It is well-known that the hidden symmetry of the $1/r$-problem $$H=\frac{{\bf p}^2}{2m}-\frac{k}{r}\tag{1}$$ is ${\rm SO(4)}$ in the sense that the components of angular momentum $L_i$ and those of the Runge-Lenz vector $A_i$ satisfy the $\mathfrak{so}(4)$ algebra, and also $[H,L_i]=[H,A_i]=0$ in quantum mechanics (QM). Therefore, in QM the invariance of the Hamiltonian under the ${\rm SO}(4)$ is trivial to check in the sense $$H'=UHU^{-1}=H\tag{2}$$ where $U$ is the unitary operator implementing the ${\rm SO}(4)$ symmetry and involves the exponential of the linear combinations of $L_i$'s and $A_i$'s. That part is fine!

Question How about the classical scenario? When we talk about the ${\rm SO}(3)$ symmetry of the Hamiltonian of a central force problem, the symmetry is manifest in both the potential and kinetic term.

${\rm SO}(4)$ transformations can be thought of transformations which leave quantities of the form $$\Delta\equiv \zeta_1^2+\zeta_2^2+\zeta_3^2+\zeta^2_4,~~ \zeta_i\in\mathbb{R}\tag{3}$$ invariant. Is there a way to reduce the Hamiltonian of $(1)$ in the form of $\Delta$ so that the ${\rm SO}(4)$ symmetry becomes manifest?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/89654/2451, physics.stackexchange.com/q/18088/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 28 '20 at 12:44
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    $\begingroup$ I have a suspicion that the answer is in this very nice article by John Baez on the RL vector and SO(4) symmetry. I believe the article shows that energy conservation requires that a quantity called the "4-velocity" (no relation to four-vectors from relativity) -- which uses slightly complicated re-parameterised coordinates -- stay on the surface of a unit 3-sphere. $\endgroup$
    – Philip
    Dec 28 '20 at 13:47
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    $\begingroup$ My teacher Bargmann, in his epochal 1936 paper introduces the coordinates ξ covering Poisson Brackets as well. $\endgroup$ Dec 28 '20 at 14:49
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Dec 30 '20 at 16:31

The classical treatment is actually simpler than the quantum one, since there are no operator ordering problems. This is what Pauli relied on in his solution of the Hydrogen atom, before Schroedinger, as every decent physicist of his generation had worked out the Kepler problem like this.

The idea is to find as many independent constants of the motion Q as possible, and break them up into independent, ideally PB-commuting, quantities.

Every such invariant must PB-commute with the Hamiltonian, since that is its time derivative, $$ \{ Q,H \}= 0.$$

As a result, the analog of your quantum (2) is $$ H'= e^{ \theta \{Q ,} H= H+\theta \{ Q,H\}+ {\theta^2 \over 2!} \{Q,\{Q,H \}\}+... =H. $$ The linear first order phase-space gradients $\{Q, \bullet \}={ \partial Q\over \partial x^i} {\partial \over \partial p^i}- { \partial Q\over\partial p^i } {\partial \over \partial x^i} $ are the generators of Lie transformations in phase space. They close into a Lie algebra purely classically, via PB commutation, $$ \{Q, \{S, - \{S, \{ Q ,= \{ \{Q,S\}, $$ by Jacobi's identity. They are thus Lie algebra elements, and can act adjointly.

For the specific Kepler problem, let me follow an older talk of mine.

$$ H = \frac{{\bf p}^2}{2}-\frac{1}{r} ~, $$ in simplified (rescaled) notation.

The invariants of the hamiltonian are the angular momentum vector, $$ {\bf L} = {\bf r} \times {\bf p}~, $$ and the Hermann-Bernoulli-Laplace vector, now usually called the Pauli-Runge-Lenz vector, $$ {\bf A}= {\bf p} \times {\bf L} -\hat{{\bf r}}~. $$ (Dotting it by $\hat{{\bf r}}$ instantly yields Kepler's elliptical orbits by inspection, $\hat{{\bf r}}\cdot {\bf A}+1 = {\bf L}^2 /r$.)

Since $~{\bf A}\cdot {\bf L}=0$, it follows that $$ H=\frac{{\bf A}^2-1}{2{\bf L}^2 } ~~. $$ However, to simplify the PB Lie-algebraic structure, $$ \{ L_i, L_j \} = \epsilon^{ijk} L_k, \qquad \{ L_i, A_j \} = \epsilon^{ijk} A_k, \qquad \{ A_i, A_j \} =-2H \epsilon^{ijk} L_k, $$ it is useful to redefine ${\bf D} \equiv \frac{{\bf A}}{\sqrt{- 2 H}}$, and further $$ {\cal R}\equiv {\bf L} + {\bf D}, \qquad {\cal L}\equiv {\bf L} - {\bf D}. $$ These six simplified invariants obey the standard $SU(2)\times SU(2) \sim SO(4)$ symmetry algebra, $$ \{ {\cal R}_i, {\cal R}_j \} = \epsilon^{ijk} {\cal R}_k, \qquad \{ {\cal R}_i, {\cal L}_j \} = 0, \qquad \{ {\cal L}_i, {\cal L}_j \} =\epsilon^{ijk} {\cal L}_k, $$ and depend on each other and the hamiltonian through
$$ \bbox[yellow]{H=\frac{-1}{2 {\cal R}^2}=\frac{-1}{2 {\cal L}^2} }~, $$ so only five of the invariants are actually algebraically independent.

It is then manifest that all six respective infinitesimal transformations leave their Casimirs invariant, the Casimirs of the other SU(2) invariant, and hence the Hamiltonian invariant; consequently the corresponding finite transformations do likewise, $$ H'= e^{ \theta^i \{ {\cal R}_i,} ~ e^{\phi^j\{ {\cal L}_j ,} ~ H= H ~. $$

You could work out these explicit infinitesimal phase-space transformations involved, if you were really inclined, and Fock and Bargmann do that, as linked in the comments, but this is the whole point of systematic formalism, that it cuts through the confusion and forestalls missing the forest for the trees. It's Lie's major contribution to geometry.

  • $\begingroup$ Thanks for the detailed answer. I did not know that the Hamiltonian transforms as $H'=e^{\theta\{Q,}H$ in classical mechanics. I never encountered your second equation in my classical mechanical courses. Where can I find this discussion? Goldstein doesn't discuss this. Right? A follow-up question: So if I want to check whether the Hamiltonian is invariant under spatial rotations, I must do $H'=e^{\theta\{{\bf L},}H$. Am I right? For the case in hand, I must separately show that $H'=e^{\theta\{{\bf L},}H=H$ and $H'=e^{\theta\{{\bf A},}H=H$. Do I get it right? @CosmasZachos $\endgroup$
    – SRS
    Dec 29 '20 at 7:23
  • $\begingroup$ Yes. This is the exponentiation of the infinitesimal transformation. Upon quantization, this becomes the adjoint action (Hadamard) lemma for commutators, showcased in the BCH article. $\endgroup$ Dec 29 '20 at 12:34
  • $\begingroup$ Appendix A of this paper might help. $\endgroup$ Dec 29 '20 at 13:43
  • $\begingroup$ Also, this answer. $\endgroup$ Dec 29 '20 at 16:22

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