Time ordering operator identity

Question

In Ref. 1, the author states that:

Making use of the fact that in a chronological product factors with different time arguments on the path $C$ may be commuted freely, application of the group property $(5.15)$ yields

$$ T \prod_{j=1}^n A_j(t_j)=S(0,t_+) \,T\left\{S(t_+,t_-)\, \prod_{j=1}^n A_j^o (t_j)\right\}\,S(t_-,0) \tag{6.15} \quad .$$

Here, $T$ denotes the time-ordering symbol, $S$ is the time-evolution operator in the interaction picture and $A$, $A^o$ are operators in the Heisenberg and interaction picture, respectively. It also holds that $t_+>t_j>t_-$ for all $j=1,2,\ldots, n$.

Other relevant equations are

$$ A(t)=S(0,t)\,A^o(t)\,S(t,0) \tag{6.14} $$

and

$$ S(t,t^\prime)\,S(t^\prime,t_0)=S(t,t_0)=S^\dagger(t_0,t)\quad,\quad S(t,t)=1\tag {5.15} \quad .$$

Question: Is equation $(6.15)$ really true without further assumptions or specifications?

Every attempt I tried to so far in proving the equation also resulted in the appearance of operators evaluated at equal times inside the time-ordering, which however is not defined.

For example, let $n=1$. Then

$$S(t_+,0)\,A(t)\,S(0,t_-)\overset{(6.14)}{=} S(t_+,0)\,S(0,t)\,A^o(t)\,S(t,0)\,S(0,t_-)\overset{(5.15)}{=}S(t_+,t)\,A^o(t)\,S(t,t_-) \quad . $$

It remains to show that the above right hand side equals $T\{S(t_+,t_-)\,A^o(t)\}$. If the time-ordering would swap $S(t,t_-)$ with $A^o(t)$, then the equality would hold. However, I don't see why this should be possible at all; for example, allowing such change of order would also imply that $$A(t)=TA(t)=T\{S(0,t)\,A^o(t)\,S(t,0)\}=T A^o(t)=A^o(t)\quad ,$$ which is obviously non-sense.

Despite that, I don't know how to understand $(6.15)$ exactly, in particular how to understand the appearance of $S(t_+,t_-)$ in the time-ordering. I guess one has to respect the fact that $S$ is a series, where each order contains operators of all times in $(t_+,t_-)$, but the author does not clarify at all.

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References:

1. A Course on Many-Body Theory Applied to Solid-State Physics. Charles P. Enz. Lecture Notes in Physics- Vol. 11. World Scientific. Chapter 2, section 6, p. 37.

Answer 1

OP presumably already know this but let us stress the following points:

1. Firstly, $S(t_2,t_1)$ is the time-evolution operator in the interaction picture. It satisfies a Schrödinger equation wrt. each end point $t_1$ and $t_2$, cf. eq. (5.20) and e.g. this related Phys.SE post.

2. Secondly, the time-ordering symbol $T$ heuristically acts on a continuum of operators inside $S(t_2,t_1)$ at various times inbetween $t_1$ and $t_2$; not just at the endpoints $t_1$ and $t_2$.

3. Thirdly, it is assumed that $S(t_2,t_1)\to {\bf 1}$ as $t_2-t_1\to 0$. Hence we should be able to ignore almost-equal-time issues.

4. Fourthly, by using the group property (5.15) we can cut $S(t_+,t_-)$ on the RHS of eq. (6.15) in so many pieces that we explicitly can arrange the RHS of eq. (6.15) in time order. E.g. for $n=2$ and $t_+>t_2>t_1>t_-$, the RHS of eq. (6.15) reads $$ S(0,t_+) S(t_+,t_2) A_2^o(t_2) S(t_2,t_1) A_2^o(t_1) S(t_1,t_-) S(t_-,0) .$$ Note that in this explicit expression the time-ordering symbol $T$ is no longer needed.

Answer 2

The comments in @Qmechanic's answer are all correct and good; I will try to provide a more concrete answer to the specific question.

Starting with your $n=1$ example, all that remains to show is $$ S(t_+, t) A^\circ(t) S(t, t_-) = \mathcal T\{ S(t_+, t_-) A^\circ(t) \} . \tag 1 $$ I would say that this equality is true by definition of the right hand side.

Time-ordering is an operation acting on formal strings of symbols. Unfortunately, this fact and its consequences are usually not made very clear. When doing algebraic manipulations withing the time ordering, one must be careful at each step to keep in mind what symbols the time ordering is acting on.

Coming back to your example, $S(t_+, t_-)$ is just a shorthand for a complicated expression involving many operators $\mathcal H_{\text{int}}^\circ(\tau)$ evaluated at some times $\tau \in [t_+, t_-]$. It is these $\mathcal H_{\text{int}}^\circ(\tau)$ that the time ordering refers to. Instead of a Dyson series, in practice I prefer thinking about an object like $S(t_+, t_-)$ in terms of a Trotter product: $$ S(t_+, t_-) = \lim_{k \to \infty} e^{-i \mathcal H_{\text{int}}^\circ(t_k)\, \Delta t} \cdots e^{-i \mathcal H_{\text{int}}^\circ(t_1)\, \Delta t}\, e^{-i \mathcal H_{\text{int}}^\circ(t_0)\, \Delta t} $$ where $\Delta t = (t_+ - t_-) / k$ and $t_i = t_- + i\Delta t$. Plugging this expression into (1), the time ordering inserts the $A^\circ(t)$ in between two of these "slices". As $k$ goes to infinity, each single slice approaches the identity; therefore it does not matter how the sorting is done if one of these $t_i$ exactly hits $t$ sometimes. (Compare QMechanic's third comment.)

In your "nonsense" example starting with $A(t) = \mathcal T A(t)$, the time ordering only acts on that time $t$ and not on those $\mathcal H_{\text{int}}^\circ(\tau)$ "hidden" inside the $S$'s that you introduce.