I have met the time ordering operator $T$ in many places, such as in the Dyson series

$$U(t) = T\exp{\left(-\dfrac{i}{\hbar}\int_0^tdt'H(t')\right)},$$

or in the definition of single particle Green's function

$$iG(t,t') = \langle T\psi(\textbf{x},t)\psi^\dagger(\textbf{x}',t')\rangle.$$

In these nice lecture notes ,written by Professor Kai Sun ,he listed some reasons to use the time ordering operator:

  • A trick to get delta functions in the equation of motion of Green's function;
  • Path integral leads to $T$ naturally;
  • The evolution operator $U(t) = T\exp[i\int dt H(t)]$(Dyson series);
  • With $T$,bosons and fermions are unified together.Same theory with two differnt boundary conditions.

$\textbf{So my question is}$ :is it just a mathematical trick for time ordering operator $T$ in quantum mehcanics ? Does it have a more profound physical meaning?For example,is there any relation to the time-reversal-symmetry?

  • 1
    $\begingroup$ Hi Jack. Your title should emphasize on what you are asking. You have provisions to add tags on the subject of your queston. $\endgroup$
    – UKH
    Commented Nov 10, 2016 at 12:15
  • $\begingroup$ Hi, Unnikrishnan.Thanks for your nice implications,I will learn and improve my skills.I am also glad to join this interesting community. $\endgroup$
    – Jack
    Commented Nov 16, 2016 at 1:20
  • $\begingroup$ Here are some related questions and answers: physics.stackexchange.com/q/173603, physics.stackexchange.com/a/56217/55689 $\endgroup$
    – Elliot Yu
    Commented Dec 1, 2016 at 0:48
  • 1
    $\begingroup$ Prof. Mark Srednicki (of the QFT textbook) once told me that in the context of QFT, he mostly views it as a technical trick that facilitates translating between the Hamiltonian operator and Lagrangian path-integral formalisms, without much physical significance. $\endgroup$
    – tparker
    Commented Dec 2, 2016 at 21:12
  • 1
    $\begingroup$ @tparker That can't be right (unless I misunderstand the question). A trick is something you don't have to use, but you do because it is convenient. I'm pretty sure that omitting the time ordering is simply wrong. For example, the time evolution operator wouldn't satisfy its defining equation without it.... $\endgroup$
    – QuantumDot
    Commented Dec 5, 2016 at 2:22

2 Answers 2


The time ordering enters as a consequence of the definition of the Hamiltonian as the generator of time translations. In the Schödinger picture: $$|\psi(t)\rangle \approx \left(1 - \frac{i}{\hbar} H(t') [t - t'] + \mathcal{O}([t-t']^2)\right) |\psi(t')\rangle,$$ where the relationship becomes exact in the limit as $t-t' \rightarrow 0^+$. It's an exercise in Lie group theory to apply multiple time translations, in order, to get: $$|\psi(t)\rangle = \lim_{N\rightarrow \infty} \prod_{j=1}^N\left(1 - \frac{i}{\hbar} H\left(t' + \frac{j}{N}[t-t']\right)\left[\frac{t-t'}{N}\right] \right) |\psi(t')\rangle,$$ which is another way of writing the Dyson series. If the Hamiltonians at different times all commute with one another, then the Dyson series becomes an ordinary exponential.

To reiterate: the time ordering just enters in naturally as a process of stepping from the start time to the end time one step at a time.

That a Green's function satisfies the equation: $$iG(t,t') = \langle T \psi(t) \psi(t') \rangle,$$ follows as a consequence of the above and actual definition of the Green's function. The Green's function is defined by the equations of motion of the free part of the theory. If the free equations of motion are: $$L \psi(t) = 0,$$ for some linear differential operator $L$, then the defintion of $G(t,t')$ is: $$ L G(t,t') = \delta(t - t').$$ In other words, it's the response of the $\psi$ field to a unit impulse (at a point) in the classical limit.

There is more than one Green's function, though, because you can add any solution to the homogeneous equation in the region of interest: $$ L G_0(t) = 0,$$ to get another Green's function. Because of that, there are multiple Greens functions possible. The most frequently used are the 'retarded' (causal) Green's function that satisfies $$ G( \mathbf{x}, t; \mathbf{x}', t') = 0$$ whenever $( \mathbf{x}, t)$ is outside of the forward directed light cone with apex at $(\mathbf{x}', t')$. The 'advanced' Green's function is identical, with the orientation of the light cone flipped to the opposite direction. The time ordered propagator is also known as the Feynman propagator, and it is symmetric under time reversal.


Time ordering is a feature of solving for $U(t)$ as an integral equation using successive iterations. It is not limited to quantum mechanics and some version of it occurs in classical Hamiltonian mechanics, for instance in the Lie transform approach to perturbation theory, as detailed in

  • J. R. Cary, Phys.Rep. 79 (1981) 129 (section 2.2)
  • Ernesto Corinaldesi's Classical Mechanics (section 9.2).

It is a trick to the extent that the solution by successive approximation is a trick, but since this is a feature of this type of approach to solving for $U(t)$ and not limited to quantum mechanics, it is unlikely to be tied at some fundamental level to any quantum symmetry. Of course the twist in QFT is that time ordering for the evolution forces an ordering of operators, thereby linking spin-statistics theorems time ordering.


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