Also at http://www.nikhef.nl/pub/theory/academiclectures/ is http://www.nikhef.nl/pub/theory/academiclectures/bs06_one.pdf, PDF pages 77 - 81, especially page 80.
Anomalies are (sum over left-handed elementary fermions) - (same for right-handed ones) of Tr(T
a.{T
b,T
c})
where the T's are gauge-group generators.
The Standard Model's symmetry is SU(3) * SU(2) * U(1) -- QCD * (weak isospin) * (weak hypercharge)
Weak hypercharge Y: average electric charge of a multiplet:
each member's charge Q = (weak isospin)
3 + Y
Some of the anomalies cancel as a result of gauge-group structure: SU(3)
2*SU(2), SU(3)*SU(2)
2, SU(2)
3, SU(3)*SU(2)*U(1), SU(3)*U(1)
2, SU(2)*U(1)
2, (gravity)
2*SU(3), (gravity)
2*SU(2), and apparently (gravity)*(nongravitational)
(2) and (gravity)
3
This leaves SU(3)
3, SU(3)
2*U(1), SU(2)
2*U(1), U(1)
3, and (gravity)
2*U(1). They mean:
All these must add up to zero, taking (sum of left-handed multiplet members) - (likewise for right-handed ones):
- Quarks 1, leptons 0
- Quark (hyper)charge
- Weak-isospin doublet (hyper)charge
- (hypercharge)3 value
- (hyper)charge
The first one is (number of left-handed quarks) = (number of right-handed quarks)
Let's work it out for the Standard Model. It has a left-handed weak-isospin doublet of quarks and one for leptons, with all the right-handed ones being weak-isospin singlets.
Anomaly #1: 1 quark doublet, 2 quark singlets. Check.
Anomaly #2: 6Y(QL) = sum of 3Y(QR's)
Anomaly #3: 6Y(QL) + 2Y(LL) = 0
Anomaly #4: 6Y(QL)
3 + 2Y(LL)
3 = sum of 3Y(QR's)
3 + sum of Y(LR's)
3
Anomaly #5: 6Y(QL) + 2Y(LL) = sum of 3Y(QR's) + sum of Y(LR's)
Let's see how much we can constrain the hypercharge values. Let Y(LL) = -1/2 for definiteness, as in the Standard Model. Then Y(QL) = 1/6, check, though Y(QR's) and the Y(LR's) are more difficult.
From #2, Y(QR1) + Y(QR2) = 1/3
From #2 and #5, sum of Y(LR's) = -1
From #4, -2/9 = 3Y(QR1)
3 + 3Y(QR2)
3 + sum of Y(LR's)
3
This is still a little ambiguous, so let us suppose that only one right-handed lepton has a nonzero hypercharge. Then Y(LR) = -1. Check.
Y(QR1) + Y(QR2) = 1/3
Y(QR1)
3 + Y(QR2)
3 = 7/27
with solution Y(QR1) = 2/3 and Y(QR2) = -1/3. Check.
So anomaly cancellation forces Y(QL) = 1/6, Y(LL) = -1/3, Y(QR1) = 2/3, Y(QR2) = -1/3, Y(LR) = -1 to within a multiplicative factor.
Electrically-neutral quarks? They'd have to be added on to the Standard Model.