If the operators $X_i$ can be written as a sum of an annihilation and a creation part

$$X_i~=~A_i + A_i^\dagger, \qquad A_i|0\rangle~=~0,$$

where

$$ [A_i(t),A_j(t^\prime)] ~=~ 0, $$

and

$$ [A_i(t),A_j^\dagger(t^\prime)] ~=~ (c~{\rm numbers}) \times {\bf 1}, $$

i.e. proportional to the identity operator ${\bf 1}$, then one may prove that

$$ T(X_i(t)X_j(t^\prime)) ~-~:X_i(t)X_j(t^\prime):
~=~\langle 0 | T(X_i(t)X_j(t^\prime))|0\rangle ~{\bf 1}. \qquad (1) $$

Proof: The time ordering is defined as

$$ T(X_i(t)X_j(t^\prime)) ~=~ \Theta(t-t^\prime) X_i(t)X_j(t^\prime)
+\Theta(t^\prime-t) X_j(t^\prime)X_i(t)$$
$$~=~X_i(t)X_j(t^\prime) -\Theta(t^\prime-t) [X_i(t),X_j(t^\prime)]$$
$$~=~X_i(t)X_j(t^\prime) -\Theta(t^\prime-t) \left([A_i(t),A_j^\dagger(t^\prime)]+[A_i^\dagger(t),A_j(t^\prime)]\right). \qquad (2)$$

The normal ordering moves the creation part to the left of the annihilation part, so

$$:X_i(t)X_j(t^\prime):~=~ X_i(t)X_j(t^\prime) - [A_i(t),A_j^\dagger(t^\prime)].\qquad (3)$$

The difference of eqs. (2) and (3) is the lhs. of eq. (1) :

$$ T(X_i(t)X_j(t^\prime)) ~-~:X_i(t)X_j(t^\prime): $$
$$~=~ \Theta(t-t^\prime)[A_i(t),A_j^\dagger(t^\prime)] + \Theta(t^\prime-t)[A_j(t^\prime),A_i^\dagger(t)],\qquad (4)$$

which is proportional to the identity operator ${\bf 1}$ by assumption. Now sandwich eq. (4) between the bra $\langle 0 |$ and the ket $|0\rangle $. Since the rhs. is proportional to the identity operator ${\bf 1}$, the unsandwiched rhs. must be equal to the sandwiched rhs. times the identity operator ${\bf 1}$. Hence also the unsandwiched lhs. must be equal to the sandwiched lhs. times the identity operator ${\bf 1}$. This yields eq. (1).

This post imported from StackExchange Physics at 2014-03-17 03:36 (UCT), posted by SE-user Qmechanic