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In quantum mechanics, when hamiltonian $H$ is constrained ($H = \sqrt{m^2 - \hbar^2 \nabla^2} $) so that it would produce simple "relativistic" model of quantum mechanics, we can show that it results in non-locality (Reference: $\nabla$ and non-locality in simple relativistic model of quantum mechanics )

The question is would Taylor-expanding every constraint equation on some quantity/operator, such as Hamiltonian, show that it will result in non-locality? Or in some case, should we check other expansions/methods?

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anyone........? – War Oct 12 '12 at 8:02
any comment...? – War Oct 13 '12 at 3:55
up vote 1 down vote accepted

In the specific case of the Hamiltonian the non-locality arises because the time evolution depends on values of the field which are arbitrary far away. In the one dimensional case we have

\begin{equation} i \frac{\partial}{\partial t}\, \psi ~~~=~~~ \tilde{H}\,\psi ~~=\ \sqrt{~m^2+\mathbf{\tilde{p}}^2_x~}\ \psi\ =\ \nonumber \end{equation}

\begin{equation} \sqrt{ ~m^2-\partial_x^2~}~~ \psi ~~=~~ \frac{m}{x} K_1\left(mx\right) ~*~ \psi ~~~~~~~ \end{equation}

( We used $\hbar=c=1$ ). In the last term $*$ denotes a convolution, in this case with a Bessel K function. It is clear that this instantaneous dependency violates the speed of light restriction.

See also my stackexchange answer here: $\nabla$ and non-locality in simple relativistic model of quantum mechanics

Now in the general case the value of $\psi(x)$ will depend on $\psi(y)$ at other locations in the past an it will depend on other fields such as $A^\mu(y)$ at other locations in the past.

Mathematically these dependencies stem from "Taylor-expanded series" of differential operators but as long as you don't violate the speed of light restriction then this is perfectly fine.


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