1. Dec 17, 2012

### tut_einstein

Hi,

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 = $\sqrt{P2c2 + m2c4}$,

schrodinger's equation in position basis becomes:

i$\hbar$$\dot{\psi}$ = $\sqrt{-\hbar2c2\nabla2 + m2c4}$$\psi$

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!

2. Dec 17, 2012

### ShayanJ

Well,that just goes back to the definition of a derivative
For a first derivative,you take two points,calculate the difference of the value of the function at those points and divide by the separation of points.So for the first derivative at a point,you need to consider two points,one the point in which you want the derivative and the other,another point at its vicinity.For second derivative you need one more point and so on,the number of points increases and the points get farther from the point you want the derivative on.So the more the order of derivatives,the behaviour of a point depends on farther and farther points which makes such theories non local.

Last edited: Dec 17, 2012
3. Dec 18, 2012

### The_Duck

Here's a suggestive argument:

By Taylor expansion, we can write

f(x + a) = f(x) + a f'(x) + (1/2)a^2 f''(x) + (1/6)a^3 f'''(x) + ...

where to get equality we need an infinite number of derivatives. The right hand side looks local (it looks like it only refers to the properties of f at the point f) but is actually nonlocal (it actually depends on the properties of f at some distance from x).

4. Dec 18, 2012

### kloptok

Maybe I misunderstood what you meant in your last post The_Duck but what about the kinetic term in the Lagrangian - $\partial_{\mu}\phi \partial^{\mu} \phi$? By the same logic, wouldn't this also be non-local? I am curious to the original question as I have thought about the same thing. I can see how a derivative could be viewed as non-local from its definition (we compare infinitesimally close points etc.), so what is the difference when it is inside a square root -- and why isn't the kinetic term non-local?

5. Dec 18, 2012

### The_Duck

I was using the Taylor expansion as something that has an *infinite* number of derivatives and is therefore nonlocal, as we can see from the fact that this infinite number of derivatives combines to give us the value of the function at a finite distance from the original point x. A standard kinetic term with only two derivatives does not give rise to anything like this.

No, comparing points at infinitesimal separations is "local" in the sense discussed here. Comparing points at *finite* separations isn't.

6. Dec 18, 2012

### kloptok

Alright, that convinces me! I suspected that infinitesimal separations would still be considered "local". Thanks!

7. Dec 19, 2012

### tut_einstein

Thanks Shyan and TheDuck, I understand that infinitesimal distances away from a given point isn't considered non-local, which is why the kinetic term isn't non-local. But if we expand the square root of the gradient, why can't we do the taylor expansion assuming that the displacement from a given point is infinitesimal?

So, I guess my point of confusion is that why is it that when there are an *infinite* number of derivatives, the theory is considered non-local, while for a finite number of derivatives, it isn't?

8. Dec 19, 2012

### ShayanJ

I think even finite order derivatives with order greater than 2 can cause non locality.But that's just what I think.Is there a proof about it?