# 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?

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## 1 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 defintion 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.

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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. – Dilaton Dec 10 '11 at 10:40
@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. – Wein Eld Dec 28 '15 at 11:30
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. – Qmechanic Dec 28 '15 at 14:17