If the operators Xi can be written as a sum of an annihilation and a creation part
Xi = Ai+A†i,Ai|0⟩ = 0,
where
[Ai(t),Aj(t′)] = 0,
and
[Ai(t),A†j(t′)] = (c numbers)×1,
i.e. proportional to the identity operator 1, then one may prove that
T(Xi(t)Xj(t′)) − :Xi(t)Xj(t′): = ⟨0|T(Xi(t)Xj(t′))|0⟩ 1.(1)
Proof: The time ordering is defined as
T(Xi(t)Xj(t′)) = Θ(t−t′)Xi(t)Xj(t′)+Θ(t′−t)Xj(t′)Xi(t)
= Xi(t)Xj(t′)−Θ(t′−t)[Xi(t),Xj(t′)]
= Xi(t)Xj(t′)−Θ(t′−t)([Ai(t),A†j(t′)]+[A†i(t),Aj(t′)]).(2)
The normal ordering moves the creation part to the left of the annihilation part, so
:Xi(t)Xj(t′): = Xi(t)Xj(t′)−[Ai(t),A†j(t′)].(3)
The difference of eqs. (2) and (3) is the lhs. of eq. (1) :
T(Xi(t)Xj(t′)) − :Xi(t)Xj(t′):
= Θ(t−t′)[Ai(t),A†j(t′)]+Θ(t′−t)[Aj(t′),A†i(t)],(4)
which is proportional to the identity operator 1 by assumption. Now sandwich eq. (4) between the bra ⟨0| and the ket |0⟩. Since the rhs. is proportional to the identity operator 1, the unsandwiched rhs. must be equal to the sandwiched rhs. times the identity operator 1. Hence also the unsandwiched lhs. must be equal to the sandwiched lhs. times the identity operator 1. This yields eq. (1).
This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user Qmechanic