This is a continuation of my question here.

Page 635 of this book by Matthew Schwartz effectively says that the $\partial_\mu J^\mu_B\neq 0$ where $J^\mu_B$ is the baryon current i.e., the baryon number is not conserved at the quantum level. However, this term can be shown to be a total derivative, and hence, any Feynman diagram with this vertex will contain a factor $\sum p_\mu=0$. Since perturbation theory is based on Feynman diagrams, such a term cannot contribute at any order in perturbation theory.

However, the term $\partial_\mu J^\mu_B\neq 0$ can itself be calculated from triangle diagrams.

$\bullet$ Does it not mean that baryon number violation is possible (or at least calculable) even at the perturbative level?

$\bullet$ If it's a non-perturbative effect why is it computable using Feynman diagram?

$\bullet$ If I understand Schwartz's argument correctly, the anomaly is derivable from Feynman diagrams but the anomaly itself does not give rise to new Feynman diagrams. Is that right?

  • $\begingroup$ One presumes you read eqn (3.70) in your reference? Is that perturbation theory? $\endgroup$ – Cosmas Zachos Nov 19 '17 at 0:46

The baryon number violation happens only when $$ \Delta N= \int d^{4}x \langle 0|\hat{\text{A}}(x)|0\rangle \neq 0, $$ where $\hat{\text{A}}(x)$ is the baryon number anomaly; this is exact relation since the chiral anomaly is one-loop exact.

The VEV of $\int d^{4}x\hat{\text{A}}(x)$ obviously depends on the vacuum you choose, and the vacuum may be "perturbative" (i.e., simply the Fock vacuum with trivial solution for the gauge fields) or "non-perturbative" - the $\theta$-vacuum. For the first case the non-perturbed value of the integral is identically zero, while the integral of any perturbative excitations vanish identically because of the reason you've given in the question. However, for the second case the VEV is in general non-zero.


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