[Physics] Defining quantum effective/proper action (Legendre transformation), existence of inverse (field-source)

1pi-effective-actionlagrangian-formalismmathematical physicspath-integralquantum-field-theory

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?

Best Answer

  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

    $$W^k_{c,1}~=~0,\tag{11}$$

    then:

    • 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.

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