How Is the Schrödinger Equation Derived from the Hamilton-Jacobi Equation?

[hunc]

Homework Statement

I am reading Mathematical Concepts of Quantum Mechanics (Stephen J. Gustafson, Israel Michael Sigal. Second edition). The book would like to find an evolution equation which would lead to the Hamilton-Jacobi equation
$$\frac{\partial S}{\partial t}=-h(x, \nabla S) $$
in the way the wave equation led to the eikonal one. The book also says that $\phi (x, t) = a(x, t) \exp( i S(x,t)/\hbar)$. So I express $S(x,t)$ using $\phi (x,t)$ and substitute back to the Hamilton-Jacobi equation, taking $h (x, \nabla S) = \frac{1}{2m}|\nabla S|^2+V(x)$.

The book means to take the leading terms when $\hbar$ small compared to a typical classical action $S$ and restore Schrodinger equation. I am kind of lost during the derivation.

Homework Equations

After the substitution, I have
$$i\hbar\partial_t \phi=-\frac{\hbar^2}{2m}[(\frac{\nabla \phi}{\phi}-\frac{\nabla a}{a})^2-\frac{2im\partial_t a}{a\hbar}]\phi+V(x)\phi.$$
Comparing with Schrodinger equation, I figure that the leading term of
$$[(\frac{\nabla \phi}{\phi}-\frac{\nabla a}{a})^2-\frac{2im\partial_t a}{a\hbar}]\phi$$
should equal to $\Delta_x \phi$, but don't know how.

The Attempt at a Solution

I am not sure what to search for the problem, but wiki have something on this. A nonlinear variant of the Schrödinger equation is expressed as
$$i\hbar\partial_t \phi=-\frac{\hbar^2}{2m}\frac{(\nabla \phi)^2}{\phi}+V(x)\phi.$$
I am not sure what a nonlinear Schrödinger equation is after realizing it's not the same thing as the Schrödinger equation.

The book's goal seems to be the linear Schrödinger equation. Even though I do see how to obtain the nonlinear Schrödinger equation, I am not sure why $(\frac{\nabla \phi}{\phi})^2$ is a leading term. Could someone help me with this?

Thanks!

 

[stevendaryl]

You can derive it in the following way:

1. Let \phi = a e^{\frac{iS}{\hbar}}
2. Compute \frac{-\hbar^2}{2m} \nabla^2 \phi in terms of a and S and only keep the lowest-order terms, in powers of \hbar
3. Similarly, compute i \hbar \frac{d}{dt} \phi in terms of a and S and keep the lowest-order terms in powers of \hbar.
4. Now, use the fact that \frac{(\nabla S)^2}{2m} + V = -\frac{d}{dt} S to show that the results of 2 and 3 are equal (to lowest order in \hbar)

 

[hunc]

You can derive it in the following way:

1. Let \phi = a e^{\frac{iS}{\hbar}}
2. Compute \frac{-\hbar^2}{2m} \nabla^2 \phi in terms of a and S and only keep the lowest-order terms, in powers of \hbar
3. Similarly, compute i \hbar \frac{d}{dt} \phi in terms of a and S and keep the lowest-order terms in powers of \hbar.
4. Now, use the fact that \frac{(\nabla S)^2}{2m} + V = -\frac{d}{dt} S to show that the results of 2 and 3 are equal (to lowest order in \hbar)

Thanks! It was my first attempt, which never really got carried out. I thought $\nabla^2\phi$ can bring in $1/\hbar^2$ and $\partial_t \phi$ only $1/\hbar$... And I just go through it and all is fine.
And now I kind of want to ask what's the story about the wiki and the equation
$$
i\hbar\partial_t \phi=-\frac{\hbar^2}{2m}\frac{(\nabla \phi)^2}{\phi}+V(x)\phi.$$

 

[stevendaryl]

Well, if you let \phi = a e^{\frac{i}{\hbar} S}, then

\phi^* (-\frac{\hbar^2}{2m} \nabla^2 \phi) = -\frac{\hbar^2}{2m} a \nabla^2 a + \frac{\hbar^2}{2m} (\nabla a)^2 - \frac{i \hbar}{2m} (\nabla^2 S) a^2 + \frac{1}{2m} (\nabla S)^2 a^2

\frac{\hbar^2}{2m} |\nabla \phi|^2 = \frac{\hbar^2}{2m} (\nabla a)^2 + \frac{1}{2m} (\nabla S)^2 a^2

So the difference between them is -\frac{\hbar^2}{2m} a \nabla^2 a - \frac{i \hbar}{2m} (\nabla^2 S) a^2.