4
$\begingroup$

Baryon number $U(1)_B$ is anomalous in the Standard Model, as can be seen by computing a $U(1)_B SU(2)_L^2$ triangle diagram. This implies that $$\partial_\mu J^{\mu B} \sim W_{\mu\nu} \tilde{W}^{\mu\nu}$$ where $W_{\mu\nu}$ is the $SU(2)_L$ field strength, which allows the nonconservation of baryon number by topologically nontrivial field configurations.

According to Schwartz's QFT textbook, all other contributions to the $U(1)_B$ anomaly vanish, but I can't see why that is. In the case of $U(1)_B U(1)_Y^2$, we should have a contribution proportional to $$\sum_{\text{LH quarks}} Y_i^2 - \sum_{\text{RH quarks}} Y_i^2 \propto 2 \left( \frac16 \right)^2 - \left(\frac23 \right)^2 - \left(-\frac13\right)^2 \neq 0.$$ What am I doing wrong in this computation?

$\endgroup$
  • $\begingroup$ Related and links thereto ... There are no U(1) instantons.... $\endgroup$ – Cosmas Zachos Jul 24 '18 at 16:06
4
$\begingroup$

Actually, there's nothing wrong with this computation; Schwartz's statement is simply incorrect. There is indeed a $U(1)_B U(1)_Y^2$ anomaly, which implies $$\partial_\mu J^{\mu B} \sim W_{\mu\nu} \tilde{W}^{\mu\nu} + B_{\mu\nu} \tilde{B}^{\mu\nu}$$ where $B_{\mu\nu}$ is the $U(1)_Y$ field strength. The reason this second term is rarely mentioned is that there are no $U(1)_Y$ instantons. That is, while there can be local violations of baryon number conservation, you can't get a global violation of baryon number from $B_{\mu\nu}$ configurations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.