in first volume of Polchinski page 39 we can read a compact formula to perform normal-order for bosonic fields $$ :\cal F:=\underbrace{\exp\left\{\frac{α'}{4}∫\mathrm{d}^2z\mathrm{d}^2w\log|z-w|^2\frac{δ}{δφ(z,\bar z)}\frac{δ}{δφ(w,\bar zw)} \right\}}_{:=\mathcal{O}}\cal F, \tag{1} $$

What I do not understand it is that I would like to have (bearing in mind the definition involving $a$ and $a^†$ $$ ::\cal F::=:\cal F:\tag{2} $$ but with this formula $$ \cal O^2\cal F≠\cal O \cal F.\tag{3} $$


$$ :φ(z)φ(w):=φ(z)φ(w)-\frac{α'}{2}\log|z-w|^2\tag{4} $$ but $$ ::φ(z)φ(w)::=:φ(z)φ(w):-\frac{α'}{2}\log|z-w|^2=φ(z)φ(w)-α'\log|z-w|^2\tag{5} $$

  • $\begingroup$ recall that $A=B$ does not imply that $:A:=:B:$. $\endgroup$ – AccidentalFourierTransform Feb 2 '17 at 23:24
  • 4
    $\begingroup$ @AccidentalFourierTransform I cannot understand. So it's also possible that $:A:≠:A:$(!?) $\endgroup$ – MaPo Feb 2 '17 at 23:27
  1. Short explanation: Polchinski's eq. (1) is not a formula that transforms no normal order into normal order: The expression ${\cal F}$ on the right-hand side of eq. (1) is implicitly assumed to be radially ordered. In fact, eq. (1) is a Wick theorem for changing radial order into normal order, cf. e.g. this Phys.SE post.

  2. Longer explanation: When dealing with non-commutative operators, say $\hat{X}$ and $\hat{P}$, the "function of operators" $f(\hat{X},\hat{P})$ does not make sense unless one specifies an operator ordering prescription (such as, e.g., radial ordering, time-ordering, Wick/normal ordering, Weyl/symmetric ordering, etc.). A more rigorous way is to introduce a correspondence map $$\begin{array}{c} \text{Symbols/Functions}\cr\cr \updownarrow\cr\cr\text{Operators}\end{array}\tag{A}$$ (E.g. the correspondence map from Weyl symbols to operators is explained in this Phys.SE post.) To define an operator $\hat{\cal O}$ on operators, one often give the corresponding operator ${\cal O}$ on symbols/functions, i.e., $$ \begin{array}{ccc} \text{Normal-Ordered Symbols/Functions}&\stackrel{\cal O}{\longrightarrow} & \text{Radial-Ordered Symbols/Functions} \cr\cr \updownarrow &&\updownarrow\cr\cr \text{Normal-Ordered Operators}&\stackrel{\hat{\cal O}}{\longrightarrow} & \text{Radial-Ordered Operators}\end{array}\tag{B}$$ E.g. Polchinski's differential operator ${\cal O}$ does strictly speaking only make sense if it acts on symbols/functions. The identification (A) of symbols and operators is implicitly implied in Polchinski.

  3. Concerning idempotency of normal ordering, see also e.g. this related Phys.SE post.

  • $\begingroup$ Thank you. I read the post about idempotency. But there is still something I cannot understand. The point is that for operator that are quadratic in fields seems to hold but not in general. For instance. If I try to use eq (1) to normal order a constant I woud get $\cal O 1 =1$ instead of $0$ as I should expect. $\endgroup$ – MaPo Feb 3 '17 at 0:21
  • $\begingroup$ @MaPo The normal order of a constant is the constant, why do you expect 0? $\endgroup$ – ACuriousMind Feb 3 '17 at 0:35
  • $\begingroup$ You are right, but i really cannot understand. For instance the example of @ACuriousMind, seems to tell us that it's impossible to really define that operation. I'm very confused and I don't know how to apply the formula. Still, if the normal order of a constant is the constant isfelf, where is the mistake in deriving eqs (4) and (5) that seems to show that idempotency fails? $\endgroup$ – MaPo Feb 3 '17 at 0:37
  • $\begingroup$ You can only operate with ${\cal O}$ on radially ordered expressions. Therefore you cannot apply ${\cal O}$ twice, since after the first application of ${\cal O}$, the expression is no longer radially ordered. $\endgroup$ – Qmechanic Feb 3 '17 at 10:21
  • $\begingroup$ Thank you so much! Now I have understood how to apply properly Polchinski (1). Apart from that what is worrying me now is the example of @ACuriousMind that seems to claim that linearity fails in normal ordering just applying definition, disregarding radial order at all. Is there a way out? $\endgroup$ – MaPo Feb 3 '17 at 10:38

Applying normal ordering twice is not a valid operation. In fact, "normal ordering" is not a proper operator at all, and only defined on a single product of operators. It is undefined when acting on sums of operators, since trying to extend it linearly fails: $$ :a a^\dagger: = : a^\dagger a + 1 : = :a^\dagger a: + :1: = a^\dagger a + 1 \neq a^\dagger a = :a a^\dagger,$$ which is a contradiction. However, $\mathcal{O}$ clearly is a linear operator, and therefore the equation $: F : = \mathcal{O} F$ is only meant to hold when $F$ is a product of operators, but not their sum.

If you examine the counterexample to linearity above, you will realize this means that normal ordering isn't defined as a function in the mathematical sense on operators at all, since in $a a^\dagger = a^\dagger a +1$ the l.h.s. and the r.h.s. are the exact same mathematical object, yet normal ordering can only be applied to one of them. It is best to think of normal ordering as acting on symbols, by which I mean strings composed of $a$ and $a^\dagger$. Every operator has, if it can be written as a product of $a$ and $a^\dagger$ without any sum, exactly one such representation, so there is a well-defined map from operators to such symbols, to which you then can apply normal ordering, and then turn them back into operators again.

  • $\begingroup$ I see. The problem is then how to trust in eq (1) by Polchinski? It seem to be useful in many application, but I cannot figure out when it gives the correct result and when it fails. $\endgroup$ – MaPo Feb 3 '17 at 0:28

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