In the book Quantum field theory and the standard model from Schwartz, it is written on page 87 some results using time ordering operator.

We have the following operator:

$$ U(t,t_0)=T \exp\biggl(-\mathrm i \int_{t_0}^t V_I(u)\, \mathrm{d}u\biggr)$$

It is said the following things:

7.2.2 $U$ relations

It is convenient to abbreviate $U$ with $$U_{21} \equiv U(t_2, t_1) = T\biggl\{\exp\biggl[-t\int_{t_1}^{t_2} \mathrm{d}t'\,V_I(t')\biggr]\biggr\}.\tag{7.46}$$ Remember that in field theory we always have later times on the left. It follows that $$\begin{align} U_{21}U_{12} &= 1, \tag{7.47} \\ U_{21}^{-1} = U_{21}^{\dagger} &= U_{12} \tag{7.48} \end{align}$$ and for $t_1 < t_2 < t_3$ $$U_{32} U_{21} = U_{31}. \tag{7.49}$$ Multiplying this by $U_{12}$ on the right, we find $$U_{31}U_{12} = U_{32}, \tag{7.50}$$

Ok, I don't understand their "proof" of (7.47) and (7.49).

Remember that in field theory we always have later times on the left. It follows that:

Is it really a proof of the equations below? I don't get it.

Also, to prove that I would write the exponential in series and reason order by order but is there a better way to prove it? Because it is not really immediate (I don't know if it is possible to find a nice way to prove it).


3 Answers 3


The time-evolution operator in the interaction picture can be written as: $$U(t,t_{0})=e^{iH_{0}t}e^{-iH(t-t_{0})}e^{-iH_{0}t_{0}}$$ Using this we can write: $$U(t_{1},t_{2})U(t_{2},t_{0})=e^{iH_{0}t_{1}}e^{-iH(t_{1}-t_{2})}e^{-iH_{0}t_{2}}e^{iH_{0}t_{2}}e^{-iH(t_{2}-t_{0})}e^{-iH_{0}t_{0}}=e^{iH_{0}t_{1}}e^{-iH(t_{1}-t_{0})}e^{-iH_{0}t_{0}}$$ So: $$U(t_{1},t_{2})U(t_{2},t_{0})=U(t_{1},t_{0})$$

From Tomonaga-Schwinger equation: $$i\partial_{t}U(t,t_{0})=H_{I}(t)U(t,t_{0})$$ We can write the time-evolution operator using the initial condition $U(t_{0},t_{0})=1$: $$U(t,t_{0})=1-i\int_{t_{0}}^{t}dt_{1}H_{I}(t_{1})U(t_{1},t_{0})$$ By iteration, we obtain that: $$U(t,t_{0})=1+(-i)\int_{t_{0}}^{t}dt_{1}H_{I}(t_{1})+(-i)^2\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}H_{I}(t_{1})H_{I}(t_{2})+(-i)^3\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}\int_{t_{0}}^{t_{2}}dt_{3}H_{I}(t_{1})H_{I}(t_{2})H_{I}(t_{3})+\dots$$ i.e., $$U(t,t_{0})=\sum_{i=0}^{\infty}(-i)^n\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}\dots \int_{t_{0}}^{t_{n-1}}dt_{n}H_{I}(t_{1})H_{I}(t_{2})\dots H_{I}(t_{n})$$

$$U(t,t_{0})=\sum_{i=0}^{\infty}\frac{(-i)^n}{n!}\int_{t_{0}}^{t}dt_{1}\int_{t_{0}}^{t_{1}}dt_{2}\dots \int_{t_{0}}^{t_{n-1}}dt_{n}\mathcal{T}\left(H_{I}(t_{1})H_{I}(t_{2})\dots H_{I}(t_{n})\right)$$ $$U(t,t_{0})=\mathcal{T}\exp\left(-i\int_{t_{0}}^{t}dt'H_{I}(t')\right)$$


A very intuitive way to think about the time-ordered exponential is $$ U_{ba} = T \exp\left(-\mathrm i \int_a^b V(t) \, \mathrm dt \right) = \lim_{N \to \infty} \mathrm e^{-\mathrm i\, V(t_N) \Delta t} \cdots\, \mathrm e^{-\mathrm i\, V(t_2) \Delta t}\, \mathrm e^{-\mathrm i\, V(t_1) \Delta t} . $$ This is valid for $b \geq a$. $\Delta t$ equals $\frac{b-a}N$ and $t_k = a + k\Delta t$. (I am not sure if this is the way it is defined in the book by Schwartz, but it makes sense that this would give the correct expression for the propagator.)

Your (7.49) is now immediately obvious (for Physicists ;)). To get (7.47), we need to understand that in $U_{ab}$ (for $b \geq a$) later times are, in fact, not on the left. Instead, $$ U_{ab} = \bar T \exp\left(-\mathrm i \int_b^a V(t) \, \mathrm dt \right) = \bar T \exp\left(\mathrm i \int_a^b V(t) \, \mathrm dt \right) = \lim_{N \to \infty} \mathrm e^{\mathrm i\, V(t_1) \Delta t} \cdots\, \mathrm e^{\mathrm i\, V(t_N) \Delta t} . $$ Here, $\bar T$ is anti-time-ordering. You immediately see $U_{ab} = U_{ba}^\dagger = U_{ba}^{-1}$. See for example here.

  • 1
    $\begingroup$ This is one of the cleanest ways to answer this question in full precision. I would only add that the reason that it is this way is that we are trying to take the Schrödinger equation $i\hbar |\partial_t\Psi\rangle=\hat H(t)|\Psi\rangle$ and solve it with $|\Psi\rangle=U(t,t_0)|\Psi_0\rangle$ so we get the explicit ordering $i\hbar\partial_tU(t,t_0)=\hat H(t)U(t,t_0)$ with new time evolution happening on the left of the operator. Maybe I'd add a statement about how in the interaction picture we have $\hat H_0,\hat H_1$ and form $U=[U_0(t,t_0)]^\dagger U_1(t,t_0).$ $\endgroup$
    – CR Drost
    Oct 6, 2017 at 1:55

Schwartz is being sloppy. Recall that the time-ordering operation is singular at coinciding space-time points, so his manipulations are, strictly speaking, far from justified. Your scepticism is not unexpected. But his equations are correct anyway, in spite of his hand-wavy proof.

A slightly more convincing reasoning is as follows:

Write $H=H_0+V$, and let $$ U(t,t_0)\equiv\mathrm e^{iH_0(t-t_0)}\mathrm e^{-iH(t-t_0)}\tag1 $$

It is now trivial to prove that $U(t,t_0)$ satisfies the same initial value problem as $$ \mathrm{Texp}\left(-i\int_{t_0}^tV_I(s)\mathrm ds\right)\tag2 $$ and therefore they agree as operators, $$ U(t,t_0)\equiv\mathrm e^{iH_0(t-t_0)}\mathrm e^{-iH(t-t_0)}\equiv \mathrm{Texp}\left(-i\int_{t_0}^tV_I(s)\mathrm ds\right)\tag3 $$

From the representation $(1)$ you should be able to prove the results claimed by Schwartz, without the need to manipulate time-ordered objects. This should allow you to prove his claims with some more rigour and confidence. I leave this to you.

Further reading: most of what you want to know is analysed in exercise 9.5 in Srednicki's book on QFT. You can find the detailed worked-out solution online.


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