# Equality of positron and proton charge problem statement

When I hear that the equality of positron and proton charge is an unsolved problem I assume that we are putting the electric charge by hand in the electroweak section of the SM Lagrangian.

Is this understanding correct?

When we make a linear combination of two elements of $SU(2)$ Lie algebra to define the electric charge operator from the hypercharge operator and the generator, $L^3+Y$, aren't we applying it to the Fermions and Leptons the same way when the symmetry is broken into $U(1)$?

I appreciate a brief technical statement of the equal charge problem in SM or if possible a reference review of attempts to solve the problem from the literature.

This is part and parcel of an SM course, e.g. try M Schwartz's QFT&SM textbook, or any other modern introduction. Charge balance is an experimental fact, as you may observe in the broad answers of the linked question, and is put into the SM by hand. So you might think, naively, that the SM would also be comfortable with an electron charge 0.7% larger than thrice the down quark charge: classical gauge invariance works just fine. But you are asking how perfect charge balance fits into the SM.

It is, in fact, very-very good for it, at the quantum level. Most would argue it makes the quantum model possible (consistent, by canceling gauge anomalies).

It basically ensures that, individually in each SM generation, the quarks and the leptons must have their charges comeasurate and perfectly balanced so as to cancel the WWB triangle anomalies.

In a notional world where you built the SM without experimental handles, you'd assign arbitrary charges to the quarks and the leptons and pursue possible solutions of the constraints imposed by anomalies on those charges. Model-builders extending the SM do this sort of thing routinely, to ensure no gauge anomalies are present to spoil renormalizability.

The Two-W-cum-Hypercharge gauge field B triangle anomaly will be proportional to $$\sum_{doublets}\operatorname{Tr} T^aT^bY \propto \delta^{ab}\sum_{doublets}Y.$$ So for this to vanish, you need $\sum_{doublets} Q =0$ for the chiral fermions, all in doublets. (The singlets are arranged to match the doublets and will never trouble you. The weak isospin is traceless and the sum over its T3 piece vanishes.)

For each generation, felicitously, nay, indeed, magically!, the lepton charge is completely offset by that of the quarks, suitably multiplexed by 3, the number of color species for them, $-1+3(2/3 -1/3)=0$.

So, adding up quark charges to get baryon charges, you get the suitable matching/balancing with lepton and antilepton charges.

You will see in learned reviews increasingly baroque arguments to undergird this fact and make it virtually necessary, so as to convince you we live in the best possible world, and that all alternatives would be pathological, but it's all experimental fact, really....

"Unsolved problem" is wishful code for "search for a deeper, transcendent, non-anthropic explanation".

Also see Foot et al 1993 .