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

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


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

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.


  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.
  • $\begingroup$ So the question is mainly about what happens at exactly equal times? $\endgroup$
    – Qmechanic
    Commented Dec 12, 2023 at 11:23
  • $\begingroup$ @Qmechanic Yes and no. The book has not discussed that; the author, a few pages earlier, also explicitly mentions that this case is undefined. (BTW: I think a very large part of the book is accessible via Google books, if someone is interested). But even if we can define this, I don't see how things like $S(t,t^\prime)$ inside the time ordering makes sense a priori, because these are functions of two time-arguments (I could imagine that one treats them as a series, where each term contains operators for all times inbetween $t,t^\prime$). $\endgroup$ Commented Dec 12, 2023 at 11:30

2 Answers 2


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.

  • $\begingroup$ But I still don't fully understand yet why we can ignore these things. As I tried to argue in my question, once we "ignore almost-equal-time issues", we can get into trouble very quickly. Can we instead of e.g. $(5.15)$ use an approximate identity, i.e. with some $\epsilon$-prescription, to circumvent these things? I tried but it did not fully work out. $\endgroup$ Commented Dec 12, 2023 at 14:33
  • $\begingroup$ Let me rephrase my previous first comment: Thanks for your answer. Could you elaborate point 4 a bit more, please? I thought about this: For an arbitrary ordering of the times such that $t_{\pi(j)}\geq t_{\pi(j+1)}$, we can write $S(t_+,t_-)=\prod_{j=0}^n S(t_{\pi(j)},t_{\pi(j+1)})$, where $t_{\pi(0)}:=t_+$ and $t_{\pi(n+1)}:=t_-$. Then $T\{S(t_+,t_-) \prod_{j=1}^n A^o(t_j)\} = \prod_{j=0}^n S(t_{\pi(j)},t_{\pi(j+1)})A^o(t_{\pi(j)})$ if we really ignore the equal time issues. This then evaluates to $S(t_+,0)T\{\prod_{j=1}^n A(t_j)\}S(0,t_-)$ and we are done. Is this what you mean with point 4? $\endgroup$ Commented Dec 12, 2023 at 15:06
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Dec 12, 2023 at 15:17
  • $\begingroup$ Thank you for your effort. Yes, that agrees with the result of my latest comment and this is also what I tried to indicate in my question (although I only considered $n=1$). But as I also tried to explain in my question and comment above, as far as I can see, the issue with equal times can lead to various wrong conclusions, and hence, at least for me, everything we are discussing here feels like cheating a bit. $\endgroup$ Commented Dec 12, 2023 at 15:32
  • $\begingroup$ Actually, a (more or less) related question came into my mind: In several posts here, you nicely pointed out that time ordering is not a super operator, but a map from symbols to operators. Then why are we actually allowed to use the group property of the time-evolution operator inside the time-ordering? Other algebraic relations, such as commutation relations, cannot be replaced. Thanks, again, in advance. $\endgroup$ Commented Dec 12, 2023 at 19:50

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.

  • $\begingroup$ Hi. Thanks for your answer. Could you elaborate your last paragraph a bit? Why does this invalidate my example? As far as I can see, the difference to eq. $(1)$ is that there in both time-evolution operators, there are operators at times in $[0,t]$ and so $A$ is evaluated at a time always equal or larger than any operator inside both $S$ (if $t>0$). Hence, it should be ordered to the left, yielding my example. What am I missing? Edit: My argument, however, then assumes that one can make the use of the group property inside the time-ordering, which I think, however, is also implicit in your ans $\endgroup$ Commented Dec 13, 2023 at 8:22
  • $\begingroup$ One way of thinking about it would be that the step $TA(t) = T\{ S(0, t) A^\circ(t) S(t, 0) \}$ is not allowed. Another way of thinking would be that it is allowed, but on the right hand side the expression "$S(0, t) A^\circ(t) S(t, 0)$" remains a "unit", the time ordering doesn't suddenly start moving the constituents of this expression around. This is what I mean when I say, one needs to keep in mind what symbols the time-ordering is acting on. Put differently, time-ordering is a very context-sensitive notation. Its definition, and what you are allowed to do, differ from case to case. $\endgroup$
    – Noiralef
    Commented Dec 13, 2023 at 9:23
  • $\begingroup$ Yes, that is also what I thought and what I wanted to illustrate with my example. OTOH, if you read the book and the other answer, it seems that we indeed make use of the group property over and over again. Hence my question: Why in this situation it is allowed and in another it is not? Formally, I think, using the Trotter expansion, you end up with the same result, no? I.e. you are using the Trotter expansion, but in the end it is really the same, i.e. you end up with different parts of the evolution operator with $A$s in between... I am very confused atm... $\endgroup$ Commented Dec 13, 2023 at 9:41
  • $\begingroup$ @TobiasFünke There may be operations that you sometimes can perform inside a time-ordering and sometimes you cannot. That is not a contradiction, since it is just a notational tool. Any expression involving $\mathcal T$ has a meaning, which is an expression without any $\mathcal T$s. --- Sometimes there are situations where I already know all there is to know and just need a few nights' sleep until I can accept that everything really works. Might be one of these situations? :-) Otherwise, try to point out a specific equation that you don't believe. $\endgroup$
    – Noiralef
    Commented Dec 13, 2023 at 11:33
  • $\begingroup$ Well, yes, I think this is such a situation. But for me it is important to see that you do not run into incompatible applications of definitions/rules/conventions... But thank you anyway so far! $\endgroup$ Commented Dec 13, 2023 at 12:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.