I'm trying to convince myself that $$\Gamma[\phi_c]=W[J]-\int d^4x\hspace{0.2cm} j(x)\phi_c(x)$$ is the effective action i.e., it contains all quantum corrections to the classical action $S[\phi]$.

Now, in quantum field theory, the Euler-Lagrange equation hold as an expectation value equation (if required, I can give the derivation of it) i.e., $$\langle 0|\frac{\delta S[\phi,J]}{\delta\phi(x)}|0\rangle=0=\int D\phi \frac{\delta S[\phi,J]}{\delta\phi(x)}\exp\Big(\frac{i}{\hbar}S[\phi,J]\Big).$$

One can calculate $\Gamma[\phi_c]$ from $S[\phi]$, and after that, treat the theory as classical theory. If this is the case, the equation $$\langle 0|\frac{\delta S[\phi,J]}{\delta\phi(x)}|0\rangle=0\tag{1}$$ should coincide with $$\frac{\delta \Gamma[\phi_c]}{\delta\phi_c(x)}=0.\tag{2}$$ How can I show this?


1 Answer 1


What you want to show is neither interesting nor true. We have generally that $$ \frac{\delta \Gamma}{\delta \phi_c(x)} = -J_\phi(x),$$ where $J_\phi$ is the source current such that $$ \langle \phi(x)\rangle_{J_\phi} = \phi_c(x).$$ Your eq. (1) is wrong, the Schwinger-Dyson equation says that $$ \left\langle \frac{\delta S}{\delta \phi} + J \right\rangle_J = 0,\tag{1'}$$ which coincides with your eq. (1) only in the case $J=0$.

The classical equation of motion of $\Gamma$ is $J_\phi = 0$, meaning $\langle \phi(x)\rangle_0 = 0 = \phi_c(x)$ in a Lorentz-invariant theory, which is not an interesting equation, and in particular does not depend on the form of $S$ so cannot be directly related with the Schwinger-Dyson equation.

The proper way to see how $\Gamma$ encodes the full quantum theory "at a classical level" is not to compute equations of motion. The quantum theory is not concerned with the time evolution of a classical field, it makes no sense to expect the classical equation of motion of $\Gamma$ to encode the dynamics of the quantum theory. Instead, the following relation holds true: $$ \frac{Z[J]}{Z[0]} = \mathrm{e}^{\mathrm{i}W[J]} = \int \mathrm{e}^{\mathrm{i}/\hbar\left(S[\phi] + J\phi\right)}\frac{\mathcal{D}\phi}{Z[0]} = \mathrm{e}^{\mathrm{i}/\hbar\left(\Gamma[\phi_c] + J\phi_c\right)}\tag{3}.$$ You might say this is somewhat trivial from the definition of $W$ and $\Gamma$, and it kind of is. The crucial point is that the r.h.s. is the classical value that dominates the partition function if one were to take $\Gamma$ as an action on its own, that is the full quantum partition function of $S$ is given by the classical limit of the quantum partition function of $\Gamma$. It is in this sense that $\Gamma$ is the "classical version" of $S$, not in the sense of equations of motion.


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