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\begin{document}
\begin{center}
{\Large \bf A Classical Approach to the Schr\H{o}dinger Equation for a
Particle in a Force Field}
\bigskip
by Frank Thomas Tveter\footnote{e-mail: f.t.tveter@met.no, address:
Norwegian Meteorological Institute, P.O.Box 43 Blindern, N-0313 Oslo,
Norway}
\end{center}
%PACS: \\
% {\bf 03.20.+i} Classical mechanics of discrete systems: general mathematical aspects,\\
% {\bf 04.20.Fy} Canonical formalism, Lagrangians, and variational principles,\\
% {\bf 11.10.Ef} Lagrangian and Hamilton approach.
% {\bf 03.20.+i} Classical mechanics of discrete systems: general mathematical aspects,\\
% {\bf 03.65.-w} Quantum mechanics,\\
% {\bf 03.75.-b} Matter waves
\begin{abstract}
A common approach to the Schr\H{o}dinger equation is to
start with a well known wave function solution to a specific problem,
apply the corresponding Schr\H{o}dinger equation and show
that the result is equivalent to the classical Hamiltonian
energy equation.
In this paper the starting point is the classical Hamiltonian
energy equation for a free particle under influence of
a force potential. An appropriate variable transformation
is applied, and the equation for the corresponding generator
function that solves the problem is identified.
The relationship between the wave function and the generator
function is found by studying the momentum in the two formalisms.
Assumptions necessary to approach the Schr\H{o}dinger
equation are identified.
The approach shown here leads to an equation that is close to the
Schr\H{o}dinger equation, differing only by the appearance
of a term containing the total time derivative of the
probability density.
\end{abstract}
\subsection*{The Schr\H{o}dinger equation}
The well known Schr\H{o}dinger equation describes the time evolution
of the wave function, $\Psi$, of a quantum mechanical system.
According to this theory, the wave function squared, $|\Psi |^2$,
describes the probability, $P$, of finding the system in a given state
at a given time.
The Schr\H{o}dinger equation may be written as
\begin{eqnarray}
\left( \bar H - \bar E\right) \Psi &=& 0 \nonumber
\end{eqnarray}
where $\bar H$ and $\bar E$ are ``operators''.
If the Schr\H{o}dinger equation is applied to a one dimensional system
consisting of a particle in a force field, we get
\begin{eqnarray}
- {\hbar^2 \over 2 m} \dert{\Psi}{q}{q t}
+ U({\bf r}) \Psi &=& i \hbar \der{\Psi}{t}{q t}
\label{schroedinger}
\end{eqnarray}
where $U({\bf r})$ is the force potential, ${\bf r}(q)$ is
the position vector,
$q$ is the (canonical) coordinate,
$i^2 = -1$ and $\hbar$ is the Planck constant (divided by $2 \pi$).
The subscript in the partial derivative expression indicates the functional
dependency of $\Psi$.
In this formalism, the momentum, $p$, is given by the equation
\begin{eqnarray}
p &=& - i \hbar {1\over \Psi} \der{\Psi}{q}{q t} \label{mama} .
\end{eqnarray}
\subsection*{A Classical Approach}
Let us study a classical system where the classical momentum,
$p$, is constrained to being a function of the coordinate, $q$.
Our system consists of a particle in a force field,
so that the state of the system is given by
$\{ q, p(q) \}$.
We know from \cite{tveter1998} that generator functions may be
assigned directly to variable systems according to the equation
\begin{eqnarray}
L - p \dot q + K - {dS\over dt} &=& 0 ,
\label{def1}
\end{eqnarray}
where $L$ is the Lagrange function for the system,
$S$ is the assigned generator function and $K$ is the Hamiltonian.
Note that the value of the Lagrange function is independent
of the variable set (unlike the value of the Hamilton function).
We also know from \cite{tveter1998} that if the assigned
generator function $S$ is a constant in time, then the position
vector can only be a function of the coordinate (i.e. ${\bf r}(q)$).
In this case we may write (\ref{def1}) as
\begin{eqnarray}
L - p \dot q + K &=& 0 .
\label{defqp}
\end{eqnarray}
A variable set where the position vector, ${\bf r}$, only is a
function of the coordinate (i.e. ${\bf r}(q)$) is usually easy
to find for a given problem.
These variable sets are the sets in which we {\em describe} the problem.
If the Hamiltonian $K$ is zero for a variable set, then
the time derivatives of the variables in the variable set is also zero
($\dot Q = \dot P = 0$, where $\{Q,P\}$ is the variable set).
For these variable sets (\ref{def1}) becomes
\begin{eqnarray}
L - {dS_2\over dt} &=& 0 ,
\label{defQP}
\end{eqnarray}
with a solution where $S_2=\int L dt$ (action integral over the
Lagrange function).
Variable sets with this generator function therefore represent
the {\em solution} to the problem.
Eliminating the Lagrange function from (\ref{defqp})
and (\ref{defQP}) gives
\begin{eqnarray}
- p \dot q + K &=& - {dS_2 \over dt}. \nonumber
\end{eqnarray}
Next we place a constraint on the system by forcing the
generator function $S_2$ to be a function of only
$\{ q t \} $ (i.e. $S_2({\bf r}(q),t)$). We may then write
\begin{eqnarray}
- p \dot q + K &=& - {dS_2 \over dt} = -\der{S_2}{q}{q t}\dot q - \der{S_2}{t}{q t}, \nonumber
\end{eqnarray}
which has a solution given by
\begin{eqnarray}
p &=& \der{S_2}{q}{q t} \label{momentum2} \\
K &=& - \der{S_2}{t}{q t} . \label{pk}
\end{eqnarray}
Note that by forcing $S_2$ to only be a function of the coordinate
(i.e. $S_2({\bf r}(q),t)$),
the momentum, $p$, is forced to also be a function
of the coordinate (i.e. $p(q,t)$).
By comparing Eqs. (\ref{mama}) and (\ref{momentum2})
we see that we must have
\begin{eqnarray}
S_2 &=& - i \hbar \ln \Psi \Psi_t \nonumber
\end{eqnarray}
where $\Psi_t$ is a function of $t$ only and determined by the constraints.
This term can be eliminated by redefining the
wave function, $\Psi$, appropriately.
Taking the derivative gives
\begin{eqnarray}
\der{S_2}{t}{q t} &=& - i \hbar {1\over \Psi} \der{\Psi}{t}{q t}
\label{tt}\\
\der{S_2}{q}{q t} &=& - i \hbar {1\over \Psi} \der{\Psi}{q}{q t}
\nonumber \\
\der{S_2}{q}{q t}^2 &=& - \hbar^2 {1\over \Psi^2}
\der{\Psi}{q}{q t}^2 \label{s20} \\
\dert{S_2}{q}{q t} &=& i \hbar {1\over \Psi^2}\der{\Psi}{q}{q
t}^2 - i \hbar {1 \over \Psi} \dert{\Psi}{q}{q t}
\label{s21}
\end{eqnarray}
Let us study a single particle in a force field given by
the force potential $U(q)$. For this system we have \cite{goldstein1981}
\begin{eqnarray}
K(q,p,t) &=& {p^2 \over 2 m} + U(q), \label{defh} \\
\dot q &=& \der{K}{p}{q p t} = {p \over m}, \nonumber
\end{eqnarray}
where $m$ is the particle mass.
Using Eqs. (\ref{momentum2}), (\ref{pk}) and (\ref{defh}) gives
\begin{eqnarray}
{1 \over 2 m} {\der{S_2}{q}{q t}}^2 + U(q) &=&
- \der{S_2}{t}{q t} .
\label{action}
\end{eqnarray}
Let us call this the ``action equation'', since a
solution to this equation gives an $S_2$ that is
equal to the Lagrange action integral (which represents
the transformation that will solve the problem).
Next we need to study the probability density function, $P$,
for the particle.
Simple particle flow density conservation considerations give us
\begin{eqnarray}
\det{\ln P} &=& - \der{\dot q}{q}{q p t}
= - {1\over m} \dert{S_2}{q}{q t} \label{defP} .
\end{eqnarray}
Using Eqs. (\ref{s20}), (\ref{s21}) and (\ref{defP}) gives
\begin{eqnarray}
{1\over 2 m}\der{S_2}{q}{q t}^2 &=& -
{1\over 2 m} \hbar^2 {1\over \Psi} \dert{\Psi}{q}{q t} -
i \hbar {1 \over 2} \det{\ln P} \label{collect}
\end{eqnarray}
Combining Eq. (\ref{tt}), (\ref{action}) and (\ref{collect}) gives
\begin{eqnarray}
- {\hbar ^2 \over 2 m} \dert{\Psi}{q}{q t} + \Psi U(q) &=&
i \hbar \der{\Psi}{t}{q t}
+ i \hbar \Psi \det{\ln \sqrt{P}}
\label{taction}
\end{eqnarray}
We observe that Eq. (\ref{taction}) differs from
Eq. (\ref{schroedinger}) on the right hand side,
by the appearance of an extra term.
The extra term involves the total time derivative of the
probability density.
When the probability density of the particle does not change
as we follow the particle (i.e. constant momentum), this term
is zero, and the resulting equation then corresponds to the
Schr\H{o}dinger Equation.
\subsection*{Conclusions}
This paper has demonstrated how it is possible to start from
the classical Hamilton energy equation and arrive at the
Schr\H{o}dinger equation when several assumptions are applied.
The greatest assumption used here is that the total
time derivative of the probability density is zero, which
is clearly not the case for most systems.
A second important assumption is that the classical
momentum is given by Eq. (\ref{mama}), and a third assumption
is that the probability density changes according to
Eq. (\ref{defP}).
For instance a free particle with uniform distribution probability
and not under the influence of any force potential, obeys all
of the assumptions above, and the probability density
derived from the Schr\H{o}dinger equation,
$\Psi \sim e^{i {p\over \hbar} x - i {E\over \hbar} t}$,
will in this case be in agreement with results from classical mechanics.
\begin{thebibliography}{}
\bibitem[Goldstein 1981]{goldstein1981}
H. Goldstein 1981, Classical Mechanics Second ed.,
Addison-Wesley, USA.
\bibitem[Tveter 1998]{tveter1998}
F. Tveter 1998, Deriving the Hamilton equations of motion for a
nonconservative system using a variational principle,
J. Math. Phys., {\bf 39}, $\#$3, 1495-1500
\end{thebibliography}
\end{document}