1
$\begingroup$

I am studying QFT from Peskin & Schroeder. There I found a physical interpretation of a 2-point correlation function. According to Peskin & Schroeder, the 2-point correlation function is nothing but a propagator between two points and I am happy with that. But the question that arises immediately is, what are the n-point correlation functions?

e.g.- does the 3-point correlation function given by, $\langle\omega|T[\phi(x)\phi(y)\phi(z)]|\omega\rangle$ imply the amplitude of propagation from $x$ to $z$ via $y$?

Is this understanding correct?

$\endgroup$
1
$\begingroup$

In a typical QFT, Wicks theorem tells you that the expectation value of all higher point correlation functions can be expressed as two point correlation functions. Essentially, a $2n$-point correlation function tells you about $n$ particles propagating from point $x_i$ to $x_j$, where $i,j = 1...n$.

For example, a 4 points look like, $$\langle\Omega|T\left\{\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4)\right\}|\Omega\rangle = \langle\Omega|T\left\{\phi(x_1)\phi(x_2)\right\}|\Omega\rangle\langle\Omega|T\left\{\phi(x_3)\phi(x_4)\right\}|\Omega\rangle + \langle\Omega|T\left\{\phi(x_1)\phi(x_3)\right\}|\Omega\rangle\langle\Omega|T\left\{\phi(x_2)\phi(x_4)\right\}|\Omega\rangle + \langle\Omega|T\left\{\phi(x_1)\phi(x_4)\right\}|\Omega\rangle\langle\Omega|T\left\{\phi(x_2)\phi(x_3)\right\}|\Omega\rangle $$

Which corresponds to one particle going from $x_1$ to $x_2$ and one going from $x_3$ to $x_4$, plus all the other possibilities. Diagramatically, this looks like

Free scalar Feynman Diagrams

You will notice that your example of a three point correlation function is zero according to this theorem.

$\endgroup$
  • $\begingroup$ Yes, I have encountered the fact that odd point correlation functions turn out to be zero. But I think that was for free field. Do odd point correlation functions vanish even for interacting fields? $\endgroup$ – Samapan Bhadury Mar 19 '18 at 17:16
  • $\begingroup$ By the way, that was a very helpful answer. You pointed out the fact that a 2n point correlation function is about n particles moving from one point to another. :) $\endgroup$ – Samapan Bhadury Mar 19 '18 at 17:18

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.