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] = ℏm δ0m+n,(1)
:αmαn: = Θ(n−m)αmαn + Θ(m−n)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): = ℏ(z−w)2.(4)
4) Notice that each of the three terms in eq. (4) are invariant under z↔w 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)] = … = ℏw2∞∑n=1n(zw)n−1 = … = ℏ(z−w)2.(8)
In the last step we will use that the sum is convergent under the assumption
|z|<|w|.
This post imported from StackExchange Physics at 2014-05-01 11:55 (UCT), posted by SE-user Qmechanic