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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?

user2723984
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  • In the example you have given, if $\phi(x)$ are free fields, $C(\phi(x) \phi(y))$ is a $c$-number and we define $N{c} = c$. – Prahar Nov 12 '18 at 16:04
  • @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 $\phi^-(y)\phi^+(x)-\phi^-(y)\phi^+(x)=0$ – user2723984 Nov 12 '18 at 16:10
  • 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. – Prahar Nov 12 '18 at 16:12
  • 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! – user2723984 Nov 12 '18 at 16:20
  • @Prahar I found this previous answer of yours that seems to contradict your first comment...I'm confused https://physics.stackexchange.com/questions/395243/what-is-the-calculation-rule-of-the-normal-ordering-operator?rq=1 – user2723984 Nov 12 '18 at 16:51
  • You are right! I should consolidate my thoughts on this. Strictly speaking P&S does not define normal ordering for the identity operator. Only for operators which contain creation and annihilation modes wherein his definition is such that $\langle N(O) \rangle = 0$. The natural extension of this definition to the identity operator is $N(I) = 0$ which is what I advocated in the answer (and which is the definition used, for instance, in CFTs). – Prahar Nov 12 '18 at 17:06
  • &user2723984 - However, it seems P&S are using the definition $N(I)=I$ (as is clear from the fact that he is putting the $N$ outside of the $C$. – Prahar Nov 12 '18 at 17:07
  • Hello, I'm reading P&S and I've got the same problem. Did you find a satisfactory answer? It seems that your first equation is only true when you calculate the contraction first before doing anything else. I'm very confused as well, this all seems very fishy to me. – AfterShave Mar 23 '19 at 04:30
  • @AfterShave hi, it's been a while, but if I remember correctly it makes sense that you do have to take the contraction first and then the normal ordering. It does come before and the two operations do not commute. – user2723984 Mar 25 '19 at 16:51

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