# Quantum radiation from a classical point source

J. Dimock
Department of Mathematics
SUNY at Buffalo

###### Abstract

We study the radiation of photons from a classical charged particle. We
particularly consider a situation where the particle has a constant velocity
in the distant past, then is accelerated, and then has a constant velocity in
the distant future. Starting with no photons in the distant past we seek to
characterize the quantum state of the photon field in the distant future.
Working in the Coulomb gauge and in a C* algebra formulation, we give sharp
conditions on whether this state is or is not in Fock space.

##  1 Introduction

We study the radiation of the quantum electromagnetic field in the presence of
a classical source. With a specified four-current consisting of functions
$(j^{0},j)$ satisfying $\partial_{\mu}j^{\mu}=0$, one seeks associated quantum
field operators $A=(A^{0},A)$. In the Lorentz gauge $\partial_{\mu}A^{\mu}=0$
these are connected by the wave equation with source $\square
A^{\mu}=j^{\mu}$. A textbook treatment of this problem can be found in
Itzykson-Zuber [4] whose sign conventions we follow. A mathematical treatment
is due to Naudts-DeRoeck [6] . A characteristic phenomena is that if one
starts in a state with no photons in the distant past, the vacuum in Fock
space, it may evolve into a state in the distant future with infinitely many
photons radiated and not in Fock space. This is sometimes known as the
infrared catastrophe.

In this paper we are interested in characterizing exactly when this happens.
We particularly study the case of a point source where the four-current has
the form

$\begin{split}j^{0}(x,t)=&\delta(x-x(t))\\\
j(x,t)=&x^{\prime}(t)\delta(x-x(t))\\\ \end{split}$ (1)

and $x(t)$ is the trajectory of the charged particle in $\mathbb{R}^{3}$. We
suppose that the particle has a velocity $v_{in}$ in the distant past, then
accelerates for a finite amount of time, and then continues with a velocity
$v_{out}$ in the distant future. We start with the Fock vacuum in the distant
past considered as a state $\omega_{in}$ on the $C^{*}$-algebra generated by
the fields. It evolves to state $\omega_{out}$ in the distant future. We
establish the following results

  * •

If $v_{out}=v_{in}=0$ and the trajectory $x(t)$ is smooth then $\omega_{out}$
is a Fock state.

  * •

If $v_{out}=v_{in}=0$ and the velocity $x^{\prime}(t)$ has a discontinuity
then $\omega_{out}$ is not a Fock state.

  * •

In general for a smooth trajectory $\omega_{out}$ is a Fock state if and only
if $v_{in}=v_{out}$.

We work in the Coulomb gauge rather than the Lorentz gauge. The advantage is
that we avoid working with indefinite metric Hilbert spaces. The disadvantage
is that relativistic properties like Lorentz invariance and locality are not
manifest, although they are still there. (See Skagerstam, Erikson, Rekdal [7]
for a discussion of locality in the Coulomb gauge.)

The Coulomb gauge is defined by the condition $\nabla\cdot A=0$ on the
magnetic potential. In this gauge Maxwells equations become

$\begin{split}-\Delta A^{0}=&j^{0}\\\ \frac{\partial^{2}A}{\partial
t^{2}}-\Delta A+\nabla\frac{\partial}{\partial t}A^{0}=&j\\\ \end{split}$ (2)

The first equation for the electrostatic potential $A^{0}$ has the solution

$A^{0}(x,t)=\int V(x-y)j^{0}(t,y)dy$ (3)

where $V(x)=(4\pi)^{-1}|x|^{-1}$ is the Coulomb potential. In the second
equation we call the term $\nabla(\partial A^{0}/\partial t)$ the longitudinal
current. Since $\partial j^{0}/\partial t+\nabla\cdot j=0$ this can be
expressed as

$j_{L}(x,t)=\nabla(\partial A^{0}/\partial t)(x,t)=-\nabla\int
V(x-y)(\nabla\cdot j)(y,t)dy$ (4)

It satisfies $\nabla\cdot j_{L}=\nabla\cdot j$. The transverse current is
defined by $j_{T}=j-j_{L}$ and it satisfies $\nabla\cdot j_{T}=0$. The second
equation in (2) and the constraint can now be written

$\frac{\partial^{2}A}{\partial t^{2}}-\Delta A=j_{T}\hskip
28.45274pt\nabla\cdot A=0$ (5)

or just $\square A=j_{T}$ with $\square=\partial_{t}^{2}-\Delta$.

The electrostatic potential $A^{0}$ is given by (3) and is not quantized. We
seek quantum field operators $A$ satisfying (5).

##  2 Preliminaries

###  2.1 fundamental solutions

We review some facts about solutions of the wave equation with source. At
first this is just classical so we seek functions $A$ satisfying $\square
A=j$. We are interested in solutions of the form $A=G_{\pm}*j$ where $G_{\pm}$
are retarded and advanced fundamental solutions which we define shortly. We
want to allow $j$ to be certain distributions so $G_{\pm}*j$ is defined as
$<G_{\pm}*j,\phi>=<j,G_{\mp}*\phi>$ for $\phi$ in (vector-valued) ${\cal
S}(\mathbb{R}^{4})$, the Schwartz space of smooth rapidly decreasing
functions. However we also want to work at sharp time so the test functions
$\phi$ will be replaced by $f\otimes\delta_{t}$ with $f$ in (vector-valued)
${\cal S}(\mathbb{R}^{3})$. The upshot is that we want to define functions
$G_{\pm}*(f\otimes\delta_{t})$, or without the convolution
$G_{\pm}(f\otimes\delta_{t})\equiv G_{\pm}(f,t)$.

The $G_{\pm}$ are given by

$G_{\pm}(x,t)=-\frac{1}{(2\pi)^{4}}\int_{p_{0}\in\mathbb{R}\pm
i\epsilon}e^{i(-p_{0}t+px)}\frac{1}{p_{0}^{2}-|p|^{2}}dp_{0}dp$ (6)

For $f\in{\cal S}(\mathbb{R}^{3})$ this is to be understood as for
$G_{\pm}(f,t)=\int f(x)G_{\pm}(x,t)dx$ given by

$G_{\pm}(f,t)=-(2\pi)^{-\frac{5}{2}}\int_{p_{0}\in\mathbb{R}\pm
i\epsilon}\frac{1}{p_{0}^{2}-|p|^{2}}e^{-ip_{0}t}\tilde{f}(-p)\ dp_{0}dp$ (7)

where $\tilde{f}(p)=(2\pi)^{-\frac{3}{2}}\int e^{-ipx}f(x)dx$ is the Fourier
transform of $f$.

Consider $G_{+}(f,t)$. We have $|e^{-ip_{0}t}|=e^{\textrm{Im}(p_{0})t}$ so if
$t<0$ we can close the $p_{0}$-contour in the upper half plane and get zero.
On the other hand If $t>0$ we we close the contour in the lower half plane,
picking up poles at $p_{0}=\pm\omega$ where $\omega(p)=|p|$. Similarly for
$G_{-}(f,t)$. We obtain

$G_{\pm}(f,t)=\pm(2\pi)^{-\frac{3}{2}}\theta(\pm t)i\int\Big{(}e^{-i\omega
t}-e^{i\omega t}\Big{)}\tilde{f}(-p)\frac{dp}{2\omega}$ (8)

where $\theta$ is the Heaviside function.

The convolution is defined by
$(G_{\pm}*(f\otimes\delta_{t_{0}}))(x,t)=G_{\pm}\Big{(}f(x-\cdot),\delta_{t_{0}}(t-\cdot)\Big{)}$.
But the Fourier transform of $f(x-\cdot)$ at $-p$ is $e^{ipx}\tilde{f}(p)$ and
$\delta_{t_{0}}(t-\cdot)=\delta_{t-t_{0}}$. Thus we have

$(G_{\pm}*(f\otimes\delta_{t_{0}}))(x,t)=\pm(2\pi)^{-\frac{3}{2}}\theta(\pm(t-t_{0}))i\int\Big{(}e^{-i\omega(t-t_{0})}-e^{i\omega(t-t_{0})}\Big{)}e^{ipx}\tilde{f}(p)\frac{dp}{2\omega}$
(9)

We also will want to consider the propagator $G=G_{+}-G_{-}$. In this
expression the Heaviside functions $\theta$ disappear and we have

$(G*(f\otimes\delta_{t_{0}}))(x,t)=u_{f}(x,t-t_{0})$ (10)

where

$u_{f}(x,t)=(2\pi)^{-\frac{3}{2}}i\int\Big{(}e^{-i\omega t}-e^{i\omega
t}\Big{)}e^{ipx}\tilde{f}(p)\frac{dp}{2\omega}$ (11)

is a smooth solution of the wave equation with data at $t=0$

$u_{f}(x,0)=0\hskip 28.45274pt\frac{\partial u_{f}}{\partial t}(x,0)=f(x)$
(12)

###  2.2 free quantum field

We review the mathematics of the quantized free field in the Coulomb gauge. In
this case we can take $A^{0}=0$, and study the field equation

$\frac{\partial^{2}A}{\partial t^{2}}-\Delta A=0$ (13)

We want to define a quantum field operator satisfying the equation, the
constraint $\nabla\cdot A=0$, and the canonical commutation relations.

It is convenient to use the relativistic Hilbert space which is (vector
valued)

${\cal H}=L^{2}(\mathbb{R}^{3},(2\omega)^{-1}dp)$ (14)

again with $\omega(p)=|p|$. On this space define a projection operator.

$(Ph)_{j}(p)=\sum^{3}_{k=1}\Big{(}\delta_{jk}-\frac{p_{j}p_{k}}{|p|^{2}}\Big{)}h_{k}(p)$
(15)

which is the orthogonal projection onto the space of transverse functions
satisfying $p\cdot h(p)=0$. The one photon Hilbert space is ${\cal
H}_{T}=P{\cal H}$, the n-photon Hilbert space ${\cal H}_{T}^{n}$ is the n-fold
symmetric tensor product of ${\cal H}_{T}$ with itself, and the Fock space
over ${\cal H}_{T}$ is

${\cal F}({\cal H}_{T})=\bigoplus_{n=0}^{\infty}{\cal H}_{T}^{n}$ (16)

For $h\in{\cal H}_{T}$ let $a^{*}(h)$ and $a(h)$ be the creation and
annihilation operators on the Fock space, with $a(h)$ anti-linear in $h$ and
$a^{*}(h)$ linear in $h$. These operators (and all operators in this section)
are defined on the dense subspace ${\cal D}$ consisting elements of Fock space
with a finite number of particles. They satisfy
$[a(h_{1}),a^{*}(h_{2})]=(h_{1},h_{2})$. We also define momentum space fields
$A_{\circ}(h)$ and $\Pi_{\circ}(h)=A_{\circ}(ih)$ by

$\begin{split}A_{\circ}(h)=&\frac{1}{\sqrt{2}}\Big{(}a^{*}(h)+a(h)\Big{)}\\\
\Pi_{\circ}(h)=&\frac{i}{\sqrt{2}}\Big{(}a^{*}(h)-a(h)\Big{)}\\\ \end{split}$
(17)

These are real linear in $h$, symmetric, and satisfy
$[A_{\circ}(h_{1}),A_{\circ}(h_{2})]=i\textrm{Im}(h_{1},h_{2})$.

The field operator is defined as a distribution. For a real test function
$f\in{\cal S}(\mathbb{R}^{3})$ define $A(f,t)$ (formally $\int f(x)A(x,t)dx$)
by

$A(f,t)=\sqrt{2}A_{\circ}\Big{(}e^{i\omega
t}\tilde{f}_{T}\Big{)}=a^{*}(e^{i\omega t}\tilde{f}_{T})+a(e^{i\omega
t}\tilde{f}_{T})$ (18)

where $\tilde{f}_{T}\equiv P\tilde{f}$. This satisfies the field equation

$\frac{\partial^{2}A}{\partial t^{2}}(f,t)-A(\Delta f,t)=0$ (19)

It also satisfies $\nabla\cdot A=0$ since for scalar $g\in{\cal
S}(\mathbb{R}^{3})$ we have $(\nabla\cdot A)(g,t)\equiv-A(\nabla g,t)$ and
$(\nabla g)_{T}^{\sim}(p)=iP(p)p\tilde{g}(p)=0$. We have the commutation
relations

$[A(f_{1},t),A(f_{2},t)]=2\textrm{Im}(\tilde{f}_{1,T},\tilde{f}_{2,T})=0$ (20)

The vanishing follows since for $f$ real
$\overline{\tilde{f}(p)}=\tilde{f}(-p)$ so $(\tilde{f}_{1,T},\tilde{f}_{2,T})$
is real.

The field has a conjugate momentum $\Pi(h,t)=\partial/\partial t\ A(h,t)$
given by

$\Pi(f,t)=\sqrt{2}\Pi_{\circ}\Big{(}e^{i\omega
t}\omega\tilde{f}_{T}\Big{)}=i\Big{(}a^{*}(e^{i\omega
t}\omega\tilde{f}_{T})-a(e^{i\omega t}\omega\tilde{f}_{T})\Big{)}$ (21)

The $\Pi(f,t)$ commute with each other and satisfy the canonical commutation
relations (CCR)

$[A(f_{1},t),\Pi(f_{2},t)]=i(\tilde{f}_{1,T},2\omega\tilde{f}_{2,T})_{{\cal
H}}=i(\tilde{f}_{1,T},\tilde{f}_{2,T})_{L^{2}(\mathbb{R}^{3},dp)}$ (22)

###  2.3 the $C^{*}$ algebra and coherent states

We want to allow states for this system which are not Fock states, and for
this we focus on the algebra of the fields. See for example Bratteli-Robinson
[1] for more details on this standard material. We consider operators
$A_{\circ}(h)$ for $h$ in a dense subspace ${\cal S}\subset{\cal H}_{T}$. The
operator is essentially self-adjoint on the domain ${\cal D}\subset{\cal
F}({\cal H}_{T})$, and so we can define

$W_{\circ}(h)=e^{iA_{\circ}(h)}\hskip 28.45274pt\hskip 28.45274pth\in{\cal S}$
(23)

This satisfies

$W_{\circ}(h_{1})W_{\circ}(h_{2})=e^{-\frac{i}{2}\textrm{Im}(h_{1},h_{2})}W_{\circ}(h_{1}+h_{2})$
(24)

This is the Weyl form of the CCR. The operators $W_{\circ}(h)$ generate a
$C^{*}$ algebra, denoted $\EuFrak{A}$, which is the closure of the set of
operators which are finite sums $\sum_{n}c_{n}W_{\circ}(h_{n})$ in the Banach
space of bounded operators on ${\cal F}({\cal H}_{T})$. Different choices of
the dense subspace ${\cal S}$ give different algebras, and we will make a
specific choice later.

A state $\omega$ on the algebra is a positive linear functional of norm one.
Positivity is the condition that $\omega(B^{*}B)\geq 0$ for any
$B\in\EuFrak{A}$. Any normalized state in Fock space defines a state on the
algebra. In particular the Fock vacuum $\Omega_{0}$ defines a state by
$\omega_{0}(\ \cdot\ )=(\Omega_{0},[\ \cdot\ ]\Omega_{0})$. It satisfies
$a(h)\Omega_{0}=0$ and is characterized by

$\omega_{0}(W_{\circ}(h))=(\Omega_{0},e^{iA_{0}(h)}\Omega_{0})=(\Omega_{0},e^{ia^{*}(h)/\sqrt{2}}e^{ia(h)/\sqrt{2}}\Omega_{0})e^{-\frac{1}{4}\|h\|^{2}}=e^{-\frac{1}{4}\|h\|^{2}}$
(25)

Another class of states are the coherent states. Given $J\in{\cal H}_{T}$ and
the Fock vacuum $\Omega_{0}$ we define the coherent state

$\Omega_{J}=e^{-\frac{1}{2}\|J\|^{2}}e^{a^{*}(J)}\Omega_{0}=e^{-\frac{1}{2}\|J\|^{2}}\sum_{n=0}^{\infty}\frac{1}{n!}(a^{*}(J))^{n}\Omega_{0}$
(26)

The series converges in Fock space since
$\|(a^{*}(J))^{n}\Omega_{0}\|=\sqrt{n!}\|J\|^{n}$. It is normalized to
$\|\Omega_{J}\|=1$ and is an eigenstate of the annihilation operator:

$a(h)\Omega_{J}=(h,J)\Omega_{J}$ (27)

As a state on the algebra $\omega_{h}=(\Omega_{h},[\ \cdot\ ]\Omega_{h})$ it
satisfies

$\begin{split}\omega_{J}(W_{\circ}(h))=&(\Omega_{J},e^{iA_{0}(h)}\Omega_{J})\\\
=&(\Omega_{J},e^{ia^{*}(h)/\sqrt{2}}e^{ia(h)/\sqrt{2}}\Omega_{J})e^{-\frac{1}{4}\|h\|^{2}}\\\
=&e^{i\overline{(h,J)}/\sqrt{2}}e^{i(h,J)/\sqrt{2}}e^{-\frac{1}{4}\|h\|^{2}}\\\
=&e^{-\frac{1}{4}\|h\|^{2}+i\sqrt{2}\textrm{Re}(J,h)}\\\ \end{split}$ (28)

More generally we define a coherent state on $\EuFrak{A}$ to be a state
satisfying

$\omega(W_{\circ}(h))=e^{-\frac{1}{4}\|h\|^{2}+i\sqrt{2}\textrm{Re}L(h)}$ (29)

for some linear function $L$ on the dense domain ${\cal S}$. It can also be
written

$\omega(W_{\circ}(h))=\omega_{0}(W_{\circ}(h))e^{i\sqrt{2}\textrm{Re}L(h)}$
(30)

The positivity condition is satisfied since for any finite sequence of complex
numbers $c_{n}$ and $h_{n}\in{\cal S}$ and $B=\sum_{n}c_{n}W_{\circ}(h_{n})$
we have

$\begin{split}\omega(B^{*}B)=&\sum_{n,m}\bar{c}_{m}c_{n}\
\omega\Big{(}W_{\circ}(h_{n}-h_{m})\Big{)}e^{\frac{i}{2}\textrm{Im}(h_{n},h_{m})}\\\
=&\sum_{n,m}\Big{(}\bar{c}_{m}e^{-i\sqrt{2}\textrm{Re}L(f_{m})}\Big{)}\Big{(}c_{n}e^{i\sqrt{2}\textrm{Re}L(f_{n})}\Big{)}\
\omega_{0}\Big{(}W_{\circ}(h_{n}-h_{m})\Big{)}e^{\frac{i}{2}\textrm{Im}(h_{n},h_{m})}\\\
=&\omega_{0}(B_{0}^{*}B_{0})\geq 0\\\ \end{split}$ (31)

where

$B_{0}=\sum_{n}c_{n}e^{i\sqrt{2}\textrm{Re}L(f_{n})}W_{\circ}(h_{n})$ (32)

The coherent state $\omega$ defined by $L$ is known to be a pure state
(Honegger-Rapp [3]). Furthermore $\omega$ is a Fock state if and only if $L$
is continuous on the pre-Hilbert space ${\cal S}$ (Honegger-Rieckers [2]).
This is the criterion we will use in the following.

##  3 Regular source

To warm up we first consider our problem with a regular source. We take
$j\in{\cal C}^{\infty}_{0}(\mathbb{R}^{4})$ and seek quantum field operators
which solve $\square A=j_{T}$ and $\nabla\cdot A=0$. Note that partial Fourier
transform $\tilde{j}(p,t)$ is bounded and rapidly decreasing in $p$ as is
$\tilde{j}_{T}(p,t)=P(p)\tilde{j}(p,t)$ The inverse Fourier transform
$j_{T}(x,t)$ then has compact support in $t$, and for each $t$ is bounded and
continuous in $x$ (since $\tilde{j}(p,t)$ is integrable in $p$) and square
integrable in $x$ (since $\tilde{j}(p,t)$ is square integrable in $p$). The
localization of $j_{T}(x,t)$ in $x$ is considerably weaker than that of
$j(x,t)$.

We want a solution which is a free field in the distant past and the distant
future. Therefore we take $A_{\textrm{in}}(f,t)$ to be a free field defined
from creation and annihilation operators
$a^{*}_{\textrm{in}}(h),a_{\textrm{in}}(h)$ as in (18). Then for real
$f\in{\cal S}(\mathbb{R}^{3})$ we define

$A(f,t)=A_{\textrm{in}}(f,t)+(G_{+}*j_{T})(f,t)$ (33)

The expression $G_{+}*j_{T}$ can be analyzed pointwise. Indeed we have
explicitly

$(G_{\pm}*j_{T})(x,t)=\pm(2\pi)^{-\frac{3}{2}}i\int\theta(\pm(t-s))i\Big{(}e^{-i\omega(t-s)}-e^{i\omega(t-s)t}\Big{)}e^{ipx}\tilde{j}_{T}(p,s)\frac{dp}{2\omega}\
ds$ (34)

Then $(G_{+}*j_{T})(x,t)$ vanishes in the distant past since $j_{T}(p,s)$
does, and hence $A(f,t)=A_{\textrm{in}}(f,t)$ in the distant past. We have
$\square(G*j_{T})=j_{T}$, also in the sense of distributions, and hence
$\square A=j_{T}$. Furthermore $\nabla\cdot(G*j_{T})$, also in the sense of
distributions, and hence $\nabla\cdot A=0$

Next we define $A_{\textrm{out}}(f,t)$ by

$A(f,t)=A_{\textrm{out}}(f,t)+(G_{-}*j_{T})(f,t)$ (35)

Then $A(f,t)=A_{\textrm{out}}(f,t)$ in the distant future since
$(G_{-}*j_{T})(f,t)$ vanishes in the distant future. Combining (33) and (35)
and identifying $G=G_{+}-G_{-}$ gives

$A_{\textrm{out}}(f,t)=A_{\textrm{in}}(f,t)+(G*j_{T})(f,t)$ (36)

where now

$\begin{split}(G*j_{T})(x,t)&=(2\pi)^{-\frac{3}{2}}i\int\Big{(}e^{-i\omega(t-s)}-e^{i\omega(t-s)}\Big{)}e^{ipx}\tilde{j}_{T}(p,s)\frac{dp}{2\omega}\
ds\\\ &=(2\pi)^{-1}\ i\int\Big{(}e^{-i\omega
t}\hat{j}_{T}(\omega,p)-e^{i\omega
t}\hat{j}_{T}(-\omega,p)\Big{)}e^{ipx}\frac{dp}{2\omega}\\\ \end{split}$ (37)

where $\hat{j}$ is the four dimensional Fourier transform with Lorentz inner
product:

$\hat{j}(p_{0},p)=(2\pi)^{-2}\int e^{i(p_{0}t-px)}j(x,t)\
dxdt=(2\pi)^{-\frac{1}{2}}\int e^{ip_{0}t}\tilde{j}(p,t)dt$ (38)

Then $G*j_{T}$ satisfies $\square(G*j_{T})=0$ and $\nabla\cdot(G*j_{T})=0$,
and hence so does $A_{\textrm{out}}(f,t)$ Furthermore together with
$\Pi_{\textrm{out}}(f,t)=\partial A_{\textrm{out}}(f,t)/\partial t$ it
satisfies the CCR. Thus $A_{\textrm{out}}(f,t)$ is a free field.

In the second term in (37) make the change of variable $p\to-p$. Since $j$ is
real we have $\overline{\hat{j}_{T}(\omega,p)}=\hat{j}_{T}(-\omega,-p)$ and
the second term is identified as the complex conjugate of the first term. Thus
if we define

$J(p)=-\sqrt{2\pi}\ i\ \hat{j}(\omega,p)$ (39)

and $J_{T}(p)=P(p)J(p)$ then we have

$(G*j_{T})(x,t)=-(2\pi)^{-\frac{3}{2}}2\ \textrm{Re}\ \int e^{-i(\omega
t-px)}J_{T}(p)\frac{dp}{2\omega}$ (40)

This gives

$(G*j_{T})(f,t)=-2\ \textrm{Re}\ \int\tilde{f}(-p)e^{-i\omega
t}J_{T}(p)\frac{dp}{2\omega}=-2\ \textrm{Re}\ (e^{i\omega
t}\tilde{f}_{T},J_{T})$ (41)

Now we have

$A_{\textrm{out}}(f,t)=A_{\textrm{in}}(f,t)-2\ \textrm{Re}\ (e^{i\omega
t}\tilde{f}_{T},J_{T})$ (42)

We define creation and annihilation operators by taking what amounts to the
positive and negative frequency parts of the last expression. We define for
$h\in{\cal H}_{T}$

$\begin{split}a^{*}_{\textrm{out}}(h)=&a^{*}_{\textrm{in}}(h)-(J_{T},h)\\\
a_{\textrm{out}}(h)=&a_{\textrm{in}}(h)-(h,J_{T})\\\ \end{split}$ (43)

These satisfy the standard
$[a_{\textrm{out}}(h_{2}),a^{*}_{\textrm{out}}(h_{2})]=(h_{1},h_{2})$. Then

$\begin{split}a^{*}_{\textrm{out}}(e^{i\omega
t}\tilde{f}_{T})=&a^{*}_{\textrm{in}}(e^{i\omega
t}\tilde{f}_{T})-(J_{T},e^{i\omega t}\tilde{f}_{T})\\\
a_{\textrm{out}}(e^{i\omega t}\tilde{f}_{T})=&a_{\textrm{in}}(e^{i\omega
t}\tilde{f}_{T})-(e^{i\omega t}\tilde{f}_{T},J_{T})\\\ \end{split}$ (44)

If we add these we get $A_{in}(f,t)-2\ \textrm{Re}\ (e^{i\omega
t}\tilde{f}_{T},J_{T})=A_{out}(f,t)$ on the right. Thus we must get the same
on the left which says

$A_{\textrm{out}}(f,t)=a^{*}_{\textrm{out}}(e^{i\omega
t}\tilde{f}_{T})+a_{\textrm{out}}(e^{i\omega t}\tilde{f}_{T})$ (45)

just as for $A_{\textrm{in}}(f,t)$ .

The out vacuum in Fock space should satisfy
$a_{\textrm{out}}(h)\Omega_{\textrm{out}}=0$. Note that $j\in{\cal
C}^{\infty}_{0}(\mathbb{R}^{4})$ implies that $\hat{j}$ is bounded and rapidly
decreasing, hence $J_{T}\in{\cal H}_{T}$. Therefore we can construct the
coherent state

$\Omega_{\textrm{out}}=\Omega_{J_{T}}=e^{-\frac{1}{2}\|J_{T}\|^{2}}\exp(a_{in}^{*}(J_{T}))\Omega_{\textrm{in}}$
(46)

This gives the the desired

$a_{\textrm{out}}(h)\Omega_{\textrm{out}}=\Big{(}a_{\textrm{in}}(h)-(h,J_{T})\Big{)}\Omega_{\textrm{out}}=0$
(47)

We summarize:

######  Proposition 1.

Let $j\in{\cal C}^{\infty}_{0}(\mathbb{R}^{4})$. Then $J_{T}(p)=-\sqrt{2\pi}\
i\ \hat{j}_{T}(\omega,p)$ is in ${\cal H}_{T}$ and the out vacuum
$\Omega_{out}$ exists in Fock space.

Remarks.

  1. 1.

To obtain this result can certainly relax the condition that $j\in{\cal
C}^{\infty}_{0}(\mathbb{R}^{3})$. The key condition is is $J_{T}\in{\cal
H}_{T}$ which follows from $J\in{\cal H}$ which says

$\int|\hat{j}(\omega,p)|^{2}\frac{dp}{2\omega}<\infty$ (48)

This is similar to a condition that Naudts - De Roeck [6] found in the Lorentz
gauge.

  2. 2.

We sketch some further developments of the scattering theory associated with a
regular source. As in (17) define
$\Pi_{\circ,\textrm{in}}(h)=\frac{i}{\sqrt{2}}\Big{(}a_{\textrm{in}}^{*}(h)-a_{\textrm{in}}(h)\Big{)}$.
This is essentially self-adjoint on ${\cal D}$ and we define a unitary
operator by

$S=\exp\Big{(}i\sqrt{2}\Pi_{\circ,\textrm{in}}(J_{T})\Big{)}=\exp\Big{(}-a_{\textrm{in}}^{*}(J_{T})+a_{\textrm{in}}(J_{T})\Big{)}$
(49)

We have

$\begin{split}S^{-1}\Omega_{\textrm{in}}=&\exp\Big{(}a^{*}_{\textrm{in}}(J_{T})-a_{\textrm{in}}(J_{T})\Big{)}\Omega_{\textrm{in}}\\\
=&e^{-\frac{1}{2}\|J_{T}\|^{2}}\exp\Big{(}a^{*}_{\textrm{in}}(J_{T})\Big{)}\exp\Big{(}-a_{\textrm{in}}(J_{T})\Big{)}\Omega_{\textrm{in}}\\\
=&e^{-\frac{1}{2}\|J_{T}\|^{2}}\exp\Big{(}a^{*}_{\textrm{in}}(J_{T})\Big{)}\Omega_{\textrm{in}}=\Omega_{\textrm{out}}\\\
\end{split}$ (50)

and

$S^{-1}a_{\textrm{in}}(h)S=a_{\textrm{in}}(h)+[a_{\textrm{in}}^{*}(J_{T}),a_{\textrm{in}}(h)]=a_{\textrm{in}}(h)-(h,J_{T})=a_{\textrm{out}}(h)$
(51)

So the in and out fields are unitarily equivalent.

One can form other asymptotic states by applying creation operators
$a^{*}_{\textrm{out}}(h)$ to $\Omega_{\textrm{out}}$ and then form scattering
amplitudes like

$\Big{(}a^{*}_{\textrm{out}}(h^{\prime}_{1})\cdots
a^{*}_{\textrm{out}}(h^{\prime}_{m})\Omega_{\textrm{out}},a^{*}_{\textrm{in}}(h_{1})\cdots
a^{*}_{\textrm{in}}(h_{n})\Omega_{\textrm{in}}\Big{)}$ (52)

This can also be written

$\Big{(}a^{*}_{\textrm{in}}(h^{\prime}_{1})\cdots
a^{*}_{\textrm{in}}(h^{\prime}_{m})\Omega_{in},S\
a^{*}_{\textrm{in}}(h_{1})\cdots
a^{*}_{\textrm{in}}(h_{n})\Omega_{\textrm{in}}\Big{)}$ (53)

and $S$ is revealed as a scattering operator.

##  4 Point source - I

Again we study the equations $\square A=j_{T}$ and $\nabla\cdot A=0$ and look
for solutions which are free in the distant past and future. We now specialize
to a point source with current $j(t,x)=x^{\prime}(t)\delta(x-x(t))$. We assume
in this section that $x^{\prime}(t)$ has compact support. So the particle is
at rest in the distant past, accelerates for finite amount of time, and then
stops in the distant future.

Again we take $A_{\textrm{in}}(f,t)$ to be a free field as defined in (18) for
real $f\in{\cal S}(\mathbb{R}^{3})$. Then we define

$A(f,t)=A_{\textrm{in}}(f,t)+(G_{+}*j_{T})(f,t)$ (54)

Now we do not define $G_{+}*j_{T}$ pointwise, but treat it as a distribution
and proceed by throwing everything onto the test function $f$. Since
$G_{+}(x,t)=G_{-}(-x,-t)$ we interpret $(G_{+}*j_{T})(f,t)$ as

$\Big{\langle}G_{+}*j_{T},f\otimes\delta_{t}\Big{\rangle}=\Big{\langle}j_{T},G_{-}*(f\otimes\delta_{t})\Big{\rangle}$
(55)

Then formally we can transfer the transverse projection from $j$ and to $f$
and get

$\Big{\langle}j_{T},G_{-}*(f\otimes\delta_{t})\Big{\rangle}=\Big{\langle}j,G_{-}*(f_{T}\otimes\delta_{t})\Big{\rangle}$
(56)

The latter is well-defined for our current since
$G_{-}*(f_{T}\otimes\delta_{t})$ as given by (9) is a bounded function
continuous in the spatial variable. With this interpretation (54) becomes

$A(f,t)=A_{\textrm{in}}(f,t)+\Big{\langle}j,G_{-}*(f_{T}\otimes\delta_{t})\Big{\rangle}$
(57)

Since $j$ has compact support and $G_{-}*(f_{T}\otimes\delta_{t})$ vanishes to
the future of $t$, we have that $<j,G_{-}*(f_{T}\otimes\delta_{t})>$ vanishes
for $t$ sufficiently negative and so $A(f,t)=A_{\textrm{in}}(f,t)$ in the
distant past. Since $A_{\textrm{in}}(f,t)$ is free and since $G_{-}$ is a
fundamental solution we have

$\begin{split}\square
A_{\textrm{out}}(f,t)=&\frac{\partial^{2}A_{\textrm{out}}}{\partial
t^{2}}(f,t)-A_{\textrm{out}}(\Delta f,t)\\\
=&\Big{\langle}j,\frac{\partial^{2}}{\partial
t^{2}}G_{-}*(f_{T}\otimes\delta_{t})-G_{-}*((\Delta
f)_{T}\otimes\delta_{t})\Big{\rangle}\\\
=&\Big{\langle}j,\square(G_{-}*(f_{T}\otimes\delta_{t}))\Big{\rangle}=\Big{\langle}j,f_{T}\otimes\delta_{t}\Big{\rangle}\equiv\Big{\langle}j_{T},f\otimes\delta_{t}\Big{\rangle}\\\
\end{split}$ (58)

So $A$ satisfies $\square A=j_{T}$ in the sense of distributions. Similarly
$\nabla\cdot A=0$.

Next we define $A_{\textrm{out}}(f,t)$ by

$A(f,t)=A_{\textrm{out}}(f,t)+G_{-}*j_{T}(f,t)$ (59)

which is interpreted as

$A(f,t)=A_{\textrm{out}}(f,t)+\Big{\langle}j,G_{+}*(f_{T}\otimes\delta_{t})\Big{\rangle}$
(60)

Then $A(f,t)=A_{\textrm{out}}(f,t)$ in the distant future.

Combining (57) and (60) gives

$A_{\textrm{out}}(f,t)=A_{\textrm{in}}(f,t)-\Big{\langle}j,G*(f_{T}\otimes\delta_{t})\Big{\rangle}$
(61)

Then $A_{\textrm{out}}$ satisfies $\square A_{\textrm{out}}=0$ and
$\nabla\cdot A_{\textrm{out}}=0$ in the sense of distributions and is a free
field.

Now take the expression $(G*(f_{T}\otimes\delta_{t}))(x,s)=u_{f_{T}}(x,s-t)$
from (10) and find

$\begin{split}\Big{\langle}j,G*(f_{T}\otimes\delta_{t})\Big{\rangle}=&\int
x^{\prime}(s)\delta(x-x(s))u_{f_{T}}(x,s-t)dxds\\\ =&\int
x^{\prime}(s)u_{f_{T}}(x(s),s-t)ds\\\ =&(2\pi)^{-\frac{3}{2}}i\int
x^{\prime}(s)\Big{(}e^{-i\omega(s-t)}-e^{i\omega(s-t)}\Big{)}e^{ipx(s)}\tilde{f}_{T}(p)\frac{dp}{2\omega}\
ds\\\ =&(2\pi)^{-\frac{3}{2}}\int
x^{\prime}(s)\Big{[}ie^{-i\omega(s-t)+ipx(s)}\tilde{f}_{T}(p)-ie^{i\omega(s-t)-ipx(s)}\tilde{f}_{T}(-p)\Big{]}\frac{dp}{2\omega}\
ds\\\ =&(2\pi)^{-\frac{3}{2}}2\ \textrm{Re}\ \int
x^{\prime}(s)\Big{[}-ie^{i\omega(s-t)-ipx(s)}\tilde{f}_{T}(-p)\Big{]}\
\frac{dp}{2\omega}ds\\\ \end{split}$ (62)

Now change the order of integration (easily justified), and define (new
definition)

$J(p)=-i(2\pi)^{-\frac{3}{2}}\int x^{\prime}(s)e^{i\omega s-ipx(s)}ds$ (63)

Then

$\Big{\langle}j,G*(f_{T}\otimes\delta_{t})\Big{\rangle}=2\ \textrm{Re}\ \int
e^{-i\omega t}\tilde{f}_{T}(-p)J(p)\frac{dp}{2\omega}=2\ \textrm{Re}\
(e^{i\omega t}\tilde{f}_{T},J_{T})$ (64)

Then integral converges since $J_{T}(p)$ is bounded and $\tilde{f}_{T}(p)$ is
bounded and rapidly decreasing. We have written it as an inner product in
${\cal H}$ but are not asserting that $J_{T}\in{\cal H}_{T}$, at least not
yet. Now we have as before

$A_{\textrm{out}}(f,t)=A_{\textrm{in}}(f,t)-2\ \textrm{Re}\ (J_{T},e^{i\omega
t}\tilde{f}_{T})$ (65)

We consider the linear function $h\to(J_{T},h)$ for $h$ in a dense domain
${\cal S}\subset{\cal H}_{T}$ containing the functions $e^{i\omega
t}\tilde{f}_{T}$ for $f\in{\cal S}(\mathbb{R}^{3})$. It is the domain of of
rapidly decreasing functions:

${\cal S}=\\{h\in{\cal H}_{T}:\textrm{ for }k\geq 0\textrm{ there is
}C\textrm{ so }|h((p)|\leq C(1+|p|^{2})^{-k}\\}$ (66)

For $h\in{\cal S}$ we define

$\begin{split}a^{*}_{\textrm{out}}(h)=&a^{*}_{\textrm{in}}(h)-(J_{T},h)\\\
a_{\textrm{out}}(h)=&a_{\textrm{in}}(h)-(h,J_{T})\\\ \end{split}$ (67)

and then as in (43) - (45)

$A_{\textrm{out}}(f,t)=a^{*}_{\textrm{out}}(e^{i\omega
t}\tilde{f}_{T})+a_{\textrm{out}}(e^{i\omega t}\tilde{f}_{T})$ (68)

just as for $A_{\textrm{in}}(f,t)$.

Now we can write
$A_{\textrm{in}/\textrm{out}}(f,t)=\sqrt{2}A_{\circ,\textrm{in}/\textrm{out}}(e^{i\omega
t}\tilde{f}_{T})$ where the momentum space fields are
$A_{\circ,\textrm{in}/\textrm{out}}(h)=\frac{1}{\sqrt{2}}(a_{\textrm{in}/\textrm{out}}^{*}(h)+a_{\textrm{in}/\textrm{out}}(h))$.
In terms of these fields the defining relation (67) can be written

$A_{\circ,\textrm{out}}(h)=A_{\circ,\textrm{in}}(h)-\sqrt{2}\ \textrm{Re}\
(J_{T},h)$ (69)

With this identity we pass to the $C^{*}$ algebra. Define

$W_{\textrm{out}}(h)=e^{iA_{\circ,\textrm{out}}(h)}\hskip
28.45274ptW_{\textrm{in}}(h)=e^{iA_{\circ,\textrm{in}}(h)}$ (70)

and then

$W_{\textrm{out}}(h)=W_{\textrm{in}}(h)e^{-i\sqrt{2}\ \textrm{Re}\ (J_{T},h)}$
(71)

These are elements of $C^{*}$ algebra $\EuFrak{A}$ generated by
$W_{\textrm{in}}(h)$, $h\in{\cal S}$. In the Fock vacuum
$\omega_{in}=(\Omega_{0,in},[\ \cdot.\ ]\Omega_{0,in})$ we have

$\omega_{in}(W_{\textrm{out}}(h))=\omega_{\textrm{in}}(W_{\textrm{in}}(h))e^{-i\sqrt{2}\
\textrm{Re}(\bar{J}_{T},h)}$ (72)

But $W_{\textrm{out}}(h)$ is again a representation of the CCR and generates
the same $C^{*}$ algebra $\EuFrak{A}$ as $W_{\textrm{in}}(h)$. Then there is a
unique $*$-automorphism $\alpha$ on $\EuFrak{A}$ such that
$\alpha(W_{\textrm{out}}(h))=W_{\textrm{in}}(h)$ (see for example [1]). This
satisfies $\alpha(W_{\textrm{in}}(h))=W_{\textrm{in}}(h)e^{i\sqrt{2}\
\textrm{Re}(\bar{J}_{T},h)}$. If we define a new state by
$\omega_{\textrm{out}}\equiv\omega_{\textrm{in}}\circ\alpha$ then

$\omega_{\textrm{out}}(W_{\textrm{in}}(h))=\omega_{\textrm{in}}(W_{\textrm{in}}(h))e^{i\sqrt{2}\
\textrm{Re}(J_{T},h)}$ (73)

Instead of changing operators we have changed states. The form of the equation
shows that $\omega_{\textrm{out}}$ is a coherent state by our ${\cal C}^{*}$
algebra definition.

The question is now whether $\omega_{\textrm{out}}$ is a Fock state. Before
answering the question we sharpen our criterion for when this is the case.

######  Proposition 2.

$\omega_{\textrm{out}}$ is a Fock state if and only if $J_{T}\in{\cal H}_{T}$.

Proof. We have already noted that $\omega_{\textrm{out}}$ is a Fock state if
and only if the linear function $h\to(J_{T},h)$ is continuous on the pre-
Hilbert space ${\cal S}$. We now claim this is equivalent to $J_{T}\in{\cal
H}_{T}$. If $J_{T}\in{\cal H}_{T}$ then continuity is immediate. If the
continuity holds then there is $J^{\prime}\in{\cal H}_{T}$ such that
$(J_{T},h)=(J^{\prime},h)$ for $h\in{\cal S}$. If $h$ rapidly decreasing then
$h_{T}\in{\cal S}$ and $(J_{T}-J^{\prime},h)=(J_{T}-J^{\prime},h_{T})=0$. Then
$(J_{T}-J^{\prime},h)=0$ for $h\in{\cal S}(\mathbb{R}^{3})$. Hence
$(J_{T}-J^{\prime})/2\omega$ as distributions and hence almost everywhere as
functions. Thus $J_{T}=J^{\prime}$ almost everywhere and so $J_{T}\in{\cal
H}_{T}$. This completes the proof.

In the following we assume that the velocity of the charged particle is less
than the speed of light which is here one.

######  Proposition 3.

Let $x(t)$ be ${\cal C}^{3}$ with $|x^{\prime}(t)|\leq v_{0}<1$, and suppose
$x^{\prime}(t)$ has compact support. Then $J_{T}\in{\cal H}_{T}$ and
$\omega_{\textrm{out}}$ is the Fock state defined by
$\Omega_{\textrm{out}}=\Omega_{J_{T}}$.

Proof. It suffices to show $J\in{\cal H}$. Ignoring the multiplicative
constants in (63) we might as well assume that

$J(p)=\int x^{\prime}(t)e^{i\psi(p,t)}dt\hskip 28.45274pt\hskip
28.45274pt\psi(p,t)=\omega t-p\cdot x(t)$ (74)

Since this is a bounded continuous function of $p$, to decide whether it is in
${\cal H}$ we need to study asymptotics as $p\to\infty$. Our first thought
might be the stationary phase method. So we look for points where

$\psi_{t}(p,t)=\omega-p\cdot x^{\prime}(t)=0$ (75)

However $|p\cdot x^{\prime}(t)|<|p|v_{0}=\omega v_{0}$ and

$\omega(1-v_{0})\leq|\psi_{t}(p,t)|\leq\omega(1+v_{0})$ (76)

There is no point of stationary phase for $p\neq 0$. Instead we can integrate
by parts and write with $\psi_{tt}=-p\cdot x^{\prime\prime}(t)$

$\begin{split}J(p)=&\int
x^{\prime}(t)\frac{1}{i\psi_{t}}\frac{\partial}{\partial t}e^{i\psi(p,t)}dt\\\
=&\frac{1}{i}\int\left[\frac{\psi_{tt}}{\psi^{2}_{t}}x^{\prime}(t)-\frac{1}{\psi_{t}}x^{\prime\prime}(t)\right]e^{i\psi(p,t)}dt\\\
\end{split}$ (77)

Now $x^{\prime},x^{\prime\prime}$ are bounded and $\psi_{t}^{-1}$ and
$\psi_{tt}/\psi^{2}_{t}$ are ${\cal O}(\omega^{-1})$ for $|p|\geq 1$. Hence
$J(p)={\cal O}(\omega^{-1})$. This is not quite enough for convergence, but we
can repeat the last step and get

$\begin{split}J(p)=&\frac{1}{i}\int\left[\frac{\psi_{tt}}{\psi^{2}_{t}}x^{\prime}(t)-\frac{1}{\psi_{t}}x^{\prime\prime}(t)\right]\frac{1}{i\psi_{t}}\frac{\partial}{\partial
t}e^{i\psi(p,t)}dt\\\
=&\int\left[\Big{(}\frac{\psi_{ttt}}{\psi^{3}_{t}}-\frac{3\psi_{tt}^{2}}{\psi_{t}^{4}}\Big{)}x^{\prime}(t)+\frac{3\psi_{tt}}{\psi^{3}_{t}}x^{\prime\prime}(t)-\frac{1}{\psi_{t}^{2}}x^{\prime\prime\prime}(t)\right]e^{i\psi(p,t)}dt\end{split}$
(78)

Now $x^{\prime},x^{\prime\prime},x^{\prime\prime\prime}$ are all bounded and
the coefficients are all ${\cal O}(\omega^{-2})$ for $|p|\geq 1$. Thus
$J(p)={\cal O}(\omega^{-2})$ and we have

$\int_{|p|\geq 1}|J(p)|^{2}\frac{dp}{2\omega}\leq\textrm{const}\int_{|p|\geq
1}\omega^{-5}dp<\infty$ (79)

Hence $J\in{\cal H}$ and the proof is complete.

Remark. We can allow jump discontinuities in $x^{\prime\prime\prime}(t)$
without changing the result. Suppose we allow a jump discontinuity in
$x^{\prime\prime}(t)$, say at $t=0$. Then in the integration by parts in (78)
we get a boundary term which is

$\left[\frac{p\cdot(x^{\prime\prime}(0^{-})-x^{\prime\prime}(0^{+}))x^{\prime}(0)}{\psi^{3}_{t}(p,0)}+\frac{x^{\prime\prime}(0^{-})-x^{\prime\prime}(0^{+})}{\psi^{2}_{t}(p,0)}\right]e^{i\psi(p,0)}$
(80)

This is still ${\cal O}(\omega^{-2})$ and our results still hold. However a
discontinuity in the velocity $x^{\prime}(t)$ is more serious as we now show.

######  Proposition 4.

Suppose $x(t)$ is continuous on $\mathbb{R}$ and ${\cal C}^{3}$ on
$(-\infty,0]$ and $[0,\infty)$ with

$x^{\prime}(0^{-})-x^{\prime}(0^{+})\neq 0$ (81)

Suppose also $x^{\prime}(t)$ has compact support and $|x^{\prime}(t)|\leq
v_{0}<1$. Then $J_{T}\notin{\cal H}_{T}$ and $\omega_{out}$ is not a Fock
state

Proof. Since $J(p)$ is still bounded, it again suffices to consider
asymptotics as $p\to\infty$. In the integration by parts in (77) there is now
a boundary term and we have

$\begin{split}J(p)=&\int
x^{\prime}(t)\frac{1}{i\psi_{t}}\frac{\partial}{\partial t}e^{i\psi(p,t)}dt\\\
=&\frac{1}{i}\int\left[\frac{\psi_{tt}}{\psi^{2}_{t}}x^{\prime}(t)-\frac{1}{\psi_{t}}x^{\prime\prime}(t)\right]e^{i\psi(p,t)}dt+\frac{1}{i}\Delta(p)e^{i\psi(p,0)}\\\
\end{split}$ (82)

where

$\Delta(p)=\frac{x^{\prime}(0^{-})}{\omega-p\cdot
x^{\prime}(0^{-})}-\frac{x^{\prime}(0^{+})}{\omega-p\cdot x^{\prime}(0^{+})}$
(83)

The first term in (82) is analyzed as in the previous proposition, unaffected
by possible discontinuities in $x^{\prime\prime},x^{\prime\prime\prime}$, and
the transverse part is in ${\cal H}_{T}$. Thus to prove our result it suffices
to show that $\Delta_{T}$ is not in ${\cal H}_{T}$.

Let $v^{-}=x^{\prime}(0^{-})$ and $v^{+}=x^{\prime}(0^{+})$. Then this can be
written

$\Delta(p)=\frac{\omega(v^{-}-v^{+})+(v^{+}p\cdot v^{-}-v^{-}p\cdot
v^{+})}{(\omega-p\cdot v^{-})(\omega-p\cdot v^{+})}$ (84)

Now $v^{+}p\cdot v^{-}-v^{-}p\cdot v^{+}=p\times(v^{+}\times v^{-})$. If we
define $\delta v=v^{-}-v^{+}\neq 0$ and $n_{p}=p/\omega=p/|p|$ then
$\Delta(p)=\Delta_{1}(p)+\Delta_{2}(p)$ where

$\begin{split}\Delta_{1}(p)=&\frac{1}{\omega}\left[\frac{\delta
v}{(1-n_{p}\cdot v^{-})(1-n_{p}\cdot v^{+})}\right]\\\
\Delta_{2}(p)=&\frac{1}{\omega}\left[\frac{n_{p}\times(v^{+}\times
v^{-})}{(1-n_{p}\cdot v^{-})(1-n_{p}\cdot v^{+})}\right]\\\ \end{split}$ (85)

Note that the denominators here are bounded below since $|v^{\pm}|<1$. Now
$\Delta_{2}(p)$ is already transverse, and the transverse part of
$\Delta_{1}(p)$ is with $n_{\delta v}=\delta v/|\delta v|$

$\Delta_{1,T}(p)=\frac{1}{\omega}\left[\frac{\delta v-n_{p}(n_{p}\cdot\delta
v)}{(1-n_{p}\cdot v^{-})(1-n_{p}\cdot v^{+})}\right]=\frac{|\delta
v|}{\omega}\left[\frac{n_{\delta v}-n_{p}(n_{p}\cdot n_{\delta
v})}{(1-n_{p}\cdot v^{-})(1-n_{p}\cdot v^{+})}\right]$ (86)

We have now $\Delta_{T}(p)=\Delta_{1,T}(p)+\Delta_{2}(p)$.

Now we divide into two cases. Either $v^{\pm}$ are colinear or not. If they
are colinear then $\Delta_{2}(p)=0$, and we compute

$\begin{split}\int_{|p|\geq
1}|\Delta_{T}(p)|^{2}\frac{dp}{2\omega}&\geq\int_{|p|\geq 1,|n_{p}\cdot
n_{\delta v}|\leq\frac{1}{2}}|\Delta_{1,T}(p)|^{2}\frac{dp}{2\omega}\\\
&\geq\textrm{const}|\delta v|^{2}\int_{|p|\geq 1,|n_{p}\cdot n_{\delta
v}|\leq\frac{1}{2}}\frac{1}{\omega^{3}}dp=\infty\\\ \end{split}$ (87)

If $v^{\pm}$ are not colinear then $v_{\times}\equiv v^{+}\times v^{-}\neq 0$.
We look at the component of $\Delta_{T}(p)$ along $v_{\times}$. Again
$\Delta_{2}(p)$ does not contribute, nor does the $n_{\delta v}$ term in
$\Delta_{1,T}(p)$. We have

$n_{v_{\times}}\cdot\Delta_{T}(p)=\frac{|\delta
v|}{\omega}\left[\frac{-(n_{p}\cdot n_{v_{\times}})(n_{p}\cdot n_{\delta
v})}{(1-n_{p}\cdot v^{-})(1-n_{p}\cdot v^{+})}\right]$ (88)

Then

$\begin{split}\int_{|p|\geq
1}|\Delta_{T}(p)|^{2}\frac{dp}{2\omega}&\geq\int_{|p|\geq 1,|n_{p}\cdot
n_{v_{\times}}|\geq\frac{1}{2},|n_{p}\cdot n_{\delta
v}|\geq\frac{1}{2}}|n_{v_{\times}}\cdot\Delta_{T}(p)|^{2}\frac{dp}{2\omega}\\\
&\geq\textrm{const}|\delta v|^{2}\int_{|p|\geq 1,|n_{p}\cdot
n_{v_{\times}}|\geq\frac{1}{2},|n_{p}\cdot n_{\delta
v}|\geq\frac{1}{2}}\frac{1}{\omega^{3}}\ dp=\infty\\\ \end{split}$ (89)

The last step follows since $\delta v$ orthogonal to $v_{\times}$ implies the
cones $|n_{v_{\times}}\cdot n_{p}|\geq\frac{1}{2}$ and $|n_{p}\cdot n_{\delta
v}|\geq\frac{1}{2}$ have a non-empty intersection. This completes the proof.

##  5 Point source - II

Now we consider a trajectory $x(t)$ which has constant velocity
$v_{\textrm{in}}$ in the distant past, and constant velocity
$v_{\textrm{out}}$ in the distant future, and is otherwise smooth. So we no
longer have compact support for $x^{\prime}(t)$, but we do have compact
support for $x^{\prime\prime}(t)$.

There is now a problem with the convergence of integrals like (63). To avoid
this we approximate the current $j$ by

$j_{R}(x,t)=\begin{cases}x^{\prime}(t)\delta(x-x(t))&\hskip 28.45274pt|t|\leq
R\\\ 0&\hskip 28.45274pt\textrm{ otherwise }\\\ \end{cases}$ (90)

and seek to take the limit $R\to\infty$. The analysis proceeds as in the the
previous section and we have
$A_{\textrm{out}}(f,t)=A_{in}(f,t)-\Big{\langle}j_{R},G*(f_{T}\otimes\delta_{t})\Big{\rangle}$
where

$\begin{split}\Big{\langle}j_{R},G*(f_{T}\otimes\delta_{t})\Big{\rangle}=&(2\pi)^{-\frac{3}{2}}2\
\textrm{Re}\
\int_{-R}^{R}x^{\prime}(s)\int\Big{[}-ie^{i\omega(s-t)-ipx(s)}\tilde{f}_{T}(-p)\Big{]}\frac{dp}{2\omega}\
ds\\\ \end{split}$ (91)

######  Proposition 5.

Let $x(t)$ be ${\cal C}^{3}$ with $|x^{\prime}(t)|\leq v_{0}<1$, and suppose
$x^{\prime\prime}(t)$ has compact support. Then the limit $R\to\infty$ exists
on the right side of (91) and defines the left side at $R=\infty$. We have as
in (64)

$\Big{\langle}j,G*(f_{T}\otimes\delta_{t})\Big{\rangle}=2\ \mathrm{Re}\ \int
e^{-i\omega t}\tilde{f}_{T}(-p)J(p)\frac{dp}{2\omega}=2\ \mathrm{Re}\
(e^{i\omega t}\tilde{f}_{T},J_{T})$ (92)

where up to a multiplicative constant

$J(p)=\frac{1}{i}\int\left[\frac{\psi_{tt}}{\psi^{2}_{t}}x^{\prime}(t)-\frac{1}{\psi_{t}}x^{\prime\prime}(t)\right]e^{i\psi(p,t)}dt$
(93)

as in (77), and the representation (78) for $J(p)$ holds as well.

Proof. Note that since $\psi_{tt}=-p\cdot x^{\prime\prime}(t)$ the integrand
on the right side of (93) has compact support in $t$ and that integral exists.
Furthermore $J(p)$ is ${\cal O}(\omega^{-1})$ everywhere and
$\tilde{f}_{T}(p)$ is bounded and rapidly decreasing so the integral (92)
converges. Once (93) is established (78) holds by a further integration by
parts.

To derive (92) it suffices to consider the case $t=0$. Then in (91) with
$\psi(p,s)=\omega s-p\cdot x(s)$ we need to find an $R\to\infty$ limit for

$\begin{split}&\int_{-R}^{R}x^{\prime}(s)\left[\int
e^{i\psi(p,s)}\tilde{f}_{T}(-p)\frac{dp}{2\omega}\right]ds\\\
=&\int_{-R}^{R}x^{\prime}(s)\frac{d}{ds}\left[\int\frac{1}{i\psi_{s}}e^{i\psi(p,s)}\tilde{f}_{T}(-p)\frac{dp}{2\omega}\right]ds+\int_{-R}^{R}x^{\prime}(s)\left[\int\frac{\psi_{ss}}{i\psi^{2}_{s}}e^{i\psi(p,s)}\tilde{f}_{T}(-p)\frac{dp}{2\omega}\right]ds\\\
=&\frac{1}{i}\int_{-R}^{R}\left[\int\left(\frac{\psi_{ss}}{\psi^{2}_{s}}x^{\prime}(s)-\frac{1}{\psi_{s}}x^{\prime\prime}(s)\right)e^{i\psi(p,s)}\tilde{f}_{T}(-p)\frac{dp}{2\omega}\right]ds\\\
&+\left[x^{\prime}(s)\int\frac{1}{i\psi_{s}}e^{i\psi(p,s)}\tilde{f}_{T}(-p)\frac{dp}{2\omega}\right]_{s=-R}^{s=R}\\\
\end{split}$ (94)

In the double integral the $R\to\infty$ limit exists by the support of
$x^{\prime\prime}(t)$. The integrand is ${\cal O}(\omega^{-2})$ as $|p|\to 0$
and rapidly decreasing as $|p|\to\infty$, so the integral is is absolutely
convergent. Changing the order of integration gives the result (92), provided
the endpoint term goes to zero.

For the endpoint term we need to show
$\int\psi_{t}^{-1}e^{i\psi(t,p)}\tilde{f}_{T}(-p)dp/2\omega$ goes to zero as
$|t|\to\infty$. As $t\to\infty$ we have $x^{\prime}(t)\to v_{\textrm{out}}$ so
it suffices to show

$\int\frac{1}{\omega-p\cdot v_{\textrm{out}}}e^{i(\omega-p\cdot
v_{\textrm{out}})t}\tilde{f}_{T}(-p)\frac{dp}{2\omega}$ (95)

goes to zero as $t\to\infty$. Let $p\to-p$ and then write this in polar
coordinates $(|p|,\theta,\phi)$ with $v_{\textrm{out}}$ as the $z$-axis. Then
$\omega+p\cdot v_{\textrm{out}}=\omega(1+\cos\theta\ v_{\textrm{out}})$ and
$(1+\cos\theta\ v_{\textrm{out}})$ is bounded above and below. The integral is
now

$\int\frac{1}{2(1+\cos\theta\ v_{\textrm{out}})}\Big{[}\int
e^{i|p|(1+\cos\theta\ v_{\textrm{out}})t}\tilde{f}_{T}(|p|,\theta,\phi)\
d|p|\Big{]}\sin\theta d\theta d\phi$ (96)

Now the interior integral goes to zero by the Riemann- Lebesgue lemma, and the
overall integral goes to zero by dominated convergence. The limit
$t\to-\infty$ is entirely similar. This completes the proof.

Having established (92), (93) we proceed as in (65) - (73) and again find

$\omega_{\textrm{out}}(W_{\textrm{in}}(h))=\omega_{\textrm{in}}(W_{\textrm{in}}(h))e^{i\sqrt{2}\
\textrm{Re}(J_{T},h)}$ (97)

######  Proposition 6.

Let $x(t)$ be ${\cal C}^{3}$ with $|x^{\prime}(t)|\leq v_{0}<1$, and suppose
$x^{\prime}(t)=v_{\textrm{in}}$ in the distant past and
$x^{\prime}(t)=v_{\textrm{out}}$ in the distant future so
$x^{\prime\prime}(t)$ has compact support. Then $\omega_{\textrm{out}}$ is a
Fock state if and only if $v_{in}=v_{\textrm{out}}$.

Proof. We must show that $J_{T}\in{\cal H}_{T}$ if and only if
$v_{in}=v_{\textrm{out}}$. We can use the representation (78) as before to get
convergence for $|p|\geq 1$. So the issue is whether the the integral for
$|p|\leq 1$ converges. (In proposition 3 convergence for $|p|\leq 1$ was
easy.) Our general bound is that $J(p)$ is ${\cal O}(\omega^{-1})$, as in
$J_{T}(p)$. We show that in case $v_{in}\neq v_{\textrm{out}}$ this is
actually sharp and gives a logarithmic divergence.

In the representation (93) for $|p|\leq 1$ we write $e^{i\psi(p,t)}=1+{\cal
O}(\omega)$ and correspondingly $J(p)=F(p)+{\cal O}(1)$. The ${\cal O}(1)$
term is in ${\cal H}$ and the transverse part in in ${\cal H}_{T}$, so it
suffices to consider $F(p)$ and show that $F_{T}(p)$ either is or is not
square integrable for $|p|\leq 1$ with respect to $(2\omega)^{-1}dp$.

We have up to a multiplicative constant

$\begin{split}F(p)=&\int\left[\frac{x^{\prime}(p\cdot
x^{\prime\prime})}{\omega-p\cdot
x^{\prime}(t))^{2}}+\frac{x^{\prime\prime}}{(\omega-p\cdot
x^{\prime}(t))}\right]dt\\\
=&\int\frac{d}{dt}\left[\frac{x^{\prime}(t)}{(\omega-p\cdot
x^{\prime}(t))}\right]dt\\\ =&\frac{v_{\textrm{out}}}{(\omega-p\cdot
v_{\textrm{out}})}-\frac{v_{\textrm{in}}}{(\omega-p\cdot v_{\textrm{in}})}\\\
=&\frac{\omega(v_{\textrm{out}}-v_{\textrm{in}})+(v_{\textrm{in}}\ p\cdot
v_{\textrm{out}}-v_{\textrm{out}}\ p\cdot v_{\textrm{in}})}{(\omega-p\cdot
v_{\textrm{out}})(\omega-p\cdot v_{\textrm{in}})}\\\ \end{split}$ (98)

where $\delta v=v_{\textrm{out}}-v_{\textrm{in}}$. We see that if
$v_{\textrm{in}}=v_{\textrm{out}}$ then $F=0$ and $F_{T}=0$ and hence
$J\in{\cal H}$ and $J_{T}\in{\cal H}_{T}$ and $\omega_{\textrm{out}}$ is Fock.

We proceed with the case $v_{\textrm{out}}\neq v_{\textrm{in}}$. We are now
almost exactly in the situation that occurred in the proof of Proposition 4.
The main difference is that we consider $|p|\leq 1$ rather than $|p|\geq 1$.
We write $F_{T}(p)=F_{1,T}(p)+F_{2}(p)$ where with $\delta v\equiv
v_{\textrm{out}}-v_{\textrm{in}}$ and $v_{\times}=v_{\textrm{in}}\times
v_{\textrm{out}}$

$\begin{split}F_{1,T}(p)=&\frac{|\delta v|}{\omega}\left[\frac{n_{\delta
v}-n_{p}(n_{p}\cdot n_{\delta v})}{(1-n_{p}\cdot
v_{\textrm{out}})(1-n_{p}\cdot v_{\textrm{in}})}\right]\\\
F_{2}(p)=&\frac{1}{\omega}\left[\frac{n_{p}\times
n_{v_{\times}}}{(1-n_{p}\cdot v_{\textrm{out}})(1-n_{p}\cdot
v_{\textrm{in}}))}\right]\\\ \end{split}$ (99)

If $v_{\textrm{in}}$ and $v_{\textrm{out}}$ are colinear (which includes the
case where one of them is zero), then only $F_{1,T}$ contributes and we have
as in (87)

$\int_{|p|\leq 1}|F_{T}(p)|^{2}\frac{dp}{2\omega}\geq\textrm{const}|\delta
v|^{2}\int_{|p|\leq 1,|n_{p}\cdot n_{\delta
v}|\leq\frac{1}{2}}\frac{1}{\omega^{3}}dp=\infty$ (100)

If $v_{\textrm{in}}$ and $v_{\textrm{out}}$ are not colinear we look at the
component along $v_{\times}$ and find as in (88)

$n_{v_{\times}}\cdot F_{T}(p)=\frac{|\delta
v|}{\omega}\left[\frac{-(n_{p}\cdot n_{v_{\times}})(n_{p}\cdot n_{\delta
v})}{(1-n_{p}\cdot v_{\textrm{out}})(1-n_{p}\cdot v_{\textrm{in}})}\right]$
(101)

Then as in (89)

$\begin{split}\int_{|p|\leq
1}|F_{T}(p)|^{2}\frac{dp}{2\omega}&\geq\textrm{const}|\delta
v|^{2}\int_{|p|\leq 1,|n_{p}\cdot n_{v_{\times}}|\geq\frac{1}{2},|n_{p}\cdot
n_{\delta v}|\geq\frac{1}{2}}\frac{1}{\omega^{3}}\ dp=\infty\\\ \end{split}$
(102)

Thus the $F_{T}$ condition fails, and $\omega_{\textrm{out}}$ is not Fock.

Remarks.

  1. 1.

Note that if the particle has constant velocity then $x^{\prime\prime}(t)=0$,
hence $J(p)=0$, and $\omega_{\textrm{out}}=\omega_{\textrm{in}}$. No photons
are radiated.

  2. 2.

The result does not depend on the complete history of the trajectory, only on
the initial and final states. Except in a the special circumstance
$v_{\textrm{in}}=v_{\textrm{out}}$ we always leave Fock space.

  3. 3.

It is noteworthy that the departure from Fock space comes from a UV divergence
in the case of of discontinuities in the velocity (proposition 4), and from an
IR divergence in the case $v_{\textrm{in}}\neq v_{\textrm{out}}$(proposition
6).

  4. 4.

It might be of interest to obtain results of this kind for the radiation of
gravitons by a classical source in linearized quantum gravity. See Skagerstam,
Eriksson, Rekdal [8], for a formulation of this model in an analog of the
Coulomb gauge. In this connection we also mention the work of Kegeles, Oriti,
Tomlin [5] who suggest non-Fock coherent states of the type used here as
appropriate for a group field theory model of quantum gravity.

## References

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