Lecture 13–14 — Dynamical Symmetry: Noether’s Theorem, the Symplectic Group, and the Kepler Problem#
Source: NPTEL Classical Physics, Mod-01 Lec-13 and Lec-14 (Dynamical symmetry, Parts 1–2), Prof. V. Balakrishnan.
Lecture 10–12 left the words “symmetry,” “invariance,” and “conservation” doing a lot of unexamined work — rotational symmetry gives conserved angular momentum, translational symmetry gives conserved linear momentum, and it was left at the level of familiar special cases. This lecture makes the relationship precise (Noether’s theorem), then asks the sharper question those examples were hiding: given a Hamiltonian, what is the entire group of transformations that preserves it, and how large is that group really? The two-dimensional isotropic oscillator and the Kepler problem turn out to hide symmetry groups — \(SU(2)\) and \(SO(4)\) — much bigger than “rotational symmetry” would suggest, and that extra size is exactly what produces the “accidental” degeneracy of the hydrogen atom in quantum mechanics.
Noether’s theorem in the Lagrangian framework#
Formalize what “symmetry” means before relating it to anything else. Consider a continuous, one-parameter family of coordinate changes \(q \to Q(\alpha)\), with \(Q(0) = q\) the identity transformation, and write the Lagrangian evaluated along the transformed coordinates as \(L' \equiv L\big(Q(\alpha), \dot Q(\alpha)\big)\). The transformation is a symmetry if \(L'=L\) for every \(\alpha\) — not merely at \(\alpha=0\) — i.e. if
Since \(\alpha\) parameterizes a change of coordinates and has nothing to do with \(t\), the derivative and \(d/d\alpha\) commute: \(d\dot Q_i/d\alpha = d/dt\,(dQ_i/d\alpha)\). Using the Euler–Lagrange equations \(\partial L/\partial Q_i = \frac{d}{dt}\partial L/\partial \dot Q_i\) turns the first term into a total time derivative too, and the whole expression collapses to
This is Noether’s theorem: every continuous symmetry of \(L\) hands over one conserved quantity, built from the momentum dotted into the direction the symmetry moves each coordinate. Rotating \(q_1, q_2\) by an angle \(\alpha\) leaves a central potential \(V(q_1^2+q_2^2)\) invariant for every \(\alpha\), and the theorem’s recipe reproduces angular momentum; shifting the origin reproduces linear momentum; shifting the origin of time (an invariance of the equations of motion, not literally of this derivation’s \(\alpha\)-family) reproduces the Hamiltonian itself. The theorem generalizes twice: to several parameters at once (three rotation angles give three conserved angular-momentum components), and to symmetries where \(L\) is invariant only up to a total time derivative, \(L' - L = df/dt\) — since adding \(df/dt\) to a Lagrangian never changes the equations of motion, this is already the most general notion of invariance the formalism admits, and it is exactly how gauge invariance of the electromagnetic field hands over conservation of charge. What is essential throughout is that the transformations be continuous and connected to the identity — parity (a right-handed frame flipping to a left-handed one) cannot be reached by any continuous \(\alpha\)-family, so it sits outside what Noether’s theorem can say anything about.
Dynamical symmetry in the Hamiltonian framework: the symplectic group#
The Hamiltonian version sharpens this considerably, because it separates two conditions that the Lagrangian derivation runs together. A transformation \((q,p) \to (Q,P)\) that is going to qualify as a dynamical symmetry must first be a canonical transformation — otherwise Hamilton’s equations don’t even keep their form — and it must leave the Hamiltonian’s functional form unchanged, \(K=H\), so that the transformed trajectories really are solutions of the same physical system rather than of some other one. The first condition alone is already a rich, well-defined group; the second cuts out a subgroup of it.
Recall \(J = \begin{pmatrix}0 & I\\ -I & 0\end{pmatrix}\) from Lecture 10–12, with \(J^2=-I\). Stacking \(q\)’s and \(p\)’s into \(\vec x\) and writing the Jacobian of a transformation as \(M \equiv \partial \vec X/\partial \vec x\), the canonical Poisson-bracket relations \(\{X_i,X_j\}=J_{ij}\) translate directly into a matrix condition on \(M\):
Any matrix satisfying this is called symplectic. It is the direct analog of the orthogonality condition \(R^TR=I\) that defines a rotation — \(J\) plays the role of the metric, so this is “orthogonality with respect to \(J\)” rather than the ordinary Euclidean one. Every symplectic matrix is automatically invertible (its inverse is built from \(M^T\) exactly as an orthogonal matrix’s inverse is its transpose, just with \(J\)’s inserted), and the product of two symplectic matrices is again symplectic — canonical transformations compose into a single canonical transformation, exactly as expected. The set of all \(2n\times 2n\) real symplectic matrices therefore forms a group, denoted \(Sp(2n,\mathbb R)\): the group of canonical transformations for an \(n\)-degree-of-freedom system.
Counting parameters. Write an infinitesimal canonical transformation as \(M = I + \epsilon G\) for a small parameter \(\epsilon\) and a matrix \(G\) called the generator. Substituting into \(M^TJM=J\) and keeping only first order in \(\epsilon\) gives
the infinitesimal (linearized) version of the symplectic condition. Counting independent solutions \(G\) of this equation gives the number of generators — the dimension of the group — and the count for \(Sp(2n,\mathbb R)\) works out to \(n(2n+1)\), i.e. \(2n^2+n\). This is worth comparing to the familiar rotation group: \(SO(n)\), the group of \(n\times n\) real orthogonal matrices with determinant \(+1\), has \(\tfrac12 n(n-1)\) generators — one per independent coordinate plane a rotation can act in, which is why \(n\) Euclidean dimensions carry \(n\) mutually orthogonal rotation planes only when \(n=3\) (the unique integer solution of \(\tfrac12n(n-1)=n\), the reason “rotation about an axis” is even a sensible phrase only in three dimensions; in general, a rotation is defined by homogeneous, orthogonal, unimodular — never by an axis). \(U(n)\), the unitary \(n\times n\) matrices, has \(n^2\) generators, and its determinant-one subgroup \(SU(n)\) has \(n^2-1\).
The simplest nontrivial symplectic matrix is \(M=J\) itself, generating exactly the \(Q=p,\,P=-q\) swap met in Lecture 10–12 — a canonical transformation, but not a symmetry of any particular \(H\) unless \(H\) itself happens to be invariant under it. That is the crux of the whole lecture: canonical transformations that leave \(H\) unchanged form a subgroup of \(Sp(2n,\mathbb R)\), and finding that subgroup for a given Hamiltonian is, in general, hard — nothing here hands it to you the way Noether’s recipe hands you a conserved quantity from a symmetry you already knew about.
Worked example: the 2D isotropic oscillator and \(SU(2)\)#
Take two decoupled oscillators of equal frequency, in units where \(m=\omega=1\):
This is separable and hence integrable, with \(F_1=\tfrac12(q_1^2+p_1^2)\) and \(F_2=\tfrac12(q_2^2+p_2^2)\) constant and in involution, and equal frequencies make every orbit periodic. But \(H\) has more structure than “two independent oscillators” suggests: it is literally the sum of the squares of all four phase-space coordinates, so any rotation of \((q_1,q_2,p_1,p_2)\) as four real numbers in \(\mathbb R^4\) — mixing coordinates and momenta freely, not just rotating \(q_1,q_2\) into each other in physical space — leaves \(H\) exactly unchanged. That symmetry group is \(SO(4)\), with \(\tfrac12\cdot4\cdot3=6\) generators, while the canonical transformations available for a two-degree-of-freedom system form \(Sp(4,\mathbb R)\), with \(2(2)^2+2=10\) generators. These are different groups — an \(SO(4)\) rotation need not be canonical, and a canonical transformation need not preserve \(H\) — so the dynamical symmetry group of the isotropic oscillator is neither of them individually, but their intersection,
isomorphic to the group of \(2\times2\) unitary matrices with unit determinant. A general element of \(SU(2)\) can always be written \(\begin{pmatrix}\alpha&\beta\\-\beta^*&\alpha^* \end{pmatrix}\) with \(|\alpha|^2+|\beta|^2=1\); writing \(\alpha,\beta\) in terms of four real numbers, this constraint is exactly the equation of the unit sphere \(S^3\) sitting inside \(\mathbb R^4\) — the parameter space of \(SU(2)\) is a three-sphere. Counting generators directly, \(U(2)\) has \(2^2=4\), and fixing \(\det=1\) removes one, leaving \(3 = 2^2-1\), matching the general count \(n^2-1\) for \(SU(n)\).
The three generators of this \(SU(2)\) turn out to be
each a constant of the motion — \(\{J_i,H\}=0\) for all three — but, unlike \(F_1,F_2\), not mutually in involution:
the same Lie algebra as angular momentum (or, equivalently, as the Pauli matrices under \(\tfrac1{2i}\) times the commutator). This is the general lesson: a two-degree-of-freedom system has phase space of dimension four, so at most three functionally independent time-independent constants of the motion exist, and integrability only asks for two of them to be in involution. \(J_1,J_2,J_3\) (together with \(H\), and with \(F_1-F_2\) expressible as a combination of them) are extra constants beyond what integrability strictly requires — their existence, and their non-commuting algebra, is precisely the fingerprint of the dynamical symmetry sitting on top of plain integrability. The Jacobi identity guarantees that whenever two components of such a vector are conserved, so is the third automatically — the same mechanism that makes all three components of ordinary angular momentum conserved together for any central force.
The three-dimensional isotropic oscillator repeats the pattern one dimension up: its canonical transformations live in \(Sp(6,\mathbb R)\), and the analogous symmetry-plus-canonical subgroup is \(SU(3)\), with \(3^2-1=8\) generators — eight constants of the motion, obeying the \(SU(3)\) algebra, that are not required by integrability alone (three degrees of freedom needs only three constants in involution) but are forced by the extra spherical symmetry of the \(r^2\) potential.
The Kepler problem: the Laplace–Runge–Lenz vector and \(SO(4)\)#
Now the physically important case, \(H = \dfrac{p^2}{2m} - \dfrac{k}{r}\). As any central-force problem, it is integrable via \(H\), \(L^2\), and one component of \(\vec L\) (say \(L_z\)) — three independent, mutually commuting constants, since \(\vec L\)’s three Cartesian components are each separately conserved but not in involution with each other, \(\{L_i,L_j\}=\epsilon_{ijk}L_k\). What is special to the exact \(1/r\) potential — not \(1/r^2\), not \(r^2\), nothing else — is a further conserved vector, the Laplace–Runge–Lenz vector
Differentiating along a trajectory, using \(\dot{\vec p} = -\nabla V = -\dfrac{k}{r^2}\hat r\) (the actual force), \(\dot{\vec L}=0\), and the vector identity \(\vec a\times(\vec b\times\vec c) = (\vec a\cdot\vec c)\vec b - (\vec a\cdot\vec b)\vec c\) applied to \(\dot{\vec p}\times\vec L\), every term cancels exactly:
using \(\vec r\cdot\dot{\vec r} = r\dot r\) to simplify the last piece. The cancellation depends on the force being exactly \(1/r^2\) — nothing else makes it work. \(\vec A\) is constant in both magnitude and direction, it lies in the orbital plane (since it’s built from \(\vec r\) and \(\vec p\), both perpendicular to \(\vec L\)), and it points along the ellipse’s major axis. That constancy is the deep reason bound Kepler orbits are exactly closed ellipses that never precess: any departure from a pure \(1/r\) force — a \(1/r^2\) correction from general relativity, a perturbing planet, tidal effects — spoils the exact cancellation above and lets the orbit precess (Mercury precesses about \(43''\) per century from general relativity alone, against \(\sim 500''\) per century from ordinary planetary perturbations; some binary pulsars, with far larger masses and tighter orbits, precess by several degrees per revolution).
Counting constants of the motion now looks like an embarrassment of riches — \(H\) (1), \(\vec L\) (3), \(\vec A\) (3) — seven numbers in a six-dimensional phase space, so they cannot all be independent. Indeed \(\vec A\cdot\vec L=0\) (removing one component: \(\vec A\) lies entirely in the plane \(\vec L\) is normal to) and
so \(A\)’s magnitude is fixed once \(H\) and \(L^2\) are known. The genuinely new content is directional: \(\vec A\)’s constant direction, unlike \(\vec L\)’s, is not forced by rotational symmetry of the potential alone — a generic central potential \(V(r)\) conserves \(\vec L\) but lets its ellipse-analog precess freely. Suitably rescaled combinations \(\vec A \pm \vec L\) close into two decoupled angular-momentum-like algebras, together isomorphic to the algebra of rotations in four dimensions — the dynamical symmetry group of the Kepler problem (for bound, \(E<0\) orbits) is
with \(\tfrac12\cdot4\cdot3=6\) generators: the three components of \(\vec L\) plus three independent combinations built from \(\vec A\). Exactly as for the oscillator, this is larger than what rotational symmetry in ordinary 3-space alone would suggest — the extra generators mix “position-like” and “momentum-like” combinations, invisible to any transformation acting on \(\vec r\) alone. This is also why the Kepler and isotropic-oscillator potentials are the only two central potentials for which every bounded orbit is exactly closed: they are exactly the two potentials whose dynamical symmetry group is strictly bigger than the \(SO(3)\) that any central force gets for free.
From classical symmetry to quantum degeneracy#
Classical dynamical symmetry translates directly into quantum-mechanical degeneracy: whatever set of transformations leaves the energy unchanged relates states with the same energy but different quantum numbers. For a generic central potential, the energy depends on the principal and orbital quantum numbers \(n,\ell\) but not on the magnetic quantum number \(m\) (the ordinary \(SO(3)\) rotational degeneracy), giving \(2\ell+1\) states per \((n,\ell)\). The hydrogen atom’s spectrum has more degeneracy than this — the energy depends on \(n\) alone, not on \(\ell\) — and summing the ordinary degeneracy over every allowed \(\ell\),
reproduces the textbook hydrogen degeneracy exactly. This “accidental” degeneracy is not accidental at all: it is the direct quantum shadow of the classical \(SO(4)\) symmetry generated by \(\vec L\) and \(\vec A\) together, and it is lifted the instant the potential departs from pure \(1/r\) — which is exactly why the fine-structure correction (among other effects) splits the hydrogen \(2s\) and \(2p\) levels that a strict \(1/r\) potential could never tell apart.
Interlude: the bead on a rotating hoop#
Switching from symmetry to a worked exercise in the Lagrangian–Hamiltonian toolkit together: a bead of mass \(m\) slides frictionlessly on a vertical circular hoop of radius \(R\), and the whole hoop is spun about its vertical diameter at constant angular speed \(\omega\). Using the angle \(\theta\) from the bottom of the hoop, the bead’s position is \(\rho = R\sin\theta\) (distance from the spin axis), \(z=R(1-\cos\theta)\), and with the rotation \(\dot\varphi=\omega\) imposed as a constraint before varying, the kinetic energy collects a term \(\tfrac12 m\rho^2\omega^2 = \tfrac12 mR^2\omega^2\sin^2\theta\) that carries no \(\dot\theta\) at all — it lands in the Lagrangian with the sign of a kinetic term, but behaves exactly like an inverted potential, pushing the bead away from the axis. This is the centrifugal term, appearing automatically, with no extra bookkeeping, simply because the constraint was substituted into \(L\) directly — the same trick Lecture 7 used to eliminate the string tension in the Atwood machine.
The reduced, one-degree-of-freedom Lagrangian and its Hamiltonian (with \(p\) the momentum conjugate to \(\rho\)) are
and Hamilton’s equations give \(\dot\rho = p(R^2-\rho^2)/mR^2\) and
Critical points need \(\dot\rho=0\), forcing \(p=0\) (away from the singular endpoint \(\rho=R\)), and then \(\dot p=0\) requires either \(\rho=0\) — the bottom of the hoop, always an equilibrium — or \(\sqrt{R^2-\rho^2} = g/\omega^2\), which has a real solution only when \(\omega\) exceeds a threshold \(\omega_c = \sqrt{g/R}\):
Below the threshold, the bottom of the hoop is the only equilibrium, and it is stable (gravity always wins). Above it, that same equilibrium goes unstable and a new stable equilibrium branches off with a square-root singularity — \(\rho_{\rm eq}\) rises from zero with infinite slope right at \(\omega_c\) and saturates toward \(R\) as \(\omega\to\infty\), matching the physical expectation that a very fast spin pins the bead near the rim. This is a pitchfork bifurcation in the sense of Lecture 5–6’s classification of how equilibria are born, destroyed, or exchange stability as a parameter is varied — reached here not by studying an abstract one-dimensional flow, but by carrying a constrained mechanical system all the way from the Lagrangian, through the Hamiltonian, to a stability analysis of its own critical points, which is exactly the combination of tools the last several lectures were assembling piece by piece.