Fujikawa Jacobian for Baryon number anomaly

Question

Reviewing the anomalies of the standard model, one knows that the Baryon number is not conserved because of an anomaly associated to the global $U(1)$ symmetry that quarks have. That is the current $$J_B^\mu = \sum_\alpha \bar{\psi}_\alpha\gamma^\mu\psi_\alpha,$$ where $\alpha$ runs over flavors and colors, is not conserved and its divergence is proportional to products of the field strength tensor times its dual. Specifically we have a symmetry transformation which is $$\psi\rightarrow e^{i\theta}\psi\qquad \text{and}\qquad \bar{\psi}\rightarrow e^{-i\theta}\psi\tag{1}\label{eq:symmetry}$$

In a path integral derivation of the anomaly through Fujikawa's method, one gauges the global symmetry and imposes that the action has a zero functional variation with respect to this "auxiliary" symmetry. Then one studies the Jacobian coming out of the field redefinition involved in this process and usually realizes that it is not trivial. However trying to do this for this case, the Jacobian obtained from the transformations in Eq. \eqref{eq:symmetry} is schematically: $$\mathcal{J} = \left[\det(e^{i\theta(x)})\right]^{-1}\left[\det(e^{-i\theta(x)})\right]^{-1}$$

which doesn't show the anomaly. Am I missing something or is this anomaly "invisible" to Fujikawa's method?

---

Reference (more details on what I am saying)

• John F. Donoghue, Dynamics of the Standard Model, 2014, Cambridge University Press, Ch. III

Answer 1

The number current for each standard model particle can be decomposed into $$ \bar \psi \gamma^\mu \psi= \frac 12 \bar \psi \gamma^\mu(1+\gamma_5)+\frac 12 \psi+\bar \psi \gamma^\mu(1-\gamma_5) \psi = J^\mu_R+J^\mu_L $$ and the left handed part has an anomaly due to its coupling to the ${\rm SU}(2)_W$ gauge field. The change in the left handed quarks contribution to he Baryon number is the same as the change in the $e_L$ $\nu_L$ contribution to Lepton number (The $u_L$ $d_L$ quarks have a factor of three for colour which is compensated for by a factor of 1/3 because each quark has $B=1/3$).

The symmetry that is being violated is not the one you wrote, but is $$ \psi\to e^{i\gamma^5\theta}\psi, \quad \bar\psi \to \bar\psi e^{i\gamma^5 \theta}. $$
which does have a non-trivial Fujikawa Jacobian.