Monday, June 9, 2014

homework and exercises - Wick Order and Radial Ordering in CFT


I am not so much familiar with the computations tools of conformal field theory, and I just run into an exercise asking to demonstrate the following formula (related to the bosonic field case):


Rj(z1)j(z2) = 1(z1z2)2 + :j(z1)j(z2):


with j defined as


j(z) = kαkzk1.


My question is should I start the calculation form the Wick ordered term and make the two others appear, because starting from the left side, I don't see how could I develop some calculus?



Answer



Here we will outline a strategy to prove the sought-for operator identity (4) from the following definitions of what the commutator and the normal order of two mode operators αm and αn mean:



[αm,αn] = m δ0m+n,(1)

:αmαn: = Θ(nm)αmαn + Θ(mn)αnαm,(2)


where Θ denote the Heaviside step function.


1) Note that the current j(z) = j(z)+j+(z) is a sum of a creation part j(z) and an annihilation part j+(z).


2) Recall that the radial order R is defined as R(j(z)j(w)) = Θ(|z||w|)j(z)j(w) + Θ(|w||z|)j(w)j(z).(3)


3) Rewrite the sought-for operator identity as R(j(z)j(w))  :j(z)j(w): = (zw)2.(4)


4) Notice that each of the three terms in eq. (4) are invariant under zw symmetry. So we may assume from now on that |z|<|w|.


5) Show that j(w)j(z)  :j(z)j(w): = [j+(w),j(z)].(5)


6) Show (under the assumption |z|<|w|) that j(w)j(z)  R(j(z)j(w)) |z|<|w|= 0.(6)


7) Subtract eq. (6) from eq. (5):


R(j(z)j(w))  :j(z)j(w): |z|<|w|= [j+(w),j(z)].(7)



8) Evaluate rhs. of eq. (7):


[j+(w),j(z)] =  = w2n=1n(zw)n1 =  = (zw)2.(8)

In the last step we will use that the sum is convergent under the assumption |z|<|w|.


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