
#1
Dec1712, 08:33 PM

P: 31

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 = [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 nonlocal (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 nonlocal 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
Dec1712, 11:19 PM

P: 735

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. 



#3
Dec1812, 07:10 AM

P: 795

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
Dec1812, 07:31 AM

P: 188

Question about expanding a square root in powers of gradient
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 nonlocal? 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 nonlocal 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 nonlocal?




#5
Dec1812, 12:19 PM

P: 795





#6
Dec1812, 12:51 PM

P: 188

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




#7
Dec1912, 07:26 PM

P: 31

Thanks Shyan and TheDuck, I understand that infinitesimal distances away from a given point isn't considered nonlocal, which is why the kinetic term isn't nonlocal. 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 nonlocal, while for a finite number of derivatives, it isn't? 


Register to reply 
Related Discussions  
F of G function question  Square root inside a square root?  Precalculus Mathematics Homework  3  
scientific notation root powers exponents question  Precalculus Mathematics Homework  1  
Determine the exact value of.... (Square Root of Square Root)  General Math  1  
Root Mean Square Error, a straight line fit, and gradient issue  Calculus & Beyond Homework  0  
square root and cube root question  Precalculus Mathematics Homework  4 