I'm learning Green's function in condensed matter. The full Green's function is defined as

$$G(k_2,t_2;k_1,t_1) = \langle\Omega |T a_{k_1}(t_1)a_{k_2}^{\dagger}(t_2) |\Omega \rangle $$

The $\Omega$ is the interaction ground state and the operators are in the Heisenberg picture.

We want $k_1,k_2$ to be the quantum number of free Hamiltonian(like free electron gas). In that case, we should impose the condition that when $a_{k_1}(t_1),a_{k_2}^{\dagger}(t_2)$ are identical to the free electron creation or annihilation operators. In literatures (e.g. Mahan), they pick $a_{k_1}(-\infty )=a_{k_1}^{free},a_{k_2}^{\dagger}(-\infty)= a_{k_2}^{\dagger,free}$.

It occurs to me that, when I learn scattering theory, the S-matrix has similar structure. For potential scattering,

$$S_{k_2,k_1}=\langle{k_2}^{out}|k_1^{in}\rangle=\langle k_2^{free}|U(\infty,-\infty)|k_1^{free}\rangle=\langle\Omega_0|a_{k_2}U(\infty,-\infty)a^\dagger_{k_1}|\Omega_0\rangle$$

In this case, the $\Omega_0$ is the ground state of non-interacting Hamiltonian. They are even more likely to each other when you pick an interacting picture.

My question is, what is the relationship of these quantities? The book of QFT discusses free Green's functions and use them to expand the S matrix, but I really need the discussion of full Green's function and S-matrix.

My naive observation tell me $G(k_2,\infty;k_1,-\infty)=S_{k_2,k_1}$, but the convention I mentioned above denied it...

  • $\begingroup$ And I also wonder what the physical meaning of $a\dagger_k(t_1)$? will it create an eigen state of full Hamiltonian when acting on $\Omega$? $\endgroup$
    – Taveren Sa
    Commented May 25, 2022 at 11:39
  • $\begingroup$ Consider to use \langle and \rangle instead of $<$ and $>$. $\endgroup$ Commented May 25, 2022 at 12:58

1 Answer 1


Condensed matter and QFT are interested in different quantities of the many-body systems they study (at least, on the level of basic textbooks).

QFT aims at calculating the cross-sections of elementary processes, which begin with a few particles added to vacuum, and end in a similar state with a few particles. One can then show that the S-matrix characterizing such a process can be expressed in terms of averages in respect to vacuum and expanded in terms of Green's functions.

In (equilibrium) condensed matter one is interested in thermodynamic properties of systems - i.e., in the averages calculated in respect to the ground state (or thermal averages at finite temperature). Inclusion of interactions then similarly leads to the an expansion in terms of averages in respect to the ground state, expressible in terms of Green's functions. (The ground state now plays the role of vacuum, although it is not real physical vacuum, but, e.g., a state filled up to the Fermi energy.) Furthermore, the expansion for the quantities of interest may be somewhat cumbersome (e.g., the one for the thermodynamic potential), which is why one prefers to express them in terms of seemingly more complex, but easier to calculate Green's function (this relations are covered in condensed matter textbooks, e.g., AGD). In many cases the scattering matrix also arises - e.g., when renormalizing vertices, obtaining Bethe-Salpeter equation or studying latter diagrams, but it is not the main object of interest.

It takes somewhat bigger role in non-equilibrium statistical physics, e.g., when setting up the collision integrals in quantum kinetic equations - see, e.g., the review by Rammer and Smith.

  • $\begingroup$ I don't think this is the answer I need. Although related to different observation variable, the basic calculation tools are identical. That question can be purely asked in high energy physics, namely the relation between Feynman propagater and S-matrix (of course you need to replace the field operator in feynman propagater with the particle creation or anhilation operator). $\endgroup$
    – Taveren Sa
    Commented May 25, 2022 at 12:39
  • $\begingroup$ @TaverenSa not sure what you exactly mean by Feynmann propagator - the retarded Green's function? $\endgroup$
    – Roger V.
    Commented May 25, 2022 at 15:35

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.