In an exercise for a course on quantum field theory, I am given the following Lagrangian: $$ \mathcal{L} = -\frac{1}{2} G_{\mu\nu}^a G^{a\mu\nu} + 2 (D_\mu \phi^\dagger)^a(D^\mu \phi)^a - 2 m^2\phi^{\dagger a}\phi^a -2\lambda (\phi^{\dagger a}\phi^a)^2, $$ where the covariant derivative and the strength tensor are given as $$ G_{\mu\nu} = \partial_\mu G_\nu - \partial_\nu G_\mu - i g [G_\mu, G_\nu], $$ $$ D_\mu = \partial_\mu - i g G_\mu \equiv \partial_\mu - K_\mu, $$ and where the scalar field is complex and decomposes under the fundamental representation of ${\rm SU}(N)$ as $\phi = \phi^a T^a$, for some given $N\geq 2$. Of course, repeated indices are summed over.

Now the first question I am asked is to calculate the propagator for the scalar field, and here a conceptual problem that has been haunting me for a while makes its appearance again: when we talk about the propagator of a given field for a specific Lagrangian, aren't we talking about the Green's function for the differential equation that gives the dynamics of the (classical) field? Applying calculus of variation to the action corresponding to this Lagrangian under a change $\phi\mapsto \phi+\delta \phi$ I reach the equations of motion $$ \Omega^{ab} \phi^b = \Omega^{ab} \phi^{\dagger b} = 0 \tag{1} $$

for an operator

$$ \Omega^{ab} = \partial_\mu K^{\mu a b} - K_\mu^{ab} \partial^\mu + K^{ca}_\mu K^{\mu cb} - (\partial^2 +m^2)\delta^{ab} $$

where, as usual, inside a linear operator terms like $\partial_\mu A$ act as $\phi\mapsto \partial_\mu(A \phi)$. Now the provided solution to the exercise says

The propagator for the scalar field is contained in the kinetic term $$\mathcal{L}_{\rm kin} = -\phi^{\dagger a}(\partial^2 + m^2)\delta^{ab}\phi^b \tag{2}$$

so according to the solution the propagator is actually the Green function for the differential equation induced by $(2)$ and not by $(1)$, but why? Extremizing the action under a variation of $\phi$ for the full Lagrangian leads to (up to possible mistakes in my development) equation $(1)$ and as such I would expect to be that what determines the propagator. Is it because the (free) propagator we are looking for here derives from the interaction-free part of the Lagrangian, which by definition only includes (for $\phi$) the last term in my $\Omega^{ab}$?

If this is the reason, I would still appreciate further information on why the propagator doesn't include the whole $\Omega$. I understand that the "practical" objective is to express through the propagator $\Delta$ a free partition function as products of terms like $$ Z_{\rm free}\propto \exp\left[i\int d^4xd^4y J(x)\Delta(x-y)J(y)\right] \qquad (3) $$ but, to my intuition, the interactions between a field and a gauge field are of different "nature" than the interactions added by hand such as the $\lambda$ term, because the former appear by the necessity of imposing gauge invariance. However, I believe these other interactions "added by hand" are also incorporated by a necessity, namely making the theory renormalizable, so perhaps they are not that different in nature after all, are they?

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    $\begingroup$ So maybe some insight on the last part of your question: Normally (if we are not talking about some effective field theory) we include all renormalizable terms containing all fields that we are interested in which are allowed under the imposed symmetries (e.g. Lorentz invariance or some global symmetry or even gauge symmetry) the reason for this is that renormalizing the lagrangian will generate all of these terms anyhow $\endgroup$ May 10, 2023 at 10:15
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    $\begingroup$ General tip: Ask only 1 question per post. $\endgroup$
    – Qmechanic
    May 10, 2023 at 11:33
  • $\begingroup$ @Qmechanic, I deleted the last paragraph as, indeed, it was note directly related to the problem here. $\endgroup$
    – Albert
    May 10, 2023 at 11:41

1 Answer 1


You are right. In perturbation theory, the propagator consists of quadratic terms about the field and the interaction terms consist of terms of third order or higher with respect to the field.

One way to understand why terms of the second order are given special treatment is that in the path integral formalism it is only the second order about a field that can do the integral exactly. (i.e., Gaussian integrals).

Let us try to understand this from a pragmatic point of view. For example, if we decide to stand on the path integral formalism, the perturbation expansion is constructed in the following form: $$Z[j,h]=\int D\phi D\varphi\ e^{iS_0+iS_I+\int_x(j\phi+h\varphi)}=\int D\phi D\varphi\ \Big(\sum_n\frac{(iS_I)^n}{n!}\Big)e^{iS_0+\int_x(j\phi+h\varphi)},$$ $$S_0+S_I=\int_x\Big\{\frac{1}{2}(\partial\phi\partial\phi-m^2\phi^2)+\frac{1}{2}(\partial\varphi\partial\varphi-m^2\varphi^2)+\frac{g}{2}\varphi\phi\phi\Big\}.$$

Here, the propagator is contained in $S_0$. Also, for the case of including fermions, see the appropriate textbook; the reason for that the second order terms are important remains the same even if fermion is included.

Now suppose that only quadratic terms for the field are included in the $S_0$: $$S_0=\int_x\Big\{\frac{1}{2}(\partial\phi\partial\phi-m^2\phi^2)+\frac{1}{2}(\partial\varphi\partial\varphi-m^2\varphi^2)\Big\}$$ We perform the path integral for the free field above and obtain $$Z[j,h]=\Big(\sum_n\frac{(iS_I[\frac{\delta}{i\delta j},\frac{\delta}{i\delta h}])^n}{n!}\Big)\int D\phi D\varphi\ e^{iS_0+\int_x(j\phi+h\varphi)}$$ $$=\Big(\sum_n\frac{(iS_I[\frac{\delta}{i\delta j},\frac{\delta}{i\delta h}])^n}{n!}\Big)\ e^{-\int_{x,y}j_x(\Box+m^2)^{-1}j_y}e^{-\int_{x,y}h_x(\Box+m^2)^{-1}h_y}.$$ Here, we did the Gauss integral. This is special property for the quadratic integral.

On the other hand, if there is a third-order term for the field; $$S_0=\int_x\Big\{\frac{1}{2}(\partial\phi\partial\phi-m^2\phi^2+g\varphi\phi\phi)+\frac{1}{2}(\partial\varphi\partial\varphi-m^2\varphi^2)\Big\}$$ it cannot be handled by the Gaussian integral, so we don't know how to handle it, in fact in this case we will obtain

\begin{align} Z[j,h]&=\Big(\sum_n\frac{(iS_I[\frac{\delta}{i\delta j},\frac{\delta}{i\delta h}])^n}{n!}\Big)\\ \times&\int D\phi D\varphi\ e^{i\int_x\{\frac{1}{2}(\partial\phi\partial\phi-m^2\phi^2+g\varphi\phi\phi)\}+i\int_x(j\phi)}\\&\times e^{-\int_x\{i\varphi(\Box+m^2)\varphi\}+i\int_x(h\varphi)}. \end{align}

We don’t know to treat this complicated integral.

This is why the propagator consists of second order terms about the field. (In operator formalism, it is difficult to construct the inverse of a differential operator with interaction terms like the one you mentioned, so we use the inverse of a free-field differential equation that can be solved.)

On the other hand, if, for example, in the example above, $\varphi$ is an external field, i.e., we do not perform path integral about $\varphi$, the story changes:

\begin{align} Z[j,h]=\Big(\sum_n\frac{(iS_I[\frac{\delta}{i\delta j},\frac{\delta}{i\delta h}])^n}{n!}\Big) \times\int D\phi \ e^{-i\int_x\{\frac{1}{2}\phi(\Box+m^2-g\varphi)\phi\}+\int_x(j\phi)}\\ \end{align}

In this case, the Gaussian integral can be performed even if $S_0$ contains terms that include the external field $\varphi$. Thus, in this case, the propagator can be defined in a form that includes $\varphi$.

Regarding the last part: the symmetry-based construction of the theory has little to do with the propagator. Concepts such as propagator and vertex are objects that are introduced to construct a perturbation theory. One reason why it makes from second order terms about fields is because, as we saw above, second order terms about fields have various useful properties, such as the possibility of Gaussian integrals. On the other hand, when defining a field theory on a lattice, it is not always necessary to introduce propagators, etc. In other words, for perturbation theory, it is convenient to keep the second-order terms of the theory separate from the rest of the theory, so we keep them separate.

  • $\begingroup$ You were right on the spot where my doubts lay! I read your answer precisely after realizing that it would be impossible to write a free partition function as a product of expressions like $(3)$, if the propagator included other fields we would be integrating over. What I did not realize---until I read you---is how dependent the notion of propagator is on the goal of applying perturbative methods, and hence on separating the exactly solvable part of a partition function. Thank you! $\endgroup$
    – Albert
    May 10, 2023 at 11:54

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