- #1
tut_einstein
- 31
- 0
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{P2c2 + m2c4}[/itex],
schrodinger's equation in position basis becomes:
i[itex]\hbar[/itex][itex]\dot{\psi}[/itex] = [itex]\sqrt{-\hbar2c2\nabla2 + m2c4}[/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!
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{P2c2 + m2c4}[/itex],
schrodinger's equation in position basis becomes:
i[itex]\hbar[/itex][itex]\dot{\psi}[/itex] = [itex]\sqrt{-\hbar2c2\nabla2 + m2c4}[/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!