I'm trying to progress towards understanding, and perhaps finding a proof for, the "nested" Wick's theorem for time-ordered products $T\{ \ldots \}$ alluded to in part (II) of this answer.

Assuming bosonic operators for now, I've noticed that

$$T\{ T\{ A(t_1)B(t_2) \} T\{ C(t_3)D(t_4) \} \} \equiv T\{ A(t_1)B(t_2)C(t_3)D(t_4) \}\tag{1}$$

through brute force calculation. Is the natural generalisation of this,

$$T\{ T\{ \ldots_1 \} \ldots_2 T\{ \ldots_3 \} \ldots_4 ~~~\ldots~~~ T\{ \ldots_{n-1} \} \ldots_n \} \equiv T\{ \ldots_1 \ldots_2 ~~~\ldots~~~ \ldots_n \},\tag{2}$$

also true?

  • $\begingroup$ Is the question mostly about the special cases where some of the times are equal? $\endgroup$
    – Qmechanic
    Feb 20, 2022 at 7:21
  • $\begingroup$ @Qmechanic not necessarily, no... I was trying out some small examples of the $T\{ N\{ \ldots \} \ldots N\{ \ldots \} \}$ expression from your answer, particularly $T\{ N\{ AB \} N\{ CD \} \}$, $T\{ AB ~ N\{ CD \} \}$, and so on, and think it's possible that if the above is true, then you can make use of the normal Wick's theorem to cancel all of the terms where two operators from the same normal-ordering are contracted. $\endgroup$ Feb 20, 2022 at 18:32

1 Answer 1

  1. Let us first define a $n$-ary Heaviside step function: $$\begin{align} \theta&(t_1\geq t_2\geq\ldots \geq t_n)\cr ~:=~&\left\{\begin{array}{rl} 0&\text{if ineq. is violated}, \cr \frac{1}{m_1!\ldots m_r!}&\text{if ineq. holds and there are $r$ sets of equal} \cr &\text{times with multiplicities } m_1, \ldots, m_r. \end{array} \right.\end{align} \tag{A}$$ It satisfies $$ \sum_{\pi\in S_n} \theta(t_{\pi(1)}\geq t_{\pi(2)}\geq\ldots \geq t_{\pi(n)})~=~1. \tag{B}$$

  2. Next define time-ordering $T$ for Grassmann-even$^1$ operators as $$ T(A_1 \ldots A_n)~:=~\sum_{\pi\in S_n} \theta\left( t(A_{\pi(1)})\geq \ldots \geq t(A_{\pi(n)})\right) A_{\pi(1)} \ldots A_{\pi(n)}. \tag{C}$$ Time-ordering $T$ is a multi-linear map and it may be viewed as a symmetric operad $$T(A_1 \ldots A_n)~=~T(A_{\pi(1)} \ldots A_{\pi(n)}), \qquad \pi~\in~ S_n.\tag{D} $$ It satisfies a generalized idempotency $$\begin{align}T&\left(T(A_1 \ldots A_r)T(B_1 \ldots B_s)\ldots Z_1 \ldots Z_u \right)\cr ~\stackrel{(C)}{=}~~&\sum_{\pi\in S_r} \theta\left( t(A_{\pi(1)})\geq \ldots \geq t(A_{\pi(r)})\right)\cr &\sum_{\sigma\in S_s} \theta\left( t(B_{\sigma(1)})\geq \ldots \geq t(B_{\sigma(s)})\right)\ldots\cr &T\left(A_{\pi(1)} \ldots A_{\pi(r)}B_{\sigma(1)} \ldots B_{\sigma(s)}\ldots Z_1 \ldots Z_u \right)\cr ~\stackrel{(B)+(D)}{=}&T(A_1 \ldots A_r B_1 \ldots B_s \ldots Z_1 \ldots Z_u ) ,\end{align}\tag{E} $$ which is OP's eq. (2).


$^1$ There is as straightforward generalization to Grassmann-graded operators.

  • $\begingroup$ With the help of a friend yesterday evening, we managed to put together a similar argument but while not considering equal times. Thanks for shoring up the result with your more general derivation! $\endgroup$ Feb 22, 2022 at 1:07

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