heuristic explanation of why quantum mechanics plus SR imply antiparticles

feynmann

I'm looking for a heuristic explanation of why quantum mechanics plus special relativity requires antiparticles, Does anybody want to take a crack at it? Or am I asking for the impossible?

TriTertButoxy

Quantum mechanics plus special relativity does not necessarily require antiparticles: although it naturally accommodates them.

----The straightforward generalization of quantum mechanics to include special relativity requires a modification of the Schrodinger equation. Modifying the Hamiltonian to reflect the energy-momentum relation, [itex]E=\sqrt{m^2c^4+p^2c^2}[/itex], for relativistic particles,


leads to a non-local theory due to the differential under the square-root.

----Modifying the entire time-dependent Schrodinger equation to reflect the squared energy-momentum relation [itex]E^2=m^2c^4+p^2c^2[/itex] gives


the Klein-Gordon equation. While this gives a local theory, it contains negative-energy states in its spectrum. This is a problem since perturbations can cause transitions indefinitely into lower states (hence, this theory is unstable).

----The modern view is to abandon any attempt to directly modify the Schrodinger equation, and instead, to quantize a continuous field, [itex]\phi(\mathbf{x},t)[/itex], using ordinary quantum mechanics. The fields that are quantized, however, have dispersion relations of the same form as the squared energy-momentum relation [itex]E^2=m^2c^4+p^2c^2[/itex], with the energy, [itex]E[/itex], identified as the frequency, [itex]\omega_p[/itex], of propagating plane waves. The resulting Schrodinger equation, has no negative energy solutions, and is local. There are, however, negative frequency solutions associated with the field's dispersion relations.

In the case that a real-valued field is quantized, negative and positive frequency solutions are identified, and there are no antiparticles. In the case that a complex-valued field is quantized, negative and positive frequency solutions are the particle and anti-particle solutions, respectively.

colorSpace

"Local" in which sense? For example, no non-local communication, or even no non-local correlation in entanglement? (If I may ask at the risk of going off-topic).

My understanding from articles and discussions of about a year ago is that (at least most) local theories require hidden variables, and that a large class of hidden-variable theories has meanwhile been disproven, that entanglement correlations are considered to demonstrate non-local correlations, even though this question is still somewhat open. "Non-local correlations" means that the effects are symmetric from each particles point of view, and therefore not usable for communication, which would require an asymmetric effect.

TriTertButoxy

I'm sorry about the confusion:
By local I meant microcausal. That is, the commutators of observables (built out of field operators) with a space-like separation vanish. This ensures that two measurements with a space-like separation do not interfere with each other (no information propagates faster than the speed of light). However, observables with space-like separations may still be correlated giving rise to entanglement.

Vanadium 50

I think it depends on what you mean by "heuristic" - if it means "I want to understand the theory without understanding the mathematics behind it", the answer is no. If you're willing to live with an idea that's not too wrong, it's because the relationship between energy, momentum and mass in SR is quadratic, and just as you get two solutions to the quadratic equation, you get two particles of identical mass.