I'm studying classical molecular dynamics and have come across an object called the classical propagator in the following context.

Let $\mathcal{A}(t) = \mathcal{A}(\vec{x}(t))$ be a function on phase-space without explicit dependency on time $t$ and $\vec{x}(t)\equiv(q_{1}(t),\dots,q_{F}(t),p_{1}(t),\dots\cdot p_{F}(t))$. The time evolution of $\mathcal{A}$ can be expressed using the poisson bracket $\lbrace\dots,\dots\rbrace$ and the Liouville operator $i\mathcal{L}$ as

$\frac{\mathrm{d}}{\mathrm{d}t}\mathcal{A}(\vec{x}(t)) = \lbrace\mathcal{A},\mathcal{H}\rbrace = i\mathcal{L} \mathcal{A}$

where $\mathcal{H}$ is the Hamiltonian of the system. Now this expression is said to have a "formal solution":


The Liouville operator is the differential operator

$i\mathcal{L}=\sum_{\alpha=1}^{F}\left[\dot{q}_{\alpha}\frac{\partial}{\partial q_{\alpha}}+\dot{p}_{\alpha}\frac{\partial}{\partial p_{\alpha}}\right]$

How can I obtain a rigorous expression for this classical propagator $e^{i\mathcal{L}t}$? What actual operations are behind the action of this propagator on $\mathcal{A}(0)$?

I have tried to find an answer for the special case of a Hamiltonian $\mathcal{H}$ not explicitly dependent on time:

Expanding $\mathcal{A}(t)$ into a Taylor series in time arount $t=0$:

$\mathcal{A}(t)=\sum_{k=0}^{\infty} \frac{\mathcal{A}^{(k)}(0)}{k!}(t-0)^k = \mathcal{A}(0)+ \frac{1}{1!} \dot{\mathcal{A}}\vert_{t=0}\,t+\frac{1}{2!} \ddot{\mathcal{A}}\vert_{t=0}\,t^{2}+\ldots$

and realizing that $\mathcal{A}^{(k)}(t)=\left\{\left\{\mathcal{A}(t),\mathcal{H}\right\},\dots,\mathcal{H}\right\} = (iL)^{k}\mathcal{A}(t)$ we have $\mathcal{A}(t)=\sum_{k=0}^{\infty}(i L)^{k}\mathcal{A}(t)\big|_{t=0}\frac{t^{k}}{k!}$

This already looks a bit similar to the exponential series $e^{i L t}(\dots)=\sum_{k=0}^{\infty}{\frac{(i L t)^{k}(\dots)}{k!}}$, but I can't find a way to match the expressions.

Bonus question: What is the usefulness of the classical propagator in classical mechanics? Are there any examples for solving an actual problem with it in a more convenient fashion?


1 Answer 1


From the definition, $$ i {\cal L} f:=\{ f, H \}= \dot f, $$ for every $f$ in the domain of the operator.

Therefore, operator $e^{it {\cal L}}$, defined by the formal series of operators $$ 1+it {\cal L} + (it {\cal L})^2 + \dots = \sum_{n=0}^{\infty}\frac{(it {\cal L})^n}{n!} $$ when applied to a function $A$ at the time $t=0$, provides its time Taylor series: $$ e^{it {\cal L}} A(0) = A(0) + t\dot A(0) + \frac{t^2\ddot A(0)}{2} + \dots = A(t).$$

Notice that the similarity of the operator $e^{it {\cal L}}$ with the unitary time operator in quantum mechanics is only formal. Actually, the Liouville operator is not a Hermitian operator, and $e^{it {\cal L}}$ is not unitary (the classical time evolution does not preserve the norm of an observable).

What is the usefulness of the classical propagator in classical mechanics? Are there any examples for solving an actual problem with it more conveniently?

There are two main fields where the classical propagator plays a role as a convenient tool for formal manipulation of time evolution.

  1. General analysis, and approximate theories, of the time evolution in the phase space and its consequences on time correlation functions. For example, the Mori analysis of the evolution of time correlations in terms of memory functions was initially based on an ingenious projection technique based on the Liouvillian of the N-body system.

  2. The Liouvillian-based time-evolution operator is a convenient starting point for developing approximate numerical methods systematically. Most of the analysis of symplectic algorithms is nowadays based on techniques to split the time operator into a product of operators (see, for instance this link).

  • $\begingroup$ Thank you, GiorgioP. I find the step from the series of operators to the Taylor series confusing as I can't see where the evaluation of $A$ or its derivatives at $t=0$ is arising from. Does it follow from some implicit conventions? Consider the quadratic term, we have $(it\mathcal{L})^2 A(t) = t \lbrace t \lbrace A(t),\mathcal{H}\rbrace , \mathcal{H} \rbrace = t \lbrace t \dot{A}(t) , \mathcal{H} \rbrace = t^2 \lbrace \dot{A}(t),\mathcal{H}\rbrace = t^2 \ddot{A}(t)$ when evaluating the quadratic term of the operator series, but not $t^2 \ddot{A}(0)$. Could you maybe clarify on this? $\endgroup$ Commented Aug 9, 2023 at 13:55
  • $\begingroup$ @Christoph90 It arises from the equation saying that the Poisson parenthesis of a function with the Hamiltonian is the time derivative. Do you need additional elaboration? $\endgroup$ Commented Aug 9, 2023 at 14:49
  • $\begingroup$ I'm afraid, yes. But only on the question of how it happens that some part of the operand of $(it\mathcal{L})^k$ (namely $A(t)$ or its derivatives) suddenly is evaluated at $t=0$. I think I understand the rest. $\endgroup$ Commented Aug 9, 2023 at 16:02
  • $\begingroup$ @Christoph90 Time $t=0$ has no special meaning. One could start the Tylor series at time $t$, writing a Taylor series in the variable $\Delta t$ to get $A(t+\Delta t )$ $\endgroup$ Commented Aug 9, 2023 at 16:07
  • 1
    $\begingroup$ @Christoph90 Finally, I got your point. Sorry. Well, the confusion originates from using the same $t$ as parameter of the translation operator and as argument of the observable $A$. In order to avoid confusion, either you work with $it{\cal L}$ and $A(0)$, to get $A(t)$, or with $i \tau{\cal L}$ and $A(t)$, to get $A(t+ \tau)$. I have clarified the notation in my answer. $\endgroup$ Commented Aug 10, 2023 at 9:50

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