5
$\begingroup$

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 $\psi, \psi^\dagger$?

$\endgroup$

2 Answers 2

3
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.