# A review of Quintessential Inflation

Jaume de Haro<EMAIL_ADDRESS>Departament de Matemàtiques, Universitat
Politècnica de Catalunya, Diagonal 647, 08028 Barcelona, Spain  Llibert Aresté
Saló<EMAIL_ADDRESS>School of Mathematical Sciences, Queen Mary
University of London, Mile End Road, London, E1 4NS, United Kingdom

###### Abstract

We compute numerically the reheating temperature due to the gravitational
production of conformally coupled superheavy particles during the phase
transition from the end of inflation to the beginning of kination in two
different Quintessential Inflation (QI) scenarios, namely Lorentzian
Quintessential Inflation (LQI) and $\alpha$-attractors in the context of
Quintessential Inflation ($\alpha$-QI). Once these superheavy particles have
been created, they must decay into lighter ones to form a relativistic plasma,
whose energy density will eventually dominate the one of the inflaton field in
order to reheat after inflation our universe with a very high temperature, in
both cases greater than $10^{7}$ GeV, contrary to the usual belief that heavy
masses suppress the particle production and, thus, lead to an inefficient
reheating temperature. Finally, we will show that the over-production of
Gravitational Waves (GWs) during this phase transition, when one deals with
our models, does not disturb the Big Bang Nucleosynthesis (BBN) success.

Gravitational Particle production; Quintessential Inflation; Reheating;
Gravitational Waves; Numerical Calculations.

###### pacs:

04.20.-q, 98.80.Jk, 98.80.Bp

Understanding the universe’s evolution is one of the greatest mysteries in the
history of humanity. It is always the primary question: “where we come from
and where we are going”. In particular, its early and late expansions have
been studied a great deal at present time. Looking at the scientific
literature, one can find two popular and well accepted theories -though they
are not observationally proved-, namely the inflation (the early surprisingly
fast accelerated expansion of our universe) and the quintessence as a form of
dark energy (the current cosmic acceleration). The inflationary paradigm guth
; linde ; albrecht  is actually a very fast accelerating phase of the early
universe that lasted for an extremely tiny time and became able to solve a
number of shortcomings associated with the standard Big Bang cosmology, such
as the horizon problem, flatness problem or the primordial monopole problem.
The predictive power of inflation was soon recognized due to its ability to
explain the origin of inhomogeneities in the universe as quantum fluctuations
during this epoch chibisov ; starobinsky ; pi ; bardeen ; Linde:1982uu ,
because such an explanation greatly matches with the recent observational data
from Planck’s team Planck . Thus, it is remarkable to note that inflation,
which appeared at the beginning of the 80’s, is still considered the best way
to explain the recent observational data, because it is nowadays the simplest
viable theory that describes almost correctly the early universe in agreement
with the recent observations Planck .

On the other hand, one of the most accepted explanations for the current
cosmic acceleration comes through the introduction of some quintessence field
Copeland:2006wr . In fact, soon after the discovery of the current cosmic
acceleration at the end of the last century riess ; perlmutter , a class of
pioneering cosmological models attempting to unify the early- and late-time
accelerating expansions were introduced. By construction, unlike the standard
quintessence models (see Tsujikawa  for a review), these models -named as
Quintessential Inflation (QI) models pv ; Spokoiny ; pr \- only contain one
classical scalar field, also named inflaton as in standard inflation guth ;
linde ; starobinsky ; albrecht , and it is shown that they succeed in
reproducing the two accelerated epochs of the universe expansion (see also
deHaro:2016hpl ; hap ; deHaro:2016hsh ; deHaro:2016ftq ; Geng:2017mic ;
AresteSalo:2017lkv ; Haro:2015ljc ; hossain1 ; hossain2 ; hossain3 ; hossain4
; guendelman1  for other interesting QI models). This idea to unify Inflation
with Quintessence was indeed a novel attempt by Peebles and Vilenkin in their
seminal paper pv , and the novelty of their proposal comes through the
introduction of a single potential that at early times allows inflation while
at late times provides quintessence. Thus, a unified picture of the universe
was effectively proposed connecting the distant early phase to the present
one. Thanks to this proposal, the origin of the scalar responsible for the
current inflation of the universe can be determined and fine-tunning problems
are reduced dimopoulos01 . In addition, the majority of models we deal with
only depend on two parameters, which are determined by observational data.
Hence, because of the behavior of the slow-roll regime as an attractor, the
dynamics of the model are obtained with the value of the scalar field and its
derivative -initial conditions- at some moment during inflation. This shows
the simplicity of Quintessential Inflation, which from our viewpoint is
simpler than standard quintessence, where a minimum of two fields are needed
to depict the evolution of the universe, the inflaton and a quintessence
field. Thus, one needs two different potentials and two different initial
conditions: one for the inflaton, which has to be fixed during inflation, and
another one for the quintessence field, whose initial conditions normally have
to be fixed at the beginning of the radiation era.

This enhanced more investigation in order to connect Quintessential Inflation
with the observational data dimopoulos1 ; Giovannini:2003jw ; hossain1 ;
hossain3 ; deHaro:2016hpl ; deHaro:2016hsh ; deHaro:2016ftq ; hap ;
Geng:2017mic ; AresteSalo:2017lkv ; Haro:2015ljc ; hyp  and, consequently,
this particular topic has become a popular area of research at the present
time. However, an important difference occurs with respect to the standard
inflationary paradigm, where the potential of the inflaton field has a local
minimum (a deep well) and, thus, the inflaton field releases its energy while
it oscillates, producing enough particles kls ; kls1 ; gkls ; stb ; Basset  to
reheat our universe. In contrast, for the ”non-oscillating” models, i.e., in
Quintessential Inflation, where the inflaton field survives to be able to
reproduce the current cosmic acceleration, the mechanism of reheating is
completely different: once the inflationary phase is completed, a reheating
mechanism keeping ”alive” the inflaton field is needed to match inflation with
the Hot Big Bang universe guth  because the particles existing before the
beginning of this period were completely diluted at the end of inflation
resulting in a very cold universe.

On this way, the most accepted idea to reheat the universe in the context of
QI comes through a phase transition of the universe from inflation to kination
(a regime where all the energy density of the inflation turns into kinetic
Joyce ) where the adiabatic regime is broken, which allows to create
particles. The mechanism to produce particles is not unique in this context
since a number of distinct ones are available and can be used. The first one
is the well-known Gravitational Particle Production studied long time ago in
Parker ; fmm ; glm ; gmm ; ford ; Zeldovich , at the end of the 90’s in Damour
; Giovannini  and more recently applied to Quintessential Inflation in
Spokoiny ; pv ; dimopoulos0 ; vardayan  for massless particles. A second
important mechanism is the so-called Instant Preheating introduced in fkl0
and applied for the first time to inflation in fkl  and recently in dimopoulos
; vardayan  in the context of $\alpha$-attractors in supergravity. Other less
popular mechanisms are the Curvaton Reheating applied to Quintessence
Inflation in FL ; ABM , the production of massive particles self-interacting
and coupled to gravity tommi  and the reheating via production of heavy
massive particles conformally coupled to gravity kolb ; kolb1 ; Birrell1 ;
hashiba ; hyp .

The production of superheavy massive particles conformally coupled to gravity
is one of the most important concerns of this review. These particles are
created during the phase transition from the end of inflation to the beginning
of kination and should decay into lighter ones to form a thermal relativistic
plasma, which eventually dominates to match this early period with the Hot Big
Bang universe. In fact, the main motivation for using a conformally-coupled
scalar field is its simplicity, which allows to employ the well-known
Hamiltonian diagonalization method (see gmmbook  for a review) to calculate
the energy density of the gravitationally produced particles, showing that
before the beginning of kination the vacuum polarization effects, which are
geometric objects associated to the creation and annihilation of the so-called
quasi-particles gmmbook , are sub-dominant and have no relevant effect in the
Friedmann equation. On the contrary, after the abrupt phase transition to
kination, heavy massive particles are produced and, since their energy density
decreases as $a^{-3}$ before decaying in lighter particles and as $a^{-4}$
after that, they will eventually dominate the energy density of the inflation
whose decrease is as $a^{-6}$. And, thus, the universe will become reheated.

Coming back to the original Peebles-Vilenkin model pv , the inflationary part
is described by a quartic potential and, according to the recent observations,
this does not suit well. To be explicit, for the quartic potential in the
inflationary part of this potential the two-dimensional contour of ($n_{s}$,
$r$), where $n_{s}$ is the scalar spectral index and $r$ is the ratio of
tensor to scalar perturbations, does not enter into the 95% confidence-level
of the Planck results Planck . However, a simple change in the inflationary
piece $-$quartic potential to a plateau one$-$ can solve this issue (see hap
for a detailed discussion and also see hyp ). On the other hand, the reheating
mechanism followed in pv  is the gravitational production of massless
particles that results in a reheating temperature of the order of $1$ TeV.
This reheating temperature is not sufficient to solve the overproduction of
the Gravitational Waves (GWs). As a result, the Big Bang Nucleosynthesis (BBN)
process could be hampered.

Dealing once again with the mechanisms to reheat the universe, the question
related to the bounds of the reheating temperature arises. Some works already
considered the constraints for reheating in Quintessential Inflation models,
both on Instant Preheating sami  and on Gravitational Particle Production
figueroa . A lower bound is obtained recalling that the radiation dominated
era is prior to the Big Bang Nucleosynthesis (BBN) epoch which occurs in the
$1$ MeV regime gkr . As a consequence, the reheating temperature has to be
greater than $1$ MeV (see also hasegawa  where the authors obtain lower limits
on the reheating temperature in the MeV regime assuming both radiative and
hadronic decays of relic particles only gravitationally interacting and taking
into account effects of neutrino self-interactions and oscillations in the
neutrino thermalization calculations). The upper bound may depend on the
theory we are dealing with; for instance, many supergravity and superstring
theories contain particles such as the gravitino or a modulus field with only
gravitational interactions and, thus, the late-time decay of these relic
products may disturb the success of the standard BBN lindley , but this
problem can be successfully removed if the reheating temperature is of the
order of $10^{9}$ GeV (see for instance eln ). This is the reason why we will
restrict the reheating temperature to remain, more or less, between $1$ MeV
and $10^{9}$ GeV.

Finally, one has to take into account that a viable reheating mechanism has to
deal with the effect of the Gravitational Waves (GWs), which are also produced
during the phase transition, in the BBN success by satisfying the
observational bounds coming from the overproduction of the GWs pv  or related
to the logarithmic spectrum of their energy density maggiore . As we will see
throughout this review, the overproduction of GWs does not disturb the BBN
success in the models studied (Lorentzian Quintessential Inflation and
$\alpha$-attractors in the context of QI), when the reheating is due to the
production of superheavy particles, which shows their viability.

In summary, due to the lack of works that explain all the outstanding
contributions made in Quintessential Inflation since the appearance of the
seminal Peebles-Vilenkin article until now, the main goal of this review is to
show the reader the most relevant aspects obtained on this topic during this
time period. To do it we have chosen the most prominent papers on the subject
from our viewpoint, and we have used the results provided by them to write
this review.

The review is organized as follows: In Section I we review, with great detail,
the seminal Peebles-Vilenkin model, calculating the value of the two
parameters on which the model depends, the evolution of the inflaton field
throughout history, the number of e-folds as a function of the reheating
temperature, and additionally introducing models that improve the original
one. Section II is devoted to the study of a class of Exponential
Quintessential Inflation models. We show that, with initial conditions when
the pivot scale leaves the Hubble radius (recall that this happens in the
slow-roll regime which is an attractor, so one can safely take initial
conditions close to the slow-roll solution) during the radiation phase, the
solution is never in the basin of attraction of the scaling solution.
Therefore, in order to match with the current observational data the potential
should be modified introducing a pure exponential term. In Section III we
review the so-called Lorentzian Quintessential Inflation (LQI), which is based
on the assumption that the main slow-roll parameter as a function of the
number of e-folds evolves as the Lorentzian distribution. The
$\alpha$-attractors in the context of Quintessential Inflation are revisited
in Section IV, showing that we obtain the same results as in the case of LQI.
In Section V we review other Quintessential Inflation models and in particular
some aspects of the work of K. Dimopoulos in QI. Next, in Section VI we study
some reheating mechanisms in Quintessential Inflation, namely via
Gravitational Particle Production of massless and superheavy particles, via
Instant Preheating and via Curvature Reheating, obtaining bounds for the
reheating temperature. In Section VII we deal with the constraints to preserve
the BBN success coming from the logarithmic spectrum of GWs and also from its
overproduction during the phase transition from the end of inflation to the
beginning of kination, showing that for the LQI model, when the reheating is
via gravitational production of superheavy particles, all these bounds are
easily overpassed. The last section is devoted to the conclusions of the
review.

The units used throughout the paper are $\hbar=c=1$, and the reduced Planck’s
mass has been denoted by $M_{pl}\equiv\frac{1}{\sqrt{8\pi G}}\cong 2.44\times
10^{18}$ GeV.

##  I The Peebles-Vilenkin model

The first Quintessential Inflation scenario was proposed by Peebles and
Vilenkin in their seminal paper pv  at the end of the last century, soon after
the discovery of the current cosmic acceleration riess ; perlmutter . The
corresponding potential is given by

$\displaystyle
V(\varphi)=\left\\{\begin{array}[]{ccc}\lambda(\varphi^{4}+M^{4})&\mbox{for}&\varphi\leq
0\\\ \lambda\frac{M^{8}}{\varphi^{4}+M^{4}}&\mbox{for}&\varphi\geq
0,\end{array}\right.$ (3)

where $\lambda$ is a dimensionless parameter and $M\ll M_{pl}$ is a very small
mass compared with the Planck’s one. At this point, note that the quartic
potential is the responsible for inflation and the inverse power law leads to
dark energy (in that case quintessence) at late times.

One can see that the model is very simple and only depends on two parameters
which can be calculated as follows: First of all, we calculate the main slow-
roll parameters

$\displaystyle\epsilon\equiv\frac{M_{pl}^{2}}{2}\left(\frac{V_{\varphi}}{V}\right)^{2}\cong\frac{8M_{pl}^{2}}{\varphi^{2}},\qquad\eta\equiv{M_{pl}^{2}}\frac{V_{\varphi\varphi}}{V}=\frac{12M_{pl}^{2}}{\varphi^{2}}.$
(4)

So, the spectral index and the ratio of tensor to scalar perturbations are
given by btw

$\displaystyle n_{s}\cong 1-6\epsilon_{*}+2\eta_{*}\cong
1-\frac{24M_{pl}^{2}}{\varphi^{2}_{*}},\qquad r\cong
16\epsilon_{*}\cong\frac{128M_{pl}^{2}}{\varphi^{2}_{*}},$ (5)

where the star denotes that the quantities are evaluated when the pivot scale
leaves the Hubble radius (at the horizon crossing). Thus, one has the
important relation $16(1-n_{s})=3r$ between both quantities.

On the other hand, using the formula of the power spectrum of scalar
perturbations (see for instance btw )

$\displaystyle P_{\zeta}=\frac{H_{*}^{2}}{8\pi^{2}\epsilon_{*}M_{pl}^{2}}\sim
2\times 10^{-9},$ (6)

for the Peebles-Vilenkin model it yields that

$\displaystyle
P_{\zeta}\cong\frac{\lambda\varphi_{*}^{4}}{24\pi^{2}\epsilon_{*}M_{pl}^{4}}\cong\frac{8\lambda}{3\pi^{2}\epsilon_{*}^{3}}\cong\frac{72\lambda}{\pi^{2}(1-n_{s})^{3}}\sim
2\times 10^{-9},$ (7)

where we have used the formulas (4) and (5). And, taking for example
$n_{s}\cong 0.96$, which is approximately its observational value planck18 ;
planck18a , we get

$\displaystyle\lambda\sim\frac{\pi^{2}}{36}(1-n_{s})^{3}\times 10^{-9}\sim
10^{-14}.$ (8)

The calculation of the other parameter is more involved and for that one needs
to know the evolution of the inflaton field up to the present time. As we will
see, the value of the inflaton at the present time is of the order of
$40M_{pl}$ (see Fig. 2). So, taking into account that in order to reproduce
the current cosmic acceleration the kinetic energy has to be negligible
compared with the potential one at the present time, in order to match with
the present energy density we will have

$\displaystyle\lambda\frac{M^{8}}{40^{4}M_{pl}^{4}+M^{4}}\cong\lambda\frac{M^{8}}{40^{4}M_{pl}^{4}}\sim
H_{0}^{2}M_{pl}^{2},$ (9)

where $H_{0}\sim 10^{-61}M_{pl}$ is the present value of the Hubble rate.
Finally, we get

$\displaystyle M\sim 40^{1/2}\lambda^{-1/8}H_{0}^{1/4}M_{pl}^{3/4}\sim
10^{-13}M_{pl}\sim 10^{5}\mbox{ GeV}.$ (10)

Unfortunately, the original Peebles-Vilenkin model provides a spectral index
and a tensor/scalar ratio that do not enter in the two dimensional
marginalized joint confidence contour at $2\sigma$ CL for the Planck TT, TE,
EE + low E+ lensing + BK14 + BAO likelihood (see Fig. 1).

Figure 1: Clearly the line $16(1-n_{s})=3r$ (black line) does not enter in the
$2\sigma$ Confidence Level, in contrast to the parabola
$(1-n_{s})^{2}=\frac{\alpha^{2}}{2}r$ (red line) with $\alpha=1$ from the
improved model studied in next subsection

(Figure courtesy of the Planck 2018 Collaboration planck18a )

For this reason, we need to improve the part of the potential which provides
inflation, that is, the quartic potential.

###  I.1 Improvements

For that purpose, one would need to consider plateau potentials Geng:2017mic
or $\alpha$-attractors attractor1 ; vardayan , such as an Exponential SUSY
Inflation type potential

$\displaystyle V^{1}_{\alpha}(\varphi)=\left\\{\begin{array}[]{ccc}\lambda
M_{pl}^{4}\left(1-e^{\alpha\varphi/M_{pl}}+\left(\frac{M}{M_{pl}}\right)^{4}\right)&\mbox{for}&\varphi\leq
0\\\ \lambda\frac{M^{8}}{\varphi^{4}+M^{4}}&\mbox{for}&\varphi\geq
0,\end{array}\right.$ (13)

or a Higgs Inflation-type potential

$\displaystyle V^{2}_{\alpha}(\varphi)=\left\\{\begin{array}[]{ccc}\lambda
M_{pl}^{4}\left(1-e^{\alpha\varphi/M_{pl}}+\left(\frac{M}{M_{pl}}\right)^{2}\right)^{2}&\mbox{for}&\varphi\leq
0\\\ \lambda\frac{M^{8}}{\varphi^{4}+M^{4}}&\mbox{for}&\varphi\geq
0.\end{array}\right.$ (16)

For the first model, the slow-roll parameters are given by

$\displaystyle\epsilon\cong\frac{\alpha^{2}}{2}\frac{e^{2\alpha\varphi/M_{pl}}}{\left(1-e^{\alpha\varphi/M_{pl}}\right)^{2}},\qquad\eta\cong-{\alpha^{2}}\frac{e^{\alpha\varphi/M_{pl}}}{1-e^{\alpha\varphi/M_{pl}}},$
(17)

and, taking into account that when the pivot scale leaves the Hubble radius
$e^{\alpha\varphi_{*}/M_{pl}}\ll 1$, one gets

$\displaystyle\epsilon_{*}\cong\frac{\alpha^{2}}{2}{e^{2\alpha\varphi_{*}/M_{pl}}},\qquad\eta_{*}\cong-{\alpha^{2}}{e^{\alpha\varphi_{*}/M_{pl}}},$
(18)

and thus, the spectral index and the tensor/scalar ratio are equal to

$\displaystyle n_{s}\cong 1-6\epsilon_{*}+2\eta_{*}\cong
1-2{\alpha^{2}}{e^{\alpha\varphi_{*}/M_{pl}}},\qquad r\cong
8{\alpha^{2}}{e^{2\alpha\varphi_{*}/M_{pl}}},$ (19)

obtaining the relation $(1-n_{s})^{2}=\frac{\alpha^{2}}{2}r$.

We can also calculate the number of e-folds from the horizon crossing to the
end of inflation, namely ${\mathcal{N}}=\int_{t_{*}}^{t_{END}}Hdt$. In the
slow-roll regime, where

$\displaystyle 3H\dot{\varphi}+V_{\varphi}\cong 0,\qquad
H^{2}\cong\frac{V}{3M_{pl}^{2}},$ (20)

it can be approximated by

$\displaystyle{\mathcal{N}}\cong\frac{1}{M_{pl}}\int_{\varphi_{*}}^{\varphi_{END}}\frac{1}{\sqrt{2\epsilon}}d\varphi,$
(21)

which for the first potential leads to

$\displaystyle{\mathcal{N}}\cong\frac{1}{\alpha
M_{pl}}\int_{\varphi_{*}}^{\varphi_{END}}\left(e^{-\alpha\varphi/M_{pl}}-1\right)d\varphi=\frac{1}{\alpha^{2}}\left(e^{-\alpha\varphi_{*}/M_{pl}}-e^{-\alpha\varphi_{END}/M_{pl}}\right)$
$\displaystyle-\frac{1}{\alpha
M_{pl}}(\varphi_{END}-\varphi_{*})\cong\frac{1}{\alpha^{2}}e^{-\alpha\varphi_{*}/M_{pl}}\cong\frac{2}{1-n_{s}}.$
(22)

In the same way, for the second potential, the slow-roll parameters are given
by

$\displaystyle\epsilon\cong
2{\alpha^{2}}\frac{e^{2\alpha\varphi/M_{pl}}}{\left(1-e^{\alpha\varphi/M_{pl}}\right)^{2}},\qquad\eta\cong-2{\alpha^{2}}\frac{e^{\alpha\varphi/M_{pl}}}{(1-e^{\alpha\varphi/M_{pl}})^{2}},$
(23)

and, taking into account again that when the pivot scale leaves the Hubble
radius $e^{\alpha\varphi_{*}/M_{pl}}\ll 1$, we get

$\displaystyle\epsilon_{*}\cong
2{\alpha^{2}}{e^{2\alpha\varphi_{*}/M_{pl}}},\qquad\eta_{*}\cong-2{\alpha^{2}}{e^{\alpha\varphi_{*}/M_{pl}}},$
(24)

and thus, the spectral index and the tensor/scalar ratio are equal to

$\displaystyle n_{s}\cong 1-6\epsilon_{*}+2\eta_{*}\cong
1-4{\alpha^{2}}{e^{\alpha\varphi_{*}/M_{pl}}},\qquad r\cong
32{\alpha^{2}}{e^{2\alpha\varphi_{*}/M_{pl}}},$ (25)

obtaining the same relation as for the other potential, namely
$(1-n_{s})^{2}=\frac{\alpha^{2}}{2}r$.

In addition, for the number of e-folds, one also obtains the same result

$\displaystyle{\mathcal{N}}\cong\frac{1}{2\alpha^{2}}e^{-\alpha\varphi_{*}/M_{pl}}\cong\frac{2}{1-n_{s}},$
(26)

and what is important to note is that for both potentials the spectral index
and the ratio of tensor to scalar perturbations have the same relation with
the number of e-folds

$\displaystyle n_{s}\cong 1-\frac{2}{{\mathcal{N}}},\qquad
r\cong\frac{8}{\alpha^{2}{\mathcal{N}}^{2}},$ (27)

which implies that for $\alpha\sim{\mathcal{O}}(1)$ and for a number of
$e$-folds greater than $60$, which is typical in quintessential inflation due
to the kination phase, the ratio of tensor to scalar perturbations is less
than $0.003$. Thus, the spectral index and the tensor/scalar ratio enter
perfectly in the two dimensional marginalized joint confidence contour at
$2\sigma$ CL for the Planck TT, TE, EE + low E+ lensing + BK14 + BAO
likelihood (see Fig. 1).

###  I.2 Dynamical evolution of the Peebles-Vilenkin model: from the beginning
of kination to the matter-radiation equality

Now we want to understand the evolution of the Quintessential Inflation models
after inflation, which only depends on the tail of the potential. We will
focus on the Peebles-Vilenkin one, and for that case, inflation ends when
$\epsilon=1$, which occurs for $\varphi_{END}=2\sqrt{2}M_{pl}$. Taking into
account that an alternative expression of this slow-roll parameter is
$\epsilon=-\frac{\dot{H}}{H^{2}}$, we conclude that at the end of inflation
the effective Equation of State (EoS) parameter, namely $w_{eff}$, satisfies

$\displaystyle
w_{eff}=\frac{P}{\rho}=\frac{\dot{\varphi}^{2}/2-V}{\dot{\varphi}^{2}/2+V}=-1-\frac{2\dot{H}}{3H^{2}}=-1/3,$
(28)

where we have used the Friedmann and Raychaudhuri equations

$\displaystyle
H^{2}=\frac{\rho}{3M_{pl}^{2}},\qquad\dot{H}=-\frac{1}{2M_{pl}^{2}}(\rho+P),$
(29)

being $\rho$ the energy density and $P$ the corresponding pressure.

Then, from (28) one gets the following relation at the end of inflation,
$\dot{\varphi}_{END}^{2}=V(\varphi_{END})$, meaning that the energy density at
the end of inflation is

$\displaystyle\rho_{END}=\frac{3}{2}V(\varphi_{END})=\frac{3}{2}\lambda(\varphi_{END}^{4}+M^{4})\cong\frac{3}{2}\lambda\varphi_{END}^{4}=96\lambda
M_{pl}^{4}.$ (30)

Next, we have to assume, as usual, that there is no drop of energy between the
end of inflation and the beginning of kination, which for these models occurs
when $\varphi_{kin}\cong 0$. So, at the beginning of kination the energy
density is approximately given by $\rho_{kin}\cong 96\lambda
M_{pl}^{4}\Longrightarrow H_{kin}\cong 4\sqrt{2\lambda}M_{pl}$ and, since all
the energy is practically kinetic, the effective EoS parameter is very close
to $w_{eff}\cong 1$. For this value, combining the Friedmann and Raychaudhuri
equations (29), one gets

$\displaystyle\dot{H}=-3H^{2},$ (31)

whose solution is given by $H(t)=\frac{1}{3t}$.

Then, coming back to the Friedmann equation and taking into account that
during kination all the energy density is kinetic, we get

$\displaystyle\frac{1}{9t^{2}}=\frac{\dot{\varphi}^{2}}{6M_{pl}^{2}}\Longrightarrow\dot{\varphi}(t)=\sqrt{\frac{2}{3}}\frac{M_{pl}}{t}\Longrightarrow\varphi(t)=\varphi_{kin}+\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{t}{t_{kin}}\right),$
(32)

which in terms of the Hubble rate could be written as follows:

$\displaystyle\dot{\varphi}(t)=\sqrt{6}H(t){M_{pl}},\qquad\varphi(t)=\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{H_{kin}}{H(t)}\right).$
(33)

At this point, we have to take into account that during the phase transition
from the end of inflation to the beginning of kination the adiabatic regime is
broken and particles are created. Here we will assume that these particles,
which we consider very massive and conformally coupled to gravity, are
gravitationally produced. So, they must decay into lighter ones in order to
form a relativistic plasma which will eventually dominate the energy density
of the universe and will become reheated. And then, since the kination regime
ends when the energy density of the inflaton field is of the same order than
the one of the produced particles, two different scenarios appear:

  1. 1.

Decay before the end of the kination period.

  2. 2.

Decay after the end of the kination period.

####  I.2.1 Decay before the end of kination

In this case, since the thermalization of the decay products is nearly
instantaneous and occurs before the end of kination, the reheating time
coincides with the end of the kination. Hence, we will have

$\displaystyle\dot{\varphi}_{reh}=\sqrt{6}H_{reh}{M_{pl}},\qquad\varphi_{reh}=\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{H_{kin}}{H_{reh}}\right).$
(34)

Taking into account that at the reheating time the energy density of the
inflaton field is the same as the one of the radiation, the Friedmann equation
will become $H_{reh}^{2}=\frac{2\rho_{reh}}{3M_{pl}^{2}}$ and, using the
Stefan-Boltzmann law $\rho_{reh}=\frac{\pi^{2}}{30}g_{reh}T_{reh}^{4}$, where
$g_{reh}$ denotes the effective number of degrees of freedom at the reheating
time, we get

$\displaystyle\varphi_{reh}=\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{12}{\pi}\sqrt{\frac{10\lambda}{g_{reh}}}\left(\frac{M_{pl}}{T_{reh}}\right)^{2}\right)\qquad\mbox{and}\qquad{\dot{\varphi}_{reh}=\sqrt{\frac{\pi^{2}g_{reh}}{15}}T_{reh}^{2}}.$
(35)

The next step is to calculate the value of the field at the beginning of the
matter-radiation equality. During the radiation domination the effective EoS
parameter is $w_{eff}=1/3$. Thus, combining once again the Friedmann and
Raychaudhuri equations we have $\dot{H}=-2H^{2}$, whose solution is
$H(t)=\frac{1}{2t}$. Then, an approximate solution of the conservation
equation

$\displaystyle\ddot{\varphi}+3H\dot{\varphi}+V_{\varphi}=0$ (36)

can be obtained disregarding once again the potential, because during the
radiation domination the potential energy is irrelevant compared with the
kinetic one. Hence, the equation

$\displaystyle\ddot{\varphi}+\frac{3}{2t}\dot{\varphi}=0$ (37)

has the following solution,

$\displaystyle\dot{\varphi}(t)=\dot{\varphi}_{reh}\left(\frac{t_{reh}}{t}\right)^{3/2}=\sqrt{\frac{\pi^{2}g_{reh}}{15}}T_{reh}^{2}\left(\frac{2H(t)}{3H_{reh}}\right)^{3/2},$
$\displaystyle\varphi(t)=\varphi_{reh}+2\dot{\varphi}_{reh}t_{reh}\left(1-\sqrt{\frac{t_{reh}}{t}}\right)=\varphi_{reh}+2\sqrt{\frac{2}{3}}M_{pl}\left(1-\sqrt{\frac{2H(t)}{3H_{reh}}}\right),$
(38)

which at the matter-radiation equality leads to

$\displaystyle\varphi_{eq}=\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{12}{\pi}\sqrt{\frac{10\lambda}{g_{reh}}}\left(\frac{M_{pl}}{T_{reh}}\right)^{2}\right)+2\sqrt{\frac{2}{3}}M_{pl}\left(1-\sqrt{\frac{2}{3}}\left(\frac{g_{eq}}{g_{rh}}\right)^{1/4}\frac{T_{eq}}{T_{rh}}\right),$
$\displaystyle\dot{\varphi}_{eq}=\frac{2\pi}{9}\sqrt{\frac{2g_{eq}}{5}}\left(\frac{g_{eq}}{g_{rh}}\right)^{1/4}\frac{T_{eq}^{3}}{T_{rh}},$
(39)

where we have used that $H_{eq}^{2}=\frac{2\rho_{eq}}{3M_{pl}^{2}}$ with
$\rho_{eq}=\frac{\pi^{2}}{30}g_{eq}T_{eq}^{4}$ being $g_{eq}\cong 3.36$ the
effective number of degrees of freedom at the matter-radiation epoch.

Finally, noticing the adiabatic evolution after the matter-radiation, we will
have $T_{eq}=\frac{a_{0}}{a_{eq}}T_{0}=(1+z_{eq})T_{0}$, where the sub-index
$"0"$ denotes the present time and $z_{eq}$ is the red-shift at the matter-
radiation equality. So,

$\displaystyle\varphi_{eq}=\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{12}{\pi}\sqrt{\frac{10\lambda}{g_{reh}}}\left(\frac{M_{pl}}{T_{reh}}\right)^{2}\right)+2\sqrt{\frac{2}{3}}M_{pl}\left(1-\sqrt{\frac{2}{3}}\left(\frac{g_{eq}}{g_{rh}}\right)^{1/4}(1+z_{eq})\frac{T_{0}}{T_{reh}}\right),$
$\displaystyle\dot{\varphi}_{eq}=\frac{2\pi}{9}\sqrt{\frac{2g_{eq}}{5}}\left(\frac{g_{eq}}{g_{rh}}\right)^{1/4}(1+z_{eq})^{3}\frac{T_{0}}{T_{reh}}T_{0}^{2},$
(40)

that is, we have obtained the values of the field at the matter-radiation
equality as a function of the reheating temperature and the observational data
$T_{0}\cong 2.73$ K and $z_{eq}\sim 3\times 10^{3}$. In fact, since $T_{0}\ll
T_{reh}$, one can safely do the approximation

$\displaystyle\varphi_{eq}\cong\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{12}{\pi}\sqrt{\frac{10\lambda}{g_{reh}}}\left(\frac{M_{pl}}{T_{reh}}\right)^{2}\right),\qquad\dot{\varphi}_{eq}\cong
0,$ (41)

which only depends on the reheating temperature.

####  I.2.2 Decay after the end of the kination period

In that scenario, at the end of kination, when the energy density of the field
is equal to the one of the produced particles, one has

$\displaystyle\varphi_{end}=\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{H_{kin}}{H_{end}}\right),\qquad\dot{\varphi}_{end}=\sqrt{6}M_{pl}H_{end}.$
(42)

Denoting by $\rho_{\varphi}(t)$ the energy density of the inflaton field and
by $\rho_{\chi}$ the energy density of the produced particles, at the end of
kination we will have

$\displaystyle\rho_{\varphi,end}=\rho_{\varphi,kin}\left(\frac{a_{kin}}{a_{end}}\right)^{6},\qquad\rho_{\chi,end}=\rho_{\chi,kin}\left(\frac{a_{kin}}{a_{end}}\right)^{3},$
(43)

and, since $\rho_{\varphi,end}=\rho_{\chi,end}$, the important parameter
defined by $\Theta\equiv\left(\frac{a_{kin}}{a_{end}}\right)^{3}$ and named
heat efficiency satisfies

$\displaystyle\Theta=\frac{\rho_{\chi,kin}}{\rho_{\varphi,kin}}=\frac{\rho_{\chi,kin}}{96\lambda
M_{pl}^{4}}.$ (44)

On the other hand, using that
$H_{end}^{2}=\frac{2\rho_{\varphi,end}}{3M_{pl}^{2}}$ one arrives at
$H_{end}=\sqrt{2}H_{kin}\Theta=8\sqrt{\lambda}M_{pl}\Theta$, and then

$\displaystyle\varphi_{end}=-\sqrt{\frac{2}{3}}M_{pl}\ln\left(\sqrt{2}\Theta\right),\qquad\dot{\varphi}_{end}=8\sqrt{6\lambda}M_{pl}^{2}\Theta.$
(45)

During the period between $t_{end}$ and $t_{reh}$ the universe is matter
dominated and, thus, the Hubble parameter evolves as $H=\frac{2}{3t}$. Since
the gradient of the potential can also be disregarded at this epoch, the
equation of the scalar field becomes
$\ddot{\varphi}+\frac{2}{t}\dot{\varphi}=0$, yielding at the reheating time

$\displaystyle\varphi_{reh}=\varphi_{end}+\sqrt{\frac{2}{3}}M_{pl}\left(1-\frac{{t}_{end}}{t_{reh}}\right)=\varphi_{end}+\sqrt{\frac{2}{3}}M_{pl}\left(1-\frac{H_{reh}}{2H_{end}}\right)$
$\displaystyle=-\sqrt{\frac{2}{3}}M_{pl}\ln\left(\sqrt{2}\Theta\right)+\sqrt{\frac{2}{3}}M_{pl}\left(1-{\frac{\pi}{48}\sqrt{\frac{g_{reh}}{5\lambda}}\frac{T_{reh}^{2}}{M_{pl}^{2}\Theta}}\right)$
(46)

and

$\displaystyle\dot{\varphi}_{reh}=\sqrt{\frac{2}{3}}\frac{M_{pl}{t}_{end}}{t_{reh}^{2}}=\frac{\sqrt{3}}{16\sqrt{2\lambda}}\frac{H_{rh}^{2}}{\Theta}=\frac{\sqrt{3}\pi^{2}}{720\sqrt{2\lambda}}\frac{g_{reh}T_{reh}^{4}}{M_{pl}^{2}\Theta}.$
(47)

Finally, in all the radiation period one can continue disregarding the
potential, obtaining

$\displaystyle\varphi(t)=\varphi_{reh}+2\dot{\varphi}_{reh}t_{ter}\left(1-\sqrt{\frac{t_{reh}}{t}}\right),$
(48)

and thus, at the matter-radiation equality one has

$\displaystyle\varphi_{eq}=\varphi_{rh}+\sqrt{\frac{2}{3}}M_{pl}\frac{H_{reh}}{H_{end}}\left(1-\sqrt{\frac{4H_{eq}}{3H_{rh}}}\right)$
$\displaystyle=\varphi_{rh}+\frac{\pi}{24}\sqrt{\frac{g_{reh}}{30\lambda}}\frac{T_{reh}^{2}}{M_{pl}\Theta}\left(1-\sqrt{\frac{4}{3}}\left(\frac{g_{eq}}{g_{reh}}\right)^{1/4}\frac{T_{eq}}{T_{reh}}\right)$
$\displaystyle=-\sqrt{\frac{2}{3}}M_{pl}\ln\left(\sqrt{2}\Theta\right)+\sqrt{\frac{2}{3}}M_{pl}\left(1-{\frac{\pi}{48}\sqrt{\frac{g_{reh}}{5\lambda}}\frac{T_{reh}^{2}}{M_{pl}^{2}\Theta}}\right)$
(49)
$\displaystyle+\frac{\pi}{24}\sqrt{\frac{g_{reh}}{30\lambda}}\frac{T_{reh}^{2}}{M_{pl}\Theta}\left(1-\sqrt{\frac{4}{3}}\left(\frac{g_{eq}}{g_{reh}}\right)^{1/4}\frac{T_{eq}}{T_{reh}}\right)$
$\displaystyle=-\sqrt{\frac{2}{3}}M_{pl}\ln\left(\sqrt{2}\Theta\right)+\sqrt{\frac{2}{3}}M_{pl}-\frac{\pi}{36}\sqrt{\frac{g_{reh}}{10\lambda}}\left(\frac{g_{eq}}{g_{reh}}\right)^{1/4}(1+z_{eq})\frac{T_{0}T_{reh}}{M_{pl}\Theta},$

and in the same way,

$\displaystyle\dot{\varphi}_{eq}=\dot{\varphi}_{reh}\left(\frac{t_{reh}}{t_{eq}}\right)^{\frac{3}{2}}=\left(\frac{16g_{eq}}{9g_{reh}}\right)^{\frac{3}{4}}\left(\frac{T_{eq}}{T_{reh}}\right)^{3}\dot{\varphi}_{reh}=\frac{\pi^{2}(1+z_{eq})^{3}}{270\sqrt{2\lambda}}\left(\frac{g_{eq}^{3}}{g_{reh}}\right)^{\frac{1}{4}}\frac{T_{0}^{3}T_{reh}}{M_{pl}^{2}\Theta}.$
(50)

As we will see, by calculating the value of $\Theta$ for several reheating
mechanisms, we can safely do the approximation

$\displaystyle\varphi_{eq}\cong-\sqrt{\frac{2}{3}}M_{pl}\ln\left(\sqrt{2}\Theta\right)+\sqrt{\frac{2}{3}}M_{pl},\qquad\dot{\varphi}_{eq}\cong
0,$ (51)

which only depends on the value of the parameter $\Theta$. Effectively, using
for example Instant Preheating (see subsection 7.2), one gets
$\Theta=\frac{g^{2}}{2\pi^{2}}$, which for the viable value of the coupling
$g\sim 10^{-4}$ yields $\Theta\sim 10^{-9}$, and one can easily check that the
approximation holds.

###  I.3 Dynamical evolution of the Peebles-Vilenkin model: from the matter-
radiation equality up to now

After the matter-radiation equality the dynamical equations cannot be solved
analytically and, thus, one needs to use numerics to compute them. In order to
do that, we need to use a ”time” variable that we choose to be the number of
$e$-folds up to the present epoch, namely
$N\equiv-\ln(1+z)=\ln\left(\frac{a}{a_{0}}\right)$. Now, using the variable
$N$, one can recast the energy density of radiation and matter respectively as

$\displaystyle\rho_{r}(a)={\rho_{r,eq}}\left(\frac{a_{eq}}{a}\right)^{4}\Longrightarrow\rho_{r}(N)=\frac{\rho_{eq}}{2}e^{4(N_{eq}-N)}$
(52)

and

$\displaystyle\rho_{m}(a)={\rho_{m,eq}}\left(\frac{a_{eq}}{a}\right)^{3}\Longrightarrow\rho_{m}(N)=\frac{\rho_{eq}}{2}e^{3(N_{eq}-N)},$
(53)

where $\rho_{eq}=\rho_{m,eq}+\rho_{r,eq}$ and the value of the energy density
of the matter or radiation $\rho_{m,eq}=\rho_{r,eq}$ at the matter-radiation
equality can be calculated as follows: Using the present value of the ratio of
the matter energy density to the critical one $\Omega_{m,0}=0.308$ and the
present value of the Hubble rate $H_{0}\cong 68\;\mbox{Km/sec/Mpc}\sim
10^{-33}$ eV, one can deduce that the present value of the matter energy
density is $\rho_{m,0}=3H_{0}^{2}M_{pl}^{2}\Omega_{m,0}\cong 3\times
10^{-121}M_{pl}^{4}$, and at the matter-radiation equality
$\rho_{m,eq}=\rho_{m,0}(1+z_{eq})^{3}$.

In order to obtain the dynamical system for this scalar field model, we
introduce the following dimensionless variables,

$\displaystyle x=\frac{\varphi}{M_{pl}},\qquad
y=\frac{\dot{\varphi}}{H_{0}M_{pl}},$ (54)

where $H_{0}\sim 1.42\times 10^{-33}$ eV denotes once again the current value
of the Hubble parameter. Now, using the variable $N=-\ln(1+z)$ defined above
and also using the conservation equation
$\ddot{\varphi}+3H\dot{\varphi}+V_{\varphi}=0$, one can construct the
following non-autonomous dynamical system,

$\displaystyle\left\\{\begin{array}[]{ccc}x^{\prime}&=&\frac{y}{\bar{H}}~{},\\\
y^{\prime}&=&-3y-\frac{\bar{V}_{x}}{\bar{H}}~{},\end{array}\right.$ (57)

where the prime represents the derivative with respect to $N$,
$\bar{H}=\frac{H}{H_{0}}$ and $\bar{V}=\frac{V}{H_{0}^{2}M_{pl}^{2}}$.
Moreover, the Friedmann equation now looks as

$\displaystyle\bar{H}(N)=\frac{1}{\sqrt{3}}\sqrt{\frac{y^{2}}{2}+\bar{V}(x)+\bar{\rho}_{r}(N)+\bar{\rho}_{m}(N)}~{},$
(58)

where we have introduced the dimensionless energy densities
$\bar{\rho}_{r}=\frac{\rho_{r}}{H_{0}^{2}M_{pl}^{2}}$ and
$\bar{\rho}_{m}=\frac{\rho_{m}}{H_{0}^{2}M_{pl}^{2}}$.

Then, we have to integrate the dynamical system, starting at
$N_{eq}=-\ln(1+z_{eq})$, with initial conditions $x_{eq}$ and $y_{eq}$ which
are obtained analytically in the previous subsection. The value of the
parameter $M$ is obtained equaling at $N=0$ the equation (58) to $1$, i.e.,
imposing $\bar{H}(0)=1$.

Numerical calculations show that $M\sim 10^{5}$ GeV for a reheating
temperature of $10^{5}$ GeV, as in our heuristic reasoning. In addition, it
remains of the same order for the other values of the reheating temperature
within the allowed range. In Figure 2 we display the evolution of the scalar
field $\varphi$ and in Figure 3 we plot the graph of the parameter
$\Omega_{i}\equiv\frac{\rho_{i}}{3H^{2}M_{pl}^{2}}$ for matter, radiation and
dark energy, and also the evolution of the effective EoS parameter.

Figure 2: Evolution of the scalar field $\varphi$ in the improved Peebles-
Vilenkin model.

Figure 3: Left: The density parameters
$\Omega_{m}=\frac{\rho_{m}}{3H^{2}M_{pl}^{2}}$ (orange curve),
$\Omega_{r}=\frac{\rho_{r}}{3H^{2}M_{pl}^{2}}$ (blue curve) and
$\Omega_{\varphi})=\frac{\rho_{\varphi}}{3H^{2}M_{pl}^{2}}$, from kination to
future times. Right: The effective Equation of State parameter $w_{eff}$, from
kination to future times. As one can see in the picture, after kination the
Universe enters in a large period of time where radiation dominates. Then,
after the matter-radiation equality, the Universe becomes matter-dominated
and, finally, near the present, it enters in a new accelerated phase where
$w_{eff}$ approaches $-1$.

###  I.4 The number of e-folds

In this subsection we aim to find the relation between the number of e-folds
from the horizon crossing to the end of inflation and the reheating
temperature $T_{reh}$. We start with the well-known formula ll

$\displaystyle\frac{k_{*}}{a_{0}H_{0}}=e^{-{\mathcal{N}}}\frac{H_{*}}{H_{0}}\frac{a_{END}}{a_{kin}}\frac{a_{kin}}{a_{reh}}\frac{a_{reh}}{a_{eq}}\frac{a_{eq}}{a_{0}},$
(59)

where, as previously, $a$ is the scale factor and $a_{END}$, $a_{kin}$,
$a_{reh}$, $a_{eq}$ and $a_{0}$ denote respectively its value at the end of
inflation, at the beginning of kination, radiation and matter domination, and
finally at the present time.

Taking into account the evolution during kination and radiation we have

$\displaystyle\left(\frac{a_{kin}}{a_{reh}}\right)^{6}=\frac{\rho_{reh}}{\rho_{kin}}\quad\mbox{and}\quad\left(\frac{a_{reh}}{a_{eq}}\right)^{4}=\frac{\rho_{eq}}{\rho_{reh}}$
(60)

and, noting that $H_{0}\sim 2\times 10^{-4}\mbox{Mpc}^{-1}\sim 6\times
10^{-61}M_{pl}$ and $k_{*}=a_{0}k_{phys}$, where we have chosen as usual
$k_{phys}=0.02\mbox{Mpc}^{-1}$, we have obtained

$\displaystyle{\mathcal{N}}\cong-5+\ln\left(\frac{H_{*}}{H_{0}}\right)+\ln\left(\frac{a_{END}}{a_{kin}}\right)+\frac{1}{4}\ln\left(\frac{g_{eq}}{g_{reh}}\right)+\frac{1}{6}\ln\left(\frac{\rho_{rh}}{\rho_{kin}}\right)+\ln\left(\frac{T_{0}}{T_{rh}}\right),$
(61)

where we have used once again that after the matter-radiation equality the
evolution is adiabatic, that is, $a_{0}T_{0}=a_{eq}T_{eq}$, as well as the
Stefan-Boltzmann law $\rho_{eq}=\frac{\pi^{2}}{30}g_{eq}T_{eq}^{4}$ and
$\rho_{reh}=\frac{\pi^{2}}{30}g_{reh}T_{reh}^{4}$ being $g_{eq}=3.36$ the
degrees of freedom at the matter-radiation equality. And we have chosen as
degrees of freedom at the reheating time the ones of the Standard Model, i.e.,
$g_{reh}=106.75$.

Now, from the formula of the power spectrum we infer that $H_{*}\sim
4\pi\times 10^{-4}\sqrt{\frac{\epsilon_{*}}{10}}M_{pl}\sim 4\times
10^{-4}\sqrt{\epsilon_{*}}M_{pl}$, obtaining

$\displaystyle{\mathcal{N}}\cong
125+\frac{1}{2}\ln\epsilon_{*}+\ln\left(\frac{a_{END}}{a_{kin}}\right)+\frac{1}{6}\ln\left(\frac{\rho_{reh}}{\rho_{kin}}\right)+\ln\left(\frac{T_{0}}{T_{reh}}\right),$
(62)

and introducing the current value of the temperature of the universe
$T_{0}\sim 10^{-31}M_{pl}$ we get

$\displaystyle{\mathcal{N}}\cong
54+\frac{1}{2}\ln\epsilon_{*}+\ln\left(\frac{a_{END}}{a_{kin}}\right)-\frac{1}{3}\ln\left(\frac{H_{kin}T_{reh}}{M_{pl}^{2}}\right).$
(63)

Taking into account that for the Peebles-Vilenkin model one has $H_{kin}\sim
4\sqrt{2\lambda}M_{pl}\sim{4\sqrt{2}\times 10^{-7}M_{pl}}$ and
$\epsilon_{*}\cong(1-n_{s})/3$, we get

$\displaystyle{\mathcal{N}}\cong
59+\frac{1}{2}\ln(1-n_{s})+\ln\left(\frac{a_{END}}{a_{kin}}\right)-\frac{1}{3}\ln\left(\frac{T_{reh}}{M_{pl}}\right)$
(64)

and, using the value of the spectral index $n_{s}\cong 0.96$ and assuming that
$\ln\left(\frac{a_{END}}{a_{kin}}\right)$ is negligible, we finally obtain

$\displaystyle{\mathcal{N}}\cong
57-\frac{1}{3}\ln\left(\frac{T_{reh}}{M_{pl}}\right).$ (65)

Since the scale of nucleosynthesis is $1$ MeV and in order to avoid the late-
time decay of gravitational relic products such as moduli fields or gravitinos
which could jeopardise the nucleosynthesis success, one needs temperatures in
the range $1\mbox{ MeV}\leq T_{reh}\leq 10^{9}\mbox{ GeV}$, which leads to
constrain the number of e-folds to $64\lesssim\mathcal{N}\lesssim 74$.

A final remark is in order: Note that in Quintessential Inflation we have
obtained a number of e-folds greater than in standard inflation, which
normally ranges between $50$ and $60$ e-folds. This fact is due to the
kination regime appearing in QI, which increases significantly the number of
e-folds.

##  II Exponential Quintessential Inflation

In this section we will consider more complex models as the Peebles-Vilenkin
one: the same Exponential Inflation-type potentials studied for the first time
in hossain2 ,

$\displaystyle V(\varphi)=\ V_{0}e^{-\gamma{\varphi}^{n}/{M_{pl}}^{n}},$ (66)

where $\gamma$ is a dimensionless parameter and $n$ is an integer.

In this case the power spectrum of scalar perturbations, its spectral index
and the ratio of tensor to scalar perturbations are given by (see for a
detailed derivation Geng:2017mic )

$\displaystyle{\mathcal{P}}_{\zeta}=\frac{V_{0}e^{-\gamma{\varphi}^{n}/{M_{pl}}^{n}}}{12\pi^{2}n^{2}\gamma^{2}M_{pl}^{6-2n}\varphi^{2n-2}}\sim
2\times 10^{-9},$ (67) $\displaystyle
n_{s}=1-n\gamma\left(\frac{\varphi}{M_{pl}}\right)^{n-2}\left(2n-2+\gamma
n\left(\frac{\varphi}{M_{pl}}\right)^{n}\right)$ (68)

and

$\displaystyle r=8n^{2}\gamma^{2}\left(\frac{\varphi}{M_{pl}}\right)^{2n-2}.$
(69)

An important relation is obtained combining the equations (68) and (69),

$\displaystyle\gamma\left(\frac{\varphi}{M_{pl}}\right)^{n}=\frac{r(2n-2)}{n(8(1-n_{s})-r)},$
(70)

which leads to the formula for the power spectrum

$\displaystyle{\mathcal{P}}_{\zeta}=\frac{2V_{0}e^{-\frac{r(2n-2)}{n(8(1-n_{s})-r)}}}{3\pi^{2}M_{pl}^{4}r}\sim
2\times 10^{-9},$ (71)

and thus,

$\displaystyle V_{0}\sim 3\pi^{2}re^{\frac{r(2n-2)}{n(8(1-n_{s})-r)}}\times
10^{-9}M_{pl}^{4},$ (72)

which, for the viable values of $n_{s}\cong 0.96$ (the central value of the
spectral index) and $r=0.02$ (whose value, as we will immediately see,
guarantees that the model provides a viable a number of e-folds), leads to

$\displaystyle V_{0}\sim 6\pi^{2}\times 10^{-11}M_{pl}^{4}.$ (73)

Next, it is important to realize that a way to find theoretically the possible
values of the parameter $\gamma$ is to combine the equations (69) and (70) to
get

$\displaystyle\gamma=\left(\frac{n(8(1-n_{s})-r)}{2r(n-1)}\right)^{n-1}\left(\frac{r}{8n^{2}}\right)^{n/2}$
(74)

and, using the theoretical values $r\leq 0.1$ and $n_{s}=0.9649\pm 0.0042$
(see for instance planck18 ; planck18a ), one can find the candidates of
$\gamma$ at $2\sigma$ C.L. These values have to be checked for the joint
contour in the plane $(n_{s},r)$ at $2\sigma$ C.L., when the number of e-folds
is approximately between 60 and 75, which is what usually happens in
Quintessential Inflation due to the kination phase ll ; deHaro:2016ftq  -the
energy density of the scalar field is only kinetic Joyce ; Spokoiny \- after
the end of the inflationary period.

For this kind of potentials, in order to compute the number of e-folds we need
to calculate the main slow-roll parameter

$\displaystyle\epsilon\equiv\frac{M_{pl}^{2}}{2}\left(\frac{V_{\varphi}}{V}\right)^{2}=\frac{\gamma^{2}n^{2}}{2}\left(\frac{\varphi}{M_{pl}}\right)^{2n-2},$
(75)

whose value at the end of inflation is $\epsilon_{END}=1$, meaning that at the
end of this epoch the field reaches the value
$\varphi_{END}=\left(\frac{2}{n^{2}\gamma^{2}}\right)^{\frac{1}{2n-2}}M_{pl}$.

Then, the number of e-folds is given by

$\displaystyle{\mathcal{N}}=\frac{1}{M_{pl}}\int_{\varphi}^{\varphi_{END}}\frac{1}{\sqrt{2\epsilon}}d\varphi=\frac{1}{n\gamma(n-2)}\left[\left(\frac{\varphi}{M_{pl}}\right)^{2-n}-\left(\frac{2}{n^{2}\gamma^{2}}\right)^{\frac{2-n}{2n-2}}\right].$
(76)

Thus, combining the equations (68), (69) and (76), one obtains the spectral
index and the tensor/scalar ratio as a function of the number of e-folds and
the parameter $\gamma$.

On the other hand, since inflation ends at
$\varphi_{END}=\left(\frac{2}{n^{2}\gamma^{2}}\right)^{\frac{1}{2n-2}}M_{pl}$,
when the effective Equation of State (EoS) parameter is equal to $-1/3$,
meaning that $\dot{\varphi}_{END}^{2}=V(\varphi_{END})$, the energy density at
the end of inflation is

$\displaystyle\rho_{\varphi,END}=\frac{3}{2}V(\varphi_{END})\cong
9\pi^{2}e^{-\gamma\left(\frac{2}{n^{2}\gamma^{2}}\right)^{\frac{n}{2n-2}}}\times
10^{-11}M_{pl}^{4}$ (77)

and the corresponding value of the Hubble rate is given by

$\displaystyle
H_{END}\cong\sqrt{\frac{3}{10}}e^{-\frac{\gamma}{2}\left(\frac{2}{n^{2}\gamma^{2}}\right)^{\frac{n}{2n-2}}}\times
10^{-5}M_{pl},$ (78)

which will constrain very much the values of the parameter $\gamma$ because in
all viable inflationary models at the end of the inflation the value of the
Hubble rate is of the order of $10^{-6}M_{pl}$ linde . In fact, when (78) is
of the order of $10^{-6}M_{pl}$ one gets

$\displaystyle\gamma\sim\frac{(2/n^{2})^{n/2}}{\ln^{n-1}(30\pi^{2})}.$ (79)

Therefore, to perform numerical calculations, we will use the values of $n=10$
and $r=0.02$ and, thus, for $\gamma\sim 10^{-16}$ we obtain approximately $67$
e-folds, which is a viable value in Quintessential Inflation.

After this, we have to find out the time when kination starts, which can be
chosen at the moment when the effective EoS parameter is very close to $1$.
Assuming for instance that kination starts at $w_{eff}\cong 0.99$, we have
numerically obtained $\varphi_{kin}\cong 47M_{pl}$ and
$\dot{\varphi}_{kin}\cong 6\times 10^{-9}M_{pl}^{2}$, hence $H_{kin}\cong
2\times 10^{-9}M_{pl}$.

Next, we want to calculate the value of the scalar field and its derivative at
the reheating time (for example, in the case that the decay was before the end
of kination). Using for our model the formulas (35)

$\displaystyle\varphi_{reh}=\varphi_{kin}+\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{H_{kin}}{\sqrt{\frac{\pi^{2}g_{reh}}{45}}\frac{T_{reh}^{2}}{M_{pl}}}\right),\qquad{\dot{\varphi}_{reh}=\sqrt{\frac{\pi^{2}g_{reh}}{15}}T_{reh}^{2}},$
(80)

and assuming that the effective number of degrees of freedom is the same as in
the Standard Model, i.e. $g_{reh}=106.75$, we get for a viable reheating
temperature $T_{reh}\sim 10^{7}$ GeV that at the beginning of the radiation
the initial conditions of the scalar field are

$\displaystyle\varphi_{reh}\cong{70}M_{pl}\qquad{\dot{\varphi}_{reh}\cong{5\times
10^{-22}}M_{pl}^{2}}.$ (81)

###  II.1 The scaling and tracker solutions

The scaling and tracker solutions appear in pure exponential potentials and
approximately in more generic exponential potentials as (66) (see for instance
btw ; Geng:2017mic ).

####  II.1.1 The scaling solution

Assuming as usual the flat Friedmann-Lemaître-Robertson-Walker (FLRW) geometry
of our universe, we consider a pure exponential potential,
$V(\varphi)=V_{0}e^{-\gamma\varphi/M_{pl}}$, and a barotropic fluid with
Equation of State (EoS) parameter $w=1/3$, i.e., a radiation fluid, whose
energy density we continue denoting by $\rho_{r}$. Then, the dynamical system
is given by the equations

$\displaystyle\left\\{\begin{array}[]{ccc}\ddot{\varphi}+3H\dot{\varphi}+V_{\varphi}&=&0\\\
\dot{\rho}_{r}+4H\rho_{r}&=&0,\end{array}\right.$ (84)

where $H=\frac{1}{\sqrt{3}M_{pl}}\sqrt{\rho_{\varphi}+\rho_{r}}$.

The scaling solution is a solution with the property that the energy density
of the scalar field scales as the one of radiation, meaning that the Hubble
parameter evolves as $H(t)=\frac{1}{2t}$. At this point, we look for a
solution of the form

$\displaystyle\varphi_{sc}(t)=\frac{M_{pl}}{\gamma}\ln\left(\frac{t^{2}}{\bar{t}^{2}}\right),\qquad\rho_{r}(t)=\frac{C^{2}}{t^{2}},$
(85)

where $C$ and $\bar{t}$ are constants. The equation
$\dot{\rho}_{r}+4H\rho_{r}=0$ is satisfied for any value of $C^{2}$ and the
equation $\ddot{\varphi}+3H\dot{\varphi}+V_{\varphi}=0$ requires that
$\bar{t}^{2}=\frac{M_{pl}^{2}}{\gamma^{2}V_{0}}$.

On the other hand, equating $H(t)=\frac{1}{2t}$ with
$H=\frac{1}{\sqrt{3}M_{pl}}\sqrt{\rho_{\varphi}+\rho_{r}}$, which yields the
relation

$\displaystyle
C^{2}=\frac{3}{4}\left(1-\frac{4}{\gamma^{2}}\right)M_{pl}^{2},$ (86)

shows that the scaling solution only exists for $\gamma>2$, and the
corresponding energy densities evolve as follows,

$\displaystyle\rho_{\varphi_{sc}}(t)=\frac{3M_{pl}^{2}}{\gamma^{2}t^{2}}\qquad\rho_{r}(t)=\frac{3}{4}\left(1-\frac{4}{\gamma^{2}}\right)\frac{M_{pl}^{2}}{t^{2}},$
(87)

and thus, one could derive the corresponding density parameters,

$\displaystyle\Omega_{\varphi_{sc}}=\frac{4}{\gamma^{2}},\qquad\Omega_{r}=1-\frac{4}{\gamma^{2}}.$
(88)

An alternative approach to get this solution goes as follows (see for instance
copeland ): one can introduce the dimensionless variables

$\displaystyle\tilde{x}\equiv\frac{\dot{\varphi}}{\sqrt{6}M_{pl}H}\quad\mbox{and}\quad\tilde{y}\equiv\frac{\sqrt{V}}{\sqrt{3}M_{pl}H},$
(89)

which enable us to write down the following autonomous dynamical system,

$\displaystyle\left\\{\begin{array}[]{ccc}\tilde{x}^{\prime}&=&-3\tilde{x}+\sqrt{\frac{3}{2}}\gamma\tilde{y}^{2}+\frac{3}{2}\tilde{x}\left[(1-w)\tilde{x}^{2}+(1+w)(1-\tilde{y}^{2})\right]\\\
\tilde{y}^{\prime}&=&-\sqrt{\frac{3}{2}}\gamma\tilde{x}\tilde{y}+\frac{3}{2}\tilde{y}\left[(1-w)\tilde{x}^{2}+(1+w)(1-\tilde{y}^{2})\right]\end{array},\right.$
(92)

where $w$ is the Equation of State (EoS) parameter, which in our case is equal
to $1/3$, together with the constraint

$\displaystyle\tilde{x}^{2}+\tilde{y}^{2}+\Omega_{r}=1.$ (93)

One can see that for $w=1/3$ the dynamical system (92) has the attractor
solution $\tilde{x}=\left(\sqrt{\frac{2}{3}}\;\right)\frac{2}{\gamma},\ \
\tilde{y}=\frac{2}{\sqrt{3}}\frac{1}{\gamma}$, whose energy density scales as
radiation, and $\Omega_{\varphi}=\frac{4}{\gamma^{2}}$.

So, since during radiation $H=\frac{1}{2t}$, thus, from
$\tilde{x}=\left(\sqrt{\frac{2}{3}}\;\right)\frac{2}{\gamma}$, we have

$\displaystyle\dot{\varphi}_{sc}=\frac{2}{\gamma
t}M_{pl}\Longrightarrow\varphi_{sc}(t)=\frac{M_{pl}}{\gamma}\ln\left(\frac{t^{2}}{\bar{t}^{2}}\right).$
(94)

Finally, in order to obtain $\bar{t}$, we use the equation
$\tilde{y}=\frac{2}{\sqrt{3}}\frac{1}{\gamma}$, which leads to
$\bar{t}^{2}=\frac{M_{pl}^{2}}{\gamma^{2}V_{0}}$, recovering the scaling
solution

$\displaystyle\varphi_{sc}(t)=\frac{M_{pl}}{\gamma}\ln\left(\frac{\gamma^{2}V_{0}t^{2}}{M_{pl}^{2}}\right).$
(95)

####  II.1.2 The tracker solution

A tracker solution Steinhardt:1999nw ; UrenaLopez:2000aj  is an attractor
solution of the field equation which describes the dark energy domination at
late times. In fact, for an exponential potential, it is the solution of
$\ddot{\varphi}+3H\dot{\varphi}+V_{\varphi}=0$ when the universe is only
filled with the scalar field. Once again, we look for a solution of the type

$\displaystyle\varphi_{tr}(t)=\frac{M_{pl}}{\gamma}\ln\left(\frac{t^{2}}{\bar{t}^{2}}\right).$
(96)

Inserting this equation into $\ddot{\varphi}+3H\dot{\varphi}+V_{\varphi}=0$,
one gets

$\displaystyle\bar{t}^{2}=\frac{2M_{pl}^{2}}{V_{0}\gamma^{4}}(6-\gamma^{2}),$
(97)

which means that $0<\gamma<\sqrt{6}$ and, thus, the tracker solution is given
by

$\displaystyle\varphi_{tr}=\frac{M_{pl}}{\gamma}\ln\left(\frac{V_{0}\gamma^{4}}{2(6-\gamma^{2})M_{pl}^{2}}t^{2}\right).$
(98)

Moreover, for this equation, it is not difficult to show that

$\displaystyle\frac{1}{2}\dot{\varphi}^{2}_{tr}=\frac{2M_{pl}^{2}}{\gamma^{2}t^{2}},\qquad
V(\varphi_{tr})=\frac{2M_{pl}^{2}}{\gamma^{2}t^{2}}(6-\gamma^{2}),$ (99)

meaning that the corresponding effective EoS parameter is given by

$\displaystyle w_{eff}=\frac{\gamma^{2}}{3}-1,$ (100)

which proves that, in order to impose the late-time acceleration of the
universe, we need to restrict $\gamma$ as $0<\gamma<\sqrt{2}$.

Note also that the system (92) when $w=0$ (in the matter domination era) has
the fixed point
$(\tilde{x},\tilde{y})=\left(\gamma/\sqrt{6},\sqrt{1-\gamma^{2}/6}\right)$,
which of course corresponds to the solution (98).

Finally, in terms of the time $N=-\ln(1+z)$, taking into account that the
relation between $N$ and the cosmic time is given by
$t=\frac{2}{\gamma^{2}H_{0}}e^{\gamma^{2}H/2}$, the tracker solution has the
following expression (see Section V of pvharo  for a detailed derivation),

$\displaystyle\varphi_{tr}(N)=\gamma
NM_{pl}+\frac{M_{pl}}{\gamma}\ln\left(\frac{2V_{0}}{(6-\gamma^{2})M_{pl}^{2}H_{0}^{2}}\right).$
(101)

###  II.2 Numerical simulation during radiation

In this subsection we will show numerically that in Quintessential Inflation,
for exponential types of potentials, during radiation the scalar field is
never in the basin of attraction of the scaling solution and, thus, does not
evolve as the scaling solution.

To prove it, first of all we heuristically calculate the value of the red-
shift at the beginning of the radiation epoch

$\displaystyle
1+z_{reh}=\frac{a_{0}}{a_{reh}}=\frac{a_{0}}{a_{eq}}\frac{a_{eq}}{a_{rh}}=(1+z_{eq})\left(\frac{\rho_{r,reh}}{\rho_{r,eq}}\right)^{1/4}=(1+z_{eq})\left(\frac{g_{reh}}{g_{eq}}\right)^{1/4}\frac{T_{reh}}{T_{eq}}$
$\displaystyle=\left(\frac{g_{reh}}{g_{eq}}\right)^{1/4}\frac{T_{reh}}{T_{0}}\sim\frac{T_{reh}}{T_{0}},$
(102)

where we have used that
$\rho_{r,eq}=\rho_{r,reh}\left(\frac{a_{reh}}{a_{eq}}\right)^{4}$ and the
adiabatic evolution after the matter-radiation equality. Then, taking
$z_{eq}\sim 3\times 10^{3}$ and using the current temperature $T_{0}\sim
10^{-31}M_{pl}$, for the reheating temperature $T_{reh}\sim 10^{9}$ GeV we get
$z_{reh}\sim 10^{22}$.

In addition, at the beginning of radiation the energy density of the matter
will be

$\displaystyle\rho_{m,reh}=\rho_{m,eq}\left(\frac{a_{eq}}{a_{reh}}\right)^{3}=\rho_{m,eq}\left(\frac{\rho_{r,reh}}{\rho_{r,eq}}\right)^{3/4}=\frac{\pi^{2}}{30}g_{eq}\left(\frac{g_{reh}}{g_{eq}}\right)^{3/4}(1+z_{eq})T^{3}_{reh}T_{0}$
$\displaystyle\sim{10^{13}}\mbox{ GeV}^{4},$ (103)

and, for radiation,

$\displaystyle\rho_{r,reh}=\frac{\pi^{2}}{30}g_{reh}T_{reh}^{4}\sim
10^{29}\mbox{GeV}^{4}.$ (104)

In this way, the dynamical equations after the beginning of radiation can be
easily obtained using once again as a time variable
$N\equiv-\ln(1+z)=\ln\left(\frac{a}{a_{0}}\right)$. Recasting the energy
density of radiation and matter respectively as functions of $N$, we get

$\displaystyle\rho_{r}(a)={\rho_{r,reh}}\left(\frac{a_{reh}}{a}\right)^{4}\Longrightarrow\rho_{r}(N)={\rho_{r,reh}}e^{4(N_{reh}-N)}$
(105)

and

$\displaystyle\rho_{m}(a)={\rho_{m,reh}}\left(\frac{a_{reh}}{a}\right)^{3}\Longrightarrow\rho_{m}(N)={\rho_{m,reh}}e^{3(N_{reh}-N)},$
(106)

where $N_{reh}=-\ln(1+z_{reh})\sim-50$ denotes the value of the time $N$ at
the beginning of radiation for a reheating temperature around $10^{9}$ GeV.

To obtain the dynamical system for this scalar field model, we will introduce
the dimensionless variables

$\displaystyle x=\frac{\varphi}{M_{pl}}\qquad\mbox{and}\qquad
y=\frac{\dot{\varphi}}{KM_{pl}},$ (107)

where $K$ is a parameter that we will choose accurately in order to facilitate
the numerical calculations. So, taking into account the conservation equation
$\ddot{\varphi}+3H\dot{\varphi}+V_{\varphi}=0$, one arrives at the following
dynamical system,

$\displaystyle\left\\{\begin{array}[]{ccc}x^{\prime}&=&\frac{y}{\bar{H}}~{},\\\
y^{\prime}&=&-3y-\frac{\bar{V}_{x}}{\bar{H}}~{},\end{array}\right.$ (110)

where the prime is the derivative with respect to $N$, $\bar{H}=\frac{H}{K}$
and $\bar{V}=\frac{V}{K^{2}M_{pl}^{2}}$. It is not difficult to see that one
can write

$\displaystyle\bar{H}=\frac{1}{\sqrt{3}}\sqrt{\frac{y^{2}}{2}+\bar{V}(x)+\bar{\rho}_{r}(N)+\bar{\rho}_{m}(N)}~{},$
(111)

where we have defined the dimensionless energy densities as
$\bar{\rho}_{r}=\frac{\rho_{r}}{K^{2}M_{pl}^{2}}$ and
$\bar{\rho}_{m}=\frac{\rho_{m}}{K^{2}M_{pl}^{2}}$.

Next, to integrate from the beginning of radiation up to the matter-radiation
equality, i.e., from $N_{reh}\sim-50$ to $N_{eq}\sim-8$, we will choose
$KM_{pl}=10^{-17}\mbox{GeV}^{2}$, yielding

$\displaystyle\bar{\rho}_{r}(N)\sim{4\times
10^{64}}e^{4(N_{reh}-N)},\qquad\bar{\rho}_{m}(N)\sim{8\times
10^{47}}e^{3(N_{reh}-N)},$ (112)

and

$\displaystyle\bar{V}(x)=\frac{V_{0}\times 10^{34}}{\mbox{GeV}^{4}}e^{-\gamma
x^{n}}.$ (113)

Finally, with the initial conditions for the field being $x_{reh}\sim{70}$ and
$y_{rh}\sim{3\times 10^{32}}$ (these initial conditions are obtained in the
equations (81)) and integrating numerically the dynamical system, we conclude
that, for values of the parameter $n$ greater than $10$, the value of the EoS
parameter $w_{\varphi}$ remains $1$ during radiation domination, namely
between $N_{rh}$ and $N_{eq}$, thus proving that the inflaton field does not
belong to the basin of attraction of the scaling solution for large values of
$n$. In addition, integrating the system up to now, we can see that the model
never matches with the current observational data.

Figure 4: Evolution of $w_{\varphi}$ during radiation domination.

###  II.3 A viable model

To depict the late-time acceleration, one has to modify the original potential
because it does not explain the current observational data (for the potential
(66), at the present time, the density parameter
$\Omega_{\varphi}=\frac{\rho_{\varphi}}{3H^{2}M_{pl}^{2}}$ is far from its
observational value, namely $0.7$). For this reason and following the spirit
of the Peebles-Vilenkin model pv , to match with the current observational
data it is mandatory to introduce a new parameter $M$ with units of mass which
must be calculated numerically. In that case, we will consider the following
modification of the potential (66) by introducing a new exponential term
containing the parameter $M$,

$\displaystyle V(\varphi)=\
V_{0}e^{-\bar{\gamma}{\varphi}^{n}/{M_{pl}}^{n}}+M^{4}e^{-\gamma\varphi/M_{pl}},$
(114)

with $0<\gamma<\sqrt{2}$ in order to obtain, as we have already seen, that at
late times the solution is in the basin of attraction of the tracker solution,
which evolves as a fluid with effective EoS parameter
$w_{eff}=\frac{\gamma^{2}}{3}-1$.

Remark.- An important remark is in order. Contrary to the potential used in
hossain2 ; Geng:2017mic

$\displaystyle V(\varphi)=\
V_{0}e^{-\bar{\gamma}{\varphi}^{n}/{M_{pl}}^{n}}+(\bar{\rho}_{\nu}-3\bar{p}_{\nu})e^{\beta\varphi/M_{pl}},$
(115)

where $\rho_{\nu}=\bar{\rho}_{\nu}e^{\beta\varphi/M_{pl}}$ is the neutrino
energy density with constant bare mass $m_{\nu}$ and $\beta>0$ is the non-
minimal coupling of neutrinos with the inflaton field (see for details
Geng:2017mic ), which has a minimum where the inflaton ends its evolution,
thus acting as an effective cosmological constant which stands for the current
cosmic acceleration, the potential considered here does not have a minimum and
the scalar field continues rolling down the potential following the spirit of
the quintessential inflation.

On the other hand, in this case one can also justify this choice considering
that the inflaton field is non-minimally coupled with neutrinos but with a
negative coupling constant, namely $\beta=-\gamma<0$. Therefore,
$M^{4}=\bar{\rho}_{\nu}-3\bar{p}_{\nu}$ and the effective mass of neutrinos is
given by $m_{\nu,eff}(\varphi)=m_{\nu}e^{-\gamma\varphi/M_{pl}}$, which tends
to zero for large values of the field, so the neutrinos become relativistic,
contrary to what happens when $\beta$ is positive, where the neutrinos acquire
a heavy mass becoming non-relativistic particles. Finally, to match with the
current observational data, $M^{4}=\bar{\rho}_{\nu}-3\bar{p}_{\nu}$ must be
very small compared with the Planck’s energy density. In fact, for $n=10$ and
$\gamma=0.1$ we have numerically obtained $M\sim 10^{-2}\mbox{ eV}$.

Now, one has to solve the dynamical system (110) with initial conditions at
the beginning of the radiation era. Choosing now $K=H_{0}$ (the present value
of the Hubble rate) we have to take into account that to obtain the parameter
$M$ one has to impose the value of $\bar{H}=H(N)/H_{0}$ to be $1$ at $N=0$.

The numerical results are presented in Figure 5, where one can see that at
very late times the energy density of the scalar field dominates and the
universe has an eternal accelerated expansion because its evolution is the
same as that of the tracker solution.

Figure 5:  Left: The density parameters
$\Omega_{m}=\frac{\rho_{m}}{3H^{2}M_{pl}^{2}}$ (orange curve),
$\Omega_{r}=\frac{\rho_{r}}{3H^{2}M_{pl}^{2}}$ (blue curve) and
$\Omega_{\varphi}=\frac{\rho_{\varphi}}{3H^{2}M_{pl}^{2}}$ (green curve), from
kination to future times, for a reheating temperature around $10^{9}$ GeV. To
perform numerical calculations we have taken $n=10$ and $\gamma=0.1$. Right:
The effective Equation of State parameter $w_{eff}$, from kination to future
times, for $n=10$ and $\gamma=0.1$. As one can see in the picture, after
kination the universe enters in a large period of time where radiation
dominate. Then, after the matter-radiation equality, the universe becomes
matter-dominated and, finally, near the present, it enters in a new
accelerated phase where $w_{eff}$ approaches $\frac{0.1^{2}}{3}-1\cong-1$

, that is, it has the same effective EoS parameter as the tracker solution,
meaning that the solution is in the basin of attraction of the tracker one.

##  III Lorentzian Quintessential Inflation

Based on the Cauchy distribution (Lorentzian in the physics language) the
authors of Benisty:2020xqm ; Benisty:2020qta  considered the following ansatz,

$\displaystyle\epsilon({N})=\frac{\xi}{\pi}\frac{\Gamma/2}{{N}^{2}+\Gamma^{2}/4},$
(116)

where $\epsilon$ is the main slow-roll parameter,
${N}=\ln\left(\frac{a}{a_{0}}\right)$ denotes the total number of e-folds (do
not confuse with $\mathcal{N}$, the number of e-folds from the horizon
crossing to the end of inflation), $\xi$ is the amplitude of the Lorentzian
distribution and $\Gamma$ is its width. From this ansatz, we can find the
exact corresponding potential of the scalar field, namely

$\displaystyle V(\varphi)=\lambda
M_{pl}^{4}\exp\left[-\frac{2\xi}{\pi}\arctan\left(\sinh\left(\gamma\varphi/M_{pl}\right)\right)\right]\bm{\cdot}\left(1-\frac{2\gamma^{2}\xi^{2}}{3\pi^{3}}\frac{1}{\cosh\left(\gamma\varphi/M_{pl}\right)}\right),$
(117)

where $\lambda$ is a dimensionless parameter and the parameter $\gamma$ is
defined by

$\gamma\equiv\sqrt{\frac{\pi}{\Gamma\xi}}.$

This potential can be derived by using equations $(37)$ in Martin:2016iqo .
However, in this work we are going to use a more simplified potential, keeping
the same properties as the original potential but not coming exactly from the
ansatz (116). However, from this potential we do recover the ansatz by using
the suitable approximations, which are valid if we properly choose the
parameters such that $\gamma\gg 1$ and $\frac{2\xi}{\Gamma\pi}>1$. For this
reason and also because the results of the original potential happen to be
exactly the same as shown in Guendelman2 , it is better to work with the
simplified potential, namely

$\displaystyle V(\varphi)=\lambda
M_{pl}^{4}\exp\left[-\frac{2\xi}{\pi}\arctan\left(\sinh\left(\gamma\varphi/M_{pl}\right)\right)\right].$
(118)

We can see the shape of the potential on Fig. 6, where the inflationary epoch
takes place on the left-hand side of the graph, while the dark energy epoch
occurs on the right-hand side.

Figure 6: The shape of the scalar potential (66) with $\xi\sim 122$ and
$\lambda\sim 10^{-69}$. The left side shows the inflationary energy density
and the right side shows the late dark energy density.

###  III.1 Calculation of the value of the parameters involved in the model

We start with the main slow-roll parameter, which is given by

$\displaystyle\epsilon\equiv\frac{M_{pl}^{2}}{2}\left(\frac{V_{\varphi}}{V}\right)^{2}=\frac{2\xi/(\Gamma\pi)}{\cosh^{2}\left(\gamma\frac{\varphi}{M_{pl}}\right)}=\frac{2\gamma^{2}\xi^{2}/\pi^{2}}{\cosh^{2}\left(\gamma\frac{\varphi}{M_{pl}}\right)}$
(119)

and, since inflation ends when $\epsilon_{END}=1$, one has to assume that
$\frac{2\xi}{\Gamma\pi}>1$ to guarantee the end of this period.

In fact, we have

$\displaystyle\varphi_{END}=\frac{M_{pl}}{\gamma}\ln\left(\sqrt{\frac{2\xi}{\Gamma\pi}}-\sqrt{\frac{2\xi}{\Gamma\pi}-1}\right)=\frac{M_{pl}}{\gamma}\ln\left[\frac{\sqrt{2}\xi}{\pi}\left(\gamma-\sqrt{\gamma^{2}-\frac{\pi^{2}}{2\xi^{2}}}\right)\right]<0,$
(120)

and we can see that, for large values of $\gamma$, one has that
$\varphi_{END}$ is close to zero. Thus, we will choose $\gamma\gg
1\Longrightarrow\Gamma\xi\ll 1$, which is completely compatible with the
condition $\frac{2\xi}{\Gamma\pi}>1$.

On the other hand, the other important slow-roll parameter is given by

$\displaystyle\eta\equiv
M_{pl}^{2}\frac{V_{\varphi\varphi}}{V}=\frac{2\xi\gamma^{2}}{\pi}\frac{\tanh\left(\gamma\frac{\varphi}{M_{pl}}\right)}{\cosh\left(\gamma\frac{\varphi}{M_{pl}}\right)}+\frac{4\gamma^{2}\xi^{2}/\pi^{2}}{\cosh^{2}\left(\gamma\frac{\varphi}{M_{pl}}\right)}.$
(121)

Both slow-roll parameters have to be evaluated when the pivot scale leaves the
Hubble radius, which will happen for large values of
$\cosh\left(\gamma\varphi/M_{pl}\right)$, obtaining

$\epsilon_{*}=\frac{2\gamma^{2}\xi^{2}/\pi^{2}}{\cosh^{2}\left(\gamma\frac{\varphi_{*}}{M_{pl}}\right)},\quad\eta_{*}\cong\frac{2\xi\gamma^{2}}{\pi}\frac{\tanh\left(\gamma\frac{\varphi_{*}}{M_{pl}}\right)}{\cosh\left(\gamma\frac{\varphi_{*}}{M_{pl}}\right)},$
(122)

with $\varphi_{*}<0$. Then, since the spectral index is given in the first
approximation by $n_{s}\cong 1-6\epsilon_{*}+2\eta_{*}$, one gets after some
algebra

$\displaystyle n_{s}\cong 1+2\eta_{*}\cong 1-\gamma\sqrt{r/2},$ (123)

where $r=16\epsilon_{*}$ is the ratio of tensor to scalar perturbations.

Now, we calculate the number of e-folds from the horizon crossing to the end
of inflation, which is given by

$\displaystyle{\mathcal{N}}=\frac{1}{M_{pl}}\int_{\varphi_{*}}^{\varphi_{END}}\frac{1}{\sqrt{2\epsilon}}d\varphi=\frac{\pi}{2\gamma^{2}\xi}\left[\sinh\left(\gamma\varphi_{END}/M_{pl}\right)-\sinh\left(\gamma\varphi_{*}/M_{pl}\right)\right]\cong\frac{\xi}{\sqrt{2\epsilon_{*}}},$

so we have that

$\displaystyle n_{s}\cong 1-\frac{2}{{\mathcal{N}}},\qquad
r\cong\frac{8}{{\mathcal{N}}^{2}\gamma^{2}},$ (124)

meaning that our model leads to the same spectral index and tensor/scalar
ratio as the $\alpha$-attractors models with $\alpha=\frac{2}{3\gamma^{2}}$
(see for instance Linde:1981mu ).

Figure 7: The Marginalized joint confidence contours for $(n_{s},r)$ at
$1\sigma$ and $2\sigma$ CL, without the presence of running of the spectral
indices. We have drawn the curve for the present model for $\gamma\gg 1$ from
${\mathcal{N}}=58$ to ${\mathcal{N}}=67$ e-folds. (Figure courtesy of the
Planck2018 Collaboration planck18a )

Finally, it is well-known that the power spectrum of scalar perturbations is
given by

$\displaystyle{\mathcal{P}}_{\zeta}=\frac{H_{*}^{2}}{8\pi^{2}\epsilon_{*}M_{pl}^{2}}\sim
2\times 10^{-9}.$ (125)

Now, since in our case $V(\varphi_{*})\cong\lambda M^{4}_{pl}e^{\xi}$, meaning
that $H_{*}^{2}\cong\frac{\lambda M^{2}_{pl}}{3}e^{\xi}$, and taking into
account that $\epsilon_{*}\cong\frac{(1-n_{s})^{2}}{8\gamma^{2}}$, one gets
the constraint

$\displaystyle\lambda\gamma^{2}e^{\xi}\sim 7\times 10^{-11},$ (126)

where we have chosen as a value of $n_{s}$ its central value $0.9649$.

Summing up, we will choose our parameters satisfying the condition (143), with
$\gamma\gg 1$ and $\xi\gg 1$, which will always fulfill the constraints
$\Gamma\xi\ll 1$ and $\frac{2\xi}{\Gamma\pi}$ that we have imposed. Then, to
find the values of the parameters one can perform the following heuristic
argument: Taking for example $\gamma=10^{2}$, the constraint (143) becomes
$\lambda e^{\xi}\sim 7\times 10^{-15}$. On the other hand, at the present time
we will have $\gamma\varphi_{0}/M_{pl}\gg 1$ where $\varphi_{0}$ denotes the
current value of the field. Thus, we will have $V(\varphi_{0})\sim\lambda
M_{pl}^{4}e^{-\xi}$, which is the dark energy at the present time, meaning
that

$\displaystyle
0.7\cong\Omega_{\varphi,0}\cong\frac{V(\varphi_{0})}{3H^{2}_{0}M_{pl}^{2}}\sim\frac{\lambda
e^{-\xi}}{3}\left(\frac{M_{pl}}{H_{0}}\right)^{2}.$ (127)

So, for the value $H_{0}=67.81\;\mbox{Km/sec/Mpc}=5.94\times 10^{-61}M_{pl}$,
we get the equations

$\displaystyle\lambda e^{\xi}\sim 7\times
10^{-15}\qquad\mbox{and}\qquad\lambda e^{-\xi}\sim 10^{-120},$ (128)

whose solution is given by $\xi\sim 122$ and $\lambda\sim 10^{-69}$.

If we choose $\gamma\sim 10^{2}$, we see that the values of $\xi$ and $\gamma$
could be set equal in order to obtain the desired results from both the early
and late inflation. From now on we will set $\xi=\gamma$, since it may help to
find a successful combination of parameters because it reduces the number of
effective parameters. As we will see later, numerical calculations show that,
in order to have $\Omega_{\varphi,0}\cong 0.7$ (observational data show that,
at the present time, the ratio of the energy density of the scalar field to
the critical one is approximately $0.7$), one has to choose $\xi=\gamma\cong
121.8$.

Next, we aim to find the relation between the number of e-folds and the
reheating temperature $T_{reh}$. We start once again with the formula

$\displaystyle\frac{k_{*}}{a_{0}H_{0}}=e^{-{\mathcal{N}}}\frac{H_{*}}{H_{0}}\frac{a_{END}}{a_{kin}}\frac{a_{kin}}{a_{reh}}\frac{a_{reh}}{a_{eq}}\frac{a_{eq}}{a_{0}}$
(129)

and, following the same steps as in the case of the Peebles-Vilenkin model, we
get

$\displaystyle{\mathcal{N}}\cong-5+\ln\left(\frac{H_{*}}{H_{0}}\right)+\ln\left(\frac{a_{END}}{a_{kin}}\right)+\frac{1}{4}\ln\left(\frac{g_{eq}}{g_{reh}}\right)+\frac{1}{6}\ln\left(\frac{\rho_{reh}}{\rho_{kin}}\right)+\ln\left(\frac{T_{0}}{T_{reh}}\right).$
(130)

Now, from the formula of the power spectrum of scalar perturbations (141) we
infer that $H_{*}\sim 4\times 10^{-4}\sqrt{\epsilon_{*}}M_{pl}$, obtaining

$\displaystyle{\mathcal{N}}\cong
125+\frac{1}{2}\ln\epsilon_{*}+\ln\left(\frac{a_{END}}{a_{kin}}\right)+\frac{1}{6}\ln\left(\frac{\rho_{rh}}{\rho_{kin}}\right)+\ln\left(\frac{T_{0}}{T_{reh}}\right),$
(131)

and introducing the current value of the temperature of the universe
$T_{0}\sim 9.6\times 10^{-32}M_{pl}$ we get

$\displaystyle{\mathcal{N}}\cong
54+\frac{1}{2}\ln\epsilon_{*}+\ln\left(\frac{a_{END}}{a_{kin}}\right)-\frac{1}{3}\ln\left(\frac{H_{kin}T_{reh}}{M_{pl}^{2}}\right).$
(132)

So, since we have numerically checked that $H_{kin}\sim 4\times
10^{-8}M_{pl}$, we get

$\displaystyle{\mathcal{N}}+\ln{\mathcal{N}}\cong
55-\frac{1}{3}\ln\left(\frac{T_{reh}}{M_{pl}}\right),$ (133)

where we have used that $\epsilon_{*}=\frac{1}{2\gamma^{2}N^{2}}$ and we have
also numerically computed that
$\ln\left(\frac{a_{END}}{a_{kin}}\right)\cong-0.068$.

Finally, taking into account that the scale of nucleosynthesis is $1$ MeV and
in order to avoid the late-time decay of gravitational relic products such as
moduli fields or gravitinos which could jeopardise the nucleosynthesis
success, one needs temperatures lower than $10^{9}$ GeV. So, we will assume
that $1\mbox{ MeV}\leq T_{reh}\leq 10^{9}\mbox{ GeV}$, which leads to
constrain the number of e-folds to $58\lesssim N\lesssim 67$. And for this
number of e-folds, $0.966\lesssim n_{s}\lesssim 0.970$, which enters within
its $2\sigma$ CL range.

###  III.2 Present and future evolution: Numerical analysis

To understand the present and future evolution of our universe, we have
integrated numerically the dynamical system (110) with $K=H_{0}$ and initial
conditions at the reheating time (obtained numerically)
$\varphi_{kin}\sim-0.003M_{pl}$ and $\dot{\varphi}_{kin}\sim 8\times
10^{-8}M_{pl}^{2}$. The results are presented in Figure 8, where one can see
that in LQI the universe accelerates forever at late times with an effective
EoS parameter equal to $-1$.

Figure 8: Left: The density parameters
$\Omega_{m}=\frac{\rho_{m}}{3H^{2}M_{pl}^{2}}$ (orange curve),
$\Omega_{r}=\frac{\rho_{r}}{3H^{2}M_{pl}^{2}}$ (blue curve) and
$\Omega_{\varphi}=\frac{\rho_{\varphi}}{3H^{2}M_{pl}^{2}}$, from kination to
future times, for a reheating temperature around $10^{9}$ GeV. Right: The
effective Equation of State parameter $w_{eff}$, from kination to future
times. As one can see in the picture, after kination the universe enters in a
large period of time where radiation dominates. Then, after the matter-
radiation equality, the universe becomes matter-dominated and, finally, near
the present, it enters in a new accelerated phase where $w_{eff}$ approaches
$-1$.

##  IV $\alpha$-attractors in Quintessential Inflation

The concept of $\alpha$-attractor, in the context of standard inflation, was
introduced for the first time in kalloslinde , obtaining models which
generalize the well-known Starobinsky model (see riotto  for a detailed
revision of the Starobinsky model).

In this section we consider $\alpha$-attractors in the context of
Quintessential Inflation. For this reason we deal with the following
Lagrangian motivated by supergravity and corresponding to a non-trivial Kähler
manifold (see for instance dimopoulos0  and the references therein), combined
with a standard exponential potential,

$\displaystyle\mathcal{L}=\frac{1}{2}\frac{\dot{\phi}^{2}}{(1-\frac{\phi^{2}}{6\alpha})^{2}}M_{pl}^{2}-\lambda
M_{pl}^{4}e^{-\kappa\phi},$ (134)

where $\phi$ is a dimensionless scalar field, and $\kappa$ and $\lambda$ are
positive dimensionless constants.

In order that the kinetic term has the canonical form, one can redefine the
scalar field as follows,

$\displaystyle\phi=\sqrt{6\alpha}\tanh\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right),$
(135)

obtaining the following potential,

$\displaystyle V(\varphi)=\lambda
M_{pl}^{4}e^{-n\tanh\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right)},$ (136)

where we have introduced the dimensionless parameter $n=\kappa\sqrt{6\alpha}$.
Similarly to Benisty:2020qta ; Benisty:2020xqm ; Guendelman2 , the potential
satisfies the cosmological seesaw mechanism, where the left side of the
potential gives a very large energy density -the inflationary side- and the
right side gives a very small energy density -the dark energy side. The
asymptotic values are $V_{\pm}=\lambda\exp(\pm n)$. The parameter $n$ is the
logarithm of the ratios between the energy densities, as $\xi$ in the earlier
versions Guendelman2 .

###  IV.1 Calculation of the value of the parameters involved in the model

Dealing with this potential at early times, the main slow-roll parameter is
given by

$\displaystyle\epsilon\equiv\frac{M_{pl}^{2}}{2}\left(\frac{V_{\varphi}}{V}\right)^{2}=\frac{n^{2}}{12\alpha}\frac{1}{\cosh^{4}\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right)},$
(137)

where we must assume that $\frac{n^{2}}{12\alpha}>1$ because inflation ends
when $\epsilon_{END}=1$, and the other slow-roll parameter is

$\displaystyle\eta\equiv
M_{pl}^{2}\frac{V_{\varphi\varphi}}{V}=\frac{n}{3\alpha}\left[\frac{\tanh\left(\frac{\varphi/M_{pl}}{\sqrt{6\alpha}}\right)}{\cosh^{2}\left(\frac{\varphi/M_{pl}}{\sqrt{6\alpha}}\right)}+\frac{n/2}{\cosh^{4}\left(\frac{\varphi/M_{pl}}{\sqrt{6\alpha}}\right)}\right].$
(138)

Both slow-roll parameters have to be evaluated at the horizon crossing, which
will happen for large values of
$\cosh\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right)$, obtaining

$\displaystyle\epsilon_{*}=\frac{n^{2}}{12\alpha}\frac{1}{\cosh^{4}\left(\frac{\varphi_{*}/M_{pl}}{\sqrt{6\alpha}}\right)},\qquad\eta_{*}\cong-\frac{n}{3\alpha}\frac{1}{\cosh^{2}\left(\frac{\varphi_{*}/M_{pl}}{\sqrt{6\alpha}}\right)},$
(139)

with $\varphi_{*}<0$.

Next, we calculate the number of e-folds from the horizon crossing to the end
of inflation, which for small values of $\alpha$ is given by

$\displaystyle{\mathcal{N}}=\frac{1}{M_{pl}}\int_{\varphi_{*}}^{\varphi_{END}}\frac{1}{\sqrt{2\epsilon}}d\varphi\cong\sqrt{\frac{3\alpha}{4\epsilon_{*}}},$
(140)

so we get the standard form of the spectral index and the tensor/scalar ratio
for an $\alpha$-attractor linde ,

$\displaystyle n_{s}\cong 1-6\epsilon_{*}+2\eta_{*}\cong
1-\frac{2}{{\mathcal{N}}},\qquad r\cong
16\epsilon_{*}\cong\frac{12\alpha}{{\mathcal{N}}^{2}}.$ (141)

Finally, it is well-known that the power spectrum of scalar perturbations is
given by

$\displaystyle{\mathcal{P}}_{\zeta}=\frac{H_{*}^{2}}{8\pi^{2}\epsilon_{*}M_{pl}^{2}}\sim
2\times 10^{-9}$ (142)

and, since in our case $V(\varphi_{*})\cong\lambda M^{4}_{pl}e^{n}$ and, thus,
$H_{*}^{2}\cong\frac{\lambda M^{2}_{pl}}{3}e^{n}$, taking into account that
$\epsilon_{*}\cong\frac{3\alpha}{16}(1-n_{s})^{2}$ one gets the constraint

$\displaystyle\lambda e^{n}/\alpha\sim 10^{-10},$ (143)

where we have chosen as the value of $n_{s}$ its central value given by the
Planck’s team, i.e., $n_{s}=0.9649$ planck18 .

Choosing for example $\alpha={10^{-2}}$, the constraint (143) becomes $\lambda
e^{n}\sim{10^{-12}}$. On the other hand, at the present time we will have
$\frac{\varphi_{0}}{\sqrt{6\alpha}M_{pl}}\gg 1$, where $\varphi_{0}$ denotes
the current value of the inflaton field. Hence, we will have
$V(\varphi_{0})\sim\lambda M_{pl}^{4}e^{-n}$, which is the dark energy at the
present time, meaning that

$\displaystyle
0.7\cong\Omega_{\varphi,0}\cong\frac{V(\varphi_{0})}{3H^{2}_{0}M_{pl}^{2}}\sim\frac{\lambda
e^{-n}}{3}\left(\frac{M_{pl}}{H_{0}}\right)^{2}.$ (144)

Thus, taking for example the value provided by the Planck’s team planck18 ;
planck18a , $H_{0}\cong 68\;\mbox{Km/sec/Mpc}\cong 6\times 10^{-61}M_{pl}$, we
get the equations

$\displaystyle\lambda e^{n}\sim 10^{-12}\qquad\mbox{and}\qquad\lambda
e^{-n}\sim 10^{-120},$ (145)

whose solution is given by $n\sim 124$ and $\lambda\sim 10^{-66}$.

To end, some remarks are in order:

  1. 1.

We have chosen $\alpha\sim 10^{-2}$, but we can safely choose $\alpha\sim 1$.
In that case the equations (145) will become

$\displaystyle\lambda e^{n}\sim 10^{10}\qquad\mbox{and}\qquad\lambda
e^{-n}\sim 10^{-120},$ (146)

whose solution is $n\sim 127$ and $\lambda\sim 10^{-65}$.

  2. 2.

If one takes $\alpha$ very small -for example of the order $10^{-2}$-, then,
as has been shown in vardayan , a much simpler model than the exponential one
is possible with a linear potential. In fact, the authors of vardayan  showed
that the potential $V(\phi)=\lambda(\phi+\sqrt{6\alpha})M_{pl}^{4}+\Lambda
M_{pl}^{4},$, which in terms of the canonically normalized field $\varphi$ has
the form

$\displaystyle
V(\varphi)=\lambda\sqrt{6\alpha}\left(\tanh\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right)+1\right)M_{pl}^{4}+\Lambda
M_{pl}^{4},$ (147)

is viable for values of $\alpha\sim 10^{-2}$ and $\Lambda\sim 10^{-120}$.

  3. 3.

Another important important application of the $\alpha$-attractors is its use
to alleviate the current Hubble tension finelli . In that work, the authors,
in the framework of $\alpha$-attractors, include to the model an Early Dark
Energy (EDE) component that adds energy to the Universe, because the success
in easing the Hubble tension crucially depends on the shape of this energy
injection. In fact, in finelli  the authors use the following potential for
EDE,

$\displaystyle
V(\varphi)=\Lambda+V_{0}\frac{\tanh^{2p}\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right)}{\left(1+\tanh\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right)\right)^{2n}},$
(148)

obtaining for $p=2$ and $n=4$ the following value of the Hubble rate at the
present time: $H_{0}=(70.9\pm 1.1)$ $\frac{\mbox{km/s}}{\mbox{Mpc}}$ at
$1\sigma$ C.L. Thus, the tension with the measurement from the SH0ES team
($H_{0}=(74.04\pm 1.42)$ $\frac{\mbox{km/s}}{\mbox{Mpc}}$) reduces to
$1.75\sigma$, while this tension with the measurement from the Planck’s team
($H_{0}=(67.36\pm 0.54)$ $\frac{\mbox{km/s}}{\mbox{Mpc}}$) was of $4.4\sigma$.

###  IV.2 Present and future evolution: Numerical analysis

Once again, we have integrated numerically the dynamical system (110) with
$K=H_{0}$ and initial conditions at the reheating time (obtained numerically)
$\varphi_{kin}\sim-0.5M_{pl}$ and $\dot{\varphi}_{kin}\sim\times
10^{-6}M_{pl}^{2}$. The results presented in Figure 9 are the same as the ones
obtained in the LQI scenario. So, both models predict the same evolution of
the universe.

Figure 9: Left: The density parameters
$\Omega_{m}=\frac{\rho_{m}}{3H^{2}M_{pl}^{2}}$ (orange curve),
$\Omega_{r}=\frac{\rho_{r}}{3H^{2}M_{pl}^{2}}$ (blue curve) and
$\Omega_{\varphi})=\frac{\rho_{\varphi}}{3H^{2}M_{pl}^{2}}$, from kination to
future times for a reheating temperature around $10^{9}$ GeV. Right: The
effective Equation of State parameter $w_{eff}$, from kination to future
times. As one can see in the picture, after kination the Universe enters in a
large period of time where radiation dominates. Then, after the matter-
radiation equality, the Universe becomes matter-dominated and, finally, near
the present, it enters in a new accelerated phase where $w_{eff}$ approaches
$-1$.

##  V  Other Quintessential Inflation models

In the present section, some QI models, different from the ones studied in the
previous sections, are presented and revisited.

###  V.1 Dimopoulos work in Quintessential Inflation

In this subsection we briefly review some points of the vast contribution of
K. Dimopoulos in this field, starting with the class of models introduced in
Dimopoulos , whose potential is given by

$\displaystyle
V_{k}(\varphi)=m^{4}\left(1-\tanh(\varphi/M_{pl})\right).\left(1-\sin\left(\frac{\pi\varphi/2}{\sqrt{\varphi^{2}+M^{2}}}\right)\right)^{k},$
(149)

where $M\ll M_{pl}$ and $m$ are two masses and $k$ is a positive integer.

It is interesting to note that the potential has the following asymptotic
behavior,

$\displaystyle
V_{k}(\varphi)=\left\\{\begin{array}[]{ccc}2^{k+1}m^{4}&\mbox{for}&\varphi/M\ll
0,\\\ &&\\\
2^{1-k}\left(\frac{\pi}{4}\right)^{2k}e^{-2\varphi/M_{pl}}\left(\frac{mM^{k}}{\varphi^{k}}\right)^{4}&\mbox{for}&\varphi/M\gg
0,\end{array}\right.$ (153)

meaning that at very early times the potential approaches to a constant non-
vanishing false vacuum responsible for inflation, and at late times it is
approximately a quasi-exponential one which leads to quintessence.

The value of the parameter $m$ is obtained, as in standard inflation, using
the power spectrum of scalar perturbation, which for $k\leq 4$ leads to the
value $m\sim 10^{15}$ GeV. And, in order to match with the current
observational data, one has to choose

$\displaystyle M\sim 4\times 10^{2}(3\times
10^{70})^{1/(4k)}10^{-30/k}M_{pl},$ (154)

which has only sub-Planckian values for $k\leq 4$. In fact, for $k=4$ one has
$M\sim 10^{18}$ GeV.

Another important contribution comes from Modular Quintessential Inflation
Dimopoulos1 , where the author introduced the following toy potential,

$\displaystyle
V_{q}(\varphi)=\frac{m^{4}}{\left[\cosh\left(\varphi/M\right)\right]^{q}},$
(155)

where $m$ and $M$ are masses and $q$ a positive integer. Once again, the
asymptotic behavior is given by

$\displaystyle
V_{q}(\varphi)=\left\\{\begin{array}[]{ccc}m^{4}-\frac{q}{2}\left(\frac{m^{2}}{M}\right)^{2}\varphi^{2}&\mbox{for}&0<\varphi\ll
M,\\\ &&\\\ 2^{q}m^{4}e^{-q\varphi/M}&\mbox{for}&\varphi\gg
M,\end{array}\right.$ (159)

showing that the inflationary piece of the potential is like a top-hill one
and the tail, which leads to quintessence, is a pure exponential one. For this
modular model the minimum value of $m$ is around $10^{7}$ GeV and, as we have
already seen, for pure exponential potentials, in order that the tracker
solution, which is an attractor, reproduces an accelerated universe, one needs
that $q$ and $M$ satisfy the constraint $0<q\frac{M_{pl}}{M}<\sqrt{2}$. In
addition, it is not difficult to show that the spectral index is given by

$\displaystyle n_{s}=1-2\epsilon_{*}-2q\left(\frac{M_{pl}}{M}\right)^{2},$
(160)

and thus, taking $n_{s}\cong 0.96$, we can deduce that
$q\left(\frac{M_{pl}}{M}\right)^{2}\leq 2\times 10^{-2}$. Finally, using the
power spectrum of scalar perturbation one gets the following important
relation between the parameters involved in the model:

$\displaystyle\left(\frac{m}{M_{pl}}\right)^{4}\sim\left(10^{-1}-5q\left(\frac{M_{pl}}{M}\right)^{2}\right)\times
10^{-7}.$ (161)

Next, we review the treatment of $\alpha$-attractor done in dimopoulos0 .
First of all, one has to introduce a negative Cosmological Constant (CC)
(recall that Einstein’s CC is positive) with the following form,
$\Lambda=-\lambda M_{pl}^{2}e^{-n}$, and thus, adding to the Lagrangian (134)
the term $\Lambda M_{pl}^{2}$ leads to the following effective potential,

$\displaystyle V(\varphi)=\lambda
M_{pl}^{4}e^{-n}\left(e^{n\left(1-\tanh\left(\frac{\varphi}{\sqrt{6\alpha}M_{pl}}\right)\right)}-1\right),$
(162)

whose main difference with the potential (136) is that it vanishes at very
late times.

During inflation $\varphi<0$ and the potential becomes as (136) because
$e^{-n}\ll 1$. So, as we have shown for the potential (136), for small values
of $\alpha$ inflation works also well when this CC is introduced in the model.

In the same way, for large values of the scalar field the potential will
become

$\displaystyle V(\varphi)=2n\lambda
e^{-n}M_{pl}^{4}e^{-\gamma\varphi/M_{pl}},$ (163)

with $\gamma\equiv\sqrt{\frac{2}{3\alpha}}$. Since for an exponential
potential a late-time eternal acceleration is achieved when $\gamma<\sqrt{2}$,
one has to choose $\alpha>1/3$. However, since in this case one has to choose
$\alpha>1/3$, the calculation of the spectral index and the ratio of tensor to
scalar perturbations changes a little bit with respect to the case $\alpha\ll
1$, obtaining (see for details Dimopoulos:2017zvq)

$\displaystyle n_{s}\cong
1-\frac{2}{{\mathcal{N}}+\frac{\sqrt{3\alpha}}{2}},\quad
r=\frac{12\alpha}{\left({\mathcal{N}}+\frac{\sqrt{3\alpha}}{2}\right)^{2}}.$
(164)

Another important contribution of Dimopoulos is the introduction of the
concept of Warm Quintessential Inflation in Dimopouloswarm , which is applied
to the original Peebles-Vilenkin model, obtaining the unification of early-
and late-time acceleration, and providing a very natural mechanism to reheat
our universe. In the same way, an interesting idea is to consider
Quintessential Inflation in the context of Palatini $R^{2}$-gravity
Dimopoulos3 , adapting the well-known Starobinsky model to QI.

About the reheating mechanism in Quintessential Inflation, in tommi  the
authors introduced a very interesting concept: an originally subdominant, non-
minimally coupled and massive scalar field with quartic self-interactions, and
without any interaction with the inflaton field. During inflation the field is
at zero vacuum, but during kination the minimum of the potential displaces
from zero because the field becomes tachyonic, and thus, it is “kicked-off” of
the origin (which becomes a potential hill) due to quantum fluctuations.
Finally, when it reaches the new minimum of potential it starts to oscillate
coherently realising its energy and creating particles as in standard
inflation. However, as it is highlighted in rubio1 ; rubio2 ; rubio3 ; rubio4
, there is an important limitation of the treatment considered in [54], namely
the fact that reheating through this mechanism is far from being an
homogeneous, coherent and perturbative process. This observation has far
reaching consequences, completely absent in the original proposal, such as the
production of topological defects, the generation of a detectable
gravitational wave background or baryogenesis. Moreover, it allows the onset
of radiation domination for arbitrary potentials, and not only for quartic
interactions, as expected from the homogeneous treatment.

Another effective reheating can also be obtained from a combination of
supergravity and string theory K1 . The model contains the inflaton and a
Peccei-Quinn field, which will be the responsible for particle production. In
fact, both fields are coupled as in the theory of Instant Preheating, which we
will study in next section, and thus, when the adiabatic evolution is broken
the total kinetic energy of the inflaton decays into radiation through
resonant production of Peccei-Quinn particles, and the residual potential
density of the inflaton field can be the responsible for dark energy.

Finally, the Curvaton rehating, which we deal with in next section, is applied
to QI in K2 , where bounds on the parameters of curvaton models are found. In
addition, using a minimal curvaton model, the authors showed that the allowed
parameter space is considerably larger than in the case of the usual
oscillatory inflation models, i.e, than in standard inflation.

###  V.2 Quintessential Inflation with non-canonical scalar fields

In hossain2  (see also hossain1 ; Geng:2017mic ) the authors study
Quintessential Inflation using the following action,

$\displaystyle S=\int
d^{4}x\sqrt{-g}\left[\frac{M_{pl}^{2}}{2}R-\frac{k^{2}(\varphi)}{2}\partial^{\mu}\varphi\partial_{\mu}\varphi-V(\varphi)\right]+S_{m}+S_{r}+S_{\nu}\left(\varphi,\Psi_{\nu}\right),$
(165)

where $S_{m}$ and $S_{r}$ are the actions for matter and radiation and
$S_{\nu}$ is the action for neutrinos (see neutrinos  and references therein
for a discussion of the coupling of neutrinos and QI). Here, the Lagrangian
density for massive neutrinos is given by Geng:2017mic

$\displaystyle{\mathcal{L}}_{\nu}=i\bar{\Psi}_{\nu}\gamma^{\lambda}\partial_{\lambda}\Psi_{\nu}-m_{\nu,0}e^{\beta\varphi/M_{pl}}\bar{\Psi}_{\nu}{\Psi}_{\nu},$
(166)

which depicts a non-minimal coupling between the inflaton and the neutrinos,
the potential is a pure exponential one,

$\displaystyle V(\varphi)=M_{pl}^{4}e^{-\alpha\varphi/M_{pl}},$ (167)

and the coupling $k^{2}$ is given by

$\displaystyle
k^{2}(\varphi)=\left(\frac{\alpha^{2}-\tilde{\alpha}^{2}}{\alpha^{2}}\right)\frac{1}{1+\beta^{2}e^{\alpha\varphi/M_{pl}}}+1,$
(168)

being $\alpha$, $\tilde{\alpha}$, $\gamma$ and $\beta$ the parameters of the
model.

For this model the slow-roll parameters as a function of the number of e-folds
${\mathcal{N}}$ are given by

$\displaystyle\epsilon_{*}=\frac{\tilde{\alpha}^{2}}{2}\frac{1}{1-e^{-\tilde{\alpha}^{2}\mathcal{N}}}\qquad\mbox{and}\qquad\eta_{*}=\epsilon_{*}+\frac{\tilde{\alpha}^{2}}{2},$
(169)

and thus, the spectral index and the tensor/scalar ratio are

$\displaystyle n_{s}\cong
1-\tilde{\alpha}^{2}\coth\left(\frac{\tilde{\alpha}^{2}\mathcal{N}}{2}\right)\qquad\mbox{and}\qquad
r=\frac{8\tilde{\alpha}^{2}}{1-e^{-\tilde{\alpha}^{2}\mathcal{N}}}.$ (170)

At first glance the model seems more complicated than the ones we have studied
previously because it contains four parameters, but it simplifies very much
because at late times, i.e, for large values of $\varphi$, the coupling
$k^{2}$ goes to $1$, recovering the canonical form of the action. In addition,
the effect of neutrinos, at late times, is the modification of the potential
$V(\varphi)$, becoming the following effective potential,

$\displaystyle
V_{eff}(\varphi)=V(\varphi)+\rho_{\nu,0}e^{\gamma(\varphi-\varphi_{0})/M_{pl}},$
(171)

where $\varphi_{0}$ and $\rho_{\nu,0}$ are the current values of the inflaton
and the energy density of the massive neutrinos.

Therefore, since the potential $V(\varphi)$ is an exponential one, we are in
the same situation studied in the subsection $3.3$, which, as we have already
seen, leads to a viable model.

###  V.3 Gauss-Bonnet Quintessential Inflation

One of the problems of Quintessential Inflation is that due to the huge
difference between the energy density scale of inflation and the current
energy density (which is over a hundred orders of magnitude), in standard
Quintessential Inflation the inflaton field typically rolls over super-
Planckian distances in field space, resulting into a multitude of problems.
Firstly, the flatness of the quintessential tail may be lifted by radiative
corrections. Also, because the associated mass is so small, the quintessence
field may give rise to a so-called 5th force problem, which can lead to a
violation of the equivalence principle. To avoid these problems, it is
desirable to keep the field variation sub-Planckian. In this case, however, to
bridge the huge difference between the inflation and dark energy density
scales, the quintessential tail must be steep. But, if the quintessential tail
is too steep, when the field becomes important today, it unfreezes and rolls
down the steep potential not leading to accelerated expansion at all. One way
to overcome this problem is to make sure the scalar field remains frozen today
even though the quintessential tail is steep. To this end, a solution (see GB1
for a more detailed discussion) is the coupling of the field with the Gauss-
Bonnet term, because such coupling impedes the variation of the field even if
the potential is steep. Thus, a reason to use the GB scalar is that in some
models the GB coupling could become important at late times making sure that
the field freezes with sub-Planckian displacement, such that it becomes the
dark energy today without the aforementioned problems. Fortunately, as we have
already shown, the potential in Lorentzian Quintessential Inflation and the
one of $\alpha$-attractors is steep enough and the field also freezes
($w_{eff}$ approaches to $-1$ at late times) at the present time leading to
the current acceleration.

Therefore, here we will analyse the paper GB1  (see also oikonomou  for
another paper about Gauss-Bonnet QI) where the authors consider the following
action,

$\displaystyle S=\int
d^{4}x\sqrt{-g}\left[\frac{M_{pl}^{2}}{2}[R-F(\varphi)\mathcal{G}]-\frac{1}{2}\partial^{\mu}\varphi\partial_{\mu}\varphi-V(\varphi)\right],$
(172)

where $F(\varphi)=F_{0}e^{-q\varphi/M_{pl}}$ ($q>0$) is the coupling with the
field and
$\mathcal{G}=R^{2}-4R^{\mu\nu}R_{\mu\nu}+R^{\rho\mu\sigma\nu}R_{\rho\mu\sigma\nu}$
is the Gauss-Bonnet scalar, which for the flat FLRW geometry has the simple
form $\mathcal{G}=24H^{2}(\dot{H}+H^{2})$.

On the other hand, the authors have chosen as a potential a mathematically
convenient prototype for situations in which an early time plateau, favoured
by Planck, as well as an exponential quintessential tail are present. For
example,

$\displaystyle
V(\varphi)=V_{0}\left[1+\tanh\left(p\varphi/M_{pl}\right)\right],\qquad p>0.$
(173)

For this potential the number of e-folds is
$\mathcal{N}\cong\frac{1}{4p^{2}}e^{2p\varphi/M_{pl}}$, and the spectral index
and the ratio of tensor to scalar perturbations as a function of $\mathcal{N}$
are given by

$\displaystyle n_{s}=1-\frac{2}{\mathcal{N}}\qquad\mbox{and}\qquad
r=\frac{2}{p^{2}\mathcal{N}^{2}},$ (174)

which have the same form of the $\alpha$-attractors for
$\alpha=\frac{1}{6p^{2}}$, so they clearly enter in the 2$\sigma$ C.L.

Finally, since our universe must be reheated by other means than inflaton
decay, the authors employ the instant preheating mechanism, in which, as we
will see in next section, the field is coupled to some other degrees of
freedom. So, as the field is rapidly rolling down the quintessential tail of
its runaway potential, it induces massive particle production, which after the
decay into lighter ones produces the radiation bath of the hot Big Bang. Soon
afterwards, the inflaton field freezes at some value with small residual
energy density, which becomes important at present, playing the role of dark
energy.

###  V.4 Simple models of Quintessential Inflation

In this subsection we will discuss very simple models of Quintessential
inflation that either do not have enough physical motivation or do not agree
with the experimental data.

We start with a model based on the $U(1)$ Peccei-Quinn symmetry proposed in
rosenfeld , where the real part of a complex field plays the role of the
inflaton whereas the imaginary part is the quintessence field.

The Lagrangian is given by

$\displaystyle{\mathcal{L}}=\partial_{\mu}\Phi\partial^{\mu}\Phi^{*}-\lambda\left(\Phi\Phi^{*}-\frac{f^{2}}{2}\right)^{2}-M^{4}\left[\cos(\arg(\Phi)-1\right]],$
(175)

where $\arg(\Phi)$ denotes the argument of the field $\Phi$. Writing the field
as follows, $\Phi=\frac{1}{\sqrt{2}}\phi e^{i\varphi/f}$, one finds the
following equations for the real and imaginary part of the field,

$\displaystyle\ddot{\phi}+3H\dot{\phi}-\frac{\dot{\varphi}^{2}}{f^{2}}\phi+V^{\prime}(\phi)=0,$
$\displaystyle\ddot{\varphi}+\left(3H+\frac{\dot{g}}{g}\right)\dot{\varphi}-\frac{1}{g(\phi)}V^{\prime}(\varphi)=0,$
(176)

where $g(\phi)=\frac{\phi^{2}}{f^{2}}$ and the corresponding potentials are
given by

$\displaystyle
V(\phi)=\frac{\lambda}{4}\left(\phi^{2}-f^{2}\right)^{2}\qquad\mbox{and}\qquad
V(\varphi)=M^{4}\left[\cos\left(\varphi/f\right)-1\right].$ (177)

The inflaton field $\phi$ moves in a potential like a “mexican-hat” and when
it arrives at one of its minima it starts to oscillate and releases its energy
producing particles as in standard inflation. On the contrary, the
quintessence field $\varphi$ rolls in the potential $V(\varphi)$ producing
enough dark energy to match with the current data provided that the parameters
involved in the model satisfy

$\displaystyle f>\frac{M_{pl}}{\sqrt{3}}\qquad\mbox{and}\qquad M\cong 3\times
10^{-3}\mbox{ eV}.$ (178)

Another simple model was built in bento1 ; bento2  in the context of brane-
world context, where the modified Friedmann equation for the flat FLRW
geometry becomes

$\displaystyle
H^{2}=\frac{1}{3M_{pl,4}^{2}}\left(1+\frac{\rho}{2\lambda}\right),$ (179)

where $M_{pl,4}$ is the 4D reduced Planck’s mass and the brane tension is
given by $\lambda=\frac{3}{32\pi^{2}}\frac{M_{pl,5}^{6}}{M_{pl,4}^{2}}$ being
$M_{pl,5}$ the 5D reduced Planck’s mass.

In this context it has been shown that a potential composed by a sum of
exponentials or a hyperbolic cosine leads to Quintessential Inflation. For
this reason the authors, adopting natural units ($M_{pl,4}=1$), choose as a
potential

$\displaystyle
V(\varphi)=e^{-\alpha\varphi}\left[A+(\varphi-\varphi_{0})^{2}\right].$ (180)

For this potential, it is not difficult to calculate the spectral index and
the tensor/scalar ratio as a function of the number of e-folds,

$\displaystyle n_{s}=1-\frac{4}{\mathcal{N}+1}\qquad\mbox{and}\qquad
r=\frac{24}{\mathcal{N}+1},$ (181)

which leads to the relation $6(n_{s}-1)=r$, which is, as in the original
Peebles-Vilenkin model, clearly incompatible with the recent Planck’s data
(see Figure 1). Anyway, we continue with the model. The chosen reheating
mechanism is the Instant Preheating, thus obtaining, as we will see in next
section, a reheating temperature around $10^{9}$ GeV. Finally, the model
exhibits transient acceleration at late times for $0.96\leq A\alpha^{2}\leq
1.26$ and $271\leq\alpha\varphi_{0}\leq 273$, while eternal acceleration is
obtained for $2.3\times 10^{-8}\leq A\alpha^{2}\leq 0.98$ and
$255\leq\alpha\varphi_{0}\leq 273$.

The last simple Quintessential Inflation model that we deal with was
introduced in deHaro:2016hpl , where the idea is to find an unstable non-
singular solution of the equations of General Relativity (the Friedmann and
Raychaudhuri equations), as in the Starobinsky model staro  or the famous
Einstein’s static model einstein , in QI. The idea to build the model goes as
follows: First of all, we look at early times for a Raychaudhuri equation of
the form $\dot{H}=F(H)$ with $F(H)=-\alpha^{2}H^{\beta}$, and note that it has
a finite time singularity for $\beta>1$. So, for large values of $H$, we
choose $F$ as a linear function of the Hubble rate. Secondly, to obtain a
phase transition, one assumes that the derivative of $F$ is discontinuous at
some point $H_{E}$ (this discontinuity enhances the particles production of
heavy massive particles obtaining a finite reheating temperature compatible
with the BBN success deHaro:2016hpl ), choosing for $H\lesssim H_{E}$ a
kination regime, which corresponds to $w_{eff}=1\Longrightarrow F(H)=-3H^{2}$.
And finally, if the model has to take into account the current cosmic
acceleration, the simplest way is to assume that our dynamical system
$\dot{H}=F(H)$ has a fixed point $H_{f}$, i.e., $F(H_{f})=0$, meaning that
$H_{f}$ is a de Sitter solution, which can be modeled for $H_{f}\lesssim H\ll
H_{E}$ by the function $F(H)=-\alpha^{2}(H-H_{f})^{2}$.

An example with all the properties mentioned above is

$\displaystyle\dot{H}=\left\\{\begin{array}[]{ccc}-3H_{e}(2H-H_{e})&\mbox{when}&H>H_{E}\\\
-3(H-H_{f})^{2}&\mbox{when}&H\leq H_{E},\end{array}\right.$ (184)

where the parameters involved satisfy $H_{e}\gg H_{f}$, and in order for $F$
to be continuous one has to choose $H_{E}=H_{e}+H_{f}+\sqrt{2H_{e}H_{f}}\cong
H_{e}.$

For that model the effective EoS parameter
$w_{eff}=-1-\frac{2\dot{H}}{3H^{2}}$ is given by

$\displaystyle
w_{eff}=\left\\{\begin{array}[]{ccc}-1+2\frac{H_{e}}{H}\left(2-\frac{H_{e}}{H}\right)&\mbox{when}&H>H_{E}\\\
&&\\\ -1+2\left(1-\frac{H_{f}}{H}\right)&\mbox{when}&H\leq
H_{E},\end{array}\right.$ (188)

which shows that for $H\gg H_{e}$ one has an early quasi de Sitter period with
$w_{eff}\cong-1$, when $H_{e}\gtrsim H\gg H_{f}$ the universe is in a kination
phase ($w_{eff}=-1$), and finally, for $H\gtrsim H_{f}$ one has a late quasi
de Sitter regime with $w_{eff}\cong-1$.

On the other hand, integrating the equation $\dot{H}=F(H)$, we can see that
the non-singular solution is

$\displaystyle
H(t)=\left\\{\begin{array}[]{ccc}\frac{H_{e}}{2}\left[\left(1+2\frac{H_{f}}{H_{e}}+\sqrt{\frac{8H_{f}}{H_{e}}}\right)e^{-6H_{e}t}+1\right]&\mbox{when}&t<0\\\
\frac{H_{e}+\sqrt{2H_{e}H_{f}}}{3(H_{e}+\sqrt{2H_{e}H_{f}})t+1}&\mbox{when}&t>0,\end{array}\right.$
(191)

which can be approximated by

$\displaystyle
H(t)\cong\left\\{\begin{array}[]{ccc}\frac{H_{e}}{2}\left[e^{-6H_{e}t}+1\right]&\mbox{when}&t<0\\\
\frac{H_{e}}{3H_{e}t+1}&\mbox{when}&t>0.\end{array}\right.$ (194)

To end, the potential can be obtained using the Raychaudhuri equation, which
leads to

$\displaystyle\varphi=M_{pl}\int\sqrt{-2\dot{H}}dt=-M_{pl}\int\sqrt{-\frac{2}{\dot{H}}}dH.$
(195)

Then, for the model studied here, when $H>H_{E}$, one gets

$\displaystyle\varphi=-\frac{2M_{pl}}{\sqrt{3}}\sqrt{\frac{H}{H_{e}}-\frac{1}{2}}\Longleftrightarrow
H=\frac{H_{e}}{2}\left(\frac{3\varphi^{2}}{2M_{pl}^{2}}+1\right),$ (196)

and for $H<H_{E}$ a simple calculation shows that

$\displaystyle\varphi=-\sqrt{\frac{2}{3}}M_{pl}\ln\left(\frac{H-H_{f}}{H_{e}+\sqrt{2H_{e}H_{f}}}\right)+\varphi_{E}$
$\displaystyle{\big{\Updownarrow}}\qquad\qquad\qquad\qquad$ (197)
$\displaystyle
H=(H_{e}+\sqrt{2H_{e}H_{f}})e^{-\sqrt{\frac{3}{2}}(\varphi-\varphi_{E})/M_{pl}}+H_{f},$

where we have introduced the notation
$\varphi_{E}\equiv-\sqrt{\frac{2}{3}}M_{pl}\sqrt{1+\frac{2H_{f}}{H_{e}}+\sqrt{\frac{8H_{f}}{H_{e}}}}\cong-\sqrt{\frac{2}{3}}M_{pl}$.

Finally, using the relation $V(H)=(3H^{2}+\dot{H})M_{pl}^{2}$, one obtains the
following potential,

$\displaystyle
V(\varphi)=\left\\{\begin{array}[]{ccc}\frac{27H^{2}_{e}M_{pl}^{2}}{16}\left(\frac{\varphi^{2}}{M_{pl}^{2}}-\frac{2}{3}\right)&\mbox{when}&\varphi<\varphi_{E}\\\
3H_{f}^{2}M_{pl}^{2}\left[2\left(\frac{H_{e}}{H_{f}}+\sqrt{\frac{2H_{e}}{H_{f}}}\right)e^{-\sqrt{\frac{3}{2}}(\varphi-\varphi_{E})/M_{pl}}+1\right]&\mbox{when}&\varphi>\varphi_{E},\end{array}\right.$
(200)

which depicts, at early times, a $1$-dimensional Higgs potential, also called
Double Well Inflationary potential, and, at late times, an exponential-
quintessence potential.

##  VI Reheating mechanisms

Different reheating mechanisms, such as gravitational particle production,
instant preheating and curvaton reheating are revisited, with all the details,
in this section.

###  VI.1 Gravitational particle production

####  VI.1.1 Massless particle production

Here, we consider a massless quantum field $\chi$ (intensively considered in
the early literature, see for instance ford ; Zeldovich ; Damour ; Giovannini
; pv ), which will be the responsible for particle production. This kind of
field only interacts with gravity and we only assume that the particles are
nearly conformally coupled with gravity, i.e., the coupling constant is
approximately $1/6$ ($\xi\cong\frac{1}{6}$) but not equal to $1/6$, because
free massless spinor and gauge fields are conformally- invariant, so they do
not contribute to the energy density of the relativistic plasma formed by the
produced particles.

Then, the modes, in Fourier space, satisfy the Klein-Gordon equation

$\displaystyle{\chi}^{\prime\prime}_{k}(\tau)+\left(k^{2}+\left(\xi-\frac{1}{6}\right)a^{2}(\tau)R(\tau)\right){\chi}_{k}(\tau)=0,$
(201)

where $\tau$ is, once again, the conformal time and $R(\tau)$ is the Ricci
scalar curvature.

To define the vacuum modes before and after the phase transition, we assume
that at early and late times the term $a^{2}R$ will vanish fast enough at
early and late times, then its behavior at early and late times is
respectively

$\displaystyle\chi_{in,k}(\tau)\simeq\frac{e^{-ik\tau}}{\sqrt{2{k}}}(\mbox{
when
}\tau\rightarrow-\infty),\quad\chi_{out,k}(\tau)\simeq\frac{e^{-ik\tau}}{\sqrt{2{k}}}(\mbox{
when }\tau\rightarrow+\infty).$ (202)

Therefore, the vacuum modes at early (“in” modes) and late times (“out” modes)
(exact solutions of (201)) will be given by ford

$\displaystyle\chi_{in,k}(\tau)=\frac{e^{-ik\tau}}{\sqrt{2{k}}}-\frac{\xi-1/6}{{k}}\int_{-\infty}^{\tau}a^{2}(\tau^{\prime})R(\tau^{\prime})\sin({k}(\tau-\tau^{\prime}))\chi_{in,k}(\tau^{\prime})d\tau^{\prime},$
$\displaystyle\chi_{out,k}(\tau)=\frac{e^{-ik\tau}}{\sqrt{2{k}}}+\frac{\xi-1/6}{{k}}\int_{\tau}^{\infty}a^{2}(\tau^{\prime})R(\tau^{\prime})\sin({k}(\tau-\tau^{\prime}))\chi_{out,k}(\tau^{\prime})d\tau^{\prime},$
(203)

and, since we are considering particles nearly conformally coupled to gravity,
we can consider the term $(\xi-1/6)a^{2}(\tau)R(\tau)$ as a perturbation, and
we can approximate the “in” and “out” modes by the first order Picard’s
iteration, i.e., inserting (202) in the right-hand side of (VI.1.1), as

$\displaystyle\chi_{in,k}(\tau)\cong\frac{e^{-ik\tau}}{\sqrt{2{k}}}-\frac{\xi-1/6}{{k}\sqrt{2{k}}}\int_{-\infty}^{\tau}a^{2}(\tau^{\prime})R(\tau^{\prime})\sin({k}(\tau-\tau^{\prime}))e^{-ik\tau^{\prime}}d\tau^{\prime},$
$\displaystyle\chi_{out,k}(\tau)\cong\frac{e^{-ik\tau}}{\sqrt{2{k}}}+\frac{\xi-1/6}{{k}\sqrt{2{k}}}\int_{\tau}^{\infty}a^{2}(\tau^{\prime})R(\tau^{\prime})\sin({k}(\tau-\tau^{\prime}))e^{-ik\tau^{\prime}}d\tau^{\prime},$
(204)

which will represent, respectively, the vacuum before and after the phase
transition.

Thus, after the phase transition, we could write the “in” mode as a linear
combination of the “out” mode and its conjugate as follows,

$\displaystyle\chi_{in,k}(\tau)=\alpha_{k}\chi_{out,k}(\tau)+\beta_{k}\chi^{*}_{out,k}(\tau).$
(205)

Imposing the continuity of $\chi$ and its first derivative at the transition
time we get, up to order $\left(\xi-1/6\right)^{2}$, the value of these
coefficients Birrell1 ; Zeldovich ,

$\displaystyle\alpha_{k}\cong
1-\frac{i({\xi}-\frac{1}{6})}{2k}\int_{-\infty}^{\infty}a^{2}(\tau)R(\tau)d\tau,\quad\beta_{k}\cong\frac{i({\xi}-\frac{1}{6})}{2k}\int_{-\infty}^{\infty}e^{-2ik\tau}a^{2}(\tau)R(\tau)d\tau.$
(206)

Finally, in a simple model of gravitational production of quanta with
negligible rest mass the energy density of the produced particles due to the
phase transition is given by Birrell

$\displaystyle\rho_{\chi}=\frac{N_{s}}{2\pi^{2}a^{4}}\int_{0}^{\infty}k^{3}|\beta_{k}|^{2}dk,$
(207)

where $N_{S}$ is the number of fermion and boson fields involved in the model,
which for simplicity we will assume of the order $1$, and the integral of the
$\beta$-Bogoliubov coefficient (206) is convergent because at early and late
times the term $a^{2}(\tau)R(\tau)$ converges fast enough to zero. In
addition, if at the transition time the first derivative of the Hubble
parameter is continuous one has $\beta_{k}\sim{\mathcal{O}}(k^{-3})$, which
means that the energy density of the produced particles is not ultraviolet
divergent. Then, the energy density of the produced massless particles
approximately becomes ford

$\displaystyle\rho_{\chi}(t)\cong\bar{\mathcal{N}}\left({\xi}-\frac{1}{6}\right)^{2}H^{4}_{kin}\left(\frac{a_{kin}}{a(t)}\right)^{4},$
(208)

where $\bar{\mathcal{N}}$ is a dimensionless numerical factor.

Remark.- The number $\bar{\mathcal{N}}$ is clearly model dependent. In the
scenario proposed by Ford in ford  where there is a transition from de Sitter
to a matter domination modeled by
$a^{2}(\tau)R(\tau)\equiv\frac{12}{\tau^{2}+\tau_{0}^{2}}$, the number
$\bar{\mathcal{N}}$ can be calculated analytically giving as a result
$\frac{9}{8}$. However, note that in this case reheating is impossible because
the energy density of the produced particles decreases faster than those of
the background. To go beyond, we have calculated numerically this factor for
some simple models that have a transition from a de Sitter regime to a
kination one, and in all cases $\bar{\mathcal{N}}$ is of the order $1$ (see
for instance ha ).

On the other hand, it is well-known that the thermalization of the produced
particles is a very fast but not instantaneous process allahverdi  (see also
the Section III of hyp ). In fact, following the work of Spokoiny Spokoiny
(see also pv ), we can consider the thermalization rate
$\Gamma_{th}=n_{\chi,kin}\sigma_{2\rightarrow 2}$, where the most important
process for kinetic equilibrium are $2\rightarrow 2$ scatterings with gauge
boson exchange, whose typical energy energy is $E\sim\rho_{\chi,kin}^{1/4}$,
in the $t$-channel, which is given by $\sigma_{2\rightarrow
2}=\frac{\alpha^{3}}{E^{2}}$ (see for instance the section IV of allahverdi ),
where, as usual, $\alpha^{2}\sim 10^{-3}$ Spokoiny .

Given that

$\displaystyle
n_{\chi}(t)=\frac{1}{2\pi^{2}a^{3}(t)}\int_{0}^{\infty}k^{2}|\beta_{k}|^{2}dk=\left({\xi}-\frac{1}{6}\right)^{2}{\mathcal{M}}H^{3}_{kin}\left(\frac{a_{kin}}{a(t)}\right)^{3},$
(209)

where for many models one finds hap1

$\displaystyle{\mathcal{M}}\equiv\frac{1}{16\pi
a^{3}_{kin}H^{3}_{kin}}\int_{-\infty}^{\infty}a^{4}(\tau)R^{2}(\tau)d\tau\sim
1,$ (210)

which is finite, because in all the considered models $R=6(\dot{H}+2H^{2})$ is
continuous and, as we have already explained, $a^{2}R$ vanishes fast enough at
early and late times. So, we get that

$\displaystyle\Gamma_{th}=\alpha^{3}\left({\xi}-\frac{1}{6}\right)\frac{{\mathcal{M}}H_{kin}}{\bar{\mathcal{N}}^{\frac{1}{2}}}\sim\alpha^{3}\left({\xi}-\frac{1}{6}\right){H_{kin}}.$
(211)

The relativistic fluid reaches the thermal equilibrium when $H(t_{th})\sim
H_{th}\sim\Gamma_{th}$ pv ; Spokoiny  and, since during kination the Hubble
rate scales as $a^{-3}$, one has

$\displaystyle
H_{th}=H_{kin}\left(\frac{a_{kin}}{a_{th}}\right)^{3}\sim\Gamma_{th}\sim\alpha^{3}\left({\xi}-\frac{1}{6}\right){H_{kin}}\Longrightarrow\left(\frac{a_{kin}}{a_{th}}\right)^{3}\sim\alpha^{3}\left({\xi}-\frac{1}{6}\right),$
(212)

meaning that the temperature when the thermalization is achieved is of the
order

$\displaystyle
T_{th}=\left(\frac{30}{\pi^{2}g_{th}}\right)^{1/4}\rho_{\chi}^{1/4}(t_{th})\sim\alpha
g_{th}^{-1/4}\left({\xi}-\frac{1}{6}\right)^{5/6}H_{kin}.$ (213)

Taking for instance $H_{kin}\sim 10^{-6}M_{pl}$ (the typical value of the
Hubble rate at the beginning of kination for many models), $g_{th}=106.75$
(the effective number of degrees of freedom of the Standard Model) and
${\xi}-\frac{1}{6}\sim 10^{-2}$, one gets a thermalization temperature of the
order of $10^{9}$ GeV.

Finally, the reheating occurs when the energy density of the background and
the one of the created particles is of the same order
($\rho_{\chi}(t_{reh})\sim\rho(t_{reh})\Longrightarrow\frac{a_{kin}}{a_{reh}}\sim\left({\xi}-\frac{1}{6}\right)\frac{H_{kin}}{M_{pl}}$),
which implies, for the typical value $H_{kin}\sim 10^{-6}M_{pl}$, a reheating
temperature of the order

$\displaystyle T_{reh}\sim g_{reh}^{-1/4}\rho_{\chi}^{1/4}(t_{reh})\sim
g_{reh}^{-1/4}\left({\xi}-\frac{1}{6}\right)^{3/2}\left(\frac{H_{kin}}{M_{pl}}\right)^{2}M_{pl}$
$\displaystyle\sim 2\times
10^{6}g_{reh}^{-1/4}\left({\xi}-\frac{1}{6}\right)^{3/2}\mbox{GeV},$ (214)

which for ${\xi}-\frac{1}{6}\sim 10^{-2}$ and $g_{reh}=106.75$ leads to a low
reheating temperature around $10^{3}$ GeV.

####  VI.1.2 Superheavy particle production: Calculations using the WKB
approximation

The study of the massive particle production in QI is a bit more involved than
for massless particles. Therefore, to warm up, we will consider, once again, a
toy model based on that of Peebles-Vilenkin, namely

$\displaystyle
V(\varphi)=\left\\{\begin{array}[]{ccc}\frac{1}{2}m^{2}\left(\varphi^{2}-M_{pl}^{2}+M^{2}\right)&\mbox{for}&\varphi\leq-
M_{pl}\\\
\frac{1}{2}m^{2}\frac{M^{6}}{(\varphi+M_{pl})^{4}+M^{4}}&\mbox{for}&\varphi\geq-
M_{pl},\end{array}\right.$ (217)

where $m$ is the inflaton mass and $M$ is another very small mass whose value
can be calculated in the same way as in the Peebles-Vilenkin model: during the
radiation and matter domination epochs the inflaton field is all the time of
the order $M_{pl}$ (see pv  for a detailed discussion). Then, in the model the
field will dominate at late times when

$\displaystyle\frac{m^{2}M^{6}}{M^{4}_{pl}+M^{4}}\sim\frac{m^{2}M^{5}}{M^{4}_{pl}}\sim
H_{0}^{2}M_{pl}^{2}\Longrightarrow
M\sim\left(\frac{H_{0}}{m}\right)^{\frac{1}{3}}M_{pl}\sim 10^{-18}M_{pl}\sim
1\mbox{ GeV},$ (218)

where we have used that the current value of the Hubble parameter is
$H_{0}\sim 10^{-61}M_{pl}$.

Remark The mass of the inflaton field $m$ can be calculated using the formula
of the power spectrum of scalar perturbations
$P_{\zeta}=\frac{H_{*}^{2}}{8\pi^{2}M_{pl}^{2}}\sim 2\times 10^{-9}$,
obtaining after a simple calculation

$\displaystyle m\sim\sqrt{5}\pi\epsilon_{*}\times 10^{-4}M_{pl}$ (219)

and, since $\epsilon_{*}\sim 10^{-2}$, one finally gets $m\sim 10^{-5}M_{pl}$.

On the other hand, regarding the creation of superheavy massive particles
conformally coupled with gravity that have no interaction with the inflaton
field, the time-dependent frequency of the $\chi$-particles in the $k$-mode is
$\omega_{k}(\tau)=\sqrt{k^{2}+m_{\chi}^{2}a^{2}(\tau)}$, where $m_{\chi}$
denotes the mass of the quantum field $\chi$, that is, the $\chi$ modes, in
Fourier space, satisfy the Klein-Gordon (KG) equation

$\displaystyle\chi_{k}^{\prime\prime}+\omega_{k}^{2}(\tau)\chi_{k}=0.$ (220)

Before continuing, the following observation should be made:

Remark.- It is well-known that at temperatures of the order of the Planck’s
mass quantum effects become very important and the classical picture of the
universe is not possible. However, at temperatures below $M_{pl}$, for example
at GUT scales (i.e., when the temperature of the universe is of the order of
$T\sim 10^{-3}M_{pl}\sim 10^{15}$ GeV), the beginning of the classical Hot Big
Bang (HBB) scenario is possible. Since for the flat FLRW universe the energy
density of the universe, namely $\rho$, and the Hubble parameter $H$ are
related through the Friedmann equation $\rho=3H^{2}M_{pl}^{2}$ and the
temperature of the universe is related to the energy density via the Stefan-
Boltzmann law $\rho=(\pi^{2}/30)g_{*}T^{4}$ (where $g_{*}=106.75$ is the
number degrees of freedom for the energy density in the Standard Model), one
can conclude that a classical picture of our universe might be possible when
$H\sim 10^{-5}M_{pl}\sim 10^{13}$ GeV. Then, if inflation starts at this
scale, i.e. taking the value of the Hubble parameter at the beginning of
inflation as $H_{B}\sim 10^{-5}M_{pl}$, we will assume as a natural initial
condition that the quantum $\chi$-field is in the vacuum at the beginning of
inflation. We will also choose the mass of the $\chi$-field two orders greater
than this value of the Hubble parameter ($m_{\chi}\sim 10^{-3}M_{pl}\sim
10^{15}$ GeV, which is a mass of the same order as those of the vector mesons
responsible for transforming quarks into leptons in simple theories with SU(5)
symmetry lindebook ) because, as we will immediately see, the polarization
terms will be sub-dominant and do not affect the dynamics of the inflaton
field. So, we will choose $10^{13}\mbox{ GeV}\sim m\sim H_{B}\ll m_{\chi}\sim
10^{15}\mbox{ GeV}\ll M_{pl}\sim 10^{18}\mbox{ GeV}$.

Thus, during the adiabatic regimes, i.e., when $H(\tau)\ll
m_{\chi}\Longrightarrow\omega_{k}^{\prime}(\tau)\ll\omega_{k}^{2}(\tau)$, one
can use the WKB approximation Haro  of the equation (220)

$\displaystyle\chi_{n,k}^{WKB}(\tau)\equiv\sqrt{\frac{1}{2W_{n,k}(\tau)}}e^{-{i}\int^{\tau}W_{n,k}(\eta)d\eta},$
(221)

where $n$ is the order of the approximation to calculate the $k$-vacuum mode.
When some high order derivative of the Hubble parameter is discontinuous, one
has to match the $k$-vacuum modes before and after this moment, and it is
precisely at this moment when one needs to use positive frequency modes after
the breakdown of the adiabatic regime in order to perform this matching,
which, following Parker’s viewpoint Parker , is the cause of the gravitational
particle production.

For our toy model, note that the derivative of the potential is discontinuous
at $\varphi=-M_{pl}$, which means, due to the conservation equation, that the
second derivative of the inflaton field is discontinuous at the transition
time and, consequently, from the Raychaudhuri equation
$\dot{H}=-\frac{\dot{\varphi}^{2}}{2M_{pl}^{2}}$ one can deduce that the
second derivative of the Hubble parameter is also discontinuous at this time.

In this case one only needs the first order WKB solution to approximate the
$k$-vacuum modes before and after the phase transition

$\displaystyle\chi_{1,k}^{WKB}(\tau)\equiv\sqrt{\frac{1}{2W_{1,k}(\tau)}}e^{-{i}\int^{\tau}W_{1,k}(\eta)d\eta},$
(222)

where Winitzki

$\displaystyle
W_{1,k}=\omega_{k}-\frac{1}{4}\frac{\omega^{\prime\prime}_{k}}{\omega^{2}_{k}}+\frac{3}{8}\frac{(\omega^{\prime}_{k})^{2}}{\omega^{3}_{k}}.$
(223)

Before the transition time, namely $\tau_{kin}$, which is very close to the
beginning of kination ($\varphi\cong-M_{pl}$), the vacuum is depicted by
$\chi_{1,k}^{WKB}(\tau)$, but after the phase transition this mode becomes a
mix of positive and negative frequencies of the form
$\alpha_{k}\chi_{1,k}^{WKB}(\tau)+\beta_{k}(\chi_{1,k}^{WKB})^{*}(\tau)$.
Then, the $\beta$-Bogoliubov coefficient could be obtained matching both
expressions at $\tau=\tau_{kin}$, obtaining

$\displaystyle\beta_{k}=\frac{{\mathcal{W}}[\chi_{1,k}^{WKB}(\tau_{kin}^{-}),\chi_{1,k}^{WKB}(\tau_{kin}^{+})]}{{\mathcal{W}}[(\chi_{1,k}^{WKB})^{*}(\tau_{kin}^{+}),\chi_{1,k}^{WKB}(\tau_{kin}^{+})]},$
(224)

where
${\mathcal{W}}[f(\tau_{kin}^{-}),g(\tau_{kin}^{+})]=f(\tau_{kin}^{+})g^{\prime}(\tau_{kin}^{-})-f^{\prime}(\tau_{kin}^{+})g(\tau_{kin}^{-})$
is the Wronskian of the functions $f$ and $g$ at the transition time, and
being $F(\tau_{kin}^{\pm})=\lim_{\tau\rightarrow\tau_{kin}^{\pm}}F(\tau)$.

The square modulus of the $\beta$-Bogoliubov coefficient will be given
approximately by hap

$\displaystyle|\beta_{k}|^{2}\cong\frac{m^{4}_{\chi}a^{10}_{kin}\left(\ddot{H}(\tau_{kin}^{+})-\ddot{H}(\tau_{kin}^{-})\right)^{2}}{256(k^{2}+m_{\chi}^{2}a^{2}_{kin})^{5}}\cong\frac{m^{4}_{\chi}m^{6}a^{10}_{kin}}{256\omega_{kin}^{10}},$
(225)

because

$\displaystyle\ddot{H}(\tau_{kin}^{+})-\ddot{H}(\tau_{kin}^{-})=-\frac{\dot{\varphi}_{kin}}{M_{pl}^{2}}(\ddot{\varphi}(\tau_{kin}^{+})-\ddot{\varphi}(\tau_{kin}^{-}))=-\frac{\dot{\varphi}_{kin}}{M_{pl}^{2}}V_{\varphi}(-M_{pl}^{-})=\frac{m^{2}\dot{\varphi}_{kin}}{M_{pl}}=m^{3},$
(226)

where we have used that at the transition time all the energy at the end of
inflation, which is approximately $\frac{1}{2}m^{2}M_{pl}^{2}$ because
$\varphi_{END}=-\sqrt{2}M_{pl}$, was converted into kinetic energy.

Thus, for our model, the number density of the produced particles and its
energy density will be

$\displaystyle n_{\chi}(t)\sim\frac{5m^{3}}{2\pi\times
16^{4}}\left(\frac{m}{m_{\chi}}\right)^{3}\left(\frac{a_{kin}}{a(t)}\right)^{3}\sim
10^{-5}\left(\frac{m}{m_{\chi}}\right)^{3}m^{3}\left(\frac{a_{kin}}{a(t)}\right)^{3},\quad\rho_{\chi}(t)\sim
m_{\chi}n_{\chi}(t).$ (227)

Now, one has to note that there are two different situations, namely, when the
superheavy massive particles decay before and after the end of the kination
regime.

  1. 1.

Decay of the superheavy particles into lighter ones before the end of
kination.

Let $\Gamma$ be the decay rate of the superheavy particles. The decay is
practically finished when $\Gamma$ is of the same order of the Hubble rate,
i.e., when ${\Gamma}\sim
H(t_{dec})=H_{kin}\left(\frac{a_{kin}}{a_{dec}}\right)^{3}\cong\frac{m}{\sqrt{6}}\left(\frac{a_{kin}}{a_{dec}}\right)^{3}$,
and thus, the corresponding energy densities will be

$\displaystyle\rho_{\varphi,dec}=3{\Gamma}^{2}M_{pl}^{2}\quad\mbox{and}\quad\rho_{\chi,dec}\sim
2\times 10^{-5}\left(\frac{m}{m_{\chi}}\right)^{2}\frac{{\Gamma}}{m}m^{4}.$
(228)

Since the decay is before the end of kination, one has $\Gamma<H_{kin}$ and
$\rho_{\chi,dec}<\rho_{\varphi,dec}$, which leads to the following bound,

$\displaystyle
10^{-5}\left(\frac{m^{2}}{m_{\chi}M_{pl}}\right)^{2}m<\Gamma<H_{kin},$ (229)

which for $m\sim 10^{13}$ GeV, $m_{\chi}\sim 10^{15}$ GeV and $H_{kin}\sim
10^{12}$ GeV becomes

$\displaystyle 10^{-6}<\Gamma/\mbox{GeV}<10^{12}.$ (230)

Then, assuming as usual nearly instantaneous thermalization, the reheating
temperature, i.e., the temperature of the universe when the relativistic
plasma in thermal equilibrium starts to dominate
($\rho_{\chi,reh}\sim\rho_{\varphi,reh}$), will be obtained taking into
account that

$\displaystyle\rho_{\chi,reh}=\rho_{\chi,dec}\left(\frac{a_{dec}}{a_{reh}}\right)^{4}\sim\rho_{\varphi,dec}\left(\frac{a_{dec}}{a_{reh}}\right)^{6}=\rho_{\varphi,reh}\Longrightarrow\rho_{\chi,reh}\sim\frac{\rho_{\chi,dec}^{3}}{\rho_{\varphi,dec}},$
(231)

and thus,

$\displaystyle
T_{reh}\sim\left(\frac{\rho_{\chi,reh}}{g_{reh}}\right)^{\frac{1}{4}}\sim\left(\frac{\rho_{\chi,reh}}{g_{reh}}\right)^{\frac{1}{4}}\sqrt{\frac{\rho_{\chi,dec}}{\rho_{\varphi,dec}}}\sim
10^{-4}\left(\frac{m}{m_{\chi}}\right)^{\frac{3}{2}}\left(\frac{m}{\Gamma}\right)^{\frac{1}{4}}\left(\frac{m}{M_{pl}}\right)^{2}M_{pl},$
(232)

which for the values of our parameters leads to a reheating temperature of the
order

$\displaystyle T_{reh}\sim
10^{-17}\left(\frac{M_{pl}}{\Gamma}\right)^{1/4}M_{pl}\sim 3\times
10^{-15}\left(\frac{\mbox{GeV}}{{\Gamma}}\right)^{1/4}\mbox{GeV}.$ (233)

Finally, using the bound of the decay rate, we deduce that the reheating
temperature is bounded by

$\displaystyle 3\times 10^{2}\mbox{ GeV}<T_{reh}<10^{6}\mbox{ GeV}.$ (234)

  2. 2.

Decay of the superheavy particles into lighter ones after the end of kination

First of all, recall that we have already seen that at the end of kination one
has

$\displaystyle H_{end}=\sqrt{2}H_{kin}\Theta\qquad\mbox{ with
}\qquad\Theta=\frac{\rho_{\chi,kin}}{\rho_{\varphi,kin}}\cong 2\times
10^{-5}\left(\frac{m^{2}}{m_{\chi}M_{pl}}\right)^{2}\sim 10^{-19}.$ (235)

Taking into account that the decay is after the end of kination we will have
the bound $\Gamma<H_{end}\sim 10^{-6}$ GeV, and thus, since the thermalization
is nearly instantaneous, the reheating time coincides when the decay is
completed, i.e, when $H\sim\Gamma$, and consequently, the reheating
temperature will be

$\displaystyle T_{reh}\sim g_{reh}^{-1/4}\sqrt{\Gamma M_{pl}},$ (236)

which is bounded by

$\displaystyle 1\mbox{ MeV}<T_{reh}<10^{6}\mbox{ GeV},$ (237)

where the lower bound is obtained taking into account that the reheating will
be before the Big Bang Nucleosynthesis which occurs around $1$ MeV.

####  VI.1.3 The diagonalization method

In the last subsection we have used the WKB method to calculate the particle
production when some derivative of the potential has a discontinuity, but when
the potential is smooth enough it does not work because one needs the exact
solution of the equation (220).

So, here we will explain the so-called diagonalization method, which is
essential to calculate the massive particle creation in more realistic
scenarios such as LQI and $\alpha$-QI.

The idea of the method goes as follows: Given the quantum scalar field $\chi$
of superheavy particles conformally coupled to gravity satisfying the KG
equation (220), the modes that define the vacuum state, at a given initial
time $\tau_{i}$, must satisfy the condition

$\displaystyle\chi_{k}(\tau_{i})=\frac{1}{\sqrt{2\omega_{k}(\tau_{i})}}e^{-i\int^{\tau_{i}}\omega_{k}(\bar{\eta})d\bar{\eta}},\quad\chi_{k}^{\prime}(\tau_{i})=-i\omega_{k}(\tau_{i})\chi_{k}(\tau_{i}),$
(238)

and the energy density of the vacuum is given by Bunch

$\displaystyle\rho_{\chi}(\tau)\equiv\langle
0|\hat{\rho}_{\chi}(\tau)|0\rangle=\frac{1}{4\pi^{2}a^{4}(\tau)}\int_{0}^{\infty}k^{2}dk\left(|\chi_{k}^{\prime}(\tau)|^{2}+\omega^{2}_{k}(\tau)|\chi_{k}(\tau)|^{2}-\omega_{k}(\tau)\right),$
(239)

where in order to obtain a finite energy density gmmbook  we have subtracted
the energy density of the zero-point oscillations of the vacuum
$\frac{1}{(2\pi)^{3}a^{4}(\tau)}\int d^{3}k\frac{1}{2}\omega_{k}(\tau)$.