Explicit independence of Hamiltonian phase-space variables from the time parameter

Question

In general, we have for a Hamiltonian flow $H$ of some "time" parameter $t$, the following relation for any function $f=f(q,p;t)$ of the phase-space generalized position ($q$) and conjugate momentum ($p$),

$$ \frac{df}{dt}=\frac{\partial f}{\partial t}+\{f,H\}, $$

where $\{\cdot,\cdot\}$ denotes Poisson's bracket on these canonical variables.

So, only if the function $f$ is explicitly independent of $t$ (meaning $\frac{\partial f}{\partial t}=0$), will we get a time derivative completely described by the Poisson bracket; i.e. we get

$$ \frac{df}{dt}=\{f,H\}. $$

If we now choose such $f$ to be one of our canonical variables themselves, $q$ or $p$, I have seen in some discussions [e.g. see eq.(2) in this paper] that it is postulated immediately (without much explanation) that $q$ and $p$ evolve as

$$ \frac{dq}{dt}=\{q,H\} , \ \ \ \frac{dp}{dt}=\{p,H\},$$

which must imply that $q,p$ are considered not explicitly dependent on the parameter $t$.

Why is that true? And is this general or only limited to particular Hamiltonian system types? (an example that comes to mind is that in the Angle-Action variables, we have the form $q=q_{0}+\omega t$, which is explicit in $t$, etc.)

Any elaboration on this would be appreciated.

Answer 1

The phase space should generically be thought of as a manifold first. The functions, such as $f$, only exist over this manifold and should be thought of a coordinate-independent objects (things get a little fuzzy when including time dependence, but then they can be imagined as 1-parameter families if functions over the manifold). As with any manifold, coordinates come second as everything is defined to exist before we think about coordinate expressions (since coordinates only exist locally anyway).

In the canonical formalism, the $p$'s and $q$'s are simply a particular coordinate system, though a special one. They are defined to be the coordinates in which the Poisson bracket takes precisely the form mentioned in OP. Though usually stated in terms of the symplectic form, it's a theorem that there always exist (at least on some local coordinate patch on phase space) a set of coordinates $p,q$ in terms of which the Poisson bracket takes the canonical form. These are known as Darboux coordinates. I'll also note that these coordinates are defined not to be part of a 1-parameter collection of coordinatizations, and hence have no time dependence by construction.

In the vast majority of commonly encountered cases, the phase space is essentially just $\mathbb{R}^{2n}$ for some $n$, so the Darboux coordinates may be used globally without any issues, hence why this point often isn't mentioned. In the very interesting case of constrained systems (including gauge theories), however, the phase space is generically not topologically trivial and hence the Darboux coordinates can only be used locally.

The long story short is that there is no loss of generality in assuming we are working in Darboux coordinates because it's a theorem that we always can. The reason the action-angle variables don't have the same property with regards to the Poisson bracket is nothing more than the statement that the action angle variables are not Darboux coordinates. They would be more akin to something like coordinatizing a manifold by the curves formed by flowing points forward by the action of some vector field (because this is, in fact, exactly what they are).