In the 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 $$\begin{align} ::φ(z)φ(w)::~=~&:φ(z)φ(w):-\frac{α'}{2}\log|z-w|^2\cr ~=~&φ(z)φ(w)-α'\log|z-w|^2.\end{align}\tag{5} $$


1 Answer 1

  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., $$ \begin{array}{ccc} \text{Radial-Ordered Symbols/Fcts}&\stackrel{\cal O}{\longrightarrow} & \text{Normal-Ordered Symbols/Fcts} \cr\cr \updownarrow &&\updownarrow\cr\cr \text{Radial-Ordered Operators}&\stackrel{\hat{\cal O}}{\longrightarrow} & \text{Normal-Ordered Operators}\end{array}\tag{B}$$ 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, 2017 at 0:21
  • $\begingroup$ @MaPo The normal order of a constant is the constant, why do you expect 0? $\endgroup$
    – ACuriousMind
    Feb 3, 2017 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, 2017 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, 2017 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, 2017 at 10:38

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