text:
\section{Why the method of images does not solve the problem}

 It is natural to expect that a Green function for a problem with a
flat, perfectly reflecting boundary can be constructed from the
Green function for all of space by subtracting the Green function
for an image source located symmetrically on the other side of the
boundary. When a potential function is involved, however, the
situation is not as simple as it seems, especially when the
derivative of the potential is not zero at the boundary.

  In our problem \eref{schrod}, it is easy to fall into either of
two traps.

 First, we know the ``free'' propagator \eref{freeprop}.   It might
seem that
 $U_\mathrm{free}(t,x,y) - U_\mathrm{free}(t,x,-y)$
 is the propagator $U(t,x,y)$ for the scenario with the ceiling
present, as it would be if the potential were absent.
 But this function does not satisfy the boundary condition,
$U(t,0,y)=0$.  A source ``uphill'' does not have the same effect at
the boundary as a source ``downhill''\negthinspace.

 The second variant of the fallacy is to write
$U_\mathrm{free}(t,x,y) - U_\mathrm{free}(t,-x,y)$.
 This time the boundary condition is satisfied, but the second term
obeys the wrong differential equation (as a function of $t$
and~$x$), because the sign of the potential has been reversed.

 The only correct way to apply the method of images is \cite{AS}
 to extend the potential to negative~$x$ as an even function:
 $V(x) = -|x|$.  With the modified potential,
$U_\mathrm{free}(t,x,y)$ will be invariant under simultaneous sign
change of $x$ and~$y$, so that the two formulas proposed previously
are equivalent and either  satisfies the propagator problem.
(The reflected term in the propagator presumably
has as its semiclassical approximation
the WKB contribution of the bounce paths.)
Although indisputably correct, this construction is useless for our
purposes, because solving the Schr\"odinger equation for the
piecewise defined potential, $|x|$, is at least as hard as solving
the original ceiling problem.
 Indeed, the standard textbook advice for solving a problem on the
whole real line with an even potential is to decompose into odd and
even modes, and to find the latter by solving the problem on the
half-line with Dirichlet and Neumann boundary condition,
respectively.  Instead of solving the ceiling problem, the gambit
has doubled~it.

\section{Soft ceilings} \label{caustic}

What is happening at the critical curve, $t = \sqrt{x}+\sqrt{y}$,
is made clearer by studying a smoother model.  We replace the hard
ceiling (Dirichlet boundary condition) by a smooth but steeply
rising potential. For algebraic convenience we place the ceiling at
$x=1$ and the barrier on the right instead of the left;
then the potential function
\begin{equation}
V(x) = x + x^n
\label{softpot}\end{equation}
for some large $n$ does what we want.
We take $n$ to be an even integer.  Then
as $n\to+\infty$, the term $x^n$ vanishes for $|x|<1$ and approaches
infinity for $|x|>1$.
The barrier at $x=-1$ was not present in our original scenario
(where it would have been at $x=2$),
but it will not affect the classical solutions in the regime where
we shall examine them.

A potential of this type creates a conventional caustic to which
the Maslov theory applies. As $n\to\infty$, part of the caustic
curve converges to the critical curve while the rest of the
caustic converges to the initial portion of the ceiling.  The
families of classical trajectories starting from $y=0$ are
displayed by {\sl Mathematica\/} in Figures \ref{pot6} and
\ref{pot30} for $n = 6$ and $30$, respectively.  The stray curves
in the upper right of the closeup plots, \ref{pot6}(b) and
\ref{pot30}(b), are artifacts of instability in the numerical
solution of the differential equation; it is not surprising that
these occurred very close to the caustic limit where various
solutions are nearly tangent. (The physical invalidity of these
curves is clear from the observation that they intersect other
trajectories that started with greater kinetic energy but now
appear to have less.) The returning trajectories in the lower
right of plots \ref{pot6}(a) and \ref{pot30}(a), on the other
hand, are artifacts of the model, representing reflection off the
gratuitous floor at $x=-1$ introduced by the potential
\eref{softpot}.

\begin{figure}
\centering\caption{A family of paths leaving the origin with various
initial momenta, for $n=6$ in \eref{softpot}.
Formation of a fold caustic at the top of the figure is clearly visible.
Horizontal axis is $t$, vertical is $x$, $y$ is fixed.}
\includegraphics{pot6a.eps} \qquad \includegraphics{pot6b.eps}
\label{pot6}\end{figure}

\begin{figure}
\centering\caption{The same for $n=30$.
The leftmost part of the caustic curve is converging to  a ``ceiling''
at $x=1$.
The trajectory in the upper right is close to the critical trajectory
\eref{criticalbounce} and can be loosely described as
``the last path that misses the caustic''\negthinspace.}
\includegraphics{pot30a.eps} \qquad
\includegraphics{pot30b.eps}
\label{pot30}\end{figure}

Disregarding these extraneous features, one clearly sees a
standard fold caustic, the envelope of the trajectories,
developing at the top of the figures. Below the caustic each
point has two trajectories through it, which we can label
``bounce'' or ``direct'' according to whether or not the path is
returning from touching the caustic. For a fixed $y$, all
sufficiently energetic paths appear, for large~$n$, to bounce off
the horizontal line $x=1$, which becomes the ceiling. Less
energetic paths do not reach all the way to the ceiling, but on
their return they cross over paths of still lower energy,
creating an envelope close to the critical trajectory of the
hard-ceiling problem.

If we were to construct the semiclassical propagator for this
system, we would presumably find that it is a poor approximation
near the caustic, in the sense that when substituted into the
time-dependent Schr\"odinger equation \eref{schrod}) it leaves a
large residual. At points well inside the caustic ($\sqrt{x} \ll
t - \sqrt{y}$) it should be a good approximation if direct and
bounce contributions are combined with the proper phase.
Nevertheless, as $n\to\infty$ the contribution of the direct
paths alone will converge to $U_{y\mathrm{d}}$ in \eref{Uyd},
which we have seen to be an exact solution of \eref{schrod}
(though not of the ceiling boundary condition).  That is, the
region where $U_{y\mathrm{d}}$ is a poor solution is pressed into
the caustic boundary as $n\to\infty$, as \eref{softpot}
increasingly well approximates a pure linear potential at
smaller~$x$.

Near a typical caustic the amplitude function \eref{wkb_amp} of
the semiclassical approximation diverges, since it is the
Jacobian determinant of the mapping from initial momentum to
final position (which is by definition singular on the caustic).
As just remarked, this is not true of the semiclassical solution
for the hard wall, \eref{Uyd} and \eref{Uyb}. The amplitude of
\eref{Uyb} actually vanishes on the critical curve.  It does,
however, have a square-root singularity there, so that its
residual in \eref{schrod} (which involves second derivatives of
the amplitude) will still blow up.  For this reason, and because
the position-space semiclassical construction yields no nonzero
prediction at all for the region of space-time beyond the
caustic, we turn in the next section to a momentum-space
construction.

\section{\label{Intro}Introduction}

The propagator of a quantum-mechanical system is the integral kernel
(Green function)
that solves the initial-value problem for its time-dependent
Schr\"odinger equation:
\begin{equation}
\rmi \hbar\,\pd{t}{\psi} = H \psi
\mathrel{\Longleftrightarrow} \psi(\mathbf{x},t) =
\int U(\mathbf{x},\mathbf{y},t)
\psi(\mathbf{y},0)\, \rmd\mathbf{y}.
\label{genprop}\end{equation}
The semiclassical or WKB ansatz, in its simplest form, is to write
\begin{equation}
U(\mathbf{x},\mathbf{y},t)
 = A(\mathbf{d}_\mathbf{y})\rme^{\rmi S(\mathbf{d}_\mathbf{y})/\hbar}
\qquad (\mathbf{d}_\mathbf{y} \equiv [(\mathbf{y},0),(\mathbf{x},t)])
\label{genWKB}\end{equation}
and to choose $S((\mathbf{y},s),(\mathbf{x},t))$ and
$A((\mathbf{y},s),(\mathbf{x},t))$
so that \eref{genprop} is satisfied up to order $\hbar^2$.
The \emph{action} $S$ and \emph{amplitude} $A$
 can be constructed by solving the classical equations of motion
and integrating
certain first-order differential equations along the resulting
paths (trajectories) in space-time.

The classical-mechanical problem that must be solved in this
construction is a two-point boundary-value problem, since the
initial point $\mathbf{y}$, the final point $\mathbf{x}$, and the
elapsed time $t-s$ (for time-independent Hamiltonians) are
prescribed.  This problem is generally not well-posed, in the
sense that for certain data there may be more than one path, for
others no path at all (``classically forbidden regions''), and at
still other points $A$ may become infinite. This last phenomenon
is called a \emph{caustic}, which technically can be defined as a
set where the mapping to $\mathbf{x}$ from the initial momentum
$\mathbf{p}$ (at~$\mathbf{y}$) fails to be a diffeomorphism.
Ordinarily a caustic represents a breakdown of the WKB
approximation (although the WKB propagator of a harmonic
oscillator is exact, despite having caustics). A semiclassical
solution that is accurate in the vicinity of a caustic often can
be found by transforming, at least temporarily, into a momentum
representation \cite{MF,delos,litjohn}.

The literature of semiclassical approximation is well developed
for Hamiltonians $H = \mathbf{p}^2/2m + V(\mathbf{x})$ (and much
more general phase-space functions) in $\mathbf{R}^n$ without
boundaries, as considered in the classic paper of Van\,Vleck
\cite{vv} and many other works, such as \cite{MF,delos,litjohn}.
The case $H= -\nabla^2$, with a reflecting boundary but no
potential, is also well studied (for example in \cite{KR}, or
indeed in all geometrical optics). But there has been little or
no attention to problems with both boundaries and potentials,
although the small-$t$ asymptotic expansion of $U$, or rather its
close relative the heat kernel, is well known for very general
operators with boundaries, potentials, Riemannian curvature and
external gauge fields \cite{BGil,kirsten}.

In this paper we construct in detail two WKB propagators for a
simple system: The configuration space is one-dimensional, the
positive $x$~axis; the potential is a decreasing, linear function
of~$x$; the boundary condition at the origin is perfect
reflection (Dirichlet). Because classically this potential
describes a force pulling a particle away from the boundary, we
think of the force as gravitational and call this boundary a
\emph{ceiling}. (Of course, it could at least equally well be a
constant electrostatic force.
 The problem also arises in the study of diffraction by a smooth convex
 obstacle \cite{SS}.) The extension to dimension 3 with a flat
ceiling is immediate, since the propagator factors and the WKB
approximation for the transverse propagator is exact. This model
already presents many of the features of the general case: Since
space-time is two-dimensional, the full multidimensional WKB
theory (using tools from Hamilton--Jacobi theory) is needed. The
WKB approximation is not exact for the wave reflected from the
ceiling, so this model is not ``trivial'' in the sense that the
WKB propagator for a linear or quadratic potential in all of
$\mathbf{R}^n$~is. The spectrum is continuous, which complicates
the solution of the Schr\"odinger equation by separation of
variables; on the other hand, no classical path strikes the
ceiling more than once, so the WKB solution is relatively simple.
(The spectrum is also unbounded below, but only because of what
happens as $x\to+\infty$, so no physical pathology results.) In
the semiclassical construction of $U(x,y,t)$ a certain region of
space-time is classically forbidden, because if $t$ is too large
compared to $x$ and~$y$, ``a ball cannot be thrown from $y$ to
$x$ in exact time $t$ because the ceiling gets in the
way''\negthinspace. The boundary of that region is not a caustic
in the usual sense, because the amplitudes do not diverge there;
however, the amplitude of the wave reflected from the ceiling
does go to $0$ there, indicating a nondiffeomorphic dynamics, and
its derivatives do become singular there. To investigate
semiclassically what happens on and beyond this critical curve,
we construct the propagator $U(x,p,t)$ that yields, in analogy to
\eref{genprop}, the wave function $\psi(x,t)$ in terms of the
initial wave function in momentum space, $\hat\psi(p,0)$. (This
step is the paper's most original contribution to the
literature.) The momentum-based calculation
 seems to be superior to the position-based one,
at least for certain initial data and times,
 because it does not falsely predict that the propagator is
identically $0$ at large~$t$.

In the remainder of this introduction we set up notation and
review some basic formalism. In \sref{position} we find all
solutions of the classical two-point boundary value problem for
initial position~$y$ (and given destination $(x,t)$).  If $t$ is
sufficiently small, there are always two solutions, one that
never reaches the ceiling and one that bounces off it.  If $t$ is
too large, there are no solutions.  There is a unique solution on
the \emph{critical curve}, \eref{criticalbounce}. We calculate
the action and amplitude for both the direct and the bounce path,
thereby obtaining an approximate propagator $U(x,y,t)$ adequate
for points inside (and not too near) the critical curve. In
\sref{caustic} we elucidate the critical curve by replacing the
ceiling by a smooth but rapidly rising potential.  Numerical
solutions for families of classical paths reveal caustics (in the
familiar sense) that converge, in the limit of an infinitely
steep potential, to the union of the critical curve and the
portion of the ceiling that gets struck by the bounce paths. In
\sref{momentum} we solve the two-point boundary value problem for
initial momentum $p$ in place of~$y$; in this case the structure
of the set of allowed paths is more complicated.  Again there is
a mild caustic-like behavior (compare \eref{Apb2} with
\eref{Aby}), but it has moved to a different part of phase space.
Again the actions and amplitudes are calculated. The resulting
approximate propagator $U(x,p,t)$, when applied to an initial
wave packet concentrated at values of $y$ for which $(x,t)$ would
be in the classically forbidden region or on the critical curve,
gives a plausible, nonvanishing value to the wave function. We
have not proved error bounds, but in \sref{numerical} we offer
some preliminary numerical calculations on Gaussian initial
packets that show the expected physical behaviors.

Although an exact analytical solution for the propagator
$U(x,y,t)$ exists as an integral of Airy functions (see
\sref{eigen}), it is difficult (for the present authors, at
least) to use in numerical work, so we have no trustworthy
comparisons to present.  In contrast, the related problem with a
\emph{floor} instead of a ceiling has been often studied (e.g.,
\cite{gea,geacom1,geacom2}); it has a discrete spectrum. In that
case it is the WKB approach that runs into complications, because
there are infinitely many classical paths to sum over.

  The detailed solution of the ceiling problem with linear
potential, presented in this paper, points the way to a
prescription for handling boundaries in problems with arbitrary
potentials (and perhaps ultimately also higher-dimensional
problems with curved boundaries).
 This ``localization'' strategy is analogous to the treatment of
diffraction from corners by applying the formulas for diffraction
by an infinite straight wedge \cite{Keller}, and it provided the
original motivation for this work.

This  paper, especially \sref{momentum} and \sref{numerical},
is based on the Master of Science thesis of the first author
\cite{zapata}.
 A  preliminary account of the conclusions in \sref{position} and
 \sref{caustic} was presented by the second author at QMath8,
 Taxco, Mexico, December 2001.

\subsection{The classical system}  \label{classicsys}
The system considered in this article is a 1-dimensional one
with an impenetrable
barrier at the origin and a linear potential,
\begin{eqnarray}
V(q)=-\alpha q.
\label{potential}\end{eqnarray}
Here $q$ represents the position of the particle, and $\alpha$
characterizes the strength and direction of the potential. For
$\alpha$ greater than zero the barrier is a ``ceiling'' and the
particle may bounce off at most once. If $\alpha$ is less than zero
then the barrier will act as a floor, and will have, in principle,
an infinite number of bounces.
This article considers the ceiling case
for two different types of data,
\begin{itemize}
  \item Initial position, $y$, and final position, $x$;
  \item Initial momentum, $p$, and final position, $x$.
\end{itemize}
The final time is $t$, and when it is necessary to consider an
initial time other than~$0$, it is denoted~$s$.  The time
parameter for a path interpolating between these data is~$\tau$,
and the corresponding position variable is~$q$; occasionally the
notation $p(\tau)$ is needed for the momentum at some point along
the path.

The equation governing the dynamics of the classical system and its
general solution are
\begin{eqnarray}
\frac{\rmd^2q}{\rmd\tau ^2}=\frac{\alpha}{m}\,, \qquad
q(\tau)=\frac{\alpha}{2m}\,\tau^2+A\tau+B.\label{classic}
\end{eqnarray}
The constants of integration, $A$ and $B$, are determined from the
initial data considered.
 For bounce trajectories two subsidiary solutions are
required, $q_1(\tau)$ and $q_2(\tau)$,
corresponding to the dynamics before and after the collision,
respectively. An extra condition must be placed on the bounce
trajectories so that the momenta of these paths at the ceiling are
equal in magnitude and opposite in direction. If $b$ is the
time at which the particle ricochets, this condition is
\begin{eqnarray}
p_1(b)=-p_2(b).
\label{pc}\end{eqnarray}

Unless otherwise stated the remainder of the paper will use the
natural units
\begin{eqnarray}
\hbar\equiv 1, \qquad
m\equiv {\textstyle\frac{1}{2}}\,, \qquad
\alpha\equiv 1.
\label{units}\end{eqnarray}
Note that when the last two of these equations are in force,
position will have the
same dimensionality as time squared, and momentum will have the same
dimensionality as time:
\begin{eqnarray}
[x]=[t]^2, \qquad
[p]=[t],
\label{unitsexamples}\end{eqnarray}
 in contrast to the more familiar situation where $\hbar$ and~$c$
(but not~$m$) are dimensionless and hence
$[x]=[t]$, $[p] = [t]^{-1}$.
Also, the velocity is twice the momentum;
although this is classically weird, it makes the quantum formulas
more elegant.

\subsection{The classical solutions}  \label{classsol}
The preliminary quest is to determine all possible trajectories
connecting the
initial data $(p,s)$  or $(y,s)$  with the final data
$(x,t)$, and the corresponding constraints. For notational simplicity
 the two-point data will be abbreviated
\[
\mathbf{x}_y\equiv[(y,s),(x,t)], \qquad
\mathbf{x}_p\equiv[(p,s),(x,t)].
\]
There are three types of non-bounce trajectories to consider:
\begin{enumerate}
  \item trajectories that initially and finally move away from
         the ceiling:
  $p\equiv p_s\in(0,\infty)$ and $p_\mathrm{f}\in(0,\infty)$;
  \item trajectories that initially and finally move toward the ceiling:
$p\equiv p_s\in(-\infty,0)$ and $p_\mathrm{f}\in(-\infty,0)$;
  \item trajectories that initially move toward the ceiling, and end
  moving away from the ceiling:
$p_0\in(-\infty,0)$ and $p_\mathrm{f}\in(0,\infty)$.
\end{enumerate}
A plot of these three types of  trajectory is in figure
\ref{direct}. The time at which the particle's momentum is zero,
$n$, governs the transition between motion toward the ceiling and
away from the ceiling for type (iii) trajectories.
\begin{figure}
\centering \caption{Three types of non-bounce trajectories with
equal initial data, $y$, and final time, $t$. The only variable is
the final position, $x$. $xd1$ represents the final position of a
type (i) trajectory; $xd2$ the final position for type (ii); and
$xt$ that for a turning trajectory, type (iii).}
\includegraphics{direct.eps}
\label{direct}
\end{figure}

There are certain boundary trajectories, which are
helpful in analyzing the more general situations. Trajectories of
type (i) are separated from trajectories of types (ii) and (iii)
 by a  trajectory whose initial
momentum is zero. It has initial position
\begin{equation}
\tilde{y}\equiv x-t^2,
\label{ytilde}\end{equation}
so that
$q(\tau)=\tau^2+x-t^2$.
In this case, if
$x>t^2$ the initial position will be on the physical side of
the ceiling, but for $x<t^2$ the initial position will not
be physical. Also, keeping the final data fixed implies that if an
initial positive momentum is given to this trajectory, then the
initial position must be moved closer to the ceiling.
Conversely, for an initial negative momentum for a trajectory of
type (ii), the initial position will be pushed away from the
ceiling. Therefore,
one  has
$y<\tilde y$ for type (i) and $y>\tilde y$ for types (ii) and (iii).

The linear potential implies that the momentum gained by the
particle is equal to the time of flight, $\tau$
(recall \eref{unitsexamples}),
 and the total
momentum of the trajectory at any time is
\begin{equation}
\tau+p.
\label{pgain}\end{equation}
Therefore,
in case (iii) the time at which the particle will turn around is
\begin{eqnarray}
n=-p.\label{n}
\end{eqnarray}
 The distance from the initial point to the turning point
 for a trajectory where $t=n$ is
\begin{eqnarray}
\Delta=-\int_0^n v(\tau)\,\rmd\tau=
-2\int_0^{-p}(p+\tau)\,\rmd\tau = p^2=t^2.
\label{delta}
\end{eqnarray}
The negative sign occurs because the
velocity is inherently negative during the initial segment of the
motion considered.
The boundary between cases (ii) and (iii) is marked by $y=\Delta$,
or $t=\sqrt{y}$.

For some initial and final data, both a direct (non-bounce) trajectory
and
a more energetic bounce trajectory exist; see figure \ref{dirbou}.
\begin{figure}
\centering \caption{General bounce trajectory (dotted line)
connecting initial data, $y$, to the final data, $x$, in the same
time, $t$, as a direct trajectory (solid line).}
\includegraphics{dirbou.eps}
\label{dirbou}
\end{figure}
The critical trajectories for the bounce paths are the type (iii)
trajectories with zero momentum at the ceiling, corresponding to
zero energy.
\begin{figure}
\centering \caption{Generic family of bounce trajectories with the
same initial position and final time. The final position, $x$, is
varying. The critical trajectory is dashed, and $bc$ denotes the
time of bounce for this trajectory, $xc$ denotes the final position.
The region above and to the left of the dashed curve is
classically inaccessible.
}
\includegraphics{bounce.eps}
\label{bounce}
\end{figure}
All other  trajectories for the bounce case  have energy
greater than zero, since those with negative energy can never
reach the ceiling.
Also, all allowed trajectories of type (iii) have negative energy,
since otherwise they would pass through the ceiling.
 A way to view the construction of bounce trajectories is to join
two trajectories at the ceiling which there obey equation
({\ref{pc}}); one trajectory  has the correct initial data, and
the other the final data. Therefore, the critical trajectory
 occurs when these two trajectories are the same;
the bounce trajectory becomes an ordinary path of type (iii). A particle
in a linear potential beginning at the ceiling with zero energy (and
therefore zero initial momentum) will end at location $x$ with
momentum $\sqrt{x}$. Conversely, if the particle
arrives at the ceiling with zero energy, its momentum at the
starting point, $y$, must have been $-\sqrt{y}$.
Therefore, by \eref{pgain} the total momentum the particle will
gain on the critical bounce trajectory is
\begin{eqnarray}
\sqrt{x}+\sqrt{y}=t.\label{criticalbounce}
\end{eqnarray}
This equation characterizes the critical trajectory.
As will be seen in detail in \sref{position},
 the critical trajectory marks
the boundary between the region of space-time points $(x,t)$
 that can be connected to $y$ by classical paths and those that cannot
(see figure \ref{bounce}).

\subsection{The quantum theory}\label{eigen}

The time-dependent Schr\"odinger equation satisfied by our particle,
in the units \eref{units}, is
\begin{equation}
-\,\pd{x^2}{^2\psi} - x\psi = \rmi \,\pd t{\psi}\,.
\label{schrod}\end{equation}
The ceiling is implemented by restricting $x$ to positive values
and imposing the boundary condition $\psi(0,t)=0$.

The solutions of the time-independent Schr\"odinger equation,
\begin{equation}
-\,\frac{\rmd^2\psi_E}{\rmd x^2} - x\psi_E = E \psi_E\,,
\label{schrodE}\end{equation}
are Airy functions, $\Ai(-x-E)$ and $\Bi(-x-E)$ and their
linear combinations.
The spectrum is continuous and unbounded above and below:
$-\infty<E<\infty$.

It is known (e.g., \cite{dean_fulling})
 that the propagator that solves the initial-value problem for
\eref{schrod} is
\begin{equation}
U(x,y,t) = \int_{-\infty}^\infty \psi_E(x)\psi_E(y) e^{-iEt}
\rho(E) \,\rmd E
\label{ceileigen}\end{equation}
with
\begin{eqnarray}
\psi_E(x) = \pi[\Bi(-E)\Ai(-x-E) -\Ai(-E) \Bi(-x-E)], \label{eigfn}\\
 \rho(E) = \pi^{-2} [\Ai(-E)^2 + \Bi(-E)^2]^{-1}.\label{specdensity}
\end{eqnarray}
The integral \eref{ceileigen} is very oscillatory and hence hard to
use in practice.

On the other hand, for a free particle (that is,
when \eref{schrod} holds on the whole real line, with no boundary),
the corresponding formula,
\begin{equation}
U_\infty(x,y,t) = \int_{-\infty}^\infty
\Ai(-x-E) \Ai(-y-E) e^{-iEt}\, \rmd E,
\label{freeeigen}\end{equation}
is known in closed form (e.g., \cite{CN},
\cite[and references therein]{hol}):
\begin{equation}
U_\mathrm{free}(t,x,y) = (4\pi \rmi t)^{-1/2} \,
e^{\rmi(x-y)^2/4t +\rmi(x+y)t/2 -\rmi
t^3/12}.
\label{freeprop}\end{equation}
We rederive this formula in passing in \eref{Uyd}.

For completeness, note that in the presence of a floor instead of
a ceiling
(that is, $\alpha=-1$),
the spectrum is discrete and the propagator is easily seen to take
the form
\begin{equation}
U_\mathrm{floor}(x,y,t) = \sum_{n=1}^\infty
\Ai'(-E_n)^{-2} \Ai(x-E_n)\Ai(y-E_n) e^{-iE_n t},
\label{flooreigen}\end{equation}
where the $E_n$ are the roots of
$ \Ai(-E_n)=0$.
The WKB solution in this case would be difficult, because of the
need to sum over many reflected  classical paths
(as the particle bounces inside the potential well).

\subsection{The WKB ansatz}

In this section we reintroduce the constants $\hbar$ and $m$ and
allow for a
higher-dimensional configuration space;
thus $x$ becomes $\mathbf{x}$, and we usually omit the arguments
$y$, $s$, and~$t$.
The mathematical correspondence between classical physics and the
quantum propagator begins with the ansatz
\begin{equation}
U(\mathbf{x})=
\e^{\frac{\rmi}{\hbar}S(\mathbf{x})}\sum_{j=0}^\infty
\left(\frac{\rmi}{\hbar}\right)^{-j}A_j(\mathbf{x})\label{wkbansatz}
\end{equation}
for a solution to the time-dependent Schr\"{o}dinger equation
\begin{equation}
-\,\frac{\hbar^2}{2m}\nabla^2\psi+V(x)\psi=\rmi\hbar\frac{\partial
\psi}{\partial t}\,.\label{schrodinger}
\end{equation}

The resulting $\emph{O}(\hbar^0)$ equation is the Hamilton--Jacobi
equation of classical physics,
\begin{equation}
\frac{1}{2m}\left(\nabla S(\textbf{x})\right)^2+V(x)=
-\,\frac{\partial S(\textbf{x})}{\partial
t},\label{wkbphase}
\end{equation}
whose solution, the action $S(\mathbf{x})$,  is the phase of the
WKB propagator and also the
generating function for the Lagrangian manifold corresponding to the
classical system \cite{litjohn}:
\begin{equation}
\mathbf{p}(\mathbf{x})\equiv\mathbf{p}(t,\mathbf{x})
=\nabla S(\mathbf{x}),  \qquad
\mathbf{p}(0,\mathbf{y}) = -\nabla_{\mathbf{y}}S.\label{lman_p}
\end{equation}

The amplitude of the propagator is the solution to the
${\emph{O}}(\hbar)$ equation resulting from (\ref{wkbansatz}),
\begin{equation}
A(\textbf{x})\nabla^2 S(\textbf{x})
+2 \nabla S(\textbf{x}) \cdot \nabla A (\textbf{x})
=-\,\frac{\partial A(\textbf{x})}{\partial t}\,.\label{wkbamp}
\end{equation}
The quantity $\rho({\textbf{x}})\equiv\abs{A(\textbf{x})}^2$
is interpreted as the probability density associated with an
ensemble of classical particles. The
initial amplitude, $\rho_0(\textbf{x})$, (i.e., the amplitude
corresponding to $t\to s=0$, $\textbf{x}\to\textbf{y})$
 should agree with the correct
initial form of the quantum propagator, and the evolution of
$\rho(\textbf{x})$ obeys
\begin{eqnarray}
m\,\pd{t}{\rho\left(\mathbf{x}\right)}=-\nabla\cdot
\left[\rho\left(\mathbf{x}\right)\,
\mathbf{p}\left(\mathbf{x}\right)\right]\,,\label{continuity_eq}
\end{eqnarray}
where the first of equations (\ref{lman_p}) defines the momentum
field, $\mathbf{p}\left(\mathbf{x}\right)$.
Equation (\ref{continuity_eq}) is a continuity equation for the
probability density function,
and it can be shown that the evolution of the density function
 from the initial configuration, $\mathbf{x}_0\equiv\mathbf{y}$,
  to the final configuration, $\mathbf{x}$, is
\begin{eqnarray}
\rho_0(\textbf{x}_0)\,\rmd \mathbf{x}_0=\rho(\textbf{x})\,\rmd \mathbf{x}
 \;\Rightarrow\;
\rho(\textbf{x})=\rho_o(\textbf{x}_0)
\abs{\frac{\partial \mathbf{x}_0}{\partial
\mathbf{x}}}\,.\label{amptra}
\end{eqnarray}
In the 1-dimensional case the Jacobian
 $\frac{\partial \mathbf{x}_o}{\partial\mathbf{x}}$ becomes $\pd{x}{y}$.

For the propagator based on trajectories with fixed initial position $y$,
 equation (\ref{amptra}) becomes inapplicable;
the trajectories are labeled by their initial
momenta, $p\equiv p(0)\equiv p(0,x)$, and the Jacobian must be
replaced by
\begin{eqnarray}\left|\frac{\partial p}{\partial x}\right| =
\left| \frac{\partial^2 S}{\partial x \,\partial y}\right| =
\left| \frac{\partial p(t)}{\partial y}\right|,
\label{wkb_amp}
\end{eqnarray}
where \eref{lman_p} has been used.
In that situation the second of equations \eref{lman_p} is replaced by
\begin{equation}
y = \frac{\partial S}{\partial p}\,.
\label{Sreversed} \end{equation}

This paper will not consider solutions of (\ref{wkbansatz}) higher
than $O(\hbar)$.
The WKB approximation for the propagator to first order in $\hbar$ is
therefore
\begin{equation}
U(\textbf{x})=A(\textbf{x})\,\rme^{\frac{\rmi}{\hbar}S(\mathbf{x})}\,,
\end{equation}
and is subject to the same initial conditions as the corresponding
quantum propagator.
For trajectories with initial position data the initial value is
(in dimension~$1$)
\begin{eqnarray}
\lim_{t\rightarrow s}U(x)=\delta(x-y)\,,
\end{eqnarray}
and for trajectories described with initial momentum data the
initial value of the propagator is
\begin{eqnarray}
\lim_{t\rightarrow s}U(p)=\frac{\rme^{\frac{\rmi}{\hbar}xp}}
{\sqrt{2\pi\hbar}}\,.
\end{eqnarray}

It is often said that the WKB approximation is exact whenever the
Hamiltonian is a polynomial of degree at most $2$ in both
position and momentum.  That statement is true, however, only for
the propagator, not for solutions of the Schr\"odinger equation
with more general initial data \cite{BS}, and only when the
Hamiltonian is globally of such a form.  In our problem we shall
find in \sref{position} that the WKB solution is exact until the
wave hits the ceiling but has
 only $O(\hbar)$ accuracy for the reflected wave.

\section{Numerical comparison of the propagators}
\label{numerical}

In this section we try out the propagators
 (\ref{Upd}), (\ref{Upb}), (\ref{Uyd}), and (\ref{Uyb})
by applying them to localized wave packets.
We take
 the initial state of the
system in position space to be a general Gaussian wave packet:
\begin{equation}
\psi\left(y\right)=\left(\frac{2}{\gamma\pi}\right)^{\frac{1}{4}}
\rme^{-\frac{(y-\bar{y})^2}{\gamma}-\rmi y\bar{p}},\label{initialq}
\end{equation}
where $y$ denotes the initial position, the average initial position
is $\bar{y}$, the average initial momentum is $\bar{p}$, and the
constant $\gamma$ prescribes the width of the packet.
Then the initial wave packet in momentum space is the
Fourier transform of \eref{initialq}:
\begin{equation}
\phi\left(p\right)=\left(\frac{\gamma}{2\pi}\right)^{\frac{1}{4}}
\rme^{-\gamma\frac{(p+\bar{p})^2}{4}-\rmi
\bar{y}(p+\bar{p})},\label{initialp}
\end{equation}
which is also Gaussian, with width $4/\gamma$.

Gaussians are chosen to make the treatment as symmetrical as
possible between position space and momentum space, thereby
minimizing any bias in numerical calculations and their
interpretation.    Unfortunately, the support of the Gaussian
$\psi(y)$ extends into the forbidden region $y<0$,
so it does not represent a state of our system, strictly
speaking.
A momentum-space counterpart of this fact  is that
the signed momentum does not exist as a legitimate
quantum observable (a self-adjoint operator) when the
configuration space is a half-line.  An exact Fourier analysis of
such a system would require Fourier sine transforms and a
``momentum'' observable that is nonnegative.  However, the whole
spirit of the semiclassical approximation requires that the
particle be thought of as approximately localized and having an
approximate classical momentum, and our foregoing calculations
have been conducted in this framework.
If
 \begin{equation}
\bar{y}\gtrsim \gamma,
\label{goodgamma}
\end{equation}
then we expect the semiclassical picture to hold.
Furthermore, because of the rapid decay of the Gaussian function,
whenever the wave packet $\psi(y)$ or $\phi(p)$ is located well
inside a classically allowed region in the sense of Table
\ref{tablex} or~\ref{tablep}, the limits of integration
can be extended to infinity without committing great error.

\begin{figure}
\centering \caption{Plot of WKB evolution of an initial wave
packet, from the point of view of a final point $(x,t)$
The solid lines are the classical trajectories to that point for
a particle initially
located at the average value of the initial wave packet;  the
dashed lines are the classical paths starting at points
rather far into the wings of the initial packet.
 The classically forbidden region for trajectories is at the
lower left of the plot, bounded by the critical trajectory (not
shown), which intersects the horizontal axis at a point $y_c$
(compare Figure \ref{bounce}, inverted).}
\includegraphics{packet.eps}
\label{packet}
\end{figure}
Figure \ref{packet} depicts the evolution of an initial Gaussian
wave function from an initial time $\tau=0$ to a final point
$(x,t)$. From each initial point to the right of a critical
value, $y_c\,$, there are two trajectories that arrive at $(x,t)$.
The contribution of each path to the wave function at $(x,t)$
is weighted by the initial wave packet and otherwise determined
by the propagator (direct or bounce, as appropriate).
The value of $y_c$ is determined from \eref{tbound} to be
\begin{equation}
y_c = (t-\sqrt{x})^2.
\label{yc}\end{equation}
Note that the roles of $x$ and $y$ in Figures \ref{bounce} and
\ref{packet} are reversed, so that in Figure \ref{packet} the
classically forbidden
region is below and to the left of the critical trajectory
starting from $y_c\,$.

 Figures \ref{dircomp} and \ref{boucomp}
show the real parts of the direct and bounce contributions
to the final wave function at the  point $(x,t)=(4,5)$.
From Table \ref{tablep} one can see that the classical limits on
the initial momentum
 data are $(-\infty,-3)$.
(The significance of $p=-3$ is that it is the critical value
where $b_p=-p\,$; see \eref{bouncetimep}.)
The classical limits on initial position are
 $(9,\infty)$, in accord with \eref{yc}.
We choose $\gamma=2$, so that  the width of the packet in
natural units is the same in both $y$ and $p$ space,
and  the effective supports of the initial
wave packets (encompassing $\sim 95\%$ of
the  packet) for the position and momentum cases are
$(\bar{y}-2,\bar{y}+2)$ and
$(\bar{p}-2,\bar{p}+2)$, respectively.
Thus (a) \eref{goodgamma} is well satisfied for $\bar{y} \ge 4$;
(b) the initial momentum-space packet is well inside the
classically allowed region if $\bar{p}=-6$ (as we arbitrarily
choose for the plots); (c) the initial position-space packet is
well inside the classically allowed region if $\bar{y} \ge 11$.
The computations indeed show that the two propagators give
essentially equal results for $\bar{y} \ge 11$ but not for
smaller $\bar{y}$.
 Our interpretation then is that the
momentum-space calculation should be preferred for
$4<\bar{y}<11$, and neither should be trusted for smaller
$\bar{y}$.
(The position-space solution might become superior when $\bar{p}$
is too close to an endpoint of a classically allowed interval of
momentum, but we have not verified that.)
In Figures \ref{dircomp} and \ref{boucomp} the two
pieces of the position-space
solution make  spurious large excursions in the critical region
$7<\bar{y}<11$ and then fall rapidly to $0$ for smaller
$\bar{y}$, as expected, since the $y$-space propagator is
identically $0$ in that region and the wave function is coming
entirely from a wing of the Gaussian packet.
The more trustworthy momentum-space solution is small but
nontrivial in that region, again as expected.

\begin{figure}
\centering \caption{Comparison of the evolution of initial wave
functions  by the WKB propagators associated with the
classical direct paths. The initial data is such that $\gamma=2$
and
$\bar{p}=-6$. The final data is $(x,t)=(4,5)$, so that $y_c=9$.
The average initial position, $\bar{y}$, is varied in the plot.}
\includegraphics{dircomp.eps}
\label{dircomp}
\end{figure}

\begin{figure}
\centering \caption{Comparison of the evolution of initial wave
functions  by the WKB propagators associated with the
classical bounce paths. Again,
$\gamma=2$,
$\bar{p}=-6$,  $(x,t)=(4,5)$, $y_c=9$, and the average
initial position, $\bar{y}$, is the independent variable.}
\includegraphics{boucomp.eps}
\label{boucomp}
\end{figure}

\section{The propagator for initial momentum data}\label{momentum}
Throughout this section the denotation of initial data, $(p,s)$,
and final data, $(x,t)$, is
\begin{eqnarray}
\mathbf{d}_p&\equiv\left[(p,0),(x,t)\right],\nonumber\\
\mathbf{b}_p&\equiv\left[(p,0),(0,b_p)\right],\nonumber\\
\mathbf{r}_p&\equiv\left[(-p-b_p,b_p),(x,t)\right].\nonumber
\end{eqnarray}

The general classical path and the corresponding momentum connecting
the initial data $\left(p,s\right)$ with final data
$\left(x,t\right)$ are\numparts
\begin{eqnarray}
q(\tau;\mathbf{x}_p)=
\left(\tau-t\right)^2+2\left(\tau-t\right)
\left(p+t-s\right)+x\label{xpqgen}\\
p(\tau;\mathbf{x}_p)= \left(\tau-s\right)+p.\label{xppgen}
\end{eqnarray}
\endnumparts
The general Hamiltonian of the classical system is therefore
\begin{eqnarray}
H(\tau;\mathbf{x}_p)= \left(t-s+p\right)^2-x.\label{xpHgen}
\end{eqnarray}
Setting $s=0$ corresponds to the direct paths under consideration
in this paper, and to the initial segments of bounce
paths:\numparts
\begin{eqnarray}
q(\tau;\mathbf{d}_p)=\left(\tau-t\right)^2+2(\tau-t)\left(p+t\right)+x,
\label{xpq}\\
p\left(\tau;\mathbf{d}_p\right)=\tau+p\label{xpp}.
\end{eqnarray}
\endnumparts
The Hamiltonian for the direct case is therefore given by
\begin{eqnarray}
H_{p,d}=\left(p+t\right)^2-x=p^2+2pt+t^2-x,\label{Hpd}
\end{eqnarray}
and the energy of the trajectory may therefore be categorized
by:\numparts
\begin{eqnarray}
H_{p,d}>0:\quad
p\in\left(-\infty,-\sqrt{x}-t\right)\cup\left(\sqrt{x}-t,\infty\right),
\label{Hpd>0}\\
H_{p,d}=0:\quad p=\pm\sqrt{x}-t,\label{Hpd=0}\\
H_{p,d}<0:\quad
p\in\left(-\sqrt{x}-t,\sqrt{x}-t\right).\label{Hpd<0}
\end{eqnarray}
\endnumparts
Using equation (\ref{xpqgen}), the bounce paths may be characterized
as:\numparts
\begin{eqnarray}
q\left(\tau;\mathbf{b}_p\right)&=
\left(\tau-b_p\right)^2+2\left(\tau-b_p\right)\left(p+b_p\right),
\label{xpbq1}\\
q\left(\tau;\mathbf{r}_p\right)&=\left(\tau-t\right)^2
+2\left(\tau-t\right)
\left(t-p-2b_p\right)+x.\label{xpbq2}
\end{eqnarray}
\endnumparts
The Hamiltonian for the bounce trajectories is thus
\begin{eqnarray}
H_{p,b}=p(b_p;\mathbf{b}_p)^2=\left(p+b_p\right)^2.
\end{eqnarray}
Note that the bounce Hamiltonian is always greater than or equal to
zero, with equality for $p=-b_p\,$.

Since all trajectories are required to stay on the positive side of
the ceiling, the initial momentum for the non-bounce trajectories is
constrained by
\begin{eqnarray}
q(0;\mathbf{d}_p)=-t^2-2pt+x\geq0\nonumber.
\end{eqnarray}
Therefore a constraint on type (i) and types (ii) and (iii)
trajectories is,
respectively,
\begin{eqnarray}
0< p\leq\frac{x}{2t}-\frac{t}{2}\,,\label{ca}\\
p\le\frac{x}{2t}-\frac{t}{2}<0\label{cb}.
\end{eqnarray}

\subsection{Trajectories of type (i)}
For the particle to initially move away from the ceiling, $p>0$.
Equation (\ref{ca}) is the only other constraint on the system
and therefore
specifies the interval of integration over $p$ for this class of
paths.
Note that (\ref{ca})
implies $\sqrt{x}>t$; therefore, since
$\sqrt{x}-t<\frac{x-t^2}{2t}\,$,
equations (\ref{Hpd>0}) and (\ref{Hpd<0}) imply that the energy of
type (i) trajectories may be either greater than or less than
zero:
\begin{eqnarray}
H<0:\quad p\in\left(0,\sqrt{x}-t\right)\nonumber\\
H>0:\quad p\in\left(\sqrt{x}-t,\frac{x-t^2}{2t}\right).\nonumber
\end{eqnarray}

\subsection{Trajectories of type (ii)} Besides the requirement
that
$p<0$, and equation (\ref{cb}), the particle must also not turn
around nor enter the forbidden region. Equation (\ref{xpp})
shows
that a turning point of the trajectory will occur at
$\tau=-p=\abs{p}$, therefore $t<\abs{p}$. Forbidding the
particle to
 turn around and requiring $x>0$ are enough to ensure that the
trajectory will not enter the forbidden region. Since
$-t<\frac{x}{2t}-\frac{t}{2}$, the second constraint is weaker
than the first and hence the operative constraint on the initial
momentum is
\begin{eqnarray}
p<-t\label{pii}
\end{eqnarray}
for trajectories of type (ii).

Finally, equations (\ref{Hpd>0}) and (\ref{Hpd<0}) reveal that the
energy of the trajectory may again be greater than or less than
zero according to:
\begin{eqnarray}
H_{p,d}>0:\quad p\in\left(-\infty,-\sqrt{x}-t\right),\nonumber\\
H_{p,d}<0:\quad p\in\left(-\sqrt{x}-t,-t\right).
\end{eqnarray}

\subsection{Trajectories of type (iii)} Using equations \eref{n}
and (\ref{Hpd}),
the position and energy of the particle at the turning point are
\begin{eqnarray}
q(n;\mathbf{d}_p)=x-(p+t)^2=-H_{p,d}.\label{quad}
\end{eqnarray}
Therefore, a turning-point trajectory implies that the energy is
less than zero; otherwise the turning point is behind the
ceiling.
Using equation (\ref{Hpd<0}), the initial momentum is therefore
constrained by:
\begin{eqnarray}
-\sqrt{x}-t<p<\sqrt{x}-t\nonumber.
\end{eqnarray}
Also, the opposite of equation (\ref{pii}) must be true;
otherwise
the turning point would not have time to occur:
\begin{eqnarray}
t>\abs {p}\Rightarrow p>-t\label{p3}.
\end{eqnarray}
If $\sqrt{x}>t\Rightarrow x>t^2$, then the upper bound in equation
(\ref{Hpd<0}) will be positive. Therefore, the regions corresponding
to the momentum of the particle and the energy being less than zero
are:
\begin{eqnarray}
x<t^2:\quad p\in\left(-\sqrt{x}-t,\sqrt{x}-t\right)\nonumber\\
x>t^2:\quad p\in\left(-\sqrt{x}-t,0\right)\nonumber.
\end{eqnarray}
However, equation (\ref{p3}) implies that the lower bounds must be
replaced by $-t$, therefore the correct constraints on the momentum
are:\numparts
\begin{eqnarray}
x<t^2:\quad p\in\left(-t,\sqrt{x}-t\right),\\
x>t^2:\quad p\in\left(-t,0\right).
\end{eqnarray}
\endnumparts

\subsection{Bounce trajectories} Since the construction of
equations \eref{xppgen}, \eref{xpbq1}, and \eref{xpbq2}
guarantees the validity of
(\ref{pc}), the final constraint to impose on the bounce
trajectories is the location of the ceiling:
\begin{eqnarray}
q(b_p;\mathbf{r}_p)=0.\label{xpbbc}
\end{eqnarray}
This yields the quadratic equation
\[
b_p^2+b_p\left(\frac{2}{3}p-\frac{4}{3}t\right)+\frac{1}{3}
\left(t^2-2pt-x\right)=0,\]
with solution
\[b_p=\frac{1}{3}\left(2t-p\pm\sqrt{\left(p+t\right)^2+3x}\right).
\]
Requiring the time of bounce, $b_p\,$, to be less than the
trajectory time, $t$, and
the initial momentum to be less than zero yields
\begin{equation}\label{bounce constraint}
\left(\abs{p}-t\right)\pm\sqrt{\left(\abs{p}-t\right)^2+3x}\,\leq0
,\end{equation}
which requires the negative root.
Therefore, independently of the relationship between the initial
momentum and the trajectory time, the correct bounce time is given
by
\begin{eqnarray}
b_p=\frac{1}{3}\left(2t-p-\sqrt{\left(p+t\right)^2+3x}\right).
\label{bouncetimep}
\end{eqnarray}
Constraining the initial position of the trajectory to be in the
classical region implies
\begin{equation}
q(0;\mathbf{b}_p)> 0\Rightarrow
0<b_p<-2p.\label{b3}\end{equation}
However, the ``critical'' trajectory for the bounce path is when the
initial momentum is equal to the time of bounce, which corresponds
to the particle just grazing the ceiling. All other initial
momenta
must be greater, in magnitude, than the time of bounce:
\begin{eqnarray}
0<b_p<-p.
\label{b4}\end{eqnarray}
This constraint is stronger than \eref{b3} and
therefore is the final constraint to impose on the bounce
trajectories.
Equality on the right side of \eref{b4} implies
\begin{equation}
2t+2p-\sqrt{\left(t+p\right)^2+3x}=0.
\label{tempconstraint}\end{equation}
This leads, on the one hand, to
\begin{equation}
p=-t\pm\sqrt{x}\label{pbounceconstraint}.
\end{equation}
But now \eref{tempconstraint} also requires that $t+p>0$, so the
correct solution is
\begin{eqnarray}
p=\sqrt{x}-t.\label{p2}
\end{eqnarray}
On the other hand, the initial momentum
corresponding to a bounce time equal to zero is the
solution of
\[
2t-p-\sqrt{\left(t+p\right)^2+3x}=0,\]
or
\begin{equation}
p=\frac{t^2-x}{2t}\,.\label{p1}
\end{equation}
The right-hand sides of \eref{p1} and \eref{p2} have opposite
signs, and only negative values of them are operative.
Therefore,
the constraints to impose on the initial momentum so
that both inequalities (\ref{b4}) hold are
\numparts
\begin{eqnarray}
x<t^2:\quad p<\sqrt{x}-t,\\
x>t^2:\quad p<\frac{t^2-x}{2t}\,.
\end{eqnarray}
\endnumparts

The  constraints on the initial momentum for given final
data $\left(x,t\right)$ are summarized in Table \ref{tablep}.
They are somewhat more complex than those for given initial
position in \sref{position}, although in compensation we did not
need to solve a cubic equation this time.

\Table{\label{tablep} Constraints on the initial momentum given
the
final data $\left(x,t\right)$. Endpoint cases are not included.}
\br $x\in$
&$p\in$&Trajectory\\
\mr
$\left(t^2,\infty\right)$&$\left(0,\frac{x-t^2}{2t}\right)$&Type
(i) (rightward)\\
$\left(0,\infty\right)$&$\left(-\infty,-t\right)$&Type (ii)
(leftward)\\
$\left(0,t^2\right)$&$\left(-t,\sqrt{x}-t\right)$&Type
(iii) (turning)\\
$\left(t^2,\infty\right)$&$\left(-t,0\right)$&Type (iii)
(turning)\\
$\left(0,t^2\right)$&$\left(-\infty,\sqrt{x}-t\right)$&Bounce\\
$\left(t^2,\infty\right)$&$\left(-\infty,\frac{t^2-x}{2t}\right)$&Bounce\\
\br
\endTable

If $x\in(t^2,\infty)$,
all four types of trajectories may occur for various initial
momenta.

If the initial momentum is
positive, only the trajectory of type (i) is allowable.
Even it does not exist if $p$ is too large for the given $(x,t)$
(or if $x<t^2$).

For $p<0$, further
analysis reveals a distinction between  the regions
  $x\in(t^2,3t^2)$ and
   $x\in(3t^2,\infty)$.
In both cases the monotonic (type (ii)) and
turning regimes of $p$ are mutually exclusive.
When $x<3t^2$,
 the bounce
regime overlaps both of those,
but when $x>3t^2$,
 the bounce regime is a subset of the monotonic one.
In either case there is a possibility of only one path --- that
is, a bounce path may not exist for the given data.

For $x\in(0,t^2)$ there are no paths corresponding to positive
initial momentum. The intervals corresponding to the always
left-moving  and the
turning trajectories are disjoint and
their union coincides with the bounce interval.
Therefore, two paths are always
possible for $x\in(0,t^2)$;
one will always be a bounce, and the other is either type (ii)
$(p<0)$ or  turning.

In short,
 there are regimes of $(p,x,t)$ data where only one path
exists and other regimes where two exist,  or none.
In contrast, for $(y,x,t)$ data, if a  path exists at all
  (i.e., \eref{tbound} is satisfied),
 then there will always be two distinct trajectories,
except in the very special case of a critical trajectory.

\subsection{The action}\label{xpaction}
Using equations (\ref{xpqgen}) and (\ref{xppgen}), the classical
Lagrangian for the general initial and final conditions is found to be
\begin{eqnarray}
L_p(\tau;\mathbf{x}_p)=
\left(\tau-s+p\right)^2
+\left(\tau-t\right)^2+2\left(\tau-t\right)\left(p+t-s\right)+x.
\label{plagr}\end{eqnarray}
Thus the general action for $\tau=t$ and arbitrary
initial and final data, $\mathbf{x}_p\,$, up to a constant of
integration, $S_0$, is
\begin{eqnarray}
S_p(\mathbf{x}_p)=
-\,\frac{\left(t-s\right)^3}{3}+\left(t-s\right)\left(x+p^2\right)
+S_0 = S+S_0\,.
\label{xpSgen}
\end{eqnarray}
The following partial derivatives of the general action will again
prove useful in determining the appropriate initial
condition:\numparts
\begin{eqnarray}
\frac{\partial S}{\partial t}(\mathbf{x}_p)
=-\left(t-s\right)^2+\left(p^2+x\right),\label{dxpSgendt}\\
\frac{\partial S}{\partial
s}(\mathbf{x}_P)=\left(t-s\right)^2-\left(p^2+x\right)
= -\,\frac{\partial S}{\partial t}\,,\label{dxpSgendt0}\\
\frac{\partial S_p}{\partial
x}(\mathbf{x}_p)=\left(t-s\right).\label{dxpSgendx}
\end{eqnarray}
\endnumparts
Using the derivatives of the general initial position from equation
 (\ref{xpqgen}),
\numparts
\begin{eqnarray}
\frac{\partial q}{\partial
t}(s;\mathbf{x}_p)=-2\left(p+t-s\right),\label{dxpqgendt}\\
\frac{\partial q}{\partial x}(s;\mathbf{x}_p)=1\label{dxpqgendx},
\end{eqnarray}
\endnumparts
one sees that the correct initial action is
\begin{eqnarray}
S_0=p(s;\mathbf{x}_p)\,q(s;\mathbf{x}_p)=p\,q(s;\mathbf{x}_p),\nonumber
\end{eqnarray}
because then (cf.~\eref{Sreversed})
\numparts\begin{eqnarray}
 \frac{\partial S_p}{\partial t} = -H(\mathbf{x}_p), \\
 \frac{\partial S_p}{\partial x} = p(t;\mathbf{x}_p), \\
 \frac{\partial S_p}{\partial p} = q(s;\mathbf{x}_p).
\end{eqnarray}\endnumparts
Then the actions for the direct and bounce cases
are respectively\numparts
\begin{eqnarray}
\fl S_{pd}&=&S_p(\mathbf{d}_p)+ p\,q(0;\mathbf{d}_p)\nonumber\\
\fl &=&-\frac{t^3}{3}-p\,t(p+t)+x(p+t),\label{xpSd}\\
\fl S_{pb}&=&S_p(\mathbf{b}_p)+S_p(\mathbf{r}_p)+
 p\,q(0;\mathbf{b}_p)\nonumber\\
\fl&=&-\frac{1}{3}\left[b_p^3+\left(t-b_p\right)^3\right]
-b_pp\left(p+b_p\right)+
\left(t-b_p\right)\left(x+\left(p+b_p\right)^2\right)
-pb_p(2p+b_p).\label{xpSb}
\end{eqnarray}
\endnumparts
In these calculations, terms resulting from the differentiation of $b_p$
play a crucial role,
whereas in the initial-position formulation they cancel each other.

\subsection{The amplitude}

The Jacobian corresponding to the amplitude function is
$\abs{\frac{\partial q\left(s\right)}{\partial q\left(t\right)}}$.
Using equation (\ref{xpqgen}) for the trajectories yields for the
direct case
\begin{eqnarray}
\frac{\partial q(0)}{\partial x}=1,\label{xpdjacobian}
\end{eqnarray}
which reveals that the amplitude for the direct case is a constant
solely depending on the initial density of particles. This initial
density must be in agreement with the the initial form of the
quantum propagator in the momentum representation, and therefore
\begin{eqnarray}
A_d=\frac{1}{\sqrt{2\pi}}\,.\label{xpampdir}
\end{eqnarray}
Using equations (\ref{xpbq1}) and (\ref{xpbq2}) for the
trajectories in the Jacobian for the bounce amplitude yields
\begin{eqnarray}
\frac{\partial q_1(0)}{\partial
x}=\frac{p+b_p}{\sqrt{\left(p+t\right)^2+3x}}\,,
\end{eqnarray}
and therefore the amplitude function for the bounce trajectory is,
up to a phase,
\begin{eqnarray}
A_b=\frac{1}{\sqrt{2\pi}}\sqrt{\abs{\frac{p+b_p}
{\sqrt{\left(p+t\right)^2+3x}}}}\,,\label{Apb}
\end{eqnarray}
where the initial (normalization) factor is carried over from the
incident direct path.

Since  the critical curve for the bounce trajectory is
equivalent to the $E=0$ type-(iii) trajectory, it is evident that
$b_p=n_p=-p$ for the critical trajectory. Hence the amplitude will
vanish for the critical path. All other values of $b_p$ are less
than $-{p}$, by \eref{b4}.
 Therefore, the numerator of equation (\ref{Apb}) will
be negative for bounce paths.
As in \sref{position}, our phase convention is
 that the direct and bounce propagators
 agree at the ceiling, $x=0$.
Therefore, the correct  bounce amplitude is
\begin{equation}
A_{pb}=\frac{1}{\sqrt{2
\pi}}\sqrt{\frac{|p+b_p|}{\sqrt{(p+t)^2+3x}}}\,.
\label{Apb2}\end{equation}
And, with this convention,
to $O(\hbar)$ the WKB approximation for the quantum
propagator is the difference,
$U_{p\mathrm{d}} - U_{p\mathrm{b}}\,$.
of the propagators constructed from
the bounce and non-bounce paths:\numparts
\begin{eqnarray}
\fl U_{p\mathrm{d}}(x,p,t)&=\frac{1}{\sqrt{2\pi}}\,
\exp\left[\rmi\left(-\frac{t^3}{3}+t\left(p^2+x\right)
-p\left(t^2+pt-x\right)\right)\right],\label{Upd} \\
\fl
U_{p\mathrm{b}}(x,p,t)&=-\,\frac{1}{\sqrt{2\pi}}\sqrt{\frac{|p+b_p|}
{\sqrt{\left(p+t\right)^2+3x}}}\,
\exp\left[i\left(-\frac{1}{3}\left[b_p^3+\left(t-b_p\right)^3\right]
-b_pp\left(3p+2b_p\right)\right)\right]\nonumber\\
&\times\exp\left[i\left(\left(t-b_p\right)
\left(x+\left(p+b_p\right)^2\right)\right)\right].\label{Upb}
\end{eqnarray}
\endnumparts
Here $b_p$ is defined by \eref{bouncetimep}.

\section{The propagator for  initial position data}\label{position}
Throughout this section
\begin{eqnarray}
\mathbf{d}_y\equiv\left[(y,0),(x,t)\right], \quad
\mathbf{b}_y\equiv\left[(y,0),(0,b_y)\right], \quad
\mathbf{r}_y\equiv\left[(0,b_y),(x,t)\right].
\label{xdatapoints}\end{eqnarray}
Here $b_y$ denotes the time of bounce for a trajectory starting at~$y$.
In our considerations $x$, $y$, and $t$ are all nonnegative.

The general trajectory and momentum connecting the initial data
$(y,s)$ to the final data $(x,t)$ for a
linear potential are
\numparts\begin{eqnarray}
q(\tau;\mathbf{x}_y)=
\left(\tau-s\right)^2+\left(\tau-s\right)
\left[\frac{x-y}{t-s}-\left(t-s\right)\right]+y,
\label{generalqy}\\
p(\tau;\mathbf{x}_y)=
\left(\tau-s\right)+\frac{1}{2}\left[\frac{x-y}{t-s}
-\left(t-s\right)\right].
\label{generalpy}
\end{eqnarray}
\endnumparts
For the direct paths, $s=0$ in $\mathbf{x}_y\,$, and the
trajectory and momentum are
\begin{eqnarray}
q(\tau;\mathbf{d}_y)=
\tau^2+\tau\left[\frac{x-y}{t}-t\right]+y\label{specificqy},\\
p(\tau;\mathbf{d}_y)=
\tau+\frac{1}{2}\left(\frac{x-y}{t}-t\right).\label{specificpy}
\end{eqnarray}
Note that equation (\ref{specificpy}) implies that a trajectory
with initial momentum $0$ will have initial position $\tilde y$,
 as predicted \eref{ytilde}.
Using equations \eref{n}, \eref{specificpy}, and \eref{specificqy},
 one sees that the time at which the particle  turns around,
$n(x,y,t)$, is
\begin{eqnarray}
n=-p=-\left(\frac{x-y}{2t}-\frac{t}{2}\right) =
\frac{1}{2}\left(\frac{y-x}{t}+t\right),
\label{n1}\end{eqnarray}
and the place where this event occurs is
\begin{equation}
q(n) = -\,\frac14\left(\frac{x-y}t -t\right)^2 +y.
\label{qb}\end{equation}

\subsection{Admissible paths}
In our system the path \eref{specificqy}  is admissible if and
only if it does not penetrate the ceiling during the time interval
$(0,t)$.
If $n\le0$, the path is admissible and of type~(i);
if $n\ge t$, it is admissible and of type~(ii).
If $0<n<t$ and $q(n)\ge0$, the path is admissible and of type (iii);
if
\begin{equation}
 0 < n < t\quad\textrm{and}\quad q(n) < 0 ,
\label{bouncecond}\end{equation}
it is forbidden.
After some calculation these two conditions  respectively translate to
\begin{equation}
t^2 > |x-y|,
\label{bbound}\end{equation}
\begin{equation}
(x+y-t^2)^2 > 4xy.
\label{qbound}\end{equation}

The solution of \eref{qbound} falls naturally into two cases:
If $t^2>x+y$, then \eref{qbound} is equivalent to
$t > \sqrt{x} +\sqrt{y}$,
and \eref{bbound} is then satisfied.
Conversely, if $t \le \sqrt{x} +\sqrt{y}$ and $t^2>x+y$, then $q(n)\ge 0$
and the trajectory is allowed.
If $t^2 \le x+y$, \eref{qbound} is equivalent to
$t^2\le |\sqrt{x}-\sqrt{y}|^2$,
which is easily shown to contradict \eref{bbound};
so all trajectories are allowed in this case.

In summary, all allowed trajectories satisfy
\begin{equation}
t\le \sqrt{x}+\sqrt{y}
\label{tbound}\end{equation}
and all forbidden trajectories violate it,
in accordance with the energy argument at the end of \sref{classsol}.
In the next subsection we verify that \eref{tbound} is also the necessary
and sufficient condition for existence of a bounce trajectory.
For given data $(y,x,t)$, therefore, there will be either one direct
path and one bounce path, or no path at all.  The only exception
is the critical case \eref{criticalbounce}, for which the direct and
bounce paths are the same.

A finer classification of paths is given in Table \ref{tablex},
which is useful for comparison with the listing of paths of
prescribed initial momentum in Table \ref{tablep} in
\sref{momentum}.  Our tables do not include paths
corresponding to endpoints of the parameter intervals listed,
partly because the classification of those can be ambivalent.
The classification is more symmetrical in $x$ and $y$ than it may
appear:
Interchanging $x$ with $y$ interchanges type~(i) with
type~(ii), and it leaves the type-(iii) region, as a whole,
invariant, its boundaries being the lines $t^2=|x-y|$ and
a segment of the parabola $(x-y)^2 -2(x+y)t^2+ t^4=0$.

\Table{\label{tablex} Constraints on the initial position given
the
final data $\left(x,t\right)$. Endpoint cases are not included.}
\br
$x\in$
&$y\in$&Trajectory\\
\mr
$\left(t^2,\infty\right)$&$\left(0,x-t^2\right)$&Type (i)
(rightward)\\
$\left(0,\infty\right)$&$\left(x+t^2,\infty\right)$&Type (ii)
(leftward)\\
$\left(0,t^2\right)$&$\left(\left(\sqrt{x}-t\right)^2,x+t^2\right)$
&Type (iii) (turning)\\
$\left(t^2,\infty\right)$&$\left(x-t^2,x+t^2\right)$&Type (iii)
(turning)\\
$\left(0,\infty\right)$&$\left(\left(t-\sqrt{x}\right)^2,\infty\right)$
&Bounce
\\\br
\endTable

\subsection{The time of bounce}  \label{bouncetime}
   From equations (\ref{generalpy}) and (\ref{generalqy}) the
trajectories and momenta for the bounce paths are
\numparts
\begin{eqnarray}
q_1(\tau;\mathbf{b}_y)=\tau^2-\tau\left(
b+\frac{y}{b}\right)+y, \label{5}\\
q_2(\tau;\mathbf{r}_y)=\left(\tau^2-t^2\right)+\frac{\tau-t}{b-t}
\left(-\left(b^2-t^2\right)-x\right)+x, \label{6}\\
p_1(\tau;\mathbf{b}_y)=\tau-\left(\frac{y}{2b}+\frac{b}{2}\right),
\label{6a}\\
p_2(\tau;\mathbf{r}_y)=\tau-\frac{1}{2}\left(\frac{x}{b-t}
+\left(b+t\right)\right).
\label{6b}
\end{eqnarray}
\endnumparts
Thus the  ceiling condition for the bounce trajectory,
$q_{1}(b)=-q_{2}(b)$, yields the
equation
\begin{equation}
f(b)\equiv b^3+a_2b^2+a_1b+a_0=0,
\label{cubic1}\end{equation}
\[
a_2\equiv-\,\frac{3t}{2}\,, \quad
a_1\equiv\frac{t^2}{2}-\frac{1}{2}\left(x+y\right), \quad
a_0\equiv\frac{yt}{2}\,.
\]

The polynomial discriminant of the cubic equation, $D$, is defined
as \cite{Cubic}
\begin{eqnarray}
D&\equiv R^2+Q^3 \label{Discriminant} \\
&=
\frac{t^2}{64}\left((x-y)^2-\frac{4}{9}(x+y)^2\right)
-\left(\frac{t^2}{12}\right)^3-\left(\frac{1}{6}(x+y)\right)^3
-\frac{t^4}{288}(x+y),  \nonumber
\end{eqnarray}
where the definitions of $R$ and $Q$ are
\numparts
\begin{eqnarray}
Q&=&\frac{3a_1-a_2^2}{9}=-\frac{t^2}{12}-\frac{1}{6}(x+y)\label{Q},\\
R&=&\frac{9a_1a_2-27a_0-2a_2^3}{54}=\frac{t}{8}(x-y)\label{R}.
\end{eqnarray}
\endnumparts
If $D>0$ then one root is real and the other two are complex
conjugates; $D=0$ if all roots are real and at least two are equal;
and $D<0$ if all roots are real and unequal. From equation
(\ref{Discriminant}),
\begin{eqnarray}
D(0)<0
\qquad\textrm{and}\qquad
\lim_{t\rightarrow\infty}D(t)=-\infty. \label{dis}
\end{eqnarray}
To analyze the discriminant's behavior as $t\rightarrow \infty$ let
$T\equiv t^2$; then
\begin{eqnarray}
\frac{dD}{dT}=\frac{1}{64}\left(\left(x-y\right)^2-\frac{4}{9}
\left(x+y\right)^2\right)
-\left(\frac{T}{24}\right)^2-\frac{T}{144}\left(x+y\right).\label{dDdt}
\end{eqnarray}
Therefore, the derivative of the discriminant will tend to $-\infty$
as $t\rightarrow\infty$. Setting \eref{dDdt} to zero reveals
that the critical points of the discriminant occur where
\begin{equation}
t = \pm\sqrt{\left(x(-2\pm3)-y(2\pm3)\right)}.
\label{discrderi} \end{equation}
The two negative roots are not physical, so the only possible
critical points are
\begin{eqnarray}
t_{\mathrm{c}+}=\sqrt{x-5y}, \qquad
t_{\mathrm{c}-}=\sqrt{y-5x}.
\label{critpts}\end{eqnarray}
If the initial and final data do not satisfy $x>5y$ or $y>5x$, then
the derivative of the discriminant will always be negative, and by
\eref{dis}  so will the discriminant. If one of these inequalities
is satisfied,
 then only
one of the  roots \eref{critpts} will be real.
Without loss of generality, assume $x>5y$,
so that the relevant root of is $t_{\mathrm{c}+}$.
Furthermore, relations \eref{dis} imply that this
root must be a maximum for $D(t)$.
Therefore, if $D(t_{\mathrm{c}+})<0$, then
the discriminant will be negative for all values of~$t$.
According to (\ref{Discriminant}),
\[
D(t_{\mathrm{c}+})=\frac{1}{64}\left(\left(x-5y\right)
\left(x-y\right)^2-\left(x-y\right)^3\right).
\]
Thus the discriminant will always be negative if
$x-5y<x-y$, or $y>0$.
Therefore, provided that $y\neq0$, $D(t)<0$ for all $t$  and all
three roots to equation (\ref{cubic1}) are real and unequal.
Similarly, if $y>5x$ then the requirement that $D(t)<0$ is $x>0$,
which is in general true except for the special case $x=0$. The
special cases $y=0$ and $x=0$ respectively imply that the initial
and final positions are at the ceiling; the respective times of bounce
are $0$ and~$t$.

The maximum and minimum of the cubic function \eref{cubic1} are
\[
b_{\mathrm{c}\pm}=\frac{t}{2}\pm\sqrt{\frac{t^2}{12}+\frac{1}{6}(x+y)}\,,
\]
and the point at which the equation changes its concavity is
$\frac{t}{2}\,$. Since $b\rightarrow\pm\infty$ implies
$f(b)\rightarrow\pm\infty$, the point $b_\mathrm{c}$ is where $f(b)$
changes
from concave to convex. Since all three roots must be real, one
root of $f(b)$ will lie between the two extrema, and the other two
must lie outside of the range:
\begin{eqnarray}
r_1\in\left(-\infty,b_{\mathrm{c}-}\right), \qquad
r_2\in\left(b_{\mathrm{c}+},\infty\right), \qquad
r_3\in\left(b_{\mathrm{c}-},b_{\mathrm{c}+}\right).
\label{r3}\end{eqnarray}
Since $r_3$ is the only root which may equal $t$ and $0$, which are
the critical values of the bounce time, and the bounce time is a
continuous function, $r_3$ is the correct root for all trajectories.
Therefore, if a root of $f(b)$ is found which agrees with the
critical  trajectories,
it will be the correct root for all the trajectories.

For the case where $D<0$ the three cubic roots may be written as
(e.g., \cite{Cubic,Namias})
\begin{equation}
r_j\left(x,y,t\right)=\frac{t}{2}+2\sqrt{-Q(x,y,t)}
\cos\left(\frac{\Theta\left(x,y,t\right)+2j\pi}{3}\right),
\label{root}
\end{equation}
where  $j=0,1,2$ and
\begin{eqnarray}
\Theta(x,y,t)=\arccos\left(\frac{R(x,y,t)}{\sqrt{-Q(x,y,t)^3}}\right).
\end{eqnarray}
Finally, for a trajectory in which $x=y$, the correct bounce time is
$b_y=\frac{t}{2}\,$, and only the last range in (\ref{root}) contains
this value. Therefore, by the considerations above
\begin{eqnarray}
b_y &= r_2(x,y,t)   \nonumber \\
&= \frac t2 + \frac1{\sqrt{3}}\sqrt{t^2 + 2(x+y)}
\sin\left[\frac13\,\sin^{-1}
{3\sqrt3 \,t(y-x) \over (t^2+ 2(x+y))^{3/2}} \right].
\label{sinroot}\end{eqnarray}

\subsection{The action}\label{xyaction}
To complete the WKB construction the action of the trajectories and
the amplitude function must be determined. From equations
(\ref{generalqy}) and (\ref{generalpy}) the general Lagrangian for
the initial-position formulation is
\begin{eqnarray}
\fl L\left[q,\dot{q},\tau\right]&=p(\tau;\mathbf{x}_y)^2+
q(\tau;\mathbf{x}_y)\nonumber\\
\fl
&=2\left(\tau-t_0\right)^2+2\left(\tau-t_0\right)
\left[\frac{x-y}{t-t_0}-\left(t-t_0\right)\right]+
\frac{1}{4}\left[\frac{x-y}{t-t_0}-\left(t-t_0\right)\right]^2
+y\label{xyLgen},
\end{eqnarray}
and the corresponding action is
\begin{eqnarray}
S_y(t;\mathbf{x}_y)&=&
\frac{2}{3}\left(t-t_0\right)^3+\left(t-t_0\right)^2
\left[\frac{x-y}{t-t_0}-\left(t-t_0\right)\right]\nonumber\\
&&{}+\left(t-t_0\right)\left(\frac{1}{4}
\left[\frac{x-y}{t-t_0}-\left(t-t_0\right)\right]^2+y\right).
\label{xySgen2}
\end{eqnarray}
Therefore, the actions for the direct and bounce trajectories are
\numparts
\begin{eqnarray}
S_{\mathbf{d}y}=S_y(t;\mathbf{d}_y),\label{xySd}\\
S_{\mathbf{b}y}=S_y(b_y;\mathbf{b}_y)+S_y(t;\mathbf{r}_y).\label{xySb}
\end{eqnarray}
\endnumparts
Note that for the direct case, as $t\rightarrow0$ the action becomes
\begin{eqnarray}
\lim_{t\rightarrow0}S_y(t;\mathbf{d}_y)\rightarrow\
frac{\left(x-y\right)^2}{4t}\,,
\end{eqnarray}
which is the free particle action.

\subsection{The amplitude}
   From equations \eref{specificpy} and \eref{wkb_amp},
 the amplitude for the direct trajectory is
\begin{eqnarray}
\sqrt{\frac{1}{2\pi\rmi}}\sqrt{\frac{\partial p(t;\mathbf{d}_y)}
{\partial y}}
=\frac{1}{\sqrt{4\pi\rmi t}}\,.
\label{directamp}\end{eqnarray}
Note that as  $(x,t)\rightarrow (x,0)$ the constructed amplitude
yields the correct initial condition, $\delta\left(y-x\right)$:
\begin{eqnarray}
\lim_{t\rightarrow
0}\frac{1}{\sqrt{4\pi\rmi t}}\,e^{\rmi S_d\left(x,y,t\right)}=
\lim_{t\rightarrow
0}\frac{1}{\sqrt{4\pi\rmi t}}e^{\frac{\left(y-x\right)^2}{4\rmi t}}
=\delta\left(y-x\right)\,.
\nonumber\end{eqnarray}

Using $r_2$ from equation (\ref{root}) and $p_1\left(0\right)$ from
equation (\ref{6a}), we find the Jacobian corresponding to the action for
the bounce case:
\begin{eqnarray}
\fl\frac{\partial p(0;\mathbf{b}_y)}{\partial
x}&=&\left(\frac{y}{2b_y^2}-\frac{1}{2}\right)\frac{\partial
b_y}{\partial x}\nonumber\\
\fl&=&\left(\frac{y}{2b_y^2}-\frac{1}{2}\right)
\frac{\cos\left(\frac{\Theta+4\pi}{3}\right)}
{6\sqrt{-Q}}\left[1-\frac{Q}{6\sqrt{-D}}
\left(\frac{t}{2}+\frac{R}{Q^4}\right)
\tan\left(\frac{\Theta+4\pi}{3}\right)\right].
\end{eqnarray}
However, linearizing the classical equations of motion for the
potential $V=-\abs{q}$, and then ``folding'' the negative half of
the plane onto the positive half produces a more explicit form of
the Jacobian. For the given potential we have
\begin{eqnarray}
-\frac{\partial V}{\partial
q}=\sgn(q)=2\theta(q)-1,\label{jacobi3}
\end{eqnarray}
where $\theta(q)$ is the Heavyside step function.
Differentiating (\ref{jacobi3}) with respect to a parameter,
$\alpha$, yields
\begin{eqnarray}
\frac{\partial}{\partial\alpha}\left[2\theta(q)-1\right]=2\delta(q)=
2\,\frac{\delta(\tau-b_y)}{\abs{\dot{q}(b_y)}}\,\frac{\partial
q}{\partial \alpha}\,.
\end{eqnarray}
where we have used the functional dependence of the trajectory
 $q(\tau)$ and the fact that
$q(b_y)=0$.  Using equation (\ref{6a}) and
the fact that $p(b_y)<0$ for the trajectory
moving toward the ceiling, we have
\begin{eqnarray}
\abs{\dot{q}(b_y)}=\frac{y}{b_y}-b_y\,.\label{jacobi4}
\end{eqnarray}
Therefore, differentiating Hamilton's equations with respect
$\alpha\equiv y$ yields
\numparts
\begin{eqnarray}
\frac{\rmd}{\rmd\tau}\frac{\partial p}{\partial
 y}=2\,\frac{\delta(\tau-b_y)}{\abs{\dot{q}(b_y)}}\frac{\partial
q}{\partial  y}\,,\label{jacobi1}\\
\frac{\rmd}{\rmd\tau}\frac{\partial q}{\partial y}=2\frac{\partial
p}{\partial y}\,.\label{jacobi2}
\end{eqnarray}
\endnumparts
For the trajectory moving toward the ceiling, $\tau<b_y\,$, equation
(\ref{jacobi1}) predicts
\begin{eqnarray}
\frac{\rmd}{\rmd\tau}\frac{\partial p}{\partial y}=0,\nonumber
\end{eqnarray}
so that $\frac{\partial p}{\partial y}$ is a constant. Therefore,
\begin{eqnarray}
\frac{\partial p}{\partial y}(\tau)=C \equiv
\frac{\partial p}{\partial y}(0)
\end{eqnarray}
for $\tau<b_y\,$, and hence the solution to (\ref{jacobi3}) for
$\tau<b_y$ is
\begin{eqnarray}
\frac{\partial q}{\partial y}=2C\tau+1,
\end{eqnarray}
since $q(0)=y$ implies $\frac{\partial q}{\partial y}(0)=1$.
Integrating (\ref{jacobi1}) for $\tau>b_y$ and using equation
(\ref{jacobi4}) yields
\begin{eqnarray}
\frac{\partial p}{\partial
y}=C+\frac{2}{\abs{\dot{q}(b_y)}}\frac{\partial
q}{\partial y}(b_y)
=\frac{C\left[y+3b_y^2\right]+2b_y}{y-b_y^2}\,.\label{dpdy>b}
\end{eqnarray}
And using this result for $\frac{\partial q}{\partial y}$ yields
\begin{eqnarray}
\frac{\partial q}{\partial y}(\tau)=\frac{\partial q}{\partial
y}(b_y)+2\frac{\partial p}{\partial
y}\int_{b_y}^t d\tau
=\frac{\partial q}{\partial y}(b_y)+2\frac{\partial
p}{\partial y}\left(\tau-b_y\right).\nonumber
\end{eqnarray}
Since $q(t)=x$ it follows that $\frac{\partial q}{\partial y}(t)=0$.
Therefore, after algebraic manipulations, the constant
$C$ is
\begin{eqnarray}
\frac{\partial p}{\partial y}(0)=
\frac{1}{2}\,\frac{5b_y^2-4tb_y-y}{\left(y-b_y^2\right)b
+(t-b_y)\left(y+3b_y^2\right)}\,.
\end{eqnarray}
Finally, putting this result into (\ref{dpdy>b}) and using the
cubic equation
to simplify the denominator, we get the desired Jacobian:
\begin{eqnarray}
-\frac{\partial p}{\partial
y}(t)=\frac{b_y^2-y}{2\left[-3tb_y^2+2(t^2-x-y)b_y+3yt\right]}\,;
\label{dpdyt}\end{eqnarray}
the minus sign comes from ``folding'' the negative half-plane over.

The initial amplitude function for the bounce trajectories must match
in magnitude the one  for the direct trajectories, \eref{directamp}.
Therefore, the final form of the
amplitude function for bounce paths is, up to a possible phase,
\begin{eqnarray}
A_{yb}=\frac{1}{\sqrt{2\pi\rmi}}
\abs{\frac{b_y^2-y}{2\left[-3tb_y^2+2(t^2-x-y)b_y+3yt\right]}}^
{\frac{1}{2}}.
\label{Aby}\end{eqnarray}

Equation (\ref{criticalbounce}) predicts that on the critical
curve, $x=t^2+y-2t\sqrt{y}$. Since the critical curve for the bounce
path is identical with the trajectory of type (iii) for $E=0$, the
time of bounce is given by $n_y$ on the critical curve. And
substituting for $x$ into equation (\ref{n1}) yields
\begin{eqnarray}
y=b_y^2
\end{eqnarray}
on the critical curve. Therefore, the amplitude for the bounce path
will vanish on the critical curve.
Below the critical curve (in the ``allowed'' region) $y>b_y^2$,
so the numerator in \eref{Aby} is negative.

On the ceiling we want the bounce propagator to agree numerically
(up to sign, see below) with the direct propagator of
which it is a continuation.  This limit corresponds to
$x=0$ and $b_y=t$.  One can check (see \eref{Uyd} and \eref{Uyb} below)
that the actions agree there, and that the denominator in the bounce
amplitude reduces as
\[
-3tb_y^2+2(t^2-x-y)b_y +3yt
=-3t^3+2t^4-2yt+3yt = t(y-b_y^2),
\]
so that the amplitudes match as well.

Putting everything together we have the WKB propagators for the direct
and bounce cases:
\numparts
\begin{eqnarray}
\fl U_{y\mathrm{d}}(x,y,t)&=\frac{1}{\sqrt{4\pi \rmi t}}
\exp\left[\rmi\left(\frac{2}{3}t^3+t^2\left[\frac{x-y}{t}-t\right]+
t\left(\frac{1}{4}\left[\frac{x-y}{t}-t\right]^2+y\right)\right)\right],
\label{Uyd}\\
\fl U_{y\mathrm{b}}\left(x,y,t\right)&=\frac{1}{\sqrt{4\pi\rmi}}
\left[{\frac{y-b_y^2}{-3tb_y^2+2(t^2-x-y)b_y+3yt}}\right]^
{\frac{1}{2}}\nonumber\\
\fl&\times\exp\left[\rmi\left(\frac{2}{3}b_y^3
-b_y^2\left[\frac{y}{t}+b\right]+
b\left(\frac{1}{4}\left[\frac{y}{t}+b_y\right]^2
+y\right)\right)\right]\nonumber\\
\fl&\times\exp\left[\rmi\left(\frac{2}{3}(t-b_y)^3+(t-b_y)^2
\left[\frac{x}{t}-(t-b_y)\right]\right)\right]\nonumber\\
\fl&\times\exp\left[\rmi\,\frac{(t-b_y)}{4}\left(\frac{x}{t}
-(t-b_y)\right)^2\right].
\label{Uyb}\end{eqnarray}
\endnumparts
Note that, with the exception of the constraints on the initial
position, the propagator for the direct paths is identical with that
for the linear potential without a ceiling, as given in \eref{freeprop}.
The complete propagator obeying the Dirichlet boundary condition,
$\psi(0,t)=0$, must vanish at the ceiling.  Therefore, its correct WKB
approximation is the \emph{difference},
$U_{y\mathrm{d}}-U_{y\mathrm{b}}\,$, if the phase convention for the
bounce propagator is that adopted in \eref{Uyb}.
The \emph{sum} solves, to lowest order in $\hbar$, the Neumann problem,
$\frac{\partial\psi(0,t)}{\partial x} =0$
(by virtue of \eref{lman_p} and \eref{pc}).
(In more technical language, \eref{Uyb} does not include the Maslov index,
or, rather, its analogue for a sharp boundary.)

redpajama_set_name: RedPajamaArXiv


very nice jquery form plugin! Emblematiq Niceforms is Wonderful. Tank u!
Excellent choice list thanks for sharing.
I just discover your site (or blog). Very interesting.
I am creating our association's WEBsite and would like to use some plugins.
How can I dowload them ?
Very nice! I'll use some ideas... especially in forms.


The State of Vermont does require Home Inspectors to be licensed.
2. Location of Property – how far does the inspector have to travel? Most do not charge for this, but go ahead and ask.
3. Additional Tests – these could include radon and water tests.
Sometimes, either before, during or after the inspection, the inspector may recommend you hire a specialist. This could include a structural engineer, a septic inspector, a heating specialist, an electrician, or a plumber.


text:
Discover and read the best of Twitter Threads about #KYsos
Stephen Wolf
@PoliticsWolf
If Democrats win the Kentucky governor's race, Andy Beshear has promised to restore voting rights to more than 100,000 citizens whom GOP Gov. Matt Bevin has kept disenfranchised #KYgov dailykos.com/stories/2019/5…
https://twitter.com/DKElections/status/1164618099863293953
Kentucky bans 1 in 11 adults from voting, including 1 in 4 black adults, the highest rate of any state in the country. This is the context in which this election is taking place, & it's profoundly undemocratic sentencingproject.org/wp-content/upl…
https://twitter.com/PoliticsWolf/status/1164618821216247809
The Democratic poll has Dems easily winning the #KYsos election. There's been very little polling of Kentucky's governor election, let alone for secretary of state, so it's hard to get a read on how competitive this race is in such a conservative state
https://twitter.com/DavidLJarman/status/1164622290929111040
#KYgov #KYsos

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: What is the proper way to script darcs? I have a few different darcs repos and I would like to write a script to automatically push changes from these repos at the end of the day.  I read at this thread that the "right" way to script darcs is using the library.  But here http://darcs.net/Library it says that "it is very much a work in progress" and lacks a stable API.  What is the "right" way to script darcs.

A: It's perfectly fine to script everyday darcs commands, like any other command-line program. It's also fine to script interactive darcs commands, but that's where it gets fragile, because those interactive prompts get refined and changed quite often in new darcs releases. So if at all possible, make the command non-interactive, eg using the -a/--all flag. Push and pull is easy, here's a regular pull that I run from /etc/crontab:
*/15 * * * * darcsden  cd /home/darcsden/darcs/darcs-reviewed; darcs pull -a http://darcs.net/reviewed >/dev/null # 2>&1

but watch out for things that can prevent a successful non-interactive push or pull:

*

*conflicts, especially with unrecorded changes

*wrong user/file ownership/file permissions

*stray _darcs/lock file left over from an abnormal termination

redpajama_set_name: RedPajamaStackExchange


My Kubo fan art using pen, markers and colored pencils.
A young Japanese boy named Kubo, with his magical shamisen and a monkey companion embark on a journey to find his father's samurai armor and defeat a vengeful spirit from the past. Kubo and the Two Strings is the latest stop-motion animated film from Laika studios and is a gorgeous work of art! Beautifully animated and surprisingly emotional, I was completely taken in by this film. So much so, that I created fan art.
My first sketches of Kubo's head. One looks like a girl and another looks too old. I had to keep trying!
After sketching several heads of Kubo for practice, I felt confident enough to try a full body pose. Using blue pencil first, I sketched out his pose, changed his feet a few times and settled on the final look. I used Pigma Micron pens, 03 and 05 sizes, to ink over my pencil. It was fun to use a combination of colored pencils and Copic markers for color.
The inked version before adding color. I really wanted the color black added throughout the drawing to add weight and balance.
Kubo plays his shamisen and makes origami paper fold on its own into animals and characters that move. My goal was to capture that first moment when he begins to play and the magic happens.
If you haven't seen Kubo and the Two Strings I strongly encourage you to see it! I wasn't sure if it was going to be interesting but I enjoy stop-motion animation so I took a chance. Within the first few minutes, I knew this movie was a masterpiece!
My only gripe was that they used white actors to play the main characters. This is supposed to be ancient Japan. I know Hollywood feels that it needs famous names to draw crowds but it's sad when you have someone as well known as George Takei playing a background character with two lines.
As an actor of Japanese descent, he really should have been given a bigger role. If they plan on distributing it in Japan, I would love to hear what it sounds like with actual Japanese actors doing voice-overs.
Even with that in mind, Kubo and the Two Strings is such a great film and the artistry of a story told with animation done by hand is too captivating to miss. Make sure you go see it and tell me what you think!


text:
Oct 28, 2019 Oct 28, 2019 ~ Peter A Bell
Nicola Sturgeon is being somewhat disingenuous when she speaks of the "SNP plan". Ian Blackford has been roped in by Jo Swinson; apparently without consulting his colleagues. The "plan" to which the First Minister refers is a Liberal Democrat plan. If it's their plan, then we have to assume that they intend it to work to their advantage. Quite why Nicola Sturgeon and Ian Blackford suppose that the SNP and/or Scotland stand to gain from serving as Jo Swinson's side-kick remains to be explained.
When any politician – Nicola Sturgeon most emphatically not excluded – says 'there's no other way' then we may assume two things. Firstly, that there almost certainly is another way. Secondly, that they don't want anybody scrutinising the way they are saying is the only way. They say 'there's no alternative' when they've made a bad choice. Or when they've made a choice for bad reasons and want to pretend that it was out of their hands. So, it pays us to look at how this choice is justified.
Only a few days ago, Ian Blackford was describing the idea of a December election as "barking mad". What changed his mind? How did this go, in the space of a couple of days, from being a totally daft idea to the only sensible course of action? Mr Blackford alone can answer that question. Until he does, we can only deduce his reasoning from his actions. And it looks very much as if he was 'got at' by the LibDems in some way. Somehow, they have persuaded him that partnering them in this little ploy was going to give him something that he wants.
According to Nicola Sturgeon, the putative gain from forcing a general election on 9 December is twofold – stopping Johnson pushing through his 'deal'; and preventing a no-deal Brexit at the end of whatever extension might be granted by the EU. But, as Angus MacNeil correctly points out, this is a bit of a non sequitur. The claimed outcome does not necessarily follow from the action taken or proposed.
We first of all have to wonder about that action. Jo Swinson and Ian Blackford have written to EU Council president Donald Tusk. I'm sure Mr Tusk will respond as politely as he may. But why should he have any regard for a letter from two of the opposition parties in the British parliament? Not even from the official opposition, but from the leaders of two 'lesser' parties. The EU deals with the elected governments of member states. They do not deal with political parties. Not even the governing party. Only with the government.
There is no reason whatever to suppose the Swinson-Blackford letter will carry any weight at all. Even if Mr Tusk was minded to be influenced by it, the decision on an Article 50 extension is made by the governments of the member nations, not the president of the EU Council.
And what does the letter ask for? Only what was all but certainly going to be granted anyway. So, what is the point of the exercise? If/when the extension is agreed, will Jo Swinson and Ian Blackford claim credit? If so, they will be roundly and deservedly mocked. That's not much of a gain.
Whether they can deliver on the 9 December election is also highly dubious. But let's suppose they can. What might this achieve? The likely outcome of an immediate election – to the extent that anything can be described as 'likely' amid the current political chaos – is a UK Parliament dominated by British Nationalist Brexiteers to an even greater extent than at present. If, as is often assumed, these forces want a no-deal Brexit, then a UK general election makes that outcome more likely, not less.
If there is no decisive win for the Mad Brexiteers, then the next most likely outcome is a less-than-decisive win for the Mad Brexiteers. If they are forced to compromise then there is only Boris Johnson's 'deal' to fall back on. There is no chance of another new 'deal'. The EU went above and beyond what was required of them when they reopened negotiations. There is no possibility that they will do so again. So, if an election means there is to be a 'deal' it has to be the one that the Scottish Government has said is unacceptable.
No doubt the LibDems are hoping that an election will put them in a position to demand a new EU referendum. This doesn't look likely, the way the polls stand. But a 'people's vote' would almost certainly require another extension. The patience of the EU member states is not infinite. It would not be at all surprising if one or two governments broke ranks and vetoed any further extension.
Even if there was a new EU referendum, what are the chances it would resolve anything? Practically non-existent. We'd all end up pretty much back where we are now.
There is another justification (rationalisation?) for Ian Blackford's action offered by an SNP spokesperson – getting rid of Boris Johnson. Is that likely to be the outcome of a general election? Even if Johnson were to be removed, would whoever replaces him be any better? What does Scotland stand to gain from a change of British Prime Minister?
Because it's not Boris Johnson that's the problem. Nor is it Brexit – with or without a deal. The problem is the Union. The problem is the grotesque constitutional anomaly which means Scotland will invariably have imposed on it British Prime Ministers and British governments and British policies that the people of Scotland did not vote for or explicitly rejected through the ballot box.
It will doubtless be argued that, in a UK general election, the SNP are likely to enjoy a landslide victory in Scotland on a scale similar to that of 2015. But what advantage did the SNP winning 56 out of 59 seats bring to Scotland? Obviously, it is better than the alternative. At least we can assume that SNP MPs will actually represent and defend Scotland's interests. But how effective can they ever be? Even if the SNP group held the balance of power in numerical terms, the British parties would never allow them to use that power in any meaningful way.
Ian Blackford is getting the SNP group at Westminster embroiled in the British political game in an effort to at least look effective. But no good ever comes of getting into bed with the treacherous Liberal Democrats, or partnering with someone as brazenly self-serving as Jo Swinson. Blackford may imagine he's formed an alliance. Swinson sees it as her using the SNP.
Angus MacNeil is right. As far as Scotland is concerned, there is absolutely nothing to be gained from a UK general election. There is nothing to be gained from Scotland's presence in the parliament of England-as-Britain. Nor will there ever be.
Posted in Politics, Uncategorized	Brexitgeneral electionIan BlackfordNicola Sturgeon
‹ PreviousThen what?
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10 thoughts on "Being used"
Papko says:
Its interesting to read someone who made their name undermining Better Together arguments with a withering analysis.
Now turning that rational on the only party who have a hope of securing Scotland's Independence.
Always find George Orwell's "Animal Farm" very prescient at times like this.
Have the pigs turned into the Farmer?
Big Jock says:
I have been very critical of the SNP tactics recently.
However on the point of an election. I can't see the SNP have anything to lose! We have a chance to remove the 12 Tory Mp's from Scotland and return 50 plus SNP mp's.
The section 30 legislation is going through in December anyway. So what is the purpose in not having an election. Bearing in mind Nicola said she wanted a referendum in 2020 in any case.
It's not as though we are going to have an indy ref vote before the Spring. So why not use the time to strengthen your party. I might not agree with the Section 30 or the tactics , but that is the chosen path. So we need to try an make sure it happens by strengthening our hand. The less unionist MP's in Scotland the better as far as I am concerned!
If your only purpose is to take seats from the Tories (or other British parties?) then a December election will suit you fine. But if you have more ambitious aims, then it achieves nothing. In terms of Scotland's cause, the practical difference between the SNP having 35 seats and 59 is zero. A landslide for the SNP may prove support for a new constitutional referendum and/or for independence, but Westminster doesn't care. If there was a 100% turnout and every single voter in Scotland voted SNP, it would make not the slightest difference to the British state's position. And the British political elite has all the power it needs to defend that position. They get that power from the Union, augmented by the No vote in 2014.
There is no Section 30 legislation going through anywhere in December. A Section 30 order hasn't even been requested… sorry! DEMANDED by Nicola Sturgeon. She may have something to say about that on Saturday. But there is absolutely no reason to suppose Boris Johnson will even respond. He can simply ignore the DEMAND. It worked for Theresa May. She never did give a formal response to Nicola Sturgeon's earlier request. And the Scottish Government was powerless to do anything about it.
Nicola Sturgeon may want a referendum in 2020, but that's not enough to make it happen. She has handed that decision to Boris Johnson. She has conceded to the British Prime Minister the authority to veto our right of self-determination. If a referendum does happen, it'll only happen when Boris Johnson says it can happen and only on his terms. That's where we are. A UK general election will not change that one iota.
Robert Hastings says:
BORING! Peter. Please give it a rest.
Feel free to fuck off at your best pace.
steelewires says:
The SNP has become a British Party that fields candidates and holds Westminster seats only in Scotland, and forms the Devolved Government in Scotland. It's focus is British politics. It talks of independence but recognises the sovereignty of the English parliament of the UK over Scotland. Only using International Law can get Scotland independence, and the SNP will not use it. WHY NOT!
I have been of the opinion we do need a General Election now.
However, I am at same time, very weary of MP Jo Swinson. She is not to be trusted.
It was her own doing, all by herself, that stopped any chance of getting some pro EU tories from getting Corbyn into Downing Street. She is was, who sowed the seeds of doubt in their minds.
And others too, have raised the concerns Peter A. Bell does, here.
But however much I would want an Election just now, it is on one point only, and that it must be SNP leadership use it as a push for Independence. Not to use it only to ask for "permission" for Section 30.
But go back to the principle that a majority of SNP MPs is grounds for Independence.
Otherwise, it is a waste of time to even bother!
Obviously, a majority of votes cast for Independence MPs is better, but as things work in UK today, a simple majority should be sufficient, if politicians use it properly.
SNP had that opportunity in 2015. They should have walked out of Westminster with those 56 MPs in June 2016.
They totally wasted that opportunity, and now look we are are!
So yes, I do want an early Election, and it is for Labour to try to make the best of it in England. The fact they seem incapable and unwilling to consider their chances is a matter for them.
But why should Scotland keep holding back, simply to suit Labour?
But it is also the case, SNP have to be more assertive, and unless they are, not much will change, as Peter, (and others too) outline above.
For example,, SNP is not making any demands, for anything in return, for this support of the Lib Dem plan.
I would opine… this may well be SNP's last chance to prove themselves for Scotland, otherwise, we really do need to consider a new pro Independence Party for this country.
The downside of that, is possible split in the Independence vote, and even if it doesn't quite split the vote i some places anyway, there is a certain timescale, to get such a thing up and running, and to be effective the way it needs to be.
But SNP has benefited from this concern too long now, and there just might come a time, when folks have become so disillusioned, they are left with no other alternative.
But as it stands, SNP are ever more beginning to look like the old Labour Party with its overwhelming majority of MPs all throughout the Thatcher/Major years, and were utterly impotent, and useless!
It is for SNP to prove us wrong with these concerns, and we would love to be proven wrong, too, in this regard.
We wait for that moment to come at this time, however.
SNP had that opportunity in 2015. They should have walked out of Westminster with those 56 MPs in June 2016. They totally wasted that opportunity, and now look we are are!"
And what if a majority of sovereign Scots had still wanted to remain in the Union at that time? As I'm sure was the case. Would that, they, not count, Gordon?
Petra, if those MPs walked out, there would have been a Constitutional Crisis, and the matter would have had to be dealt with.
Instead, this play it by the London rules, play it nicely policy was the one they went with… and gotten us nowhere!
The Sovereign Scots have to choose between London, and EU, and by the looks of it, the majority want EU.
It is clear, they can't have both.


text:
Q: put each word in a span with jQuery I have a text that looks like this ->
car, building, money, computer, new phone, media chat, google plus, desk, gym

Is there a way of putting each of these words in a  tag with jQuery..?
Can't seem to find something like this in internet...

A: From your string, you'll probably want to start by splitting it on the commas and creating an array:
var sentence = "car, building, money";
var words = sentence.split(',');

Then we'll need a destination for the spans:
var $container = $('#container'); // something in your UI, a div maybe

Then you can wrap each one in a span and add it to the container:
$.each(words, function() {
     // create a span whose content is the result of trimming
     // the current array item
     $container.append($('').html($.trim(this)));
});

jsFiddle Demo

A: I believe what you are looking for is:

var myString = "car, building, money, computer, new phone, media chat, google plus, desk, gym";
var result = myString.split(', ');

result now references an array of all the names in your string. You can iterate through the array and dynamically create html tags and put the values in.

A: var str = 'car, building, money, computer, new phone, media chat, google plus, desk, gym'
''+str.replace(/, /g,"")+''

Or
''+str.replace(/, /g,", ")+''

depening on what is the outcome you want to achieve (you're not very good at formulating tasks)

A: Try following :
   var string1 = "car, building, money, computer, new phone, media chat, google plus, desk, gym";
    var str = string1.split(",");
    var container = jQuery('
');
    jQuery.each(str,function(k,v){
        container.append(jQuery('').html(v));
    });
    console.log(container);

Here is the demo :http://jsfiddle.net/dCYyY/


My writing doesn't often stray from design, tech or publishing-related news but as I find myself so deeply immersed in social media throughout my work and personal life, recent news events have forced my hand.
The awful attack on a British serviceman in Woolwich, South-east London yesterday was relayed to the world as the tragedy unfolded via Twitter. Through a combination of words, photos and videos, the terrorists, murderers (or whatever title they eventually fall under) were not only caught in the violent act but their admission of guilt and their misguided view of the people of this great country were recorded. Their fates were sealed.
I say misguided as they clearly haven't observed our resolve under previous threats and atrocities by mindless terrorist organisations. We don't bow to pressure, we don't like being told what to do and we certainly don't give in to fanaticism.
I have been critical of news organisations in the past for responding too slowly to major events. I'm not unreasonably expecting camera teams and reporters to rush to every false alarm, I'm just suggesting they give more power to the people. In much the same way a sherif deputises any number of law enforcement helpers, so the news networks should be quicker to jump on social networks and deputise the men and women on the street to open channels before they arrive on the scene.
The hunting of the Boston Marathon suspects was the perfect example of Twitter providing not only the up-to-the-minute coverage of the hunt for the Tsarnaev brothers but also the background and names of the suspects before any news networks. I was viewing Dzhokhar Tsarnaev's social network profile page before many 'official' organisations had even identified him.
Understandably, criticism can be levelled at social reporting when this coverage lacks the filtering mechanism of good journalism. Fact-checking and the immediate threat of legal action are disregarded in favour of apparent sensationalism and personal opinion – in many cases inflammatory and counter-productive. However, Following the right people at the right time can make a real difference. As an event unfolds, take a (virtual) step backwards and try not to be swept up it the moment. Find the person on the scene, not the loudest voice.
Coverage via social channels not only uncovers the gruesome details but also highlights the incredible spontaneous bravery of the individuals we live amongst. Cub Scout leader Ingrid Loyau-Kennett approached the Woolwich murderers and kept them distracted as the time approached when local school children would begin to file into the crowds on the high street. This took incredible courage and this one act highlights the resolve these senselessly violent acts, from Woolwich to Boston, face – and why they achieve nothing more that strengthen our resolve.


text:
Q: Why is my IDE telling me my if statement is redundant? Why is NetBeans with TMC (IDE recommended by MOOC fi Java) telling my that my final if statement is redundant?
@Override
    public boolean equals(Object object) {
        if (this == object) {
            return true;
        }
        if (object == null) {
            return false;
        }
        if (this.getClass() != object.getClass()) {
            return false;
        }
        LicensePlate compared = (LicensePlate) object;

        if (!this.country.equals(compared.country)) {
            return false;
        }
        if (!this.liNumber.equals(compared.liNumber)) {
            return false;
        }
        return true;

    }

A: I would assume that you can simplify it using the return statement.
Example:
return this.liNumber.equals(compared.liNumber);