# Topological terms VEVs and ghosts

Suppose we have the Standard model, and we want to calculate with VEVs of topological susceptibilities of $SU_{L}(2), U_{Y}(1)$ and $SU_{c}(3)$ fields, which have the form $$\tag 1 \kappa \equiv \lim_{p \to 0} p_{\mu}p_{\nu}\int d^{4}xe^{ipx} \langle \text{vac}|T\left( K^{\mu}(x)K^{\nu}(0)\right) |\text{vac}\rangle \equiv \lim_{p \to 0}p_{\mu}p_{\nu}\Pi^{\mu \nu}(p)$$ In principle, there exist topological configurations for all of groups (we need to take into an accounts that instanton configurations are forbidden for electroweak sector, but there are exist instanton-like configurations for which an action is finite), however, existence of nonzero VEV is more complicated question.

From the other way, existence of nonzero topological susceptibility implies the pole structure of $\Pi^{\mu \nu}$: for being nonzero, the $\Pi^{\mu \nu}$ in rhs has to be $$\lim_{p \to 0}\Pi^{\mu \nu}(p) = A\frac{g^{\mu \nu}}{p^{2}} + B\frac{p^{\mu}p^{\nu}}{p^{4}},$$ i.e., there must be the ghost state which couples to gauge variant Chern-Simons form $K^{\mu}$.

The situation is known in pure QCD, for which $$\lim_{p \to 0}p_{\mu}p_{\nu}\Pi^{\mu \nu}(p) \sim -\langle| \bar{\tilde{u}}\tilde{u} |\rangle \neq 0$$ in the broken phase. This can be undestood as the fact that the gluons are confined in the broken phase.

What to do in the case of $SU_{L}(2)\times U_{Y}(1)$ group? How to calculate topological susceptibilities?

• How is this different from this old question of yours? – ACuriousMind Feb 5 '16 at 21:08
• @ACuriousMind : now I relate nonzero value of topological susceptibility to the presence of the ghost. – Name YYY Feb 5 '16 at 21:11