We want to calculate the path integral
$$ Z = \int \mathcal{D}{\phi}\, e^{i \hbar^{-1} S[\phi]} $$
which encodes a transition amplitude between initial and final quantum states.

If we had the effective action $\Gamma[\phi]$ at our disposal, we would have calculated *the same* result by solving for
$$ \phi_c(x):\quad \left. \frac{\delta \Gamma}{\delta \phi} \right|_{\phi=\phi_c} = 0 $$
and plugging it back in the effective action:
$$ Z = e^{i \hbar^{-1} \Gamma[\phi_c]}. $$

This is the definition of $\Gamma$.

Note that no path integrals are required at this point. Boundary conditions are implicitly present throughout this answer, encoding the exact states between which the quantum transition occurs. Their existence ensures that there is only one solution $\phi_c$.

Now to why $\phi_c$ is called *classical*: it solves the e.o.m. given by the action $\Gamma$.

Think of $\Gamma$ as of an object in which all the short-scale properties of the integration measure $\mathcal{D}\phi$ (including renormalization-related issues) are already accounted for. You simply solve the e.o.m. and plug the solution in the exponential and you are done: here is your transition amplitude.

That being said, $\Gamma$ is not classical in the sense that it still describes dynamics of a quantum theory. Only in a different fashion. Simple algebraic manipulations instead of path integrals.

Finally, note how if the path integral is Gaussian,
$$\Gamma[\phi] = S[\phi] + \text{const},$$
where $\text{const}$ accounts for the path integral normalization constant. There are no quantum corrections.

In classical theory, however, we solve the e.o.m. w.r.t. $\phi = \phi_c$ for $S[\phi]$, not $\Gamma[\phi]$. Plugging it back into $S[\phi_c]$ gives us the Hamilton function. When the path integral is Gaussian, it doesn't matter if we use $S$ or $\Gamma$, and exponentiating the Hamilton function gives you the transition amplitude. However if we are dealing with an interacting theory, the correct way to do this would be to use $\Gamma$ instead of $S$. In this sense, $\Gamma$ is the quantum-corrected version of $S$.

And yes, it is always true (can be shown using the saddle point approximation formula) that
$$ \Gamma[\phi] = S[\phi] + \mathcal{O}(\hbar). $$

Why wouldn't we just use $\Gamma[\phi]$ to *define* the quantum theory and forget about $S[\phi]$ alltogether? Because $\Gamma$ is *non-local* and contains *infinitely many adjustable parameters*. These can be determined from the form of $S[\phi]$ by, well, quantization. That's why it is $S[\phi]$ which defines the theory, not $\Gamma$. $\Gamma$ is to be calculated via path integrals.

**UPDATE:**
It is also important to understand that in naive QFT $\Gamma$ contains divergences, while $S$ doesn't. However, the actual situation is opposite. It is $S$ which contains divergences (divergent bare couplings), which cancel out against the divergences coming from the path integral, rendering a finite (i.e. renormalized) $\Gamma$. That $\Gamma$ should be finite is evident from how we use it to calculate physical properties: we only solve the e.o.m. and plug the result back in $\Gamma$.

Actually, the whole point of renormalization is to make $\Gamma$ finite and well-defined while adjusting only a finite number of diverging couplings in the bare action $S$.

This post imported from StackExchange Physics at 2017-09-17 12:58 (UTC), posted by SE-user Solenodon Paradoxus