When doing QFT or statistical field theory of, say $N$ scalar fields $\varphi_i$, we consider the the generating functional $$ W[J] = - \ln Z[J], \quad Z[J] = \int \mathcal D \varphi \, e^{-S[\varphi] + J\cdot \varphi}. $$ Here, I denote both integration and summation over indices by a dot product. This is analogous to Helmholtz free energy in statistical field theory. The analogous quantity to the Gibbs free energy is the 1PI generating functional $$ \Gamma[\Phi] = W[J] + J\cdot\Phi, \quad \Phi = \langle \varphi \rangle_J. $$ The two potentials and the variables are connected by $\frac{\delta W}{\delta J} = - \Phi$ and $\frac{\delta \Gamma}{\delta \Phi} = J$. What is calculated in perturbative QFT is the 1PI generating funcitonal, as it is given by the sum of one-particle-irreducible (1PI) diagrams. Let $$ \Gamma^{(N)}(x_1...x_N) = \frac{\delta^N \Gamma[\Phi]}{\delta \Phi(x_1)...\delta\Phi(x_N)}\bigg|_{\langle \phi\rangle_{J = 0}}. $$ $\Gamma^{(N)}$ is then given by all 1PI diagrams with $N$ external legs. Using the same notaiton for $W$, $W^{(N)}$ are the sum of all connected diagrams. Taking derivatives of $\frac{\delta \Gamma}{\delta \Phi} = J$, we may show that these are related by (in very schematic notation) $$ W^{(2)} = \left( \Gamma^{(2)} \right)^{-1}, \quad W^{(3)} = \left(W^{(2)}\right)^3 \Gamma^{(3)},\quad W^{(4)} = \left(W^{(2)}\right)^3 \Gamma^{(4)} + 3 \left(W^{(2)}\right)^2 \Gamma^{(3)} W^{(2)} \Gamma^{(3)} \left(W^{(2)}\right)^2, $$ and so on (modulo some possible minus signs hear and there). The connected correlation functions are given by 1PI verteces, connected by the connected propagators. (See section 11.5 in Peskin & Schröder.)

My question is, can we write down a similar relation for $W^{(1)} = \Phi$? My first intuition, and by looking at the diagramatics, would be to write $$ W^{(1)} = W^{(2)} \Gamma^{(1)}, $$ or $$ \Phi = D J, $$ where $D = W^{(2)}$ is the connected propagator. This, however, would force us to make the conclusion that $J = 0 \implies \Phi = 0$. That is, $\Phi$ must have a vanishing vacuum expectation value. Is this right? Are we not allowed to use these tools unless we choose $\varphi$ such that the expectation value vanish? This seems to be in contradiction with the discussion of Peskin & Schröder, p.355, where they define the renormalization conditions for the linear sigma model. They say "such a shift [of the expectation value to be non-zero] is acceptable.". I interpret that we are free to choose $\Phi|_{J = 0} \neq 0$. This seems to me to be a contradiction, what is the resolution?

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    $\begingroup$ You are tacitly assuming that there is no solution of $D^{-1}\Phi=0$. $\endgroup$
    – mike stone
    Nov 12, 2023 at 14:02
  • $\begingroup$ As @Qmechanic shows here, the equation $\Phi = DJ$ is wrong. Rather, it should be $\Phi - \langle \varphi \rangle_{J = 0} = D J + \mathcal O (J^2)...$ $\endgroup$ Nov 12, 2023 at 16:23

1 Answer 1


The Legendre transformation between the generating functional $W_c[J]$ for connected diagrams and the the effective/proper action $\Gamma[\phi_{\rm cl}]$ also works for non-zero vacuum expectation value $\langle \phi \rangle_{J=0}$. The first few orders are e.g. worked out in my Phys.SE answer here.

  • $\begingroup$ Thank you, I see that $\Phi = DJ$ is in fact wrong, as the expansion is in terms of $\Delta \Phi = \Phi - \langle \varphi \rangle_{J = 0}$, the expression should be $\Delta \Phi = DJ + \mathcal O (J^2)...$ With this, there is no contradiction. $\endgroup$ Nov 12, 2023 at 18:00

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