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?

  • 2
    $\begingroup$ More on classical limit of path integrals. $\endgroup$
    – Qmechanic
    Commented Feb 11, 2018 at 12:32
  • 2
    $\begingroup$ the density matrix formalism is better for large systems. $\endgroup$
    – anna v
    Commented Feb 11, 2018 at 15:28

1 Answer 1


Suppose I have an action functional $S[\phi]$ in dependence of some fields $\phi$. We can also express the action in the form

$S[\phi] = \int d^4x \mathcal{L}[\phi]$. The one-loop quantum correction is determined by the quadratic term obtained by Taylor expansion to second order in field; this generates also (one can show e.g. through perturbation theory) the higher order Loop corrections. The leading order term in the semiclassical expansion is

$S[\phi'] = \frac{1}{2} \int d^4x \delta_\phi^2 \mathcal{L}[\phi]|_{\phi_0} \phi'^2$.

Now we can reformulate this term in dimensionless variables (denoted by bars). We set

$x = L \bar{x}$ ($L$ is characteristic length scale; and $T = \frac{L}{c}$ the corresponding characteristic time scale for e.g. an electromagnetic process)

$\phi' = \eta \bar{\phi'}$ ($\eta$ is characteristic magnitude of the field)

and $\int d^4x = \frac{L^4}{c}\int d^4 \bar{x}$.

Thus the exponent of the path integral phase factor scales as

$\frac{S}{\hbar} = \frac{L^4\eta^2}{2 \hbar c}\int d^4 \bar{x} \delta_{\phi}^2 \mathcal{L}[\phi]|_{\phi_0}\bar \phi'^2$.

( It is typical, that the dominant contributions of the action are the quadratic terms. )

Now we see that we have a Gaussian integral with variation of $\sigma^2 = \frac{\hbar c}{L^4 \eta^2}$. For large $L$ the variation is extremely small. But also if the characteristic field value (wave function magnitudes are proportional to the square root of number density) is large, this variation becomes small. This explains also why at a high number of particles quantum fluctuations can be neglected.


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