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Particles as Force Carriers - How Does It Work? |
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| Jun22-12, 05:09 PM | #18 |
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Particles as Force Carriers - How Does It Work?
Dirac is RQFT, of course. And it's all 4-momentum. Where did you see any 3-momentum operators in that theory?
Where non-relativistic QFT shows up is condensed matter. There, you would often just use Shroedinger equation, but with second-quantized fields. I don't deal with these, though. So I'm a bad person to ask about any details. |
| Jun22-12, 06:07 PM | #19 |
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(i suspect it's generally assumed that everything is "R", so there's no need to say it) ![]() i see creation and annihilation operators which are functions of the 3-momentum p i see them, for example, as a(p,σ,n) throughout §6.1 of weinberg (pp.259-274) i do not see any creation and annihilation operators in the form a(q,σ,n), as functions of the 4-momentum q nor do i see any creation and annihilation operators at all in §6.3 (which is why i say there are no creation or annihilation operators for off-mass-shell virtual particles in the maths) §6.1 deals with the coordinate-space representation of the S-matrix, in which only 3-momentum is conserved at each vertex … by the end of §6.1, the creation and annihilation operators have been eliminated, resulting in a non-operator formula which, however, is a function of ps, and is not lorentz invariant§6.3 deals with the momentum-space representation of the S-matrix, in which 4-momentum is conserved at each vertex … by defining an extra variable, the non-operator formula from §6.1 is made lorentz invariant, resulting in a non-operator formula which is a function of qs, and is lorentz invariantso i say that coordinate-space (ie feynman diagrams of the 1st type) has on-mass-shell 4-momentums, and creation and annihilation operators for (on-mass-shell virtual) particles with those 4-momentums, but momentum-space (ie feynman diagrams of the 2nd type) has off-mass-shell 4-momentums, but no creation or annihilation operators at all ie off-mass-shell virtual particles don't even exist in the maths, because the maths doesn't include any creation or annihilation operators for them |
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