# Naïve relativistic schrodinger equation [duplicate]

Possible Duplicate:
Why are higher order Lagrangians called 'non-local'?

Bjorken and Drell presents the equation:

$$i\hbar\frac{d\psi}{dt}=H\psi=\sqrt{p^2 c^2+m^2 c^4}\psi=\sqrt{-\hbar^2 c^2 \nabla^2+m^2 c^4}\psi$$

The squareroot can be expanded to obtain an equation with all powers of the derivative operator. What do they mean when they say this leads to a non-local theory?

And is this equation incorrect or just impractical?

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## marked as duplicate by David Z♦Sep 19 '11 at 16:30

Not naïve? (more characters) –  mbq Sep 19 '11 at 14:24
Would you mind $\TeX$ing this equation so that it's readable? –  leftaroundabout Sep 19 '11 at 14:25
I do believe there is even a meta policy to not answer ill formated (not in $\TeX$) questions. –  AdamRedwine Sep 19 '11 at 14:32
The question has already been answered: physics.stackexchange.com/questions/13624/… –  FranceV Sep 19 '11 at 14:42
There is also a mathematically precise way to interpret locality. See Peetre theorem. In particular, an operator of the form $\sqrt{-\nabla^2 + k^2}$ cannot be represented as a partial differential operator, but only as a pseudo-differential one. –  Willie Wong Sep 19 '11 at 16:13

To see why the theory is nonlocal, consider the effect of the derivative operator... I like to put things on a lattice, so I will: $\psi_i=\psi(x_i)$, then the derivatives (in 1D, for simplicity) become $$\nabla^2 \psi_i \propto (\psi_{i-1}-2\psi_i + \psi_{i+1})$$ Now, you can see what happens as you continue to apply derivatives (as you must, in the expansion of the square root) -- For high order derivatives, the time-derivative of $\psi$ at a lattice site will depend on the instantaneous spatial values of the whole field!