I'm reading chapter 4 of Peskin & Schröder, and I'm confused how they express the time ordering of two fields: let $T$ denote time ordering, $N$ the normal ordering and I use $C$ for the contraction of the two fields (as I couldn't find a way to write it better in mathjax)

$$T\{\phi(x)\phi(y)\}=N\{ \phi(x)\phi(y)+ C({\phi(x)\phi(y)})\} $$

My problem is that I think it should be:

$$T\{\phi(x)\phi(y)\}=N\{ \phi(x)\phi(y)\}+ C({\phi(x)\phi(y)}) $$

As $C({\phi(x)\phi(y)})$ is a commutator, and normal ordering a commutator will just result in $0$, more specifically, if for instance $y_0<x_0$

$$ C({\phi(x)\phi(y)})=\phi^+(x)\phi^-(y)-\phi^-(y)\phi^+(x)$$

where $\phi^\pm$ is the positive/negative frequency part of the field.

What am I missing? Moreover, they state that this $\phi^\pm$ decomposition is always possible for free fields, but I thought that the whole point of this was to deal with interacting fields, is it possible in general?

  • $\begingroup$ In the example you have given, if $\phi(x)$ are free fields, $C(\phi(x) \phi(y))$ is a $c$-number and we define $N\{c\} = c$. $\endgroup$ – Prahar Nov 12 '18 at 16:04
  • $\begingroup$ @Prahar I see this $c$-number everywhere, do you mean it's a number in $\mathbb{C}$? And while I agree that definition makes sense, does this mean that the normal ordering operator is not linear? Because if you normal order the terms in the contraction one at the time it seems to me that we'd get $\phi^-(y)\phi^+(x)-\phi^-(y)\phi^+(x)=0$ $\endgroup$ – user2723984 Nov 12 '18 at 16:10
  • $\begingroup$ You don't normal order terms in the contraction. You first perform the contraction, then do the normal ordering. The reason he is putting the contraction inside the normal ordering symbol is because you need it to be that way when there are more than two fields. With just two fields, it doesn't matter. Also, by $c$-number I do mean a complex number. $\endgroup$ – Prahar Nov 12 '18 at 16:12
  • $\begingroup$ I think I understand, does this amount to saying, as I was suggesting, that $N(AB+CD)\neq N(AB)+N(CD)$ in general? Also do you have a quick reference or justification on why the commutator of free fields is a $c$-number? Thanks for the help! $\endgroup$ – user2723984 Nov 12 '18 at 16:20
  • $\begingroup$ @Prahar I found this previous answer of yours that seems to contradict your first comment...I'm confused physics.stackexchange.com/questions/395243/… $\endgroup$ – user2723984 Nov 12 '18 at 16:51

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