Physical interpretations of non-equilibrium Green's functions $G^{<}(p, \omega)$ and $G^{>}(p, \omega)$

Question

The greater and lesser non-equilibrium Green's function is defined as $$ G^<(r, t) =\pm \frac{1}{i}\langle\psi^\dagger(0, 0)\psi(r, t)\rangle,\qquad G^>(r, t) =\frac{1}{i}\langle\psi(r, t)\psi^\dagger(0, 0)\rangle, $$ where $+$ is for bosons and $-$ is for fermions, and we assume that the system is translationally invariant. The Fourier transforms of the two functions are defined as $$ G^{<}(p, \omega) = \pm i \int dr dt e^{-ip\cdot r + i\omega t} G^{<}(r, t),\\ G^{>}(p, \omega) = i \int dr dt e^{-ip\cdot r + i\omega t} G^{>}(r, t). $$

In Kadanoff&Baym's book, $G^{<}(p, \omega)$ is interpreted as the average density of particles in the system with momentum $p$ and energy $\omega$: $$ G^{<}(p, \omega) = \langle n(p, \omega)\rangle = A(p, \omega)f(\omega), $$ where $A(p, \omega)$ is the spectral density $A(p, \omega)\equiv G^{>}(p, \omega)\mp G^{<}(p, \omega)$ and $f(\omega) = 1/(e^{\beta(\omega - \mu)}\pm1)$.

What's physical intuition for this interpretation? Is there a way to explicitly express $n(p, \omega)$ in terms of $\psi, \psi^\dagger$?

Answer 1

To expand on the answer by @Galilean:

• $G^<(\omega)$ is usually interpreted as the average occupancy of states at energy $\omega$. This interpretation is more transparent when we talk about a discrete spectrum, but this is what Kadanoff&Baym mean.
• Likewise, $G^>(\omega)$ is the density of vacancies. In some contexts this may correspond to the density of holes.
• $G^>(\omega)-G^<(\omega)=G^r(\omega)-G^a(\omega)$ is just the density-of-states
• Keldysh Green's function $G^K=G^>(\omega)+G^<(\omega)$ usually becomes the classical $f(\mathbf{p}, \mathbf{q},t)$ in the Boltzmann/kinetic equation. (Kadanoff&Baym approach was developed in parallel to Keldysh, so they use different terminology, but there is an equivalent object in their derivations.)

For more modern discussion see

• Quantum field-theoretical methods in transport theory of metals by Rammer and Smith
• Quantum Kinetics in Transport and Optics of Semiconductors by Haug and Jauho

See also some basic derivations in these threads:
How to dress free Green functions with constant broadening?
Fluctuation-dissipation theorem in the Keldysh formalism

Answer 2

Lets try to understand this for non-interacting Fermions. Say you have a single band and the Hamiltonian reads $H = \sum_k \epsilon_k c_k^\dagger c_k$. Then the momentum resolved occupation is given by the Fermi distribution, i.e., $n(k) = \langle c_k^\dagger c_k \rangle = f(\epsilon _k)$. This same occupation can be written as: \begin{align} n(k) &= \frac{1}{2\pi}\int d\omega\; 2\pi \delta(\omega - \epsilon_k)f(\omega) \\ & = \frac{1}{2\pi}\int d\omega\; A(k,\omega)f(\omega) \label{eq1}\tag{1} \end{align} Here, $A(k,\omega) = 2\pi \delta(\omega - \epsilon_k)$ is the spectral function for non-interacting fermions. The integrand of \eqref{eq1} can be interpret as an occupation of momentum and frequency occupation number $n(k,\omega)=A(k,\omega)f(\omega)$. This is the probability of finding a particle at momentum and frequency $(q,\omega)$ with temperature $1/\beta$ and chemical potential $\mu$, which is same as $G^<(k,\omega)$.

When you turn on interaction, \eqref{eq1} is still valid as long as the spectral function is quite sharp. One of the things that interaction does to is introducing a width to the spectral function because of a finite scattering rate.