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I already know that $T\,:\,\,:\,=T$ and $\,:\,\,:\,T=\,:\,\,:\,$ simply because when one operator acts after the other, play a bit with what the previous has done and do what he wants to actually do, or at least that's my interpretation, probably wrong. I also always considered that time ordering and normal ordering acts linearly on operators (acting strange when there are things like $\hat{a}\hat{a}^\dagger+1$) and also act as idempotent operators, meaning that $T^2=T$ and $\,:\,(\,:\,\,:\,)\,:\,=\,:\,\,:\,$ .

But now I'm stuck with this apparently paradoxical result using Wick's theorem in a composition of different field operators considered at equal time \begin{gather*} \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \equiv T\left( \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \right) \\ \,:\! \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \!:\, \equiv \,:\! \left( \,:\! \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \!:\, + \sum_{j=1}^{\left\lfloor\frac{k}{2}\right\rfloor} \sum\limits_{\bullet\times j} \,:\! \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \!:\, \right) \!:\, \\ \,:\!\left( \,:\! \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \!:\,\right) \!:\, \overset{?}{=} \,:\! \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \!:\, \\ \Rightarrow 0 = \sum_{j=1}^{\left\lfloor\frac{k}{2}\right\rfloor} \sum\limits_{\bullet\times j} \,:\! \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \!:\, \\ \,:\! \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \!:\, = T\left( \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \right) = \overset{k}{\underset{i=1}{\circ}} \hat{\phi}_i(t,\boldsymbol{r}_i) \end{gather*} The last equation seems very wrong to me and for that I can suspect the hypothesis that normal ordering was idempotent. Where is the issue?

Qmechanic
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Rob Tan
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    normal ordering and time ordering have multiple argument slots, not a single one. So $T(\phi_1\phi_2\cdots\phi_n)$ is bad notation, the correct one is $T(\phi_1,\phi_2,\dots,\phi_n)$. Now it is clear that $T\circ T=T$: the first one eats a tuple and spits out a singlet. The second one takes this singlet and returns it unchanged. Same for normal ordering. – AccidentalFourierTransform Feb 10 '21 at 16:17
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    Related: https://physics.stackexchange.com/q/162339/2451 – Qmechanic Feb 10 '21 at 16:19
  • Thanks @AccidentalFourierTransform and @Qmechanics; some days ago I just saw the related question you are now posting I just didn't remember it! Using $NT=N$ solves the issue, but at the same time I cannot recognise were the error is in the second equation! Thank you very much – Rob Tan Feb 10 '21 at 16:36
  • @Qmechanic I was doing these passages as an introduction just to nested Wick's theorem, that you cite in the related question. Attempting the demonstration I completely canceled the passages in this question and I tried to arrive at the result that all contractions of an equal time normal ordered composition of field operators, don't contribute with theirs contractions to Wick's theorem when we have to time order something like $,:!\overset{l}{\underset{j=1}{\circ}}\hat{\phi}_j(t,\boldsymbol{r}_j)!:,\overset{k}{\underset{i=1}{\circ}}\hat{\phi}_i(\boldsymbol{x}_i)$. – Rob Tan Feb 11 '21 at 09:07
  • The problem is I can suppose the $i$-indexed composition to be already normal ordered, but I'm not authorized to suppose $t<t_i,,\forall,i$ or $t>t_i,,\forall,i$, so time ordering the expression before is a true nightmare! I don't really know how to do this, so have you some sources to suggest how to demonstrate nested Wick's theorem? – Rob Tan Feb 11 '21 at 09:09

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