Derivation of Electro-Weak Unification and Final Form of Standard Model with QCD and Gluons  1W1+  2W2 +  3W3

Substitute B = cosW A  + sin W Z0  Sum over first generation particles. text error up down Left handed only text error Flavor up Flavor down Flavor changing interactions.

Weak interaction terms flavor changing: leptons flavor changing: quarks

We want the coefficient for the electron-photon term to be -e f = 0 for neutrino and = 1 for others A -e A Z0 Z0

Consider only the A term: gives agreement with experiment. Cf = 2T3  = -1

The following values for the constants give the correct charge for all the particles.

Coefficients for the Z0 term g1 = g2 sinW/cos W YfL = 2 [ Q f – T3] f YfR = 2 Q f Z0 Z0 Cf cos W = 2T3 [1-sin2 W]/ cos W  = -1

Final Form for Electro-Weak Interaction Terms (the QCD terms have yet to be written in terms of the color/anti-color gluons) A Z0 In the text by Gordon Kane, p.92, Eq. 7.32, there appears to be a typo in the sign of the third term.

- The Standard Model Interaction Lagrangian for the 1st generation + (E & M) QED interactions weak neutral current interactions - weak flavor changing interactions + QCD color interactions

Weak neutral current interactions Z0 Z0 Z0 Z0 note: no flavor changes

Weak charged flavor changing interactions quarks -g2 leptons

Quantum Chromodynamics (QCD): color forces Only non-zero components of  contribute.

To find the final form of the QCD terms, we rewrite the above sum, collecting similar quark “color” combinations.

The QCD interaction Lagrangian density

grg - - ggb Note that there are only 8 possibilities: r g The red, anti-green gluon - ggb The green, anti-blue gluon

At any time the proton is color neutral. That is, The gluon forces hold the proton together At any time the proton is color neutral. That is, it contains one red, one blue and one green quark. proton

beta decay u u d d d u W- proton neutron W doesn’t see color

decay of - - u d -

- W production from - p p p d u p u W+ - d - u - p p - u

n p p n The nuclear force u u d d d u u W- u d d d u u Note that W-  d + u = - In older theories, one would consider rather the exchange of a - between the n and p. -

Cross sections and Feynman diagrams everything happens here transition probability amplitude must sum over all possible Feynman diagram amplitudes with the same initial and final states .

Feynman rules applied to a 2-vertex electron positron scattering diagram Note that each vertex is generated by the interaction Lagrangian density. time spin spin metric tensor Mfi = left vertex function right vertex function coupling constant – one for each vertex propagator The next steps are to do the sum over  and  and carry out the matrix multiplications. Note that  is a 4x4 matrix and the spinors are 4-component vectors. The result is a a function of the momenta only, and the four spin (helicity) states.

Confinement of quarks  free quark terms free gluon terms quark- gluon interactions The free gluon terms have products of 2, 3 and 4 gluon field operators. These terms lead to the interaction of gluons with other gluons.

G G Nf= # flavors Nc= # colors Nc Nf Note sign normal free gluon term 3-gluon vertex Nf= # flavors Nc= # colors Nc Nf quark loop gluon loop

The terms given explicitly in M are only those loops shown in the previous diagram. Higher order terms are indicated by “ + … “ momentum squared of exchanged gluon Nf Nc  M2quark Nf Nc -7 In QED one has no terms corresponding to the number of colors (the 3-gluon) vertex. This term has a negative sign.

Quark confinement arises from the increasing strength of the interaction at long range. At short range the gluon force is weak; at long range it is strong. This confinement arises from the SU(3) symmetry – with it’s non-commuting (non-abelian) group elements. This non-commuting property generates terms in the Lagrangian density which produce 3-gluon vertices – and gluon loops in the exchanged gluon “propagator”.

Conditions on the SU(2) Gauge Particle Fields which complete the Invariance of the Lagrangian Density

= D’ ’ = D’ [ei/2 ] = ei/2 D  We want to find  W such that the following is satisfied: D’ ’ = D’ [ei/2 ] = ei/2 D  cancel =

 0 and the k are linearly independent so [ …] = 0 and the following is the expression for  W

The Higgs Lagrangian Contribution