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 ..


1 Answer 1


When calculating the tree propagator it's not matter whether You field is abelian or not.

You can obtain the expression of the propagator in the following equivalent way:

  • 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.

  • Path integral approach. In this approach

$$ (1) \equiv \frac{\delta^{2}Z[J] }{\delta W^{\mu,\dagger}(x)\delta W^{\nu}(y)}\bigg|_{J = 0}, $$ 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, $D_{\mu\nu}(p) = K_{\mu\nu}^{-1}(x)$.

It is not matter in which approach - path integral or operator formalism - you work. They give the same answer, by definition


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.