# Is time ordering defined for a single operator depending of two time variables?

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?

-
There is a clear definition of the time-ordering operator given in the wikipedia link you have provided. It is just a permutation operator. If you want a clear picture of regularization from the causal point of view, check out this link arxiv.org/pdf/0906.1952v2.pdf – dj_mummy Sep 13 '13 at 18:55
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. – NikolajK 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. – NikolajK 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. physics.stackexchange.com/questions/44455/… – joshphysics Sep 14 '13 at 19:14