- #1
Identity
- 152
- 0
The Hamilton-Jacobi equation
[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
[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