The background field method of deriving a 2PI effective action. Calzetta and Hu book

Question

I am going through "Nonequilibrium Quantum Field Theory" by Calzetta and Hu right now and it seems that I cannot fully understand the derivations in chapter 6.5. There, they consider the derivation of a 2PI effective action. It is defined as a double Legendre transform of the functional $W[J,R]$ (I use slightly different notations): \begin{equation} e^{\frac{i}{\hbar} W[J,R]} = \int D \Phi_a \exp{\left\lbrace \frac{i}{\hbar} \left[ S[\Phi] + J_a \Phi_a + \frac{1}{2} R_{ab} \Phi_a \Phi_b \right] \right\rbrace}\tag{6.144} \end{equation}

such that

\begin{equation} \Gamma[\phi,G] = W[J,R] - J_a \phi_a - \frac{1}{2} R_{ab} \left[\phi_a \phi_b + G_{ab} \right],\tag{6.146} \end{equation}

where

\begin{equation} \frac{\delta W}{\delta J_a} = \phi_a, \qquad \frac{\delta W}{\delta R_{ab}} = \frac{1}{2} \left[\phi_a \phi_b + G_{ab} \right].\tag{6.145} \end{equation}

The equations of motion read

\begin{equation} \frac{\delta \Gamma}{\delta \phi_a} = -J_a - R_{ab} \phi_b, \qquad \frac{\delta \Gamma}{\delta G_{ab}} = -\frac{1}{2} R_{ab}.\tag{6.147} \end{equation}

We can use the above to get

\begin{equation} e^{i\Gamma/\hbar} = \int D \Phi_a \, e^{\frac{i}{\hbar} \left[S[\Phi] + J_a (\Phi_a - \phi_a) + \frac{1}{2} R_{ab} (\Phi_a \Phi_b - \phi_a \phi_b - G_{ab}) \right]}.\tag{6.148} \end{equation}

We can now decompose the field into classical part and fluctuations, i.e. $$\Phi_a = \phi_a + \varphi_a,\tag{6.149b}$$ and using the quantum equations of motion rewrite the exponent as

\begin{equation} S[\phi_a + \varphi_a] - \frac{\delta \Gamma}{\delta \phi_a} \varphi_a - \frac{\delta \Gamma}{\delta G_{ab}} \left[\varphi_a \varphi_b - G_{ab}\right]. \tag{6.149} \end{equation}

Therefore, the result is

\begin{equation} \Gamma [\phi,G] = S[\phi] - i \hbar \log \int D \varphi_a \, e^{\frac{i}{\hbar} \left\lbrace (S[\phi_a + \varphi_a] - S[\phi_a]) - \frac{\delta \Gamma}{\delta \phi_a} \varphi_a - \frac{\delta \Gamma}{\delta G_{ab}} \left[\varphi_a \varphi_b - G_{ab}\right] \right\rbrace} \end{equation}

We now expand the classical action as

\begin{equation} S[\phi_a + \varphi_a] - S[\phi_a] = \frac{\delta S[\phi]}{\delta \phi_a} \varphi_a + \frac{1}{2} \frac{\delta^2 S[\phi]}{\delta \phi_a \delta \phi_b} \varphi_a \varphi_b + S_Q,\tag{6.150} \end{equation}

where $S_Q$ denotes the part containing higher order terms with respect to $\varphi$. Substituting, we get

\begin{equation} \Gamma [\phi,G] = S[\phi] + \frac{\delta \Gamma}{\delta G_{ab}} G_{ab} - i \hbar \log \int D \varphi_a \, e^{\frac{i}{\hbar} \left\lbrace \left(\frac{\delta S}{\delta \phi_a} - \frac{\delta \Gamma}{\delta \phi_a}\right) \varphi_a + \frac{1}{2}\left(\frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2 \frac{\delta \Gamma}{\delta G_{ab}} \right) \varphi_a \varphi_b + S_Q\right\rbrace} \end{equation}

After that they use some magic words and get the final result, but I would like to actually obtain it by using mathematics.

Let us forget about $S_Q$ for now, i.e. make computations on a Gaussian level. Then the above integral can be evaluated

\begin{align} &\int D \varphi_a \, e^{\frac{i}{\hbar} \left\lbrace \left(\frac{\delta S}{\delta \phi_a} - \frac{\delta \Gamma}{\delta \phi_a}\right) \varphi_a + \frac{1}{2}\left(\frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2 \frac{\delta \Gamma}{\delta G_{ab}} \right) \varphi_a \varphi_b \right\rbrace}\\ &= \mathrm{const} \cdot \left[\det \left(\frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2 \frac{\delta \Gamma}{\delta G_{ab}} \right) \right]^{-1/2} e^{\left\lbrace \frac{-i}{2 \hbar} \left(\frac{\delta S}{\delta \phi_a} - \frac{\delta \Gamma}{\delta \phi_a}\right) \left(\frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2 \frac{\delta \Gamma}{\delta G_{ab}} \right) \left(\frac{\delta S}{\delta \phi_b} - \frac{\delta \Gamma}{\delta \phi_b}\right)\right\rbrace}. \end{align}

Thus,

\begin{align} \Gamma[\phi,G] &= S[\phi] + \frac{\delta \Gamma}{\delta G_{ab}} G_{ab} + \frac{i \hbar}{2} \mathrm{Tr} \log \left(\frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2 \frac{\delta \Gamma}{\delta G_{ab}} \right)\\ &+ \frac{1}{2} \left(\frac{\delta S}{\delta \phi_a} - \frac{\delta \Gamma}{\delta \phi_a}\right) \left(\frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2 \frac{\delta \Gamma}{\delta G_{ab}} \right) \left(\frac{\delta S}{\delta \phi_b} - \frac{\delta \Gamma}{\delta \phi_b}\right) \end{align}

I can kinda see how to get $-\frac{i\hbar}{2}\mathrm{Tr} \log G$. From the $1$-loop $1$PI result:

\begin{equation} i G_{0,ab}^{-1} + R = \frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2\frac{\delta \Gamma}{\delta G_{ab}} \approx i G^{-1}_{ab}, \end{equation}

which yields

\begin{equation} \frac{i \hbar}{2} \mathrm{Tr} \log \left(\frac{\delta^2 S}{\delta \phi_a \delta \phi_b} - 2 \frac{\delta \Gamma}{\delta G_{ab}} \right) \approx -\frac{i \hbar}{2} \mathrm{Tr} \log G + \mathrm{const}. \end{equation}

But what should I do with the second piece to get $\frac{i}{2} G_{0,ab}^{-1} G_{ab}$? I can't see it.

References:

1. E. A. Calzetta and B.-L. B. Hu, Nonequilibrium quantum Field theory, Cambridge Monographs on Mathematical Physics (Cambridge University Press, 2008). Chapter 6.5. The two-particle irreducible (2PI) effective action.

Answer 1

You need to start with the expansion for effective action $\Gamma$ ($\hbar=1$): $$\Gamma[ϕ,G]=S[ϕ]−i\ln\int Dφ\cdot e^{i(S[ϕ+φ]−S[ϕ])−i\frac{\delta Γ}{\delta \phi^a}φ^a−i\frac{\delta Γ}{\delta G^{ab}}(φ^aφ^b−G^{ab})}$$ and then introduce $\Gamma_Q$ via $$\Gamma[\phi, G] = S[\phi] +\frac{1}{2}G_{0ab}^{-1}[\phi]G^{ba}-\frac{i}{2}\ln\text{sdet}G + \Gamma_Q[\phi,G] -\frac{i}{2}\delta^a_a $$ You put this formula to the former one and after some algebra get expansion for $\Gamma_Q$ (formula 6.152 in the book). The idea is to cancel quadratic term $-\frac{1}{2}φG^{-1}_0φ$ in $S[\phi+φ]-S[\phi]$ by term $\frac{1}{2}G_{0}^{-1}G$ in $\Gamma$, and substitute it by $-\frac{1}{2}φG^{-1}φ$ using $\frac{i}{2}\ln\text{sdet}G$ in $\Gamma$. So now all gaussian integrations in expansion for $\Gamma_Q$ are done with full propagator $G$.