Given a Quantum field theory, for a scalar field $\phi$ with generic action $S[\phi]$, we have the generating functional $$Z[J] = e^{iW[J]} = \frac{\int \mathcal{D}\phi e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi e^{iS[\phi]}}.$$

The one-point function in the presence of a source $J$ is.

$$\phi_{cl}(x) = \langle \Omega | \phi(x) | \Omega \rangle_J = {\delta\over\delta J}W[J] = \frac{\int \mathcal{D}\phi \ \phi(x)e^{i(S[\phi]+\int d^4x J(x)\phi(x))}} {\int \mathcal{D}\phi \ e^{i(S[\phi]+\int d^4x J(x)\phi(x))}}.$$

The effective Action is defined as the Legendre transform of $W$

$$\Gamma[\phi_{cl}]= W[J] -\int d^4y J(y)\phi_{cl}(y),$$ where $J$ is understood as a function of $\phi_{cl}$.

That means we have to invert the relation $$\phi_{cl}(x) = {\delta\over\delta J}W[J]$$ to $J = J(\phi_{cl})$.

How do we know that the inverse $J = J(\phi_{cl})$ exists? And does the inverse exist for every $\phi_{cl}$? Why?


2 Answers 2

  1. If we treat the generating functional $W_c[J]$ for connected diagrams as a formal power series in the sources $J_i$, and if the connected propagator$^1$ $$\begin{align} \langle \phi^k\phi^{\ell} \rangle^c_J~=~& \frac{\hbar}{i}\frac{\delta^2 W_c[J]}{\delta J_k\delta J_{\ell} }\cr ~=~&\langle \phi^k\phi^{\ell} \rangle_J-\langle \phi^k \rangle_J \langle \phi^{\ell} \rangle_J\end{align} \tag{1}$$ is invertible at $J=0$, then the effective/proper action $$ \Gamma[\phi_{\rm cl}]~=~W_c[J]-J_k \phi^k_{\rm cl} \tag{2}$$ exists as a formal power series in the Legendre transformed variable $\phi_{\rm cl}$. In particular, the inversion of the formal power series $$\phi^k_{\rm cl}~=~\frac{\delta W_c[J]}{\delta J_k} ~=~ \langle \phi^k \rangle_J\tag{3}$$ then follows from a multi-variable generalization of Lagrange inversion theorem.

  2. Concretely, to lowest orders, if we expand $$\begin{align} W_c[J]~=~&W_{c,0} ~+~J_k W_{c,1}^k ~+~\frac{1}{2}J_k W_{c,2}^{k\ell}J_{\ell}\cr ~+~&\frac{1}{6} W_{c,3}^{k\ell m}J_k J_{\ell} J_m ~+~{\cal O}(J^4), \end{align}\tag{4}$$ we calculate $$\begin{align} \Delta\phi^k_{\rm cl} ~:=~& \phi^k_{\rm cl}~-~ W_{c,1}^k\cr ~\stackrel{(3)+(4)}{=}&~W_{c,2}^{k\ell}J_{\ell}~+~\frac{1}{2} W_{c,3}^{k\ell m} J_{\ell} J_m ~+~{\cal O}(J^3),\end{align} \tag{5}$$ so that $$\begin{align} J_k~\stackrel{(5)}{=}~&(W^{-1}_{c,2})_{k\ell}\left(\Delta\phi^{\ell}_{\rm cl}~-~\frac{1}{2}\Delta\phi^p_{\rm cl} (W^{-1}_{c,2})_{pm} W_{c,3}^{m\ell n}(W^{-1}_{c,2})_{nq}\Delta\phi^q_{\rm cl} \right)\cr ~+~&{\cal O}(\Delta\phi^3_{\rm cl}).\end{align}\tag{6}$$ Perturbatively, the Legendre transform becomes $$ \begin{align} \Gamma[\phi_{\rm cl}]~\stackrel{(2)+(4)+(6)}{=}& W_{c,0} ~-~\frac{1}{2}\Delta\phi^k_{\rm cl} (W^{-1}_{c,2})_{k\ell}\Delta\phi^{\ell}_{\rm cl}\cr ~+~&\frac{1}{6} W_{c,3}^{k\ell m}(W^{-1}_{c,2})_{kp}(W^{-1}_{c,2})_{\ell q}(W^{-1}_{c,2})_{mr}\Delta\phi^p_{\rm cl}\Delta\phi^q_{\rm cl}\Delta\phi^r_{\rm cl} \cr ~+~&{\cal O}(\Delta\phi^4_{\rm cl}),\end{align}\tag{7}$$ and so forth.

  3. Similarly, perturbatively, the inverse Legendre transform becomes $$ \begin{align} W_c[J]~\stackrel{(2)+(9)}{=}& \Gamma_0 ~-~\frac{1}{2}\Delta J_k (\Gamma^{-1}_2)^{k\ell}\Delta J_{\ell}\cr ~-~&\frac{1}{6} \Gamma_{3,k\ell m}(\Gamma^{-1}_2)^{kp}(\Gamma^{-1}_2)^{\ell q}(\Gamma^{-1}_2)^{mr}\Delta J_p\Delta J_q\Delta J_r\cr ~+~&{\cal O}(\Delta J^4),\end{align}\tag{8}$$ and so forth, where $$ \Delta J_k~:=~ J_k~+~ \Gamma_{1,k}.\tag{9}$$

  4. At this point it seems natural to finish with the following useful Proposition.

    Proposition. If$^2$

    $$\langle \phi^k \rangle_{J=0}~=~0,\tag{10}$$

    or equivalently, if



    • The full 2-pt function is equal to the full connected 2-pt function: $$\langle \phi^k\phi^{\ell} \rangle_{J=0}~=~\langle \phi^k\phi^{\ell} \rangle^c_{J=0}~=~\frac{\hbar}{i}G_c^{k\ell},\tag{12}$$ cf. eq. (1).

    • $$\Gamma_{1,k}~=~0,\tag{13}$$ cf. eq. (7).

    • $$-(\Gamma_2^{-1})^{k\ell}~=~(W_{c,2})^{k\ell}~=~G_c^{k\ell}\tag{14}$$ is the full connected propagator, cf. eq. (8).

    • There are no tadpoles$^3$ in the sense that if a single cut cuts a connected diagram in 2 parts, then both parts contain $J$-sources, cf. e.g. Srednicki, QFT, chapter 9, p. 67. This follows from the fact that (a sum of all possible) connected diagrams is (a sum of all possible) trees of full propagators and (amputated) 1PI vertices, cf. this Phys.SE post.

    • In particular, the connected vacuum diagrams $W_{c,0}=\Gamma_0$ are all 1PI diagrams, cf. eq. (8).

    • In particular, the self-energy $$ \Sigma~=~G_0^{-1}-G_c^{-1},\tag{15}$$ [who in general consists of connected diagrams with 2 amputated legs such that the 2 legs cannot be disconnected by cutting a single internal line] now only consists of 1PI diagrams.

    • The Wilsonian effective action $W_{c,\rm int}[J\!=\!0,\phi_L]$ consists$^4$ of only 1PI action terms.


$^1$ We use DeWitt condensed notation to not clutter the notation. See also e.g. this related Phys.SE post.

$^2$ This is a standard renormalization condition. Due to momentum conservation, $$ \langle \widetilde{\phi}(k) \rangle_{J=0}~=~\widetilde{W}_{c,1}(k)~~\propto~~\delta^4(k) \tag{16}$$ is automatically satisfied.

$^3$ Be aware that the above notion of tadpole diagrams is not the same as self-loop diagrams, cf. Wikipedia.

$^4$ The renormalization condition (11) reads in this situation $$0~=~\langle \phi^k_H \rangle_{J=0,\phi_L}~=~\left. \frac{\delta W_c[J,\phi_L]}{\delta J_k}\right|_{J=0}.\tag{17}$$ The renormalization condition (17) should be satisfied for all values of the background field $\phi_L$. Note that $\phi_H$-tadpole terms $\phi_L\ldots\phi_L\phi_H$ are kinematically suppressed in the Wilsonian effective action. The renormalization condition (17) is consistent with the renormalization group flow. This is due to the Polchinski exact renormalization group flow equation (ERGE), $$\frac{d W^k_{1,c}}{d\Lambda}~\propto~\ldots \underbrace{\frac{\delta W^k_{1,c}}{\delta\phi_L}}_{=0} +\ldots \underbrace{\frac{\delta^2 W^k_{1,c}}{\delta\phi_L^2}}_{=0}~=~0,\tag{18}$$ cf. e.g. my Phys.SE answer here.

  • $\begingroup$ An $L$-loop diagram inside the $n$-point functions $W^{k_1\ldots k_n}_{n,c}$ or $\Gamma_{n, k_1\ldots k_n}$ is ${\cal O}(\hbar^L)$, cf. the $\hbar$/loop-expansion. $\endgroup$
    – Qmechanic
    Commented Jul 4, 2018 at 13:24
  • $\begingroup$ Very minor nitpicking: Lagrange inversion is a little bit different. Here it's just the inverse mapping theorem for formal power series, ie, solution to the reversion problem which one can find in Bourbaki, Algebra II, Chapters 4-7, Berlin, Springer-Verlag, 1990 and in any case is easy to prove by recursion. For explicit formulas in terms of tree diagrams see emis.de/journals/SLC/wpapers/s49abdess.html $\endgroup$ Commented Nov 4, 2021 at 18:34

This is an interesting question, and although I don't know a rigorous answer, we can discuss some typical cases.

Usually, the inverse exists, but the cases where this inverse does not exist are not necessarily pathological (sound models can have the problem that the inverse does not exist).

For standard field theories (say, $\phi^4$, O(N) models, classical spins models, ...), generically the inverse exists, and this can be shown order by order in a loop expansion (I don't know if this has been proven at all order, but in standard textbooks, this is shown to order 1 or 2). However, the inverse will not exist necessarily for all $\phi_{cl}$, especially in broken symmetry phases. Indeed, an ordered phase is characterized by $$\bar\phi_{cl}=\lim_{J\to 0 } \phi_{cl}[J]=\lim_{J\to 0 }W'[J]\neq 0 ,$$ where $\bar \phi_{cl}$ is the equilibrium value of the order parameter. Therefore, you cannot inverse the relationship $\phi_{cl}[J]$ for $\phi_{cl}\in [0,\bar \phi_{cl}]$ ($\phi_{cl}[J]$ generically increases when $J$ increases).

Furthermore, there are cases where the inverse is simply not defined, because $\phi_{cl}[J]={\rm const}$ for all $J$. This is usually the case when the field has no independent dynamics without a source. For instance, if you take a single quantum spin at zero temperature, the only dynamics is given by external magnetic field (here in the $z$ direction) $$\hat H= -h.\sigma_z.$$ With $h>0$, the ground state is always $|+\rangle$, and the "classical field" $\phi_{cl}(h)=\langle \sigma_z\rangle=1/2$ for all $h$, and the Gibbs free energy (the Legendre transform of the free energy with respect to $h$, which is essentially the effective action) does not exist.


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