That's due to anomaly cancellation where the anomalies are certain diagrams that cannot be evaluated self-consistently. The triangle anomaly features a fermion loop with three gauge particles coming out of it. It cannot be evaluated consistently if the gauge-fermion interactions violate parity. The only way to get rid of them is to make them cancel each other out, and that supplies constraints on the fields.
Anomalies and the Standard Model
For gauge particles a, b, and c, with operators on the fermions T
a, T
b, T
c, operators like electric charge.
Anomaly ~ Tr(T
a.{T
b,T
c})
left - Tr(T
a.{T
b,T
c})
right
If all the interactions conserve parity, then the triangle anomaly will cancel out. So the interesting case is when they don't, like the electroweak interactions.
Electroweak multiplets have two quantum numbers, weak isospin and weak hypercharge
Weak isospin (WIS) is like 3D angular momentum. Members have (WIS)
3 form - (WIS) to + (WIS) in integer steps
Weak hypercharge (WHC) is like electric charge
Electric charge Q = (WIS)
3 + (WHC)
Elementary fermions come in left-handed doublets and right-handed singlets. The right-handed singlets must have hypercharge values that enable the electromagnetic interaction to conserve parity. A left-handed multiplet with hypercharge Y has charges
up: Y + 1/2
dn: Y - 1/2
Right-handed:
up: WHC = Y + 1/2
dn: WHC = Y - 1/2
Right-handed neutrinos are electroweak and Standard-Model singlets, so their presence or absence does not affect this argument.Let's see if we can get constraints on possible hypercharge values.
QCD-QCD-QCD - cancels, from # left-handed degrees of freedom equaling # right-handed ones
QCD-QCD-WIS - cancels, from Tr(WIS operator) = 0
QCD-QCD-WHC - cancels, from left-handed hypercharges 2Y equaling right-handed ones (Y+1/2)+(Y-1/2) = 2Y
QCD-(WIS/WHC)-(WIS/WHC) - cancels, from Tr(QCD operator) = 0
WIS-WIS-WIS - cancels, since it's zero for (WIS) = 0 and 1/2
WIS-WIS-WHC - gives 2Y
WIS-WHC-WHC - cancels, since Tr(WIS operator) = 0
WHC-WHC-WHC - gives 2Y
3 - (Y+1/2)
3 - (Y-1/2)
3 = -(3/2)Y
So the sum of the hypercharges of the left-handed multiplets must equal zero, with the sums being weighted by the multiplets' multiplicities.
Let's now look at lepton and quark triangle anomalies.
Leptons first. Electrons have charge -1 and neutrinos 0, giving their left-handed multiplets a WHC of -1/2
2*that = -1 -- anomaly!
Now quarks. Up-like ones have charge 2/3 and down-like ones charge -1/3, giving their left-handed multiplets a WHC of +1/6
Being sure to count colors,
3*2*that = +1 -- anomaly also!
But the lepton and quark ones cancel out.
Thus, quarks' fractional charges are due to anomaly cancellation and their coming in 3 colors.