- #1

Identity

- 152

- 0

[tex]\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[/tex]

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

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

[tex]dn = \frac{dW_0}{\sqrt{2m(E-V)}}[/tex]

(In other words, [tex]\frac{dW_0}{dn} = |\nabla W|[/tex])

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

thanks