Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In quantum scattering theory, Green's Function is defined as [1] $$G_0(z)=(z-H_0)^{-1},$$ $$G(z)=(z-H)^{-1},$$ where $H_0$ and $H=H_0+V$ are separately non-interacting and interacting Hamiltonian. $V$ is interaction.
One can then use the identity $$\tag{1}V=G_0^{-1}-G^{-1},$$ to obtain Lippmann-Schwinger equation $$\tag{2}G=G_0 + G_0 V G. $$ However, on the other hand, in quantum field theory(QFT), Green's function is defined as correlation function. For 2-point Green's function, we have Dyson equation $$\tag{3} G=G_0+ G_0 \Sigma G, $$ where $\Sigma$ is here defined as self-energy. Equivalently $$\tag{4} \Sigma:=G_0^{-1}-G^{-1}.$$ My questions are
Are the two Green's functions the same? What's the relation between the two formalisms? And the relation between Lippmann-Schwinger equation and Dyson equation? If they are actually the same thing, then does it mean $V=\Sigma$(this sounds very stupid)? Are the possible differences relating to the discrepancy between S-matrix theory and QFT?

[1]: John R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions.

share|cite|improve this question
up vote 4 down vote accepted

There are very important differences between this two approaches, that can be summarized by noting that the Lippmann-Schwinger is the (formal) solution of a one-body problem (scattering of a particle by an external potential) whereas the Dyson equation gives the solution of a many-body problem. I focus here on the non-relativistic many-body case (it is also the case of the scattering problem).

It is only in the case where the system is in an external potential and non-interacting (or empty,assuming conservation of the number of particle) that the two approaches are equivalent (or more precisely the Dyson equation gives back the Lippmann-Schwinger equation).

share|cite|improve this answer
According to wikipedia, Lippmann-Schwinger equation still could be used for other few-body problem, say for there-body. Maybe for many-body problem it is very difficult to be applied. For the late approach: QFT could describe both interacting few-body and many-body problem, right? If so, the Dyson equation in QFT frame, still could be applied. The difference, i think, is in bare propagator for any Feynman diagram, right? – Mathieu Mar 26 '14 at 5:51
@Mathieu: you are right, you can use the LS equation for more than one body, but it conserves the number of particles (the ground state of the system is trivial: it is the vacuum). Stated otherwise, it is a reformulation of the Schrodinger equation. Of course, in principle, if you can solve the N-body ($N\to\infty$) SL equation, you could describe a many-body state (say, a supraconductor), but that's definitely not doable in practice. And yes, as I said in the answer, Dyson equation is more general and give the SL equation in the appropriate limit. – Adam Mar 26 '14 at 13:34
To make it clear, allow me to ask alternative questions: do you mean that the generalization of LS equation for many-body problem is not equivalent to Dyson equation? But they are both correct and precise? Could Dyson equation for two-body system with interaction be reduced to LS equation for two-body case? – Mathieu Mar 28 '14 at 8:43
The main difference between both equation is that they do not give access to the same physics (even when you treat the same problem). The object you work with in the SL eq. is an operator, the Green function of the D eq. is a function (the average of an operator). For the two body problem, D eq. is not useful (the green function is unchanged by the other particle) and the correct object to work with is the 4-point function, which equation is given by the Bethe-Salpeter equation. – Adam Mar 28 '14 at 13:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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