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Hamilton's equation can be written in terms of Poisson brackets, as follows: $$\dot{q} = \{q,H\}$$ $$\dot{p} = \{p,H\}$$ where $H$ is the Hamiltonian of the system. Now, wikipedia says that the solution of these equations can be written as: $$q(t) = \exp(-t\{H,\cdot\})q(0)$$ $$p(t) = \exp(-t\{H,\cdot\})p(0)$$ What does these exponentials mean? I mean, is $\{H,\cdot\}$ some sort of operator? And can we prove rigorously that the latter is the solution of the former? Or is it just formal?

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    $\begingroup$ I believe the Wikipedia entry says that the Poisson bracket here is to be read as an operator, but you are correct, from a practical perspective this is mostly a formal relationship. It is, in general, not sufficient to find actual solutions of the equations of motion. It turns out that only about a dozen Hamiltonians are integrable, i.e. we can only calculate long term solutions for a tiny subset of all possible Hamiltonian systems. See e.g. en.wikipedia.org/wiki/Integrable_system for details. $\endgroup$ Commented Jun 8, 2023 at 21:42

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The formula is formal, and isn’t very useful for actual computation. You have to view $\{\cdot,H\}$ as a linear operator acting on the vector space of observables, i.e. real functions defined on phase space: $O(q,p)$.

From the equation of motion: $$ \dot O=\{O,H\} $$ You can view it as an autonomous linear equation which is solved as usual by exponentiation: $$ \begin{align} O(t)&=e^{t\{\cdot,H\}}O(0)\\ &= \sum_{n=0}^N\frac{t^n}{n!} \{\{…O(0)…H,\},H\}+o(t^N) \end{align} $$ which can be obtained directly from Taylor expanding in time and applying the time derivative multiple time on the equation of motion: $$ \frac{d^nO}{dt^n}=\{\cdot,H\}^nO $$

Mathematically, it is better to view it as the Picard contraction by reformulating the equation of motion as: $$ O(t)=O(0)+\int_0^tds\{O(s),H\} $$ By successive application of the “contraction”, you recover the previous power series.

This latter derivation is better suited for answering questions of convergence to a true solution. You can use the fixed point theorem, as long as you can prove that using a suitable topology the above functional is indeed a contraction. Things are a bit more tricky compared to the usual Cauchy Lipschitz equation since the space is infinite dimensional and you’ll need functional analysis.

Hope this helps.

Answer to comments

The convergence of the Taylor series is not as simple as in the finite dimensional case.

Unlike the usual exponential, you wouldn't expect the radius of convergence to be infinite. Take for example: $$ H = \frac{p^2}{2}+\frac{q^3}{3} $$ which exhibits finite time blowup (in $(t_b-t)^{-1/2}$). Since the blowup time depends on the initial condition in phase space, it is arbitrarily short and you can already anticipate major issues.

This is why you should rather substitute such uniform convergence to weaker form. Take for example a slight modification of the previous example: $$ H = \frac{p^2}{2}+\frac{\epsilon q^2}{2}+\frac{q^3}{3} $$ Now, you have a stable basin of attraction around the origin. You'd expect some form of uniform convergence inside this basin. More formally, You shouldn't try to prove that: $O_N(t)\to O(t)$ with $O_N(t)$ being the truncated Taylor series of $N$ terms and the convergence being some kind of strong convergence with a given norm. Instead, you should rather prove a weak convergence $(\rho,O_N(t))\to (\rho,O(t))$ with $\rho$ a density function (physically, a convergence in expected values). For the convergence to be uniform, you'd expect that $\rho$ needs to be supported in the basin. Furthermore, by imposing $\rho$ to be sufficiently smooth, you can define the powers of $\{\cdot,H\}$ by duality so that the smoothness of the observable is not necessary.

In general, things are more intuitive if you look at the equation of variations, ie the one of the flow: $$ \dot \phi=\{\phi,H\} \\ \phi(0) = Id $$ so that: $$ O(t)=O(0)\circ \phi(t) $$ For a quantum mechanical analogy, physically, the analogue of the equation of motion would be the Heisenberg picture. However the mathematical structure is actually closer to the Schrödinger picture with $\phi$ being the analogue of the evolution operator. This allows to bypass the necessity of smooth observables as well. Once again, you wouldn't expect the convergence of the Taylor series for $\phi$ to be uniform, and you'll need to weaken the convergence in a similar manner.

For the perturbation analysis, I guess so, the usual methods of quantum mechanics have natural analogues in classical mechanics. CosmasZachos' comment can be translated in more familiar terms such as the interaction picture (Duhamel's formula giving the leading order correction). By adding a perturbation $H=H_0+V$, then you similarly get: $$ \begin{align} e^{t\{\cdot,H\}} &= e^{t\{\cdot,H_0\}}\mathcal T \exp\left(\int_0^t \{\cdot, V\}_I(s)ds\right) \\ \{\cdot, V\}_I(s) &= e^{s\{\cdot,H_0\}}\{\cdot, V_I\}e^{-s\{\cdot,H_0\}}\\ &= \{\cdot \circ \phi_0(-s), V\}\circ \phi_0(s) \end{align} $$

which is a formal way of writing out what you naively do in perturbation calculations.

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  • $\begingroup$ I was wondering if the formal relationship can be used as a starting point for classical perturbation analysis? The few (very old) books about the topic that I have seen seemed a bit of a hodgepodge of methods without strong mathematical motivation (but that's mostly due to my lack of understanding). The exponential suggests that it's already in a form that allows us to analyze stability more easily. $\endgroup$ Commented Jun 8, 2023 at 22:38
  • $\begingroup$ Thanks for your answer! I think my comment to the other answer also applies to your answer, tho. $\endgroup$
    – MathMath
    Commented Jun 8, 2023 at 23:12
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    $\begingroup$ @FlatterMann PBs have the formal structure of commutators, and Duhamel's formula is rigorous and incontrovertibly dispositive here. The OP might not be conversant in it. $\endgroup$ Commented Jun 8, 2023 at 23:14
  • $\begingroup$ @CosmasZachos I am not very familiar with differential topology and for this reason I don't know much about Lie algebras and groups. I understand it in the present context, however. What is not clear to me is the topology which one should equip $C^{\infty}(\mathbb{R}^{2n})$, the space of observables, so to ensure the expomential is well defined in terms of its Taylor expansion. $\endgroup$
    – MathMath
    Commented Jun 9, 2023 at 0:20
  • $\begingroup$ @MathMath My apologies for not knowing the waterfront well enough to send you to suitable references; but I do know such exist! In principle, you don't need the Taylor expansion; physicists normally use it as a conversation stopper to get ornery mathematicians off their back... I am too tired to slug through Arnold and Marsden & Ratiu to locate the relevant Chapter & verses... The operator $\{ H, \bullet = \partial _x H \partial_p -\partial_p H \partial_x $ and its higher-dimensional generalizations... $\endgroup$ Commented Jun 9, 2023 at 0:27
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If $X$ is a smooth ($=C^\infty$ henceforth) vector field on a smooth manifold $M$, the local flow (also known as the one-parameter group of diffeomorphisms generated by $X$) is the map defined as $$\Phi_t^X(p) := \gamma(t|p)\in M$$ where $I_p\ni t \mapsto \gamma(t|p)$ is the (unique) maximal solution of the differential equation $$\frac{d \gamma}{dt} = X(\gamma(t))$$ with initial condition $$\gamma(0) = p\:.$$ Above $I_p\subset \mathbb{R}$ is an open interval (possibly the whole real line) that includes $0$, completely and uniquely determined by $p$ under the requirement that the domain of the solution is the largest possible. Varying $p\in M$, $$A:= \bigcup_{p\in M} I_p \times \{p\} \subset \mathbb{R}\times M$$ turns out to be an open set and the map $A \ni (t,p)\mapsto \Phi^X_t(p) \in M$ turns out to be smooth.

For various reasons the following notation is used $$\Phi_t^X(p) = \exp(tX)(p)\:.$$ The most important reason is that the left-hand side has the same properties of the standard exponential: $$\exp(0X)= id\:, \quad \exp(tX)\exp(t'X)(p)= \exp((t+t')X)(p)$$ the second identity is valid for $p\in M$ and $t,t' \in \mathbb{R}$ such that both sides are defined. An important corollary is that $$(\exp(tX))^{-1}= \exp(-tX)\:.$$ It is also true that, under suitable real-analyticity hypotheses (generally false), the Taylor expansion theorem furnishes $$f\left(\Phi_t^X(p)\right) = \sum_{n=0}^{+\infty} \frac{t^n}{n!} (X)^n(f)|_p =: (e^{t X} f)(p)\:,$$ where we have also used the fact that vector fields are first-order differential operators $$X(f) = \sum_{k=1}^n X^k(p) \frac{\partial f}{\partial x^k}|_p\:.$$

Differently from the exponential of numbers however, if $X\neq Y$, generally: $$\exp(tX) \exp(sY)\neq \exp(sY)\exp(tX)$$ (where both sides are defined). However the following crucial result is true $$\left(\exp(tX) \exp(sY)\right)(p)= \left(\exp(sY)\exp(tX)\right)(p)\quad \mbox{for $s,t,p$ such that both sides are defined}$$ if and only if $$[X,Y](p) =0 \quad \forall p \in M\:.$$ Above $[\cdot,\cdot]$ is the Lie commutator of smooth vector fields. That is the starting point of the various ``infinitiesimal formulations'' of the Noether theorem in Hamiltonian mechanics (see my answer here and my book quoted below).

If we are dealing with an Hamiltonian theory (a symplectic manifold) $(M,\omega)$, then $X_g:= \{g, \cdot\}$ is a vector field for every smooth function $g: M \to \mathbb{R}$. The flow $\{\Phi_s^{X_g}\}_{s\in \mathbb{R}}$ is made of active canonical transformations.

Time evolution is a special case when $-g=H$ is the Hamiltonian function of the theory. For instance, if $q,p$ are canonical coordinates we can view $q$ and $p$ as maps on $M$ (at least defined locally). Since $\mathbb{R}\ni t \mapsto \exp(-t\{H,\cdot\})(x)$ is the solution of Hamilton equations with initial condition $x=(q,p)$, $$q(t) = e^{-t\{H, \cdot\}}(q)\:,$$ $$p(t) = e^{-t\{H, \cdot\}}(p)\:.$$


Refernces. V. Moretti, Analytical Mechanics: Classical, Lagrangian and Hamiltonian Mechanics, Stability Theory, Special Relativity, Springer 2023

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    $\begingroup$ Valter, this is an amazing answer! Thank you very much. I would like to ask you what you mean by "real-analyticity conditions"? As far as I understood from you answer, the Taylor expansion is not central, however. It can be proven that under these conditions its Taylor expansion is well-defined, but the focus the exponential is more like a notation which follows from the properties you listed, is that right? $\endgroup$
    – MathMath
    Commented Jun 9, 2023 at 12:44
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    $\begingroup$ You are right, in the general case, it is nothing but a notation: the properties actually matter which are similar to the ones of the exponential. The analyticity condition is that $f$ and the components of $X$ in the considered local chart on $M$ are real analytic functions. $\endgroup$ Commented Jun 9, 2023 at 13:15
  • $\begingroup$ Right! Thank you again. I wonder why these expressions in terms of $\exp(t\{H,\cdot\})$ are not usually found in the mathematical physics literature, at least not explicitly. To me, this is the perfect motivation for introducing the time evolution $e^{-it H}$ in quantum mechanics. $\endgroup$
    – MathMath
    Commented Jun 9, 2023 at 14:54
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It can indeed be seen as an operator. The object $\{H,\cdot\}$ is a linear map that maps functions on phase space $f(p,q,t)$ to functions on phase space. Because it is linear, it behaves similar to matrices.

The exponent is defined in terms of its power series. Let's look at an example with a square matrix $A$. You can easily replace every $A$ with $\{H,\cdot\}$ later on. $$\exp(A)\equiv\sum_{n=0}^\infty\frac{1}{n!}A^n$$ The matrix exponential can be used to solve the following matrix differential equation. \begin{align} \dot q&=Aq\\ \implies q(t)&=\exp(At)q(0) \end{align} With some work you can prove that $$\frac{d}{dt}\exp(At)=\frac{d}{dt}\sum_{n=0}^\infty\frac{1}{n!}t^nA^n=A\exp(At),$$

which indeed solves the differential equation.

Note that powers of $\{H,\cdot\}$ are well defined: $$\{H,\cdot\}^2f=\{H,\cdot\}\Big(\{H,f\}\Big)=\{H,\{H,f\}\}$$

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  • $\begingroup$ I like your approach here. However, some things still need clarification. To use Poisson brackets one needs to assume observables are differentiable. It is usually assumed that the space of observables is $C^{\infty}(\mathbb{R}^{2n})$. You defined $\exp(At)$ by means of its Taylor series, but this assumes that the underlying topological space is Banach and $A$ to be bounded. However, it is not clear to me what is the appropriate norm on $C^{\infty}(\mathbb{R}^{2n})$ which makes it a Banach space. Moreover, is $\{H,\cdot\}$ a bounded operator on this space? $\endgroup$
    – MathMath
    Commented Jun 8, 2023 at 23:11
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    $\begingroup$ @MathMath sorry I can't help you further, my education is not that mathematically inclined. I hope my answer could at least help you a little bit $\endgroup$ Commented Jun 9, 2023 at 10:42
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Another way of looking at it is via Liouville equation: $$ \frac{\partial \rho(q,p,t)}{\partial t} + \left\{\rho(q,p,t), H(q,p,t)\right\}=0 $$ This is a linear differential equation and we can solve this for any initial condition $\rho(q,p,t_0)=\rho_0(q,p)$, which will give us the Green's function $$ \rho(q,p,t)=\int dq'\int dp' G(q,p,t|q',p',t_0)\rho(q',p',t_0). $$ This Green's function is the evolution operator.

Remarks:

  • For well-defined initial position and momentum we can always use $\rho(q,p,t_0)=\delta(q-q(t_0))\delta(p-p(t_0))$
  • This also can be viewed as a product of matrices in continuous Hilbert space, as we often do in QM (e.g.,w hen working in position or momentum representation.)
  • (Retarded) one-particle Green's functions of quantum theory can also be represented as matrix elements of the evolution operator.
  • In a somewhat hidden form this operator is ubiquitous in the theory of canonical transformations in classical mechanics. See motion as a canonical trasformation.
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    $\begingroup$ Roger, thank you very much for the answer. It is a very interesting point of view! $\endgroup$
    – MathMath
    Commented Jun 9, 2023 at 12:45

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