2
$\begingroup$

What does the classical vacuum of the $SU(2)$ Yang-Mills action correspond to? Does it correspond to $F_{\mu\nu}=0$ everywhere or just at the spatial infinity? In Srednicki’s book, he has shown that, there are infinite number of classical field configurations of zero energy, each charecterized by a winding number $n$. I believe, instantons doesn’t require $F_{\mu\nu}^a=0$ everywhere but only at the boundary. Am I right?

In Ryder's book, he talks about instantons but does he really shows that how SU(2) Yang-Mills theory has an infinite number of degenerate vacua?

$\endgroup$
  • $\begingroup$ Are you sure those instanton configurations have zero energy? As far as I know, the classical action of the one-instanton solution in $\text{SU}(2)$ is $8\pi^2/g^2$. Furthermore instantons are not vacua but (quantum mechanically interpreted) vacuum tunneling configurations, for which the only requirement is that you get the vacuum ($F_{\mu\nu}=0$ everywhere as you say) only for $t\to\pm\infty$. $\endgroup$ – David Vercauteren Jan 13 '15 at 7:46
1
$\begingroup$

A vacuum is a field configuration that is pure gauge, i.e. $A = g^{-1}\mathrm{d}g$ for a gauge transformation $g$, and hence $F = \mathrm{d}_A A = 0$ (for $\mathrm{d}_A$ the gauge covariant derivative).

An instanton is a local minimum of the action, which is given by an (anti-)self-dual configuration $F = \pm \star F$. It is not a vacuum for non-zero winding number $k$, because the value of a $k$-instanton in the Yang-Mills action is

$$ S_{YM}[F] = \int\mathrm{tr}(F \wedge \star F) = \int\mathrm{tr}(F \wedge F) = 8\pi k^2$$

since the winding number is also the second Chern class, and this unequivocally shows that $F \neq 0$ for non-zero $k$. Hence, an instanton is not a vacuum. It is, however, as a local minimum of the action, a natural start for perturbation theory, where it then takes the place of the vacuum for all purposes.

However, only fields with finite value of the action should be considered physical fields. Hence, if we have a non-compact spacetime, we must require that $F(x)\to 0$ as $\lvert x\rvert \to \infty$.

If we put the instanton on a 4-cylinder $(-\infty,\infty)\times S^3$ (compactifying the spatial part of our spacetime for this argument), then, naturally, $F(t = \pm \infty) = 0$, and the partition function of this naturally represents the propagator between the Hilbert spaces associated to the ends of the cylinder - hence instantons are said to mediate vacuum transititon, because one can show (see my answer here) that the vacua at the end differ by $k$ in their Chern classes, and hence are different vacua.

You may ask how they can be different, when only field configurations with $F = 0$, or rather $A = g^{-1}\mathrm{d}g$, are vacua, and hence pure gauge. The reason is that there is a subtle distinction between small and large gauge transformations, and only the small ones (which are homotopic to the identity transformation) do not change the Chern class of the theory. Since they thus differ in global structure, the vacua with different winding numbers are different states.

Any non-abelian gauge theory with a simple Lie group as gauge group has its third homotopy group as $\mathbb{Z}$, and it is the third fundamental group that tells you which Chern classes are allowed (see instantons and the three-sphere in my answer linked above). Since $\mathbb{Z}$ is infinite, there is thus an infinite number of different vacua between which instantons may mediate.

$\endgroup$
  • $\begingroup$ I think you're being a bit sloppy about the distinction between the second Chern class/number $\int F \wedge F$ and the Chern-Simons form/action $\int A \wedge \mathrm d A + \frac{2}{3} A\wedge A \wedge A$. The former is always invariant under (large) gauge transformation, but the latter is not. Moreover the instanton doesn't couple vacua with different Chern numbers but rather with different value of the Chern-Simons action. $\endgroup$ – Ruben Verresen Feb 13 '16 at 16:23

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.