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In quantum field theory, the time ordering operator (TOO) appears in the formal expressions for the scattering amplitudes. It acts upon a product of operators that each depends on time, and returns the product of these operators sorted by time.

But, there is a problem with this definition. Since if the argument of the TOO is a Hilbert space operator (HSO), then the HSO itself "has no memory" of the time of which its terms were taken. In more mathematical words, if $T$ is the TOO and $A(t_1), B(t_2), C(t_3), D(t_4)$ are four different HSOs at times $t_1 < t_2 < t_3 < t_4$, and assume that $B(t_2)*A(t_1) = C(t_3)*D(t_4) = E$, and also that $D(t_4)*C(t_3) = F$ which is different from $E$. Then:

$$T\left\lbrace B(t_2)*A(t_1)\right\rbrace = B(t_2)*A(t_1) = E \Rightarrow T\left\lbrace E\right\rbrace = E $$

But on the other hand:

$$T\left\lbrace C(t_3)*D(t_4)\right\rbrace = D(t_4)*C(t_3) = F \Rightarrow T\left\lbrace E\right\rbrace = F $$

So there is a contradiction.

This can't be solved by making $T$ be a function not of a single HSO, but of $n$ HSOs and $n\!-\!1$ times. This is because then you won't be able to do any algebraic manipulations inside the argument of T (like performing the products of the HSOs) as is done in QFT, since each HSO would be in a different "slot" of the function $T$.

So my questions are: is this really a problem? is there a more rigorous definition of the TOO?

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    $\begingroup$ The time ordering is surely not a linear operator on a Hilbert space, nevertheless is, in its action, well-defined. $\endgroup$ – yuggib May 1 '15 at 7:04
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The time-ordered product is not an operation on operators. It is an operation on time-dependent operators. Let $\cal{O}$ denote the algebra of linear operators on a Hilbert space $\cal{H}$. If $A, B: [0, T] \to \cal{O}$ are two time-dependent operators, we can define an operator $$ AB: [0,T]^2 \to \mathcal{O}, \ \ (AB)(t_1, t_2) = A(t_1) B(t_2) $$ which depends on two times $t_1, t_2$. Similarly we can form the operator $BA$, $$ (BA)(t_1, t_2) = B(t_1) A(t_2) $$ which depends on two times as well. In general these don't agree. Now, there is a third product we can form, the time-ordered product $T(AB)$, defined by $$ (T(AB))(t_1, t_2) \left\{ \begin{array}{rr} (AB)(t_1, t_2), & t_1 > t_2 \\ (BA)(t_2, t_1), & t_1 < t_2 \end{array} \right.$$ which is again a function of two times $t_1, t_2$ (but in principle is undefined for $t_1=t_2$ unless $A$ and $B$ commute at equal times). The extension to products of $n$ time-dependent operators is straightforward (it will be an operator depending on $n$ times $t_1, \dots, t_n$).

The point is, both the input and the output of the time-ordering operation are time-dependent operators. The variables $t_1, t_2$, etc. should be treated as indeterminates. Failing to do so is exactly the mistake made in deriving the "contradiction" in your question.

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  • $\begingroup$ but this is not a complete definition as it doesn't specify the corresponding distribution in time. it is the latter which causes problems with the naive applications of time ordering. The question asked for a rigorous definition. $\endgroup$ – Arnold Neumaier May 7 '17 at 13:34
  • $\begingroup$ What's the definition of $[0,T]$? $\endgroup$ – WillG Apr 12 '19 at 3:12
  • $\begingroup$ Is that the interval of real numbers between $0$ and some time $T$, or some sort of bracket notation involving $T$ as the time ordering symbol? Confused. $\endgroup$ – WillG Apr 12 '19 at 3:20
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A rigorous version of the time-ordered product is given in the Epstein-Glaser causal approach to quantum field theory. The key to getting rid of difficulties is to make sure that distributions are always multiplied only with sufficiently regular expressions to make sure that the product is well-defined. See, e.g., Pinter's article Finite renormalizations in the Epstein Glaser framework and renormalization of the S-matrix of $\Phi^4$-theory.

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