How to describe time-shifts in Noether's theorem in Hamiltonian formalism

Question

As was described in, for example, this post, one can formulate Noether's Theorem also in Hamiltonian Mechanics. Symmetries are then represented by vector fields generated by observables whose Poisson brackets vanish with the Hamiltonian of the system.

My question is: How do you describe transformations of time in this formalism? Flows in the phase space only represent active transformations of the phase space, but they don't change anything about the time variable. It is not possible to derive, for example, conservation of energy as the consequence of a symmetry regarding the system's time-shifts.

Answer 1

Yes, it is possible, same as for Lagrangian action. Taking the Hamiltonian action as $$ S_H\{q, p\} = \int_{t_i}^{t_f}{dt\; \left[p\dot{q} - H(q, p, t) \right]} $$ consider an infinitesimal time shift $t' = t +\delta t$, with $q'(t') = q(t) +\delta q$, $p'(t') = p(t) +\delta p$, and assume invariance under time translations: $$ S_H\{q, p\} = \int_{t_i}^{t_f}{dt\; \left[p\dot{q} - H(q, p, t) \right]} = \int_{t'_i}^{t'_f}{dt\; \left[p'\dot{q'} - H(q', p', t) \right]} $$ The 2nd equality above gives $$ \int_{t_i}^{t_f}{dt\; \left[p\dot{q} - H(q, p, t) \right]} = \int_{t_i + \delta t_i}^{t_f + \delta t_f}{dt\; \left[p\dot{q} - H(q, p, t) + p\delta \dot{q} + \dot{q}\delta p - \frac{\partial H}{\partial q}(q, p, t)\delta q - \frac{\partial H}{\partial p}(q, p, t)\delta p \right]} $$ or after slight rearrangement, $$ \left[p\dot{q} - H \right]\Big|_{t_f}\delta t_f - \left[p\dot{q} - H \right]\Big|_{t_i}\delta t_i + \int_{t_i + \delta t_i}^{t_f + \delta t_f}{dt\;\left[p\delta \dot{q} + \dot{q}\delta p + \dot{p}\delta q - \dot{q}\delta p \right]} = \\ = \int_{t_i}^{t_f}{dt\; \frac{d}{dt}\left[ p\dot{q}\delta t - H(q,p,t)\delta t + p\delta q\right]} = \int_{t_i}^{t_f}{dt\; \frac{d}{dt}\left[ 2 p\delta q - H(q,p,t)\delta t \right]} = 0 $$ where use is made of $\dot{q}\delta t = \delta q$ and the Hamiltonian EOMs, $ \dot{q}=\frac{\partial H}{\partial p}$, $\dot{p}=-\frac{\partial H}{\partial q}$. For a time translation that leaves end points invariant such that $\delta q\Big|_{t_i} = \delta q\Big|_{t_f} = 0$, the last equality above reduces to $$ \frac{dH}{dt}(q, p, t) = 0 $$ or $$ H(q, p, t) = const. $$

Answer 2

Comments to the question (v2):

1. It should first of all be mentioned that the Hamiltonian generates time-evolution, although I'm sure OP already knows this.

2. The Hamiltonian Lagrangian $$ \tag{1} L_H(q,\dot{q},p,t) ~:=~\sum_{i=1}^n p_i \dot{q}^i - H(q,p,t)$$
   may be viewed as a first-order Lagrangian system $L_H(z,\dot{z},t)$ in twice as many variable $$\tag{2} (z^1,\ldots,z^{2n}) ~=~ (q^1, \ldots, q^n;p_1,\ldots, p_n).$$

3. How to apply Noether's theorem for energy conservation in a Lagrangian system is e.g. discussed in my Phys.SE answer here. None of the suggested infinitesimal transformations in my Phys.SE answer are of the form of a vertical vector field $$\tag{3} \delta z^I~=~\varepsilon V^I, \qquad \delta t~=~0, $$
   in such a way that the vector field components $V^I$ do not depend on $\dot{z}$, which seems to be at the core of OP's question.

4. The corresponding Lagrangian energy function $$\tag{4} h_H(z,\dot{z},t)~:=~ \sum_{I=1}^{2n}\dot{z}^I\frac{\partial L_H(z,\dot{z},t)}{\partial \dot{z}^I} - L_H(z,\dot{z},t)~=~H(z,t) $$ is unsurprisingly the Hamiltonian itself. It will play the role of Noether charge for time translations.

5. Energy is conserved on-shell if the Hamiltonian Lagrangian $L_H$ (or equivalently, the Hamiltonian $H$), has no explicit time dependence.