# Why do we only have complete particle generations?

There are 3 generations of fermions in the standard model. I know that there is a theorem that states, that only complete generations are allowed. This means that there have to be quarks with three colors. With this, if we sum over all leptons and quarks in a generation the overall charge is zero.

Why is it assumed that this must hold?

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It is not assumed, it is a fact that having an incomplete generation of standard model particles is inconsistent. It leads to an inconsistency known as an "anomaly", specifically a gauge anomaly. This leads to the creation of unphysical polarization states of the gauge bosons, which end up ruining either Lorentz invariance or unitarity - the condition that total probabilities have to add up to 100%.

In the standard model the dangerous anomaly affects the hypercharge gauge boson (which becomes part of the photon and $Z^0$ after the Higgs mechanism kicks in). The condition that the total anomaly vanishes is

$$\sum_\text{all doublets} Y_L = 0,$$

and

$$\sum_\text{left handed} Y_L^3 - \sum_\text{right handed} Y_R^3 = 0,$$

where the sums run over all electroweak doublets and left/right handed particles respectively. If you plug in the standard model quantum numbers (from wiki for example) you should find that these conditions hold. (Note that wiki only lists left handed particles so instead of subtracting right handed terms you add the left handed ones. The string of numbers you should get for the second equation is $-1+8-1+0+\frac{1}{9}-\frac{64}{9}+\frac{1}{9}+\frac{8}{9}=0$.) It should also be easy to see that getting rid of any particle or adding a new particle willy-nilly will upset the delicate cancellation that takes place. Because of this anomalies place strong constraints on building models beyond the standard model.

There is a reasonably pedagogical account of anomalies here:

I do not understand, in the (numerical) second equation, why, for quarks part, there is not a $3$ term (because there are $3$ colors for quarks). –  Trimok Jul 26 '13 at 12:30
@Trimok Oh I already included that. E.g. $1/9 = 3 \times \left( \frac{1}{3} \right)^3$ etc. I didn't want to display all of the arithmetic for the OP. Sorry for the confusion. :) –  Michael Brown Jul 26 '13 at 14:00