# Are Hamilton's equations reversible?

Say I define a time dependent vector field $$\Psi(t):\mathbb{R}^d\to \mathbb{R}^d$$ as reversible (also here) if, for $$f(x,y)=(x,-y)$$, we have:

$$f\circ \Psi \circ f =\Psi(-t)=\Psi^{-1}(t).$$

Just to clarify $$\Psi:[0,T]\times \mathbb{R}^d\to \mathbb{R}^d$$. Now let $$\Psi$$ be some Hamiltonian dynamics $$\Psi=(q,p)$$, where:

$$\frac{d q}{dt}=\nabla_p H (q,p), \ \ \ ~q(0)=x,$$ $$\frac{d p}{dt}=-\nabla_q H (q,p), \ \ \ ~p(0)=y,$$ where $$H(q,p)=H(q,-p)$$. Is it obvious that $$\Psi$$ is reversible? Can I use Reversibility of Hamiltonian dynamics somehow?

(Note the 2nd equality $$\Psi(-t)=\Psi^{-t}(t)$$ follows since $$(q,p)$$ is a flow. https://en.wikipedia.org/wiki/Flow_(mathematics) )

• I think Hamiltonian may have time-reversal symmetry but not necessarily for the wave function. Jul 9 at 17:59
• @TanmoyPati this is a problem in classical (not quantum) mechanics. Jul 9 at 18:06

## 1 Answer

1. A dynamical equation system with dynamical variables $$z$$ is called reversible if it is invariant under the combination $$(t\to -t, z\to I(z))$$ where $$I$$ is an involution$$^1$$, cf. Ref. 1.

2. Main example: Any autonomous Hamiltonian $$H(q,p)=H(q,-p)$$ that is even in momenta describes a reversible system. Here the involution is $$I(q,p)=(q,-p)$$.

3. Concerning OP's title question, it is possible to find non-reversible non-autonomous Hamiltonian systems. Think e.g. on a Hamiltonian of the form $$H(q,p,t)=f(t)q+g(t)p$$ for 2 appropriate functions $$f,g$$.

4. More interestingly, Ref. 1 claims [e.g. around eq. (1.29)] that there exist non-reversible autonomous Hamiltonian systems.

References:

1. J.A.G. Roberts & G.R.W. Quispel, Chaos and time-reversal symmetry. Order and chaos in reversible dynamical systems. Phys. Rep. 216 (1992) 63.

$$^1$$ More generally in a geometric language: If $$M$$ denotes the manifold of dynamical variables (sans time $$t$$), then the map $$I:\Gamma(TM)\to \Gamma(TM)$$ is a (possibly time-dependent, not necessarily integrable) mixed tensor field, such that it is pointwise an involution $$\forall z\in M: I_z^2={\bf 1}_{TM}$$. This is often called an almost product structure in the literature.

• For a system to be reversible, the involution in question need not be $\vec{p} \to -\vec{p}$, right? I'm thinking of (for example) a particle in a magnetic field, which is not even in $\vec{p}$ but might have an involution of $\vec{p} \to - \vec{p} + 2 q \vec{A}$ or something like that. Jul 22 at 12:04
• $\uparrow$ Right. Jul 22 at 12:10