# Non Abelian-Abelian gauge field propagator via a Fourier transformation

Can any one explain how to get the propagator:

\begin{align} i \Delta_{\mu\nu} = - i \frac{g_{\mu\nu - k_\mu k_\nu/M^2_W}}{k^2-M^2_W} \end{align}

from the Lagrangian:

\begin{align} \mathcal{L}_W = - \frac{1}{4} (\partial_\mu W^\dagger_\nu - \partial_\nu W_\mu^\dagger ) (\partial^\mu W^\nu - \partial^\nu W^\mu ) + M^2_W W^\dagger_\mu W^\mu \end{align}

Actually this topic seems difficult for me when I read in QFT books. I learnt that the propagator is two point function , put how did we get from the action ? and then how to make the proper Fourier transformation into the momentum space to get the final formula of the propagator . Also it's not clear for me here weather will we work by path integral or canonical quantization .

Any help will be appreciated ..

• Operator approach. You start from the definition of a propagator: $$\tag 1 D_{\mu\nu}(x - y) = \langle 0| T(W_{\mu}^{\dagger}(x)W_{\nu}(y))|0\rangle$$ After straightforward calculation (see, for example, Weinberg's QFT Vol. 1) it can be shown that this expression is nothing but $$D_{\mu\nu}(x - y) \simeq \int \frac{d^{4}p}{(2\pi)^{4}} e^{ip(x-y)}D_{\mu\nu}(p), \quad D_{\mu\nu}(p) = \frac{\sum_{\sigma}\epsilon^{\dagger}_{\sigma,\mu}(p)\epsilon_{\sigma, \nu}(p)}{p^{2} - m^{2}}$$ The expression $\sum_{\sigma}\epsilon^{\dagger}_{\sigma,\mu}(p)\epsilon_{\sigma, \nu}(p)$ is the polarization sum rule, which can be derived from the lagrangian.
$$(1) \equiv \frac{\delta^{2}Z[J] }{\delta W^{\mu,\dagger}(x)\delta W^{\nu}(y)}\bigg|_{J = 0},$$ where $$Z[J] = \frac{1}{\int DW_{\mu}^{\dagger}DW_{\mu}e^{iS[W_{\mu},W_{\mu}^{\dagger}]}}\int DW^{\dagger}_{\mu}DW_{\mu} e^{iS[W_{\mu},W^{\dagger}_{\mu}] + i\int d^{4}xW_{\mu}J^{\mu,\dagger} - i\int d^{4}xW_{\mu}^{\dagger}J^{\mu}}$$ is generating functional. In our case it is Gaussian integral, and it can be shown that for bose variables $W_{\mu}, W_{\mu}^{\dagger}$ it is equal to $$Z[J] = e^{i\int d^{4}x J_{\mu}^{\dagger}(x)\big(K^{\mu\nu}(x-y)\big)^{-1}J_{\nu}(y)},$$ where $K^{\mu\nu}(x)$ is the action quadratic form operator $$S = \int d^{4}x W_{\mu}^{\dagger}(x)\big(K^{\mu\nu}(x)\big)^{-1}W_{\nu}(x)$$ So, $D_{\mu\nu}(p) = K_{\mu\nu}^{-1}(x)$.