I'm trying to work out the 2PI-effective action for complex scalar fields.
Introducing a multi field index $(a,b,c...)$ the complex conjugation and all other degrees of freedoms are suppressed, and field vectors are used $$\phi=(\phi_a,\phi_b,...)^T=(\phi_1,\phi_1^*,....)^T .$$
The generating functional $Z[J,R]$ and $W[J,R]$ is then given by;
$$ Z[J,R]= \int \mathcal{D} \phi \, \exp (i \left[ S[\phi] + J_a \phi_a + \frac{1}{2} \phi_a R_{a,b} \phi_b \right] ) = \exp(i W[J,R]). $$
We can now define the connected two-points function $G_{a,b}$ and the macroscopic field $\Phi_a$ and find for the derivative of $W$:
$$ \qquad \Phi_a = \frac{\delta W}{\delta J_a} \qquad G_{a,b} = \frac{\delta^2 W}{\delta J_a \delta J_b} \qquad \Rightarrow \qquad \frac{\delta W}{\delta R_{a,b}} = \frac{1}{2} (\Phi_a \Phi_b + i G_{a,b}).$$
We can now look for the Legendre transformation of $W$ w.r.t. the two fields.
$$ \Gamma[\Phi,G] = W[J,R] - \Phi_a J_a - \frac{1}{2} (\Phi_a \Phi_b + i G_{a,b}) R_{b,a}. $$
From this we find for the sources;
$$ \frac{\delta \Gamma}{\delta \Phi_a} = - J_a - \frac{1}{2} \Phi_a (R_{a,b}+R_{b,a})= - J_a - \Phi_a R_{a,b}$$ $$ \frac{\delta \Gamma}{\delta G_{a,b}} =- \frac{i}{2} R_{b,a} = - \frac{i}{2} R_{a,b}. $$
- And here is my first question. I can see that $R$ must be symmetric by definition, therefore the expression does not depend on taking the derivative with $G_{a,b}$ or $G_{b,a}$. But how to take the derivative in general?
$$ \qquad \qquad \frac{\delta G_{d,c}}{\delta G_{a,b}} = \delta_{a,d} \delta_{c,b} \qquad or \qquad \frac{\delta G_{d,c} }{\delta G_{a,b}} =\frac{1}{2} ( \delta_{a,d} \delta_{c,b} +\delta_{c,d} \delta_{a,b} ). $$
- The second question is if that statement also implies that $G_{a,b}$ must by symmetric?
Following further the development the effective action is then found as;
$$ \Gamma[\Phi,G]= S[\Phi] +\frac{i}{2} \text{tr} \log G^{-1} + \frac{i}{2} \text{tr} G_{0}^{-1}G + \Gamma^2. $$
Where $\Gamma^2$ contains all 2-PI diagrams and will contribute the self-energy. The effective action has boundary conditions at $J=0, R=0$, from where the Greens function can be found. Therefore we just have to take the functional derivative w.r.t. $G$ of the upper equation.
$$ \frac{\delta \Gamma}{\delta G_{a,b}}=0 = \frac{-i}{2} G^{-1}_{b,a} + \frac{i}{2} (G_{b,a}^0)^{-1} + \frac{\delta \Gamma}{\delta G_{a,b}}. $$
Does this also implies that the self-energy $$\Sigma_{b,a} = 2i \frac{\delta \Gamma}{\delta G_{a,b}} $$ is symmetric in $a,b$?
