# Generating functional for fields with non-zero expectation value

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?

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

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.
• 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. Nov 12, 2023 at 18:00