Generators and Conserved Charges
In the preceding sections, we reformulated classical dynamics in terms of the geometry of symplectic manifolds. We will now use the same geometrical framework to discuss symmetries of dynamical systems and their corresponding conserved charges.
Consider a continuous transformation that leaves a classical dynamical system unchanged — a symmetry transformation. In the Lagrangian formalism, Noether’s theorem tells us that such symmetry transformations are associated with conserved quantities. In the Hamiltonian formalism, a symmetry is a canonical transformation generated by a function on phase space, and that same function is its conserved charge. We will reframe these results geometrically, and for symmetries that form a group (for example, translations), the corresponding charges combine into a single geometric object called a moment map.
Generators
We have already seen that, given a smooth function $f\in C^{\infty}(M)$, there is a unique vector field $X _ {f}$ satisfying \begin{equation} \iota _ {X _ {f}}\omega=-\mathrm{d}f \ , \label{eq:hamiltonian-vector-field-definition} \end{equation} called a Hamiltonian vector field. The one-parameter flow of $X _ {f}$ moves points through phase space. This is clear from its component form: \begin{equation} X _ {f} =\frac{\partial f}{\partial p^{a}} \frac{\partial}{\partial q _ {a}} -\frac{\partial f}{\partial q _ {a}} \frac{\partial}{\partial p^{a}} \ . \label{eq:hamiltonian-vector-field-darboux} \end{equation} Consequently, the infinitesimal transformation generated by $f$ is \begin{equation} \delta _ {f}q _ {a} =\epsilon\frac{\partial f}{\partial p^{a}} =\epsilon\left\lbrace q _ {a},f\right\rbrace \ , \qquad \delta _ {f}p^{a} =-\epsilon\frac{\partial f}{\partial q _ {a}} =\epsilon\left\lbrace p^{a},f\right\rbrace \ . \label{eq:canonical-transformation-from-generator} \end{equation} More generally, an observable $g\in C^{\infty}(M)$ changes infinitesimally as \begin{equation} \delta _ {f} g = \epsilon X _ {f}[g] = \epsilon \left\lbrace g,f\right\rbrace \ . \label{eq:hamiltonian-vector-field-action} \end{equation}
Let’s consider some examples of this. On $T^{\star}\mathbb{R}$, the function $p$ generates translations (of position), since $X _ {p}=\partial/\partial q$. The function $q$ generates translations of momentum (in the opposite direction), since $X _ {q}=-\partial/\partial p$. The Hamiltonian $H$ generates time evolution: \begin{equation} \frac{\mathrm{d}g}{\mathrm{d}t} =X _ {H}[g] =\left\lbrace g,H\right\rbrace \ . \label{eq:observable-time-evolution} \end{equation} This restates something we already know: the paths traced by dynamical systems in phase space are integral curves of $X _ {H}$.
Before continuing, we wish to also revisit a distinction we made earlier, between Hamiltonian and symplectic vector fields, and consider a concrete example where a global, topological distinction exists between them. (The reader will recall that we had already established that locally, these are equivalent.) Consider the $ 2 $-torus $ \mathbb{T}^ {2} $ with periodic coordinates $(x,y)$ as a symplectic manifold. Let the symplectic $ 2 $-form be $\omega=\mathrm{d}x\wedge\mathrm{d}y$. Notice that \begin{equation} \iota _ {\partial _ {x} }\omega=\mathrm{d}y \end{equation} is certainly closed — this is the definition of a symplectic vector. However, it is not the exterior differential of a globally defined real-valued function on the torus, due to the periodicity of the $ y $ coordinate. Therefore, it is not Hamiltonian. This fact — that $ \iota _ {\partial x} \omega $ is closed but not exact — is a signal of nontrivial first de Rham cohomology class due to the existence of non-contractible $ 1 $-cycles on $ \mathbb{T}^ {2} $.
Conserved Charges
Consider a function $Q\in C^{\infty}(M)$ with no explicit time dependence. Equation \eqref{eq:observable-time-evolution} gives \begin{equation} \frac{\mathrm{d}Q}{\mathrm{d}t}=\left\lbrace Q,H\right\rbrace = -X _ {Q} [H] \ . \label{eq:charge-conservation-condition} \end{equation} We conclude from this that when $ Q $ Poisson commutes with the Hamiltonian, it is conserved. Alternatively, from the second equality, we may say for conserved $ Q $, the Hamiltonian is invariant under the transformation generated by $ Q $.
At the risk of being pedantic, let us stress one last time that there are two directions to this statement: the first is from conserved quantity to symmetry, and the second is from symmetry to conserved quantity. Let’s deal with the first one: if $ Q $ is a conserved quantity , then the canonical transformations it generates leave the Hamiltonian invariant: \begin{equation} \frac{\text{d} }{\text{d} t} Q = 0 \Rightarrow \delta _ {Q} H = \epsilon \left\lbrace H,Q \right\rbrace = - \epsilon \left\lbrace Q, H \right\rbrace = \frac{\text{d} }{\text{d} t} Q = 0 \ . \end{equation} We can say this another way: the functions $ H $ and $ Q $ generate time evolution and symmetry transformations respectively, whose flows are determined by $ X _ {H} $ and $ X _ {Q} $. It is easy to check that \begin{equation} \left[ X _ {H} , X _ {Q} \right] = - X _ {\left\lbrace Q,H \right\rbrace} = 0 \ , \end{equation} so the flows commute, i.e.~it doesn’t matter if we evolve in time first and then perform the symmetry transformation or vice versa.
Now let’s tackle the opposite direction. Given a symmetry transformation generated by a function $ Q $, it can be associated to a Hamiltonian vector field $ X _ {Q} $ such that \begin{equation} \iota _ {X _ {Q} } \omega = - \text{d} Q \ . \end{equation} Now, if it is a symmetry, it leaves $ H $ invariant, so \begin{equation} X _ {Q} [H] = 0 \Rightarrow X _ {Q} [H] = - \left\lbrace Q, H \right\rbrace = - \frac{\text{d} }{\text{d} t} Q = 0 \ , \end{equation} and so $ Q $ is conserved. The point of the $ \mathbb{T}^ {2} $ example discussed above is to highlight that one can have symplectic (as opposed to Hamiltonian) vector fields that are not associated to a globally defined conserved charge.
Let us now consider some examples. For a particle moving in $\mathbb{R}^{N}$ under the influence of some potential $ V(q) $ we have the Hamiltonian \begin{equation} H=\frac{1}{2m}\delta _ {ab}p^{a}p^{b}+V(q) \ , \end{equation} and invariance under translations in the direction of a constant vector $u _ {a}$ means we must have \begin{equation} u _ {a}\frac{\partial V}{\partial q _ {a}}=0 \ . \end{equation} The function \begin{equation} Q _ {u}=u _ {a}p^{a} \end{equation} generates $\delta q _ {a}=\epsilon u _ {a}$ and $\delta p^{a}=0$, and \begin{equation} \left\lbrace Q _ {u},H\right\rbrace=-u _ {a}\frac{\partial V}{\partial q _ {a}}=0 \ . \end{equation} Translation invariance along some direction and the conservation of momentum along that direction are therefore equivalent, a statement that is familiar to us from our study of Noether’s theorem.
As a second example, let a particle move in $ \mathbb{R}^ {3} $ in a central potential, so that \begin{equation} H=\frac{\boldsymbol{p}^{2}}{2m}+V(|\boldsymbol{q}|) \ . \end{equation} For a fixed infinitesimal rotation vector $\boldsymbol{\xi}$, define \begin{equation} Q _ {\boldsymbol{\xi}} =\boldsymbol{\xi}\mathbin{\boldsymbol{\cdot}} \bigl(\boldsymbol{q}\mathbin{\boldsymbol{\times}}\boldsymbol{p}\bigr) = \epsilon _ {abc} \xi ^ {a} q ^ {b} p ^ {c} \ , \label{eq:angular-momentum-smeared-charge} \end{equation} where $ \epsilon _ {abc} $ is the Levi-Civita symbol. Its Hamiltonian vector field produces \begin{equation} \delta\boldsymbol{q} =\epsilon\boldsymbol{\xi}\mathbin{\boldsymbol{\times}}\boldsymbol{q} \ , \qquad \delta\boldsymbol{p} =\epsilon\boldsymbol{\xi}\mathbin{\boldsymbol{\times}}\boldsymbol{p} \ . \end{equation} Since the Hamiltonian is rotationally invariant, $\left\lbrace Q _ {\boldsymbol{\xi}},H\right\rbrace=0$. The three components of $\boldsymbol{q}\mathbin{\boldsymbol{\times}}\boldsymbol{p}$ are therefore conserved. This is simply the familiar fact of conservation of angular momentum in a central potential.