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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

$$\tag{2} L_H(q,\dot{q},p,t) ~:=~\sum_{i=1}^n p_i \dot{q}^i - H(q,p,t). $$

One may prove that Euler-Lagrange (EL) 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 $L_H(z,\dot{z},t)$ in twice as many variable

$$\tag{5} (z^1,\ldots,z^{2n}) ~=~ (q^1, \ldots, q^n;p_1,\ldots, p_n).$$

Returning to OP's question, the Noether theorem may then be applied to the Hamiltonian action (1) to investigate symmetries and conservation laws.

Example 1: Let there be given a quantity $Q=Q(q,p,t)$ such that the infinitesimal transformation

$$ \delta z^I~=~ \{z^I,Q\}\epsilon,\qquad I~\in~\{1, \ldots, 2n\}, \qquad \delta t~=~0,$$ $$\tag{6} \delta q^i~=~\frac{\partial Q}{\partial p_i}\epsilon, \qquad \delta p_i~=~ -\frac{\partial Q}{\partial q^i}\epsilon, \qquad i~\in~\{1, \ldots, n\},$$

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

$$\tag{7} \delta L_H~=~\epsilon \frac{d f^0}{dt},$$

where $f^0=f^0(q,p,t)$ is some function. The bare Noether charge is by definition

$$\tag{8} Q^0~:=~ \sum_{I=1}^{2n}\frac{\partial L_H}{\partial \dot{z}^I} \{z^I,Q\} ~\stackrel{(2)}{=}~ \sum_{i=1}^n p_i \frac{\partial Q}{\partial p_i}.$$

Noether's theorem then guarantees an off-shell Noether identity

$$ \frac{d (Q^0-f^0)}{dt} ~=~-\sum_{I=1}^{2n}\frac{\delta S_H}{\delta z^I} \{z^I,Q\} $$ $$\tag{9}~\stackrel{(2)}{=}~ \sum_{I=1}^{2n}\dot{z}^I \frac{\partial Q}{\partial z^I} +\{H,Q\} ~=~\frac{dQ}{dt}-\frac{\partial Q}{\partial t} +\{H,Q\}. $$

Firstly, the off-shell Noether identity (9) implies that the corresponding full Noether charge $Q^0-f^0$ is conserved on-shell

$$\tag{10} \frac{d(Q^0-f^0)}{dt}~\approx~0.$$

Secondly, the off-shell Noether identity (9) can be rewritten as

$$\tag{11} \{Q,H\}+\frac{\partial Q}{\partial t}~=~\frac{dg^0}{dt}~=~\sum_{I=1}^{2n}\dot{z}^I \frac{\partial g^0}{\partial z^I}+\frac{\partial g^0}{\partial t}, $$

where we have defined the quantity

$$\tag{12} g^0~:=~Q+f^0-Q^0.$$

We conclude from the off-shell identity (11) that (i) $g^0=g^0(t)$ is a function of time only,

$$\tag{13} \frac{\partial g^0}{\partial z^I}~=~0;$$

and that (ii) the following off-shell identity holds

$$\tag{14} \{Q,H\} +\frac{\partial Q}{\partial t} ~=~\frac{\partial g^0}{\partial t}.$$

Note that the quasisymmetry and the eqs. (6)-(10) are invariant if we redefine the generator

$$\tag{15} Q ~~\longrightarrow~~ \tilde{Q}~:=~Q-g^0 . $$

Then the new $\tilde{g}^0=0$ vanishes. Dropping the tilde from the notation, the off-shell identity (14) simplifies to

$$ \tag{16} \{Q,H\} +\frac{\partial Q}{\partial t}~=~0.$$

Example 2: Conversely, if there is given a quantity $Q=Q(q,p,t)$ such that (16) holds off-shell, then the infinitesimal transformation (6) generated by $Q$ is a quasisymmetry of the Hamiltonian Lagrangian

$$ \delta L_H ~\stackrel{(2)}{=}~\sum_{i=1}^n\dot{q}^i \delta p_i -\sum_{i=1}^n\dot{p}_i \delta q^i -\delta H +\frac{d}{dt}\sum_{i=1}^np_i \delta q^i \qquad$$
$$~\stackrel{(6)+(8)}{=}~ -\sum_{I=1}^{2n}\dot{z}^I \frac{\partial Q}{\partial z^I}\epsilon -\{H,Q\}\epsilon + \epsilon \frac{d Q^0}{dt}$$ $$\tag{17}~\stackrel{(16)}{=}~ \epsilon \frac{d (Q^0-Q)}{dt}~\stackrel{(18)}{=}~ \epsilon \frac{d f^0}{dt},$$

because $\delta L_H$ is a total time derivative. Here we have defined $$\tag{18} f^0~=~ Q^0-Q .$$

The corresponding full Noether charge

$$\tag{19} Q^0-f^0~\stackrel{(18)}{=}~Q $$

is just the generator $Q$ we started with! Finally, Noether's theorem states the full Noether charge is conserved on-shell

$$\tag{20} \frac{dQ}{dt}~\approx~0.$$

Note that it is overkill to use Noether's theorem to deduce eq. (20) from eq. (16). In fact, eq. (20) follows directly from the starting assumption (16) by use of Hamilton's eoms (4) without the use of Noether's theorem! For the above reasons, as purists, we disapprove of the common praxis to refer to the implication (16)$\Rightarrow$(20) 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 (20) leads to a quasisymmetry (6) of the action (1), cf. e.g. my Phys.SE answer here.

Example 3: 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

$$\tag{21} L~=~ \frac{m}{2}\dot{q}^2 + \frac{k}{q},$$

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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