# A kind of Noether's theorem for the Hamiltonian

How can I (conveniently?) show that an invariance of the Lagrangian and Hamiltonian (i.e. the kinetic as well as the potential energy are independently invariant) will lead to a conservation law using only the Hamiltonian formalism? Am I correct, that the invariance of a Hamiltonian does not need to lead to a conservation law, if the corresponding Lagrangian is not invariant (i.e. only $T + V$ is invariant, but $T - V$ is not)?

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It should be stressed that Noether's theorem is a statement about consequences of symmetries of an action functional (as opposed to, e.g., symmetries of equations of motion, or solutions thereof). So to use Noether's theorem, we first of all need an action formulation. How do we get an action for a Hamiltonian theory? Well, let us for simplicity consider point mechanics (as opposed to field theory, which is a straightforward generalization). Then the Hamiltonian action reads

$$\tag{1} S_H[q,p] ~:= \int \! dt ~ L_H(q,\dot{q},p,t).$$

Here $L_H$ is the so-called Hamiltonian Lagrangian

$$\tag{2} L_H(q,\dot{q},p,t) ~:=~ p_i \dot{q}^i - H(q,p,t).$$

One may prove that Euler-Lagrange equations for the Hamiltonian action (1) leads to Hamilton's equations of motion.

In turn, we may view the action (1) as a first-order Lagrangian system in twice as many variable $(q^i,p_j)$. Returning to OP's question: Noether's theorem may then be applied to the Hamiltonian action (1) to investigate symmetries and conservation laws.

Example: The symmetries associated with conservation of the Laplace-Runge-Lenz vector in the Kepler problem is difficult to understand via a purely Lagrangian formulation in configuration space

$$\tag{3} L~=~ \frac{m}{2}\dot{q}^2 + \frac{k}{q},$$

but may easily be described in the corresponding Hamiltonian formulation in phase space, cf. Wikipedia and this Phys.SE post.

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If your Hamiltonian is invariant, that means there should be a vanishing Poisson bracket for some function $F(q,p)$ of your canonical coordinates so that $$\{ H(q,p), F(q,p)\} = 0$$ Since the Poisson bracket with the Hamiltonian also gives the time derivative, you automatically have your conservation law.

One thing to note: The Lagrangian is a function of position and velocity, whereas the Hamiltonian is a function of position and momentum. Thus, your $T$ and $V$ in $L = T - V$ and $H = T + V$ are not the same functions.

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