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Hamilton-Jacobi Equation

There is nothing stopping us from transforming variables on the phase space from $(q_i, p_i), i = 1, 2, \ldots, n$ to some new $(Q_i, P_i), i = 1, 2, \ldots, n$. Normally, if you change variables this way the functional form of a Hamiltonian is going to change too. In general it is not the case that if (qi, pi) and H satisfy Hamilton equations then the new (Qi, Pi) and $\bar{H}$ will satisfy Hamilton equations too.

But it turns out that if the transformation $(q_i, p_i) \rightarrow
(Q_i, P_i)$ satisfies certain conditions then Hamilton equations are preserved. Such transformations are called canonical transformations and the condition, it turns out, is as follows: for a transformation to be canonical there must exist a function

\begin{displaymath}S(q_1, \ldots, q_n, P_1, \ldots, P_n, t),
\end{displaymath}

called a forming or a generating function, such that
pi = $\displaystyle \frac{\partial S(\boldsymbol{q}, \boldsymbol{P}, t)}
{\partial q_i}$ (4.26)
Qi = $\displaystyle \frac{\partial S(\boldsymbol{q}, \boldsymbol{P}, t)}
{\partial P_i}$ (4.27)
$\displaystyle \bar{H}(\boldsymbol{Q}, \boldsymbol{P}, t)$ = $\displaystyle H(\boldsymbol{q}, \boldsymbol{p}, t) +
\frac{\partial S}{\partial t}$ (4.28)

This observation provides us with the means to solve Hamilton equations and find quite easily all constants of motion. Imagine that we have found a canonical transformation, given by some forming function S, such that the new Hamiltonian is zero. Then from Hamilton equations it follows that:

\begin{eqnarray*}\dot{Q}_i &=& \frac{\partial\bar{H}}{\partial P_i} = 0 \\
\dot{P}_i &=& - \frac{\partial\bar{H}}{\partial Q_i} = 0
\end{eqnarray*}


therefore Qi and Pi must be constants of motion and function S(q, P, t) = PS(q, t) is parametrised by P only, and, furthermore,

\begin{eqnarray*}q_i &=& q_i(\boldsymbol{Q}, \boldsymbol{P}, t) =
{}^{\boldsym...
...\boldsymbol{P}, t) =
{}^{\boldsymbol{Q},\boldsymbol{P}}p_i(t)
\end{eqnarray*}


If S is to be a forming function of a canonical transformation then we must have first that:

\begin{displaymath}p_i = \frac{\partial S(\boldsymbol{q}, \boldsymbol{P}, t)}
{\partial q_i}
\end{displaymath}

and then substituting this condition into H(q, p, t), we obtain:

 \begin{displaymath}
H\left(q_1, \ldots, q_n,
\frac{\partial S}{\partial q_1},...
...{\partial q_n}, t\right)
+ \frac{\partial S}{\partial t} = 0
\end{displaymath} (4.29)

This is the Hamilton-Jacobi equation.

The Hamilton-Jacobi equation is enormously useful in solving analytically and numerically equations of motion for classical particles. The main reason for its usefulness is that it yields all constants of motion automatically, and the solution itself becomes formulated in terms of those constants of motion.

Another interesting feature of this equation is that the forming function S behaves a little like a wave. It can be shown that particle trajectories pierce surfaces of constant S.

A yet another interesting feature of the Hamilton-Jacobi equation is that it can be easily derived from the Schrödinger equation of Quantum Mechanics by representing the wave function $\Psi(\boldsymbol{r}, t)$ in the polar form

 \begin{displaymath}
\Psi(\boldsymbol{r}, t) = {\cal A}(\boldsymbol{r}, t) e^{i S(\boldsymbol{r}, t) / \hbar}
\end{displaymath} (4.30)

where ${\cal A}$ and S are both real.

Here is how this comes about.

Start from the Schrödinger equation for a single quantum ``particle'':

\begin{displaymath}i\hbar \frac{\partial \Psi}{\partial t}
= - \frac{\hbar^2}{2 m} \nabla^2 \Psi + U \Psi,
\end{displaymath} (4.31)

where $\hbar = h/(2\pi)$, and h is the Planck constant, $\Psi$ is the wave function of the particle, and U is a potential. For example we can have U(r) = q V(r), where q is an electric charge of the particle. Now substitute equation (4.34) in place of $\Psi$. This yields for $\partial \Psi/\partial t$:

 \begin{displaymath}
\frac{\partial\Psi}{\partial t} =
e^{iS/\hbar}
\left(
\...
...rac{i}{\hbar}{\cal A}
\frac{\partial S}{\partial t}
\right)
\end{displaymath} (4.32)

There is a little more work that we have to do with the Laplacian on the right hand side of the Schrödinger equation. Let us evaluate $\boldsymbol{\nabla}\Psi$ first:

\begin{displaymath}\boldsymbol{\nabla}\Psi =
e^{iS/\hbar}\left(
\boldsymbol{\n...
...l A}
+ \frac{i}{\hbar}{\cal A}\boldsymbol{\nabla}S
\right)
\end{displaymath} (4.33)

Acting with $\boldsymbol{\nabla}$ again yields:

 \begin{displaymath}
\nabla^2\Psi =
e^{iS/\hbar}
\left(
\nabla^2{\cal A} - \...
... \boldsymbol{\nabla}S
+ {\cal A}\nabla^2 S
\right)
\right)
\end{displaymath} (4.34)

Now we have to substitute equations (4.36) and (4.38) into the Schrödinger equation, divide both sides by $e^{iS/\hbar}$, this term will accompany all other terms, and collect separately real and imaginary parts of the resulting equation.

Collecting the real part yields:

\begin{displaymath}- {\cal A}\frac{\partial S}{\partial t}
= - \frac{\hbar^2}{2...
...^2}
\left(\boldsymbol{\nabla}S\right)^2
\right) + U{\cal A}
\end{displaymath} (4.35)

Dividing this equation by $-{\cal A}$ and grouping all terms where $\hbar$ cancels out on the left hand side yields:

 \begin{displaymath}
\frac{\partial S}{\partial t}
+ \frac{\left(\boldsymbol{\n...
...m}
+ U = \frac{\hbar^2}{2m}\frac{\nabla^2{\cal A}}{{\cal A}}
\end{displaymath} (4.36)

The imaginary part, in turn, yields

\begin{displaymath}\hbar\frac{\partial A}{\partial t}
=
-\frac{\hbar^2}{2m}\fr...
... A} \cdot \boldsymbol{\nabla}S
+ {\cal A}\nabla^2 S
\right)
\end{displaymath} (4.37)

Multiplying both sides of this equation by $2{\cal A}/ \hbar$ lets us rewrite it finally as:

 \begin{displaymath}
\frac{\partial {\cal A}^2}{\partial t}
= -
\boldsymbol{\n...
... \left(
{\cal A}^2 \frac{\boldsymbol{\nabla} S}{m}
\right)
\end{displaymath} (4.38)

If you look at the real part of the Schrödinger equation, equation (4.40), the one that gives $\partial S/\partial t$, you can see that in the limit $\hbar \rightarrow 0$ it turns into:

\begin{displaymath}\frac{\partial S}{\partial t}
+ \frac{\left(\boldsymbol{\nabla}S\right)^2}{2m}
+ U = 0
\end{displaymath} (4.39)

which is the Hamilton-Jacobi equation for a classical particle moving in a potential field U. The neglected term proportional to Planck constant

\begin{displaymath}\frac{\hbar^2}{2m} \frac{\nabla^2{\cal A}}{{\cal A}}
\doteq - Q
\end{displaymath} (4.40)

is called a quantum potential. For wave functions that are valid solutions of the Schrödinger equation, this quantum potential does not vanish as we move away from a quantum ``particle''. It is the source of non-locality of Quantum Mechanics. Using this non-vanishing quantum potential particles can feel their way around as they travel through space. Or, since the quantum potential is actually made of the particle's wave function, we can just as well say, that a quantum particle spreads as far as its quantum potential, the potential itself being an integral component of what a quantum particle is.

But returning to the Hamilton-Jacobi equation, in summary, what we have just demonstrated is that:

The Hamilton-Jacobi function S is, on the one hand, the forming function of a canonical transformation that annihilates the Hamiltonian, and, on the other, it is the phase of the quantum mechanical wave function that represents a quantum particle.

next up previous index
Next: Solving the Hamilton-Jacobi Equation Up: Kinematics and Dynamics of Previous: Poisson Brackets
Zdzislaw Meglicki
2001-02-26