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Klein-Gordon Approximation Question

  1. May 9, 2008 #1

    div curl F= 0

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    I'd be greatful for a bit of help on this question, can't seem to get the answer to pop out:

    A particle moving in a potential V is described by the Klein-Gordon equation:

    [tex] \left[-(E-V)^2 -\nabla^2 + m^2 \right] \psi = 0 [/tex]

    Consider the limit where the potential is weak and the energy is low:
    [tex] |V| << m \;;\; |\epsilon| << m \;;\; \epsilon = E - m [/tex]

    Show that in this limit the KG equation can be approximated by the Schrodinger equation:

    [tex] \left[-\nabla^2 + 2mV \right] \psi = 2m\epsilon \psi [/tex]

    --------------


    It seems that a taylor expansion is required or other approximation is needed to get the factors of 2 but I've tried several now and can't seem to make the answer.

    Any help much appreciated.
     
    div curl F= 0, May 9, 2008
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  3. May 9, 2008 #2

    lbrits

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    Divide the original equation by [tex]m^2[/tex], and Taylor expand. Of course [tex]E[/tex] is of order [tex]m[/tex], so keep that in mind when you divide by "large" things.
     
    Last edited: May 9, 2008
    lbrits, May 9, 2008
  4. May 9, 2008 #3

    lbrits

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    Sorry, divide everything through by [tex]E[/tex], then [tex]V/E[/tex] is small and you can Taylor expand the square and drop the term [tex]V^2/E^2[/tex]. Now multiply through by E again and divide by m?
     
    lbrits, May 9, 2008
  5. May 9, 2008 #4

    ironhill

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    No need for Taylor expansions, just algebra, keep stuff of 1st order of smallness, expand the square;

    E^2 -> m^2 + 2m[tex]\epsilon[/tex]
    2EV -> 2mV
    V^2 -> 0

    Stick that in and you get it.
     
    Last edited: May 9, 2008
    ironhill, May 9, 2008
  6. May 9, 2008 #5

    lbrits

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    Uhm, heh, about the Taylor expand thing... "If all you have is a hammer then everything looks like a nail", it doesn't bother me that it's already a polynomial =)
     
    lbrits, May 9, 2008
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