Confusion about baryon number violation in the standard model

Question

I'm reading Gauge Theory by David Tong and not understanding the concept of baryon number violation. I understand that the massless Dirac field has two symmetries, an $e^{i\theta}$ $U(1)$ symmetry and $e^{i\theta\gamma^5}$. I understand that the former corresponds to the current $\bar\psi\gamma^\mu\psi$ and the latter to $\bar\psi\gamma^\mu\gamma^5\psi$. I understand that these amount $n-\bar n$ and $n+\bar n$ respectively for $n$ particles and $\bar n$ antiparticles. I understand the 2nd symmetry is anomalous, and I understand in detail why. Now, how exactly does any of this show baryon number violation? Isn't $B=\frac{n-\bar n}3$ essentially? I see the example of $\pi^0\to\gamma\gamma$ is used as baryon number violation, but clearly $1-1=0-0$. What am I not understanding? Is everyone implicitly renaming the baryon number to $n$ and calling that baryon number violation, or what is going on?

Answer 1

Yeah, the U(1) vector global symmetry is almost not anomalous, but this is marred by the aggressively chiral weak interactions: there is a very weak nonperturbative possibility for an anomaly, in principle, because of the lopsided couplings of the Ws and Zs to the left and right fermions.

A baryon current divergence may pick up a nonperturbative piece bilinear in the EW gauge fields analogous to the two-photon operator in neutral pion decay. This violation has not been observed, in practice, but it is there in principle, for model building and cosmology, and your text covers it, as mentioned, on p 173.

The point is that, within a family, all quarks, Left ($q_L$) and Right ($u_R,d_R$) have B=1/3, as per your correct expression for quarks, and antiquarks minus that (with their chiralities reversed).

However, both the SU(2) and hypercharge U(1) gauge fields couple to them very differently! For instance, the respective weak hypercharges are 1/3,1/3,4/3,-2/3, for $u_L,d_L,u_R,d_R$, respectively. So the sum of the L quarks' $BY^2$s is 2 (I've tripled everything by the number of color species), from which we subtract that for the R quarks, -20, totaling 18, non-vanishing! The weak isospin gauge fields will not cancel that.

Within a fermion family, this is exactly equal for the lepton number current, so B-L is not anomalous, a good SM symmetry. However, B in isolation can be violated in principle, if only you can imagine a nonperturbative process, however unlikely, doing the job ('t Hooft did). Unless you do model building or cosmology, it is unlikely you'll stumble onto this in the lab, needless to say: the effective suppression factor is $\exp(-16\pi^2/g_2^2)\sim 10^{-160}$...

PS. You might be shocked that a vector current such as B's has a triangle chiral anomaly. However, because of the chirality-defying hypercharge couplings, above, you may have a VVA triangle just fine... Same holds for the weak isospin part. The weak interactions violate parity, and the hypercharge, in particular, smashes it in magnificently chaotic ways!

Answer 2

Baryon number is not $B=\frac{1}{3}(n-\bar{n})$ for the number of fermions $n$ minus the number antifermions $\bar{n}$; it is actually $B=\frac{1}{3}(n_{q}-\bar{n}_{q})$, where $n_{q}$ is the number of quarks. However, any anomalies that might potentially lead to $B$ nonconservation must be related to chiral gague theories; they are thus related to the physics of the electroweak sector, not the strong sector [in which the $SU(3)_{c}$ gauge symmetry is a vector, not chiral, symmetry].

So the natural fermionic conserved quantities need to include contributions from both the quark and lepton sectors. In fact, the key quantities turn out to be $B\pm L$, where $B$ is the baryon number (defined in terms of the quarks) and $L$ is the analogous lepton number. It turns out that in the standard model, both $B+L$ and $B-L$ are conserved at tree level. However, anomalous processes (sphalerons) lead to nonconservation of $B+L$, although $B-L$ remains conserved. Therefore, in electroweak baryogenesis scenarios, it is possible to create baryon-lepton pairs, like $p^{+}+e^{-}$; the simultaneous creation of a proton and electron leave the charge $Q$ and $B-L$ unchanged. In larger grand unified theories like the $SO(10)$ GUT, the $B-L$ theory can also be broken.