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?


1 Answer 1


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

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

$$\begin{align} A_i|\Omega\rangle~=~0, \qquad & \langle \Omega |A^{\dagger}_i~=~0, \cr \qquad i~\in~&I,\end{align}\tag{2}$$


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


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

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

$$\begin{align} T(&X_i(t)X_j(t^{\prime})) ~-~:X_i(t)X_j(t^{\prime}):\cr ~=~&\langle \Omega | T(X_i(t)X_j(t^{\prime}))|\Omega\rangle ~{\bf 1}.\end{align} \tag{5}$$

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

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

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

$$\begin{align}:X_i(t)X_j(t^\prime):~\stackrel{(1)}{=}~& X_i(t)X_j(t^{\prime}) \cr ~-~& [A_i(t),A^{\dagger}_j(t^{\prime})], \end{align}\tag{7}$$

$$ \langle \Omega | :X_i(t)X_j(t^{\prime}):|\Omega\rangle~\stackrel{(1)+(2)}{=}~0.\tag{8}$$

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

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

which is proportional to the identity operator ${\bf 1}$ by assumption (4). Now sandwich eq. (9) between the bra $\langle \Omega |$ and the ket $|\Omega\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). $\Box$

A similar argument applied to 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. (5) without the time-order $T$.


$^1$ 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: $\langle \Omega | \Omega\rangle=1$.

  • 1
    $\begingroup$ Thanks a lot @Qmechanic, this clear proof is exactly what I needed. The only thing left to do for me now is to check that the preconditions are valid in my specific case. $\endgroup$
    – Dilaton
    Dec 10, 2011 at 10:40
  • $\begingroup$ @Qmechanic, for the operators $A$ and $B$, is the equation $\langle 0|A+B|0\rangle=\langle 0|A|0\rangle+\langle 0|B|0\rangle$ satisfied? If it is satisfied, then we can sandwich the lhs. of eq. (1) and the equation simply becomes $\langle 0|:X_i(t)X_j(t'):|0\rangle=0$. But such equation sometimes fails. $\endgroup$
    – Wein Eld
    Dec 28, 2015 at 11:30
  • $\begingroup$ I updated the answer. It seems you are talking about situations where one uses definitions of Fock vacuum and normal order that are not properly compatible/adjusted to each other. $\endgroup$
    – Qmechanic
    Dec 28, 2015 at 14:17

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