Background 1. In classical SU(N) Yang-Mills theories, there are a countably infinite number of homotopically inequivalent gauge field configurations of zero energy labelled by a winding number $n\in \mathbb{Z}$. In the corresponding quantum theory, the states $|n\rangle$ labelled by the quantum number $n$ are not the true vacua of the theory. This is because due to instanton effects, there is a nonzero tunnelling amplitude between the states $|n\rangle$. The true vacua of such theories are not the states $|n\rangle$ but given by a superposition $$|\theta\rangle=\sum\limits_{n=-\infty}^{\infty}e^{in\theta}|n\rangle\tag{1}$$ called the $\theta-$vacua. These states do not have a definite winding number.

Background 2. In the Standard Model, the baryon current $J^\mu_B$ is anomalous $$\partial_\mu J^\mu_B=\frac{N_f}{32\pi^2}(g^2W_{\mu\nu}^a \tilde{W}^{\mu\nu a}-g^{\prime 2} B_{\mu\nu}\tilde{B}^{\mu\nu})\tag{2}$$ where $N_f$ is the number of fermion flavours, $W_{\mu\nu}^a,B_{\mu\nu}^a$ are the SU(2) and U(1) field strengths.

One says that when there is a transition from a state of $n=1$ to $n=2$ (say, for example) there is a baryon number violation $\Delta B$ is given by $$\Delta B=2N_f(2-1)=2N_f.\tag{3}$$

This implies that, initially, the Universe is in a state with a definite winding number and finally it can make a transition to another state with definite (but different) winding number. This suggests to me that one considers states $|n\rangle$ as the vacua of the theory (instead the $\theta$ vacua) and somehow baryon numbers are related to the winding number. The states $|n\rangle$ can be considered as vacua only when the tunneling amplitudes are negligible. Another possibility is that, one considers the theory to be classical. Classical vacua are gauge field configurations of zero energy separated by finte energy barriers and having definite $n$.


$\bullet$ Therefore, my question is why are we calling $|n\rangle$ to be the vacua in the Standard model rather than $|\theta\rangle$?

$\bullet$ How is the winding number related to the baryon number?


1 Answer 1


There is a definite fermion number only before and after the transition, as one can see from your equation for divergence of baryon current. Analogously, the system is in the definite vacuum state only at $t= \pm \infty$.

  • 1
    $\begingroup$ If the system is in $\theta$-vacuum, which is an eigenstate of the Hamiltonian, there shouldn't be any jump. Am I wrong? $\endgroup$
    – SRS
    Commented Dec 21, 2016 at 18:38
  • $\begingroup$ @SRS Yes, I think so. $\endgroup$ Commented Dec 21, 2016 at 18:39
  • $\begingroup$ Do you agree or you think I'm wrong? I didn't get your point. If the system is in $\theta-$vacuum, why should we talk about, for example, evolution from $n=1$ to $n=2$? $\endgroup$
    – SRS
    Commented Dec 21, 2016 at 18:43
  • $\begingroup$ @SRS As far as I understood, you considered transition from vacuum labeled by $n_1$ to some other one with $n_2$. Am I right? In any case, there is a very detailed discussion of the subject in Ch. 23 of Weinberg. Perhaps you should consult it? $\endgroup$ Commented Dec 21, 2016 at 18:49
  • $\begingroup$ My first question is whether the fermions are in the $\theta$-vacuum or in one of the vacua labelled by $|0,n\rangle$? @AndreyFeldman $\endgroup$
    – SRS
    Commented Jul 18, 2017 at 17:05

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