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The time ordering for the purpose of quantum mechanics is e.g. given by

$${\mathcal T} \left[A(x) B(y)\right] := \begin{matrix} A(x) B(y) & \textrm{ if } & x_0 > y_0 \\ \pm B(y)A(x) & \textrm{ if } & x_0 < y_0, \end{matrix}$$

where $x_0$ and $y_0$ denote the time-coordinates of the points x and y.

Now strictly speaking, the above definition works by pattern recognition - I can tell the operators $A(x)$ and $B(y)$ appart. I could either imagine one could write down the quantum mechanical theory by repacing all occurences of $\mathcal T \left[A(x) B(y)\right]$ by function notation $\mathcal T \left[A(x), B(y)\right]$ or that there is a computable procedure to make sense of $\mathcal T$ whenever I'm given an operator $F(x,y)$, i.e. $\mathcal T \left[F(x,y)\right]$.

Which is the case?

Nikolaj-K
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  • What about viewing time ordering as a mapping from $2n$-tuples $(A_1, \dots, A_n, x_1, \dots, x_n)$ of $n$ operator-valued distributions and $n$ spacetime points? It seems to me that this would properly formalize the notion of pattern-recognition to which you refer. – joshphysics Sep 13 '13 at 20:25
  • @dj_mummy: Which definition do you mean? At first sight, I see no proper functional definition there. E.g. writing $\mathrm{add}(x\cdot y):=x+y$ or $\mathrm{perm}(f(t_a,t_b)):=f(t_b,t_a)$ doesn't properly define $\mathrm{add}$ or $\mathrm{perm}$ in a classical way as maps as they rely on pattern recognition. – Nikolaj-K Sep 14 '13 at 12:55
  • @joshphysics: Yeah, I just wanted to know in which sense it's usually ment or a reference to a mathematical treatment of it. – Nikolaj-K Sep 14 '13 at 13:00
  • @NickKidman I think this essentially asks and answers your question. In fact, I think it's essentially a duplicate. http://physics.stackexchange.com/q/44455/ – joshphysics Sep 14 '13 at 19:14
  • @dj_mummy: I'm not sure you understand what I mean. I wrote $\mathrm{add}(x\cdot y)$ (analogous to the time ordering operation ${\mathcal T} \left[A(x) B(y)\right]$) and not $\mathrm{add}(x,y)$. – Nikolaj-K Sep 15 '13 at 15:16
  • @dj_mummy: Example: Each function has a domain and if the images of maps from time to $\mathcal T$'s domain are three pairwise non-commuting operators ($A,B,C\in\mathrm{dom(\mathcal T)}$) then with $x,y\in \mathbb R$ and because of the operators "definition", it's not clear $\mathcal T[A(x)\ C\ B(y)]$ maps to. Second exmaple from the original post, given a function $F(x,y)$ which maps into that domain, we don't know how to evaluate $\mathcal T[F(x,y)]$. – Nikolaj-K Sep 16 '13 at 07:44
  • @dj_mummy: For given $x,y$, having the image of operators as domain doesn't wor. If for $x,y\in \mathbb R, x<y$ the operators $A(x), B(y)$ are in the domain, and if $C$ is another operator in that domain where we don't have an argument to scan for, how does $\mathcal T[A(x)\ C\ B(y)]$ evaluate. If $\mathcal T$ is a function with domain and all, then there must be a unique output to this. Is it e.g. $B(y)\ A(x)\ C$ or $C\ B(y)\ A(x)$, or something else. – Nikolaj-K Sep 16 '13 at 11:51

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