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I have a quick question about making quantum mechanics relativistic by simply replacing the hamiltonian by a relativistic hamiltonian. If we write the hamiltonian operator as:

H = [itex]\sqrt{P^{2}c^{2}+ m^{2}c^{4}}[/itex],

schrodinger's equation in position basis becomes:

i[itex]\hbar[/itex][itex]\dot{\psi}[/itex] = [itex]\sqrt{-\hbar^{2}c^{2}\nabla^{2}+ m^{2}c^{4}}[/itex][itex]\psi[/itex]

If you expand the square root in powers of nabla, you get an infinite number of gradients. I remember reading that an infinite number of spatial gradients acting on psi implies that the theory is non-local (I don't recall where I read this, but it might be in Mark Srednicki's QFT textbook.) I don't get the jump of logic in saying that an infinite number of gradient operators implies a non-local theory. I think I've come across similar arguments in other contexts in QFT (I'm sorry, I don't recall specifically which ones).

Could someone please explain to me what I am missing here?

Thanks!

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# Question about expanding a square root in powers of gradient

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