Path integral of vector bosons My question is about the path integral formulation of QFT. Specifically about vector/gauge bosons
I understand that I have measures $\mathcal D \phi$, $\mathcal D\psi$, $\mathcal D \overline\psi$ for scalar and Dirac fields. I have $\mathcal D A$ for the vector field. Now there is a partition function (only scalar and Dirac)
$$
\tilde{Z}[\eta, \overline{\eta}, J]
=
\frac{\int\mathscr{D}\psi\mathscr{D}\overline{\psi}\mathscr{D}\phi\exp\left({i\mathcal{S}[\psi,\overline{\psi},\phi]+i\int\mathrm{d}^4x[\overline{\psi}\eta+\overline{\eta}\psi+\phi J]}\right)}{\int\mathscr{D}\psi\mathscr{D}\overline{\psi}\mathscr{D}\phi\exp\left(i\mathcal{S}[\psi,\overline{\psi},\phi]\right)},
$$
and (2m+n)-point Green function
$$
\langle x_1...y_mz_1... z_{n}\rangle=
\frac{\int\mathscr{D}\psi\mathscr{D}\overline{\psi}\mathscr{D}\phi\left[
\psi(x_1)...\psi(x_m)\overline{\psi}(y_1)...\overline{\psi}(y_n)\phi(z_1)...\phi(z_n)
\exp\left({i\mathcal{S}[\psi,\overline{\psi},\phi]}\right)\right]}{\int\mathscr{D}\psi\mathscr{D}\overline{\psi}\mathscr{D}\phi\exp\left(i\mathcal{S}[\psi,\overline{\psi},\phi]\right)}.
$$
Now I'd like to include vector field $A^\mu$. Here's the partition function (?):
$$
\tilde{Z}\left[\eta, \overline{\eta}, J, j^\mu\right]
=
\frac{\int\mathscr{D}\psi\mathscr{D}\overline{\psi}\mathscr{D}\phi\mathscr{D}A\exp\left({i\mathcal{S}[\psi,\overline{\psi},\phi,A^\mu]+i\int\mathrm{d}^4x[\overline{\psi}\eta+\overline{\eta}\psi+\phi J+ j^\mu A_\mu]}\right)}
{\int\mathscr{D}\psi\mathscr{D}\overline{\psi}\mathscr{D}\phi\mathscr{D}A\exp\left(i\mathcal{S}[\psi,\overline{\psi},\phi,A^\mu]\right)},
$$
I have 2 questions:

*

*How measure $\mathcal DA$ work? $A$ is 4-vector. Is $\mathcal D$ a 4-vector too? (if yes I don't need much more explanation)


*What is the expression for Green function with $A$? I can't have $k$ 4-vector fields without coupling them to something, right?
 A: For your first question, please check out this for the definition of path-integral of gauge fields. Roughly speaking, the measure $\mathcal{D}[A]$ is defined in the following way: $$\mathcal{D}[A]\equiv\prod_{\mu,a,x}dA^{a}_{\mu}(x),$$
where $A^{\mu}(x)=A^{\mu}_{a}(x)T^{a}$, with $[T^{a},T^{b}]=f^{ab}_{\,\,\,\,\,c}T^{c}$ being the commutator relation of the Lie algebra of the gauge group.
But one must be very careful with the above infinite dimensional integral measure. The action $S[A]$ and the integral measure are both gauge invariant under $$A[U]=U^{-1}dU+U^{-1}AU\quad\mathrm{or}\quad\delta_{U}A=d\,\delta U+[A,\delta U].$$
This means that the naive integral $\int\mathcal{D}[A]$ actually integrates out an infinite much redundant degrees of freedom. One must isolate these redundant gauge degrees of freedom so that the partition function makes sense. This procedure is known as Faddeev-Popov procedure, which is introduced in the link above.
For your second question, please check out this link. In the Feynman-'t Hooft gauge, the two-point function is given by $$\langle A^{a}_{\mu}(x)A^{b}_{\nu}(y)\rangle=\int\frac{d^{d}k}{(2\pi)^{d}}\frac{-i}{k^{2}+i\epsilon}g_{\mu\nu}\delta^{ab}e^{-ik\cdot(x-y)}.$$
This can be obtained from taking functional derivatives of the partition function.
