Let $T$ be the time-ordering operator which orders operators $A_1(t_1), A_2(t_2), \ldots$ such that the time parameter decreases from left to right:

$$T[A_1(t_1) A_2(t_2)] = A_2(t_2) A_1(t_1) \text{ if } t_2 > t_1 \text{ and }= A_1(t_1)A_2(t_2) \text{ otherwise. } $$

The time $t_i$ does not have to be a physical time, it can also be an imaginary time, etc.

Question: I would like to know why the following equation holds: for $t_i \leq t_1, t_2 \leq t_f$ it holds that $$T\left[A_1(t_1) A_2(t_2) \exp\left(-i\int_{t_i}^{t_f}H(t) dt\right)\right] \\ = T\left[\exp\left(-i\int_{t_{\pi_1}}^{t_f}H(t) dt\right)\right] A_{\pi_1}(t_{\pi_1}) \cdot T\left[\exp\left(-i\int_{t_{\pi_2}}^{t_{\pi_1}}H(t) dt\right)\right] A_{\pi_2}(t_{\pi_2}) \cdot T\left[\exp\left(-i\int_{t_i}^{t_{\pi_2}}H(t) dt\right)\right] ,$$ where $\pi$ is a permutation such that the times are ordered.

I encountered this equation in Negele & Orland (1998) in eq. (2.49) on p. 63 and in eq. (2.67b) on p. 70, where they split the integral

$$\int_{t_i}^{t_f} dt = \int_{t_i}^{t_{\pi_2}} dt + \int_{t_{\pi_2}}^{t_{\pi_1}} dt + \int_{t_{\pi_1}}^{t_f} dt$$

and used the time-ordering. It appears in calculations of greens functions respectively correlation functions.

I tried to prove this equation in an elementary way by using

$$ T\left[\exp\left(-i\int_{t_i}^{t_f}H(t) dt\right)\right] = 1 + \sum_{n=1}^\infty \frac{(-i)^n}{n!} \int_{t_i}^{t_f} d\tau_1 \ldots \int_{t_i}^{t_f} d\tau_n T \left[ H(\tau_1) \ldots H(\tau_n) \right]$$

[cf. eq. (2.10) on p. 50] and applying the $T$-operator on the expression, but I did not succeed yet. If someone can show me a valid proof or point out some literature where it is proven, I'd be thankful.


3 Answers 3



$\int_0^t \int_0^{t_1} dt_1 dt_2 \, a(t_1) a(t_2) = \frac{1}{2!} \int_0^t\int_0^t dt_1 dt_2\, \mathcal{T}\{ a(t_1) a(t_2) \}$

and so forth. You can see this by noting that the (square) integration region in the second integral can be split up into two triangular integration regions like in the first integral. This is one way to define the time-ordered integration.

Edit: A bit more to your question. You can consider

$\mathcal{T} \left\{ e^{\int_0^t\!dt\, H(t)} \right \}$

to be the product

$\mathcal{T} \left\{ e^{H(0) \Delta t} \, e^{H(\Delta t) \Delta t} \, e^{H(2\Delta t) \Delta t} \dots e^{H(t) \Delta t}\right \}$ where we have turned the integral into a Riemann sum and factored exponential of a sum into a product of exponentials. Now, once you add another object into the time ordering, e.g. $A(t_1)$, it just slides past as many of these items in the product until it finds it's right spot. E.g.

$\mathcal{T} \left\{ e^{H(0) \Delta t} \, e^{H(\Delta t) \Delta t} \, e^{H(2\Delta t) \Delta t} \dots A(t_1) \dots e^{H((N-1)\Delta t)} e^{H(t) \Delta t}\right \}$

Then everything that's left (everything on either side of $A(t_1)$ can again be factored into exponentials of sums, i.e., exponentials of integrals. Repeat as necessary.

I hope that helps.


Hints to the question (v1):

  1. Recall that the operator time ordering is symmetric $$\tag{1}T[A_1(t_1)\ldots A_n(t_n)]~=~T[A_{\pi(1)}(t_{\pi(1)})\ldots A_{\pi(n)}(t_{\pi(n)})], $$ where $\pi\in S_n$ is a permutation. (Here we assume for simplicity that all operators are Grassmann-even. Else there will be additional sign factors.)

  2. Recall that if $t_1> \ldots >t_n$, then the operator time ordering is defined $$\tag{2}T[A_1(t_1)\ldots A_n(t_n)]~:=~A_1(t_1)\ldots A_n(t_n).$$

  3. It becomes a bit technical to explain and work with the time ordering rule $T[A_1(t_1)\ldots A_n(t_n)]$ when a subset of the times $t_1,\ldots, t_n$, happens to be exactly equal. Of course in this case, the operators $A_1(t_1)\ldots A_n(t_n)$ should be symmetrized in an appropriate sense.

  4. Extend the definition of time-ordering $T[A_1(t_1)\ldots A_n(t_n)]$ by multi-linearity.

  5. To avoid that technical point 3, let us discretize time. More precisely, let us assume that OP's three different operators $H(t)$, $A(t)$, and $B(t)$ live on three different time discretizations, such that two different operators are never taken at exactly the same instant. (That e.g. a power of $H(t)$ appears in the same time $t$ does not matter, since $H(t)$ commutes with itself, so we can ignore the symmetrization procedure from point 3 without introducing errors.) In that way we can easily time-order any operator expression of $H(t)$, $A(t)$, and $B(t)$, only knowing the un-equal-time ordering rules (1)-(2).

  6. In OP's sought-for identity, replace time integrations $\int\! dt$ with appropriate discrete sums $\sum$ of operators that live on their respective sub-lattices of time. The corresponding discretized version of OP's sought-for identity becomes a trivial identity, since the LHS is defined as the RHS.

  7. At the end of the calculation, take the continuum limit where the lattice constant goes to zero, and summations becomes integrals again. Argue that the identity continues to hold.


I accepted lionelbrits' answer because I've already figured it out with his/her hint to "slide the operator past". I will write now a more detailed version for others who are interested in the identy but can't figure it out.

At first, I advice the reader to read Galindo & Pascual, Volume I, p. 70ff (2012) to see how the time-ordering is introduced and how lionelbrits arrives at his first equation (or general versions thereof).

Now, we will look at the expression with only one operator, $T\left[\exp\left(-i \int_{0}^{t}H(t') dt' \right) A(\tau)\right]$ and $t_i=0, t_f=t$. I think the argument also works for more operators, but it will become more opaque and cumbersome to write it in an online forum. The n-th order term of the above expression is of the form $$ \frac{(-i)^n}{n!} \int_0^t d\tau_1 \ldots \int_0^t d\tau_n T[H(\tau_1)\cdots H(\tau_n) A(\tau)] .$$

In order to apply the $T$-operator, we define $t_1 =\tau_1, \ldots, t_n=\tau_n, t_{n+1}=\tau$ and $A_1 = H(t_1), \ldots, A_n =H(t_n), A_{n+1}=A(t_{n+1})$. Then by the definition of $T$ we have $$T[A_1 \ldots A_{n+1}] = \sum_{\pi} \theta(t_{\pi_1}-t_{\pi_2}) \theta(t_{\pi_2}-t_{\pi_3})\ldots \theta(t_{\pi_n}-t_{\pi_{n+1}}) A_{\pi_1}\ldots A_{\pi_{n+1}} .$$ We do not care for the case of equal times, because the corresponding regions of integration have measure 0 (typical physicist argument). In the case that $t_{\pi_k} = \tau$, the integration is only non-vanishing in the region $t_{\pi_1} > t_{\pi_2} > \ldots > \tau > t_{\pi_{k+1}} > \ldots > t_{\pi_{n+1}}$. There are $n!$ such permutations for fixed $k$. Furthermore, for every addend in the sum (i.e. for every permutation $\pi$), we change the order of integration accordingly (we assume we can) and relabel the integration variables $d\tau_{\pi_1} \rightarrow d\tau_1$, etc. Thus we arrive at the following result for the n-th order term $$(-i)^n \left[ A(\tau) \int_0^t d\tau_1 \int_0^{\tau_1}d\tau_2\ldots \int_0^{\tau_{n-1}}d\tau_n H(\tau_1)\ldots H(\tau_n) \\ + \int_\tau^t d\tau_1 H(\tau_1) A(\tau) \int_0^{\tau}d\tau_2 \int_0^{\tau_2}d\tau_3 \ldots \int_0^{\tau_{n-1}}d\tau_n H(\tau_2)\ldots H(\tau_n) \\ + \int_{\tau}^t d\tau_1 \int_{\tau}^{\tau_1}d\tau_2 H(\tau_1) H(\tau_2) A(\tau) \int_0^{\tau}d\tau_3 \ldots \int_0^{\tau_{n-1}}d\tau_n H(\tau_3)\ldots H(\tau_n)\\ + \ldots \right] ~.$$ The factor $1/n!$ cancels because for each position of $A(\tau)$ there are $n!$ integrals which give the same result. We can simplify the expression using the time-ordering again and get $$ A(\tau) \frac{(-i)^n}{n!} \int_0^t d\tau_1 \ldots \int_0^t d\tau_n T[H(\tau_1) \ldots H(\tau_n)] \\ + \frac{-i}{1!} \int_\tau^t d\tau_1 H(\tau_1) A(\tau) \frac{(-i)^{n-1}}{(n-1)!} \int_0^\tau d\tau_2 \ldots \int_0^\tau d\tau_n T[H(\tau_2)\ldots H(\tau_n)] \\ + \frac{(-i)^2}{2!} \int_\tau^t d\tau_1 \int_\tau^t d\tau_2 T[H(\tau_1) H(\tau_2)] A(\tau) \frac{(-i)^{n-2}}{(n-2)!} \int_0^\tau d\tau_3 \ldots \int_0^\tau d\tau_nT[H(\tau_3)\ldots H(\tau_n)] \\ + \ldots .$$ Now, I think it is clear that this is one of the terms one gets if one calculates $T \left[\exp\left(-i \int_\tau^t H(t')dt'\right)\right] A(\tau) T \left[ \exp\left(-i \int_0^\tau H(t')dt'\right) \right]$. So if we sum the correct terms of this expression up, we get the n-th order term of our first expression. Summing everthing up gives the sought identy.


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