# Schrodinger solution to hamilton-jacobi

1. Apr 6, 2010

### Identity

The Hamilton-Jacobi equation

$$\frac{\partial W}{\partial t}+\frac{1}{2m}\left[\left(\frac{\partial W}{\partial x}\right)^2+\left(\frac{\partial W}{\partial y}\right)^2+\left(\frac{\partial W}{\partial z}\right)^2\right] + V(x,y,z) = 0$$

Can be re-expressed as $$|\nabla W| = \sqrt{2m(E-V)}$$ by taking $$W = -Et+S(x,y,z)$$

Schrodinger says that if we think of the level curves of W, and assign an arbitrary curve the value $$W_0$$, that we can take a normal to that paticular level curve (spanning $$W_0+dW$$) to be":

$$dn = \frac{dW_0}{\sqrt{2m(E-V)}}$$

(In other words, $$\frac{dW_0}{dn} = |\nabla W|$$)

Where does this come from? How do we know the normal differential has this value?

thanks

2. Apr 6, 2010

### Ben Niehoff

This is just from the definition of the gradient.

3. Apr 6, 2010

### Identity

Isn't it just $$|n| = |\nabla W|$$??