The original Noether's theorem assumes a Lagrangian formulation. Is there a kind of Noether's theorem for the Hamiltonian formalism?

  • 1
    $\begingroup$ The Hamiltonian formalism is the one chosen by the author John M. Lee to introduce the theorem. Check Theorem 22.22 at his Introduction to Smooth Manifolds. You may have an institutional access to the book: link.springer.com/book/10.1007/978-1-4419-9982-5 $\endgroup$
    – Andrestand
    Commented Nov 13, 2022 at 19:45

2 Answers 2


Action formulation. 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

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

Here $L_H$ is the so-called Hamiltonian Lagrangian

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

We may view the action (1) as a first-order Lagrangian system $L_H(z,\dot{z},t)$ in twice as many variable

$$ (z^1,\ldots,z^{2n}) ~=~ (q^1, \ldots, q^n;p_1,\ldots, p_n).\tag{3}$$

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

$$\begin{align} 0~\approx~&\frac{\partial S_H}{\partial z^I} ~=~\sum_{J=1}^{2n}\omega_{IJ}\dot{z}^J -\frac{\partial H}{\partial z^I}\cr\cr \qquad\Updownarrow&\qquad\cr\cr \dot{z}^I~\approx~&\{z^I,H\}\cr\cr \qquad\Updownarrow&\qquad\cr\cr \dot{q}^i~\approx~& \{q^i,H\} ~=~\frac{\partial H}{\partial p_i}\cr \qquad \wedge&\qquad\cr \dot{p}_i~\approx~& \{p_i,H\}~=~-\frac{\partial H}{\partial q^i}. \end{align}\tag{4}$$

[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 (4) as

$$ \frac{dQ}{dt}~\approx~ \{Q,H\}+\frac{\partial Q}{\partial t}.\tag{5}$$

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

Statement 1: "A symmetry is generated by its own Noether charge."

Sketched proof: Let there be given an infinitesimal (vertical) transformation

$$\begin{align} \delta z^I~=~& \epsilon Y^I(q,p,t), \cr &I~\in~\{1, \ldots, 2n\}, \cr \delta t~=~&0,\end{align}\tag{6}$$

where $Y^I=Y^I(q,p,t)$ are (vertical) generators, and $\epsilon$ is an infinitesimal parameter. Let the transformation (6) be a quasisymmetry of the Hamiltonian Lagrangian

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

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

$$ Q^0~:=~ \sum_{I=1}^{2n}\frac{\partial L_H}{\partial \dot{z}^I} Y^I \tag{8}$$

while the full Noether charge is

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

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

$$\begin{align} \sum_{I=1}^{2n}\dot{z}^I \frac{\partial Q}{\partial z^I} &+\frac{\partial Q}{\partial t}\cr ~=~& \frac{dQ}{dt}\cr ~\stackrel{\text{NI}}{=}~& -\sum_{I=1}^{2n} \frac{\delta S_H}{\delta z^I}Y^I \cr ~\stackrel{(4)}{=}~& \sum_{I,J=1}^{2n}\dot{z}^I\omega_{IJ}Y^J\cr &+\sum_{I=1}^{2n} \frac{\partial H}{\partial z^I}Y^I . \end{align} \tag{10}$$

By comparing coefficient functions of $\dot{z}^I$ on the 2 sides of eq. (10), we conclude that the full Noether charge $Q$ generates the quasisymmetry transformation

$$ Y^I~=~\{z^I,Q\}.\tag{11}$$ $\Box$

Statement 2: "A generator of symmetry is essentially a constant of motion."

Sketched proof: Let there be given a quantity $Q=Q(q,p,t)$ (a priori not necessarily the Noether charge) such that the infinitesimal transformation

$$\begin{align} \delta z^I~=~& \{z^I,Q\}\epsilon,\cr &I~\in~\{1, \ldots, 2n\}, \cr \delta t~=~&0,\cr \delta q^i~=~&\frac{\partial Q}{\partial p_i}\epsilon, \cr \delta p_i~=~& -\frac{\partial Q}{\partial q^i}\epsilon, \cr &i~\in~\{1, \ldots, n\},\end{align}\tag{12}$$

generated by $Q$, and with infinitesimal parameter $\epsilon$, is a quasisymmetry (7) of the Hamiltonian Lagrangian. The bare Noether charge is by definition

$$\begin{align} Q^0~:=~& \sum_{I=1}^{2n}\frac{\partial L_H}{\partial \dot{z}^I} \{z^I,Q\}\cr ~\stackrel{(2)}{=}~& \sum_{i=1}^n p_i \frac{\partial Q}{\partial p_i}.\end{align}\tag{13}$$

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

$$\begin{align} \frac{d (Q^0-f^0)}{dt} ~\stackrel{\text{NI}}{=}~& -\sum_{I=1}^{2n}\frac{\delta S_H}{\delta z^I} \{z^I,Q\}\cr ~\stackrel{(2)}{=}~& \sum_{I=1}^{2n}\dot{z}^I \frac{\partial Q}{\partial z^I} +\{H,Q\}\cr ~=~&\frac{dQ}{dt}-\frac{\partial Q}{\partial t} +\{H,Q\}. \end{align}\tag{14}$$

Firstly, Noether theorem implies that the corresponding full Noether charge $Q^0-f^0$ is conserved on-shell

$$ \frac{d(Q^0-f^0)}{dt}~\approx~0,\tag{15}$$

which can also be directly inferred from eqs. (5) and (14). Secondly, the off-shell Noether identity (14) can be rewritten as

$$\begin{align} \{Q,H\}+\frac{\partial Q}{\partial t} ~\stackrel{(14)+(17)}{=}&~\frac{dg^0}{dt}\cr ~=~~~&\sum_{I=1}^{2n}\dot{z}^I \frac{\partial g^0}{\partial z^I}+\frac{\partial g^0}{\partial t},\end{align}\tag{16} $$

where we have defined the quantity

$$ g^0~:=~Q+f^0-Q^0.\tag{17}$$

We conclude from the off-shell identity (16) that (i) $g^0=g^0(t)$ is a function of time only,

$$ \frac{\partial g^0}{\partial z^I}~=~0\tag{18}$$

[because $\dot{z}$ does not appear on the lhs. of eq. (16)]; and (ii) that the following off-shell identity holds

$$ \{Q,H\} +\frac{\partial Q}{\partial t} ~=~\frac{\partial g^0}{\partial t}.\tag{19}$$

Note that the quasisymmetry and the eqs. (12)-(15) are invariant if we redefine the generator

$$ Q ~~\longrightarrow~~ \tilde{Q}~:=~Q-g^0 .\tag{20} $$

Then the new $\tilde{g}^0=0$ vanishes. Dropping the tilde from the notation, the off-shell identity (19) simplifies to

$$ \{Q,H\} +\frac{\partial Q}{\partial t}~=~0.\tag{21}$$

Eq. (21) is the defining equation for an off-shell constant of motion $Q$.


Statement 3: "A constant of motion generates a symmetry and is its own Noether charge."

Sketched proof: Conversely, if there is given a quantity $Q=Q(q,p,t)$ such that eq. (21) holds off-shell, then the infinitesimal transformation (12) generated by $Q$ is a quasisymmetry of the Hamiltonian Lagrangian

$$\begin{align} \delta L_H ~\stackrel{(2)}{=}~~& \sum_{i=1}^n\dot{q}^i \delta p_i -\sum_{i=1}^n\dot{p}_i \delta q^i \cr &-\delta H +\frac{d}{dt}\sum_{i=1}^np_i \delta q^i \cr ~\stackrel{(12)+(13)}{=}&~ -\sum_{I=1}^{2n}\dot{z}^I \frac{\partial Q}{\partial z^I}\epsilon\cr &-\{H,Q\}\epsilon + \epsilon \frac{d Q^0}{dt}\cr ~\stackrel{(21)}{=}~~& \epsilon \frac{d (Q^0-Q)}{dt}\cr ~\stackrel{(23)}{=}~~& \epsilon \frac{d f^0}{dt},\end{align}\tag{22}$$

because $\delta L_H$ is a total time derivative. Here we have defined

$$ f^0~=~ Q^0-Q .\tag{23}$$

The corresponding full Noether charge

$$ Q^0-f^0~\stackrel{(23)}{=}~Q \tag{24}$$

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

$$ \frac{dQ}{dt}~\approx~0.\tag{25}$$

Eq. (25) is the defining equation for an on-shell constant of motion $Q$.


Discussion. Note that it is overkill to use Noether's theorem to deduce eq. (25) from eq. (21). In fact, eq. (25) follows directly from the starting assumption (21) by use of Hamilton's eoms (5) without the use of Noether's theorem! For the above reasons, as purists, we disapprove of the common praxis to refer to the implication (21)$\Rightarrow$(25) as a 'Hamiltonian version of Noether's theorem'.

Interestingly, an inverse Noether's theorem works for the Hamiltonian action (1), i.e. a on-shell conservation law (25) leads to an off-shell quasisymmetry (12) of the action (1), cf. e.g. my Phys.SE answer here.

In fact, one may show that (21)$\Leftrightarrow$(25), cf. my Phys.SE answer here.

Example 4: The Kepler problem: 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

$$ L~=~ \frac{m}{2}\dot{q}^2 + \frac{k}{q},\tag{26}$$

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

  • $\begingroup$ One question:Give conserved quantity $Q(p,q,t)$, I can use your formula to compute $\delta q, \delta p$. How to compute $\delta t$? Or there is no $\delta t$ in phase space variation? $\endgroup$
    – maplemaple
    Commented Dec 18, 2017 at 4:53
  • $\begingroup$ $\delta t$ is assumed zero in the answer for simplicity. It would be interesting to extend the analysis to non-zero $\delta t$. $\endgroup$
    – Qmechanic
    Commented Dec 19, 2017 at 20:29
  • 1
    $\begingroup$ @Qmechanic Very beautiful answer (+1). But it seems to me that you omitted to mention a fact. The notion of symmetry can be defined completerly in the Hamiltonian setup independently form the action principle. Symmetries are those (one-parameter groups of) active canonical transformations that leave invariant in form the Hamitonian. This requirment is equivanent to (21) for instance. But I guess you know very well these things (better than me). $\endgroup$ Commented Dec 5, 2020 at 9:50

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.

  • $\begingroup$ This doesn't work for e.g. Galilean symmetry, where the generators of the symmetry transformations have explicit time dependence. Instead $\dot{F} = 0$ implies $\{ H ,F\} + \partial_t F = 0$ $\endgroup$ Commented Jan 26, 2022 at 16:00

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