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Here $\phi_I$ is just the free Klein-Gordon field. So, this field is decomposed of two components shown above. Now let $N$ be the normal ordering operator. Then, I think that $N(\phi_I^+(x)\phi_I^-(y))=N(\phi_I^-(y)\phi_I^+(x))=\phi_I^-(y)\phi_I^+(x)$. Is this true? If so, then $N([\phi_I^+(x), \phi_I^-(y)])=0$ must hold. Is this right?

!!!Addtional question!!!

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This is page 89 of Peskin and Schroeder's QFT book. First, from the definition (4.35) and my original question, $N(\text{contraction of}\; \phi(x)\; \text{and} \; \phi(y))$ must be zero. However, from (4.36) and the definition of the Feynman propagator (which seems to contain no operators $a_\textbf{p}$, $N(\text{contraction of}\; \phi(x)\; \text{and} \; \phi(y))=N(D_F(x-y))=D_F(x-y)$. So I am extremely confused... Also, (4.37) equation seems to be wrong. The right side seems to be just equal to $N(\phi(x)\phi(y))$ and it does not match with the left side. For the last, what exactly is the definition of the contraction of non-adjacent fields? For example what exactly is $\phi_1\phi_2\phi_3\phi_4$ with $\phi_1$ and $\phi_3$ contracted? I mean the thing inside $N$ of the left side of the last line of the picture I posted.

Qmechanic
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Keith
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    Related: https://physics.stackexchange.com/q/345898/2451 , https://physics.stackexchange.com/q/323801/2451 and links therein. – Qmechanic Mar 23 '18 at 20:43

1 Answers1

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Normal-ordering of the identity operator is zero, i.e. $$ N(I) = 0. $$ The idea of normal-ordering is introduced so that a normal-ordered operator has vanishing vacuum expectation value $$ \langle 0| N({\cal O}) |0 \rangle = 0 \qquad \qquad (1) $$ The way this is done is follows. Let ${\cal O}(x)$ be an operator with $\langle 0| {\cal O}(x) |0 \rangle = a\neq0$.

Then, define the normal ordering of the operator by $$ N({\cal O}(x)) = {\cal O}(x) - a I $$ Then, by construction (1) is true. The following are now exercises left to the reader -

  1. Convince yourself that this definition is in fact the same as the one defined in Peskin and Schroeder, which defines the operation in terms of creation and annihilation operators. It asks us to move all annihilation operators to the right and all creation operators to the left. That is the same as the procedure above.

  2. By the above definition $N(I) = 0$.

Prahar
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  • Yes I understand this construction. Then, is the equation (4.37) in my picture true? What exactly is the left side of equation (4.37)? – Keith Mar 23 '18 at 20:45
  • (4.37) should read $T(\phi(x) \phi(y)) = N(\phi(x) \phi(y) ) + contraction ( \phi (x) \phi(y) )$. – Prahar Mar 23 '18 at 20:50
  • So is it a typo? Then what becomes (4.38) and (4.39)? The book seems to be clearly wrong then... – Keith Mar 23 '18 at 20:52
  • (4.38) has the normal-ordering sign on all operators on the RHS, but not the identity operator. Strictly speaking, you are right of course, the book is wrong. However, I rather think of it as abuse of notation. Its more cumbersome to write out the exact result as is and so PS abuse notation and write it in a short but loose way. – Prahar Mar 23 '18 at 20:55
  • Where is an identity operator in the RHS of (4.38)? As a double major in math, I just cannot tolerate anything that is not in the most exact form....Is this attitude an impediment to studying physics? – Keith Mar 23 '18 at 21:07
  • @Keith To your first question - Hidden inside the phrase "all possible contractions". Let me denote a contraction by $c(AB)$. For instance all possible contractions of a product of 3 operators s $c(ABC) = c(AB)C + c(AC)B + c (BC)A$. In this case, there is no identity operator in c(ABC) and everything is normal ordered in (4.38). On the other hand, $c(ABCD) = c(AB)CD+c(AC)BD+c(AD)BC+c(BC)AD+c(BD)AC+c(CD)AB + c(AB)c(CD)+c(AC)c(BD)+c(AD)c(BC)$. – Prahar Mar 23 '18 at 21:30
  • @Keith - In this case, the last three terms correspond to the identity operator and these are not normal ordered in (4.38). – Prahar Mar 23 '18 at 21:32
  • @Keith - In general, with odd number of fields, there is no identity operator and everything is normal ordered. Thus, the vacuum expectation value of a time-ordered product of odd number of fields is zero. With even number of fields, there are several terms corresponding to the identity operator and none of those are normal ordered. – Prahar Mar 23 '18 at 21:32
  • @Keith - To your second question - it is not. In fact, it is necessary to get to the physics part of the problem and not be bogged down by lot of mathematical details. To be clear, every physicist does one of three things - (1) all the math is cleanly performed and everything is explicitly written down (in books or papers), (2) all the math is cleanly performed but they abuse notation or do not write things down cleanly for purposes of abbreviation or (3) the math is not cleanly performed but several checks are done to ensure the accuracy of the final result. – Prahar Mar 23 '18 at 21:35
  • @Keith - As a student, you will wish everyone was of type (1). As you get more advanced you will yourself see the usefulness of (2) and probably transform into that type. A great physicist is often of type (3), i.e. they have sufficient intuition about the problem to know what the final answer should look like despite not being backed up by rigorous mathematics along the way. Of course, one's intuition is not enough to convince others so you do have to prove that it is correct by doing lots of checks so that part is crucial. – Prahar Mar 23 '18 at 21:37
  • The each of last three terms you said seem to correspond to the product of Feynman propagators I think, not the identity operator. – Keith Mar 24 '18 at 04:35
  • Or do you call Feynman propagators the identity operator? – Keith Mar 24 '18 at 04:37
  • The Feynman propagator is a c-number (in this case a function) so yes they multiply the identity operator on the Hilbert space. – Prahar Mar 24 '18 at 04:42
  • Now come to think about it. The PS definition of normal orderong seems to leave the identity operator as it is because there is no operators to move in the identity operator. But you said $N(I)=0$. Why do they agree? – Keith Mar 24 '18 at 04:47
  • PS has not defined normal ordering for identity operator. I have done so in my answer above. – Prahar Mar 24 '18 at 04:49
  • This answer suggests that it's more useful to go with a definition of normal ordering in which $\mathbf{:} \mathbf{1} \mathbf{:} = \mathbf{1}$, and that also allows P&S to pull the contraction into the normal ordering due to linearity. – tomdodd4598 Feb 19 '22 at 03:35
  • It is more natural to consider the normal ordering symbol $N$ is only defined on products of creation/annihilation operators (no sum. in other words, monimials). Then we can extend the definition to include any other operators by assuming that $N$ does nothing on them. Also, $N$ will be assumed to possess linearity. – Keyflux Dec 08 '22 at 07:57