# Scalar field with a time-independent classical source, not trivial after
all:
from vacuum decay to scattering

Leonardo Tinti111<EMAIL_ADDRESS>Institute of Physics, Jan
Kochanowski University, ul. Uniwersytecka 7, 25-406, Kielce, Poland  Arthur
Vereijken222<EMAIL_ADDRESS>Institute of Physics, Jan Kochanowski
University, ul. Uniwersytecka 7, 25-406, Kielce, Poland  Shahriyar
Jafarzade333<EMAIL_ADDRESS>Institute of Physics, Jan Kochanowski
University, ul. Uniwersytecka 7, 25-406, Kielce, Poland  Institute of
Radiation Problems, Ministry of Science and Education, B.Vahabzade 9 Street,
AZ1143, Baku, Azerbaijan  Center for Theoretical Physics, Khazar University,
Mehseti 41 Street, Baku, AZ1096, Azerbaijan  Francesco
<EMAIL_ADDRESS>Institute of Physics, Jan Kochanowski
University, ul. Uniwersytecka 7, 25-406, Kielce, Poland  Institute for
Theoretical Physics, J. W. Goethe University,
Max-von-Laue-Straße. 1, 60438 Frankfurt am Main, Germany

###### Abstract

Historically it has been believed that a time-independent classical source has
no effect on the scattering of relativistic uncharged field, in contrast with
single particle quantum mechanics. In this work we show that the dynamics is
not trivial. We solve exactly for the scattering amplitudes and find that a
key ingredient is the production of particles from the unstable vacuum,
conceptually similar to the Schwinger mechanism. We compute exactly the
probabilities for the vacuum to decay in $n$ particles. The time dependence of
such probabilities displays interesting properties such as the quantum Zeno
effect and in particular has no regime where the exponential decay law is a
good approximation. We show that the trivial scattering found in the past is
the byproduct of the adiabatic switching of the interaction. In fact, it is
not possible to switch off the interaction (adiabatically or otherwise) at
distant times and recover the exact results. Finally, this non trivial vacuum
behavior is a source of particle production. We argue that such non-
perturbative calculations can be phenomenologically relevant for the
production processes that are suppressed at the lower orders in perturbation
theory, for instance dilaton production in a medium.

##  I Introduction

The quantum field theory of an uncharged free scalar particle coupled to an
external classical source has received much attention during the development
of quantum field theory. This is very natural given its simplicity, and it has
become a textbook example illustrating concepts in quantum field theory [1, 2,
3, 4]. The claim is usually that if the classical source is static in time,
the theory has no scattering and the dynamics is trivial. We will take a close
look at the arguments used for this claim and give reasoning why it can be
disputed. Namely, in this work we compute exactly the time evolution of
quantum states, and it turns out that in the specific case of a time-
independent source the scattering is not trivial, the vacuum decays in a
similar fashion than in the Schwinger mechanism. In fact, this vacuum decay is
the key element to solve all the dynamics of the system, whatever the initial
state.
Vacuum decay of scalar fields has been studied extensively by using, for
example, thin wall approximations [5, 6, 7, 8, 9, 10, 11]. It is also closely
related to quantum mechanical tunneling [12], in particular when it involves
the decay of a metastable vacuum. A possible experimental test of false vacuum
decay is proposed in [13, 14]. One important application is the study of the
electroweak vacuum of the Higgs potential [15, 16, 17, 18, 19, 20]. Namely,
due to quantum corrections to the Higgs self-interaction, the minimum, which
classically is global, has the possibility of becoming only a local minimum.
Whether or not the electroweak vacuum is only a local metastable vacuum
depends on the parameters, in particular, the Higgs mass and the top quark
mass. Current knowledge states that it is indeed metastable [16, 17, 18] with
a lifetime longer than the age of the universe, but close enough to stability
that further studies and precise measurements are necessary. The inclusion of
gravity also has interesting effects on vacuum stability and vice versa, see
e.g. [6, 21, 22, 23]. Recently it has been shown that also in the non-
interacting case non trivial vacua can be necessary to properly renormalize
the expectation values of observables in flat [24, 25], as well as non-flat
spacetime and cosmology [26, 27]. Such non-trivial vacuum effects have been
shown to be a source of particles to be taken into account in heavy-ion
collisions [24, 28].
Furthermore, scalar fields are plentiful in meson physics as well as BSM
physics, examples of such being the dilaton, closely related to the scalar
condensate in QCD [29, 30], and the possible inflaton in cosmology.
Perturbation of the ground state by a temporary classical external field is
also a convenient way to approximately model a physical phenomenon such as the
quark-gluon plasma and the dilaton production from its decay.

The vacuum decay process for scalar fields is conceptually analogous to the
nonperturbative Schwinger mechanism [31, 32, 33, 12], in which particles are
created from the vacuum when there exists a strong static electric field. Past
the conceptual similarity, the results and how they are reached are different.
In the Schwinger mechanism there is particle-antiparticle pair production, in
contrast to our case, in which a single particle (and indeed an arbitrary
number of particles) can be emitted. The particle production does not follow
an exponential and at infinite times the particle content is stable. In
particular, the survival probability is quadratic at short times, thus
allowing for the quantum Zeno effect if a sequence of measurements is
performed. In fact, the quantum Zeno effect is realized when the time
evolution of the system can be slowed down or frozen by repeated measurements
at short time intervals and occurs when the short decay at short time is
slower than exponential, see e.g.[34, 35, 36, 37, 38]. It is an important
effect in cavity QED [39, 40] and potentially even in large systems when
gravity is involved [41]. In the case of the anti Zeno effect [38, 42, 43] the
decay is accelerated by repeated measurements with longer time intervals. They
are both generic results of the time evolution of quantum mechanics, but in
most cases - for example the decay of muons - an exponential decay is an
extremely good approximation experimentally de facto indistinguishable from
the full decay law. In the vacuum decay studied in this work this is not the
case, as will be shown the exponential decay is at no time scale a good
approximation for the decay law.
We proceed as follows. In section II we describe the system. In section III we
compute exact relations for the scattering amplitudes and the probabilities of
the vacuum decaying to $n$ particles, and in section IV we analyse the vacuum
decay, looking at the small time and large time behaviour, compute the
probability distribution for specific external sources, and find the behaviour
for large volume. We compare findings to previous work and discuss
phenomenological applications in V. Finally, we conclude in section VI. Some
details of the calculations and formal arguments are left in the appendices.

##  II Setup of the model

We examine a free scalar field coupled to a static external field. The
simplicity of this model allows us to do exact calculations, with no need for
an expansion in Wick or Feynman diagrams. Unless explicitly stated otherwise,
everything will be done in the Heisenberg picture. The action of the system
can be written in the following way

${\cal S}=\int d^{4}x\;{\cal L}=\int
d^{4}x\left\\{\frac{1}{2}\partial_{\mu}\hat{\phi}(x)\partial^{\mu}\hat{\phi}(x)-\frac{1}{2}m^{2}\hat{\phi}^{2}(x)+g\rho({\bf
x})\hat{\phi}(x)\right\\}.$ (1)

Here $\rho(\mathbf{x})$ is a time independent, classical source of the field
operator $\hat{\phi}$ that carries dimensions Energy-3. The Euler-Lagrange
equation of motion for the system reads

$\Box\,\hat{\phi}(x)+m^{2}\hat{\phi}(x)=g\rho(\bf x).$ (2)

Solutions to this equation of motion are given by

$\hat{\phi}(x)=\hat{\phi}_{0}(x)+i\,g\int_{0}^{t}dy_{0}\int
d^{3}y\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf
p}}\left[\vphantom{\frac{}{}}e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}\right]\,\rho(\bf
y)\text{ ,}$ (3)

where $\hat{\phi}_{0}(x)$ is a solution to the homogeneous equation of motion,
i.e. $\Box\,\hat{\phi}_{0}(x)+m^{2}\hat{\phi}_{0}(x)=0$, decomposed into
positive and negative energy plane waves as:

$\hat{\phi}_{0}(x)=\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\left[\vphantom{\frac{}{}}a_{\bf p}e^{-ip\cdot x}+a_{\bf
p}^{\dagger}e^{ip\cdot x}\right]\text{ ,}$ (4)

where $a_{\mathbf{p}},a^{\dagger}_{\mathbf{p}}$ are the particle creation and
annihilation operators at an initial time, thus the particle vacuum $\ket{0}$
is such that $a_{\mathbf{p}}\ket{0}=0$ for each $\mathbf{p}$. The $y_{0}$
integral in Eq. (3) vanishes for $t=0$, so the homogeneous solution
corresponds to the full solution at time 0 and can be seen as the “in-field”
in the usual $S$-matrix language. Contrary to standard scattering theory, we
do not assume asymptotically free fields for $t\to\pm\infty$. Instead, we will
assume that the initial condition at $t=0$ is the particle vacuum $\ket{0}$.
This vacuum is unstable due to the external source $\rho(\mathbf{x})$, and we
make a later measurement of the particle content at some time $t>0$.
Evaluating the $y_{0}$ integral in (3) and using the homogeneous solution (4),
the exact form of the field for $t>0$ can be written in the explicit form

$\begin{split}\hat{\phi}(x)&=\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\left[\vphantom{\frac{}{}}\left(a_{\bf p}+g\frac{e^{iE_{\bf
p}t}-1}{\sqrt{2E_{\bf p}^{3}}}\int d^{3}y\;e^{-i{\bf p}\cdot{\bf y}}\rho(\bf
y)\right)e^{-ip\cdot x}\right.\\\ &\qquad\left.+\left(a^{\dagger}_{\bf
p}+g\frac{e^{-iE_{\bf p}t}-1}{\sqrt{2E_{\bf p}^{3}}}\int d^{3}y\;e^{i{\bf
p}\cdot{\bf y}}\rho(\bf y)\right)e^{ip\cdot x}\right].\end{split}$ (5)

Note that the time dependence of the in-homogeneous part is proportional to
$\sin(E_{\mathbf{p}}t)/E_{\mathbf{p}}$, which is a nascent delta function. For
$t\to\infty$ it approaches the Dirac delta function $\delta(E_{\mathbf{p}})$,
which might hint towards a trivial infinite time behaviour, because the mass
sets the on-shell energy strictly larger than zero. However for any finite $t$
we can calculate this integral explicitly and find that it is not simply a
peak around $E_{\mathbf{p}}=0,$ but rather it oscillates and is nonzero even
for the massive case when $E_{\mathbf{p}}>0$. We must be careful treating this
expression and take the infinite time limit at a later point, which we will
see will leave us with non-trivial dynamics.
Calling $h$ and $f$ the functions

$\begin{split}h({\bf p})&=g\,\frac{-1}{\sqrt{2E_{\bf p}^{3}}}\int
d^{3}y\;e^{-i{\bf p}\cdot{\bf y}}\rho({\bf y}),\\\ f(t,{\bf
p})&=g\,\frac{e^{iE_{\bf p}t}-1}{\sqrt{2E_{\bf p}^{3}}}\int d^{3}y\;e^{-i{\bf
p}\cdot{\bf y}}\rho({\bf y}),\end{split}$ (6)

equation (5) can be written in the compact way

$\hat{\phi}(x)=\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\left[\left(\vphantom{\frac{}{}}a_{\bf p}+f(t,{\bf p})\right)e^{-ip\cdot
x}+\left(\vphantom{\frac{}{}}a^{\dagger}_{\bf p}+f^{*}(t,{\bf
p})\right)e^{ip\cdot x}\right].$ (7)

The field can be written in a form similar to (4), but with time dependent
creation and annihilation operators. From (7) we can immediately see the form
of the time dependent ladder operators

$a_{\bf p}(t)=a_{\bf p}+f(t,\mathbf{p})=a_{\mathbf{p}}+(1-e^{iE_{\bf
p}t})h(\mathbf{p})\text{ .}$ (8)

From this relation we have, by construction, that $a_{\mathbf{p}}(0)$ is just
$a_{\mathbf{p}}$ which is the initial condition, so we will omit the time
argument at $t=0$. The field, even in an interacting theory, can always be
written in terms of time dependent creation and annihilation operators, see
appendix A. The free field operators are shifted by a function, and so the
expectation value of the particle number operator with respect to the particle
vacuum $\ket{0}$ is

$N_{0}(t)=\bra{0}\hat{N}(t)\ket{0}=\int\frac{d^{3}p}{(2\pi)^{3}}\langle
a^{\dagger}_{\bf p}(t)a_{\bf
p}(t)\rangle=\int\frac{d^{3}p}{(2\pi)^{3}}\absolutevalue{f(t,\bf
p)}^{2}=2\int\frac{d^{3}p}{(2\pi)^{3}}\left(\vphantom{\frac{}{}}1-{\rm
cos}(E_{\bf p}t)\right)\left|h({\bf p})\right|^{2}.$ (9)

The normal-ordered Hamiltonian is computed in Appendix A for a general source;
if the source is time independent it turns out the Hamiltonian is also time
independent:

$\hat{H}=\int\frac{d^{3}p}{(2\pi)^{3}}\,E_{\bf
p}\,\left[\left(a^{\dagger}_{\bf p}+h^{*}({\bf
p})\vphantom{\frac{}{}}\right)\left(a_{\bf p}+h({\bf
p})\vphantom{\frac{}{}}\right)-|h({\bf p})|^{2}\right]\text{ .}$ (10)

As long as the integral

$\int\frac{d^{3}p}{(2\pi)^{3}}\,E_{\bf p}|h({\bf
p})|^{2}=\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf p}^{2}}\,\left|\int
d^{3}x\;e^{i{\bf p}\cdot{\bf x}}\rho({\bf x})\right|^{2}\in(0,\infty)$ (11)

converges, the positive time Hamiltonian corresponds to a finite (and
positive) off-set plus a “free-like” term, having the same spectrum as the
free Hamiltonian, but with the creation (and annihilation) operators shifted
by a function. The Hamiltonian, unsurprisingly, has the same spectrum as a
free Hamiltonian. The final Hamiltonian is, however, not diagonal for the
basis that diagonalizes the free one, since the fields themselves are shifted.
This is necessary to have any evolution at all of the observables in the
Heisenberg picture of quantum mechanics, since the operators change in time,
not the states (except for measurements). The creation and annihilation
operators present in the Hamiltonian are different from those in the field as
they are shifted by a different factor. After all, the fields are time
dependent through the creation and annihilation operators $a_{\mathbf{p}}(t)$,
while the Hamiltonian is time independent. This is already a first hint that
the Hamiltonian does not have a basis of mutual eigenstates with the momentum
or the particle number operator. In appendix A more details as well as the
general form of the Hamiltonian for a time dependent source are given.

##  III Scattering amplitudes

###  III.1 General expressions

Given a generic initial state of the system $|\psi_{i}\rangle$
measured/prepared at time $t_{i}=0$, the probability amplitude and probability
to measure the state $|\psi_{f}\rangle$ at a later time $t>0$ are given by the
time-evolution operator of the theory $U(t,0)$ by

$a_{i\to f}(t)=\langle\psi_{f}|U(t,0)|\psi_{i}\rangle\text{
,}\phantom{....}P_{i\to f}(t)=|a_{i\to f}(t)|^{2}\text{ .}$ (12)

The operator $U(t,0)$ is also referred to as the ‘propagator operator’ [1, 44]
(but clearly this expression should not be confused with the Feynman
propagator or similar quantities). We will make use of the expression $U(t,0)$
as the evolution-time propagator or propagator ‘tout court’ below.

The probability amplitude simplifies significantly when the Hamiltonian is
time independent. For a time independent source, the Hamiltonian is also time
independent. When the source depends on time then so will the Hamiltonian, and
solving (12) becomes a non-trivial task. Before focusing on the case for the
static source, we will solve the dynamics for an arbitrary source, and
simplify to a static source later on. One way to tackle the problem is to
solve exactly for the Dyson series. However, since there are the full
solutions of the Euler-Lagrange equation of motion at all time (7), it is
possible to avoid altogether solving for the operator $U(t,0)$. It requires a
small detour into some unphysical states and some conceptual considerations
over the states and how to recognize them in measurements. As it is explained
in Appendix A, any uncharged scalar field can be decomposed in terms of time-
dependent ladder operators, and the explicit form is taken from (7). This
allows us to use all the free-field concepts of normal ordering and build the
particles state. In the appendix A, it has been used to build a free field
$\phi_{0}$ that coincides with the interacting one at the time $t_{0}=0$.
Since the fields coincide, the particles states built with the free field are
the particle states of the interacting scalar field at $t=0$. In the
Heisenberg picture the physical states do not evolve in time, therefore the
states at time $t_{0}=0$ are the physical states at all times. On the other
hand, there is nothing special about fixing $t_{0}=0$. The very same procedure
could have been done at a different time, providing an equally valid base.
As will be shown next, it is possible to use this freedom of choice of the
base to exactly solve the dynamics of the system, while bypassing the need to
solve for the time-evolution propagator $U(t,0)$. The main ingredient is to
notice that the time-dependent ladder operator evolve with the interaction
propagator. In general for quantum field theories of scalar fields we can
write the annihilation operator as (see appendix A for details):

$a_{\bf p}(t)=\frac{i}{\sqrt{2E_{\bf p}}}\int d^{3}x\;e^{iE_{\bf
p}t}\,e^{-i{\bf p}\cdot{\bf x}}\left[\vphantom{\frac{1}{1}}{\hat{\Pi}}(t,{\bf
x})-iE_{\bf p}\,{\hat{\phi}}(t,{\bf x})\right],$ (13)

where $\hat{\Pi}(t,x)=\partial\cal L/\partial\dot{\phi}$. The hermitian
conjugate gives the expression for the creation operator.

The interacting (Heisenberg) fields evolve according with the full propagator
operator $U(t,0)$ of the theory, therefore

$\begin{split}a_{\bf p}(t)&=\frac{i}{\sqrt{2E_{\bf p}}}\int d^{3}x\;e^{iE_{\bf
p}t}\,e^{-i{\bf p}\cdot{\bf
x}}\left[\vphantom{\frac{1}{1}}U^{-1}(t,0)\,{\hat{\Pi}}(0,{\bf
x})\,U(t,0)-iE_{\bf p}\,U^{-1}(t,0)\,{\hat{\phi}}(0,{\bf x})\,U(t,0)\right]\\\
&=U^{-1}(t,0)\left\\{\frac{i}{\sqrt{2E_{\bf p}}}\int d^{3}x\;e^{iE_{\bf
p}t}\,e^{-i{\bf p}\cdot{\bf x}}\left[\vphantom{\frac{1}{1}}{\hat{\Pi}}(0,{\bf
x})-iE_{\bf p}\,{\hat{\phi}}(0,{\bf x})\right]\right\\}U(t,0)\\\
&=U^{-1}(t,0)\left\\{\vphantom{\frac{1}{2}}e^{iE_{\bf p}t}\;a_{\bf
p}\right\\}U(t,0)\text{ .}\end{split}$ (14)

On the other hand, the ladder operators have the same commutation relations
with $\hat{H}_{0}$ as the free field. That is, with the normal ordered Klein-
Gordon Hamiltonian, built with the fields at the same time as the ladder
operators. Thus the phase $e^{iE_{\bf p}t}$ can be written in terms of the
free propagator $U_{0}(t,0)={\rm exp}\\{-i\hat{H}_{0}t\\}$, leading to

$\begin{split}a_{\bf
p}(t)&=U^{-1}(t,0)\left\\{\vphantom{\frac{1}{2}}e^{iE_{\bf p}t}\;a_{\bf
p}\right\\}U(t,0)\\\
&=U^{-1}(t,0)\left\\{\vphantom{\frac{1}{2}}U_{0}(t,0)\,a_{\bf
p}\,U_{0}^{-1}(t,0)\right\\}U(t,0)=U_{I}^{-1}(t,0)\,a_{\bf
p}\,U_{I}(t,0)\text{ ,}\end{split}$ (15)

where we recognize the propagator in the interaction picture
$U_{I}=U_{0}^{-1}U$. The very same arguments hold for the creation operator,
since the numerical phase in the time has the opposite sign, as it is required
for the opposite sign of the commutation relations with $\hat{H}_{0}$ compared
to the annihilation (or destruction) operator. Using this fact that the
creation and annihilation operators evolve with the interaction picture
propagator, it is possible to recognize a continuous set of states labeled by
$t$:

$\begin{split}|0\rangle_{t}=U_{I}^{-1}(t,0)|0\rangle\quad\Rightarrow\quad
a_{\bf p}(t)|0\rangle_{t}&=\left(\vphantom{\frac{}{}}U_{I}^{-1}(t,0)\,a_{\bf
p}\,U_{I}(t,0)\right)U_{I}^{-1}(t,0)|0\rangle\\\
&=U_{I}^{-1}(t,0)\left(\vphantom{\frac{}{}}a_{\bf p}|0\rangle\right)=0,\\\
|{\bf p_{1}},\cdots,{\bf p}_{n}\rangle_{t}=U_{I}^{-1}(t,0)|{\bf
p_{1}},\cdots,{\bf p}_{n}\rangle\quad\Rightarrow&\\\ \quad a^{\dagger}_{{\bf
p}}(t)\,a_{{\bf p}}(t)\,|{\bf p_{1}},\cdots,{\bf
p}_{n}\rangle_{t}&=U_{I}^{-1}(t,0)\left(\vphantom{\frac{}{}}a^{\dagger}_{{\bf
p}}\,a_{{\bf p}}\,|{\bf p_{1}},\cdots,{\bf p}_{n}\rangle\right).\end{split}$
(16)

They are eigenstates of the number operator at time $t$, and the
$|\cdots\rangle$ are just the regular Heisenberg states, eigenstates of the
number operator at time $t=0$. It is important to note that these states are
not physical states555They are not normalized to one. However they are a basis
for the physical states as usual., nor are they states in the Interaction
representation, since they are linked by $U_{I}^{-1}$ instead of the regular
$U_{I}$. They are also not energy eigenstates, as there are no simultaneous
eigenstates of the Hamiltonian and the number operator (or equivalently of the
Hamiltonian and the momentum operator, see appendix A for details). We are
still working completely in the Heisenberg picture, and these states are just
a useful tool for the calculation. By construction we have

${}_{t}\langle{\bf p}_{1},\cdots,{\bf p}_{n}|{\bf k}_{1},\cdots{\bf
k}_{n}\rangle=\langle{\bf p}_{1},\cdots,{\bf p}_{n}|\,U_{I}(t,0)\,|{\bf
k}_{1},\cdots{\bf k}_{n}\rangle.$ (17)

A short note on which states to project on is in order: it might be intuitive
to think that, since the basis of particle number eigenstates changes over
time, one has to sandwich the propagator between the initial time $n$ particle
state and an $n$ particle state at final time. The reason why this should not
be done is most easily seen in the Schrödinger picture, where the basis of
eigenstates of an operator does not change over time, and one would sandwich
in between two states of the same basis. Matrix elements are equivalent
between pictures, so it is correct to sandwich the operator between two $t=0$
eigenstates666Or between two eigenstates for some other time $t=t^{\prime}$ as
long as they are taken at the same time, since the basis spans the Hilbert
space for all times all the information is still present in these inner
products.. Thus the regular matrix elements of the full propagator read

$\begin{split}&\langle{\bf p}_{1},\cdots,{\bf p}_{n}|\,U(t,0)\,|{\bf
k}_{1},\cdots{\bf k}_{l}\rangle=\langle{\bf p}_{1},\cdots,{\bf
p}_{n}|\,U_{0}(t,0)\,U_{0}^{-1}(t,0)\,U(t,0)\,|{\bf k}_{1},\cdots{\bf
k}_{l}\rangle\\\ &\qquad=\langle{\bf p}_{1},\cdots,{\bf
p}_{n}|\,U_{0}(t,0)\,U_{I}(t,0)\,|{\bf k}_{1},\cdots{\bf
k}_{l}\rangle=\left[\prod_{j=1}^{n}e^{-iE_{{\bf
p}_{j}}t}\right]\;{}_{t}\langle{\bf p}_{1},\cdots,{\bf p}_{n}|{\bf
k}_{1},\cdots{\bf k}_{l}\rangle.\end{split}$ (18)

Here we see more precisely the idea of using the time dependent particle
number eigenstates; one projects it without the propagator and with some
additional phases. It is possible to use this result to solve for the
scattering exactly. Namely, from the definition (16), we can recognize that
the $t$ states are built just like the regular $n$ particle states, but with
the $t$ vacuum and time dependent creation operators:

$|{\bf p}_{1},\cdots,{\bf
p}_{n}\rangle_{t}=\sqrt{\frac{\prod_{j=1}^{n}(2E_{{\bf
p}_{j}})}{n!}}\;a^{\dagger}_{{\bf p}_{1}}(t)\cdots a^{\dagger}_{{\bf
p}_{n}}(t)\,|0\rangle_{t}\text{ .}$ (19)

The explicit form of the $a_{\bf p}(t)$ in terms of the $a_{\bf p}$ is known
exactly from Eq. (7), consisting in a shift. In particular, the mixed
commutators still give a Dirac delta

$a_{\bf p}(t)=a_{\bf p}+f(t,{\bf p})\qquad\Rightarrow\qquad\left[a_{\bf
p}(t),a^{\dagger}_{\bf p}\right]=(2\pi)^{3}\delta^{3}({\bf p}-{\bf q}),$ (20)

and the vacua are proportional to a coherent state of each other (a quick
overview of coherent states is given in appendix D)

$\begin{split}&a_{\bf p}(t)\,|0\rangle_{t}=0\quad\Rightarrow\quad
0=\left(\vphantom{\frac{}{}}a_{\bf p}+f(t,{\bf
p})\right)|0\rangle_{t}\quad\Rightarrow\quad a_{\bf p}|0\rangle_{t}=-f(t,{\bf
p})|0\rangle_{t},\\\ &a_{\bf p}\,|0\rangle=0\quad\Rightarrow\quad
0=\left(\vphantom{\frac{}{}}a_{\bf p}(t)-f(t,{\bf
p})\right)|0\rangle\quad\Rightarrow\quad a_{\bf p}(t)|0\rangle=f(t,{\bf
p})|0\rangle.\end{split}$ (21)

Applying the latter (21) to (18) one almost has the propagator matrix element
already

$\begin{split}&\langle{\bf p}_{1},\cdots,{\bf p}_{n}|\,U(t,0)\,|{\bf
k}_{1},\cdots{\bf k}_{l}\rangle=\left[\prod_{j=1}^{n}e^{-iE_{{\bf
p}_{j}}t}\right]\;{}_{t}\langle{\bf p}_{1},\cdots,{\bf p}_{n}|{\bf
k}_{1},\cdots{\bf k}_{l}\rangle\\\ \\\
&=\sqrt{\frac{\prod_{j=1}^{n}\left(2E_{{\bf
p}_{j}}\right)}{n!}}\sqrt{\frac{\prod_{s=1}^{l}\left(2E_{{\bf
k}_{s}}\right)}{l!}}\left[\prod_{j=1}^{n}e^{-iE_{{\bf
p}_{j}}t}\right]\;{}_{t}\langle 0|\,a_{{\bf p}_{1}}(t)\cdots a_{{\bf
p}_{n}}(t)\;a^{\dagger}_{{\bf k}_{1}}\cdots a^{\dagger}_{{\bf
k}_{l}}|0\rangle\\\ \\\ &=\frac{{}_{t}\langle
0|0\rangle}{\sqrt{n!l!}}\left\\{\prod_{j=1}^{n}\left[e^{-iE_{{\bf
p}_{j}}t}f(t,{\bf p}_{j})\sqrt{2E_{{\bf
p}_{j}}}\right]\prod_{s=1}^{l}\left[-f^{*}(t,{\bf k}_{s})\sqrt{2E_{{\bf
k}_{s}}}\right]\right.\\\ &+(2\pi)^{3}\sum_{j=1}^{n}\sum_{s=1}^{l}(2E_{{\bf
k}_{s}})\,e^{-iE_{{\bf k}_{s}}t}\,\delta^{3}({\bf p}_{j}-{\bf
k}_{s})\prod_{j^{\prime}\neq j}\left[e^{-iE_{{\bf p}_{j^{\prime}}}t}f(t,{\bf
p}_{j^{\prime}})\sqrt{2E_{{\bf
p}_{j^{\prime}}}}\vphantom{\int}\right]\prod_{s^{\prime}\neq
s}\left[\vphantom{\int}-f^{*}(t,{\bf k}_{s^{\prime}})\sqrt{2E_{{\bf
k}_{s^{\prime}}}}\right]\\\ &\quad+\quad\left.\vphantom{\int}\mbox{ all
combinations with two and more Dirac deltas}\quad\right\\}.\end{split}$ (22)

The second equality is gained using the commutation relation (20) and the
actions on the vacuum (21), it gives one term without any Dirac Delta, $nl$
(all the combinations between $p$’s and $k^{\prime}s$) with a single delta. If
both $n$ and $l$ are larger than $1$, then there are $n(n-1)l(l-1)$ terms with
two deltas (all the combinations for two contractions), and so on until either
$n-n$ or $l-l$ is hit. The numerical factor is the same in all instances
except that the functions for the indices that appear in the deltas are
omitted.
To finish the calculation, one has to compute the overlap ${}_{t}\langle
0|0\rangle$. The modulus can be obtained by the normalization of the physical
states, for instance $|0\rangle$:

$\begin{split}1&=\langle 0|0\rangle=\langle
0|\,U^{\dagger}(t,0)\,\,U(t,0)\,|0\rangle=\sum_{n=0}^{\infty}\left[\prod_{j=1}^{n}\int\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right]\left|\langle{\bf p}_{1},\cdots,{\bf
p}_{n}|\,U(t,0)\,|0\rangle\right|^{2}\\\
&=\sum_{n=0}^{\infty}\frac{1}{n!}\left[\prod_{j=1}^{n}\int\frac{d^{3}p_{j}}{(2\pi)^{3}}\right]\left|{}_{t}\langle
0|\,a_{{\bf p}_{1}}(t)\cdots a_{{\bf p}_{n}}(t)|0\rangle\right|^{2}\\\
&=\sum_{n=1}^{\infty}\frac{\left|{}_{t}\langle
0|0\rangle\right|^{2}}{n!}\left(\int\frac{d^{3}p}{(2\pi)^{3}}\;\left|f(t,{\bf
p})\right|^{2}\right)^{n}=\left|{}_{t}\langle
0|0\rangle\right|^{2}e^{N_{0}(t)}\text{ ,}\end{split}$ (23)

Where we recall $N_{0}(t)$ is the expectation value of the particle number
operator given in (9), leading to

$\left|{}_{t}\langle 0|0\rangle\right|=e^{-\frac{1}{2}N_{0}(t)}\text{ .}$ (24)

In Eq. (23) we recognize the probabilities for the vacuum to decay into
$n$-particle states after a time $t$. The use of $t$ states simplifies the
calculation of these to a few lines, compared to a more standard procedure of
projecting onto eigenstates of the Hamiltonian. It is given by a Poissonian
distribution with expectation value $N_{0}(t)$ (as expected), and the
normalization of the probabilities is guaranteed by the normalization of the
initial time vacuum state $\ket{0}$, i.e. (23). What remains is to calculate
the phase of the overlap ${}_{t}\langle 0|0\rangle$. We leave the calculation
of the phase to appendix B. If the overlap is written as ${}_{t}\langle
0|0\rangle=\left|{}_{t}\langle 0|0\rangle\right|e^{i\varphi(t)}$, the phase
$\varphi(t)$ is given by:

$\varphi(t)=\frac{i}{2}\int_{0}^{t}dx^{0}\int\frac{d^{3}p}{(2\pi)^{3}}\left[\partial_{x^{0}}f^{*}(x^{0},{\bf
p})\,f(x^{0},{\bf p})-f^{*}(x^{0},{\bf p})\,\partial_{x^{0}}f(x^{0},{\bf
p})\vphantom{\frac{}{}}\right]\text{ .}$ (25)

###  III.2 One particle scattering

We consider here the scattering of a single particle against the classical
static potential generated by $g\rho({\bf x})$ and show in which sense we
recover the intuitive notion of “the quantum particle seems free as long as it
misses the target”.

We take a single particle initial state

$|\psi_{1}\rangle=\int\frac{d^{3}k}{(2\pi)^{3}2E_{\bf k}}\;\psi_{1}({\bf
k})\;|{\bf k}\rangle,\text{ .}$ (26)

Moreover, we call $\psi_{0}(t;{\bf p}_{1},\cdots,{\bf p}_{n})$ as the
$n$-particle partial wave of the time evolving vacuum

$\psi_{0}(t;{\bf p}_{1},\cdots,{\bf p}_{n})=\langle{\bf p}_{1},\cdots,{\bf
p}_{n}|\,U(t,0)\,|0\rangle=\frac{{}_{t}\langle
0|0\rangle}{\sqrt{n!}}\prod_{j=1}^{n}\left[e^{-iE_{{\bf p}_{j}}t}f(t,{\bf
p}_{j})\sqrt{2E_{{\bf p}_{j}}}\right],$ (27)

which is computed from (22). We know its absolute value squared from (23), but
this intermediate result is convenient to recognize in the scattering
amplitude formula. For $n=0$ this is the survival amplitude of the particle
vacuum which is given by $\psi_{0}(t)={}_{t}\langle 0|0\rangle$, see Eq. (24)
for its modulus. Only two terms in (22) are nonzero in the one-particle
scattering and they are given by:

$\begin{split}\langle{\bf p}_{1}\cdots{\bf
p}_{n}|\,U(t,0)\,|\psi_{1}\rangle&=-\psi_{0}(t;{\bf p}_{1}\cdots{\bf
p}_{n})\int\frac{d^{3}k}{(2\pi)^{3}\sqrt{2E_{\bf k}}}\;f^{*}(t,{\bf
k})\,\psi_{1}({\bf k})\\\
&+\frac{1}{\sqrt{n}}\sum_{j=1}^{n}\left[\psi_{1}({\bf p}_{j})\,e^{-iE_{{\bf
p}_{j}}t}\;\psi_{0}(t;{\bf p}_{1},\cdots{\bf p}_{j-1},{\bf
p}_{j+1},\cdots,{\bf p}_{n})\right].\end{split}$ (28)

The simplest cases to look at are for $n=0$ or $n=1$. In the $n=0$ case the
only term is the absorption of the incoming particle by the source, but
nothing is emitted as a result. For $n=1$ there is a contribution where the
particle is absorbed and re-emitted, and a contribution where it does not
interact with the source and propagates freely. Diagrammatically, for $n=1$
the amplitude can be understood as

$\includegraphics[width=130.08731pt,valign={c}]{feynman-7.pdf}+\includegraphics[width=86.72267pt,valign={c}]{feynman-8.pdf},$
(29)

corresponding to the amplitude given by

$\langle{\bf p}|\,U(t,0)\,|\psi_{1}\rangle=-\psi_{0}(t;{\bf
p})\int\frac{d^{3}k}{(2\pi)^{3}\sqrt{2E_{\bf k}}}\;f^{*}(t,{\bf
k})\,\psi_{1}({\bf k})\\\ +\left[\psi_{1}({\bf p})\,e^{-iE_{{\bf
p}}t}\;\psi_{0}(t)\right]\text{ ,}$ (30)

where: (i) the first term on the r.h.s. contains the amplitude for the
emission of a particle with momentum ${\bf p}$ out of the vacuum,
$\psi_{0}(t;{\bf p})$ (that includes all vacuum contributions with creation
and annihilation of an arbitrary number of particles) multiplied by the
amplitude for the annihilation of the original state $|\psi_{1}\rangle$; (ii)
the second term on the r.h.s. contains the amplitude for the free propagation
of $|\psi_{1}\rangle$ multiplied by the vacuum amplitude $\psi_{0}(t)$, that
includes the creation and subsequent annihilation of any number of particle.

While the pictures above are, strictly speaking, not Feynman diagrams, they
offer an intuitive picture of the process. For general $n$ the structure is
largely similar. In general the amplitude can be diagrammatically written as

$\includegraphics[width=130.08731pt,valign={c}]{feynman-13.pdf}+\includegraphics[width=130.08731pt,valign={c}]{feynman-12.pdf}\,.$
(31)

The first term can be understood as the probability amplitude that the
starting particle gets absorbed, times the $n$-particle wave-function of the
vacuum. The second part is the free-particle wave-function $\psi_{i}({\bf
p})e^{-iE_{\bf p}t}$ times the $(n-1)$-particles vacuum contribution, where
the $\mathbf{p}_{j}$ is not in the vacuum contribution. The sum guarantees
that this contribution is manifestly symmetric in the exchange of any
particle, and the $1/{\sqrt{n}}$ makes sure it is normalized to $1$, as is
mandatory for physical states. The entire dynamics of the system can be summed
up as an interference pattern between the decaying vacuum, a probability
amplitude of absorbing the original particle and a free-like evolution of the
original state. The vacuum decay is a key component of the scattering; if the
vacuum did not decay, there would have been no scattering.
Focusing on the absorption amplitude, calling $\psi_{1}(x)$

$\psi_{1}(x)=\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf p}}\;e^{-ip\cdot
x}\psi_{1}({\bf p}),$ (32)

the wave function in space-time that the original state would have if it were
a free system, making use of the definition of $f(t,{\bf k})$ in (6), one has

$\begin{split}{\cal A}(t)&=-\int\frac{d^{3}k}{(2\pi)^{3}\sqrt{2E_{\bf
k}}}\;f^{*}(t,{\bf k})\,\psi_{1}({\bf k})=-i\,g\int_{0}^{t}dy^{0}\int
d^{3}y\int\frac{d^{3}k}{(2\pi)^{3}2E_{\bf k}}\;\rho({\bf y})\,e^{-ip\cdot
y}\psi_{1}({\bf k})\\\ &=-i\int_{0}^{t}dy^{0}\int d^{3}y\;g\,\rho({\bf
y})\;\psi_{1}(y).\end{split}$ (33)

The absorption amplitude can also be seen as a functional ${\cal
A}[g\rho,\psi_{1}]$. This amplitude vanishes if the time integral of the
overlap between the free-like wave-function $\psi_{1}(x)$ and the classical
source is vanishing. The absorption amplitude could be zero for another
reason, for example if $\psi(x)$ is an even function of space and
$\rho(\mathbf{x})$ is an odd function of space. The simplest case is, of
course, that the support of $\rho$ and $\psi_{1}$ is mutually exclusive during
the whole time of the evolution

$\rho({\bf x})\psi_{1}(x)=0,\qquad\forall\,x^{0}\in[0,t],\;\forall\,{\bf
x}\in{\mathbb{R}}.$ (34)

This corresponds to the intuitive notion of “the wave-packet misses the
classical source”. Because of the exact and general (28), if the amplitude
${\cal A}$ is vanishing the only remaining part of the dynamics is the
eventual decay of the vacuum, in the form of the $\psi_{0}$’s, with an
appropriate totally symmetric combination of the free-like part and the vacuum
contribution. It is not possible to remove this vacuum decay, but we can look
at the conditional probability that no extra particles come from the vacuum
decay. That is, we normalize the probability amplitudes by dividing by
$\int\frac{d^{3}\mathbf{p}}{(2\pi)^{3}2E_{\mathbf{p}}}|\langle{\bf
p}|U|\psi_{1}\rangle|^{2}$, so the probabilities are conditional on the
observation of ‘one particle goes in, one particle comes out’, while keeping
the original phase in the wave functions. (In turn, this operation also
corresponds to the ‘collapse’ of the wave function if the final state is
measured to contain one particle only). By construction, the resulting wave
functions are normalized to 1. When the absorption amplitude is zero, the
resulting one particle wave function after scattering is

$e^{i\varphi(t)}\;e^{-iE_{\bf p}t}\psi_{1}(\bf p)\text{ .}$ (35)

This is ‘almost’ the wave-function of a free particle in the Shrödinger
representation. In fact, the energy phase $e^{-iEt}$ is responsible of the
time evolution of the states, even for free fields. The phase $e^{i\varphi}$
is not, but it can be removed from the theory without changing the Euler-
Lagrange equations of motion since it corresponds to a (time-dependent) phase-
convention of the basis, rather than any physical observable.
Since the $\varphi(t)$ itself is given as a time integral (25), it is possible
to add to the Lagrangian of the system the function $-\partial_{t}\varphi(t)$
to compensate exactly in the final formulas.
In this sense then, for a vanishing overlap ${\cal A}(t)=0$, the elastic
scattering with the classical source $g\rho$ is trivial and indistinguishable
from a free particle. The non-trivial dynamics involve the creation of and
interference with the eventual extra particles from the vacuum decay.
It goes without saying that the elastic scattering is not trivial for a non-
vanishing ${\cal A}$. In that case the normalized partial wave is given by:

$\left[{\cal A}(t)\,\psi_{0}(t;{\bf p})\;+\;\psi_{0}(t)\,e^{-iE_{\bf
p}t}\psi_{i}({\bf
p})\vphantom{\int}\right]\frac{\exp[\frac{1}{2}N_{0}(t)]}{\sqrt{|{\cal
A}(t)|^{2}\left[\vphantom{\frac{}{}}N_{0}(t)-2\right]+1}},$ (36)

or just $0$ if $\sqrt{2E_{\bf p}}f(t,{\bf p})$ happens to be, at time $t$,
exactly equal to the starting $\psi_{1}$, $f(t,{\bf p})\sqrt{2E_{\bf
p}}=\psi_{1}(\bf p)$.
All of this can be extended to the case of many particles initial states.
Instead of one absorption amplitude there are amplitudes for the absorption of
one or more of the starting particles, the vacuum decays as in the one
particle case, and (parts of 777In the case of some of the particles being
absorbed, but not all off them, only the remaining contribution of the initial
state evolves just with the phases $e^{-iE_{{\bf p}_{j}}t}$. There is not only
the case of no-particle absorbed, and a free evolution of the initial state. )
the initial state that evolves freely.

##  IV Vacuum decay and Schwinger mechanism

In the previous section we have seen that the initial particle vacuum
$\ket{0}$ is not the lowest energy eigenstate (it is not even an energy
eigenstate) and so it will decay and produce particles. We have a situation
conceptually quite similar to the Schwinger mechanism, but there are some
interesting differences in the result. In (23) we can see the probability
distribution for the particle vacuum to decay into $n$ particle after time $t$
has elapsed. It is a Poisson distribution with the expectation value found
earlier in (9), that is

$\begin{split}&P_{0\to
n}(t)=\frac{e^{-N_{0}(t)}}{n!}\left[\vphantom{\frac{}{}}N_{0}(t)\right]^{n},\\\
&N_{0}(t)=\bra{0}\hat{N}(t)\ket{0}=2\int\frac{d^{3}p}{(2\pi)^{3}}\left(\vphantom{\frac{}{}}1-{\rm
cos}(E_{\bf p}t)\right)\left|h({\bf p})\right|^{2}.\end{split}$ (37)

The function $h(\mathbf{p})$ is a weighted Fourier transform of the source
$g\rho(\mathbf{x})$, given in (6). As expected, in the limit of $t\to 0$ the
expectation value is 0 and so the vacuum is stable. The limit $t\to\infty$
depends on the form of the source, but we can deduce it is finite. The
integral can be re-written as an integral over energy and angle (which is
bounded), and is absolutely convergent. Therefore the Riemann-Lebesgue lemma
applies which says that in the limit of $t\to\infty$ the oscillatory term goes
to 0, meaning the expectation value goes to the finite nonzero positive value:

$\lim_{t\to\infty}N_{0}(t)=2\int\frac{d^{3}p}{(2\pi)^{3}}\left|h({\bf
p})\right|^{2}<\infty\text{ ,}$ (38)

implying that the particle production rate goes to zero for large times,
leaving us with a finite amount of particles (or particle density for large
volumes), without any need to turn off the source. Since the derivation of
section III was valid even for time dependent sources, the source can of
course be chosen so that it turns off after some time.

###  IV.1 Quantum Zeno effect

For short times there is a quadratic behavior that allows for the quantum Zeno
effect. Namely, the vacuum survival probability

$P_{0\to 0}(t)={\rm exp}\left\\{-2\int\frac{d^{3}q}{(2\pi)^{3}}\left(1-{\rm
cos}(E_{\bf q}t)\vphantom{\frac{}{}}\right)\left|h({\bf
q})\right|^{2}\right\\}$ (39)

can be expanded for short times:

${\rm exp}\left\\{-\int\frac{d^{3}q}{(2\pi)^{3}}E_{\bf
q}^{2}t^{2}\vphantom{\frac{}{}}\left|h({\bf
q})\right|^{2}\right\\}=e^{-\frac{t^{2}}{\tau_{Z}^{2}}}\text{ ,}$ (40)

where we define a ‘Zeno time’ $\tau_{Z}$ by

$\frac{1}{\tau_{Z}^{2}}\equiv\int\frac{d^{3}q}{(2\pi)^{3}}E_{\bf
q}^{2}\vphantom{\frac{}{}}\left|h({\bf q})\right|^{2}.$ (41)

We may look at the situation where we have repeated measurements of the vacuum
and whether or not it survived. We let the system evolve over a large time $T$
with $N$ measurements separated by equally spaced (small) time intervals
$t=T/N$. For the vacuum to survive at time $T$, it needs to survive at each
measurement, meaning

$P_{0\to 0}(T)=(P_{0\to
0}(t))^{N}\approx\exp\left[-N\frac{t^{2}}{\tau_{Z}^{2}}\right]=\exp\left[-\frac{1}{N}\frac{T^{2}}{\tau_{Z}^{2}}\right].$
(42)

The small time behaviour of the survival probability now also holds for
arbitrarily long times $T$, given $N$ is chosen sufficiently large so that $t$
is sufficiently small. Furthermore, for a given time $T$, $N$ can be chosen
arbitrarily large, which makes the survival probability approach 1. This is an
explicit example of the quantum Zeno effect [34, 35, 36, 37, 38]: by
repeatedly measuring the system we can preserve the initial state which would
normally be prone to change over time. It is a consequence of a decay that is
slower than an exponential decay for small times. The quantum Zeno effect is a
generic feature of theories obeying quantum mechanics. It is found for example
in cavity QED [39, 40] and may even affect Hawking radiation in large black
holes [41]. Moreover, upon inspection of (39), we can reason there is no
regime where an exponential decay law is a good approximation. For small times
it is given by (40), after which the oscillatory behaviour becomes relevant.
At large time the oscillation dies out and the particle number expectation
value becomes a non-vanishing constant.

###  IV.2 Numerical examples

As a computational example, and to illustrate the time dependent behaviour, we
shall compute the probability distribution for two different forms of the
external source, a Gaussian and a flat source. We will take them of the forms:

$\displaystyle g\rho_{1}({\bf x})$
$\displaystyle=g\,m^{3}\,\exp\left(-\,\pi\frac{\bf x^{2}}{L^{2}}\right),$ (43)
$\displaystyle g\rho_{2}(\mathbf{x})$
$\displaystyle=g\,m^{3}\,\theta\left(\frac{L}{2}-x\right)\theta\left(\frac{L}{2}+x\right)\theta\left(\frac{L}{2}-y\right)\theta\left(\frac{L}{2}+y\right)\theta\left(\frac{L}{2}-z\right)\theta\left(\frac{L}{2}+z\right)\text{
,}$ (44)

where $g$ plays the role of a dimensionless coupling constant determining the
strength of the interaction, and $L$ is a parameter that determines the volume
of the sources. The sources are normalized such that their peaks and total
charges are equal $\int d^{3}\mathbf{x}g\rho_{1}(\mathbf{x})=\int
d^{3}\mathbf{x}g\rho_{2}(\mathbf{x})=g\,m^{3}L^{3}$, and
$\rho_{1}(0)=\rho_{2}(0)=g\,m^{3}$. For the expectation value we need the
weighted Fourier transforms of the sources (see Eq. (6)), which are given by

$\displaystyle h_{1}(\textbf{p})$
$\displaystyle=-\frac{g}{(\mathbf{p^{2}}+m^{2})^{3/4}}\,m^{3}L^{3}\,e^{-\frac{p^{2}L^{2}}{4\pi}},$
(45) $\displaystyle h_{2}(\mathbf{p})$
$\displaystyle=-\frac{g}{(\mathbf{p^{2}}+m^{2})^{3/4}}\frac{8m^{3}\sin\left(\frac{Lp_{x}}{2}\right)\sin\left(\frac{Lp_{y}}{2}\right)\sin\left(\frac{Lp_{z}}{2}\right)}{p_{x}p_{y}p_{z}},$
(46)

where $p_{x},p_{y},p_{z}$ are the components of the 3-momentum $\mathbf{p}$.
This allows us to compute the integral (LABEL:expectation_value) numerically.
Since the mass is the only parameter with dimension of energy for this form of
the sources, we will set the mass $m=1$. Consequently the time $t$ and length-
scale $L$ will be in units $1/m$. The dependence on $g$ is simple: $N_{0}(t)$
scales with $g^{2}$ (even though the result is non-perturbative), and the
dependence on $L$ has no simple form except in the large L limit, which we
look at shortly. The result of this with the parameters $g$ and $L$ set to 1
is shown in figure 1, where we can see that saturation of the particle
expectation value sets in quite quickly. The vacuum survival probability for
the Gaussian source, together with an exponential decay and the small time
behaviour are shown in figure 2.
Taking, as an illustrative example, the case of the dilaton [29], assuming a
mass of about 1.7 GeV, and using $\hbar=6.6\cdot 10^{-25}$ GeV s = 1, units of
time are of approximately $4\cdot 10^{-25}$ seconds. From figure 1 the
particle content is close to saturation after 10 units of $m\,t$ (or earlier),
a typical timescale for the dilaton in this case is in the order of $10^{-24}$
seconds, or $\sim 1\,\text{fm}/c$.
In figure 3 a comparison of the vacuum survival probability with an
exponential approximation of the form $z_{i}+(1-z_{i})\exp{-\Gamma t}$, where
$z_{i}$ is the saturation value found by exponentiating (38), the index
$i=1,2$ denotes to which shape of the source i.e. (45) or (46), it belongs to,
and where we have taken $\Gamma\sim m$. We have also taken $g=1$, but scaling
to $g\neq 1$ is done simply by taking the probability to the power of $g^{2}$.
Note, the choice $\Gamma=m$ here is arbitrary and is chosen just to ‘guide the
eye’ in order to manifestly show that it does not lead to a good description
of the actual survival probability. The only regime where the exponential
approximation is close is at large times when the fluctuations around the
saturation value are too small. A better description of the theoretical curve
could be obtained by an exponential with superimposed oscillations as well as
an offset, but we leave this point for future phenomenological investigations.

It is interesting to observe that similar oscillations (but not decays) arise
in the case of mixing of bosonic fields [45, 46], which in turn might be
relevant for neutrino oscillations [47].

Figure 1: Expectation value for the number of particles produced for the
different sources as a function of time. The parameters $g$ and $m$ are
absorbed in the expectation value and time units respectively, and we use
$L=1$ (in $1/m$ units). The grey lines are the infinite time asymptotes found
by setting the oscillatory term to 0. Figure 2: Vacuum survival probability
with a Gaussian source, compared with the small time behaviour found in (40).
The parameter $m$ is absorbed in the time units, and $g=1$ and $L=1$ are used.
Figure 3: Vacuum survival probability with both sources with a longer
timescale, compared with a exponential decay which saturates to a non-zero
value at infinity. The parameter $m$ is absorbed in the time units, and we set
$L=1$ in units of $1/m$, and $g=1$. The grey lines indicate the asymptotic
values for the survival probability for the two different sources.

###  IV.3 Large volume limit

The scenario where the size of the classical source is much larger than the
scales probed by the system is of course a simplification which is however
interesting for practical purposes. In our system, this large volume limit can
be explicitly computed for both sources. Starting with the Gaussian shaped
source, the expectation value per volume element $L^{3}$ is found using
(LABEL:expectation_value):

$\displaystyle\frac{1}{L^{3}}N_{0}(t)=2\int\frac{d^{3}\mathbf{p}}{(2\pi)^{3}}\left(1-\cos(E_{\mathbf{p}}t)\right)\frac{g^{2}}{(\mathbf{p}^{2}+m^{2})^{3/2}}m^{6}L^{3}\,e^{-\frac{p^{2}L^{2}}{2\pi}}.$
(47)

For taking the large volume limit, that is $L\to\infty$, we will want to
identify representations of the Dirac delta function. In this case for the
Gaussian source, the relevant representation of the Dirac delta is

$\delta(x)=\lim_{\epsilon\to
0^{+}}\frac{e^{-x^{2}/4\epsilon}}{2\sqrt{\pi\epsilon}}$ (48)

Renaming $\frac{L^{2}}{2\pi}=\frac{1}{4\epsilon}$ on the RHS of (47) this form
of the Dirac deltas will be more easily identifiable:

$\displaystyle\frac{1}{L^{3}}N_{0}(t)=2\int\frac{d^{3}\mathbf{p}}{(2\pi)^{3}}\left(1-\cos(E_{\mathbf{p}}t)\right)\frac{g^{2}m^{6}}{(\mathbf{p}^{2}+m^{2})^{3/2}}\pi^{3}\sqrt{8}\left(\frac{e^{-p_{x}^{2}/4\epsilon}}{2\sqrt{\pi\epsilon}}\right)\left(\frac{e^{-p_{y}^{2}/4\epsilon}}{2\sqrt{\pi\epsilon}}\right)\left(\frac{e^{-p_{z}^{2}/4\epsilon}}{2\sqrt{\pi\epsilon}}\right).$
(49)

When we take the large volume limit $L\to\infty$ which implies $\epsilon\to
0^{+}$ the last 3 factors approach delta functions of the components of
$\mathbf{p}$. Evaluating the momentum integral with these delta functions
gives

$\displaystyle\frac{1}{L^{3}}N_{0}(t)=\frac{g^{2}}{\sqrt{2}}m^{3}(1-\cos(mt)).$
(50)

Similarly, in the computation for the square source we will need the following
representation of the Dirac delta:

$\delta(x)=\lim_{L\to\infty}\frac{\sin^{2}(L\,x)}{\pi L\,x^{2}}$ (51)

Using the weighted Fourier transform of the flat source (46) in the
expectation value (LABEL:expectation_value) gives

$\displaystyle\frac{1}{L^{3}}N_{0}(t)=2\int\frac{d^{3}\mathbf{p}}{(2\pi)^{3}}\left(1-\cos(E_{\mathbf{p}}t)\right)\frac{g^{2}}{(\mathbf{p}^{2}+m^{2})^{3/2}}\frac{64m^{6}}{L^{3}}\left[\frac{\sin^{2}(Lp_{x}/2)}{p_{x}^{2}}\right]\left[\frac{\sin^{2}(Lp_{y}/2)}{p_{y}^{2}}\right]\left[\frac{\sin^{2}(Lp_{z}/2)}{p_{z}^{2}}\right].$
(52)

Once again, we bring it into the form for Dirac deltas:

$\displaystyle\frac{1}{L^{3}}N_{0}(t)=2\int\frac{d^{3}\mathbf{p}}{(2\pi)^{3}}\left(1-\cos(E_{\mathbf{p}}t)\right)\frac{g^{2}m^{6}}{(\mathbf{p}^{2}+m^{2})^{3/2}}\pi^{3}\left[\frac{\sin^{2}(Lp_{x}/2)}{\pi
L(p_{x}/2)^{2}}\right]\left[\frac{\sin^{2}(Lp_{y}/2)}{\pi
L(p_{y}/2)^{2}}\right]\left[\frac{\sin^{2}(Lp_{z}/2)}{\pi
L(p_{z}/2)^{2}}\right]$ (53)

In the limit $L\to\infty$ the sinc functions become delta functions of the
components of $\mathbf{p}/2$ which equal $8$ times the delta function of
$\mathbf{p}$ and we find

$\displaystyle\frac{1}{L^{3}}N_{0}(t)=2g^{2}m^{3}(1-\cos(mt)).$ (54)

Note that taking this limit does not commute with taking the $t\to\infty$
limit, as here the oscillation does not die down anymore since the integral is
already evaluated. Taking the infinite time limit before the volume limit, the
oscillatory term vanishes while leaving the rest of the derivation entirely
the same. Then the large volume and infinite time limit is

$\begin{split}\lim_{L\to\infty}\lim_{t\to\infty}\frac{1}{L^{3}}N_{0}(t)&=2g^{2}m^{3}\qquad\phantom{\frac{1}{\sqrt{2}}}(\text{flat
source}),\\\
\lim_{L\to\infty}\lim_{t\to\infty}\frac{1}{L^{3}}N_{0}(t)&=\frac{g^{2}}{\sqrt{2}}m^{3}\qquad\phantom{2}(\text{Gaussian
source}).\end{split}$ (55)

Most of the form can be found just using dimensional arguments, since both
sides are dimensionless and the dependence on $g$ is just $g^{2}$, but the
prefactor is dependent on the shape of the source and on how the volume is
defined. In this sense the flat source gives the same result using the
informal statement ‘$\delta^{3}(0)\to(2\pi)^{3}V$’, but for the Gaussian
source it does not.

These results are differs from the Schwinger mechanism, where the particle
production rate is constant, so the number of particles is linear in time and
diverges for $t\to\infty$. This divergence is of course unphysical, and it is
tamed by the adiabatic hypothesis which turns off the source for large times.
In our situation the particle production rate goes to zero for large times,
leaving us with a finite amount of particles (or particle density for large
volumes), without any need to turn off the source. Since the derivation of
section III was valid even for time dependent sources, the source can of
course be turned off, even in a non-adiabatic way.

##  V Discussion

The quantum field theory of a scalar field coupled to some external source is
one of the simplest QFTs one can write down, it has been considered in the
past and it is not surprising that it has been mentioned or discussed in
textbooks, see e.g. [1, 2, 3, 4]. It is commonly accepted that the dynamics
can be non-trivial for a classical source $\rho(t,{\bf x})$ which is Fourier
transformable in spacetime, hence time dependent. The case of a time
independent source is treated as a limiting case of the former. In particular,
the originally time-independent classical source is multiplied by an adiabatic
function that switches off the interaction in the long past and future. The
scattering matrix in the infinite time limit of persistence of the interaction
is essentially the identity matrix, and this is usually assumed to imply an
equivalence between the interacting system (without the adiabatic function)
and a free quantum field. Since some of the intuitive statements commonly used
are incomplete, and it is easy to miss some fundamental conceptual steps
looking just at the scattering matrix, it is convenient to have a more
detailed discussion.

This section is dedicated to the paradoxical comparison between the results of
this paper, in which the dynamics is not trivial, and the past literature. We
will refer especially to the seminal lectures on QFT by Coleman [2], since the
treatment is rather extensive and it inspired some of the other books, e.g.
[4], especially regarding the triviality of a time independent source. The
starting point is the exact solution for the time dependent case, obtained
resumming all the terms of the Dyson series [2]. Using the notation in this
work, the scattering matrix reads

$\begin{split}S&=U_{I}(\infty,-\infty)=:{\rm exp}\left\\{-\frac{g^{2}}{2}\int
d^{4}x\int d^{4}y\,\rho(x)\rho(y)\,\Delta_{F}(x-y)\right.\\\ &\left.-ig\int
d^{4}x\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf p}}}\left[\rho(x)e^{-ip\cdot
x}\,a_{\bf p}+\rho(x)e^{ip\cdot x}\,a^{\dagger}_{\bf
p}\vphantom{\frac{}{}}\right]\right\\}:\,.\end{split}$ (56)

This formula makes sense for a fast decreasing (and hence Fourier
transformable in spacetime) classical source $\rho(x)$. In particular the
field is asymptotically free in the far past and far future
$(\Box+m^{2})\hat{\phi}\to 0$ as $t\to\pm\infty$ by the fast decreasing (in
time too) assumption on the source. Making use of the energy integrated form
of the Feynman propagator

$\Delta_{F}(x-y)=\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf
p}}\left[\vphantom{\frac{}{}}\theta(x^{0}-y^{0})e^{-ip\cdot(x-y)}+\theta(y^{0}-x^{0})e^{ip\cdot(x-y)}\right],$
(57)

it is rather straightforward888The fastest way is to notice that the relations
(21) between the basis $|\cdots\rangle_{t}$ and $|\cdots\rangle$ at $t=0$ can
be generalized to the case
$|\cdots\rangle_{t}\leftrightarrow|\cdots\rangle_{t_{0}}$ using the very same
arguments as before, the difference being the function $f$ is integrated from
$t_{0}$ in the lower bound instead of $0$ as in (93). Therefore the formulas
in (16) generalize to the $t_{0}$ basis using $U(t,t_{0})$ instead of $U(t,0)$
as it could have been expected. to show that the first term in the right hand
side of (56) is just ${}_{t}\langle 0|0\rangle_{t_{0}}$ in the $t\to\infty$
and $t_{0}\to-\infty$ limit. The term in the creation operator reads

${\rm exp}\left\\{-\int\frac{d^{3}p}{(2\pi)^{3}}\,f(\infty,-\infty;E_{\bf
p},{\bf p})\,a^{\dagger}_{\bf p}\vphantom{\frac{}{}}\right\\},$ (58)

and the analogous for the term in the destruction operators is

${\rm exp}\left\\{\int\frac{d^{3}p}{(2\pi)^{3}}\,f^{*}(\infty,-\infty;E_{\bf
p},{\bf p})\,a_{\bf p}\vphantom{\frac{}{}}\right\\}\text{ .}$ (59)

Because of the normal ordered product the two terms factorize, with the
creation term on the left and the destruction one on the right999Within normal
ordered products the creation and destruction operators commute. This is the
usual case of factorization of the exponential of the sum of two commuting
operators. The normal ordering symbol can be removed as long as the exponent
on the creation operators is on the right.. The notation used for the $f$
function is the general (that admits time dependent sources) one as in (93) in
Appendix A. The one used in the main text (6) is thus a specific case. One can
verify that the matrix elements of the scattering matrix (56) coincide with
the limit

$\langle{\bf p}_{1},\cdots,{\bf p}_{n}|S|{\bf q}_{1},\cdots,{\bf
q}_{m}\rangle=\lim_{{t_{-}\to-\infty}}\left[\lim_{{t_{+}\to\infty}}\left(\vphantom{\frac{}{}}{}_{t_{+}}\langle{\bf
p}_{1},\cdots,{\bf p}_{n}|{\bf q}_{1},\cdots,{\bf
q}_{m}\rangle_{t_{-}}\right)\right],$ (60)

using the algebra of the time-dependent bases of Section III. The usual
approach to treat the time-independent case is to assume that the scattering
matrix can be obtained from (56). Since the expression in (56) requires a
classical source which is Fourier transformable in time, and not only in the
space coordinates, one considers the particular case of $\rho(t,{\bf
x})=s(t)\rho(\bf x)$. The (smooth) adiabatic function $s(t)$

$0<s(t)<1,\qquad
s(t)=1\,\forall|t|<\frac{T}{2},\qquad\lim_{t\to\pm\infty}s(t)t^{n}=0\quad\forall
n\in{\mathbb{Z}}^{+},$ (61)

effectively switches on(off) the interaction in the far past(future) in a
Fourier transformable way and keeping the interaction the wanted (time-
independent) one in the bulk $t\in\\{-T/2,T/2\\}$. The intuitive idea is that
in the limit of $T\to\infty$ one has to recover the full time-independent
theory, after all, in other cases, this procedure works. We show in this
section that this is not the case, despite the fact that for a sudden switch
$s(t)=\theta(t+T/2)\theta(T/2-t)$ equation (60) is fulfilled in the
$T\to\infty$ limit101010Technically, both in the left hand side and in the
righ hand side there is an undefined phase that diverges. It can be treated
adding a constant in the Lagrangian (not the Lagrangian density) to off-set
the minimum of the Hamiltonian and have a well-defined phase. This is
completely analogous to the treatment in [2] of the phase in “model II”..
Already in the lectures [2] it is recognized that the conditions in (61)
should be supplemented. Indeed, in order to recover the correct imaginary part
of the function in (56) in the $T\to\infty$ limit, it is necessary to add

$s(t)=0\quad\forall|t|\geq T/2+\Delta(T),\qquad\Delta(T)\geq
0,\qquad\lim_{T\to\infty}\frac{\Delta(T)}{T}=0\text{ .}$ (62)

Later on, we shall simply write $\Delta$ for $\Delta(T)$. In this way the
Fourier transform of the adiabatic function fulfills the following relations

$\int dt\,e^{\pm
iEt}s(t)\xrightarrow{T\to\infty}(2\pi)\,\delta(E),\quad\frac{1}{T}\left|\int
dt\,e^{\pm iEt}s(t)\right|^{2}\xrightarrow{T\to\infty}(2\pi)\,\delta(E).$ (63)

The conditions (62) are not really necessary for the first limit. Any
adiabatic function would suffice. As it can be seen with a direct
calculation111111The function $s$ must be suppressed fast enough for $|t|>T/2$
to be able to use minorants of the modulus square outside of
$t\in\\{-T/2,T/2\\}$ to show that they do not contribute. The rest of the
proof consists just of the properties of the sinc delta family $[{\rm
sin}(x/\epsilon)]/(\pi x)\xrightarrow{\epsilon\to 0}\delta(x)$ and the
$\epsilon[{\rm sin}^{2}(x/\epsilon)]/(\pi x)\xrightarrow{\epsilon\to
0}\delta(x)$ one., it is needed for the second one.

At this point in Ref [2] the operatorial parts (58) and (59) are dismissed
with an intuitive but deeply misleading argument. Since the function $f$ in
(58) reads, in the presence of the adiabatic function

$f(\infty,-\infty;E_{\bf p},{\bf p})=\int dt\;e^{iE_{\bf
p}t}\,s(t)\;\left[i\frac{g}{\sqrt{2E_{\bf p}}}\int d^{3}x\,\rho({\bf
x})\,e^{-i{\bf p}\cdot{\bf x}}\right],$ (64)

thus it approaches a Dirac delta as $T\to\infty$ according to the first
equation in (63). It is assumed that it goes to zero in this limit, the
support of the Dirac Delta is just the origin, and the on-shell energy $E_{\bf
p}\neq 0$ is never in the support for a massive field after all. This argument
would be valid if $a^{\dagger}_{\bf p}$ were a test function rather than an
operator. In the latter case the convergence properties in the space of
generalized functions or distributions are irrelevant. In the lectures [2] it
is presented an incorrect argument to argue that in the function space the
integral of the adiabatic function is going to be vanishing for on-shell
energies. It is based on the common misconception that, since some successions
of functions converging to a Dirac delta are point wise vanishing outside of
$0$ and sharply peaked in $0$, for instance a succession of Gaussians with a
decreasing width $\sigma\to 0$, this must happen for any succession. This is
clearly not the case for the family

$\delta_{\varepsilon}(x)=\sqrt{\frac{2}{\pi}}\,\frac{1}{\varepsilon}\,{\rm
sin}\left(\frac{x^{2}}{\varepsilon^{2}}\right)\xrightarrow{\varepsilon\to
0}\delta(x),$ (65)

which is converging to a Dirac delta121212This can be seen from the general
construction of a family of regular functions converging to a delta. Any
function $g(x)$ normalized to $1=\int dx\,g(x)$, that doesn’t spoil the
uniform convergence of the integral $\int
dy\,g(y)\phi(y\varepsilon)\xrightarrow{\varepsilon\to 0}\int
dy\,g(y)\phi(0)=\phi(0)$ with the test functions, produces a family of
functions $1/\varepsilon\,g(x/\varepsilon)\xrightarrow{\varepsilon\to
0}\delta(x)$. Namely, their action converges to the action of a Dirac delta
$\int dx/\varepsilon\,g(x/\varepsilon)\phi(x)=\int
dy\,g(y)\phi(y\varepsilon)\xrightarrow{\varepsilon\to 0}\phi(0)$ for any test
function $\phi(x)$.. All the functions are vanishing in $x=0$, the infinite
peaks (and valleys) are just less dense around $x=0$; for any $x\neq 0$ the
function $\delta_{\varepsilon}(x)$ does not converge to anything as
$\varepsilon\to 0$.

In the sudden limit $s(t)\sim\theta(t+T/2)\theta(t-T/2)$ the time integral can
be computed exactly, and it reads

$\frac{2\,{\rm sin}\left(\tfrac{E_{\bf p}T}{2}\right)}{E_{\bf
p}}\xrightarrow{T\to\infty}(2\pi)\delta(E_{\bf p}).$ (66)

It still converges, distribution-wise, to a Dirac delta, it has a peak
$\propto T$ in $0$, but the functions for the on-shell energy $E_{\bf p}>0$
are not converging to anything in the $T\to\infty$ limit, certainly not $0$.
They keep bouncing periodically between the extremes $\pm 2/E_{\bf p}$. The
(fast, in the $T\to\infty$ limit) oscillations from the sin function do not
grant any convergence to $0$: The Riemann-Lebesgue lemma works for functions
not operators. The correct procedure is to apply the operators to a physical
states, pick the partial waves amplitude (which in general are also not
defined in the $T\to\infty$ limit), compute the expectation values of the
observable operators (modulus squares of the partial wave functions) and see
if these integrals corresponding to physical observations are well defined in
the large time limit. It is interesting to note that tweaking the calculations
in Section III, namely from time $-T/2$ to $T/2$ rather than from $0$ to $t$,
one obtains the same amplitudes (hence the very same, well defined, time
limits for the probabilities) as using the scattering matrix in the sudden
limit. This must not be interpreted though as proof that the sudden switch is
the “correct way to insert the interaction for a finite time”. In fact, as we
will show at the end of this section, there is no such a thing as a correct
way to switch off the interaction asymptotically to recover the full results
of the theory, not even when restricting to the long time limit behavior. The
analysis of the sudden switch is to point out that if the adiabatic function
$s(t)$ fulfilling (61) and (62) “converges” to a sudden switch (the precise
meaning will be apparent later) the result of a trivial scattering matrix as
in Ref. [2] can’t be recovered. In order to recover the trivial scattering,
one has to add to (61) and (62) the additional conditions

$\Delta\xrightarrow{T\to\infty}\infty,\quad\sup\left|\partial_{t}s(t)\right|\leq\frac{M}{\Delta},\qquad\forall\,T>\bar{T}>0\text{
.}$ (67)

In other words the modulus of the derivative of the (smooth) adiabatic
function $|\dot{s}|$ must be smaller than a maximum that decreases at least
linearly with $\Delta$ in the limit $T\to\infty$. Of course it is inessential
if this bound is violated at finite times and it is required only in the large
$T$ limit. One can see that manipulating directly the integral

$\int dt\,s(t)\,e^{iE_{\bf p}t}=\frac{i}{E_{\bf p}}\int
dt\;\partial_{t}s(t)\,e^{iE_{\bf p}t}\text{ .}$ (68)

The derivation by parts is granted by the smoothness of $s$ and the positivity
of the on-shell energy $E_{\bf p}\geq m>0$. Thanks to the requirements (61),
the time derivative of the adiabatic function is vanishing everywhere except
in the intervals $t\in(-T/2-\Delta,-T/2)$ and $t\in(T/2,T/2+\Delta)$. It is
convenient then to define the the functions $s_{\pm}$

$s_{\pm}(\lambda)=s\left(\pm\lambda\,\Delta\pm\tfrac{T+\Delta}{2}\right),\qquad\lambda\in(-\tfrac{1}{2},\tfrac{1}{2})$
(69)

in term of the dimensionless variable $\lambda$. The integral (68) then reads

$\frac{i}{E_{\bf p}}\,{\rm exp}\left\\{-iE_{\bf
p}\left(\frac{T+\Delta}{2}\right)\right\\}\int
d\lambda\;s^{\prime}_{-}(\lambda)\,e^{-i(E_{\bf
p}\Delta)\lambda}-\frac{i}{E_{\bf p}}\,{\rm exp}\left\\{iE_{\bf
p}\left(\frac{T+\Delta}{2}\right)\right\\}\int
d\lambda\;s^{\prime}_{+}(\lambda)\,e^{i(E_{\bf p}\Delta)\lambda},$ (70)

with $s^{\prime}_{\pm}=\partial_{\lambda}s_{\pm}$. The condition (67) then
implies $|s^{\prime}_{\pm}|\leq M$ in the large $T$ limit. It is possible,
then, to prove that both the $s_{-}^{\prime}$ and the $s^{\prime}_{-}$
integrals vanish in the large-$T$ (hence large $\Delta$) limit131313For
instance using the fact that any Fourier transform $\tilde{f}(\xi)=\int
dx\,e^{ix\xi}f(x)$ can be written as $\tilde{f}(\xi)=\int
dx\,e^{ix\xi}[f(x)-f(x+\pi/\xi)]/2$, and use the absolute convergence of the
integral, the continuity and bound of $s_{\pm}^{\prime}$ at any $T>\bar{T}$ to
prove that it can always be found a large enough $T$ such that the integral is
arbitrary small, regardless of the specific function $s_{\pm}$ at said time.
In other words, it vanishes in the large $T$ limit.. The simplest and most
intuitive case would be $s_{-}(\lambda)=s_{+}(-\lambda)$ independent of $T$ in
the first place, them the highly oscillatory integrals vanish as a consequence
of the Riemann-Lebesgue lemma. It can be seen also that, removing the bound on
gradient $s_{\pm}^{\prime}$, one could have a family of functions converging
to a Dirac delta141414Any delta family $\delta_{\varepsilon}(\lambda)$ can be
used as a start. It bust be multiplied by a smooth drop to the border, like
$e^{-4\lambda^{2}/(1-4\lambda^{2})}$ and properly normalized to one for all
$\varepsilon$. As long as $\varepsilon\Delta\xrightarrow{T\to\infty}0$ the
integrals in $s^{\prime}_{\pm}$ converge to just $1$. $\delta(\lambda)$,
substantially recovering the results of the sudden limit. Of course other
outcomes are possible for adiabatic functions with unbounded derivatives in
the $T\to\infty$ limit.

It is possible then to force the system to have a trivial scattering
matrix151515Besides a diverging phase that can be removed with a counterterm
in the Lagrangian as discussed in [2]. even if the full interaction is on for
a long time. However it is not the case for any function (pointwise)
converging $s(t)\to 1$. The smoothness of the adiabatic function can be
relaxed to just continuous and smooth except a finite (constant) number of
points, for instance $\\{-\Delta-T/2,-T/2,T/2,\Delta+T/2\\}$. One has just to
be careful with the integration by parts and add “when defined” in every
sentence containing the derivatives of $s$. On the other hand all the
conditions in (61), (62) and (67) are crucial. Since the long permanence of
the interaction doesn’t grant a unique limit, but the behavior on the distant
past and future drastically change the final outcome, one should realize that
none of them in particular is supposed to reproduce the full theory. In fact
none of them are.

It is possible to rework the arguments presented in this section161616The
calculations in Ref. [2] regarding “Model I” and “Model II” are essentially a
way to resum the Dyson series. In no instance there is a switch to the
momentum space, and all the calculations can be repeated without much effort
integrating time in a compact domain, rather than the full real axis. In this
way one obtains the $U_{I}(t,-T/2-\Delta)$. The formulas are the same as the
one presented at the beginning of the section, except that the time integral
has an upper bound in $t$, not at the end of the support of the adiabatic
function $s$. to prove the analogue of the Adiabatic theorem

$U(t,-\tfrac{T}{2}-\Delta)|{\bf p}_{1},\cdots,{\bf p}_{n}\rangle\sim{\rm
exp}\left\\{-i\left[E_{0}+\sum_{j=1}^{n}E_{{\bf
p}_{j}}\right]\left(t+\frac{T}{2}+\Delta\right)\right\\}|{\bf
p}_{1},\cdots,{\bf p}_{n}\rangle_{\Omega}$ (71)

The states $|{\bf p}_{1},\cdots,{\bf p}_{n}\rangle_{\Omega}$ build energy
eigenstates basis used in the appendix B. The $\sim$ instead of an equality
represents two facts. The first one is that it is valid in the large $T$ limit
and only if all the requirements (61), (62) and (67) are fulfilled. The second
is that there is an additional phase making the effective time $\Delta t_{\rm
eff}\neq\Delta t=t+T/2+\Delta$ for the $E_{0}$ coefficient different than the
one for the $E_{{\bf p}_{j}}$’s; however, in the large $T$ limit $\Delta
t_{\rm eff}/(t+T/2+\Delta)\to 1$. Namely, the phase is asymptotic to the
correct phase.

On the other hand in the sudden case
$s(t)=\theta(T/2+\Delta+t)\theta(T/2+\Delta-t)$ one has, calling $U_{S}$ the
time evolution here to distinguish it from the previous case

$\begin{split}&\langle{\bf q}_{1},\cdots,{\bf
q}_{m}|U_{S}(t,-\tfrac{T}{2}-\Delta)|{\bf p}_{1},\cdots,{\bf p}_{n}\rangle\\\
&\quad={\rm exp}\left\\{-i\left(\sum_{j=1}^{n}E_{{\bf p}_{j}}\right)\Delta
t\right\\}{}_{\Delta t}\langle{\bf q}_{1},\cdots,{\bf q}_{m}|{\bf
p}_{1},\cdots,{\bf p}_{n}\rangle,\qquad
t\in[-\tfrac{T}{2}-\Delta,\tfrac{T}{2}+\Delta]\end{split}$ (72)

as it is rather obvious since the system is the exact one for
$t\in(-\tfrac{T}{2}-\Delta,\tfrac{T}{2}+\Delta)$, hence time translation
invariant. The results of the previous section must be recovered in this case
as long as $t$ remains in its bound. needless to say, for initial conditions
before $-T/2-\Delta$ or final $t>T/2+\Delta$ the phases keep growing but the
$\Delta t$ in the $t$ dependent basis is stuck and the results become
different from the exact ones. The difference between the two cases stems from
the fact that

${}_{\Delta t}\langle{\bf q}_{1},\cdots,{\bf q}_{m}|{\bf p}_{1},\cdots,{\bf
p}_{n}\rangle\neq e^{-iE_{0}\Delta t}\langle{\bf q}_{1},\cdots,{\bf
q}_{m}|{\bf p}_{1},\cdots,{\bf p}_{n}\rangle_{\Omega}\text{ ,}$ (73)

something which is rather immediate to see if one works out these bases
solving exactly the system, but it is rather obscure if one looks only at the
Dyson series for the $U_{I}$: for free fields (including fields in the
interaction picture) there is no difference between the basis $|{\bf
p}_{1},\cdots,{\bf p}_{n}\rangle$ at different times and the basis of the
energy excitations. They are all the same basis. This is related to the fact
that the regular basis of in and out states (usually assumed to exist, rather
than proved, in practical applications) simply does not exist for a time
independent classical source171717In general, it does not exist also for time-
dependent but not Fourier transformable in time sources. . Mostly due to the
fact that the state corresponding to the minimum of the Hamiltonian
$\Omega\rangle$ is not a momentum eigenstate. More information about this in
the Appendix E. This difference between bases is crucial from the physical
point of view. In the full theory, the initial and final states can be written
in any basis, possibly different ones. The matrix elements of $U$ with respect
to the regular $|{\bf p}_{1},\cdots,{\bf p}_{n}\rangle$ basis are enough to
write the matrix elements in all bases, one has to just use the ordinary
change of basis rules. Conceptually it is fine, at worst it can be complicated
numerically for inconvenient bases. If one artificially switches off the
interaction in the far past and future it is not possible to distinguish, eg,
if the initial or final conditions were given in the energy basis or in the
particle momentum one, as they are the same. One has to “chose a different
method to insert the interaction for a long time” to recover the exact
results, if any. For instance if one has the initial and final condition in
the energy basis one should use the proper adiabatic case. Because of (71)
this approach smoothly connects the free states to the energy eigenstates,
then back to the corresponding free states in the far future. Seen from this
perspective it is obvious that the scattering is trivial. By definition of the
energy basis the only evolution of its eigenstates is a (time and energy
dependent) phase. The non trivial part is that the spectrum of the Hamiltonian
(besides an off-set) is the same as the free one, and with no additional
degeneracy. Something that can be proved only with the ful exact solutions, as
done in this paper, and cannot be obtained by looking at the Dyson series. If
the initial states are given in terms of momentum (and particle number)
eigenstates, the proper switch is the sudden one. If the former sends “free
states to energy eigenstates” (which have an undefined particle number) the
latter keeps the momentum and particle eigenstates (which are not energy
eigenstates). It is not the purpose of this paper to consider other bases. At
the moment it is also possible that for some, more complicated, bases there is
not a “corresponding way to switch off the interaction at asymptotically large
times”. In any case, there is not an universal “correct way” to keep the
interaction for a finite, but arbitrarily large time.

##  VI Conclusion

An external classical source causes the particle vacuum to be unstable, and
its decay into $n$ particles is given by a Poisson distribution, which is
well-defined whenever the source has a Fourier transform. The time dependence
of the vacuum survival probability oscillates and exhibits a quadratic
behavior at short times that allows for the quantum Zeno effect. The
oscillation of the survival probability dies out at large times, and it tends
to a nonvanishing constant, at odds with standard particle decay. In the large
volume limit the particle density saturates, and the particle production never
displays a constant rate, in contrast with the Schwinger mechanism. In
addition, the precise shape of the source, and not just its effective volume,
is relevant for the particle production.
We find that the scattering is non trivial also for a time-independent source,
the vacuum decay playing the most important role because the full dynamics can
be reduced to an absorption probability amplitude of the initial particles and
an interference pattern between the remaining particles and the ones produced
by the vacuum. The result of trivial scattering was previously obtained in the
framework of adiabatic switching off the interaction and under the assumption
that in the limit of long persistence of the interacting Hamiltonian one can
recover the full results of the unmodified theory. This claim does not hold
after a closer inspection because removing the interaction changes the algebra
of the physical states. In particular, the full treatment reveals the
existence of a particle vacuum and an energy vacuum. As a consequence particle
number eigenstates are never energy eigenstates and vice-versa. This exact
aspect of the theory cannot be recovered in the long-time limit if one
switches off the interaction, no mater how it is performed.
In high energy experiments, a possible use of this specific model would be to
simulate (a part of) the dilaton production out of the medium approximated as
a classical source. The effect due to the vacuum decay is fully
nonperturbative and could give a relevant contribution to dilaton production,
since perturbative production mechanisms are suppressed. The produced dilatons
(i.e. predominantly scalar glueballs) then decay and produce mesons that can
be measured, see for instance Ref. [48] where a similar process is discussed.
Any generalization of the main results of this paper to higher spin and/or
charged particles would be interesting and eventually prove useful in modeling
emission particles other than the dilaton, but it remains a question whether
the same techniques can be used.

In low energy experiments, the Schwinger mechanism and Zeno effect are more
easily accessible, e.g. [49] for the Schwinger mechanism and [39, 40] for the
Zeno effect. Therefore (appropriate extension of) the model might work as a
description of those phenomena at low energies.

## Acknowledgement

F. G. and A. V. acknowledge financial support from from the Polish National
Science Centre (NCN) via the OPUS project No. 2019/33/B/ST2/00613. L. T. and
A.V. have been supported by the Polish National Science Centre Grant SONATA
project No. 2020/39/D/ST2/02054.

##  Appendix A Exact solutions of the field equations and the generators of
space-time translations

In this appendix we show the explicit calculations for a generic classical
source $g\rho(x)$, regardless of the time dependence (or independence) of
$\rho$. In particular the exact solution of the field equation, and we perform
the calculations for the total four-momentum operators, the Hamiltonian
$\hat{H}$ and $\hat{P}$ in the Heisenberg picture.
The starting Lagrangian reads

${\cal
L}=\frac{1}{2}\partial_{\mu}\hat{\phi}\;\partial^{\mu}\hat{\phi}-\frac{1}{2}m^{2}\hat{\phi}^{2}+g\rho\;\hat{\phi},$
(74)

whose Euler-Lagrange equation of motion then reads

$\Box\hat{\phi}+m^{2}\hat{\phi}=g\,\rho\text{ .}$ (75)

By hypothesis, the classical source $\rho(x)$ can be Fourier transformed with
respect to the space variables, but we do not assume anything about the time
dependence, beside a generic smoothness (continuous, differentiable and lack
of singularities) of the real function $\rho$. In particular, we are
interested in the case of a time-independent $\rho(x)=\rho({\bf x})$. An
integral from $y^{0}=-\infty$ to a finite $t$ of $e^{\pm ip(x-y)}\rho(y)$ is
not defined in general. It is surely not defined for a time-independent
source. The same can be said for integrals from a finite $y^{0}=t$ to
$+\infty$. Therefore, neither the advanced, nor the retarded green functions
can be used to obtain a solution in term of a solution $\hat{\phi}_{0}$ of the
homogeneous equation

$\Box\hat{\phi}_{0}+m^{2}\hat{\phi}_{0}=0.$ (76)

It is possible, however, to split the time domain in two components. Without
loss of generality, given a finite $t_{0}\in{\mathbb{R}}$

$(-\infty,\infty)=(-\infty,t_{0}]\cup[t_{0},\infty).$ (77)

One can safely use the advanced Green’s function in the $y^{0}<t_{0}$ region,
and the retarded ones in the $y^{0}>t_{0}$ region. It is immediate then to
obtain the exact solutions in the the intervals $(-\infty,t_{0})$ and
$(t_{0},\infty)$. A special solution can be written exactly in the same way
(for a $t$ larger or smaller than $t_{0}$)

$ig\int_{t_{0}}^{t}dy^{0}\int
d^{3}y\;\rho(y)\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf
p}}\left(\vphantom{\frac{}{}}e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}\right)\text{
,}$ (78)

which is manifestly continuous and differentiable in $t_{0}$. Therefore, the
general solution reads

$\hat{\phi}(x)=\hat{\phi}_{0}(x)+ig\int_{t_{0}}^{t}dy^{0}\int
d^{3}y\;\rho(y)\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf
p}}\left(\vphantom{\frac{}{}}e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}\right).$ (79)

Alternatively, one can just check directly that (79) is a solution of (75).
Clearly the “initial” condition is that the free field $\hat{\phi}_{0}$ at
$t=t_{0}$ is the full interacting field $\hat{\phi}(t_{0},{\bf
x})=\hat{\phi}_{0}(t_{0},{\bf x})$. This implies that, at least for the time
$t_{0}$, the physical field must coincide with some free field, regardless of
the details of the classical source. In other words both a free field and an
interacting one must have “the same initial conditions at $t=t_{0}$”.

In general, it is not granted that two fields following different equation of
motions can coincide over an extended region. Namely, the Klein-Gordon
equation with different sources, and the whole space at fixed time, as it is
needed in Eq. (79).

The key to prove that (79) is an acceptable solution is the canonical
quantization rules. Indeed, they are enough to ensure that any interacting
field (regardless of the equation of motion, which may be much more
complicated than (75)) coincide with some free field at fixed time in an
inertial reference frame. In particular, the quantization relations ensure
that, for any fixed time $\bar{t}$ it can be built canonically a field
$\bar{\phi}$ which coincides with the starting field at $t=\bar{t}$ and
fulfills the free Klein-Gordon equation. In particular at the starting time
$t_{0}$ the field must be coinciding with some free field and $\phi_{0}$ can
be taken as an initial condition without loss of generality.

This result, beside justifying the use of the solution (79) in this work, is
essential to use the interaction picture in the general case. More
importantly, it shows that there are always some (time-dependent) ladder
operators that one can use to span the particles’ states for the interacting
field, not only for the Lagrangian density (74), but definitely also for the
system studied in the present work. This lesser known time dependent basis is
the main tool to solve exactly for the transition amplitudes in Section III.
Therefore, as a technical reminder, we show a brief proof for the construction
of such a free field and we highlight some of the consequences of this
formulation.

In the case of a Hermitian scalar field $\hat{\phi}^{\dagger}=\hat{\phi}$, one
can call ${\tilde{\alpha}}(x^{0};p)$ and ${\tilde{\alpha}}^{\dagger}(x^{0};p)$
the volume integrals

${\tilde{\alpha}}(x^{0};p)=i\int d^{3}x\;e^{ip\cdot
x}\,\left[\vphantom{\frac{}{}}\hat{\Pi}(x)-iE\,\hat{\phi}(x)\right],$ (80)

and its hermitian conjugate

${\tilde{\alpha}}^{\dagger}(x^{0};p)=-i\int d^{3}x\;e^{-ip\cdot
x}\left[\vphantom{\frac{}{}}\hat{\Pi}(x)+iE\,\hat{\phi}(x)\right].$ (81)

The (Hermitian) field $\hat{\Pi}$ is the canonical conjugate of the
$\hat{\phi}$

$\hat{\Pi}(t,{\bf x})=\frac{\partial{\cal
L}}{\partial(\partial_{t}\hat{\phi})},$ (82)

which is, in turn, equal to $\hat{\Pi}=\partial_{t}\hat{\phi}$ in the absence
of a derivative interaction. The four momentum $p=(E,{\bf p})$ at this point
is generic and does not have to be on-shell. Making use of the equal-time
commutators

$[\hat{\phi}(t,{\bf x}),\hat{\phi}(t,{\bf y})]=[\hat{\Pi}(t,{\bf
x}),\hat{\Pi}(t,{\bf y})]=0\text{ ,}\qquad[\hat{\phi}(t,{\bf
x}),\hat{\Pi}(t,{\bf y})]=i\delta^{3}({\bf x}-{\bf y})\text{ ,}$ (83)

it is straightforward to check that

$\begin{split}&[\tilde{\alpha}(x^{0};p_{1}),\tilde{\alpha}(x^{0};p_{2})]=[\tilde{\alpha}^{\dagger}(x^{0};p_{1}),\tilde{\alpha}^{\dagger}(x^{0};p_{2})]=0,\\\
&[\tilde{\alpha}(x^{0};p_{1}),\tilde{\alpha}^{\dagger}(x^{0};p_{2})]=e^{i(E_{1}-E_{2})x^{0}}\left(\vphantom{\frac{}{}}E_{1}+E_{2}\right)(2\pi)^{3}\delta^{3}({\bf
p}_{1}-{\bf p}_{2}).\end{split}$ (84)

It is immediate to note that for on-shell energies $E_{1}=E_{{\bf
p}_{1}}=\sqrt{m^{2}+{\bf p}_{1}\cdot{\bf p}_{1}}$, $E_{2}=E_{{\bf
p}_{2}}=\sqrt{m^{2}+{\bf p}_{2}\cdot{\bf p}_{2}}$ this is a generalization of
the commutation relations of the ladder operators. More precisely, for an
uncharged free field $\hat{\phi}_{f}$ with mass $m$ and normalization

$\hat{\phi}_{f}(x)=\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\left[\vphantom{\frac{}{}}a_{f}({\bf p})e^{-ip\cdot
x}+a^{\dagger}_{f}({\bf p})e^{ip\cdot x}\right],$ (85)

one has

$[a_{f}({\bf p}),a_{f}({\bf q})]=[a^{\dagger}_{f}({\bf
p}),a^{\dagger}_{f}({\bf q})]=0,\qquad[a_{f}({\bf p}),a^{\dagger}_{f}({\bf
q})]=(2\pi)^{3}\delta^{3}({\bf p}-{\bf q})\text{ ,}$ (86)

therefore, the on-shell $\alpha(x^{0};m,{\bf{p}})=\tilde{\alpha}(x^{0};E_{\bf
p},{\bf p})$ is a time dependent generalization of the free $\sqrt{2E_{\bf
p}}a_{f}({\bf p})$ to the interacting case. In other words

$a_{\bf p}(x^{0},m)=\frac{\alpha(x^{0};m,{\bf p})}{\sqrt{2E_{\bf
p}}}=\frac{\tilde{\alpha}(x^{0};E_{\bf p},{\bf p})}{\sqrt{2E_{\bf p}}},\qquad
a^{\dagger}_{\bf
p}(x^{0},m)=\frac{\alpha^{\dagger}(x^{0};m,{\bf{p}})}{\sqrt{2E_{\bf
p}}}=\frac{\tilde{\alpha}^{\dagger}(x^{0};E_{\bf p},{\bf p})}{\sqrt{2E_{\bf
p}}},$ (87)

are the generalization, respectively, of the destruction and creation
operators. They have the exactly the same commutation relations as the free
ones (86). Therefore, if one has an uncharged interacting field $\phi$ and its
canonical conjugate $\Pi$ at the same time, it is possible to write both of
them in terms of the $a_{\bf p}(x^{0},m)$ and $a_{\bf p}^{\dagger}(x^{0},m)$
operators

$\begin{split}\hat{\phi}(x^{0},{\bf
x})&=\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\left[\vphantom{\frac{1}{2}}a_{\bf p}(x^{0},m)\;e^{-iE_{\bf
p}x^{0}}e^{i{\bf p}\cdot{\bf x}}+a^{\dagger}_{\bf p}(x^{0},m)\;e^{iE_{\bf
p}x^{0}}e^{-i{\bf p}\cdot{\bf x}}\right],\\\ \hat{\Pi}(x^{0},{\bf
x})&=-i\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf p}}}\,E_{\bf
p}\left[\vphantom{\frac{1}{2}}a_{\bf p}(x^{0},m)\;e^{-iE_{\bf p}x^{0}}e^{i{\bf
p}\cdot{\bf x}}-a^{\dagger}_{\bf p}(x^{0},m)\;e^{iE_{\bf p}x^{0}}e^{-i{\bf
p}\cdot{\bf x}}\right],\end{split}$ (88)

which can be directly verified by simply plugging the definitions (87) and
then (80) and (81) in the integrals. The last equations clearly work
exclusively for the same time $x^{0}$ in the exponential and in the time-
dependent ladder operators.

Finally, if one builds the (normal ordered181818The normal ordering is defined
in the same way as for the free fields. This definition holds in the
interacting case thanks to (7).) Klein-Gordon Hamiltonian with the fields at
time $x_{0}$ and the same mass as the one in the definition of $a_{\bf
p}(x^{0},m)$

$\hat{H}_{0}=\frac{1}{2}:\\!\\!\\!\int d^{3}x\left[\hat{\Pi}^{2}(x^{0},{\bf
x})+\nabla_{\bf x}\hat{\phi}(x^{0},{\bf x})\cdot\nabla_{\bf
x}\hat{\phi}(x^{0},{\bf x})+m^{2}\hat{\phi}(x^{0},{\bf
x})\hat{\phi}(x^{0},{\bf x})\right]\\!\\!:\text{ ,}$ (89)

it is possible to prove, making use of the commutation relations (83) and the
definitions (88), (87), (80) and (81), that the commutators of the generalized
ladder operators with $\hat{H}_{0}$ are the same as the commutators of a free
field

$[\,\hat{H}_{0}\;,\;a_{\bf p}(x^{0},m)\,]=-E_{\bf p}\;a_{\bf p}(x^{0},m)\text{
,}\qquad[\,\hat{H}_{0}\;,\;a^{\dagger}_{\bf p}(x^{0},m)\,]=E_{\bf
p}\;a^{\dagger}_{\bf p}(x^{0},m)\text{ .}$ (90)

Therefore, using the free propagator ${\rm exp}\\{-i\hat{H}_{0}\Delta
x^{0}\\}$, it is possible to write a free field $\bar{\phi}$ that follows the
free Klein-Gordon equation with mass $m$ and coincides with the physical field
at a specific time $\bar{t}$, $\bar{\phi}_{0}({\bar{t}},{\bf
x})=\bar{\phi}({\bar{t}},{\bf x})$ :

$\begin{split}{\bar{\phi}}(t,{\bf
x})&=e^{i\hat{H}_{0}(t-{\bar{t}})}\;\hat{\phi}({\bar{t}},{\bf
x})\;e^{-i\hat{H}_{0}(t-{\bar{t}})}\\\
&=\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\left[\vphantom{\frac{1}{2}}a_{\bf p}({\bar{t}},m)\;e^{-iE_{\bf
p}t}e^{i{\bf p}\cdot{\bf x}}+a^{\dagger}_{\bf p}({\bar{t}},m)\;e^{iE_{\bf
p}t}e^{-i{\bf p}\cdot{\bf x}}\right],\end{split}$ (91)

as it was stated before. The same result holds for the analogous $\bar{\Pi}$
that coincides with $\hat{\Pi}$ at ${\bar{t}}$ and evolves according to the
free Klein-Gordon equation.

These arguments are rather general, so they work, in particular, for the
simple Lagrangian density in (74). Without loss of generality, for any $t_{0}$
one can build a free field that coincides with the full interacting field at
$t_{0}$. The solution in (79) is thus general.

It is possible now to plug (91), with $t=t_{0}$, in (79). Writing all in therm
of the same $d^{3}p$ integral and factoring the ${\rm exp}\\{\pm ip\cdot x\\}$
phases from both of the terms

$\hat{\phi}(x)=\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\left[\vphantom{\frac{1}{2}}\left(\vphantom{\frac{}{}}a_{\bf
p}(t_{0},m)+f(t,t_{0};E_{\bf p},{\bf p})\right)e^{-ip\cdot
x}+\left(\vphantom{\frac{}{}}a^{\dagger}_{\bf p}(t_{0},m)+f^{*}(t,t_{0};E_{\bf
p},{\bf p})\right)e^{ip\cdot x}\right],$ (92)

the function $f$ being the integral of the classical source

$f(t,t_{0};p)=i\,g\int_{t_{0}}^{t}dy^{0}\int
d^{3}y\;\frac{\rho(y)}{\sqrt{2E_{\bf p}}}\;e^{ip\cdot y},$ (93)

and $f^{*}$ its complex conjugate. In almost all the cases in the manuscript,
including (92), only the on-shell integral is needed. For the most part of the
paper, also, only a time-independent $\rho$ is considered and the $y^{0}$
integral can be performed exactly. For ease of reading in the main text the
the on-shellness is understood and the $t_{0}$ dependence (taken as $t_{0}=0$)
is omitted. In a similar manner the formal notation $a_{\bf p}(t_{0}=0,m)$ is
left out in favor of the simpler $a_{\bf p}$. The exact field
$\hat{\Pi}=\partial_{t}\hat{\phi}$ can be computed directly from from (92).

$\begin{split}\hat{\Pi}(x)=\partial_{t}\hat{\phi}(x)=-i\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}E_{\bf p}&\left[\left(\vphantom{\frac{}{}}a_{\bf
p}(t_{0},m)+f(t,t_{0};E_{\bf p},{\bf p})+\frac{i}{E_{\bf
p}}\partial_{t}f(t,t_{0};E_{\bf p},{\bf p})\right)e^{-ip\cdot x}\right.\\\
&\left.-\left(\vphantom{\frac{}{}}a^{\dagger}_{\bf
p}(t_{0},m)+f^{*}(t,t_{0};E_{\bf p},{\bf p})-\frac{i}{E_{\bf
p}}\partial_{t}f^{*}(t,t_{0};E_{\bf p},{\bf p})\right)e^{ip\cdot
x}\right].\end{split}$ (94)

Making use of the definition (93) it is immediate to show that, for any on-
shell and positive frequency four-momentum $p^{\mu}$

$\begin{split}\partial_{t}f^{*}(t,t_{0};E_{\bf p},{\bf p})\;e^{i(E_{\bf
p})t}\,e^{-i{\bf p}\cdot{\bf x}}&=-i\,g\int d^{3}y\;\frac{\rho(t,{\bf
y})}{\sqrt{2E_{\bf p}}}\;e^{-i{\bf p}\cdot({\bf x}-{\bf y})}=-\left(i\,g\int
d^{3}y\;\frac{\rho(t,{\bf y})}{\sqrt{2E_{\bf p}}}\;e^{i(-{\bf p})\cdot({\bf
x}-{\bf y})}\right)\\\
&=-\partial_{t}f(t,t_{0};E_{\bf-p},{\bf-p})\;e^{-i(E_{-{\bf p}})t}\,e^{i(-{\bf
p})\cdot{x}}.\end{split}$ (95)

Within the minimal integrability hypothesis for the classical source (and its
partial Fourier transform), the two terms in the time derivatives of $f$ in
(94) converge separately, and they have an opposite sign, as it is apparent
performing a ${\bf p}\to-{\bf p}$ in one of them and not the other. Their
contribution to the total field simplifies

$\begin{split}\hat{\Pi}(x)=-i\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}E_{\bf p}&\left[\left(\vphantom{\frac{}{}}a_{\bf
p}(t_{0},m)+f(t,t_{0};E_{\bf p},{\bf p})\right)e^{-ip\cdot
x}-\left(\vphantom{\frac{}{}}a^{\dagger}_{\bf p}(t_{0},m)+f^{*}(t,t_{0};E_{\bf
p},{\bf p})\right)e^{ip\cdot x}\right].\end{split}$ (96)

Confronting the results (92) and (96) with the general formulas (88), it is
immediate to recognize

$a_{\bf p}(t_{0},m)+f(t,t_{0};E_{\bf p},{\bf p})=a_{\bf p}(t,m),\qquad
a^{\dagger}_{\bf p}(t_{0},m)+f^{*}(t,t_{0};E_{\bf p},{\bf p})=a^{\dagger}_{\bf
p}(t,m),$ (97)

as the on-shell ladder operators for the generic time $t$. Hence, the number
operator in the Heisenberg picture reads, in term of the ladder operators at
$t_{0}$, and the integrals of the classical source

$\hat{N}(t)=\int\frac{d^{3}p}{(2\pi)^{3}}\;a^{\dagger}_{\bf p}(t,m)a_{\bf
p}(t,m)=\int\frac{d^{3}p}{(2\pi)^{3}}\left(\vphantom{\frac{}{}}a^{\dagger}_{\bf
p}(t_{0},m)+f^{*}(t,t_{0};E_{\bf p},{\bf
p})\right)\left(\vphantom{\frac{}{}}a_{\bf p}(t_{0},m)+f(t,t_{0};E_{\bf
p},{\bf p})\right).$ (98)

It is possible then to compute now the total four-momentum operators. The
system is, in general, not space-time translation invariant, but $\hat{H}(t)$
and $\hat{\bf P}(t)$ are still the generators of space-time translations in
the lab frame at time $t$. They are the regular volume integral of the
(contribution of the quantum field to the total) stress-energy tensor
$\hat{T}^{\mu\nu}$, for a non-derivative interaction term like (74) this has
the standard form

$\hat{T}^{\mu\nu}(x)=\frac{1}{2}\partial^{\mu}\hat{\phi}(x)\partial^{\nu}\hat{\phi}(x)+\frac{1}{2}\partial^{\nu}\hat{\phi}(x)\partial^{\mu}\hat{\phi}(x)-g^{\mu\nu}{\cal
L},$ (99)

and thus

$\begin{split}\hat{H}(t)&=\int d^{3}x\;\hat{T}^{00}=\frac{1}{2}\int
d^{3}x\left[\hat{\Pi}(x)^{2}+\nabla_{\bf x}\hat{\phi}\cdot\nabla_{\bf
x}\hat{\phi}(x)+m^{2}\hat{\phi}(x)^{2}-2\,g\,\rho(x)\,\hat{\phi}(x)\right]\\\
\hat{P}^{i}(t)&=\int d^{3}x\;\hat{T}^{0i}=\frac{1}{2}\int
d^{3}x\left[\hat{\Pi}(x)\;\partial^{i}\hat{\phi}(x)\,+\,\partial^{i}\hat{\phi}(x)\;\hat{\Pi}(x)\right].\end{split}$
(100)

The derivation is usually mentioned in textbook [50, 2]. Also, the
computations of the volume integrals follows the steps of the textbook
examples for the free scalar field since the (100) involves only $\hat{\phi}$,
$\hat{\Pi}$ and the spatial gradients $\nabla_{\bf x}\hat{\phi}$ at equal
limes, and they all can be written using the same algebra of the free fields
(88), including the normal ordering. The notable difference is the additional
term due to the interaction with the classical source. The time-dependent
momentum operator in the Heisenberg picture then reads

$:\\!\\!\hat{\bf P}(t)\\!\\!:=\int\frac{d^{3}p}{(2\pi)^{3}}\;{\bf
p}\;\left(\vphantom{\frac{}{}}a^{\dagger}_{\bf
p}(t_{0},m)+f^{*}(t,t_{0};E_{\bf p},{\bf
p})\right)\left(\vphantom{\frac{}{}}a_{\bf p}(t_{0},m)+f(t,t_{0};E_{\bf
p},{\bf p})\right),$ (101)

it is immediate to verify then, since the commutators of the time-dependent
ladder operator are the same as the free ones, that this is a valid generator
of the space-translations

$[:\\!\\!\hat{P}^{j}(t)\\!\\!:,\hat{\phi}(t,{\bf
x})]=-\int\frac{d^{3}p}{(2\pi)^{3}\sqrt{2E_{\bf
p}}}\;p^{j}\,\left[\vphantom{\frac{}{}}a_{\bf p}(t,m)\,e^{-ip\cdot
x}-a^{\dagger}_{\bf p}(t,m)\,e^{ip\cdot
x}\right]=i\partial^{j}\hat{\phi}(t,{\bf x})=-i\partial_{j}\hat{\phi}(t,{\bf
x}).$ (102)

In other words, the fields at time $t$ translate in space with the unitary
operator ${\rm exp}\\{i:\hat{\bf P}(t):\;\cdot\;\Delta{\bf x}\\}$. Since the
interaction in the Lagrangian density (74) is so simple, the additional term
in the Hamiltonian can be easily computed in terms of the on-shell $f$
function (93). The normal ordering is also trivial

$\begin{split}:\\!\\!\hat{H}(t)\\!\\!:&=\int\frac{d^{3}p}{(2\pi)^{3}}\;E_{\bf
p}\;a^{\dagger}_{\bf p}(t,m)\,a_{\bf p}(t,m)\\\
&\qquad+\int\frac{d^{3}p}{(2\pi)^{3}}\left[\vphantom{\frac{}{}}-i\partial_{t}f^{*}(t,t_{0};E_{\bf
p},{\bf p})\;a_{\bf p}(t,m)+i\partial_{t}f(t,t_{0};E_{\bf p},{\bf
p})\;a^{\dagger}_{\bf p}(t,m)\right],\end{split}$ (103)

which is convenient to reorganize in a free-like operatorial part, and a time-
dependent offset (function, not an operator)

$\begin{split}:\\!\\!\hat{H}(t)\\!\\!:&=\int\frac{d^{3}p}{(2\pi)^{3}}\;E_{\bf
p}\,\left(a^{\dagger}_{\bf p}(t,m)-\frac{i}{E_{\bf
p}}\partial_{t}f^{*}(t,t_{0};E_{\bf p},{\bf p})\right)\left(a_{\bf
p}(t,m)+\frac{i}{E_{\bf p}}\partial_{t}f(t,t_{0};E_{\bf p},{\bf p})\right)\\\
&\qquad-\int\frac{d^{3}p}{(2\pi)^{3}}\;E_{\bf
p}\,\left|\frac{\partial_{t}f(t,t_{0};E_{\bf p},{\bf p})}{E_{\bf
p}}\right|^{2}.\end{split}$ (104)

The commutation relations of the $a_{\bf p}(t,m)$ and $a^{\dagger}_{\bf
p}(t,m)$ are, by construction, the same as the free ones (86), however the
shift in the operators by a function in (104) produces a change in the
commutation relations with the ladder operators. The Hamiltonian is no longer
an integral of $Ea^{\dagger}a$ at the operator level, but of
$Ea^{\dagger}a-if^{*}a+ifa^{\dagger}$ plus functions (that commute with
everything). Then

$\begin{split}[:\\!\\!\hat{H}(t)\\!\\!:,a_{\bf p}(t,m)]&=-E_{\bf p}a_{\bf
p}(t,m)-i\partial_{t}f(t,t_{0};E_{\bf p},{\bf p}),\\\
[:\\!\\!\hat{H}(t)\\!\\!:,a^{\dagger}_{\bf p}(t,m)]&=E_{\bf p}a^{\dagger}_{\bf
p}(t,m)-i\partial_{t}f^{*}(t,t_{0};E_{\bf p},{\bf p}).\end{split}$ (105)

If one takes the commutator of $i[:\\!\\!H(t)\\!\\!:,\hat{\phi}(t,{\bf x})]$,
the extra functions reconstruct the the field $\hat{\Pi}(t,{\bf x})$ as in
(94), which can be simplified thanks to (95) in (96) as shown before. The
extra functions do not simplify if one takes the commutator with
$\hat{\Pi}(t,{\bf x})$, but they are needed to obtain the correct time
derivative $\partial_{t}\hat{\Pi}(t,{\bf x})$. The higher order derivatives
need the second (and higher) time derivatives of $f$. They are given not only
by the nested commutators with $:\\!\\!H(t)\\!\\!:$, but they involve also
commutators with the time derivatives of $:\\!\\!H(t)\\!\\!:$. Rather than
making some specific examples it is easier to sum all up and and remind that
the fields evolve in time with the unitary propagator $U(t^{\prime},t)$ that
solves the time-dependent Shrödinger equation w.r.t. the variable
$t^{\prime}$:

$\begin{split}i\partial_{t^{\prime}}\;U(t^{\prime},t)&=\;:\\!\\!\hat{H}_{S}(t^{\prime})\\!\\!:\;U(t^{\prime},t)=U(t^{\prime},t)U^{-1}(t^{\prime},t)\;:\\!\\!\hat{H}_{S}(t^{\prime})\\!\\!:\;U(t^{\prime},t)\\\
&=U(t^{\prime},t){U}^{\dagger}(t^{\prime},t)\;:\\!\\!\hat{H}_{S}(t^{\prime})\\!\\!:\;U(t^{\prime},t)={U}(t^{\prime},t)\;:\\!\\!\hat{H}(t^{\prime})\\!\\!:\;\text{
.}\end{split}$ (106)

Some comments about Eq. (106) are in order. Since it is standard textbook
material, we don’t think it is necessary to give too many explanations. It is
important to note though, since it is a common source of confusion, that the
Hamiltonian is on the left if one uses the Shrödinger time-dependent operator
and, by construction, on the right if one uses (as it is common in quantum
field theories) the Hamiltonian in the Heisenberg picture. The propagator is
the same in both pictures, however their Dyson series needs the regular time
ordering if the equation is solved for the $\hat{H}_{S}(t)$ and anti-time-
ordering if solved for the Heisenberg one $\hat{H}(t)$. In the case of time
independent operators the time ordering is irrelevant $\hat{H}_{S}=\hat{H}$,
the propagator is a regular exponential and it commutes with the Hamiltonian,
it doesn’t matter if it is on the right or on the left. It bears mentioning
that in the interaction picture one starts with the operators in the
Shrödinger picture (including the Hamiltonian) and then evolves them with the
free Hamiltonian to get their interaction picture counterpart.

Since the field can be written in the form

$\hat{\phi}(t^{\prime},{\bf x})=U^{\dagger}(t^{\prime},t)\;\hat{\phi}(t,{\bf
x})\;U(t^{\prime},t),$ (107)

the $n-$th time derivative in $t$ read

$\left(\partial_{t}\right)^{n}\hat{\phi}(t,{\bf x})=\lim_{t^{\prime}\to
t}\left[\left(\partial_{t^{\prime}}\right)^{n}\hat{\phi}(t^{\prime},{\bf
x})\right]=\lim_{t^{\prime}\to
t}\left\\{\left(\partial_{t^{\prime}}\right)^{n}\left[U^{\dagger}(t^{\prime},t)\;\hat{\phi}(t,{\bf
x})\;U(t^{\prime},t)\right]\right\\}.$ (108)

The dependence in $t^{\prime}$ in the right hand side is only on the
propagators. Therefore

$\partial_{t}\hat{\phi}(t,{\bf
x})=i[:\\!\\!\hat{H}(t)\\!\\!:\,,\,\hat{\phi}(t,{\bf x})],$ (109)

the second time derivative reads

$\partial^{2}_{t}\hat{\phi}(t,{\bf
x})=i[\partial_{t}:\\!\\!\hat{H}(t)\\!\\!:\,,\,\hat{\phi}(t,{\bf
x})]-\left[\,:\\!\\!\hat{H}(t)\\!\\!:\,,[:\\!\\!\hat{H}(t)\\!\\!:\,,\,\hat{\phi}(t,{\bf
x})]\,\right]\text{ .}$ (110)

As it can be shown using (93), the first commutator vanishes by construction.
But for the higher order derivatives, nested commutators involving the
Hamiltonian and its partial derivatives, the formula cannot be reduced to the
one valid in the time-independent case that involves only
$:\\!\\!\hat{H}\\!\\!:$. It is possible also to verify order by order that
these formulas are equivalent to just taking the time derivatives of the
explicit solution (92).
The last worthwhile calculation here is the commutators between the time-
dependent Hamiltonian and momentum operators. Making direct use of (105)

$\begin{split}-i\partial_{t}:\\!\\!\hat{\bf
P}(t)\\!\\!:\;=[:\\!\\!\hat{H}(t)\\!\\!:\,,\,:\\!\\!\hat{\bf
P}(t)\\!\\!:]\,=-i\int\frac{d^{3}p}{(2\pi)^{3}}\,{\bf
p}\left[\vphantom{\frac{}{}}\partial_{t}f^{*}(t,t_{0};E_{\bf p},{\bf
p})\,a_{\bf p}(t,m)\;+\;\partial_{t}f(t,t_{0};E_{\bf p},{\bf
p})\,a^{\dagger}_{\bf p}(t,m)\right]\text{ .}\end{split}$ (111)

This is in general different from $0$. For a generic $\rho$ it might be
possible to have commuting operators at some times. However in the time-
independent case $\rho=\rho({\bf x})$, it can never happen. The function

$f(t,t_{0};E_{\bf p},{\bf p})+\frac{i}{E_{\bf p}}\partial_{t}f(t,t_{0};E_{\bf
p},{\bf p})=-\frac{g\;e^{iE_{\bf p}t_{0}}}{\sqrt{2E_{\bf p}^{3}}}\int
d^{3}x\;\rho({\bf x})e^{-i{\bf p}\cdot x},$ (112)

as well as the modulus square of

$\partial_{t}f(t,t_{0};E_{\bf p},{\bf p})=\frac{ig\;e^{iE_{\bf
p}t}}{\sqrt{2E_{\bf p}}}\int d^{3}x\;\rho({\bf x})e^{-i{\bf p}\cdot x},$ (113)

are time-independent, making the Hamiltonian time-independent too. As well as
making the operator in the right-hand side of (111) never trivial. In the rest
of the manuscript, the time ordering is understood, and the $:\cdots:$ is
omitted.

##  Appendix B Phase of the vacuum overlap

In (24) we have found the modulus of the vacuum overlap. It remains to verify
that the phase of the overlap ${}_{t}\langle 0|0\rangle$ is given by (25). In
order to do so, it is useful to introduce a basis of energy eigenstates. The
procedure is substantially the same as the one employed in Section III, for
the time-dependent basis (16). Looking at the generic, time-dependent
Hamiltonian (104); one recognizes the “energy ladder operators” $b_{\bf p}(t)$
and $b^{\dagger}_{\bf p}(t)$

$b_{\bf p}(t)=a_{\bf p}(t)+\frac{i}{E_{\bf p}}\dot{f}(t,{\bf p})=a_{\bf
p}+f(t,{\bf p})+\frac{i}{E_{\bf p}}\dot{f}(t,{\bf p}),$ (114)

and their Hermitian conjugate for the creation operators. In the latter
formula it has been used the compact notation

$\dot{f}(t,{\bf p})=\partial_{t}f(t,0;E_{\bf p},{\bf p}).$ (115)

In particular, for a time-independent source $b_{\bf p}=a_{\bf p}+h({\bf p})$
which is time independent. For the rest of this Appendix we will consider the
general case. The time independent one being just a particular subset.

The complete basis of energy eigenstates can be obtained as it has been done
in the previous case. The energy vacuum reads

$\begin{split}|\Omega(t)\rangle&={\rm
exp}\left\\{-\frac{1}{2}\int\frac{d^{3}p}{(2\pi)^{3}}\left|f(t,{\bf
p})+\frac{i}{E_{\bf p}}\dot{f}(t,{\bf p})\right|^{2}\right\\}{\rm
exp}\left\\{-\int\frac{d^{3}p}{(2\pi)^{3}}\left(f(t,{\bf p})+\frac{i}{E_{\bf
p}}\dot{f}(t,{\bf p})\right)b^{\dagger}_{\bf p}(t)\right\\}|0\rangle\\\
&\Rightarrow b_{\bf
p}(t)\,|\Omega(t)\rangle=0,\quad\hat{H}(t)|\Omega(t)\rangle=E_{0}(t)|\Omega(t)\rangle,\qquad
E_{0}(t)=-\int\frac{d^{3}p}{(2\pi)^{3}}\,E_{\bf p}\,\left|\frac{\dot{f}(t,{\bf
p})}{E_{\bf p}}\right|^{2}<0,\end{split}$ (116)

and therefore

$\begin{split}&a_{\bf p}(t)\,|\Omega(t)\rangle=-\frac{i}{E_{\bf
p}}\dot{f}(t,{\bf p})\,|\Omega(t)\rangle,\qquad a_{\bf
p}\,|\Omega(t)\rangle=-\left(f(t,{\bf p})+\frac{i}{E_{\bf p}}\dot{f}(t,{\bf
p})\right)|\Omega(t)\rangle\\\ &b_{\bf p}(t)\,|0\rangle_{t}=\frac{i}{E_{\bf
p}}\dot{f}(t,{\bf p})\,|0\rangle_{t},\qquad b_{\bf
p}(t)\,|0\rangle=\left(f(t,{\bf p})+\frac{i}{E_{\bf p}}\dot{f}(t,{\bf
p})\right)|0\rangle.\end{split}$ (117)

The excited states are build in the usual way

$\begin{split}|{\bf p}_{1}\,\cdots\,{\bf
p}_{n}\rangle_{\Omega(t)}&=\sqrt{\frac{2E_{{\bf p}_{1}}\cdots 2E_{{\bf
p}_{n}}}{n!}}\;b^{\dagger}_{{\bf p}_{1}}(t)\cdots b^{\dagger}_{{\bf
p}_{n}}(t)\,|\Omega(t)\rangle,\\\ &\Rightarrow\hat{H}(t)\,|{\bf
p}_{1}\,\cdots\,{\bf
p}_{n}\rangle_{\Omega(t)}=\left(E_{0}(t)+\sum_{j=1}^{n}E_{{\bf
p}_{j}}\right)|{\bf p}_{1}\,\cdots\,{\bf
p}_{n}\rangle_{\Omega(t)}\;.\end{split}$ (118)

Inserting the identity in terms of the energy eigenstates at time $t$, it is
possible to find an identity that will be useful later

$\begin{split}{}_{t}\langle 0|0\rangle&={}_{t}\langle
0|\,{\mathds{1}}\,|0\rangle=\sum_{n=0}^{\infty}\left[\prod_{j=1}^{n}\int\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right]{}_{t}\langle 0|{\bf p}_{1}\,\cdots\,{\bf
p}_{n}\rangle_{\Omega(t)}\,{}_{\Omega(t)}\langle{\bf p}_{1}\,\cdots\,{\bf
p}_{n}|0\rangle\\\
&=\sum_{n=0}^{\infty}\frac{1}{n!}\left[\prod_{j=1}^{n}\int\frac{d^{3}p_{j}}{(2\pi)^{3}}\right]{}_{t}\langle
0|\,b^{\dagger}_{{\bf p}_{1}}(t)\cdots b^{\dagger}_{{\bf
p}_{n}}(t)\,|\Omega(t)\rangle\langle\Omega(t)|\,b_{{\bf p}_{1}}(t)\cdots
b_{{\bf p}_{n}}(t)\,|0\rangle\\\ &={}_{t}\langle
0|\Omega(t)\rangle\,\langle\Omega(t)|0\rangle\sum_{n=0}^{\infty}\frac{1}{n!}\left[\int\frac{d^{3}p}{(2\pi)^{3}}\left(-\frac{i}{E_{\bf
p}}\dot{f}^{*}(t,{\bf p})\right)\left(f(t,{\bf p})+\frac{i}{E_{\bf
p}}\dot{f}(t,{\bf p})\right)\right]^{n}\\\ &={}_{t}\langle
0|\Omega(t)\rangle\,\langle\Omega(t)|0\rangle\;{\rm
exp}\left\\{\int\frac{d^{3}p}{(2\pi)^{3}}\left(-\frac{i}{E_{\bf
p}}\dot{f}^{*}(t,{\bf p})\right)\left(f(t,{\bf p})+\frac{i}{E_{\bf
p}}\dot{f}(t,{\bf p})\right)\right\\}.\end{split}$ (119)

On the other hand, making use of (24), without loss of generality

${}_{t}\langle
0|0\rangle=e^{-\frac{1}{2}N_{0}(t)+i\varphi(t)},\qquad\Rightarrow
e^{i\varphi(t)}=e^{\frac{1}{2}N_{0}(t)}{}_{t}\langle 0|0\rangle,$ (120)

and making use of this, the definition (16) and the Shrödinger equation in the
Heisenberg representation (106)

$\begin{split}{}_{t}\langle 0|0\rangle&=\langle
0|\,U_{I}(t,0)\,|0\rangle=\langle 0|\,U(t,0)\,|0\rangle\\\
\Rightarrow&i\,\dot{\varphi}(t)\;e^{i\varphi(t)}=\frac{1}{2}\dot{N}_{0}(t)\;e^{\frac{1}{2}N_{0}(t)}{}_{t}\langle
0|0\rangle-ie^{\frac{1}{2}N_{0}(t)}\langle
0|\,U(t,0)\;\hat{H}(t)\,|0\rangle.\end{split}$ (121)

Calling ${\cal I}(t)$ the integral

${\cal I}(t)=\int\frac{d^{3}p}{(2\pi)^{3}}\left(-\frac{i}{E_{\bf
p}}\dot{f}^{*}(t,{\bf p})\right)\left(f(t,{\bf p})+\frac{i}{E_{\bf
p}}\dot{f}(t,{\bf p})\right),$ (122)

and making use of the identity (119) and of (116), (118), (117) one can write

$\begin{split}&\langle 0|\,U(t,0)\;\hat{H}(t)\,|0\rangle={}_{t}\langle
0|\,\hat{H}(t)\,|0\rangle=\sum_{n=0}^{\infty}\left[\prod_{j=1}^{n}\int\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right]{}_{t}\langle 0|{\bf p}_{1},\cdots{\bf
p}_{n}\rangle_{\Omega(t)}\;{}_{\Omega(t)}\langle{\bf p}_{1},\cdots{\bf
p}_{n}|\,\hat{H}(t)\,|0\rangle\\\ &={}_{t}\langle
0|\Omega(t)\rangle\,\langle\Omega(t)|0\rangle\sum_{n=0}^{\infty}\frac{1}{n!}\left[\prod_{j=1}^{n}\int\frac{d^{3}p_{j}}{(2\pi)^{3}}\left(-\frac{i}{E_{{\bf
p}_{j}}}\dot{f}^{*}(t,{\bf p}_{j})\right)\left(f(t,{\bf
p}_{j})+\frac{i}{E_{{\bf p}_{j}}}\dot{f}(t,{\bf
p}_{j})\right)\right]\left(E_{0}(t)+\sum_{j=1}^{n}E_{{\bf p}_{j}}\right)\\\
&={}_{t}\langle 0|\Omega(t)\rangle\,\langle\Omega(t)|0\rangle\,{\rm
exp}\left\\{{\cal
I}(t)\vphantom{\frac{}{}}\right\\}\,\left[E_{0}(t)+\int\frac{d^{3}p}{(2\pi)^{3}}\left(-\frac{i}{E_{{\bf
p}}}\dot{f}^{*}(t,{\bf p})\right)\left(f(t,{\bf p})+\frac{i}{E_{{\bf
p}}}\dot{f}(t,{\bf p})\right)E_{\bf p}\right]\\\ &={}_{t}\langle
0|0\rangle\,\int\frac{d^{3}p}{(2\pi)^{3}}\,\left[-i\vphantom{\frac{}{}}\dot{f}^{*}(t,{\bf
p})\,f(t,{\bf p})\right].\end{split}$ (123)

Plugging into Eq.(121), and making use of (120) to simplify the $e^{i\varphi}$
phase

$\dot{\varphi}(t)=-\frac{i}{2}\dot{N}_{0}(t)+i\int\frac{d^{3}p}{(2\pi)^{3}}\,\dot{f}^{*}(t,{\bf
p})\,f(t,{\bf
p})=\frac{i}{2}\int\frac{d^{3}p}{(2\pi)^{3}}\left[\dot{f}^{*}(t,{\bf
p})\,f(t,{\bf p})-f^{*}(t,{\bf p})\,\dot{f}(t,{\bf
p})\vphantom{\frac{}{}}\right].$ (124)

Since the phase of $\langle 0|0\rangle$ is $0$ by construction, the latter is
enough to prove that Eq. (25) is the correct solution, for time-dependent
classical source in general, and in particular the time-independent source
too. The last step is to verify that is a real number as expected. This can be
done making direct use of the definition of the $f$ function (93)

$\begin{split}\varphi(t)&=\frac{i}{2}\int_{0}^{t}dx^{0}\int\frac{d^{3}p}{(2\pi)^{3}}\left[\partial_{x^{0}}f^{*}(x^{0},{\bf
p})\,f(x^{0},{\bf p})-f^{*}(x^{0},{\bf p})\,\partial_{x^{0}}f(x^{0},{\bf
p})\vphantom{\frac{}{}}\right]\\\
&=\frac{i\,g^{2}}{2}\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf
p}}\int_{0}^{t}dx^{0}\int_{0}^{x_{0}}dy^{0}\int d^{3}x\int
d^{3}y\,\rho(x)\,\rho(y)\left[e^{-ip\cdot(x-y)}-e^{ip\cdot(x-y)}\right]\\\
&=g^{2}\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf
p}}\int_{0}^{t}dx^{0}\int_{0}^{x_{0}}dy^{0}\int d^{3}x\int
d^{3}y\,\rho(x)\,\rho(y)\,{\rm
sin}\left[p\cdot(x-y)\vphantom{\frac{}{}}\right]\in{\mathds{R}}.\end{split}$
(125)

##  Appendix C Decay in n particles

In this appendix we give a reminder in which sense the following formulas

$P_{\psi_{i}\to 0}(t_{f},t_{i})=\left|\vphantom{\frac{}{}}\langle
0|U(t_{f},t_{i})|\psi_{i}\rangle\right|^{2},\qquad P_{\psi_{i}\to
n}(t_{f},t_{i})=\int\frac{d^{3}p_{1}\cdots d^{3}p_{n}}{(2\pi)^{3n}2E_{{\bf
p}_{1}}\cdots 2E_{{\bf p}_{n}}}\left|\vphantom{\frac{}{}}\langle{\bf
p}_{1}\cdots{\bf p}_{n}|U(t_{f},t_{i})|\psi_{i}\rangle\right|^{2},$ (126)

represent the probability that a system measured in the state
$|\psi_{i}\rangle$ at the initial time $t_{i}$, is measured to be a
$n$-particles state at time $t_{f}$, regardless of the details of the
$n$-particle part of the wave function.
That is the probability reads

$P_{\psi_{i}\to 0}(t_{f},t_{i})=\left|\vphantom{\frac{}{}}\langle
0|U(t_{f},t_{i})|\psi_{i}\rangle\right|^{2},\qquad P_{\psi_{i}\to
n}(t_{f},t_{i})=\int\frac{d^{3}p_{1}\cdots d^{3}p_{n}}{(2\pi)^{3n}2E_{{\bf
p}_{1}}\cdots 2E_{{\bf p}_{n}}}\left|\vphantom{\frac{}{}}\langle{\bf
p}_{1}\cdots{\bf p}_{n}|U(t_{f},t_{i})|\psi_{i}\rangle\right|^{2},$ (127)

The amplitude, which gives the transition probability, in the Heisenberg
picture reads

$a_{i\to f}=\langle\psi_{f}|U(t_{f},t_{i})|\psi_{i}\rangle,$ (128)

with an explicit mention of the final state $|\psi_{f}\rangle$, but no
explicit concept of particle number.
Assuming that the eigenstates of all observables at the time $t_{f}$ can be
decomposed exclusively with the basis $|0\rangle$, $|{\bf p}_{1}\cdots{\bf
p}_{n}\rangle$, the final state $|\psi_{f}\rangle$ can be expressed as

$\begin{split}|\psi_{f}\rangle&={\mathds{1}}|\psi_{f}\rangle=\langle
0|\psi_{f}\rangle|0\rangle+\sum_{n=1}^{\infty}\int\frac{d^{3}p_{1}\cdots
d^{3}p_{n}}{(2\pi)^{3n}2E_{{\bf p}_{1}}\cdots 2E_{{\bf p}_{n}}}\langle{\bf
p}_{1},\cdots,{\bf p}_{n}|\psi_{f}\rangle|{\bf p}_{1},\cdots,{\bf
p}_{n}\rangle\equiv\\\
&\equiv\psi_{f}^{(0)}|0\rangle+\sum_{n=1}^{\infty}\int\frac{d^{3}p_{1}\cdots
d^{3}p_{n}}{(2\pi)^{3n}2E_{{\bf p}_{1}}\cdots 2E_{{\bf
p}_{n}}}\;\psi^{(n)}_{f}({\bf p}_{1},\cdots,{\bf p}_{n})|{\bf
p}_{1},\cdots,{\bf p}_{n}\rangle.\end{split}$ (129)

One has a clear understanding what it means to have an $\tilde{n}$-particle
final state. The partial wave for all $n\neq\tilde{n}$ is zero,
$\psi_{f}^{(n\neq\tilde{n})}=0$.

Making a similar decomposition for $U|\psi_{i}\rangle$, which is the time-
dependent state in the Shrödinger picture, one has

$U(t_{f},t_{i})|\psi_{i}\rangle=\sum_{n=0}^{\infty}\int\frac{d^{3}p_{1}\cdots
d^{3}p_{n}}{(2\pi)^{3n}2E_{{\bf p}_{1}}\cdots 2E_{{\bf p}_{n}}}\langle{\bf
p}_{1},\cdots,{\bf p}_{n}|U(t_{f},t_{i})|\psi_{i}\rangle|{\bf
p}_{1},\cdots,{\bf p}_{n}\rangle.$ (130)

Calling $|\psi_{i}^{(n)}(t_{f},t_{i})\rangle$ the $n$-particle partial wave,
after normalizing it to $1$

$\begin{split}|\psi_{i}^{(n)}(t_{f},t_{i})\rangle&=\frac{1}{N^{(n)}_{\psi_{i}}(t_{f},t_{i})}\int\frac{d^{3}p_{1}\cdots
d^{3}p_{n}}{(2\pi)^{3n}2E_{{\bf p}_{1}}\cdots 2E_{{\bf p}_{n}}}\langle{\bf
p}_{1},\cdots,{\bf p}_{n}|U(t_{f},t_{i})|\psi_{i}\rangle|{\bf
p}_{1},\cdots,{\bf p}_{n}\rangle,\\\
\left(N^{(n)}_{\psi_{i}}(t_{f},t_{i})\right)^{2}&=\int\frac{d^{3}p_{1}\cdots
d^{3}p_{n}}{(2\pi)^{3n}2E_{{\bf p}_{1}}\cdots 2E_{{\bf
p}_{n}}}\left|\langle{\bf p}_{1},\cdots,{\bf
p}_{n}|U(t_{f},t_{i})|\psi_{i}\rangle\vphantom{\frac{}{}}\right|^{2}=P_{\psi_{i}\to
n}(t_{f},t_{i}).\end{split}$ (131)

In other words

$U(t_{f},t_{i})|\psi_{i}\rangle=\sum_{n=0}^{\infty}\sqrt{P_{\psi_{i}\to
n}(t_{f},t_{i})}\,|\psi_{i}^{(n)}(t_{f},t_{i})\rangle,$ (132)

that is $P_{\psi_{i}\to n}(t_{f},t_{i})$ is the largest probability to measure
an $n$-particles state, which is realized if and only if
$|\psi_{f}\rangle=|\psi_{i}^{(n)}(t_{f},t_{i})\rangle$.
Another way to put it, $P_{\psi_{i}\to n}(t_{f},t_{i})$ is the probability to
decay in $n$-particles in the sense that it follows the usual rules of
probability decomposition, despite quantum mechanics revolving around
amplitudes. To be more precise, being $|\psi^{(n)}_{f}\rangle$ a purely
$n$-particles final state, the probability to measure it at time $t_{f}$ reads

$\left|\langle\psi^{(n)}_{f}|U(t_{f},t_{i})|\psi_{i}\rangle\vphantom{\frac{}{}}\right|^{2}=P_{\psi_{i}\to
n}(t_{f},t_{i})\left|\langle\psi^{(n)}_{f}|\psi_{i}^{(n)}(t_{f},t_{i})\rangle\vphantom{\frac{}{}}\right|^{2}.$
(133)

That is, the probability to decay into an $n$-particle state times the
probability that the $n$-particles state is actually $|\psi^{(n)}_{f}\rangle$
(the overlap between the two states). This interpretation of $P_{\psi_{i}\to
n}(t_{f},t_{i})$ is the only one compatible with both the regular
factorization of constrained probabilities, and the superposition rules in the
states of quantum mechanics.

##  Appendix D Coherent states

Eigenvectors $|c\rangle$ of the destruction operators are called coherent
states

$a|c\rangle=c|c\rangle.$ (134)

The coherent state itself can be built noting that, as a consequence of the
commutation relations

$\begin{split}[a,a^{\dagger}]=1,\quad\Rightarrow\quad
a|n\rangle=n|n-1\rangle,\end{split}$ (135)

the formal series

$|0\rangle+\sum_{n=1}^{\infty}\frac{c^{n}}{\sqrt{n!}}|n\rangle=\left[\sum_{n=0}^{\infty}\frac{\left(c\,a^{\dagger}\vphantom{\frac{}{}}\right)^{n}}{n!}\right]|0\rangle=e^{c\,a^{\dagger}}|0\rangle,\qquad\Rightarrow\qquad
a\left(e^{c\,a^{\dagger}}|0\rangle\right)=c\left(e^{c\,a^{\dagger}}|0\rangle\right),$
(136)

is an eigenvector of $a$ with with eigenvalue $c$, an arbitrary complex
number. In order to be an eigenstate it must be normalized to $1$.

The normalization can be computed in a straightforward manner using (136)

$\begin{split}\left|e^{c\,a^{\dagger}}|0\rangle\right|^{2}&=\left(\langle
0|e^{c^{*}\,a}\right)\left(e^{c\,a^{\dagger}}|0\rangle\right)=\langle
0|\left[\sum_{n=0}^{\infty}\frac{\left(c^{*}\,a\vphantom{\frac{}{}}\right)^{n}}{n!}\right]\left(e^{c\,a^{\dagger}}|0\rangle\right)=\langle
0|0\rangle\sum_{n=0}^{\infty}\frac{\left|\vphantom{\frac{}{}}\;c\;\right|^{2n}}{n!}=e^{\left|\vphantom{\frac{}{}}\;c\;\right|^{2}}.\end{split}$
(137)

Therefore the physical coherent state is given by

$|c\rangle=e^{-\frac{1}{2}|c|^{2}}\,e^{c\,a^{\dagger}}|0\rangle.$ (138)

The generalization to the uncharged scalar field situation

$a_{\bf p}|\psi\rangle=-h({\bf p})|\psi\rangle,$ (139)

is rather straightforward. One has to take into account that there are
multiple creation-annihilation operators (for each momentum ${\bf p}$), that
each annihilation operator has a different eigenvalue, and that the
normalization is not to the identity but to a distribution

$[a_{\bf p},a^{\dagger}_{\bf q}]=(2\pi^{3})\delta({\bf p}-{\bf q}).$ (140)

Instead of a straight multiplication of the creation operators times the
eigenvalue, one needs the multiplication of the momentum integral of them

$\int\frac{d^{3}p}{(2\pi)^{3}}\,\left(\vphantom{\frac{}{}}-h(\bf
p)\right)\;a^{\dagger}_{\bf p}.$ (141)

Therefore, without loss of generality, an eigenvector for each $a_{\bf p}$
corresponding to the eigenvalue $-h({\bf p})$ is

$\begin{split}&|0\rangle+\left[\sum_{n=1}^{\infty}\frac{1}{n!}\int\frac{d^{2}p_{1}d^{3}p_{2}\cdots
d^{3}p_{n}}{(2\pi)^{3n}}\left(\vphantom{\frac{}{}}-h({\bf
p}_{1})a^{\dagger}_{{\bf p}_{1}}\right)\left(\vphantom{\frac{}{}}-h({\bf
p}_{2})a^{\dagger}_{{\bf p}_{2}}\right)\cdots\left(\vphantom{\frac{}{}}-h({\bf
p}_{n})a^{\dagger}_{{\bf p}_{n}}\right)\right]|0\rangle=\\\
&=\left[\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\int\frac{d^{3}p}{(2\pi)^{3}}\,h({\bf
p})\,a^{\dagger}_{\bf p}\right)^{n}\right]|0\rangle={\rm
exp}\left\\{-\int\frac{d^{3}p}{(2\pi)^{3}}\,h({\bf p})\,a_{\bf
p}^{\dagger}\right\\}|0\rangle,\qquad\Rightarrow\\\ \\\ &\Rightarrow\qquad
a_{\bf p}\;{\rm exp}\left\\{-\int\frac{d^{3}p}{(2\pi)^{3}}\,h({\bf p})\,a_{\bf
p}^{\dagger}\right\\}|0\rangle=-h({\bf p})\;{\rm
exp}\left\\{-\int\frac{d^{3}p}{(2\pi)^{3}}\,h({\bf p})\,a_{\bf
p}^{\dagger}\right\\}|0\rangle.\end{split}$ (142)

The normalization follows from the commutation relations in a very similar way
as in the discrete case. Therefore

$\begin{split}&|\psi\rangle\quad{\rm s.t.}\quad a_{\bf p}|\psi\rangle=-h({\bf
p})|\psi\rangle,\quad\Rightarrow\\\ &\qquad\Rightarrow\quad|\psi\rangle={\rm
exp}\left\\{-\frac{1}{2}\int\frac{d^{3}p}{(2\pi)^{3}}\,\left|h({\bf
p})\right|^{2}\right\\}{\rm exp}\left\\{-\int\frac{d^{3}p}{(2\pi)^{3}}\,h({\bf
p})\,a_{\bf p}^{\dagger}\right\\}|0\rangle.\end{split}$ (143)

##  Appendix E Scattering states

Given the important role that the definition of the scattering states hold in
this work, in this section there is a reminder of the most common concept.
Within the Heisenberg representation, the in and out states

$|S_{\rm in}\rangle,\qquad|S_{\rm out}\rangle,$ (144)

are states that in the distant past (for the in-states) and in the distant
future (out-states) resemble some free states. In the Heisenberg picture, of
course, the states do not depend on time, the former statements mean that the
partial waves can be approximated by some free wave-function

$\langle{\bf p}_{1}\cdots{\bf p}_{n}|U(t,0)|S_{{\rm in}/{\rm
out}}\rangle\simeq e^{-i\sum_{j}E_{{\bf p}_{j}}t}\;\psi_{{\rm in}/{\rm
out}}({\bf p}_{1},\cdots{\bf p}_{n}),$ (145)

and that the approximation becomes more accurate the smaller(larger) time $t$
for the in(out) case. The basis $|{\bf p}_{1}\cdots{\bf p}_{n}\rangle$ is the
regular one for the free state. This is built from interaction picture fields
$\hat{\phi}_{I}(x)$ which coincide with the free fields at time $t=0$ and
fulfill the free Klein-Gordon equation. The overlap between the states

$\langle S_{\rm out}|S_{\rm in}\rangle,$ (146)

has the immediate interpretation of the probability amplitude of a state
(arbitrarily close to a free state in the distant past) that evolves into some
final state (also free-like) in the distant future.

The simpler way to isolate such states is to modify the theory, having the
interaction (and eventual counterterms) only for a limited time, from $t_{-}$
to $t_{+}>t_{-}$. In this way the states do not resemble free states in the
distant past or future, they become free states. In the sense that, in the
Shrödinger picture, they evolve according to the free propagator $U_{0}$.
Namely, for the in states

$\begin{split}U(t<t_{-},0)|S_{\rm
in}\rangle&=U(t,t_{-})\left[U(t_{-},0)\vphantom{\frac{}{}}|S_{\rm
in}\rangle\right]=U_{0}(t,t_{-})\left[U(t_{-},0)\vphantom{\frac{}{}}|S_{\rm
in}\rangle\right]\\\
&=\sum_{n=0}^{\infty}\int\prod_{j=1}^{n}\left\\{\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right\\}e^{-i\sum_{j}E_{{\bf p}_{j}}t}\;\psi_{\rm in}({\bf
p}_{1},\cdots{\bf p}_{n})|{\bf p}_{1},\cdots{\bf p}_{n}\rangle.\end{split}$
(147)

For $t<t_{-}$ the system is free and the decomposition of the state at time
$t_{-}$ in the partial wavefucntions is not ambiguous. The phase convention
for the $\psi_{\rm in}$ is, clearly

$\langle{\bf p}_{1},\cdots{\bf p}_{n}|U(t_{-},0)|S_{\rm
in}\rangle=e^{-i\sum_{j}E_{{\bf p}_{j}}t_{-}}\psi_{\rm in}({\bf
p}_{1},\cdots{\bf p}_{n}),$ (148)

since the evolution is free $e^{-i{\hat{H}}_{0}\,\Delta t}$ only before
$t_{-}$ (and after $t_{+}$). The analogue holds for the out states

$\begin{split}U(t>t_{+},0)|S_{\rm
out}\rangle&=U(t,t_{+})\left[U(t_{+},0)\vphantom{\frac{}{}}|S_{\rm
out}\rangle\right]=U_{0}(t,t_{+})\left[U(t_{+},0)\vphantom{\frac{}{}}|S_{\rm
out}\rangle\right]\\\
&=\sum_{n=0}^{\infty}\int\prod_{j=1}^{n}\left\\{\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right\\}e^{-i\sum_{j}E_{{\bf p}_{j}}t}\;\psi_{\rm out}({\bf
p}_{1},\cdots{\bf p}_{n})|{\bf p}_{1},\cdots{\bf p}_{n}\rangle.\end{split}$
(149)

The last formulas (147) and (149) can be inverted, in order write the states
$|S_{{\rm in}/{\rm out}}\rangle$ according to something recognizable $|S_{{\rm
in}/{\rm out}}\rangle=U^{-1}\left(\;\right.$ free-state decomposition
$\left.\;\right)$. In this way the overlap reads

$\begin{split}&\langle S_{\rm out}|S_{\rm
in}\rangle=\sum_{m,n}\int\prod_{j=1}^{n}\prod_{k=1}^{m}\left\\{\frac{d^{3}\tilde{p}_{j}}{(2\pi)^{3}2E_{\tilde{\bf
p}_{j}}}\frac{d^{3}p_{k}}{(2\pi)^{3}2E_{{\bf p}_{k}}}e^{iE_{\tilde{\bf
p}_{j}}t_{f}-iE_{{\bf p}_{k}}t_{i}}\right\\}\\\ &\qquad\times\psi_{\rm
out}^{*}(\tilde{\bf p}_{1},\cdots\tilde{\bf p}_{n})\psi_{\rm in}({\bf
p}_{1},\cdots{\bf p}_{m})\;\langle\tilde{\bf p}_{1},\cdots\tilde{\bf
p}_{n}|U(t_{f},0)U^{-1}(t_{i},0)|{\bf p}_{1},\cdots{\bf p}_{m}\rangle=\\\
&=\sum_{m,n}\int\prod_{j=1}^{n}\prod_{k=1}^{m}\left\\{\frac{d^{3}\tilde{p}_{j}}{(2\pi)^{3}2E_{\tilde{\bf
p}_{j}}}\frac{d^{3}p_{k}}{(2\pi)^{3}2E_{{\bf p}_{k}}}e^{iE_{\tilde{\bf
p}_{j}}t_{f}-iE_{{\bf p}_{k}}t_{i}}\right\\}\psi_{\rm out}^{*}(\tilde{\bf
p}_{1},\cdots\tilde{\bf p}_{n})\psi_{\rm in}({\bf p}_{1},\cdots{\bf p}_{m})\\\
&\qquad\qquad\qquad\times\langle\tilde{\bf p}_{1},\cdots\tilde{\bf
p}_{n}|U(t_{f},t_{i})|{\bf p}_{1},\cdots{\bf p}_{m}\rangle,\end{split}$ (150)

for any $t_{f}\geq t+$ and $t_{i}\leq t_{-}$. The problem is reformulated in
terms of the matrix elements of the propagator $U$. This formula can be
significantly simplified, removing the phases on the energies and removing the
explicit dependence on $t_{f}$ and i. Instead of directly inverting (147) and
(149), first applying $U_{0}^{-1}$

$\begin{split}&U_{I}(t<t_{-},0)|S_{\rm
out}\rangle=U_{0}^{-1}(t<t_{-},0)U(t<t_{-},0)|S_{\rm out}\rangle\\\
&\quad=U_{0}^{-1}(t<t_{-},0)\sum_{n=0}^{\infty}\int\prod_{j=1}^{n}\left\\{\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right\\}e^{-i\sum_{j}E_{{\bf p}_{j}}t}\;\psi_{\rm in}({\bf
p}_{1},\cdots{\bf p}_{n})|{\bf p}_{1},\cdots{\bf p}_{n}\rangle\\\
&\qquad=\sum_{n=0}^{\infty}\int\prod_{j=1}^{n}\left\\{\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right\\}\;\psi_{\rm in}({\bf p}_{1},\cdots{\bf p}_{n})|{\bf
p}_{1},\cdots{\bf p}_{n}\rangle,\\\ \\\ &U_{I}(t>t_{+},0)|S_{\rm
out}\rangle=U_{0}^{-1}(t>t_{+},0)U(t>t_{+},0)|S_{\rm out}\rangle\\\
&\quad=U_{0}^{-1}(t>t_{+},0)\sum_{n=0}^{\infty}\int\prod_{j=1}^{n}\left\\{\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right\\}e^{-i\sum_{j}E_{{\bf p}_{j}}t}\;\psi_{\rm out}({\bf
p}_{1},\cdots{\bf p}_{n})|{\bf p}_{1},\cdots{\bf p}_{n}\rangle\\\
&\qquad=\sum_{n=0}^{\infty}\int\prod_{j=1}^{n}\left\\{\frac{d^{3}p_{j}}{(2\pi)^{3}2E_{{\bf
p}_{j}}}\right\\}\;\psi_{\rm out}({\bf p}_{1},\cdots{\bf p}_{n})|{\bf
p}_{1},\cdots{\bf p}_{n}\rangle.\end{split}$ (151)

Then, inverting these formulas with the interaction picture propagator on the
left hand side, and following the same passages that brought (150), one has
the alternative formula

$\begin{split}&\langle S_{\rm out}|S_{\rm
in}\rangle=\sum_{m,n}\int\prod_{j=1}^{n}\prod_{k=1}^{m}\left\\{\frac{d^{3}\tilde{p}_{j}}{(2\pi)^{3}2E_{\tilde{\bf
p}_{j}}}\frac{d^{3}p_{k}}{(2\pi)^{3}2E_{{\bf p}_{k}}}\right\\}\psi_{\rm
out}^{*}(\tilde{\bf p}_{1},\cdots\tilde{\bf p}_{n})\psi_{\rm in}({\bf
p}_{1},\cdots{\bf p}_{m})\\\
&\qquad\qquad\qquad\qquad\qquad\times\langle\tilde{\bf p}_{1},\cdots\tilde{\bf
p}_{n}|U_{I}(t_{+},t_{-})|{\bf p}_{1},\cdots{\bf p}_{m}\rangle.\end{split}$
(152)

The problem is reduced to the calculation of the matrix elements of the
interaction picture propagator $U_{I}=U_{0}^{\dagger}U$. The explicit
dependence on $t_{-}$ and $t_{+}$ is still there. The hope is that, having the
regular (unaltered) interaction for the bulk of $\\{t_{-},t_{+}\\}$ plus some
reconnection to zero close to the border of this integral, if both the limits

$\lim_{t_{-}\to-\infty}\lim_{t_{+}\to\infty}\langle\tilde{\bf
p}_{1},\cdots\tilde{\bf p}_{n}|U_{I}(t_{+},t_{-})|{\bf p}_{1},\cdots{\bf
p}_{m}\rangle,\qquad\lim_{t_{+}\to\infty}\lim_{t_{-}\to-\infty}\langle\tilde{\bf
p}_{1},\cdots\tilde{\bf p}_{n}|U_{I}(t_{+},t_{-})|{\bf p}_{1},\cdots{\bf
p}_{m}\rangle,$ (153)

exist and are equal, then the results of the full theory will be recovered.
These limits are important since the $d^{4}x$ integrals produce the four-
momentum conservation deltas at the vertices in the Feynman rules. It is
immediate to notice that, within this framework, stable bound states like the
Hydrogen atom in its fundamental state cannot be treated. One needs to use
some effective Lagrangian that includes all stable states in the
$\hat{H}_{0}$.

Besides this, in order to check if this procedure of switching the interaction
reproduces the results of the full theory, at least regarding the stable
states, an alternative treatment is necessary. In axiomatic field theory [4]
one assumes (or proves, if possible) that the four-momentum operators
$\hat{P}^{\mu}$ have a (physical) vacuum eigenstate191919The Hamiltonian
operator in quantum field theory usually has to be just bounded from below. A
constant (and finite) counterterm is necessary, in general, to ensure that the
minimum is exactly zero.  $|0\rangle_{P}$

$\hat{P}^{\mu}|0\rangle_{P}=0,\qquad{}_{P}\langle 0|0\rangle_{P}=1,$ (154)

which is orthogonal to all the other physical states. In addition it is
required that a subspace of one-particle states is spanned by the formal one-
particle states

$\hat{P}^{\mu}|{\bf p}\rangle=p^{\mu}|{\bf p}\rangle,\quad\langle{\bf p}|{\bf
q}\rangle=(2\pi)^{3}2E_{\bf p}\delta^{3}({\bf p}-{\bf q}),\quad{}_{P}\langle
0|{\bf p}\rangle=0,$ (155)

for some physical mass $m\to p^{0}=E_{\bf p}=\sqrt{m^{2}+{\bf p}\cdot{\bf
p}}$. In the case of scalar particles, there must be a scalar
field202020Usually it is necessary to renormalize the fields, at least remove
the vacuum expectation value and rescale the field with a positive constant.
See, eg, the book [4]. $\hat{\phi}^{\prime}$, which does not have to be
fundamental, such that

${}_{P}\langle 0|\hat{\phi}^{\prime}(x)|0\rangle_{P}=0,\qquad{}_{P}\langle
0|\hat{\phi}^{\prime}(x)|{\bf p}\rangle=e^{-ip\cdot x}.$ (156)

These are the generalization to the interacting theory of the analogous
formulas fulfilled by free fields. The one-particle physical states can be
built just like in the free case

$|\psi_{(1)}\rangle=\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf p}}\;\psi_{(1)}({\bf
p})\,|{\bf p}\rangle,\qquad\langle\psi_{(1)}|\psi_{(1)}\rangle=1,$ (157)

for some wavefunction (in the Heisenberg representation) $\psi_{(1)}({\bf
p})$. These states correspond to the free wave-functions (in the Shrödinger
representation)

$\psi_{(1)}(t,{\bf x})=\int\frac{d^{3}p}{(2\pi)^{3}2E_{\bf p}}\;e^{-iE_{\bf
p}t+i{\bf p}\cdot{\bf x}}\,\psi_{(1)}({\bf p}).$ (158)

Following [4], it is possible then to build an operator that behaves like a
creation operator of a physical one-particle state
$\hat{\phi}^{\prime}[\psi_{(1)}]$

$\hat{\phi}^{\prime}[\psi_{(1)}](t)=i\int
d^{3}x\left[\hat{\phi}^{\prime}(x)\,\partial_{t}\psi_{(1)}(x)-\psi_{(1)}(x)\,\partial_{t}\hat{\phi}^{\prime}(x)\right].$
(159)

Thanks to (156) it is immediate to prove that

$\langle{\bf
p}|\;\hat{\phi}^{\prime}[\psi_{(1)}](t)\,|0\rangle_{P}=\psi_{(1)}(p),\qquad{}_{P}\langle
0|\;\hat{\phi}^{\prime}[\psi_{(1)}](t)\,|{\bf p}\rangle=0.$ (160)

Differently from the free case, the overlap of the “one-particle” state
$\hat{\phi}^{\prime}[\psi_{(1)}]|0\rangle_{P}$ with a generic state
$|\Psi\rangle$ is not the overlap of the one-particle partial wave
$\langle{\bf p}|\Psi\rangle$ with $\psi_{(1)}(p)$. However it can be shown
that this expected overlap is the only remaining term both in the $t\to\infty$
limit, and in the $t\to-\infty$ one. The asymptotic multiparticle-states are
obtained as the limits

$|\psi_{(1)}^{a}\,\psi_{(1)}^{b},\cdots\rangle_{{\rm in}/{\rm
out}}=\lim_{t\to\pm\infty}\left\\{\left(\hat{\phi}^{\prime}[\psi_{(1)}^{a}](t)\;\hat{\phi}^{\prime}[\psi_{(1)}^{b}](t)\cdots\right)|0\rangle_{P}\right\\}.$
(161)

Either one accept this as a definition, possibly recurring to some intuitive
localization arguments as in [2, 4], otherwise it is necessary to delve deeply
in the mathematical structure of axiomatic field theory [51]. The results of
scattering theory obtained from the switch on approach can be recovered, for
instance the Feynmann rules. There is the additional advantage to include in
the scattering theory the stable bound states like the hydrogen atom.

In any case this approach cannot be used in this particular case. From the
very start, the vacuum state in (154) does not exist. The Hamiltonian (10) is
bounded from below. It is not difficult to set the minimum to $0$ as required.
It is enough a constant off-set in the Lagrangian (not in the Lagrangian
density). However the minimum of the Hamiltonian is an eigenstate only of the
Hamiltonian operator itself. It is not an eigenstate of the momentum $\hat{\bf
P}(t)$ at any time (not even $t=0$), as shown in a appendix B: The energy
vacuum $|\Omega(t)\rangle$ is a coherent state of the momentum eigenstates at