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Intiutive approach to Green's function for SE

  1. Nov 10, 2007 #1
    Griffiths develops an intelgral equation for Scrödinger equation in his QM book. As doing so, he requires Green's function for Helmholtz equation

    (k^2 + \nabla^2) G( \mathbf r) = \delta^3(\mathbf r)

    A rigourious series of steps, including Fourier transforms and residue integrals follow immidiately. As it turns out, the Green's function is independent of direction of [itex]\mathbf r[/itex], i.e. [itex]G(\mathbf r) = G(r)[/itex]. Does anyone know a trick deducing this condition without actually finding the Green's function. The importance to prove is that, it turns out that there's a relatively simple solution if we assume [itex]G(\mathbf r) = G(r)[/itex])
  2. jcsd
  3. Nov 11, 2007 #2


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    This simpler form of the GF is caused by a space-invariance of the potential.
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