quantum mechanics - Time-ordering vs normal-ordering and the two-point function/propagator

I don't understand how to calculate this generalized two-point function or propagator, used in some advanced topics in quantum field theory, a normal ordered product (denoted between $::$) is subtracted from the usual time ordered product (denoted $T$):

$$\langle X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau')\rangle ~=~ T ( X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau')) ~-~ : X^{\mu}(\sigma,\tau)X^{\nu}(\sigma',\tau'):$$

My question is can the rhs of this propagator be derived or the meaning of the subtraction of the time ordered product explained and motivated in simple words?

Answer

If the operators $X_i$ can be written as a sum of an annihilation and a creation part$^1$

$$X_i~=~A_i + A^{\dagger}_i, \qquad i~\in ~I, \tag{1}$$ $$ A_i|0\rangle~=~0, \qquad \langle 0 |A^{\dagger}_i~=~0, \qquad i~\in ~I,\tag{2}$$

where

$$ [A_i(t),A_j(t^{\prime})] ~=~ 0, \qquad [A^{\dagger}_i(t),A^{\dagger}_j(t^{\prime})] ~=~ 0, \qquad i,j~\in ~I,\tag{3} $$

and

$$ [A_i(t),A_j^\dagger(t^{\prime})] ~=~ (c~{\rm number}) \times {\bf 1},\qquad i,j~\in ~I,\tag{4} $$

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

$$ T(X_i(t)X_j(t^{\prime})) ~-~:X_i(t)X_j(t^{\prime}): ~=~\langle 0 | T(X_i(t)X_j(t^{\prime}))|0\rangle ~{\bf 1}. \tag{5}$$

Proof of eq. (5): On one hand, the time ordering $T$ is defined as

$$ T(X_i(t)X_j(t^{\prime})) ~=~ \Theta(t-t^{\prime}) X_i(t)X_j(t^{\prime}) +\Theta(t^{\prime}-t) X_j(t^{\prime})X_i(t)$$ $$~=~X_i(t)X_j(t^{\prime}) -\Theta(t^{\prime}-t) [X_i(t),X_j(t^{\prime})]$$ $$~\stackrel{(1)+(3)}{=}~X_i(t)X_j(t^{\prime}) -\Theta(t^{\prime}-t) \left([A_i(t),A^{\dagger}_j(t^{\prime})]+[A^{\dagger}_i(t),A_j(t^{\prime})]\right). \tag{6}$$

On the other hand, the normal ordering $::$ moves by definition the creation part to the left of the annihilation part, so that

$$:X_i(t)X_j(t^\prime):~\stackrel{(1)}{=}~ X_i(t)X_j(t^{\prime}) - [A_i(t),A^{\dagger}_j(t^{\prime})], \tag{7}$$ $$ \langle 0 | :X_i(t)X_j(t^{\prime}):|0\rangle~\stackrel{(1)+(2)}{=}~0.\tag{8}$$

The difference of eqs. (6) and (7) is the lhs. of eq. (5):

$$ T(X_i(t)X_j(t^{\prime})) ~-~:X_i(t)X_j(t^{\prime}): $$ $$~\stackrel{(6)+(7)}{=}~ \Theta(t-t^{\prime})[A_i(t),A^{\dagger}_j(t^{\prime})] + \Theta(t^{\prime}-t)[A_j(t^{\prime}),A^{\dagger}_i(t)],\tag{9}$$

which is proportional to the identity operator ${\bf 1}$ by assumption (4). Now sandwich eq. (9) between the bra $\langle 0 |$ and the ket $|0\rangle $. Since the rhs. of eq. (9) 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. of eq. (9) must also be equal to the sandwiched lhs. times the identity operator ${\bf 1}$. This yields eq. (5).

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$^1$ The operators $A_i$ and $A^{\dagger}_i$ need not be Hermitian conjugates in what follows.