1  Note on sea corrections

Given the gluonic observable W(t)=t∂(t2⟨E⟩)∂t=t2(2⟨E⟩+t∂⟨E⟩∂t) W(t)=t\frac{\partial (t^2\langle E\rangle)}{\partial t}=t^2\left( 2\langle E\rangle +t \frac{\partial\langle E\rangle}{\partial t}\right)

For the δμ\delta \mu correction to a gluonic observable WW, where the counterterm lagrangian Lc=δμuˉu+δμdˉd{\cal L}_c= \delta \mu \bar u u + \delta \mu \bar d d with ru=+1r_u=+1 and rd=−1r_d=-1. Expanding the interaction e−Lce^{-{\cal L}_c} we have ⟨W⟩=⟨W⟩0−⟨W∫dx′(δμuˉu+δμdˉd)(x′)⟩0=⟨W⟩0−⟨W∫dx(δμχˉuiγ5χu−δμχˉdiγ5χd)(x)⟩0=⟨W⟩0+iδμ⟨W(Tr[∫dxSu(x,x)γ5]−Tr[∫dxSd(x,x)γ5])⟩\begin{gather} \langle W \rangle = \langle W \rangle_{0}- \langle W \int dx' (\delta\mu\bar u u+\delta\mu \bar d d)(x') \rangle_{0} \\ = \langle W \rangle_{0} - \langle W \int dx (\delta\mu\bar \chi_u i\gamma_5 \chi_u- \delta\mu\bar \chi_d i\gamma_5 \chi_d)(x) \rangle_{0} \\ = \langle W \rangle_{0} +i\delta\mu \langle W \left(\text{Tr}[ \int dx S_u(x,x)\gamma_5 ] -\text{Tr}[ \int dx S_d(x,x)\gamma_5 ]\right)\rangle \end{gather} Where we got an extra minus sign from the fermion loop