Tuesday, 29 May 2018

quantum field theory - Normal Order of Normal Order



in first volume of Polchinski page 39 we can read a compact formula to perform normal-order for bosonic fields :F:=exp{α4d2zd2wlog|zw|2δδφ(z,ˉz)δδφ(w,ˉzw)}:=OF,


What I do not understand it is that I would like to have (bearing in mind the definition involving a and a ::F::=:F:

but with this formula O2FOF.


EXAMPLE


:φ(z)φ(w):=φ(z)φ(w)α2log|zw|2

but ::φ(z)φ(w)::=:φ(z)φ(w):α2log|zw|2=φ(z)φ(w)αlog|zw|2



Answer





  1. Short explanation: Polchinski's eq. (1) is not a formula that transforms no normal order into normal order: The expression 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 ˆX and ˆP, the "function of operators" f(ˆX,ˆP) 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 Symbols/FunctionsOperators

    (E.g. the correspondence map from Weyl symbols to operators is explained in this Phys.SE post.) To define an operator ˆO on operators, one often give the corresponding operator O on symbols/functions, i.e., Normal-Ordered Symbols/FunctionsORadial-Ordered Symbols/FunctionsNormal-Ordered OperatorsˆORadial-Ordered Operators
    E.g. Polchinski's differential operator 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.




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