A Green functions and n-point correlation functions

1. Nov 20, 2016

ShayanJ

Green functions are defined in mathematics as solutions of inhomogeneous differential equations with a dirac delta as the right hand side and are used for solving such equations with a generic right hand side.
But in QFT, n-point correlation functions are also called Green functions. Why is that?

Thanks

2. Nov 20, 2016

Orodruin

Staff Emeritus
Because, from a mathematician's point of view, physicists are notoriously sloppy with what they call a Green function.

Green functions in mathematics are solutions to linear differential equations with a delta inhomogeneity. In a linear setting, this tells you all you need to know about your system given any inhomogeneity. In interacting QFT, this is really no longer the case and instead the information can be summarised in terms of the n-point correlation functions, if you know them you know the system so in some sense it is a broader use of the term.

3. Nov 20, 2016

MathematicalPhysicist

Physicists are notoriously sloppy period. :-)

4. Nov 20, 2016

dextercioby

It takes a few pages of Bailin and Love (start with page 38 and go through page 46) to see that the name "Green's functions" for the correlation functions is well-justified.

5. Nov 20, 2016

nrqed

They are actually Green functions in the mathematical sense! For a simple example, look at Peskin and Schroeder, (page 30, Equation 2.56 in my edition). There they show explicitly that the two point function of a spin 0 field is a Green function (more specifically a retarded Green function) for the Klein-Gordon operator.

6. Nov 20, 2016

samalkhaiat

$$\langle 0 |T \left( \varphi(x) \varphi^{\dagger}(y) \right) |0 \rangle = i \int \frac{d^{4}p}{(2\pi)^{4}} \ \frac{e^{-ip(x-y)}}{p^{2} - m^{2}} ,$$ $$\left( \partial^{2}_{(x)} + m^{2}\right) \langle 0 |T \left( \varphi(x) \varphi^{\dagger}(y) \right) |0 \rangle = -i \delta^{4}(x-y).$$

7. Nov 21, 2016

Orodruin

Staff Emeritus
This answers the question "why are 2-point correlation functions in a free theory called Green's functions?" I do not think this was ever doubt. The bigger question is why we call the n-point correlation functions Green's functions as well, which is what I discussed in #2. Of course, the root of the terminology is that the 2-point function in the free theory is a Green's function even in the strictly mathematical sense as a solution to a linear differential equation with a delta inhomogeneity (i.e., the Green's function of the Klein-Gordon operator).

8. Nov 21, 2016

vanhees71

If you refer to propagators used in perturbative vacuum Feynman rules, it's the time-ordered, not the retarded one. See #6.

9. Nov 21, 2016

vanhees71

Here an utmost important $\mathrm{i}0^+$ is missing in the denominator of the integral is missing!

10. Nov 21, 2016

nrqed

yes, you are absolutely right. I mentioned the retarded propagator because this is what Peskin and Schroeder consider in the equation I referred to. Sorry if my comment was misleading and thanks for the correction.

11. Nov 22, 2016

samalkhaiat

When rigor does not add anything new, we think of it as a total waste of time and choose to be "sloppy". And, when rigorous technique does not exist (as in the QFT-case), we do anything to make progress. So, in this case, sloppiness is forced upon us, and it is exactly this "sloppiness" that led us to the Standard Model.

12. Nov 22, 2016

samalkhaiat

Not at all my friend, you can do the integration over $p_{0}$ first: choose your contour $C$, in the complex $p_{0}$ plane, to go below the singularity at $p_{0} = - \omega_{\mathbf{p}}$, and above the one at $p_{0} = + \omega_{\mathbf{p}}$, and close it by a large semicircle, in the lower half plane (if $x_{0} > y_{0}$) or in the upper half plane (if $x_{0} < y_{0}$). The same result, as integrating over $C$, can be obtained if you integrate along the real $p_{0}$ axis instead and avoid hitting the poles by shifting them by an infinitesimal amount into the complex plane. In this case, as you pointed out, the denominator becomes $p^{2} - m^{2} + i \epsilon$.

Last edited: Nov 23, 2016
13. Nov 22, 2016

samalkhaiat

First of all, we call $\langle 0 |T \left( \phi(x_{1}) \cdots \phi(x_{n} \right)|0\rangle$ the interacting Green’s function, the general Green’s function or the complete n-particle Green’s function. To understand the reason for that you need to look at the original form of the LSZ reduction formulas, the cornerstone of all calculations in QFT. The remarkable feature of those expressions is the relation they provide between the on-shell transition amplitudes and the Green functions of the interacting theory. The latter is nothing but the vacuum expectation values of time-ordered field products. Further insight can be obtained by comparing the modern form of the LSZ-formula to the S-matrix (for motion in external field) constructed in the (non-relativistic) propagator theory. The S-matrix element is nothing but the residue of the (muti-)pole structure of the general Green’s function.

14. Nov 23, 2016

vanhees71

Ok, but then you have to specify the path clearly, i.e., how you run around the poles on the real axis. This is the point where you should avoid to be sloppy. Nothing contributes to confusion more in not being specific about which propagator you are talking. In vacuum theory it's usually the time-ordered. In equilibrium theory you can do with the Feynman or retareded propagator (at finite temperature there's a difference between the Feynman and the time-ordered) or use the imaginary-time formalism and use a careful analytic continuation to get the retarded of Feynman propgator. For linear-response theory (e.g., Kubo formalism to evaluate transport coefficients) you need the retarded one. In non-equilibrium many-body theory you have matrix-valued operators in the Schwinger-Keldysh-contour formalism using either the +- or the ra convention.

15. Nov 23, 2016

vanhees71

What do you mean by "the modern form of the LSZ formalism"? Is there progress compared to the standard operator treatment (as, e.g., detailed in Itzykson, Zuber, QFT) or path-integral treatment (as, e.g., in Bailin, Love, Gauge Theory).

16. Nov 23, 2016

MathematicalPhysicist

If it will be found in the future that this slopiness isn't justified mathematically, what will this have on physics enterprise?

I mean the predictions might be postdictions and no predictions at all.

17. Nov 24, 2016

samalkhaiat

No, I don’t have to because #6 is self-explanatory:
1) $\varphi$ is the common symbol for scalar field.
2) The Feynman $\Delta$-function $$\Delta_{F}(x) \equiv - i \langle 0 |T \left( \varphi (x) \varphi (0) \right) | 0 \rangle ,$$ has two and only two integral representations, either $$\Delta_{F}(x) = \frac{1}{(2\pi)^{4}} \int d^{4}p \frac{e^{-ipx}}{p^{2}-m^{2}} ,$$ or $$\Delta_{F}(x) = \frac{1}{(2\pi)^{4}} \int d^{4}p \frac{e^{-ipx}}{p^{2}-m^{2} + i\epsilon} .$$
3) All elementary textbooks on QFT explain both contours used to write $\Delta_{F}(x)$.
4) Both contours lead to the same relevant fact in this thread, that is $$(\partial^{2} + m^{2}) \Delta_{F}(x) = - \delta^{4}(x) .$$
5) This is an A-type thread, so it is natural to assume that the OP has read some elementary texts on QFT and, therefore, knows about Wick’s theorem.
If you call $\int d^{4}p \frac{e^{-ipx}}{p^{2}-m^{2}}$ “sloppy”, then $\int d^{4}p \frac{e^{-ipx}}{p^{2}-m^{2} + i\epsilon}$ is equally “sloppy” because there is no third option.

18. Nov 24, 2016

samalkhaiat

By the “original form”, I meant the LSZ-I which is the content of their paper in:
Nuovo Cimento, 1, 1425 (1955), and Nuovo Cimento, 2, 425 (1955).
And, by the “modern form” I meant the LSZ-II which is the content of their paper in: Nuovo Cimento, 6, 319 (1957). Most QFT textbooks describe LSZ-II.

19. Nov 24, 2016

vanhees71

Well, I don't want to be picky, but (1) isn't even defined, because in the usual notation it means the integral over $\mathbb{R}^4$, and one doesn't know what to do with the "on-shell poles". You have to specify the path in complex $p^0$ plane as you explained earlier how to circumwent these poles to get the correct time-ordered propagator. (2) is fine, because you know how you have to close the contour in the $p^0$ plane at infinity (depending on the sign of $t=x^0$), and then the $\mathrm{i} \epsilon$ ($\epsilon \rightarrow 0^+$) specifies which pole contributes for the two cases. Both, the specification of the contour as this $\mathrm{i} \epsilon$ prescription, lead to the same result for the time-ordered propagator as expected.

The defining equation for any (unspecified) Green's function of the d'Alembert operator doesn't determine it uniquely. You can get any Green's function, instead of the here wanted time-ordered one, among them, e.g., the retarded, advanced Green's functions.

20. Nov 24, 2016

vanhees71

Thanks, very interesting, in my paper archive, I've only the 1957 paper. So it's the standard textbook LSZ reduction formalism.