%&latex
%% 25 Sep. 2005 (14 Feb. 2006: final version)
%% 5 mars: More detailed explanations for the proof of
%% Theorem 4 on pages 15 and 16; improved last paragraph of
%% section 2 on page 9; signs corrected on page 13; and, on
%% the bottom half of page 8 and top of page 9, the proof
%% has been corrected
%% to appear in ANNALES HENRI POINCARE
%% Spacetime causality in the study of the Hankel transform
%% par Jean-Fran\c cois Burnol
%% Universit\'e Lille 1, UFR de Math\'ematiques
%% burnol@math.univ-lille1.fr
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\begin{document}
\title{Spacetime causality in the study of the Hankel transform}
\author{Jean-Fran\c cois Burnol}
\date{\ }
\maketitle
\medskip
\begin{abstract}
We study Hilbert space aspects of the Klein-Gordon equation
in two-dimensional spacetime. We associate to its
restriction to a spacelike wedge a scattering from the past
light cone to the future light cone, which is then shown to
be (essentially) the Hankel transform of order zero. We
apply this to give a novel proof, solely based on the
causality of this spatio-temporal wave propagation, of the
theorem of de~Branges and V.~Rovnyak concerning Hankel pairs
with a support property. We recover their isometric
expansion as an application of Riemann's general method for
solving Cauchy-Goursat problems of hyperbolic type.
\end{abstract}
\medskip
\begin{footnotesize}
\begin{quote}
keywords:
Klein-Gordon equation; Hankel and Fourier transforms; Scattering.
\medskip
Universit\'e Lille 1\\
UFR de Math\'ematiques\\
Cit\'e scientifique M2\\
F-59655 Villeneuve d'Ascq\\
France\\
burnol@math.univ-lille1.fr\\
\bigskip
v1: 25 Sept. 2005; Final: 5 March 2006\\
%%\ \\
\end{quote}
\end{footnotesize}
\medskip
\setlength{\normalbaselineskip}{18pt}
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\section{Introduction}
We work in two-dimensional spacetime with metric $c^2dt^2 -
dx^2$. We shall use units such
that $c=1$. Points are denoted $P = (t,x)$. And the
d'Alembertian operator $\square$ is
$\frac{\partial^2}{\partial t^2} -
\frac{\partial^2}{\partial x^2}$. We consider the
Klein-Gordon equation (with $m=1$, $\hbar = 1$; actually
we shall only study the classical wave field, no
quantization is involved in this paper):
\begin{equation}
\square \phi + \phi = 0 \end{equation}
We have an energy density:\footnote{as this paper is
principally of a mathematical nature, we do not worry
about an overall $\frac12$ factor.}
\begin{equation} \cal E = |\phi|^2 + \left|\dphidx\right|^2 +
\left|\dphidt\right|^2\end{equation}
which gives a conserved quantity:
\begin{equation} E = \frac1{2\pi} \int_{-\infty}^{+\infty} \cal
E(\phi)(t,x)\;dx\;,\end{equation}
in the sense that if the Cauchy data at time $t=0$ has
$E<\infty$ then $E$ is finite (and constant\dots) at all
times (past and future). We shall mainly work with such
finite energy solutions.
Although we failed in locating a reference for the following
basic observation, we can not imagine it to be novel:
\begin{theorem}\label{theo:1}
If $\phi$ is a finite energy solution to the
Klein-Gordon equation then:
\[ \lim_{t\to+\infty} \int_{|x|>t} \cal
E(\phi)(t,x)\;dx = 0\;.\]
\end{theorem}
Obviously this would be completely wrong for the zero mass
equation. We shall give a (simple) self-contained proof,
because it is the starting point of all that we do
here. Let us nevertheless state that the result follows
immediately from Hörmander's fine pointwise estimates
(\cite{hor1, hor2}; see also the paper of S.~Klainerman
\cite{klain} and the older papers of S.~Nelson \cite{nel1,
nel2}.) I shall not reproduce the strong pointwise
results of Hörmander, as they require notations and
preliminaries. Let me simply mention that Hörmander's
Theorem 2.1 from \cite{hor1} can be applied to the
positive and negative frequency parts of a solution with
Cauchy data which is gaussian times polynomial. So theorem
\ref{theo:1} holds for them, and it holds then in general,
by an approximation argument.
The energy conservation follows from:
\begin{equation} \dpt \cal E + \dpx \cal P = 0
\qquad\text{with}\qquad \cal P = - \dphidx \overline{\dphidt} -
\overline{\dphidx} \dphidt \end{equation}
If we apply Gauss' theorem to the triangle with vertices $O =
(0,0)$, $A = (t,t)$, $B = (t,-t)$, we obtain ($t>0$):
\[ \int_{|x|\leq t} \cal E(\phi)(t,x) dx = \int_{-t}^0 (|\phi(|x|,x)|^2 +
\left| \ddx \phi(|x|,x)\right|^2) \,dx + \int_{0}^t (|\phi(x,x)|^2 +
\left| \ddx \phi(x,x)\right|^2) \,dx\]
This proves that $\int_{|x|>t} \cal
E(\phi)(t,x)\;dx$ decreases as $t\to+\infty$. It shows
also that
theorem \ref{theo:1} is equivalent to:
\begin{equation}\label{eq:123}
E = \frac1{2\pi} \int_{-\infty}^0 (|\phi(|x|,x)|^2 +
\left| \ddx \phi(|x|,x)\right|^2 ) \,dx + \frac1{2\pi}
\int_{0}^\infty ( |\phi(x,x)|^2 +
\left| \ddx \phi(x,x)\right|^2 ) \,dx
\end{equation}
Otherwise stated, there is a unitary representation of
$\phi$ on the future light cone.
%%
%% \footnote{as there is
%% something in physics called ``light-cone quantization''
%% this unitary equivalence must be a very well-known
%% fact. I don't know if the physics literature contains the
%% rigorous proof given here.}
%%
Here
is now the basic idea: as
solutions to hyperbolic equations propagate causally,
equation \eqref{eq:123}
gives a unitary representation from the Hilbert space of
Cauchy data at time $t=0$ with support in $x\geq0$ to the
Hilbert space of functions $p(v) = \phi(v,v)$ on
$[0,+\infty[$ with squared norm $ \frac1{2\pi} \int_{0}^\infty (
|p(v)|^2 + |p'(v)|^2 ) \,dv$. Instead of Cauchy data
vanishing for $x<0$, it will be useful to use Cauchy data
invariant under $(t,x)\to(-t,-x)$. Then $p$ will be
considered as an even, and $p'$ as an odd, function, and $
\frac1{2\pi} \int_{0}^\infty ( |p(v)|^2 + |p'(v)|^2 )
\,dv$ will be $\frac12 E(\phi)$, for $\phi(t,x) =
\phi(-t,-x)$.
We can also consider the
past values $g(u) = \phi(-u,u)$, $t = -u$, $x =u$, $0\leq
u <\infty$. So there is a unitary map from such $g$'s to
the $p$'s:
\begin{theorem}\label{theo:2}
Let $g(u)$, $u>0$, and $p(v)$, $v>0$ be such that
$\int_0^\infty
|g(u)|^2+|g'(u)|^2 du < \infty$, $\int_0^\infty
|p(v)|^2+|p'(v)|^2 dv < \infty$. The necessary and
sufficient condition for $A(r) = \sqrt{r}\,
g(\frac{r^2}2)$ and $B(s) = - \sqrt{s}\, p'(\frac{s^2}2)$
to be Hankel transforms of order zero of one another
($A(r) = \int_0^\infty \sqrt{rs} J_0(rs) B(s) \,ds$) is
for $g$ and $p$ to be the values on the past and future
boundaries of the Rindler wedge $0<|t|0$
the vanishing on $00$
and $p(v)= \phi(v,v)$ for $v>0$ be the values taken by
$\phi$ on the past, respectively future, boundaries of the
Rindler wedge $0<|t|0$ the vanishing for $00$,
\[ h(x) = \sqrt x\; k(\frac{x^2}2),\qquad f(x) = \sqrt x
F(x^2), \qquad g(x) = \sqrt x G(x^2) \]
Then the equations above become:
\begin{subequations}
\begin{align}\label{eq:100a}
F(x) &=
\int_{x/2}^\infty J_0(\sqrt{x(2v-x)})k(v)\,dv\\
\label{eq:100b}
G(x) &= k(\frac{x}2) -
\int_{x/2}^\infty x\,
\frac{J_1(\sqrt{x(2v-x)})}{\sqrt{x(2v-x)}}\,
k(v)\,dv\\
\label{eq:100c}
k(v) &= G(2v) + \frac12\int_0^{2v}
J_0(\sqrt{x(2v - x)}) F(x) \,dx - \frac12\int_0^{2v}
x\, \frac{J_1(\sqrt{x(2v - x)})}{\sqrt{x(2v - x)}} G(x)
\,dx\\
\label{eq:100d}
\int_0^\infty 2&|k(v)|^2\,dv = \int_0^\infty (|F(x)|^2 +
|G(x)|^2)\,dx
\end{align}
\end{subequations}
The de~Branges Rovnyak theorem is thus the equivalence between
equations \eqref{eq:100a}, \eqref{eq:100b} and
\eqref{eq:100c}, the validity of \eqref{eq:100d}, the
fact that the pair $(F,G)$ is identically zero on $(0,2a)$
if and only if both $k$ and $\cH(k)$ vanish identically on
$(0,a)$, and finally the fact that permuting $F$ and $G$ is equivalent to
$k\leftrightarrow \cH(k)$.
It proves convenient to work with the first order
``Dirac'' system:
\begin{subequations}
\begin{align}
\dpsidt - \dpsidx = + \phi\\
\dphidt + \dphidx = - \psi
\end{align}
\end{subequations}
Let us write $\left[\begin{smallmatrix} \psi(0,x)\\
\phi(0,x)\end{smallmatrix}\right] = \left[\begin{smallmatrix} G(x)\\
F(x)\end{smallmatrix}\right]$. We shall use $K(\psi,\phi) = \frac1{2\pi}
\int_{-\infty}^{+\infty}
(|F(x)|^2 + |G(x)|^2)\,dx$ as the Hilbert space (squared)
norm. We shall require $\dphidx$ and $\dpsidx$ to be
in $L^2$ at $t=0$ (then $\phi$ and $\psi$ are continuous
on space-time). Our previous $E(\phi)$ is not invariant
under Lorentz boosts:
%%
%% \footnote{this is elementary knowledge from any
%% physics textbook, except that we need in our setting to
%% rigorously justify the dropping of boundary terms when
%% applying the divergence or Stokes or Green theorem.}
%%
it is only the first component of a Lorentz vector
$(E(\phi),P(\phi))$ (see equation \eqref{eq:P} for the
expression of $P$). And it turns out that in fact $K(\psi,\phi) =
E(\phi) - P(\phi) = E(\psi) + P(\psi)$. The point
%%
%% \footnote{again this is an
%% elementary fact from any textbook on relativistic
%% physics.}
%%
is that in order to define an action of the
Lorentz group on the solutions of the Dirac system it is
necessary to rescale in opposite ways $\psi$
and $\phi$. When done symmetrically, $K$ then becomes an
invariant under the Lorentz boosts. This relativistic
covariance of the spinorial quantity
$\left[\begin{smallmatrix} \psi \\
\phi \end{smallmatrix}\right]$ is important for the proof
of the next theorem:
\begin{subequations}
\begin{theorem}\label{theo:4}
Let $F$ and $G$ be two functions with $\int_0^\infty
|F|^2+|F'|^2+|G|^2+|G'|^2 dx < \infty$.
Let $\left[{\psi\atop \phi}\right]$ be the unique solution
in the Rindler wedge $x>|t|>0$ of the
first order system:
\begin{align}
\dpsidt - \dpsidx &= + \phi\\
\dphidt + \dphidx &= - \psi
\end{align}
with Cauchy data $\phi(0,x) = F(x)$, $\psi(0,x) =
G(x)$. The boundary values:
\[ g(u) = \phi(-u,u) \quad (u>0),\qquad\text{and}\qquad k(v) =
\psi(v,v) \quad (v>0),\]
verify $\int_0^\infty |g(u)|^2 + |g'(u)|^2 du
<\infty$, $\int_0^\infty |k(v)|^2 + |k'(v)|^2 dv < \infty$ and are a
$\cH$ transform pair. For any $a>0$ the identical
vanishing of $F(x)$ and $G(x)$ for $00$,
$\omega\geq1$), and ``negative frequency'' ($\lambda<0$,
$\omega\leq-1$) parts.
At first we only take $\alpha$ to be a smooth, compactly
supported function of $\lambda$, vanishing identically in
a neighborhood of $\lambda=0$. Then the corresponding
$\phi$ is a smooth, finite energy solution of the Klein
Gordon equation. Let us compute this energy. At $t=0$ we
have
\[ \phi(0,x) = \int_{-\infty}^{+\infty} e^{+i \mu x} \alpha(\lambda)\,d\lambda
\qquad \dphidt(0,x) = - i \int_{-\infty}^{+\infty} e^{+i\mu x}
\frac12(\lambda+\frac1\lambda) \alpha(\lambda)\,d\lambda\]
So we will apply Plancherel's theorem, after the change of
variable $\lambda\to \mu$. We must be careful that if
$\lambda$ is sent to $\mu$, then $\lambda' =
-\frac1\lambda$, is too. Let $\lambda_1 >0$ and $\lambda_2
<0$ be the ones being sent to $\mu$. Let us also define:
\[ a(\mu) =
\frac{\alpha(\lambda_1)}{\frac12(1+\frac1{\lambda_1^2})},
\qquad
b(\mu) =
\frac{\alpha(\lambda_2)}{\frac12(1+\frac1{\lambda_2^2})}\]
Then:
\[ \phi(0,x) = \int_{-\infty}^{+\infty} e^{+i \mu x} ( a(\mu) + b(\mu) ) d\mu
\qquad \dphidt(0,x) = - i \int_{-\infty}^{+\infty} e^{+i\mu x}
\frac12(\lambda_1+\frac1{\lambda_1}) ( a(\mu) - b(\mu) )d\mu\]
\[ \frac1{2\pi} \int_{-\infty}^{+\infty} (|\phi|^2 + |\dpx \phi|^2) dx = \int_{-\infty}^{+\infty} | a(\mu) + b(\mu)
|^2 (1 + \mu^2)\, d\mu \]
\[ \frac1{2\pi} \int_{-\infty}^{+\infty} |\dpt \phi|^2 dx = \int_{-\infty}^{+\infty} | a(\mu) - b(\mu)
|^2 \left(\frac12(\lambda_1+\frac1{\lambda_1})\right)^2\,
d\mu \]
Observing that $1+\mu^2 =
\left(\frac12(\lambda_1+\frac1{\lambda_1})\right)^2 =
\left(\frac12(\lambda_2+\frac1{\lambda_2})\right)^2 $, this gives
\[ E(\phi) = 2 \int_{-\infty}^{+\infty} (|a(\mu)|^2 + |b(\mu)|^2) \lambda_1^2
\left(\frac12(1+\frac1{\lambda_1^2})\right)^2 d\mu\]
\[ = 2\int_0^\infty |a(\mu)|^2 \lambda_1^2
\left(\frac12(1+\frac1{\lambda_1^2})\right)^3 \,d\lambda_1
+ 2\int_{-\infty}^0 |b(\mu)|^2 \lambda_2^2
\left(\frac12(1+\frac1{\lambda_2^2})\right)^3\,d\lambda_2\]
\[ = 2\int_0^\infty |\alpha(\lambda)|^2 \frac12 (1+\lambda^2) \, d\lambda
+2\int_{-\infty}^0 |\alpha(\lambda)|^2 \frac12 (1+\lambda^2)\, d\lambda\]
\begin{equation} E(\phi) = \int_{-\infty}^{+\infty} |\alpha(\lambda)|^2 (1 +
\lambda^2)\,d\lambda\end{equation}
Let us now compute the energy on the future light cone. We write
$g(u) = \phi(-u,u)$, $p(v) = \phi(+v,v)$. We
have:
\begin{equation}
\label{eq:5}
g(u) = \int_{-\infty}^{+\infty} e^{+ i \lambda u}
\alpha(\lambda)\,d\lambda
\end{equation}
Let $\alpha = \alpha_+ + \alpha_-$ be the decomposition of
$\alpha$ as the sum of $\alpha_+$,
belonging to the
Hardy space of the upper half-plane $\Im(\lambda)>0$ and
of $\alpha_-$, belonging to the Hardy space of the
lower half-plane. We have:
\begin{subequations}
\begin{equation}\label{eq:gandgprime}
\frac1{2\pi} \int_{-\infty}^0 |g(u)|^2 \,du = \int_{-\infty}^{+\infty}
|\alpha_+(\lambda)|^2 \,d\lambda
\end{equation}
We must be careful regarding $\frac1{2\pi} \int_{-\infty}^0
|g'(u)|^2 \,du$. We have
$g'(u)\Un_{u<0}(u) - g(0)\delta(u) =
\int_{-\infty}^{+\infty} e^{+ i \lambda u}
i\lambda \alpha_+(\lambda)\,d\lambda$ as a distribution identity
so
\[ \frac1{2\pi} \int_{-\infty}^0
|g'(u)|^2 \,du = \int_{-\infty}^{+\infty} \left|
\lambda\alpha_+(\lambda) +
\frac{g(0)}{2\pi i}\right|^2\,d\lambda\]
On the other
hand $(t\alpha(t))_+(\lambda) = \frac1{2\pi
i}\int_{-\infty}^\infty \frac{t\alpha(t)\,dt}{t-\lambda}
= \lambda\alpha_+(\lambda) + \frac{g(0)}{2\pi i}$ so the
formula is:
\begin{equation}
\label{eq:11}
\frac1{2\pi} \int_{-\infty}^0
|g'(u)|^2 \,du = \int_{-\infty}^{+\infty}
|(t\alpha)_+(\lambda)|^2 \,d\lambda
\end{equation}
\end{subequations}
Similarly, as $p(v) = \int_{-\infty}^{+\infty} e^{+ i \lambda v}
\alpha(-\frac1\lambda)\frac1{\lambda^2}\,d\lambda$, and
defining $\beta(\lambda) = \alpha(-
\frac1\lambda)\frac1{\lambda^2}$ we obtain:
\[ \frac1{2\pi} \int_0^\infty (|p(v)|^2 + |p'(v)|^2) dv =
\int_{-\infty}^{+\infty}
(|\beta_-(\lambda)|^2 + |(t\beta)_-(\lambda)|^2)\,d\lambda\]
Now, from $t\beta(t) = \frac1t \alpha(\frac{-1}t)$ it is
seen that $(t\beta)_-(\lambda) = \frac1\lambda
\alpha_-(\frac{-1}\lambda)$, so $\int_{-\infty}^{+\infty}
|(t\beta)_-(\lambda)|^2\,d\lambda = \int_{-\infty}^{+\infty}
|\alpha_-(\lambda)|^2\,d\lambda$. And, as $t\alpha(t) =
\frac1t \beta(\frac{-1}t)$ one has in a similar manner
$\int_{-\infty}^{+\infty}
|(t\alpha)_+(\lambda)|^2\,d\lambda = \int_{-\infty}^{+\infty}
|\beta_+(\lambda)|^2\,d\lambda$.
Combining, we get
\begin{align*}
&\frac1{2\pi} \int_{-\infty}^0 (|g(u)|^2 +
|g'(u)|^2) \,du + \frac1{2\pi} \int_0^\infty (|p(v)|^2 +
|p'(v)|^2) dv\\
&= \int_{-\infty}^{+\infty}
(|\alpha_+(\lambda)|^2 + |\beta_+(\lambda)|^2 +
|\beta_-(\lambda)|^2 +|\alpha_-(\lambda)|^2) \,d\lambda
\end{align*}
and
as the two Hardy spaces are mutually perpendicular in
$L^2(-\infty,+\infty;d\lambda)$ we finally obtain:
\[ \int_{-\infty}^{+\infty} |\alpha(\lambda)|^2 (1+ \lambda^2)
\,d\lambda\]
as the energy on the future light cone.
So, with this, the theorem that $E(\phi)$ is entirely on the
future light cone is proven for the $\phi$'s corresponding
to $\alpha$'s which are smooth and compactly supported
away from $\lambda = 0$. Obviously the Cauchy data for
such $\phi$'s is a dense subspace of the full initial data
Hilbert space. As energy is conserved as $t\to\infty$, the
fact that $\lim_{t\to\infty} \int_{|x|>t} \cal E(\phi) dx
=0$ holds for all finite energy $\phi$'s then follows by
approximation. Furthermore we see that a finite energy
solution is uniquely written as a wave packet:
\begin{equation} \phi(t,x) = \int_{-\infty}^{+\infty}
e^{+i(\lambda u - \frac1\lambda v)}
\alpha(\lambda)\,d\lambda\qquad E(\phi) = \int_{-\infty}^{+\infty} (1+
\lambda^2)|\alpha(\lambda)|^2\,d\lambda < \infty
\end{equation}
At this stage Theorem \ref{theo:1} is established.
When studying the Klein-Gordon equation in the right wedge
$x>0$, $|t|0$:
\[ \int_{|x|\leq t} \cal P(\phi)(t,x) dx = \int_{-t}^0 (- |\phi(-x,x)|^2 +
\left| \ddx \phi(-x,x)\right|^2) \,dx + \int_{0}^t (|\phi(x,x)|^2 -
\left| \ddx \phi(x,x)\right|^2) \,dx\]
The integral of $| \cal P |$ for $|x|>t$ tends to zero for
$t\to+\infty$ as it is bounded above by the one for $\cal
E$. So:
\begin{equation}
\label{eq:7}
P = \frac1{2\pi} \int_{-\infty}^0 ( - |g(u)|^2 +
|g'(u)|^2)\,du + \frac1{2\pi} \int_{0}^\infty ( |p(v)|^2 -
|p'(v)|^2)\,dv
\end{equation}
with, again, $g(u) = \phi(-u,u)$, $p(v) =
\phi(+v,v)$. Hence:
\begin{subequations}
\begin{align}\label{eq:1}
E - P &= \frac1{\pi} \int_{-\infty}^0
|g(u)|^2\,du + \frac1{\pi} \int_{0}^\infty |p'(v)|^2\,dv\\
E + P &= \frac1{\pi} \int_{-\infty}^0
|g'(u)|^2\,du + \frac1{\pi} \int_{0}^\infty |p(v)|^2\,dv
\end{align}
\end{subequations}
\From\ \eqref{eq:gandgprime} and the similar formulas relative
to $p$ we can express all four integrals in terms of
$\alpha(\lambda)$. Doing so we find after elementary steps:
\begin{equation}
\label{eq:9}
E - P = 2 \int_{-\infty}^{+\infty} |\alpha(\lambda)|^2\,d\lambda\qquad E + P
= 2 \int_{-\infty}^{+\infty} \lambda^2\,|\alpha(\lambda)|^2\,d\lambda
\end{equation}
So:
\begin{equation} P = \int_{-\infty}^{+\infty} (\lambda^2 -
1)|\alpha(\lambda)|^2\,d\lambda\end{equation}
This confirms that a $\lambda$ with $|\lambda|\geq1$
gives a ``right-moving''
component of the wave packet (its phase is constant for
$\omega t - \mu x = C$, $\omega =
\frac12(\lambda+\frac1\lambda)$, $\mu = \frac12(\lambda-
\frac1\lambda)$.) The values of $\lambda$ with
$|\lambda|\leq1$ give ``left-moving'' wave components. As a check,
we can observe that it is impossible to have a purely
right-moving packet with vanishing Cauchy data for $t=0$,
$x<0$, because as we saw above, for such Cauchy data
$\alpha$ has to belong to the Hardy space of the lower
half-plane and can thus (by a theorem of Wiener) not
vanish identically on $(-1,1)$. A purely right-moving
packet starting entirely on $x>0$ would have a hard time
hitting the light cone, and this would emperil Theorem
\ref{theo:1}. Such wave-packets exist for the zero-mass
equation, one way of reading Theorem \ref{theo:1} is to
say that they don't exist for non-vanishing real mass.
Let us consider the effect of a Lorentz boost on $E$ and
$P$. We take $\Lambda = e^\xi$ ($\xi\in\RR$) and replace
$\phi$ by:
\begin{subequations}
\begin{gather}
\phi_\Lambda(t,x) = \phi( \cosh(\xi) t +
\sinh(\xi) x, \sinh(\xi) t + \cosh(\xi) x )\\
\phi_\Lambda[u,v] = \phi[\frac1\Lambda u, \Lambda v]\\
g_\Lambda(u) = \phi_\Lambda[u,0] = g(\frac1\Lambda
u)\qquad p_\Lambda(v) = p(\Lambda v)\label{eq:gscale}\\
\alpha(\lambda) \mapsto \alpha_\Lambda(\lambda) = \Lambda
\alpha(\Lambda \lambda) \\
E_\Lambda - P_\Lambda = \Lambda\cdot(E -
P)\qquad\qquad E_\Lambda + P_\Lambda = \frac1\Lambda (E +
P)\\
E_\Lambda = \cosh(\xi) E - \sinh(\xi) P\\
P_\Lambda = -\sinh(\xi) E + \cosh(\xi) P
\end{gather}
\end{subequations}
So the conserved quantities $E$ and $P$ are not Lorentz
invariant but the Einstein rest mass squared $E^2 - P^2$
is.
\section{Scale reversing operators}
We begin the proof of Theorem \ref{theo:2}.
Let us consider the manner in which the function $g(u)$
for $u>0$ is related to the function $p(v)>0$. We know
that they are in unitary correspondence for the norms
$\int_{u>0} |g|^2+|g'|^2 \,du$ and $\int_{v>0}
|p|^2+|p'|^2\,dv$, and the formulas \eqref{eq:1} for $E-P$
and $E+P$ suggest that one should pair $g$ with $p'$ and
$g'$ with $p$. In fact if we take into consideration the
wave which has values $\phi(t,x) = e^{-|x|}$
for space-like points, we are rather led to pair $g$ with
$-p'$ and $g'$ with $-p$ (the values of $\phi$ at
time-like points are more involved and we don't need to
know about them here; suffice it to say that certainly
$e^{-x}$ solves Klein-Gordon, so it gives the unique
solution in the right wedge with $\phi(0,x) = e^{-x}$,
$\dphidt(0,x) = 0$.)
Let us denote by $\cH$ the operator which acts as $g\mapsto
-p'$,
on even $g$'s. Under a Lorentz boost:
$g\mapsto g_\Lambda(u) = g(\frac1\Lambda
u)$, $ -p'\mapsto -\Lambda p'(\Lambda v) $
and also the assignment $g\mapsto -p'$ is unitary for the
$L^2$ norm:
\[ g(u) = \int_{-\infty}^{+\infty} e^{iu\lambda}
\alpha(\lambda)\,d\lambda\qquad
p(v) = \int_{-\infty}^{+\infty} e^{i \lambda v}
\alpha(-\frac1\lambda)\frac1{\lambda^2}\,d\lambda
\]
\[ -p'(v) = -i \int_{-\infty}^{+\infty} e^{i \lambda v}
\alpha(-\frac1\lambda)\frac1{\lambda}\,d\lambda \]
Going from $g$ to $\alpha$ is unitary, from $\alpha$ to
$-i\alpha(-\frac1\lambda)\frac1{\lambda}$ also, and back
to $-p'$ also, in the various $L^2$ norms. So the assignment from
$g$ to $-p'$ is unitary.
Identifiying the $L^2$ space on $u>0$ with
the $L^2$ space on $v>0$, through $v=u$, $\cH$ is a
unitary operator on $L^2(0,+\infty;\,du)$. Furthermore it
is ``scale reversing'': we say that an operator $\cK$
(bounded, more generally, closed) is scale reversing if
its composition $\cK I$ with $I: g(u)\mapsto
\frac{g(1/u)}u$ commutes with the unitary group of scale
changes $g\mapsto \sqrt \Lambda g(\Lambda u)$. The Mellin
transform $g\mapsto\wh g(s) = \int_0^\infty g(u)
u^{-s}\,du$, for $s= \frac12+i\tau$, $\tau\in\RR$, is the
additive Fourier transform of $e^{t/2}g(e^{t})\in
L^2(-\infty,+\infty;dt)$. The operator $\cK I$
commutes with multiplicative translations hence is
diagonalized by the Mellin transform: we have a certain
(bounded for $\cK$ bounded) measurable function $\chi$
on the critical line $\Re(s)= \frac12$ such that for any
$g(u) \in L^2(0,\infty;du)$, and almost everywhere on the
critical line:
\[ (\cK g)^\wedge(s) = (\cK I(Ig))^\wedge(s) = \chi(s)
(Ig)^\wedge(s) = \chi(s) \wh g(1-s) \]
Let us imagine for a minute that we know a $g$ which is
invariant under $\cK$ and which, furthermore has $\wh
g(s)$ almost everywhere non vanishing (by a theorem of
Wiener, this means exactly that the linear span of its orbit under the
unitary group of scale changes is dense in $L^2$). Then we
know $\chi(s)$ hence, we know $\cK$. So $\cK$ is
uniquely determined by the knowledge of one such invariant
function.
In the case of our operator $\cH$ which goes from the
data of $g(u)$, $u>0$, to the data of $k(v) = -p'(v)$,
$v>0$, where $g$ and $p$ are the boundary values of a
finite energy solution of the Klein-Gordon equation in the
right wedge, we know that it is indeed unitary, scale
reversing, and has $e^{-u}$ as a self-reciprocal function
(so, here, $\chi(s) = \frac{\Gamma(1-s)}{\Gamma(s)}$).
On the other hand the Hankel transform of order zero is
unitary, scale reversing, and has $\sqrt u e^{-u^2/2}$ as
self-reciprocal invariant function. So we find that the
assignment of $-\sqrt v\;k'(\frac{v^2}2)$ to $\sqrt u\;
g(\frac{u^2}2)$ is exactly the Hankel transform of order
zero. This may also be proven directly by the method we
will employ in section \ref{sec:riemann}.
\section{Causality and support conditions}
The Theorem \ref{theo:2} is almost entirely proven: if the
Cauchy data vanishes identically for $00$ which gives under the $\cH$ transform the
function $k(v)$ for $v>0$ which must be considered odd and
correspond to the $PT$ anti-invariant $\psi$. We note that
if $k(0^+)\neq0$ then this $\psi$ is not of finite
energy. Using only that $\phi$ is finite energy, we have
from equation \eqref{eq:1}:
\[E(\phi) - P(\phi) = \frac1{\pi} \int_{-\infty}^0
|g(u)|^2\,du + \frac1{\pi} \int_{0}^\infty |k(v)|^2\,dv\]
\[ \frac1{2\pi} \int_{-\infty}^{+\infty} \left(|\phi(0,x)|^2 + |\psi(0,x)|^2
\right)\; dx = E(\phi) - P(\phi) = \frac1{\pi} \int_0^{\infty}
|g(u)|^2\,du + \frac1{\pi} \int_{0}^\infty |k(v)|^2\,dv\]
\begin{subequations}
\begin{align}\label{eq:13a}
\int_0^\infty \left(|\phi(0,x)|^2 + |\psi(0,x)|^2
\right)\; dx &= 2 \int_0^{\infty}
|g(u)|^2\,du \\ \label{eq:13b}
\int_0^\infty \left(|\phi(0,x)|^2 + |\psi(0,x)|^2
\right)\; dx &= 2 \int_{0}^\infty |k(v)|^2\,dv
\end{align}
\end{subequations}
We now begin the proof of Theorem \ref{theo:4}. To prove
that $\int_0^\infty |F(x)|^2+|G(x)|^2\,dx = 2\int_0^\infty
|g(u)|^2\,du = 2\int_0^\infty |k(v)|^2\,dv$, we extend $F$
to be even and $G$ to be odd. Then $\phi$ is $PT$ even of
finite energy, and $\psi$ is $PT$ odd and equations
\eqref{eq:13a} and \eqref{eq:13b} apply. Note that if
$G(0^+)\neq0$ then $\psi$ is not of finite energy but only
the fact that $\phi$ is of finite energy was used for
\eqref{eq:13a} and \eqref{eq:13b}. That $k=\cH(g)$ and
$\int_0^\infty |g(u)|^2
+ |g'(u)|^2 du <\infty$ hold are among our previous
results. If we choose $G$ to be even and $F$ to be odd, then
it is $\psi$ which is of finite energy and so $\int_0^\infty
|k(v)|^2 + |k'(v)|^2 dv < \infty$ holds true. We can also
prove $\int_0^\infty |g|^2
+ |g'|^2 du <\infty$, $\int_0^\infty |k|^2 + |k'|^2 dv <
\infty$ after extending $F$ and $G$ such that
$\int_{-\infty}^\infty
|F|^2+|F'|^2+|G|^2+|G'|^2 dx < \infty$ so that both
$\phi$ and $\psi$ are then of finite energy. The boundary
values $g(u)$, $u>0$, and $k(v)$, $v>0$ do not depend on
choices. Furthermore the vanishing of $F$ and $G$ on
$(0,2a)$ at $t=0$ is equivalent by our previous arguments to
the vanishing of $g$ and $k$ on $(0,a)$. To show that all
$\cH$ pairs with $\int_0^\infty |g|^2 + |g'|^2 du <\infty$,
$\int_0^\infty |k|^2 + |k'|^2 dv < \infty$ are obtained, let
$k_1$ be the odd function with $k_1(v) = k(v) -
k(0^+)e^{-v}$ for $v>0$ and let $g_1$ be the even function
with $g_1(u) = g(u) - k(0^+)e^{-u}$ for $u\geq 0$. Then $k_1
= \cH(g_1)$ and $\int_{-\infty}^\infty |g_1|^2 + |g_1'|^2 du
<\infty$ and $\int_{-\infty}^\infty |k_1|^2 + |k_1'|^2 dv <
\infty$. They thus correspond to $\phi_1$ and $\psi_1$ both
of finite energy. We define for $x>0$: $F(x) = \phi_1(0,x) +
k(0^+) e^{-x}$ and $G(x) = \psi_1(0,x) + k(0^+) e^{-x}$, it
then holds that $\int_{0}^\infty
|F|^2+|F'|^2+|G|^2+|G'|^2 dx < \infty$ and
$\left[\begin{smallmatrix} \psi \\ \phi
\end{smallmatrix}\right] = \left[\begin{smallmatrix} \psi_1
+ k(0^+) e^{-x}\\
\phi_1 + k(0^+)e^{-x} \end{smallmatrix}\right]$ is
the unique solution in the Rindler wedge of the Dirac system
with Cauchy data $\left[\begin{smallmatrix} G \\ F
\end{smallmatrix}\right]$ on $x>0$, $t=0$, and it has $g(u)$
and $k(v)$ as boundary values. To complete the proof of
Theorem \ref{theo:4} there only remains to show the formulas
relating $F$, $G$, $g$, and $k$ and this will be done in the
next section.
On the Hilbert space $L^2(0,\infty;\,dx)\oplus
L^2(0,\infty;\,dx)$ of the pairs $(F,G)$, we can define
a unitary group $U(\xi)$, $-\infty<\xi<\infty$, as
follows: we define its action at first for $(F,G)$ with
$F', G' \in L^2$. Let $\left[{\psi\atop
\phi}\right]$ be the solution of first order system
\eqref{eq:dirac} such that $\phi(0,x) = F(x)$, $\psi(0,x) =
G(x)$. Then we take:
\begin{equation}
U(\xi)(F,G) = (\left.\phi_{-\xi}\right|_{t=0}, \left.\psi_{-\xi}\right|_{t=0})
\end{equation}
where \eqref{eq:spinor} has been used. As
$\xi$ increases from $-\infty$ to $+\infty$ this has the
effect of transporting $\phi$ and $\psi$ forward along the
Lorentz boosts trajectories. We can also implement
$U(\xi)$ as a unitary group acting on the $L^2$ space of
the $g(u) = \phi(-u,u)$ functions, or on the space of the
$k(v) = \psi(v,v)$ functions. We then have,
taking into account \eqref{eq:spinor} (and $-\xi$):
\begin{equation}
g_\xi(u) = e^{\frac\xi2} g(e^{\xi} u)\qquad\qquad
k_\xi(v) = e^{-\frac\xi2} k(e^{-\xi} v)
\end{equation}
Following the terminology of Lax-Phillips
\cite{laxphillips} (the change of variable $u\to \log(u)$
would reduce to the additive language of
\cite{laxphillips}) we shall say that $(F,G)\mapsto I(g)$
provides an incoming (multiplicative) translation
representation ($U(\xi)$ moves the graph of
$e^{y/2}I(g)(e^{y}) = e^{-y/2}g(e^{-y})$ to the right by
an amount of additive time $\xi$) and $(F,G)\mapsto k$ is
an outgoing translation representation. We use $(Ig)(u) =
\frac1u g(\frac1u)$ as it is translated by $U(\xi)$ in the
same direction as $k$. The assignment $Ig\to k$ will be
called the ``scattering matrix'' $\cal S$ (it is canonical
only up to a translation in ``time'', which means here
only up to a scale change in $u$). With our previous
notation it is $S = \cH I$. Let us give a ``spectral''
representation of $S$. For this we represent $g$ as a
superposition of (multiplicative) harmonics, $g(u) =
\frac1{2\pi} \int_{\Re(s)=\frac12} \wh g(s)
u^{s-1}\,|ds|$, with $\wh g(s) = \int_0^\infty g(u)\,
u^{-s}\,du$, $s=\frac12+i \tau$. Then the unitary operator
$S$ will be represented as multiplication by a unit
modulus function $\chi(s)$. Multiplication by $\chi(s)$
must send the Mellin transform $\Gamma(s)$ of $I(e^{-u})$
to the Mellin transform $\Gamma(1-s)$ of $e^{-u}$, in
other words:
\begin{equation}\label{eq:chi}
\chi(s) = \frac{\Gamma(1-s)}{\Gamma(s)}
\end{equation}
We thus see that the first order system in the wedge of two
dimensional space-time provides an interpretation of this
function (for $\Re(s) = \frac12$) as a scattering matrix.
To obtain the Hankel transform of order zero, and not its
succ\'edan\'e $\cH$, one writes $s = \frac14 + \frac
w2$, where again $\Re(w) = \frac12$. In fact, with our
normalizations, the scattering matrix corresponding to the
tranform $g(t)\mapsto f(u)=\int_0^\infty
\sqrt{ut}J_0(ut)g(t)\,dt$ is
the function $2^{\frac12 -w}\frac{\Gamma(\frac34 - \frac
w2)}{\Gamma(\frac14+\frac w2)}$ on the critical line
$\Re(w) = \frac12$.
\section{Application of Riemann's method}\label{sec:riemann}
The completion of the proof of Theorem \ref{theo:4} will now
be provided.
I need to briefly review
Riemann's method (\cite[IV\textsection 1]{john},
\cite[VI\textsection5]{courant}), although it is such a
classical thing, as I will use it in a special manner
later. In the case of the (self-adjoint) Klein-Gordon
equation $\frac{\partial^2 \phi}{\partial u\partial v} =
+\phi$, $t^2 - x^2 = 4(-u)v$, Riemann's method combines:
\begin{enumerate}
\item whenever $\phi$ and $\psi$ are two solutions, the
differential form
$ \omega = \phi\dpsidu\,du + \psi\dphidv\,dv$
is
closed,
\item it is advantageous to use either for $\phi$ or for
$\psi$ the special solution (Riemann's function) $R(P,Q)$
which reduces to the constant value $1$ on
each of characteristics issued from a given point
$P$. Here $R(P,Q) = R(P-Q,0) = R(Q-P,0)$, $R((t,x),0) =
J_0(\sqrt{t^2 - x^2}) = J_0(2\sqrt{-uv})$.
\end{enumerate}
Usually one uses Riemann's method to solve for $\phi$ when
its Cauchy data is given on a
curve transversal to the characteristics. But one can
also use it when the data is on the characteristics
(Goursat problem). Also,
one usually symmetrizes the formulas obtained in combining
the information from using $\phi\frac{\partial R}{\partial
u}\,du + R\dphidv\,dv$ with the information from using
$R\dphidu\,du + \phi\frac{\partial R}{\partial v}\,dv
$. For our goal it will be better not to symmetrize in
this manner. Let us recall as a warming-up how one can use
Riemann's method to find $\phi(t,x)$ for $t>0$ when $\phi$
and $\dphidt$ are known for $t=0$. Let $P = (t,x)$, $A =
(0,x-t)$, $B = (0,x+t)$, and $R(Q) = R(P-Q)$.
\[ \phi(P) - \phi(A) = \int_{A\to P} \dphidv dv = \int_{A\to
P} R \dphidv dv + \phi \frac{\partial R}{\partial
u} du = \int_{A\to B} + \int_{B\to P} = \int_{A\to B}\]
Hence:
\[ \phi(P) = \phi(A) + \int_{A\to B} ( R \dphidv
+ \phi \frac{\partial R}{\partial u}) \frac{dx}2 \]
Using $R\dphidu\,du + \phi\frac{\partial R}{\partial v}\,dv$ we get
in the same manner:
\begin{equation}\label{eq:2} \phi(P) = \phi(B) - \int_{A\to B} ( R \dphidu
+ \phi \frac{\partial R}{\partial v}) \frac{dx}2 \end{equation}
After averaging:
\[ \phi(P) = \frac{\phi(A) + \phi(B)}2 + \frac12 \int_{A\to
B} ( R \dphidt - \phi \frac{\partial R}{\partial t}) dx\]
This gives the classical formula ($t>0$):
\begin{equation}\label{eq:classic}
\begin{split} \phi(t,x) = \frac{\phi(0,x-t) +
\phi(0,x+t)}2
- \frac12 \int_{x-t}^{x+t} t\; \frac{J_1(\sqrt{t^2 -
(x-x')^2})}{\sqrt{t^2 - (x-x')^2}}\phi(0,x')dx'\\
+ \frac12 \int_{x-t}^{x+t} J_0(\sqrt{t^2 -
(x-x')^2})\dphidt(0,x')dx'
\end{split}\end{equation}
I have not tried to use it to establish theorem
\ref{theo:1}. Anyway, when $\phi$,
$\dphidx$, $\dphidt$ all belong to $L^2$ at $t=0$, this
formula shows that $\phi(P)$ is continuous in $P$
for $t>0$. Replacing $t=0$ with $t=-T$, we find that
$\phi$ is continuous on spacetime.
Let us now consider the problem, with the notations of
Theorem \ref{theo:4}, of determining $k(v) = \psi(v,v)$
for $v>0$
when $F(x) = \phi(0,x) = -\dpsidu(0,x)$ and $G(x) =
\psi(0,x) = - \dphidv(0,x)$ are known for $x>0$. We use $P
= (v_0,v_0)$, $A = (0,0)$, $B= (0,2v_0)$. We then have:
\[ R(t,x) = J_0(\sqrt{(v_0-t)^2 - (v_0-x)^2}) =
J_0(2\sqrt{u(v_0-v)})\qquad R(0,x) = J_0(\sqrt{x(2v_0 - x)})\]
\[ \frac{\partial R}{\partial v} =
\frac{J_1(2\sqrt{u(v_0-v)})}{2\sqrt{u(v_0-v)}} 2u \qquad
\frac{\partial R}{\partial v}(0,x) =
\frac{J_1(\sqrt{x(2v_0 - x)})}{\sqrt{x(2v_0 - x)}} x\]
Hence, using \eqref{eq:2} (for $\psi$):
\begin{equation} \psi(v,v) = G(2v) + \frac12\int_0^{2v} (
J_0(\sqrt{x(2v_0 - x)}) F(x) - x\, \frac{J_1(\sqrt{x(2v_0
- x)})}{\sqrt{x(2v_0 - x)}} G(x)) \,dx\end{equation}
We then consider the converse problem of expressing $G(x) =
\psi(0,x)$ in terms of $k(v) = \psi(v,v)$. We choose
$x_0>0$, and consider the rectangle with vertices $P =
(\frac12 x_0, \frac12 x_0)$, $Q = (0, x_0)$, $Q' = (X,
X+x_0)$, $P' = (X+ \frac12x_0, X+\frac12 x_0)$ for
$X\gg0$. We take Riemann's function $S$ to be $1$ on the
edges $P\to Q$ and $Q\to Q'$. We then write:
\[ \psi(Q) - \psi(P) = \int_{P\to Q}\dpsidu \,du =
\int_{P\to Q}(S \dpsidu \,du + \psi \frac{\partial
S}{\partial v} dv)\]
\[ = \int_{P\to P'} + \int_{P'\to Q'} + \int_{Q'\to Q} =
\int_{P\to P'} \psi \frac{\partial
S}{\partial v}\, dv + \int_{P'\to Q'} S\dpsidu\,du\]
\begin{equation} G(x_0) = \psi(\frac{x_0}2,\frac{x_0}2) + \int_{P\to P'}
\psi \frac{\partial
S}{\partial v}\, dv - \int_{P'\to Q'} S\phi \,du\end{equation}
Now, $|S|\leq 1$ on the segment leading from $P'$ to $Q'$,
so we can bound the last integral, using Cauchy-Schwarz,
then the energy integral, and finally the theorem
\ref{theo:1}. So this term goes to $0$. On the light cone
half line from $P$ to $\infty$
we have:
\[ S(v,v) = J_0(\sqrt{x_0(2v-x_0)})\qquad \frac{\partial
S}{\partial v} = -
\frac{J_1(\sqrt{x_0(2v-x_0)})}{\sqrt{x_0(2v-x_0)}}\,x_0\]
\begin{equation} G(x_0) = \psi(\frac{x_0}2,\frac{x_0}2) -
\int_{x_0/2}^\infty
\frac{J_1(\sqrt{x_0(2v-x_0)})}{\sqrt{x_0(2v-x_0)}}\,x_0
\psi(v,v)\,dv
\end{equation}
Our last task is to obtain the formula for $F(x_0)$. We use
the same rectangle and same function $S$.
\[ \phi(Q) - \phi(Q') = \int_{Q'\to Q} \dphidv dv =
\int_{Q'\to Q} S\dphidv dv + \phi \frac{\partial
S}{\partial u} du = \int_{Q'\to P'} \phi \frac{\partial
S}{\partial u} du + \int_{P'\to P} S\dphidv dv + 0 \]
On the segment
$Q'\to P'$ we integrate by parts to get:
\[ \int_{Q'\to P'} \phi \frac{\partial
S}{\partial u} du = \phi(P')S(P') - \phi(Q') - \int_{Q'\to
P'} \dphidu S\,du\]
Again we can bound $S$ by $1$ and apply Cauchy-Schwarz to
$\int_{Q'\to
P'} \dphidu S\,du$. Then we observe that $\int_{Q'\to P'}
|\dphidu|^2 |du|$ is bounded above by the energy integral,
which itself is bounded above by the energy integral on
the horizontal line having $P'$ as its left end. By
Theorem \ref{theo:1} this goes to $0$. And regarding
$\phi(P')$ one has $\lim_{v\to+\infty} \phi(v,v)=0$ as
$\phi(v,v)$ and its derivative belong to
$L^2(0,+\infty;\,dv)$. We cancel the $\phi(Q')$'s on both
sides of our equations and obtain:
\[ \phi(Q) = - \int_{P\to(\infty,\infty)} S \dphidv dv =
+\int_{P\to(\infty,\infty)} S \psi dv \]
Hence
\begin{equation} F(x_0) =
\int_{x_0/2}^\infty J_0(\sqrt{x_0(2v-x_0)})\psi(v,v)\,dv\end{equation}
In conclusion: the functions $F(x) = \phi(0,x)$, $G(x) =
\psi(0,x)$, and $k(v) = \psi(v,v)$ of Theorem
\ref{theo:4} are related by the following formulas:
\begin{subequations}
\begin{align}
F(x) &=
\int_{x/2}^\infty J_0(\sqrt{x(2v-x)})k(v)\,dv\\
G(x) &= k(\frac{x}2) -
\int_{x/2}^\infty x\,
\frac{J_1(\sqrt{x(2v-x)})}{\sqrt{x(2v-x)}}\,
k(v)\,dv\\
k(v) &= G(2v) + \frac12\int_0^{2v}
J_0(\sqrt{x(2v - x)}) F(x) \,dx - \frac12\int_0^{2v}
x\, \frac{J_1(\sqrt{x(2v
- x)})}{\sqrt{x(2v - x)}} G(x) \,dx
\end{align}
\end{subequations}
Exchanging $F$ and $G$ is like applying a time reversal so
it corresponds exactly to exchanging $k(v) = \psi(v,v)$
with $g(u) = \phi(-u,u)$. So the proof of Theorem
\ref{theo:4} is complete.
\section{Conformal coordinates and concluding remarks}
The Rindler coordinates $(\xi,\eta)$ in the right wedge are defined by
the equations $x =
\eta \cosh\xi$, $t = \eta \sinh\xi$. Let us use the
conformal coordinate system:
\[
\xi = \frac12 \log\frac{x+t}{x-t}\qquad\qquad \zeta =
\frac12 \log(x^2 - t^2) - \log2 = \log\frac\eta2
\]
where $-\infty<\xi<+\infty$, $-\infty<\zeta<+\infty$. The
variable $\xi$
plays the r\^ole of time for our scattering. The reason
for $-\log2$ in $\zeta$ is the following: at $t=0$ this
gives $e^\zeta = \frac12 x = u = v$. The differential
equations we shall write are related to the understanding
of the vanishing condition for an $\cH$ pair on an
interval $(0,a)$. And $a = \frac12 (2a)$ hence the
$-\log2$ (to have equations identical with those in
\cite{hankel}.) The Klein-Gordon equation becomes:
\begin{equation}
\label{eq:6}
\frac{\partial^2\phi}{\partial\xi^2} -
\frac{\partial^2\phi}{\partial\zeta^2} + 4e^{2\zeta} \phi
= 0
\end{equation}
If we now look for ``eigenfunctions'', oscillating
harmonically in time, $\phi = e^{-i\gamma\xi}\, \Phi(\zeta)$,
$\gamma\in\RR$, we obtain a Schr\"odinger eigenvalue
equation:
\begin{equation}
\label{eq:4}
- \Phi''(\zeta) + 4e^{2\zeta} \Phi(\zeta) = \gamma^2 \Phi(\zeta)
\end{equation}
This Schrödinger operator has a potential function which can
be conceived of as acting as a repulsive exponential
barrier for the de~Broglie wave function of a quantum
mechanical particle coming from $-\infty$ and being
ultimately bounced back to $-\infty$. The solutions of
\eqref{eq:4} are the modified Bessel functions
(\cite{watson}) of imaginary argument $i\gamma$ in the
variable $2e^{\zeta}$. For each $\gamma\in\CC$ the unique
(up to a constant factor) solution
of \eqref{eq:4} which is square integrable at $+\infty$ is
$K_{i\gamma}(2e^{\zeta})$.
\From\ Theorem
\ref{theo:4} it is more convenient to express the $\cH$
transform as a scattering for the two-component,
``Dirac'', differential system. The spinorial nature of
$\left[\begin{smallmatrix} \psi\\
\phi\end{smallmatrix}\right]$ leads under the change of
coordinates $(t,x)\mapsto (\xi,\zeta)$ to
$e^{\frac\zeta2}e^{-\frac\xi2}\phi$ rather than $\phi$,
and to $e^{\frac\zeta2}e^{+\frac\xi2}\psi$ rather than
$\psi$. In order to get quantities which, in the past at
$\xi\to-\infty$, look like $\phi$ and, in the future at
$\xi\to+\infty$, look like $\psi$ we consider the linear
combinations:
\begin{subequations}
\begin{align}\label{eq:ABdefa}
\cA &= \frac12 \,e^{\frac\zeta2}\,(
+e^{-\frac\xi2}\phi + e^{\frac\xi2} \psi)\\
\label{eq:ABdefb}
\cB &= \frac i2 \,e^{\frac\zeta2}\,(
-e^{-\frac\xi2}\phi + e^{\frac\xi2} \psi)
\end{align}
\end{subequations}
Their differential system is:
\begin{subequations}
\begin{align}
+i\frac{\partial\cA}{\partial\xi} &=
+\left(\frac\partial{\partial\zeta} - 2e^\zeta\right)\cB
\\ +i\frac{\partial\cB}{\partial\xi} &=
-\left(\frac\partial{\partial\zeta} + 2e^\zeta\right)\cA
\end{align}
\end{subequations}
Or, if we look for solutions oscillating in time as $e^{-i\gamma\xi}$:
\begin{subequations}\label{eq:ABsys}
\begin{align}\label{eq:ABsysa}
\left(\frac\partial{\partial\zeta} - 2e^\zeta\right)\cB &= \gamma\cA
\\ \label{eq:ABsysb}
\left(-\frac\partial{\partial\zeta} - 2e^\zeta\right)\cA &= \gamma\cB
\end{align}
\end{subequations}
and this gives Schrödinger equations:
\begin{subequations}
\begin{align}
\label{eq:ABschroda}
- \frac{\partial^2\cA}{\partial\zeta^2} + (4e^{2\zeta} -
2e^\zeta) \cA &= \gamma^2 \cA \\
\label{eq:ABschrodb}
- \frac{\partial^2\cB}{\partial\zeta^2} + (4e^{2\zeta} +
2e^\zeta) \cB &= \gamma^2 \cB
\end{align}
\end{subequations}
So we have two exponential barriers, and two associated
``scattering functions''
giving the induced phase shifts. From our previous
discussion of the scattering in the Lax-Phillips formalism
we can expect from equation \eqref{eq:chi} that a
formalism of Jost functions will confirm these functions to be
\begin{equation}
\label{eq:S}
\cal S(\gamma) = \frac{\Gamma(\frac12 -
i\gamma)}{\Gamma(\frac12 + i \gamma)}\qquad
(\gamma\in\RR)\;,
\end{equation}
for the equation associated with $\cA$ and $-\cal
S(\gamma)$ for the equation associated with $\cB$. And
indeed the solution $\left[\begin{smallmatrix} \cA_\gamma\\
\cB_\gamma\end{smallmatrix}\right]$ of the system
\eqref{eq:ABsys} which is square-integrable at $+\infty$
is given by the formula
\begin{equation}
\label{eq:10}
\begin{bmatrix} \cA_\gamma(\zeta)\\
\cB_\gamma(\zeta)\end{bmatrix} = \begin{bmatrix}
e^{\frac\zeta2}\left(K_s(2e^{\zeta}) +
K_{1-s}(2e^\zeta)\right)\\ i\,
e^{\frac\zeta2}\left(K_s(2e^{\zeta}) -
K_{1-s}(2e^\zeta)\right)\end{bmatrix}\qquad(s=\frac12+i\gamma)
\end{equation}
Let $j_\gamma(\zeta)$ be the
solution of \eqref{eq:ABschroda} which satisfies the Jost
condition $j_\gamma(\zeta)\sim
e^{-i\gamma\zeta}$ as $\zeta\to-\infty$. Then the exact
relation holds (a detailed
treatment is given in \cite{hankel}):
\begin{equation}
\label{eq:8}
\cA_\gamma(\zeta) = \frac12\left(\Gamma(s) j_\gamma(\zeta)
+ \Gamma(1-s) j_{-\gamma}(\zeta)\right)\qquad(s=\frac12+i\gamma)
\end{equation}
We interpret this as saying that the $\cA$-wave comes
from $-\infty$ and is bounced back with a phase-shift
which at frequency $\gamma$ equals
$\arg\frac{\Gamma(\frac12
-i\gamma)}{\Gamma(\frac12+i\gamma)} = \arg\cS(\gamma)
$. For the $\cB$ equation one obtains $-\cS(\gamma)$ as
the phase shift function.
We have
associated in \cite{cras4} Schr\"odinger equations to the
cosine and sine kernels whose potential functions also
have exponential vanishing at $-\infty$ and exponential
increase at $+\infty$, and whose associated scattering
functions are the functions
arising in the functional equations of the Riemann and
Dirichlet L-functions. The equations (13a), (13b) of
\cite{cras4} are analogous to \eqref{eq:ABdefa},
\eqref{eq:ABdefb} above, and (14a), (14b) of \cite{cras4}
are analogous to \eqref{eq:ABsysa} and \eqref{eq:ABsysb}
above. The analogy is no accident. The reasoning of
\cite{cras4} leading to the consideration of Fredholm
determinants when trying to understand self- and
skew-reciprocal functions under a scale reversing operator
on $L^2(0,+\infty;\,dx)$ is quite general. The (very
simple) potential functions in the equations
\eqref{eq:ABschroda} and \eqref{eq:ABschrodb} can be
written in terms of Fredholm determinants associated
with the $\cH$ transform. The detailed treatment is given in
\cite{hankel}.
The function $\cal S(\gamma)$ arises in
number theoretical
functional equations (for the Dedekind zeta functions of
imaginary quadratic fields). We don't know if its
interpretation obtained here in terms of the Klein-Gordon
equation may lead us to legitimately hope for number
theoretical applications. An interesting physical context
where $S(\gamma)$ has appeared is the method of angular
quantization in integrable quantum field theory
\cite[App. B]{lukzak}. And, of course the group
of Lorentz boosts and the Rindler wedge are
connected by the Bisognano-Wichman theorem
\cite{biswich1,biswich2,haag}.
The potentials associated in \cite{cras4} to the cosine and
sine kernels are, contrarily to the simple-minded
potentials obtained here, mainly known through their
expressions as Fredholm determinants, and these are
intimately related to the Fredholm determinant of the
Dirichlet kernel, which has been found to be so important
in random matrix theory.
It is thus legitimately considered an important problem to
try to acquire for
the cosine and sine kernels the kind of understanding
which has been achieved here for the $\cH$ transform. Will
it prove possible to achieve this on (a subset, with
suitable conformal coordinates) of (possibly higher
dimensional) Minkowski space?
We feel that some kind of
non-linearity should be at work.
A tantalizing thought presents itself: perhaps the
kind of understanding of the Fourier transform which is
hoped for will arise from the study of the causal
propagation and scattering of (quantum mechanical?) waves
on a certain curved Einsteinian spacetime.
\setlength{\parskip}{\the\baselineskip}
\addtolength{\baselineskip}{-6pt}
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\end{document}