How to avoid paradoxes about time-ordering operation?

Question

(Original title: is time-odering operator a linear operator?)

I'm confused with two formulas, one of which is $$ \mathcal{T} \exp \left [-\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t' \hat{H}_I(t') \right] =\sum_{N=0}^\infty\frac{1}{N!} \left (-\frac{\mathrm{i}}{\hbar} \right)^N \int_{t_0}^t \mathrm{d} t_N \int_{t_0}^t \mathrm{d} t_{N-1} \cdots \int_{t_0}^t \mathrm{d} t_1 \mathcal{T} \hat{H}_I (t_N) \hat{H}_I(t_{N-1}) \cdots \hat{H}(t_1). $$ I'm clear about how the right hand side is derived following this page, but I don't understand how it goes to the left. Is that just a definition? Otherwise I have to expand the exponential to series and admit the time ordering operator is linear so that $\mathcal T(A+B)=\mathcal T(A)+\mathcal T(B)$. But I haven't seen any description on that.

The other one is $$ \mathcal T([A,B])=\mathcal T(AB-BA)=\mathcal T(AB)-\mathcal T(BA)=0 $$ whether $[A,B]$ is $0$ or not. To make the second equal sign hold, $\mathcal T$ is again required to be linear. But if it is linear, and $[A,B]$ is a constant, say $c\not =0$, $$ \mathcal T([A,B])=\mathcal T(c)=T(cI^2)=c\mathcal T(I^2)=c\not=0, $$ which is a contradiction.

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This answer points out that $\mathcal T$ maps on an operator-valued function of time in a space with more than one dimension, but no information on the linear property is mentioned.

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Later edited:

$[A,B]$ doesn't have to be a constant, in fact it's impossible because $A,B$ are operators dependent on different time variables. But the paradox survives if $[A,B] = c1\!\!1$, where $c=c(t,t')$ is a function of the time variables and $1\!\!1$ denotes the identity.

For example, commutator of the position and momentum in Heisenberg picture writes, $$ [x(t),p(t')]=\cos(t-t')=\cos t\cos t'+\sin t\sin t'=C(t)C(t')+S(t)S(t') $$ except the coefficients, where $$ C(t)=\cos t1\!\!1,\quad S(t)=\sin t1\!\!1. $$ Therefore (suppose $t>t'$) $$ 0=\mathcal T([x(t),p(t')]) =\mathcal T\left[C(t)C(t')+S(t)S(t')\right]= C(t)C(t')+S(t)S(t')=\cos(t-t'). $$ Contradiction again.

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I thought the point is, time-ordering operation on product of operators--namely, monomial--is well-defined, but operation on sum of these products, which is polynomial of operators, is not.

I thougth the polynomial can be devided into monomials that yields the same result, but according to the above example, that's not true.

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Thanks for answering! I'm not content with just regarding it as a notation, since I can't tell where paradoxes like above lie.

Answer 1

1. Time ordering $T$ (like any other operator ordering, such as, normal ordering $: ~:$, radial ordering ${\cal R}$, etc.) is technically speaking a linear map from symbols to operators, not a linear map from operators to operators, cf. e.g. this, this and this Phys.SE posts.

2. In particular, it is not allowed to use an operator relation (such as, e.g. a CCR) before applying the time ordering, cf. OP's apparent contradiction/paradox.

Answer 2

I think it's best to regard the time-ordering symbol as notation, rather than as an operator. When you see the time ordering symbol in time-ordered correlation functions, for example $\langle 0 | T \big[ \phi(x_1) \phi(x_2) \big] | 0 \rangle$, you should take this simply to be short-hand for $$ \langle 0 | T \big[ \phi(x_1) \phi(x_2) \big] | 0 \rangle \triangleq \Theta(x_1^0 - x_2^0) \langle 0 | \phi(x_1) \phi(x_2) |0 \rangle + \Theta(x_2^0 - x_1^0) \langle 0 | \phi(x_2) \phi(x_1) |0 \rangle $$ Similarly, in the time-ordered exponential, you extend $T$ to each element of the Taylor series by definition. This point is best understood by working through the derivation for the Dyson series, where you see that the time-ordering symbol is just short-hand for an expression involving lots of Theta functions.

The best way to avoid ambiguities or paradoxes is to not take the $T$ symbol too literally. Usually it's easy to guess in context what an author means when they write it, and you yourself should have a concrete expression in terms of Theta functions in mind before you use it. Then ambiguities like the commutator issue you raised never appear.

Answer 3

$$ \mathcal{T} \exp \left [-\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t' \hat{H}_I(t') \right] =... $$

Is that just a definition?

Seems like it must be, yes. Or perhaps better thought of as an abuse of notation.

Such an exponential formula never appears in Dyson's original 1948 paper, where he introduced (or popularized) the concept of the time ordering of operators. In his words:

"If $F_1(x_1), \ldots, F_n(x_n)$ are any operators defined, respectively, at the points x_1, \ldots x_n of space-time, then $P(F_1(x_1), \ldots, F_n(x_n))$ will denote the product of these operators, taken in the order, reading from right to left, in which the surfaces $\sigma(x_1),\ldots, \sigma(x_n)$ occur in time."

(He uses the symbol $P$ rather than $T$)

Otherwise I have to expand the exponential to series and admit the time ordering operator is linear...

But the example you gave shows that doesn't work to treat time-ordering as some ordinary linear function of operator arguments. Rather it is just a shorthand notation for "manhandling" (this is the language from Schwartz's QFT book) the operators into time order.