How does one think about, and apply, Noether's theorem in the classical mechanical Hamiltonian formalism?

From the Lagrangian perspective, Noether's theorem (in 1-D) states that the quantity

$$\sum_{i=1}^n \frac{\partial \mathcal{L}}{\partial ( \frac{d y_i}{dx})} \frac{\partial y_i^*}{\partial \varepsilon} - \left[\sum_{j=1}^n \frac{\partial \mathcal{L}}{\partial ( \frac{d y_j}{dx})} \frac{d y_j }{\partial x} - \mathcal{L}\right]\frac{\partial x^*}{\partial \varepsilon}$$

is conserved if the Lagrangian $\mathcal{L}(x,y_i,y_i')$ is invariant under a continuous one-parameter group of infinitesimal transformations of the form

$$T(x,y_i,\varepsilon) = (x^*,y_i^*) = (x^*(x,y_i,\varepsilon),y_i^*(x,y_i,\varepsilon).$$

From the action perspective, Noether's theorem states the equality of the 1-forms:

$$\mathcal{L}(x,y_i,y_i')dx = \sum_{j=1}^n p_i d y_j - \mathcal{H}dx = \mathcal{L}(x^*,y_i^*,y_i'^*)dx^* = \sum_{i=1}^n p_i d y_i^* - \mathcal{H}dx^*$$

which can be used to determine (additive) symmetries nicely.

How do I use this formalism to understand the Hamiltonian Noether theorem in a general context? I'll usually see a claim that $dA/dt = [H,A]$ is the Hamiltonian Noether's theorem, and I can't make sense of this in the context of my description of Noether above. This appears to derive the Poisson brackets as part of Noether from what I've developed above, but I can't make much sense of it to be honest I'm sure the answer is supposed to link the local Lie algebra tangent vector structure to the global Lie group transformation in the Lagrangian, but saying that in words is one thing, in math it's another.

Note, these (and other) old stack posts do not answer the question:


  1. Bergmann - Introduction to the Theory of Relativity, appendix
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    $\begingroup$ the key word to look for would be 'moment map'; see eg ncatlab.org/nlab/show/… $\endgroup$ – Christoph Jul 16 '14 at 14:57
  • $\begingroup$ This question is essentially a duplicate of physics.stackexchange.com/q/69271/2451 $\endgroup$ – Qmechanic Jul 16 '14 at 14:59
  • $\begingroup$ The answer I was looking for is contained in Greiner's Field Quantization book if anybody is interested. $\endgroup$ – bolbteppa Oct 12 '14 at 15:57
  • $\begingroup$ Could you post your answer then? I'm not even sure what your question here really is - since, indeed, $\partial_t A = \{H,A\}$, the Noether charge $A$ generating the symmetry is conserved almost by definition of a symmetry. For Noether's theorem as a Hamiltonian action principle, Qmechanic has linked a suitable duplicate. $\endgroup$ – ACuriousMind Jan 6 '15 at 21:23
  • $\begingroup$ Which page in Field Quantization by Greiner et. al.? $\endgroup$ – Qmechanic Jan 6 '15 at 21:32

Noether's theorem in Hamiltonian mechanics is saying the same thing as Noether's theorem in the Lagrangian setting, under the Legendre transform.

A Hamiltonian system is a triple $(M,\omega, H)$ where $(M,\omega)$ is a symplectic manifold and $H$ is the Hamiltonian. You define a continuous symmetry in the Hamiltonian setting to be a vector field $V$ that preserves both $\omega$ and $H$. That is if $\theta_t$ is the flow of $V$, then $\theta_t^\ast\omega=\omega$ and $\theta_t^\ast H=H$.

A conserved quantity is just a smooth function that Poisson commutes with $H$. That is a function $G:M\to\mathbb{R}$ such that $\{G,H\}=0=\{H,G\}$.

Noether's theorem says these are in one-to-one correspondence.

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  • $\begingroup$ I'm sorry but (entirely my fault) I see absolutely no relationship between this and how I've described Noether in my post, other than that I think you are saying that the tangent vector is the derivative of what I'm calling $T$ is the generator of a symmetry of $H$ (for some reason, I don't see why). In my (and Gelfand's) statement of Noether the symmetry doesn't even touch the Hamiltonian, and Poisson brackets just appear out of nowhere. I think there might be a nice link if there's some general version of Heisenberg's equations in the Calculus of Variations: $\endgroup$ – bolbteppa Jul 30 '14 at 1:16
  • $\begingroup$ books.google.ie/… $\endgroup$ – bolbteppa Jul 30 '14 at 1:16

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