# Fluctuation-dissipation theorem in QFT

If I understand correctly, the fluctuation-dissipation theorem (fdt) in QFT technically arises because of $\pm i\epsilon$ - infinitesimally small summand in the denominator of spectral representation of Green functions. By comparing with the derivation of fdt for oscillator with friction and external force, I see that $i\epsilon$ causes dissipation as well as the friction term for harmonic oscillator.

What is the physical reasoning of such phenomena (there are dissipations in almost free quantum systems)?

• I don't think that any sort of regularization for the purposes of integration can be used to derive the fluctuation-dissipation theorem. The physical reasons for it are thermodynamic and relativistic in nature: we have $kT>0$ even in quantum systems and for an open system there is always a dissipative solution because we are losing energy towards infinity in form of radiation. – CuriousOne Jun 1 '16 at 21:00
• Here is a relevant post. – DanielSank Jun 2 '16 at 21:00

The fluctuation-dissipation theorem is a relation between the symmetrized and retarded correlation functions $$G_s(\omega,k)=\coth\left(\frac{\omega}{2T}\right) \,{\rm Im}\, G_R(\omega,k).$$ I would not say that it is due to the "$i\epsilon$" terms. The $i\epsilon$ simply reflects the different analyticity properties of $G_{R,A}$ etc. Physically, the symmetrized correlator is a measure of fluctuations, and the retarded correlator is a measure of dissipation. For example, the retarded correlator of the stress tensor determines viscosity $$\eta(\omega) =\frac{1}{\omega}{\rm Im}\, G_R(\omega,0)\, .$$ Why should this be true? Imagine a particle, a harmonic oscillator, or a quantum field coupled to a reservoir at some temperature $T$. In the equations of motion, this is reflected by a noise term. Without dissipation, the noise would continue to pump energy into the system, so in thermal equilibrium the noise must be balanced by a dissipative term in the equations of motion.