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?
If the operators $X_i$ can be written as a sum of an annihilation and a creation part$^1$
$$X_i~=~A_i + A^{\dagger}_i, \qquad i~\in ~I, \tag{1}\label{eq:1}$$
$$\begin{align} A_i|\Omega\rangle~=~0, \qquad & \langle \Omega |A^{\dagger}_i~=~0, \cr \qquad i~\in~&I,\end{align}\tag{2}\label{eq:2}$$
where
$$\begin{align} [A_i(t),A_j(t^{\prime})] ~=~& 0, \cr [A^{\dagger}_i(t),A^{\dagger}_j(t^{\prime})] ~=~& 0, \cr i,j~\in ~&I,\end{align}\tag{3}\label{eq:3} $$
and
$$\begin{align} [A_i(t),A_j^\dagger(t^{\prime})] ~=~& (c~{\rm number}) \times {\bf 1},\cr i,j~\in~&I,\end{align}\tag{4}\label{eq:4}$$
i.e. proportional to the identity operator ${\bf 1}$, then one may prove that
$$\begin{align} T(&X_i(t)X_j(t^{\prime})) ~-~:X_i(t)X_j(t^{\prime}):\cr ~=~&\langle \Omega | T(X_i(t)X_j(t^{\prime}))|\Omega\rangle ~{\bf 1}.\end{align} \tag{5}\label{eq:5}$$
Proof of eq. $\eqref{eq:5}$: On one hand, the time ordering $T$ is defined as
$$\begin{align} T(&X_i(t)X_j(t^{\prime}))\cr ~=~& \Theta(t-t^{\prime}) X_i(t)X_j(t^{\prime}) +\Theta(t^{\prime}-t) X_j(t^{\prime})X_i(t)\cr ~=~&X_i(t)X_j(t^{\prime}) -\Theta(t^{\prime}-t) [X_i(t),X_j(t^{\prime})]\cr ~\stackrel{\eqref{eq:1}+\eqref{eq:3}}{=}&~X_i(t)X_j(t^{\prime})\cr &-\Theta(t^{\prime}-t) \left([A_i(t),A^{\dagger}_j(t^{\prime})]+[A^{\dagger}_i(t),A_j(t^{\prime})]\right).\end{align} \tag{6}\label{eq:6}$$
On the other hand, the normal ordering $::$ moves by definition the creation part to the left of the annihilation part, so that
$$\begin{align}:X_i(t)X_j(t^\prime):~\stackrel{(1)}{=}~& X_i(t)X_j(t^{\prime}) \cr ~-~& [A_i(t),A^{\dagger}_j(t^{\prime})], \end{align}\tag{7}\label{eq:7}$$
$$ \langle \Omega | :X_i(t)X_j(t^{\prime}):|\Omega\rangle~\stackrel{$\eqref{eq:1}$+$\eqref{eq:2}$}{=}~0.\tag{8}$$
The difference of eqs. $\eqref{eq:6}$ and $\eqref{eq:7}$ is the lhs. of eq. $\eqref{eq:5}$:
$$\begin{align} T(& X_i(t)X_j(t^{\prime})) ~-~:X_i(t)X_j(t^{\prime}): \cr ~\stackrel{\eqref{eq:6}+\eqref{eq:7}}{=}& \Theta(t-t^{\prime})[A_i(t),A^{\dagger}_j(t^{\prime})] \cr ~+~& \Theta(t^{\prime}-t)[A_j(t^{\prime}),A^{\dagger}_i(t)],\end{align}\tag{9}\label{eq:9}$$
which is proportional to the identity operator ${\bf 1}$ by assumption $\eqref{eq:4}$. Now sandwich eq. $\eqref{eq:9}$ between the bra $\langle \Omega |$ and the ket $|\Omega\rangle $. Since the rhs. of eq. $\eqref{eq:9}$ 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. $\eqref{eq:9}$ must also be equal to the sandwiched lhs. times the identity operator ${\bf 1}$. This yields eq. $\eqref{eq:5}$. $\Box$
A similar argument applied to eq. $\eqref{eq:7}$ yields that
$$\begin{align} X_i(t)&X_j(t^{\prime}) ~-~:X_i(t)X_j(t^{\prime}):\cr ~=~&\langle \Omega | X_i(t)X_j(t^{\prime})|\Omega\rangle ~{\bf 1} \end{align}\tag{10}$$
i.e. a version of eq. $\eqref{eq:5}$ without the time-order $T$.
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$^1$ The operators $A_i$ and $A^{\dagger}_i$ need not be Hermitian conjugates in what follows. We implicitly assume that the vacuum $|\Omega\rangle$ is normalized: $\langle \Omega | \Omega\rangle=1$.