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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)?

[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)?

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