Distributive property of the time-ordering symbol

Most derivations of the LSZ reduction formula, e.g. Srednicki (equations 5.13, 5.14, 5.15), Schwartz (equations 6.17, 6.18, 6.19), Wikipedia use a property of the time-ordering symbol that looks like the distributive property.

The relevant steps are (I'm using a single in and out particle to simplify the expressions): \begin{align} \langle 0|a_{2}(+\infty)a_1^\dagger(-\infty)|0\rangle &= \langle 0|{\rm T} a_{2}(+\infty)a_1^\dagger(-\infty)|0\rangle\\ &= \langle 0|{\rm T} \left(a_{2} (-\infty)+\int dt_{2} \partial_0 a_{2}(t_{2})\right) \left(a_{1}^\dagger (+\infty)+\int dt_{1} \partial_0 a_{1}^\dagger(t_{1})\right)|0\rangle \\ &= \langle 0|{\rm T} \left(\int dt_{2} \partial_0 a_{2}(t_{2})\right) \left(\int dt_{1} \partial_0 a_{1}^\dagger(t_{1})\right)|0\rangle \\ &= \int dt_{1}dt_{2} \langle 0| {\rm T}a_{2}(t_{2})a_{1}^\dagger(t_{1})|0\rangle \\ \end{align} Here the distributive property is used twice: first to make the ladder operators annihilate the vacuum states (line 3) and then to exchange the time ordering symbol with the integral operator (line 4).

The final expression is they way I expect it: it is only natural that operators should be applied in the order given by their time argument. However this property is far from trivial, since it works even when an operator is decomposed into a sum of operators at different times.

So my questions are: what are the conditions (e.g. which operators, how should they be decomposed..) for this property to hold? And, most important, is there a proof of it?

Edit: To clarify what I mean by "distributive property": \begin{align} \langle 0|{\rm T} \left(a_{2} (-\infty)+\int dt_{2} \partial_0 a_{2}(t_{2})\right) &\left(a_{1}^\dagger (+\infty)+\int dt_{1} \partial_0 a_{1}^\dagger(t_{1})\right)|0\rangle =\\ &\langle 0|{\rm T} a_{2} (-\infty) a_{1}^\dagger (+\infty)|0\rangle \\ +&\langle 0|{\rm T} a_{2} (-\infty)\left(\int dt_{1} \partial_0 a_{1}^\dagger(t_{1})\right)|0\rangle\\ +&\langle 0|{\rm T} \left(\int dt_{2} \partial_0 a_{2}(t_{2})\right) a_{1}^\dagger (+\infty)|0\rangle\\ +&\langle 0|{\rm T} \left(\int dt_{2} \partial_0 a_{2}(t_{2})\right) \left(\int dt_{1} \partial_0 a_{1}^\dagger(t_{1})\right)|0\rangle \end{align}

1 Answer

Comments to the question (v3):

1. The time-ordered product of operators $$T[A_1(t_1)\ldots A_n(t_n)] ~:=~\sum_{\pi\in S_n} \theta \left(t_{\pi(1)} > \ldots > t_{\pi(n)} \right) (-1)^{\varepsilon_{\pi,A}} A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)}) \tag{1}$$ is graded symmetric. [Here $(-1)^{\varepsilon_{\pi,A}}$ is a sign factor in the case of Grassmann-odd operators.] Note that the time-ordered product (1) is only defined for mono-local operators $A_i(t_i)$, i.e. when each operator $A_i(t_i)$ depends on a single time $t_i$ each.

2. Formula (1) does not make sense for bi-local operators $B(t_1,t_2)$ (and more generally multi-local operators $M(t_1,\ldots t_m)$), unless they can be decomposed in mono-local operators. We next extend definition (1) via multi-linarity. This means in particular that time-ordered product $T$ satisfies the distributive law by construction, e.g. $$T[A_1(t_1)\ldots A_i(t_i) \left\{B(t_b)+C(t_c) \right\}A_{i+1}(t_{i+1})\ldots A_n(t_n)]$$ $$~:=~T[A_1(t_1)\ldots A_i(t_i) B(t_b)A_{i+1}(t_{i+1})\ldots A_n(t_n)]$$ $$+T[A_1(t_1)\ldots A_i(t_i)C(t_c)A_{i+1}(t_{i+1})\ldots A_n(t_n)]\tag{2}$$ If $t_b\neq t_c$, then the left-hand side of eq. (2) does not have an independent meaning. If $t_b=t_c$, then eq. (2) follows from definition (1).

3. Using the fundamental theorem of calculus, we can consider a telescoping sum: $$T[A_1(t_1)\ldots A_i(t_i) \left\{B(t\!=\!\infty)- B(t\!=\!-\infty) \right\} A_{i+1}(t_{i+1})\ldots A_n(t_n)]$$ $$~=~ T\left[A_1(t_1)\ldots A_i(t_i) \left\{\int_{\mathbb{R}}\! dt~\frac{dB(t)}{dt} \right\} A_{i+1}(t_{i+1})\ldots A_n(t_n)\right]$$ $$~=~ \int_{\mathbb{R}}\! dt~T\left[A_1(t_1)\ldots A_i(t_i) \frac{dB(t)}{dt} A_{i+1}(t_{i+1})\ldots A_n(t_n)\right]$$ $$+\sum_{i=1}^n\int_{\mathbb{R}}\! dt~\delta(t-t_i)~T[A_1(t_1)\ldots [A_i(t_i), B(t_i)] A_{i+1}(t_{i+1})\ldots A_n(t_n)]$$ $$~=~\int_{\mathbb{R}}\! dt\frac{d}{dt} T[A_1(t_1)\ldots A_i(t_i) B(t) A_{i+1}(t_{i+1})\ldots A_n(t_n)], \tag{3}$$ which yields a distributive-like law.

4. Now for the mentioned step in the proof of the LSZ reduction formula, the calculation can be organized such that the distributive properties (2) and (3) suffice. Besides the distributive law, there are several other subtle points in the derivation, such as, e.g. contact terms. See e.g. this related Phys.SE post.

• Thank you for the detailed answer. You say that "the calculation can be organized such that the distributive properties (2) and (3) suffice": could you provide a reference to a book or a web resource where this is done in detail? The books I mentioned above do not even mention this issue, and Weinberg arrives at the formula using a different method. However, I'm accepting it as it fully answers my question (which deals with T, not LSZ). – agc Jul 17 '16 at 9:26