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Extra resources for Discrete quantum electrodynamics

Example text

This is the U ( 1 ) subgroup. There is also an SU ( 3 ) subgroup for states consisting of three quarks. 11 Because these currents mix flavours the same coupling constant is needed for all quarks, and the interaction density takes the form H ( x ) = gj a ( x ) ⋅ A a ( x ) where A a ( x ) are the gluon fields, which are identical to the photon, up to interactions. 12, can be read as an eigenvalue equation for the gauge covariant derivative showing the origin of gauge invariance. 12. 7 for intermediate vector bosons.

1. 4 ∑∫ 1 α |x, ν〉 〈 x , µ ˆ | = -------d 3 p d 3 qivˆ ( q, r ) ( p ⋅ γ – q ⋅ γ )v ( p, s )e ix ⋅ ( p – q ) |p, r〉 〈q, s| ∂αγ µν 8π 3 r, s Μ ∫ Μ Here v and vˆ have been ordered so that the spin index can be unambiguously omitted. 11. 12. 7. e. 5, so 〈 j 〉 is conserved. e. 2 that j is additive for multiparticle states, so it is sufficient to show the theorem for a one particle state | f 〉 ∈ * . 4 0 〈x, ν| f 〉 – γ 0 〈 f |x, ν〉 〈 x , µ ˆ| f 〉 〈 j 0 ( x ) 〉 = 〈 f | x , µˆ 〉γ µν µν = 〈 f | x 〉γ 0 γ 0 〈 x | f 〉 – 〈 x | f 〉γ 0 γ 0 〈 f | x 〉 by ordering terms so that the spinor indices can be suppressed.

2 has a co-ordinate x i for the position of each quark. 1. So physical states are either directly created from the creation operators appearing in H ( x ) , or evolve from them from the corresponding non-interacting a wave equation. The spin statistics theorem forbids even multiplets but it is possible to build a model in which the field operators create or annihilate triplets of quarks. There is no way physically to distinguish the three quarks in a hadron, so the creation operator must be (anti)symmetric under permutations of the three quarks.

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