Noether's theorem

In Hamiltonian mechanics

In the context of classical Hamiltonian systems, Noether's theorem asserts that if a Hamiltonian system $(M, ω, H)$ has a 1-parameter group of Hamiltonian symmetrys ${φ_{s}}$ generated by a Hamiltonian vector field $X_{A}$ , then the quantity $A$ is conserved along the solutions of the Hamilton's equations (the flow of $X_{H}$ ), i.e., $A$ is a first integral. Conversely, if a smooth function $A$ is conserved along the solutions of the Hamilton's equations, then there exists a Hamiltonian symmetry generated by a Hamiltonian vector field $X_{A}$ . This result shows a deep connection between the symmetries of a Hamiltonian system and its conserved quantities.

Proof
Indeed, since $ϕ_{s}$ are Hamiltonian symmetries, $X_{A} (H) = 0$ . But we know that

X_{A} (H) = d H (X_{A}) = ω (X_{H}, X_{A}) =

= - ω (X_{A}, X_{H}) = - d A (X_{H}) = - X_{H} (A)

from where we conclude the result. $◼$

In @olver86 we find the interesting Proposition 6.28
Proposition. In a Poisson manifold, a function $P (x, t)$ is a first integral of the Hamiltonian system of ODEs:

\frac{d x}{d t} = J (x) \nabla H (x, t)

if and only if

\frac{\partial P}{\partial t} + {P, H} = 0.

$◼$

In Lagrangian mechanics

Poor's man version.
One of the great advantages of Lagrangian formulation is the application of this theorem. It lets us reduce variables if we know a symmetry of the system.

Given a Lie group acting over the configuration space, we can considerate the infinitesimal action $q \mapsto q + δ X_{V}$ (see Lie algebra action) where $X_{V} = (f_{i} (q))$ in a particular coordinate system. If $p_{i}$ are the generalized momenta in this coordinates $q_{i}$ , Noether's theorem says that if the infinitesimal action doesn't change the Lagrangian then the quantity

Q = \sum_{i} p_{i} \cdot f_{i}

is conserved (see The theoretical minimum 1, page 139).

Full version.
A vector field $η$ on a manifold $M$ is said to be an infinitesimal symmetry for a tensor field $Q$ (maybe a function) defined on $M$ if

L_{η} Q = 0

This is equivalent to saying that $Q$ is invariant under the flow $g_{η}^{t}$ associated to $η$ , i.e.,

(g_{η}^{t})^{*} (Q) = Q

On the other hand, a vector field $η$ on $M$ can be lifted to a unique vector field $\tilde{η}$ on $T M$ in such a way that its flow is the corresponding lifted flow. I think it should have something to do with the prolongation of vector fields...

This means that for $(m, v) \in T M$ and $t \in R$ , the lifted flow takes $m$ to $g_{η}^{t} (m)$ and $v$ to the Lie transport of $v$ under $g_{η}^{t}$ .

It is easy to show that, in coordinates, if $m = (q_{i})$ , $v = ({\dot{q}}_{i})$ and $η = (a_{i})$ :

\tilde{η} = \sum_{i = 1}^{n} (a_{i} \frac{\partial}{\partial q_{i}} + \sum_{j = 1}^{n} {\dot{q}}_{j} \frac{\partial a_{i}}{\partial q_{j}} \frac{\partial}{\partial {\dot{q}}_{i}})

Theorem:
Let $L : T M \mapsto R$ be a Lagrangian function, and $η = (a_{i})$ a vector field such that $\tilde{η}$ is an infinitesimal symmetry for $L$ . Then for any critical path $γ$ for $L$ , the function

\sum_{i = 1}^{n} a_{i} \frac{\partial L}{\partial {\dot{q}}_{i}}

is constant along the lift of $γ$ on $T M$ .

Proof: A brief introduction to physics for mathematicians, page 14.

In QM

It is almost trivial. See formulation of QM#Symmetries.

In Lagrangian field theory

See @olver86 Theorem 4.29.
Suppose $G$ is a one-parameter local group of transformations of variational symmetry of the variational problem

J [u] = \int_{Ω} L (x, u^{(n)}) d x

Let

v = \sum_{i = 1}^{p} ξ^{i} (x, u) \frac{\partial}{\partial x^{i}} + \sum_{α = 1}^{q} ϕ_{α} (x, u) \frac{\partial}{\partial u^{α}}

be the infinitesimal generator of $G$ , and $Q = (Q_{1}, \dots, Q_{q})$ its characteristic. Then $Q$ is also the characteristic of a conservation law for the Euler-Lagrange equations.