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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:

  1. 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)

  2. What is the expression for Green function with $A$? I can't have $k$ 4-vector fields without coupling them to something, right?

Qmechanic
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Josef Bobek
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  • $\cal D$ by itself does not mean anything, so it is no meaningful to try charactering it as a four-vector. On the other hand ${\cal D}A$ is the fictitious translation invariant measure in field space analogous to $d^nx$, and used in formal manipulations. The meaningful thing is ${\cal D}A e^{- S}$ which can be defined after resorting to the Faddeev-Popov prescrpition. – Gold Dec 20 '22 at 15:46
  • $\psi$ and $\bar\psi$ are vectors too, so if you are happy with $\mathcal D\psi\mathcal D\bar\psi$, then you should also be happy with $\mathcal DA$... – AccidentalFourierTransform Dec 21 '22 at 23:37

1 Answers1

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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.

Valac
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  • Thank you so much! This is very helpful.

    1. Would be Faddeev–Popov procedure necessary if $A$ wouldn't be the gauge field? Just a vector meson for example.

    2. I'm familiar with that you can get a Green function from taking functional derivatives but is there some nice expression for the Green function like the one above (for scalar and Dirac field)?

     – Josef Bobek Dec 20 '22 at 17:47
  • @JosefBobek You are welcome. 1) If $A$ is not a gauge field, then there's no need for gauge fixing and thus no Faddeev-Popov procedure. 2) You can find the Green's functions of scalar and fermions from any QFT textbook. – Valac Dec 21 '22 at 08:54
  • Thanks! That bracket "(for scalar and Dirac field)" was referring to what I've written in the original post. My question was if there is something analogical for the vector field. Something that you could write the Green function with functional integral for the vector field. In the case of Dirac/scalar, there are either $\overline{\psi}\psi$ of $\phi$ in the numerator. Is there something like $\prod_\mu A_\mu$ instead? – Josef Bobek Dec 21 '22 at 17:18
  • Why do you want to write down a product of different components of a field $A$? Two points function is supposed to be containing a product of $A$ valuated at different spacetime points. i.e. something like $A_{\mu}(x)A_{\nu}(y)$. – Valac Dec 21 '22 at 17:35