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Spinless Salpeter equation non-locality

  1. Dec 11, 2014 #1

    ChrisVer

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    A straightforward attempt to generalize the Schrodinger equation to a relativistic equation, is obtained by taking the Eigenvector/value equation:
    [itex]H |a_k> = E_k |a_k>[/itex]

    with a hamiltonian of the form [itex]H= T+V [/itex],

    In the relativistic case we have [itex]T=\sqrt{p^2+m^2}[/itex] , and taking also (for my purposes) the case that [itex]V(\vec{x})=0[/itex] we get the Spinless Salpeter equation:

    [itex] \sqrt{p^2 +m^2} |a_k> = E_k |a_k> [/itex]

    (http://arxiv.org/abs/hep-ph/9807342 , below eq.1)

    I was wondering, how can someone see immediately that this form of the Hamiltonian is non-local? I think that someone sees this by looking at the propagator? Or does someone get this by writting the Lagrangian and seeing it's non-local (contains terms that are not functions of the fields or their derivatives)?
     
    Last edited: Dec 11, 2014
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  3. Dec 11, 2014 #2

    ShayanJ

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    The point is that you can taylor expand the kinetic energy operator w.r.t. p and it will be an infinite series. But the momentum operator in that series is a derivative operator and that means that operator(T) is in fact differentiating to infinite order. But the higher the derivative, the more points you need and those points get farther from the point your considering which, for an infinite order differentiation, means the points very very far from a point affect the properties of that point and so the theory is non-local.
     
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