathscr{N}_{i},\mathscr{P}_{0}]=-i\mathscr{P}_{i}$
(354)
$\displaystyle[\mathscr{M}_{i},\mathscr{M}_{j}]=i\epsilon_{ijk}\mathscr{M}_{k}\qquad[\mathscr{M}_{i},\mathscr{N}_{j}]=i\epsilon_{ijk}\mathscr{N}_{k}\qquad[\mathscr{N}_{i},\mathscr{N}_{j}]=-i\epsilon_{ijk}\mathscr{M}_{k}$
(355)

Recall that we are using the convention $\eta=diag(+---)$.

Since the computations are long but straightforward, we will only explicitly
verify two of these commutators, but the rest of them can be verified using
the same exact procedures.

##### $[\mathscr{N}_{i},\mathscr{P}_{j}]=-i\eta_{ij}\mathscr{P}_{0}$

The commutator $[\mathscr{N}_{i},\mathscr{P}_{j}]$ contains two components,
i.e.

$\displaystyle-i\alpha(p)\alpha(q)S(\omega_{p})S(\mathbf{q})_{j}\Bigg{\\{}\left[\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}a_{\mathbf{q}}\right]-\left[a_{\mathbf{p}}^{\dagger}\frac{\partial
a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}},a_{\mathbf{q}}^{\dagger}a_{\mathbf{q}}\right]\Bigg{\\}}$
(356) $\displaystyle
i\,\alpha(p)\alpha(q)\omega_{p}\mathbf{q}_{j}\Bigg{\\{}\left[b_{\mathbf{p}}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}},b_{\mathbf{q}}b_{\mathbf{q}}^{\dagger}\right]-\left[\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}^{\dagger},b_{\mathbf{q}}b_{\mathbf{q}}^{\dagger}\right]\Bigg{\\}}$
(357)

Starting from the first one, we have

$\displaystyle-i\alpha(p)\alpha(q)S(\omega_{p})S(\mathbf{q})_{j}\left[\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}a_{\mathbf{q}}\right]$
(358)
$\displaystyle=-i\alpha(p)\alpha(q)S(\omega_{p})S(\mathbf{q})_{j}\Bigg{\\{}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}\underbrace{[a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}]}_{=\frac{1}{\alpha}\delta(\mathbf{p}-\mathbf{q})}a_{\mathbf{q}}+a_{\mathbf{q}}^{\dagger}\underbrace{\left[\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}},a_{\mathbf{q}}\right]}_{=-\frac{1}{\alpha}\frac{\partial}{\partial
S(\mathbf{p})^{i}}\delta(\mathbf{q}-\mathbf{p})}a_{\mathbf{p}}\Bigg{\\}}$
(359) $\displaystyle=-i\alpha(p)S(\omega_{p})S(\mathbf{q})_{j}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}\delta(\mathbf{p}-\mathbf{q})a_{\mathbf{q}}$
$\displaystyle+i\alpha(p)S(\omega_{p})S(\mathbf{q})_{j}a_{\mathbf{q}}^{\dagger}\left(\frac{\partial}{\partial
S(\mathbf{p})^{i}}\delta(\mathbf{q}-\mathbf{p})\right)a_{\mathbf{p}}$ (360)

Integrating by parts the last term in the last passage above we get

$\displaystyle-i\alpha(p)S(\omega_{p})S(\mathbf{q})_{j}a_{\mathbf{q}}^{\dagger}\delta(\mathbf{q}-\mathbf{p})\frac{\partial}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}-i\frac{\partial(\alpha(p)S(\omega_{p}))}{\partial
S(\mathbf{p})^{i}}S(\mathbf{q})_{j}a_{\mathbf{q}}^{\dagger}\delta(\mathbf{q}-\mathbf{p})a_{\mathbf{p}}$
(361)

and substituting back in eq. (360) we get (we can now safely apply the Dirac
delta since no derivative acts on it)

$\displaystyle\left(-i\alpha(p)S(\omega_{p})S(\mathbf{p})_{j}\frac{\partial}{\partial
S(\mathbf{p})^{i}}(a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}})-i\frac{\partial(\alpha(p)S(\omega_{p}))}{\partial
S(\mathbf{p})^{i}}S(\mathbf{p})_{j}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}\right)\delta(\mathbf{q}-\mathbf{p})$
(362) $\displaystyle=\Bigg{(}-i\frac{\partial}{\partial
S(\mathbf{p})^{i}}(\alpha(p)S(\omega_{p})S(\mathbf{p})_{j}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}})+iS(\mathbf{p})_{j}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}\frac{\partial}{\partial
S(\mathbf{p})^{i}}(\alpha(p)S(\omega_{p}))$
$\displaystyle+i\alpha(p)S(\omega_{p})\eta_{ij}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}-i\frac{\partial(\alpha(p)S(\omega_{p}))}{\partial
S(\mathbf{p})^{i}}S(\mathbf{p})_{j}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}\Bigg{)}\delta(\mathbf{q}-\mathbf{p})$
(363)
$\displaystyle=i\alpha(p)S(\omega_{p})\eta_{ij}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}\delta(\mathbf{q}-\mathbf{p})$
(364)

where in the last passage we ignored $\frac{\partial}{\partial
S(\mathbf{p})^{i}}(\alpha(p)S(\omega_{p})S(\mathbf{p})_{j}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}})$
since it is a surface term after the application of the Dirac delta.

We can analogously proceed for the second term, as follows

$\displaystyle
i\,\alpha(p)\alpha(q)\omega_{p}\mathbf{q}_{j}\left[b_{\mathbf{p}}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}},b_{\mathbf{q}}b_{\mathbf{q}}^{\dagger}\right]$
(365)
$\displaystyle=i\,\alpha(p)\alpha(q)\omega_{p}\mathbf{q}_{j}\Bigg{\\{}b_{\mathbf{p}}\underbrace{\left[\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}},b_{\mathbf{q}}\right]}_{=-\frac{1}{\alpha}\frac{\partial}{\partial\mathbf{p}^{i}}\delta(\mathbf{q}-\mathbf{p})}b_{\mathbf{q}}^{\dagger}+0+0+b_{\mathbf{q}}\underbrace{[b_{\mathbf{p}},b_{\mathbf{q}}^{\dagger}]}_{=\frac{1}{\alpha}\delta(\mathbf{p}-\mathbf{q})}\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}\Bigg{\\}}$ (366)
$\displaystyle=+i\,\alpha(p)\omega_{p}\mathbf{q}_{j}b_{\mathbf{q}}\delta(\mathbf{p}-\mathbf{q})\frac{\partial
b_{\mathbf{p}}^{\dagger}}{\partial\mathbf{p}^{i}}+i\,\alpha(p)\omega_{p}\mathbf{q}_{j}\frac{\partial
b_{\mathbf{p}}}{\partial\mathbf{p}^{i}}\delta(\mathbf{q}-\mathbf{p})b_{\mathbf{q}}^{\dagger}$
$\displaystyle+i\,\mathbf{q}_{j}\frac{\partial(\alpha(p)\omega_{p})}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}\delta(\mathbf{q}-\mathbf{p})b_{\mathbf{q}}^{\dagger}$
(367)
$\displaystyle=\left(i\,\mathbf{p}_{j}\frac{\partial(\alpha(p)\omega_{p})}{\partial\mathbf{p}^{i}}b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}+i\,\alpha(p)\omega_{p}\mathbf{q}_{j}\frac{\partial}{\partial\mathbf{p}^{i}}(b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger})\right)\delta(\mathbf{p}-\mathbf{q})$
(368)
$\displaystyle=\Big{(}i\,\frac{\partial}{\partial\mathbf{p}^{i}}(\mathbf{p}_{j}\alpha(p)\omega_{p}b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger})-i\,\mathbf{p}_{j}\alpha(p)\omega_{p}\frac{\partial}{\partial\mathbf{p}^{i}}(b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger})-i\,\eta_{ij}\alpha(p)\omega_{p}b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}$
$\displaystyle+i\,\alpha(p)\omega_{p}\mathbf{q}_{j}\frac{\partial}{\partial\mathbf{p}^{i}}(b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger})\Big{)}\delta(\mathbf{p}-\mathbf{q})$
(369)
$\displaystyle=-i\,\eta_{ij}\alpha(p)\omega_{p}b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}\delta(\mathbf{p}-\mathbf{q}).$
(370)

In summary, we have shown that

$\displaystyle[N_{i},\mathscr{P}_{j}]$ $\displaystyle=\int
d^{3}p\,d^{3}q\,\left(i\alpha(p)S(\omega_{p})\eta_{ij}a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}-i\,\eta_{ij}\alpha(p)\omega_{p}b_{\mathbf{p}}b_{\mathbf{p}}^{\dagger}\right)\delta(\mathbf{q}-\mathbf{p})$
(371) $\displaystyle=-i\eta_{ij}\mathscr{P}_{0}$ (372)

which is what we wanted to show.

##### $[\mathscr{N}_{i},\mathscr{N}_{j}]=-i\epsilon_{ijk}\mathscr{M}_{k}$

The commutator $[\mathscr{N}_{i},\mathscr{N}_{j}]$ contains two parts, the
$a,a^{\dagger}$ part and the $b,b^{\dagger}$ part. Starting with the one
concerning the $a$’s we have

$\displaystyle-\alpha(p)\alpha(q)S(\omega_{p})S(\omega_{q})\Bigg{[}\Bigg{\\{}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}}-a_{\mathbf{p}}^{\dagger}\frac{\partial
a_{\mathbf{p}}}{\partial S(\mathbf{p})^{i}}\Bigg{\\}},\Bigg{\\{}\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{j}}a_{\mathbf{q}}-a_{\mathbf{q}}^{\dagger}\frac{\partial
a_{\mathbf{q}}}{\partial S(\mathbf{q})^{j}}\Bigg{\\}}\Bigg{]}$ (373)
$\displaystyle=$
$\displaystyle-\alpha(p)\alpha(q)S(\omega_{p})S(\omega_{q})\Bigg{\\{}\Bigg{[}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{j}}a_{\mathbf{q}}\Bigg{]}+\Bigg{[}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}},-a_{\mathbf{q}}^{\dagger}\frac{\partial
a_{\mathbf{q}}}{\partial S(\mathbf{q})^{j}}\Bigg{]}$
$\displaystyle+\Bigg{[}-a_{\mathbf{p}}^{\dagger}\frac{\partial
a_{\mathbf{p}}}{\partial S(\mathbf{p})^{i}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{j}}a_{\mathbf{q}}\Bigg{]}+\Bigg{[}-a_{\mathbf{p}}^{\dagger}\frac{\partial
a_{\mathbf{p}}}{\partial
S(\mathbf{p})^{i}},-a_{\mathbf{q}}^{\dagger}\frac{\partial
a_{\mathbf{q}}}{\partial S(\mathbf{q})^{j}}\Bigg{]}\Bigg{\\}}$ (374)

We will compute these commutators one by one starting from the first. Once
again, we use

$\displaystyle[AB,CD]=A[B,C]D+[A,C]BD+CA[B,D]+C[A,D]B$ (375)

and we get

$\displaystyle-\alpha(p)\alpha(q)S(\omega_{p})S(\omega_{q})\Bigg{[}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}a_{\mathbf{p}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial S(\mathbf{q})^{j}}a_{\mathbf{q}}\Bigg{]}$
(376)
$\displaystyle=-\alpha(p)\alpha(q)S(\omega_{p})S(\omega_{q})\Bigg{\\{}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}\left[a_{\mathbf{p}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{j}}\right]a_{\mathbf{q}}+\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial S(\mathbf{q})^{j}}\left[\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}},a_{\mathbf{q}}\right]a_{\mathbf{p}}\Bigg{\\}}$ (377)
$\displaystyle=-\alpha(p)\alpha(q)S(\omega_{p})S(\omega_{q})\Bigg{\\{}\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}\left[a_{\mathbf{p}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{j}}\right]a_{\mathbf{q}}-\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{j}}\left[a_{\mathbf{q}},\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{i}}\right]a_{\mathbf{p}}\Bigg{\\}}$ (378)

Now we notice that the second term is just the first one with
$\mathbf{p}\leftrightarrow\mathbf{q}$ and $i\leftrightarrow j$, but we can
just rename the integration variables in such a way that the second one and
the first are the same except with $i\leftrightarrow j$, and therefore the
above integral becomes

$\displaystyle-\alpha(p)\alpha(q)S(\omega_{p})S(\omega_{q})\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{[i}}\left[a_{\mathbf{p}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial S(\mathbf{q})^{j]}}\right]a_{\mathbf{q}}$
(379)

where we just used the antisymmetrized index to shorten the notation. Now we
have a small subtlety related to the presence of the $\frac{1}{\alpha}$ in
front of the delta coming from the commutator. In fact, recall from eq. (208)
that

$\displaystyle\left[a_{\mathbf{p}},a_{\mathbf{q}}^{\dagger}\right]$
$\displaystyle=\frac{1}{\alpha}\delta(\mathbf{p}-\mathbf{q}).$ (380)

However we have two choices, one where the $\frac{1}{\alpha}$ on the RHS
depends on $p$ and the other where it depends on $q$. We are in the following
situation

$\displaystyle f(p,q)\alpha(p)\alpha(q)\left[a_{\mathbf{p}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})}\right]=f(p,q)\alpha(p)\alpha(q)\frac{\partial}{\partial
S(\mathbf{q})}\left(\frac{1}{\alpha}\delta(\mathbf{p}-\mathbf{q})\right)$
(381)

Now, if the $\alpha$ in the round brackets depends on $p$, then the above
formula reduces to

$\displaystyle f(p,q)\alpha(q)\frac{\partial}{\partial
S(\mathbf{q})}\delta(\mathbf{p}-\mathbf{q})\xrightarrow{\text{int. by parts
and appl.}\delta}-\left(\frac{\partial}{\partial
S(\mathbf{q})}f(q)\alpha(q)\right)\delta(\mathbf{p}-\mathbf{q})$ (382)

However, if the $\alpha$ in the round brackets depends on $q$, then we have

$\displaystyle f(p,q)\alpha(p)\alpha(q)\frac{\partial}{\partial
S(\mathbf{q})}\left(\frac{1}{\alpha}\delta(\mathbf{p}-\mathbf{q})\right)$
(383)
$\displaystyle=-f(p,q)\alpha(p)\alpha(q)\frac{1}{\alpha^{2}(q)}\frac{\partial\alpha(q)}{\partial
S(\mathbf{q})}\delta(\mathbf{p}-\mathbf{q})+f(p,q)\alpha(p)\frac{\partial}{\partial
S(\mathbf{q})}\delta(\mathbf{p}-\mathbf{q})$ (384)
$\displaystyle=-f(q)\frac{\partial\alpha(q)}{\partial
S(\mathbf{q})}\delta(\mathbf{p}-\mathbf{q})+f(p,q)\alpha(p)\frac{\partial}{\partial
S(\mathbf{q})}\delta(\mathbf{p}-\mathbf{q})$ (385)
$\displaystyle=-f(q)\frac{\partial\alpha(q)}{\partial
S(\mathbf{q})}\delta(\mathbf{p}-\mathbf{q})-\left(\frac{\partial}{\partial
S(\mathbf{q})}f(p,q)\right)\alpha(p)\delta(\mathbf{p}-\mathbf{q})$ (386)
$\displaystyle=-\left(\frac{\partial}{\partial
S(\mathbf{q})}f(q)\alpha(q)\right)\delta(\mathbf{p}-\mathbf{q})$ (387)

where in eq. (385) we applied the Dirac delta in the first term (because it is
not acted upon by any derivatives) and in eq. (386) we integrated by parts the
second term on the RHS. One sees that both approaches lead to the same final
result. Therefore for simplicity we will always consider the
$\frac{1}{\alpha}$ in the definition of $[a,a^{\dagger}]$ to depend on the
variable that is not acted upon by the derivation inside the commutator.

Coming back to eq. (379), we have

$\displaystyle-\alpha(p)\alpha(q)S(\omega_{p})S(\omega_{q})\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{[i}}\left[a_{\mathbf{p}},\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial S(\mathbf{q})^{j]}}\right]a_{\mathbf{q}}$
(388) $\displaystyle=-\alpha(q)S(\omega_{p})S(\omega_{q})\frac{\partial
a_{\mathbf{p}}^{\dagger}}{\partial
S(\mathbf{p})^{[i}}a_{\mathbf{q}}\frac{\partial}{\partial
S(\mathbf{q})^{j]}}\delta(\mathbf{p}-\mathbf{q})$ (389)
$\displaystyle=\frac{\partial\alpha(q)}{\partial
S(\mathbf{q})^{[j}}S(\omega_{q})^{2}\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{i]}}a_{\mathbf{q}}+\alpha(q)S(\mathbf{q})_{[j}\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial
S(\mathbf{q})^{i]}}a_{\mathbf{q}}+\alpha(q)S(\omega_{q})^{2}\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial S(\mathbf{q})^{[i}}\frac{\partial
a_{\mathbf{q}}}{\partial S(\mathbf{q})^{j]}}$ (390)

We can now integrate by parts the first term of the last step, which (ignoring
surface terms) will give us

$\displaystyle\frac{\partial\alpha(q)}{\partial
S(\mathbf{q})^{[j}}S(\omega_{q})^{2}\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial S(\mathbf{q})^{i]}}a_{\mathbf{q}}$ (391)
$\displaystyle=-\alpha(q)2\frac{\partial S(\omega_{q})}{\partial
S(\mathbf{q})^{[j}}S(\omega_{q})\frac{\partial
a_{\mathbf{q}}^{\dagger}}{\partial