# Derive Schrodinger equation

1. Jan 16, 2015

### hunc

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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!

2. Jan 16, 2015

### stevendaryl

Staff Emeritus
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 $\hbar$)

3. Jan 16, 2015

### hunc

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
$$i\hbar\partial_t \phi=-\frac{\hbar^2}{2m}\frac{(\nabla \phi)^2}{\phi}+V(x)\phi.$$

4. Jan 17, 2015

### stevendaryl

Staff Emeritus
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 $-\frac{\hbar^2}{2m} a \nabla^2 a - \frac{i \hbar}{2m} (\nabla^2 S) a^2$.