Consider $SU(2)$ gauge theory. The classical ground state is $F^a_{μν}=0$ . This implies that the vector potential $A^a_μ=U∂_μU^†$. Here $U(x)$ is an element of the gauge group. Why can we impose the boundary condition that $U(x)=1$ as $|x| → ∞$?. This is done in Srednicki 571. Thanks in advance
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Related question by OP: http://physics.stackexchange.com/q/317165/2451 – Qmechanic Mar 10 '17 at 12:19
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I recently gave an answer here that contains the answer to this question. – ACuriousMind Mar 10 '17 at 13:01