1
$\begingroup$

Let's have the S-matrix: $$ S_{\beta \alpha} = \langle \beta | \hat{S} | \alpha\rangle . $$ Here $|\alpha \rangle , | \beta \rangle$ are $t \to \mp \infty$ limit of the free states, $\hat {S} = \hat{T}e^{-i\int \hat{L}_{\int}d^{4}x}$, $\hat{L}_{\int}$ refers to the operator in the interaction picture. When we decide to get the matrix element of some process we will get $$ \int d^{4}x_{1}...d^{4}x_{n}\langle \beta |\hat{T}(\hat{\varphi}_{1_{int}}(x_{1})...\hat{\varphi}_{m_{int}}(x_{n})) | \alpha \rangle . $$ So it's convenient to introduce n-point Green function, $$ \tag 1 G_{n}(x_{1},...x_{n}) = \langle 0| \hat{T}(\hat {\varphi}_{1_{int}}(x_{1})...\hat{\varphi}_{n_{int}}(x_{n}))| 0\rangle $$ and generation functionals for it.

But recently I have read anywhere that as n-point Green function people use expression $$ \tag 2 G^{H}_{n}(x_{1},...x_{n}) = \langle 0| \hat{T}(\hat {\varphi}_{1}(x_{1})...\hat{\varphi}_{n}(x_{n}))| 0\rangle , $$ where the operators of fields are in the Heisenberg picture. So they need to rewrite the operators into interaction picture: $$ \tag 3 G^{H}_{n}(x_{1},...x_{n}) = \langle 0| \hat{T}\left( \hat {\varphi}_{1_{int}}(x_{1})...\hat{\varphi}_{n_{int}}(x_{n})\hat{S}\right)|0\rangle . $$ I don't understand why we need the Green function $(2)$ where the fields operators are in the Heisenberg picture if $S$-matrix "generates" rather Green functions with operators in interaction picture.

Can you explain it why we don't use $(1)$ when talk about the Green functions?

$\endgroup$
  • 1
    $\begingroup$ Could you add the sources that use these functions? I've seen many weird definitions of n-point function, generating functionals and the like, and often they have tailored it to their specific context (sometimes they're just confused, though). $\endgroup$ – ACuriousMind Aug 7 '14 at 15:59
  • $\begingroup$ @ACuriousMind : for example, here, en.wikipedia.org/wiki/…, is used generating functional for $(3)$. $\endgroup$ – Andrew McAddams Aug 7 '14 at 16:49
1
$\begingroup$

The quantum fields in the interaction picture evolve according to the free field equations – the evolution is given just by the quadratic part of the Hamiltonian, without the interactions – so Green's functions constructed from these interaction-picture field operators would be those of the free field theory, too. It wouldn't be terribly interesting. We would only "learn" the Wick's theorem and things about the free field theory.

The normal Green's ($n$-point) functions are supposed to include all the interactions given by Feynman's vertices etc., so they need to be evaluated from the operators in the normal picture, i.e. the Heisenberg picture. The interaction picture is just a "fudged" compromise between the Heisenberg picture and the Schrödinger picture – a compromise that is useful and convenient but in no way fundamental. It's the Heisenberg picture and correlators in it that reduce to classical physics in the $\hbar\to 0$ classical limit.

$\endgroup$
  • $\begingroup$ Thank you. But where these normal Green functions arise naturally (where we must introduce $(2)$, not $(1)$, for applying to the theory)? In the S-matrix there are only the Green function like my $(1)$, and nonperturbative results (like LSZ-theorem, Ward equalities etc.) also don't use $(2)$, if I don't confuse. $\endgroup$ – Andrew McAddams Aug 7 '14 at 16:26
  • $\begingroup$ Hmm, maybe, the last statement (about nonperturbative results) is incorrect, because they are derived for Heisenberg operators. $\endgroup$ – Andrew McAddams Aug 7 '14 at 18:26
  • 1
    $\begingroup$ Dear Andrew, the LSZ theorem and all related things uses the Heisenberg Green's functions. Just look at it rationally. The nontrivial, interaction- or Feynman-diagrams-related part of the S-matrix is only carried by the Green's functions, that's what the LSZ theorem really says. And it's exactly the S-matrix that appears in your formula written in terms of the interaction-picture field operators. The field operators that appear in the product only give the external propagators etc. $\endgroup$ – Luboš Motl Aug 7 '14 at 19:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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