actionconservation-lawslagrangian-formalismnoethers-theorem
From the book Introduction to Classical Mechanics With Problems and Solutions by David Morin, page 236 states:
Noether's Theorem: For each symmetry of the Lagrangian, there is a conserved quantity.
Whereas the Wikipedia page states:
Noether's (first) theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law
Does the action and Lagrangian have identical symmetries and conserved quantities?
There are already several good answers. However, the off-shell aspect related to Noether Theorem has not been addressed so far. (The words on-shell and off-shell refer to whether the equations of motion (e.o.m.) are satisfied or not.) Let us rephrase the problem as follows.
Consider a (not necessarily isolated) Hamiltonian system with $N$ degrees of freedom (d.o.f.). The phase space has $2N$ coordinates, which we denote $(z^1, \ldots, z^{2N})$.
(We shall have nothing to say about the corresponding Lagrangian problem.)
Symplectic structure. Usually, we work in Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, with the canonical symplectic potential one-form
$$\vartheta=\sum_{i=1}^N p_i dq^i.$$
However, it turns out to be more efficient in later calculations, if we instead from the beginning consider general coordinates $(z^1, \ldots, z^{2N})$ and a general (globally defined) symplectic potential one-form
$$\vartheta=\sum_{I=1}^{2N} \vartheta_I(z;t) dz^I,$$
with non-degenerate (=invertible) symplectic two-form
$$\omega = \frac{1}{2}\sum_{I,J=1}^{2N} \omega_{IJ} \ dz^I \wedge dz^J = d\vartheta,\qquad\omega_{IJ} =\partial_{[I}\vartheta_{J]}=\partial_{I}\vartheta_{J}-\partial_{J}\vartheta_{I}. $$
The corresponding Poisson bracket is
$$\{f,g\} = \sum_{I,J=1}^{2N} (\partial_I f) \omega^{IJ} (\partial_J g), \qquad \sum_{J=1}^{2N} \omega_{IJ}\omega^{JK}= \delta_I^K. $$
Action. The Hamiltonian action $S$ reads $$ S[z]= \int dt\ L_H(z^1, \ldots, z^{2N};\dot{z}^1, \ldots, \dot{z}^{2N};t),$$ where $$ L_H(z;\dot{z};t)= \sum_{I=1}^{2N} \vartheta_I(z;t) \dot{z}^I- H(z;t) $$ is the Hamiltonian Lagrangian. By infinitesimal variation $$\delta S = \int dt\sum_{I=1}^{2N}\delta z^I \left( \sum_{J=1}^{2N}\omega_{IJ} \dot{z}^J-\partial_I H - \partial_0\vartheta_I\right)+ \int dt \frac{d}{dt}\sum_{I=1}^{2N}\vartheta_I \delta z^I, \qquad \partial_0 \equiv\frac{\partial }{\partial t},$$ of the action $S$, we find the Hamilton e.o.m. $$ \dot{z}^I \approx \sum_{J=1}^{2N}\omega^{IJ}\left(\partial_J H + \partial_0\vartheta_J\right) = \{z^I,H\} + \sum_{J=1}^{2N}\omega^{IJ}\partial_0\vartheta_J. $$ (We will use the $\approx$ sign to stress that an equation is an on-shell equation.)
Constants of motion. The solution $$z^I = Z^I(a^1, \ldots, a^{2N};t)$$ to the first-order Hamilton e.o.m. depends on $2N$ constants of integration $(a^1, \ldots, a^{2N})$. Assuming appropriate regularity conditions, it is in principle possible to invert locally this relation such that the constants of integration $$a^I=A^I(z^1, \ldots, z^{2N};t)$$ are expressed in terms of the $(z^1, \ldots, z^{2N})$ variables and time $t$. These functions $A^I$ are $2N$ (locally defined) constants of motion (c.o.m.), i.e., constant in time $\frac{dA^I}{dt}\approx0$. Any function $B(A^1, \ldots, A^{2N})$ of the $A$'s, but without explicit time dependence, will again be a c.o.m. In particular, we may express the initial values $(z^1_0, \ldots, z^{2N}_0)$ at time $t=0$ as functions $$Z^J_0(z;t)=Z^J(A^1(z;t), \ldots, A^{2N}(z;t); t=0)$$ of the $A$'s, so that $Z^J_0$ become c.o.m.
Now, let $$b^I=B^I(z^1, \ldots, z^{2N};t)$$ be $2N$ independent (locally defined) c.o.m., which we have argued above must exist. OP's title question in this formulation then becomes if there exist $2N$ off-shell symmetries of the (locally defined) action $S$, such that the corresponding Noether currents are on-shell c.o.m.?
Remark. It should be stressed that an on-shell symmetry is a vacuous notion, because if we vary the action $\delta S$ and apply e.o.m., then $\delta S\approx 0$ vanishes by definition (modulo boundary terms), independent of what the variation $\delta$ consists of. For this reason we often just shorten off-shell symmetry into symmetry. On the other hand, when speaking of c.o.m., we always assume e.o.m.
Change of coordinates. Since the action $S$ is invariant under change of coordinates, we may simply change coordinates $z\to b = B(z;t)$ to the $2N$ c.o.m., and use the $b$'s as coordinates (which we will just call $z$ from now on). Then the e.o.m. in these coordinates are just $$\frac{dz^I}{dt}\approx0,$$ so we conclude that in these coordinates, we have $$ \partial_J H + \partial_0 \vartheta_J=0$$ as an off-shell equation. [An aside: This implies that the symplectic matrix $\omega_{IJ}$ does not depend explicitly on time, $$\partial_0\omega_{IJ} =\partial_0\partial_{[I}\vartheta_{J]}=\partial_{[I} \partial_0\vartheta_{J]}=-\partial_{[I}\partial_{J]} H=0.$$ Hence the Poisson matrix $\{z^I,z^J\}=\omega^{IJ}$ does not depend explicitly on time. By Darboux Theorem, we may locally find Darboux coordinates $(q^1, \ldots, q^N; p_1, \ldots, p_N)$, which are also c.o.m.]
Variation. We now perform an infinitesimal variation $\delta= \varepsilon\{z^{I_0}, \cdot \}$, $$\delta z^J = \varepsilon\{z^{I_0}, z^J\}=\varepsilon \omega^{I_0 J},$$ with Hamiltonian generator $z^{I_0}$, where $I_0\in\{1, \ldots, 2N\}$. It is straightforward to check that the infinitesimal variation $\delta= \varepsilon\{z^{I_0}, \cdot \}$ is an off-shell symmetry of the action (modulo boundary terms) $$\delta S = \varepsilon\int dt \frac{d f^0}{dt}, $$ where $$f^0 = z^{I_0}+ \sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J.$$ The bare Noether current is $$j^0 = \sum_{J=1}^{2N}\frac{\partial L_H}{\partial \dot{z}^J} \omega^{I_0 J}=\sum_{J=1}^{2N}\omega^{I_0J}\vartheta_J,$$ so that the full Noether current $$ J^0=j^0-f^0=-z^{I_0} $$ becomes just (minus) the Hamiltonian generator $z^{I_0}$, which is conserved on-shell $\frac{dJ^0}{dt}\approx 0$ by definition.
So the answer to OP's title question is Yes in the Hamiltonian case.
I) For a mathematical precise treatment of an inverse Noether's Theorem, one should consult e.g. Olver's book (Ref. 1, Thm. 5.58), as user orbifold also writes in his answer (v2). Here we would like give a heuristic and less technical discussion, to convey the heart of the matter, and try to avoid the language of jets and prolongations as much as possible.
In popular terms, we would like to formulate an "inverse Noether machine"
$$ \text{Input: Lagrangian system with known conservation laws} $$ $$ \Downarrow $$ $$ \text{[inverse Noether machine]} $$ $$ \Downarrow $$ $$ \text{Output: (quasi)symmetries of action functional} $$
Since this "machine" is supposed to be a mathematical theorem that should succeed everytime without exceptions (else it is by definition not a theorem!), we might have to narrow down the set/class/category of inputs that we allow into the machine in order not to have halting errors/breakdowns in the machinery.
II) Let us make the following non-necessary restrictions for simplicity:
Let us focus on point mechanics with a local action functional $$ S[q] ~=~ \int\! dt~ L(q(t), \frac{dq(t)}{dt}, \ldots,\frac{d^Nq(t)}{dt^N} ;t), \tag{1} $$ where $N\in\mathbb{N}_0$ is some finite order. Generalization to classical local field theory is straightforward.
Let us restrict to only vertical transformations $\delta q^i$, i.e., any horizontal transformation $\delta t=0$ vanishes. (Olver essentially calls these evolutionary vector fields, and he mentions that it is effectively enough to study those (Ref. 1, Prop. 5.52).)
Let us assume, as Olver also does, that the Lagrangian $L$ and the transformations are real analytic$^{\dagger}$.
The following technical restrictions/extensions are absolutely necessary:
The notion of symmetry $\delta S=0$ should be relaxed to quasisymmetry (QS). By definition a QS of the action $S$ only has to hold modulo boundary terms. (NB: Olver uses a different terminology: He calls a symmetry for a strict symmetry, and a quasisymmetry for a symmetry.)
The notion of QS transformations might only make sense infinitesimally/as a vector field/Lie algebra. There might not exist corresponding finite QS transformations/Lie group. In particular, the QS transformations are allowed to depend on the velocities $\dot{q}$. (Olver refers to this as generalized vector fields (Ref. 1, Def. 5.1).)
III) Noether's Theorem provides a canonical recipe of how to turn a QS of the action $S$ into a conservation law (CL),
$$ \frac{dQ}{dt}~\approx~0,\tag{2}$$
where $Q$ is the full Noether charge. (Here the $\approx$ symbol means equality on-shell, i.e. modulo the equations of motion (eom).)
Remark 1: Apart from time $t$, the QS transformations are only allowed to act on the variables $q^i$ that actively participate in the action principle. If there are passive external parameters, say, coupling constants, etc, the fact that they are constant in the model are just trivial CLs, which should obviously not count as genuine CLs. In particular, $\frac{d1}{dt}=0$ is just a trivial CL.
Remark 2: A CL should by definition hold for all solutions, not just for a particular solution.
Remark 3: A QS of the action $S$ is always implicitly assumed to hold off-shell. (It should be stressed that an on-shell QS of the action
$$ \delta S \approx \text{boundary terms}\tag{3} $$
is a vacuous notion, as the Euler-Lagrange equations remove any bulk term on-shell.)
Remark 4: It should be emphasized that a symmetry of eoms does not always lead to a QS of the Lagrangian, cf. e.g. Ref. 2, Example 1 below, and this Phys.SE post. Hence it is important to trace the off-shell aspects of Noether's Theorem.
Example 1: A symmetry of the eoms is not necessarily a QS of the Lagrangian. Let the Lagrangian be $L=\frac{1}{2}\sum_{i=1}^n \dot{q}^i g_{ij} \dot{q}^j$, where $g_{ij}$ is a constant non-degenerate metric. The eoms $\ddot{q}^i\approx 0$ have a $gl(n,\mathbb{R})$ symmetry $\delta q^{i}=\epsilon^i{}_j~q^{i}$, but only an $o(n,\mathbb{R})$ Lie subalgebra of the $gl(n,\mathbb{R})$ Lie algebra is a QS of the Lagrangian.
IV) Without further assumptions, there is a priori no guarantee that the Noether recipe will turn a QS into a non-trivial CL.
Example 2: Let the Lagrangian $L(q)=0$ be the trivial Lagrangian. The variable $q$ is pure gauge. Then the local gauge symmetry $\delta q(t)=\epsilon(t)$ is a symmetry, although the corresponding CL is trivial.
Example 3: Let the Lagrangian be $L=\frac{1}{2}\sum_{i=1}^3(q^i)^2-q^1q^2q^3$. The eom are $q_1\approx q_2q_3$ and cyclic permutations. It follows that the positions $q^i\in\{ 0,\pm 1\}$ are constant. (Only $1+1+3=5$ out of the $3^3=27$ branches are consistent.) Any function $Q=Q(q)$ is a conserved quantity. The transformation $\delta q^i=\epsilon \dot{q}^i$ is a QS of the action $S$.
If we want to formulate a bijection between QSs and CLs, we must consider equivalence classes of QSs and CLs modulo trivial QSs and CLs, respectively.
A CL is called
trivial of first kind if the Noether current $Q$ vanishes on-shell.
trivial of second kind if CL vanishes off-shell.
trivial if it is a linear combination of CLs of first and second kinds (Ref. 1, p.264-265).
V) The most crucial assumption is that the eoms are assumed to be (totally) non-degenerate. Olver writes (Ref. 1, Def. 2.83.): A system of differential equations is called totally non-degenerate if it and all its prolongations are both of maximal rank and locally solvable$^{\ddagger}$.
The non-degeneracy assumption exclude that the action $S$ has a local gauge symmetry. If $N=1$, i.e. $L=L(q,\dot{q},t)$, the non-degeneracy assumption means that the Legendre transformation is regular, so that we may easily construct a corresponding Hamiltonian formulation $H=H(q,p,t)$. The Hamiltonian Lagrangian reads
$$ L_H~=~p_i \dot{q}^i-H.\tag{4} $$
VI) For a Hamiltonian action functional $S_H[p,q] = \int\! dt~ L_H$, there is a canonical way to define an inverse map from a conserved quantity $Q=Q(q,p,t)$ to a transformation of $q^i$ and $p_i$ by using the Noether charge $Q$ as Hamiltonian generator for the transformations, as also explained in e.g. my Phys.SE answer here. Here we briefly recall the proof. The on-shell CL (2) implies
$$ \{Q,H\}+\frac{\partial Q}{\partial t}~=~0\tag{5} $$
off-shell, cf. Remark 2 and this Phys.SE post. The corresponding transformation
$$ \delta q^i~=~ \{q^i,Q\}\epsilon~=~\frac{\partial Q}{\partial p_i}\epsilon\qquad \text{and}\qquad \delta p_i~=~ \{p_i,Q\}\epsilon~=~-\frac{\partial Q}{\partial q^i}\epsilon\tag{6} $$
is a QS of the Hamiltonian Lagrangian
$$\begin{align} \delta L_H ~\stackrel{(4)}{=}~&\dot{q}^i \delta p_i -\dot{p}_i \delta q^i -\delta H+\frac{d}{dt}(p_i \delta q^i)\cr ~\stackrel{(6)+(8)}{=}& -\dot{q}^i\frac{\partial Q}{\partial q^i}\epsilon -\dot{p}_i\frac{\partial Q}{\partial p_i}\epsilon -\{H,Q\}\epsilon + \epsilon \frac{d Q^0}{dt}\cr ~\stackrel{(5)}{=}~& \epsilon \frac{d (Q^0-Q)}{dt}~\stackrel{(9)}{=}~ \epsilon \frac{d f^0}{dt},\end{align}\tag{7} $$
because $\delta L_H$ is a total time derivative. Here $Q^0$ is the bare Noether charge
$$ Q^0~=~ \frac{\partial L_H}{\partial \dot{q}^i} \{q^i,Q\} + \frac{\partial L_H}{\partial \dot{p}_i} \{p_i,Q\} ~=~ p_i \frac{\partial Q}{\partial p_i},\tag{8} $$
and
$$ f^0~=~ Q^0-Q .\tag{9} $$
Hence the corresponding full Noether charge
$$ Q~=~Q^0-f^0\tag{10} $$
is precisely the conserved quantity $Q$ that we began with. Therefore the inverse map works in the Hamiltonian case.
Example 4: The non-relativistic free particle $L_H=p\dot{q}-\frac{p^2}{2m}$ has e.g. the two conserved charges $Q_1=p$ and $Q_2=q-\frac{pt}{m}$.
The inverse Noether Theorem for non-degenerate systems (Ref. 1, Thm. 5.58) can intuitively be understood from the fact, that:
Firstly, there exists an underlying Hamiltonian system $S_H[p,q]$, where the bijective correspondence between QS and CL is evident.
Secondly, by integrating out the momenta $p_i$ we may argue that the same bijective correspondence holds for the original Lagrangian system.
VII) Finally, Ref. 3 lists KdV and sine-Gordon as counterexamples to an inverse Noether Theorem. KdV and sine-Gordon are integrable systems with infinitely many conserved charges $Q_n$, and one can introduce infinitely many corresponding commuting Hamiltonians $\hat{H}_n$ and times $t_n$. According to Olver, KdV and sine-Gordon are not really counterexamples, but just a result of a failure to properly distinguishing between non-trivial and trivial CL. See also Ref. 4.
References:
P.J. Olver, Applications of Lie Groups to Differential Equations, 1993.
V.I. Arnold, Mathematical methods of Classical Mechanics, 2nd eds., 1989, footnote 38 on p. 88.
H. Goldstein, Classical Mechanics; 2nd eds., 1980, p. 594; or 3rd eds., 2001, p. 596.
L.H. Ryder, Quantum Field Theory, 2nd eds., 1996, p. 395.
$^{\dagger}$ Note that if one abandons real analyticity, say for $C^k$ differentiability instead, the analysis may become very technical and cumbersome. Even if one works with the category of smooth $C^\infty$ functions rather than the category of real analytic functions, one could encounter the Lewy phenomenon, where the equations of motion (eom) have no solutions at all! Such situation would render the notion of a conservation law (CL) a bit academic! However, even without solutions, a CL may formally still exists as a formal consequence of eoms. Finally, let us add that if one is only interested in a particular action functional $S$ (as opposed to all action functionals within some class) most often, much less differentiability is usually needed to ensure regularity.
$^{\ddagger}$ Maximal rank is crucial, while locally solvable may not be necessary, cf. previous footnote.
Best Answer
Yes, provided one uses the correct notions of symmetry for the action and the lagrangian.
The setup.
We assume throughout that the action can be written as the integral of a local Lagrangian. Namely, let $\mathcal C$ be the configuration space of the system, then for any admissible path $q:[t_a, t_b]\to \mathcal C$, there exists a local function $L$ of paths such that \begin{align} S[q] = \int_{t_a}^{t_b} dt \,L_q(t). \end{align} Let a smooth, $\epsilon$-deformation $q(t) \to \hat q(t, \epsilon)$ of paths be given. We will use the $\delta$ notation for first order changes in quantities under such a deformation.
Symmetry defined.
We say that this deformation is a symmetry of the action $S$ provided there exists a local function of paths $B_q$ such that \begin{align} \delta S[q] = B_q(t_b) - B_q(t_a) \end{align} for all admissible paths $q:[t_a, t_b]\to \mathcal C$. In other words, the action only changes to first order by a boundary term. We say that this deformation is a symmetry (or what Qmechanic calls a quasisymmetry in his response) of the Lagrangian $L$ provided there exists a local function $\Lambda_q$ of paths such that \begin{align} \delta L_q(t) = \frac{d\Lambda_q}{dt}(t) \end{align} for all admissible paths $q:[t_a, t_b]\to \mathcal C$. In other words, the lagrangian only changes to first order up to a total derivative.
Equivalence of notions of symmetry.
Using these definitions, one can show that a given deformation is a symmetry of $S$ if and only if it is a symmetry of $L$.
Notice that for any deformation, and for any admissible path $q:[t_a, t_b]\to \mathcal C$, one has \begin{align} \delta S[q] = \int_{t_a}^{t_b} dt\,\delta L_q(t) \end{align} Suppose now, that a given deformation is a symmetry of $S$, and let a path $q:[t_a, t_b]\to\mathcal C$ be given. For each $t\in [t_a, t_b]$ we have \begin{align} \int_{t_a}^{t} dt'\,\delta L_{q}(t') = B_{q}(t) -B_{q}(t_a), \end{align} Since the deformation is a symmetry of $S$. Taking the derivative of both sides with respect to $t$, and using the fundamental theorem of calculus on the left, we obtain \begin{align} \delta L_q(t) = \dot B_q(t) \end{align} for all $t\in[t_a, t_b]$. Identifying $B$ with $\Lambda$, we find that the deformation is a symmetry of the lagrangian.
I'll leave the converse to you.