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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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2 Answers 2

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, cf. this Phys.SE post). 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

$$\tag{3} \dot{q}^i~\approx~ \{q^i,H\}~=~\frac{\partial H}{\partial p_i}\qquad \text{and}\qquad \dot{p}_i~\approx~ \{p_i,H\}~=~-\frac{\partial H}{\partial q^i}. $$

[Here the $\approx$ symbol means equality on-shell, i.e. modulo the equations of motion (eom).] Equivalently, for an arbitrary quantity $Q=Q(q,p,t)$ we may collectively write the Hamilton's eoms (3) as

$$\tag{4} \frac{dQ}{dt}~\approx~ \{Q,H\}+\frac{\partial Q}{\partial t}.$$

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 1: If there is given a quantity $Q=Q(q,p,t)$ such that

$$ \tag{5} \{Q,H\}+\frac{\partial Q}{\partial t}~=~0$$

holds off-shell, then the infinitesimal transformation

$$\tag{6} \delta q^i~=~ \{q^i,Q\}\epsilon~=~\frac{\partial Q}{\partial p_i}\epsilon\qquad \text{and}\qquad \delta p_i~=~ \{p_i,Q\}\epsilon~=~-\frac{\partial Q}{\partial q^i}\epsilon$$

with generator $Q$ and infinitesimal parameter $\epsilon$, is a quasisymmetry of the Hamiltonian Lagrangian

$$ \delta L_H ~\stackrel{(2)}{=}~\dot{q}^i \delta p_i -\dot{p}_i \delta q^i -\delta H+\frac{d}{dt}(p_i \delta q^i)\qquad $$
$$~\stackrel{(6)+(8)}{=}~ -\dot{q}^i\frac{\partial Q}{\partial q^i}\epsilon -\dot{p}_i\frac{\partial Q}{\partial p_i}\epsilon -\{H,Q\}\epsilon + \epsilon \frac{d Q^0}{dt}$$ $$\tag{7}~\stackrel{(5)}{=}~ \epsilon \frac{d (Q^0-Q)}{dt}~\stackrel{(9)}{=}~ \epsilon \frac{d f^0}{dt},$$

because $\delta L_H$ is a total time derivative. Here $Q^0$ is the bare Noether charge

$$\tag{8} Q^0~=~ \frac{\partial L_H}{\partial \dot{q}^i} \{q^i,Q\} + \frac{\partial L_H}{\partial \dot{p}_i} \{p_i,Q\} ~=~ p_i \frac{\partial Q}{\partial p_i},$$


$$\tag{9} f^0~=~ Q^0-Q .$$

Then Noether's theorem states that the corresponding full Noether charge

$$\tag{10} Q~=~Q^0-f^0 $$

is conserved on-shell

$$\tag{11} \frac{dQ}{dt}~\approx~0.$$

But it is overkill to use Noether's theorem to deduce eq. (11). In fact, eq. (11) follows directly from the starting assumption (5) by use of Hamilton's eoms (4) without the use of Noether's theorem! For the above reasons, as purists, we disprove of the common praxis to refer to the implication (5)$\Rightarrow$(11) as a 'Hamiltonian version of Noether's theorem'.

Interestingly, an inverse Noether's theorem works for the Hamiltonian action (1), i.e. a conservation law (11) leads to a quasisymmetry (6) of the action (1), cf. e.g. my Phys.SE answer here.

Example 2: 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

$$\tag{12} L~=~ \frac{m}{2}\dot{q}^2 + \frac{k}{q},$$

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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