What is the vacuum of classical Yang Mills theory $$\mathcal{L} = - \frac14 F^{a \mu \nu} F^a_{\mu \nu}~?$$ Is it simply $A^a_\mu=0$ for all its components?


1 Answer 1


It depends on what you mean by vacuum.

If you mean a field configuration that has $F=0$, then the gauge potential is locally pure gauge (cf. this answer by Qmechanic), so there can only be global obstructions to the gauge equivalence class of the $A$ corresponding to $F=0$ being $A=0$ everywhere. On $\mathbb{R}^{1,3}$, there is no such obstruction.

If you mean a solution of the classical equations of motion in the absence of sources, but not necessarily $F=0$, then it gets a bit more interesting:

The Minkowskian $\mathbb{R}^{1,3}$ theory clearly, at least for the $\mathrm{U}(1)$ case, has non-trivial solutions - the electromagnetic waves.

The Euclidean theory in $4d,d\in\mathbb{N}$ dimensions has so-called instantons, which are (anti-)self-dual configurations of $F$, and thus local minima of the classical action. One example is given by the BPST instanton for $\mathrm{SU}(2)$ on $\mathbb{R}^4$, or the Harrington-Shepard caloron on $\mathbb{R}^3\times S^1$.

  • $\begingroup$ with vacuum I would mean the configuration that minimizes the potential of the lagrangian. Is that also $F=0$? Or is it not meaningful to divide YM fields into kinetic and potential terms? $\endgroup$
    – dan-ros
    May 10, 2016 at 15:48
  • $\begingroup$ @tonydo: Usually, one calls the entire $F\wedge{\star} F$ term (or $F^{a\mu\nu} F^a_{\mu\nu}$ in components) the "kinetic" term of a Yang-Mills field. $\endgroup$
    – ACuriousMind
    May 10, 2016 at 16:04
  • $\begingroup$ why is that? Even though the $A^4$ term doesn't come with any derivatives? $\endgroup$
    – dan-ros
    May 10, 2016 at 16:22
  • 1
    $\begingroup$ @tonydo: The $A^4$ term doesn't have any meaning on its own as it is not gauge invariant. $\endgroup$
    – ACuriousMind
    May 10, 2016 at 16:49

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