The composition property of the time-evolution operators

Question

For time-dependent Hamiltonians in Quantum Mechanics and QFT, we define the time-evolution operator as a unitary operator $U (t_2, t_1)$ such that $$ \tag{1} |\psi (t_2) \rangle = U (t_2, t_1) |\psi (t_1) \rangle. $$ From this definition and the time-dependent Schrodinger equation, it follows that \begin{align} \tag{2} U (t_2, t_1) = \begin{cases} \mathcal{T} \exp \Biggr[-i \int_{t_1}^{t_2} d\tau H (\tau) \Biggr], & t_2 \geq t_1 \\ \mathcal{AT} \exp \Biggr[-i \int_{t_1}^{t_2} d\tau H (\tau) \Biggr], & t_1 \geq t_2 \end{cases} \end{align} where $\mathcal{T}$ and $\mathcal{AT}$ are symbols for time-ordering and anti-time-ordering respectively.

Now, intuitively (and also directly from (1)), the time-evolution operators must satisfy the following composition property: $$ \tag{3} U (t_3, t_2) U (t_2, t_1) = U (t_3, t_1). $$ where no ordering has been chosen for $t_1, t_2$ and $t_3$.

Still in almost all the references that I have referred to, it is said that (3) is only valid when we choose $t_3 \geq t_2 \geq t_1$. So, is it not the case that (3) holds in general for any three times?

Also, is there an explicit proof of (3) which is just based on (2) and doesn't depend on the defining property of $U (t_2, t_1)$ in (1)?

EDIT: After reading a few other answers on this site, I have tried to provide a detailed answer to these questions below.

Answer 1

1. Yes, the group property (3) holds for any order of $t_1,t_2,t_3$.

2. The proof follows from $$U(t_2,t_1)^{-1}~=~ U(t_1,t_2) \tag{A}$$ and the time-ordered version of eq. (3).

3. Proof of eq. (A): We may wlog. assume $t_2\geq t_1$. Then $$\begin{align}U(t_2,t_1)^{-1}~\stackrel{(2)}{=}~&\left[T\exp\left[-\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right]\right]^{-1}\cr ~=~&AT\exp\left[\frac{i}{\hbar}\int_{t_1}^{t_2}\! dt~H(t)\right]\cr ~=~&AT\exp\left[-\frac{i}{\hbar}\int_{t_2}^{t_1}\! dt~H(t)\right]\cr ~\stackrel{(2)}{=}~&U(t_1,t_2). \end{align}\tag{B}$$ $\Box$

4. The reason many textbooks only list the time-ordered version of eq. (3) is presumably because they don't bother to provide eq. (2) in the anti-time-order case.

5. Related: The formal solution of the time-dependent Schrödinger equation

Answer 2

Since proving it requires knowing facts that have been mentioned in answers to some other questions, I decided to write a single answer which includes everything relevant to the proof.

For any three times $\{t_1, t_2, t_3\}$, we can convert $U (t_3, t_2) U (t_2, t_1) = U (t_3, t_1)$ to an equivalent equation such that the first argument of each $U$ is greater than the second. For example, if $t_2 \geq t_1 \geq t_3$, then $U (t_3, t_2) U (t_2, t_1) = U (t_3, t_1) \iff (U (t_3, t_2) U (t_2, t_1))^\dagger = U^\dagger (t_3, t_1) \iff U (t_1, t_2) U (t_2, t_3) = U (t_1, t_3) \iff U (t_2, t_3) = U (t_2, t_1) U (t_1, t_3)$.

So, w.l.o.g. we can restrict our attention to the expression \begin{equation*} U (t_3, t_2) U (t_2, t_1) = U (t_3, t_1) \end{equation*} where $t_3 \geq t_2 \geq t_1$. Because of this ordering in time, \begin{equation*} \begin{aligned} U (t_3, t_2) U (t_2, t_1) &= \mathcal{T} \Biggl[\exp \Biggl(-i \int_{t_2}^{t_3} d \tau \ H (\tau) \Biggl) \Biggr] \mathcal{T} \Biggl[\exp \Biggl(-i \int_{t_1}^{t_2} d \tau^\prime \ H (\tau^\prime) \Biggr) \Biggr] \\ &= \sum_{n, m} \frac{(-i)^{n+m}}{n! m!} \int_{t_2}^{t_3} d \tau_1 \cdots \int_{t_2}^{t_3} d \tau_n \int_{t_1}^{t_2} d \tau^\prime_1 \cdots \int_{t_1}^{t_2} d \tau^\prime_m \ \mathcal{T} [H (\tau_1) \cdots H (\tau_n)] \\ & \hspace{10cm} \mathcal{T} [H (\tau^\prime_1) \cdots H (\tau^\prime_m)] \\ &= \sum_{n, m} \frac{(-i)^{n+m}}{n! m!} \int_{t_2}^{t_3} d \tau_1 \cdots \int_{t_2}^{t_3} d \tau_n \int_{t_1}^{t_2} d \tau^\prime_1 \cdots \int_{t_1}^{t_2} d \tau^\prime_m \\ & \hspace{7cm} \mathcal{T} [H (\tau_1) \cdots H (\tau_n) H (\tau^\prime_1) \cdots H (\tau^\prime_m)] \\ &= \mathcal{T} \Biggl[ \sum_{n, m} \frac{(-i)^{n+m}}{n! m!} \int_{t_2}^{t_3} d \tau_1 \cdots \int_{t_2}^{t_3} d \tau_n \int_{t_1}^{t_2} d \tau^\prime_1 \cdots \int_{t_1}^{t_2} d \tau^\prime_m \\ & \hspace{7cm} H (\tau_1) \cdots H (\tau_n) H (\tau^\prime_1) \cdots H (\tau^\prime_m) \Biggr] \\ &= \mathcal{T} \Biggl[\exp \Biggl(-i \int_{t_2}^{t_3} d \tau \ H (\tau) \Biggl) \exp \Biggl(-i \int_{t_1}^{t_2} d \tau^\prime \ H (\tau^\prime) \Biggr) \Biggr] \end{aligned} \end{equation*} where in the third equation we have used a property of the time-ordering operation discussed here.

Now, we can use the BCH formula to write \begin{align*} \exp \Biggl(-i \int_{t_2}^{t_3} d \tau \ H (\tau) \Biggl) \exp \Biggl(-i \int_{t_1}^{t_2} d \tau^\prime \ H (\tau^\prime) \Biggr) \\ &= \exp \Biggl(-i \int_{t_2}^{t_3} d \tau \ H (\tau) -i \int_{t_1}^{t_2} d \tau^\prime \ H (\tau^\prime) + C (t_1, t_2, t_3) \Biggr) \\ &= \exp \Biggl(-i \int_{t_1}^{t_3} d \tau \ H (\tau) + C (t_1, t_2, t_3) \Biggr) \end{align*} where $C (t_1, t_2, t_3) \equiv -\frac{1}{2} \int_{t_2}^{t_3} d \tau_1 \int_{t_1}^{t_2} d \tau_2 [H (\tau_1), H (\tau_2)] + \text{terms with nested commutators}$.

Therefore, \begin{align*} U (t_3, t_2) U (t_2, t_1) &= \mathcal{T} \Biggl[\exp \Biggl(-i \int_{t_1}^{t_3} d \tau \ H (\tau) + C (t_1, t_2, t_3) \Biggr) \Biggr] \\ &= \sum_n \frac{1}{n!} \mathcal{T} \Biggl[ \Biggl(-i \int_{t_1}^{t_3} d \tau \ H (\tau) + C (t_1, t_2, t_3) \Biggr)^n \Biggr] \end{align*} The time-ordering operation on a product of operators which involves a commutator makes it vanish. Therefore, the only terms in the above equation which survive don't involve any commutators (See this question for more details). This means that \begin{align*} U (t_3, t_2) U (t_2, t_1) &= \sum_n \frac{1}{n!} \mathcal{T} \Biggl[ \Biggl(-i \int_{t_1}^{t_3} d \tau \ H (\tau) \Biggr)^n \Biggr] \\ &= \mathcal{T} \Biggl[\sum_n \frac{1}{n!} \Biggl(-i \int_{t_1}^{t_3} d \tau \ H (\tau) \Biggr)^n \Biggr] \\ &= \mathcal{T} \Biggl[\exp \Biggl(-i \int_{t_1}^{t_3} d \tau \ H (\tau) \Biggr) \Biggr] \\ &= U (t_3, t_1) \end{align*}