Schrodinger solution to hamilton-jacobi

  • Thread starter Identity
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  • #1
Identity
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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
 

Answers and Replies

  • #2
Ben Niehoff
Science Advisor
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This is just from the definition of the gradient.
 
  • #3
Identity
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Isn't it just [tex]|n| = |\nabla W|[/tex]??
 

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