Is a non-abelian gauge field's strength observable?

For an abelian gauge field, the field strength $G_{\mu \nu}$ is gauge-invariant. This means it is a physically observable quantity, e.g. we can build an apparatus to measure electromagnetic field strength.

For a non-abelian gauge field, $G_{\mu \nu}$ transforms non-trivially under infinitesimal gauge transformations $\xi^a$:

$G_{\mu \nu} \to G'_{\mu \nu} = G_{\mu \nu} + [\xi^a t_a, G_{\mu \nu}]$

What are the implications for the observability of $G_{\mu \nu}$? Could we in principle build a physical apparatus to measure the field strength of a non-abelian field? Let's ignore practical difficulties such as short range of non-abelian fields actually found in nature.

• it seems that you kind of answered your own question, right? it is not gauge invariant, ergo it cannot be observable. QED. ($\leftarrow$ pun intended) – AccidentalFourierTransform Nov 29 '16 at 13:27
• So what is the best we can do? If $G_{\mu \nu}$ is not observable, maybe some function of $G_{\mu \nu}$ is observable? – Sergei Patiakin Nov 29 '16 at 13:33
• hint: the Lagrangian is gauge invariant. What function of $G_{\mu\nu}$ appears in the Lagrangian? that object is in principle observable. – AccidentalFourierTransform Nov 29 '16 at 13:34
• Actually not even field strengths are observables: propagators (i. e. transition amplitudes) are, that are path integrals (Taylor expasions thereof) of the action. – gented Nov 29 '16 at 13:40
• $Tr(G_{\mu \nu} G^{\mu \nu})$ appears in the Lagrangian, but this is just one scalar observable. Compare this with the EM field, which has 3+3 observables. Informally, a non-abelian field somehow feels "richer" than an abelian field, so I expected more observables - it seems this is incorrect intuition. – Sergei Patiakin Nov 29 '16 at 13:41