$\displaystyle\boxed{\Im\Pi^{(2)}(p^{2})=\Im\Pi^{U}(p^{2})\left\\{1+\frac{1}{\kappa}\left(\frac{3}{2}p_{0}\right)+\frac{1}{\kappa^{2}}\left[p_{0}\left(-\frac{1}{6}+\frac{2}{3}\frac{m^{2}}{p^{2}}\right)+\frac{1}{24}p^{2}+\frac{1}{2}m^{2}\right]\right\\}}$
(177)

We now pass to the third option. In this case we have

$i\Pi(p^{2})=$$S(k)$$k\oplus p$ $\displaystyle
i\Pi(p^{2})=(ig)^{2}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}-m^{2}+i\epsilon}\frac{i}{(k\oplus
p)^{2}-m^{2}+i\epsilon}$ (178)

which expanded to the first two non-trivial orders in powers of $1/\kappa$ is

$\displaystyle i\Pi(p^{2})$ $\displaystyle=i\Pi^{U}(p^{2})$ (179)
$\displaystyle-\frac{g^{2}}{\kappa}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}-m^{2}+i\epsilon}\frac{\Delta_{1}[(k\oplus
p)^{2}]}{[(p+k)^{2}-m^{2}+i\epsilon]^{2}}$ (180)
$\displaystyle-\frac{g^{2}}{\kappa^{2}}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}-m^{2}+i\epsilon}\frac{\Delta_{2}[(k\oplus
p)^{2}]}{[(p+k)^{2}-m^{2}+i\epsilon]^{2}}$ (181)

Notice the presence of $(p+k)^{2}$ in the denominator. This has no impact on
the non-deformed amplitude, because we can simply switch $k\mapsto-k$, the
metric remains invariant as well as the integration domain, and we recover the
usual integral. In the deformed terms, however, we have to keep in mind that
we must send $k\mapsto-k$ also in the numerator before doing the Feynman
trick, so that we can once again use the same steps af before.

Apart from this, every consideration goes exactly as before, and this time we
need to compute

$\displaystyle\Delta_{1}^{shif+Wick}[(k\oplus
p)^{2}],\qquad\Delta_{2}^{shif+Wick}[(k\oplus p)^{2}]$ (182)

once again. We have

$\displaystyle\Delta_{1}[(k\oplus
p)^{2}]=2\mathbf{k}(k_{0}\mathbf{p}-p_{0}\mathbf{k})$ (183)
$\displaystyle\Delta_{2}[(k\oplus p)^{2}]$
$\displaystyle=(k_{0}+p_{0})\left(p_{0}\left(k_{0}(k_{0}+p_{0})+\mathbf{k}^{2}\right)-2k_{0}\mathbf{k}\mathbf{p}-k_{0}\mathbf{p}^{2}\right)$
$\displaystyle-
p_{0}^{2}\mathbf{k}^{2}+\mathbf{k}\left(\mathbf{p}^{2}-p_{0}^{2}\right)(\mathbf{k}+\mathbf{p})+\frac{\mathbf{k}^{2}\mathbf{p}^{2}}{3}$
(184)

where we used the identity

$\displaystyle(\mathbf{k}\mathbf{p})^{2}=\sum_{i,j}\mathbf{k}_{i}\mathbf{p}^{i}\mathbf{k}_{j}\mathbf{p}^{j}\mapsto\sum_{i}\mathbf{k}_{i}^{2}(\mathbf{p}^{i})^{2}\mapsto
k_{0}^{2}\mathbf{p}^{2}\mapsto\frac{1}{3}\mathbf{k}^{2}\mathbf{p}^{2}.$ (185)

We now send $k\mapsto-k$. It is obvious that $\Delta_{1}[(k\oplus p)^{2}]$ is
not affected by such a change. On the other hand $\Delta_{2}[(k\oplus p)^{2}]$
is modified. We have

$\displaystyle\Delta_{1}[(k\oplus
p)^{2}]_{new}=2\mathbf{k}(k_{0}\mathbf{p}-p_{0}\mathbf{k})$ (186)
$\displaystyle\Delta_{2}[(k\oplus p)^{2}]_{new}$
$\displaystyle=(p_{0}-k_{0})\left(p_{0}\left(\mathbf{k}^{2}-k_{0}(p_{0}-k_{0})\right)-2k_{0}\mathbf{k}\mathbf{p}+k_{0}\mathbf{p}^{2}\right)$
$\displaystyle-
p_{0}^{2}\mathbf{k}^{2}-\mathbf{k}\left(\mathbf{p}^{2}-p_{0}^{2}\right)(\mathbf{p}-\mathbf{k})+\frac{\mathbf{k}^{2}\mathbf{p}^{2}}{3}$
(187)

Since now the denominators are again the same as before, we can treat them in
the same way, obtaining

$\displaystyle-\frac{g^{2}}{\kappa}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{1}^{shift}[(k\oplus
p)^{2}]_{new}}{[k^{2}-\Delta(x)+i\epsilon]^{3}}$ (188)
$\displaystyle-\frac{g^{2}}{\kappa^{2}}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{2}^{shift}[(k\oplus
p)^{2}]_{new}}{[k^{2}-\Delta(x)+i\epsilon]^{3}}$ (189)

The shift is again $k\mapsto k+p(1-x)$, and performing the Wick rotation the
amplitude becomes

$\displaystyle
i\frac{g^{2}}{\kappa}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{1}^{shift+Wick}[(k\oplus
p)^{2}]_{new}}{[k^{2}+\Delta(x)+i\epsilon]^{3}}$
$\displaystyle+i\frac{g^{2}}{\kappa^{2}}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{2}^{shift+Wick}[(k\oplus
p)^{2}]_{new}}{[k^{2}+\Delta(x)+i\epsilon]^{3}}$ (190)

The numerators are

$\displaystyle\Delta_{1}^{shift+Wick}[(k\oplus
p)^{2}]_{new}=-2p_{0}\mathbf{k}^{2}$ (191)
$\displaystyle\Delta_{2}^{shift+Wick}[(k\oplus p)^{2}]_{new}$
$\displaystyle=-3k_{0}^{2}p_{0}^{2}x+k_{0}^{2}p_{0}^{2}+2k_{0}^{2}\mathbf{p}^{2}x-k_{0}^{2}\mathbf{p}^{2}$
$\displaystyle+p_{0}^{4}x^{3}-p_{0}^{4}x^{2}+p_{0}^{2}\mathbf{k}^{2}x-2p_{0}^{2}\mathbf{k}^{2}$
$\displaystyle-
p_{0}^{2}\mathbf{p}^{2}x^{3}-p_{0}^{2}\mathbf{p}^{2}x^{2}+3p_{0}^{2}\mathbf{p}^{2}x-p_{0}^{2}\mathbf{p}^{2}+\frac{4\mathbf{k}^{2}\mathbf{p}^{2}}{3}$
$\displaystyle+\frac{4\mathbf{p}^{4}x^{2}}{3}-\frac{5\mathbf{p}^{4}x}{3}+\frac{\mathbf{p}^{4}}{3}$
(192)

Once again, being in the Eulidean space, we can switch any isolated
$\mathbf{k}^{2}$ with $3k_{0}^{2}$, obtaining

$\displaystyle\Delta_{1}^{shift+Wick}[(k\oplus p)^{2}]_{new}=-6p_{0}k_{0}^{2}$
(193) $\displaystyle\Delta_{2}^{shift+Wick}[(k\oplus p)^{2}]_{new}$
$\displaystyle=k_{0}^{2}[-5p_{0}^{2}+\mathbf{p}^{2}(2x+3)]$
$\displaystyle+p_{0}^{4}x^{3}-p_{0}^{4}x^{2}-p_{0}^{2}\mathbf{p}^{2}x^{3}-p_{0}^{2}\mathbf{p}^{2}x^{2}+3p_{0}^{2}\mathbf{p}^{2}x-p_{0}^{2}\mathbf{p}^{2}$
$\displaystyle+\frac{4\mathbf{p}^{4}x^{2}}{3}-\frac{5\mathbf{p}^{4}x}{3}+\frac{\mathbf{p}^{4}}{3}$
(194)

Notice that the $1/\kappa$ contribution is numerically equivalent to the
previous case, but with a sign difference, so we don’t need to compute it. For
what remains, since we are only interested in the logarithmic contribution,
using eq. (150), (151), (152), since each $\log$ contributes a $-\pi$, and
since

$\displaystyle\int_{L_{1}}^{L_{2}}dx[-5p_{0}^{2}+\mathbf{p}^{2}(2x+3)]=\sqrt{1-4\frac{m^{2}}{p^{2}}}\left(p_{0}^{2}+4p^{2}\right)$
(195)

we end up with

$\displaystyle\Im\Pi^{(3)}=\Im\Pi^{U}(p^{2})-\frac{g^{2}}{\pi\kappa}\sqrt{1-4\frac{m^{2}}{p^{2}}}\frac{3p_{0}}{32}+\frac{g^{2}}{64\pi\kappa^{2}}\sqrt{1-4\frac{m^{2}}{p^{2}}}\left(p_{0}^{2}+4p^{2}\right)$
(196)

Recalling from eq. (43) that
$\Im\Pi^{U}(p^{2})=\frac{g^{2}}{16\pi}\sqrt{1-4\frac{m^{2}}{p^{2}}}$ we
finally obtain

$\displaystyle\boxed{\Im\Pi^{(3)}=\Im\Pi^{U}(p^{2})\left[1-\frac{1}{\kappa}\left(\frac{3}{2}p_{0}\right)+\frac{1}{\kappa^{2}}\left(\frac{1}{4}p_{0}^{2}+p^{2}\right)\right]}$
(197)

Finally, we consider the last graph.

$i\Pi(p^{2})=$$S(k)$$p\oplus k$ $\displaystyle
i\Pi(p^{2})=(ig)^{2}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i}{k^{2}-m^{2}+i\epsilon}\frac{i}{(p\oplus
k)^{2}-m^{2}+i\epsilon}$ (198)

The computations for this graph are exactly the same with the previous with
$k\leftrightarrow p$ up to the shift, so we have

$\displaystyle i\Pi(p^{2})$ $\displaystyle=i\Pi^{U}(p^{2})$ (199)
$\displaystyle-\frac{g^{2}}{\kappa}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}-m^{2}+i\epsilon}\frac{\Delta_{1}[(p\oplus
k)^{2}]}{[(p+k)^{2}-m^{2}+i\epsilon]^{2}}$ (200)
$\displaystyle-\frac{g^{2}}{\kappa^{2}}\int\frac{d^{4}k}{(2\pi)^{4}}\frac{1}{k^{2}-m^{2}+i\epsilon}\frac{\Delta_{2}[(p\oplus
k)^{2}]}{[(p+k)^{2}-m^{2}+i\epsilon]^{2}}$ (201)

and

$\displaystyle\Delta_{1}[(k\oplus
p)^{2}]_{new}=-2\mathbf{p}(k_{0}\mathbf{p}-p_{0}\mathbf{k})$ (202)
$\displaystyle\Delta_{2}[(k\oplus p)^{2}]_{new}$
$\displaystyle=-\mathbf{p}\left(\mathbf{k}^{2}-k_{0}^{2}\right)(\mathbf{k}-\mathbf{p})-k_{0}^{2}\mathbf{p}^{2}$
$\displaystyle+(k_{0}-p_{0})\left(k_{0}\left(\mathbf{p}^{2}-p_{0}(k_{0}-p_{0})\right)+p_{0}\mathbf{k}^{2}-2p_{0}\mathbf{k}\mathbf{p}\right)+\frac{\mathbf{k}^{2}\mathbf{p}^{2}}{3}$
(203)

Since now the denominators are again the same as before, we can treat them in
the same way, obtaining

$\displaystyle-\frac{g^{2}}{\kappa}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{1}^{shift}[(k\oplus
p)^{2}]_{new}}{[k^{2}-\Delta(x)+i\epsilon]^{3}}$ (204)
$\displaystyle-\frac{g^{2}}{\kappa^{2}}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{2}^{shift}[(k\oplus
p)^{2}]_{new}}{[k^{2}-\Delta(x)+i\epsilon]^{3}}$ (205)

The shift is again $k\mapsto k+p(1-x)$, and performing then the Wick rotations
the amplitude becomes

$\displaystyle
i\frac{g^{2}}{\kappa}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{1}^{shift+Wick}[(p\oplus
k)^{2}]_{new}}{[k^{2}+\Delta(x)+i\epsilon]^{3}}$ (206)
$\displaystyle+i\frac{g^{2}}{\kappa^{2}}\int_{0}^{1}dx\int\frac{d^{4}k}{(2\pi)^{4}}\frac{\Delta_{2}^{shift+Wick}[(p\oplus
k)^{2}]_{new}}{[k^{2}+\Delta(x)+i\epsilon]^{3}}$ (207)

and the numerators (after the same simplifications) become

$\displaystyle\Delta_{1}^{shift+Wick}[(k\oplus p)^{2}]_{new}=0$ (208)
$\displaystyle\Delta_{2}^{shift+Wick}[(k\oplus p)^{2}]_{new}$
$\displaystyle=k_{0}^{2}[p_{0}^{2}(-6x+1)+\mathbf{p}^{2}(10x-5)]$
$\displaystyle+p_{0}^{4}x^{3}-p_{0}^{4}x^{2}-2p_{0}^{2}\mathbf{p}^{2}x^{3}+2p_{0}^{2}\mathbf{p}^{2}x^{2}+p_{0}^{2}\mathbf{p}^{2}x$
$\displaystyle-
p_{0}^{2}\mathbf{p}^{2}+\mathbf{p}^{4}x^{3}-\frac{5\mathbf{p}^{4}x^{2}}{3}+\frac{\mathbf{p}^{4}x}{3}+\frac{\mathbf{p}^{4}}{3}$
(209)

Recalling once more eq. (150), (151), (152), since each $\log$ contributes a
$-\pi$ and since

$\displaystyle\int_{L_{1}}^{L_{2}}dx[p_{0}^{2}(-6x+1)+\mathbf{p}^{2}(10x-5)]=2\sqrt{1-4\frac{m^{2}}{p^{2}}}p_{0}^{2}$
(210)

we end up with

$\displaystyle\Im\Pi^{(4)}(p^{2})=\Im\Pi^{U}(p^{2})+\frac{g^{2}}{\pi\kappa^{2}}\sqrt{1-4\frac{m^{2}}{p^{2}}}\left(\frac{1}{32}p_{0}\right)$
(211)

Recalling from eq. (43) that
$\Im\Pi^{U}(p^{2})=\frac{g^{2}}{16\pi}\sqrt{1-4\frac{m^{2}}{p^{2}}}$ we
finally obtain

$\displaystyle\boxed{\Im\Pi^{(4)}=\Im\Pi^{U}(p^{2})\left[1+\frac{1}{\kappa^{2}}\left(\frac{1}{2}p_{0}^{2}\right)\right]}$
(212)

##### Summary of the individual diagrams and total 1-loop amplitude

We considered the following diagrams

$i\Pi^{(1)}(p^{2})=$$k$$S(k)\oplus p$$i\Pi^{(2)}(p^{2})=$$k$$p\oplus
S(k)$$i\Pi^{(3)}(p^{2})=$$S(k)$$k\oplus p$$i\Pi^{(4)}(p^{2})=$$S(k)$$p\oplus
k$

Each of them gives the imaginary part to the amplitude respectively in eq.
(141), (177), (197), (212). To get the complete contribution, we just sum the
four different results

$\displaystyle\Im\Pi^{TOT}(p^{2})=\frac{1}{4}\left(\Im\Pi^{(1)}(p^{2})+\Im\Pi^{(2)}(p^{2})+\Im\Pi^{(3)}(p^{2})+\Im\Pi^{(4)}(p^{2})\right)$
(213)

where the factor $1/4$ is necessary to get the correct
$\kappa\rightarrow\infty$ limit. We get

$\displaystyle\boxed{\Im\Pi^{TOT}(p^{2})=\Im\Pi^{U}(p^{2})\left\\{1-\frac{1}{\kappa^{2}}\left[p_{0}^{2}\left(\frac{1}{48}+\frac{1}{12}\frac{m^{2}}{p^{2}}\right)-p^{2}\frac{19}{48}-\frac{5}{24}m^{2}\right]\right\\}}$
(214)

Notice the remarkable fact that, although some of the diagrams contain a
contribution proportional to $1/\kappa$, this contribution goes away after we
sum all the contributions from every diagram, leaving a leading contribution
proportional to $1/\kappa^{2}$.

Notice the important point that we chose our initial particle to have momentum
$S(p)$, but we may just as well have decided to take a particle with initial
momentum $-p$. In this case momentum conservation would amount to

$\displaystyle(-p)\oplus p_{2}\oplus p_{3}$ (215)

which is immediately satisfied if $p_{2}\oplus p_{3}=S(-p)$ (in order to avoid
confusion, it is understood that the equivalent computations using the momenta
$p^{*}$ would amount to having a particle with momentum $p^{*}$, so that
$p_{2}\oplus p_{3}=S(p^{*})$). It is therefore sufficient to switch every $p$
in eq. (45), (46), (47), (48) with $S(-p)$ and perform the same computations
again. Since we are expanding up to second order in $1/\kappa$, the final
result will be the same as before plus some additional contribution. Since we
have already shown in detail the previous computations, here we limit
ourselves to present the additional contribution to the numerator of the sum
of the four diagrams after all the shifts, Wick rotation, and Euclidean
simplifications have been performed. We call $\delta\Delta_{2}^{shift+Wick}$
the additional contribution to the sum of the numerators. We have

$\displaystyle\delta\Delta_{2}^{shift+Wick}=\frac{8k_{0}^{2}\mathbf{p}^{2}}{\kappa^{2}}+\frac{8\mathbf{p}^{2}\left(p_{0}^{2}\left(-x^{2}+x-2\right)+\mathbf{p}^{2}(x-1)x\right)}{\kappa^{2}}$
(216)

where we included that $1/\kappa$ powers to show that also in this case we
have no $1/\kappa$ contribution after summing all four diagrams contributions.
Notice that, because of eq. (151), the term with no $k$ in the numerator does
not contribute to the imaginary part, since it does not generate a $\log$ term
upon integrating over $k$. The only term which contributes is the term
$\frac{8k_{0}^{2}\mathbf{p}^{2}}{\kappa^{2}}$, and we know from eq. (150) that
it contributes to the total amplitude as

$\displaystyle-8\mathbf{p}^{2}\frac{1}{64\pi^{2}}\log\Delta$ (217)

and since the $\log$ contributes as $-\pi$ to the imaginary part we only need
to compute the quantity

$\displaystyle-\frac{g^{2}}{\kappa^{2}}\frac{1}{64\pi}\int_{L_{1}}^{L_{2}}8\mathbf{p}^{2}dx=-\Im\Pi^{U}(p^{2})\left(2\frac{p_{0}^{2}}{\kappa^{2}}-2\frac{p^{2}}{\kappa^{2}}\right).$
(218)

In other words, calling $\Im\Pi_{S(p)}$ the total amplitude
$\Im\Pi^{TOT}(p^{2})$ in eq. (214) corresponding to an initial on-shell
particle of momentum $S(p)$, and calling $\Im\Pi_{-p}$ the equivalent quantity
with momentum $-p$, we have

$\displaystyle\Im\Pi_{S(p)}-\Im\Pi_{-p}=\Im\Pi^{U}(p^{2})\left(2\frac{p_{0}^{2}}{\kappa^{2}}-2\frac{p^{2}}{\kappa^{2}}\right).$
(219)

The quantity $\frac{p^{2}}{\kappa^{2}}=\frac{m_{\psi}^{2}}{\kappa^{2}}$ is
invariant, but the quantity $\frac{p_{0}^{2}}{\kappa^{2}}$ is sensible to the
particles energy, which means that this difference can be experimentally
highlighted by going to higher energies.

###  5 Mass distribution width

In section 18 in chapter 1 we addressed possible phenomenological consequences
of the deformation of $CPT$ symmetry, but we did not consider for the moment
possible contributions coming from the higher order corrections to the
propagator of the decaying particle. Such corrections would induce a widening
of the mass distribution, and such contributions need to be taken into
account.

More formally, given a mass distribution function $\omega(m;M,\Gamma)$, where
$m$ is the resonant mass, and with mean mass $M$ and decay width $\Gamma$, one
can describe the time and momentum dependent decay amplitude with the
expression

$\displaystyle
a(t,\mathbf{p})=\int_{-\infty}^{\infty}dm\,\omega(m;M,\Gamma)\,e^{-it\sqrt{m^{2}+\mathbf{p}^{2}}}.$
(220)

The decay probability is then given by the usual formula
$\mathscr{P}=|a(t,\mathbf{p})|^{2}$. Since we are interested in ultra-
relativistic particles (we need high momenta and energies in order to
counterbalance the large value of $\kappa$ in the denominator), we consider
the relativistic Breit-Wigner distribution

$\displaystyle\omega(m;M,\Gamma)$
$\displaystyle=\frac{f(M,\Gamma)}{(m^{2}-M^{2})^{2}+M^{2}\Gamma^{2}},$ (221)
$\displaystyle f(M,\Gamma)$
$\displaystyle=\frac{2\sqrt{2}}{\pi}\frac{M\Gamma\sqrt{M^{2}(M^{2}+\Gamma^{2})}}{\big{[}M^{2}+\sqrt{M^{2}(M^{2}+\Gamma^{2})}\big{]}^{1/2}}.$
(222)

Notice that $f(M,\Gamma)$ is independent of $m$. Since we want to understand
the effect of deformation, we now switch the energy
$E=\sqrt{m^{2}+\mathbf{p}^{2}}$ at the exponent for its antipode $S(E)$ (see
eq. (533)). Furthermore, one can show that

$\displaystyle a(t,\mathbf{p})$ $\displaystyle=$ $\displaystyle\int
dm\,\frac{f(M,\Gamma)}{(m^{2}-M^{2})^{2}+M^{2}\Gamma^{2}}e^{-it(\sqrt{m^{2}+\mathbf{p}^{2}}-\mathbf{p}^{2}/\kappa)}$
(223) $\displaystyle=$ $\displaystyle
e^{-it(\sqrt{M^{2}+\mathbf{p}^{2}+iM\Gamma}-\mathbf{p}^{2}/\kappa)}.$

The new, corrected decay width $\tilde{\Gamma}$ (which takes into
consideration the non-trivial mass distribution width) can be now computed by
taking into account the real part of the exponent in eq. (223) (which will
describe the exponential decay of the amplitude). However, the only effect of
deformation comes into play in the term $-it\frac{\mathbf{p}^{2}}{\kappa}$,
i.e. in the imaginary part, so no correction to $\Gamma$ can be obtained from
a non-trivial mass distribution width. Indeed, one explicitly has

$\displaystyle\tilde{\Gamma}$ $\displaystyle=$ $\displaystyle
2\,\Im\Big{(}\sqrt{M^{2}+\mathbf{p}^{2}+iM\Gamma}-\frac{\mathbf{p}^{2}}{\kappa}\Big{)}$
(224) $\displaystyle=$
$\displaystyle\sqrt{2}\Big{[}\sqrt{(M^{2}+\mathbf{p}^{2})^{2}+M^{2}\Gamma^{2}}-(M^{2}+\Gamma^{2})\Big{]}^{1/2}.$

so that the only corrections are the same that would be present in the
canonical, non-deformed context.

###  6 Summary

In this work, we have presented our model of scalar field theory, and analysed
its features and the possible phenomenological consequences coming from our
results.

We first introduced the framework of $\kappa$-Minkowski spacetime and
$\kappa$-Poincaré Hopf algebra, which are the fundamental ingredients for any
other construction. Before going to the complex scalar field, we tackled the
interesting problem of the boost of a two-particle state system. The
determination of the finite boost of a two-particle system allows us to draw
interesting phenomenological consequences for any type of particle, since the
results are purely of kinematical nature.

We then passed to the complex scalar field. After some preliminary
mathematical work concerning the Weyl maps, the action of derivatives on star
products, and the integration-by-parts relations, we introduced the complex
scalar field action and computed the EoM (which turned out to be the canonical
Klein-Gordon equations). This allowed us to derive an explicit formula for the
on-shel fields $\phi$, $\phi^{\dagger}$. Furthermore, using the surface terms
obtained during the integration-by-parts, we obtained the Noether charges
coming from the continuous translation symmetry of the action.

To compute the remaining Noether charges we used the symplectic form, which
was also important for the fact that it allowed us to obtain the
creation/annihilation operators commutation relation. Together with the
charges, these allowed us to check that the algebra satisfied by the charges
is the canonical Poincaré algebra. Notice that, despite the fact that the EoM
are the canonical Klein-Gordon equations, and the fact that the charges
satisfy the canonical Poincaré algebra, there are still effect of deformation
on single-particle states which are made manifest by looking at the non-
trivial interaction between discrete and continuous transformations. Here we
explicitly checked that, in the deformed case, we have
$[\mathscr{N}_{i},C]\neq 0$. Furthermore, $\kappa$-deformation affects in non-
trivial way the features vacuum expectation values of fields. In this work, we
discussed in detail Greenberg’s theorem, showing that it does not hold in
general in the $\kappa$-deformed context. We then explored some
phenomenological consequences of the fact that $[\mathscr{N}_{i},C]\neq 0$, by
boosting a particle and an antiparticle, both originally at rest.

We then proceeded to compute the propagator and the imaginary part of the
1-loop correction to it. To do so, we exploit the path integral formalism for
the determination of the tree-level propagation, and we assume a deformed
conservation of momenta at each vertex. Because of this, a single diagram in
the non-deformed context is translated into four different diagrams in the
deformed context. We computed all of them for initial momentum of a particle
given by $S(p)$ and $-p$, obtaining a difference in the imaginary part
proportional to $p_{0}^{2}/\kappa^{2}$. We conclude with some comments on the
relevance of a non-trivial mass distribution width on the decay times of
boosted particles and antiparticles.

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