If I would try to treat macroscopic systems consisting of a super-large number of particles (also when environment is included), I have to compute $2N$-point correlation functions with very large particle number $N$. These are given by
$$A(x_1',\dots,x_N';x_1,\dots,x_N) = \langle\prod_{i=1}^N\psi^\dagger(x_i') \psi(x_i)\rangle$$
The primed variables denote final states and $\psi(x)$ generate one fermion at point $x$. Transition probabilities are then proportional to $|A|^2$. For a classical behavior, these transition probabilities have to be deterministic, i.e. must be in the form of a delta distribution
$$\propto \prod_i \delta(x_i'-f_i(x_1,\dots,x_N))$$
for the (differential equation) solution function $f_i$. How can I derive that deterministic transition amplitudes arise in everyday life scenarios from the path integral average?
I know that I can make a decomposition $\psi(x) = \psi_0(x)+\psi'(x)$ with classical solutions of the equation of motion $\psi_0$ and quantum fluctuations $\psi'$ that obey $\langle\psi'\rangle=0$. Taylor expansion until second order in fluctuations of the action functional leads to the Gaussian distribution similar to
$$\exp\left(\frac{i}{2}\frac{\delta^2}{\delta \psi^2}S|_{\psi_0}\psi'^2\right)$$
This distribution becomes sharper peaked as the second functional derivative of the action functional gets larger.
I can introduce dimensionless variables and see that this Gaussian standard deviation depends on characteristic length scales and with higher length scales, standard deviations will get less.
That is what I know.
But by considering above multiparticle scattering amplitudes, making the expansion of quantum fields around classical solution I will pick up also LOTS of quantum corrections $\langle\psi'^2\rangle,\langle\psi'^3\rangle,...$. There will be $\frac{N(N-1)}{2}$ quadratic fluctuation terms which is a huge number. But why these are irrelevant in macroscopic systems?
Is there a detailed answer?