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Schrodinger solution to hamilton-jacobi

  1. Apr 6, 2010 #1
    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?

  2. jcsd
  3. Apr 6, 2010 #2

    Ben Niehoff

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    Gold Member

    This is just from the definition of the gradient.
  4. Apr 6, 2010 #3
    Isn't it just [tex]|n| = |\nabla W|[/tex]??
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