Symmetry and Symplectic Group of Hydrogenic Atom

Question

New version of the question: A simmetry needs to be canonical, following the first answer of this post which states:

the symmetry requirement is not necessary in the definition of canonical transformation, whereas a symmetry is by definition an (active) canonical transformation.

The geometric Simmetry generated by the Angular Momentum is $SO(3)$. By definition I need to intersect it with a Symplectic group $Sp(2n,R)$. I don't know how to do handle this.

The simmetry group generated by LRL vector $+$ rotational invariance is $SO(4)$. We can generate the $su(2)\times su(2)$ algebra without any effort and it gives us the right energy levels. This is the standard procedure when dealing with this problem, however we need to have a canonical transformation by definition. this post states that the intersection: \begin{equation} SO(4)\cap Sp(4,R) \simeq SU(2) \end{equation} Because an orthogonal and symplectic transformation in $R^{2n}$ must be unitary. However this leaves us with a trouble: we don't have the right algebra to build 2 different Casimir Operators and generate the right degeneracy of the system. As we know from Noether Theorem, there is a connection between degeneracy and symmetries and we looked for a different conserved quantity from L, exactly for this discrepancy.

This leads us to my issue. If the symmetry - the right one - of the system, generates the right degeneracy: why the intersection of momentum $+$ LRL vector with the symplectic group leads us to the wrong degeneracy?

Text before the edit: As far as I know the Hamiltonian is preserved under canonical transformations. These can be viewed as elements of the symplectic group $\operatorname{Sp}(2n,R)$.

Picking the hydrogenic Hamiltonian we spot a geometric symmetry $\operatorname{SO}(3)\simeq\operatorname{SU}(2)$ and a dynamic symmetry $\operatorname{SO}(4)\simeq\operatorname{SU}(2)\times\operatorname{SU}(2)$.

The latter gives us 2 different Casimir operators whose relations offer the right energy levels.

However I just read not all $\operatorname{SO}(4)$ transformation are canonical (I wonder why), so we must intersect as follows: \begin{equation} \operatorname{SO}(4)\cap\operatorname{Sp}(4,R) \simeq\operatorname{SU}(2) \end{equation} The reason lies in the fact every orthogonal and symplectic group is unitary.

However this falls in the geometric symmetry. Is it wrong? Were is the issue?

Answer 1

Even though the intro to your question conjures up Pauli's legendary quantization of the Hydrogen atom using the rotational so(3) symmetry and the suitably normalized LRL vector which can be combined with the above to yield the celebrated so(3)⊕so(3)~so(4) (note I am skipping groups in favor of Lie algebras, which are all that matter here!), your question is a thoroughly classical one.

Since I failed to understand your question (on account of the last two lines of it), I'll just jot down the straightforward part, the algebraic structure of the generators for the 3D canonical transformation generators, an sp(6), and their overlap with the above symmetry so(4).

For simplicity, I'll consider the so(3) rotation subalgebra; and the three peculiar components of the LRL vector symmetry underlain by the astounding Lie scaling symmetry in which the coordinates r and the time t are scaled by different powers of a parameter λ,

$$ t \rightarrow \lambda^{3}t , \qquad \mathbf{r} \rightarrow \lambda^{2}\mathbf{r} , \qquad\mathbf{p} \rightarrow \frac{1}{\lambda}\mathbf{p}. $$ This transformation changes the total angular momentum L and energy E, $ L \rightarrow \lambda L, \qquad E \rightarrow \frac{1}{\lambda^{2}} E, $ but preserves their product $EL^2$. Therefore, the eccentricity e and the magnitude A are preserved... $$ A^2 = m^2 k^2 e^{2} = m^2 k^2 + 2 m E L^2.$$ The direction of A is preserved as well, since the semiaxes are not altered by a global scaling. This transformation also preserves Kepler's third law,...

Now you may easily convince yourself that rotations (of r and p in the same sense!) preserve Poisson brackets, so provide a canonical transformation; but, visibly from the above, the Lie tfmations scale PBs by λ, so they are not canonical transformations. So these three transformations are not canonical, indicating, straightforwardly, that only rotations are canonical transformations, so they overlap with only 3 of the 21 generators of the sp(6), as indicated in your formula (as I imagined it corrected; are you faked out by sp(4)~so(5)?).

I have a sense that, by failing to provide a reference, you have withheld relevant context.

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Since you appear to be confused about the symplectic group, I'll briefly review it for you. The 6×6 matrix group elements $e^M$ act on the 6-vectors $(x_1,x_2,x_3,p_1,p_2,p_3)^T$ in such a way as to leave the symplectic metric $$ J=\begin{pmatrix} 0&0&0&0&0&1 \\ 0&0&0&0&1&0 \\ 0&0&0&1&0&0 \\ 0&0&-1&0&0&0 \\ 0&-1&0&0&0&0 \\ -1&0&0&0&0&0 \\ \end{pmatrix} $$ invariant, $$ e^{M^T} J e^{M} =J ~~~ \leadsto ~~~JM= (JM)^T\equiv S,$$ where S is an arbitrary 6×6 symmetric matrix, of which there are 3×7=21 independent ones. So the sp(6) generators are these 21 matrices $M=JS$. You may easily prove the rank is 3; the algebra is $C_3$, after all. Show it easily contains rotations!

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Edit on the edited question:

Phew... your dotty Physics Forums link at last explains your wrong minded attachment to Sp(4,R). The answerer is expounding the symmetry group of harmonic oscillators in 2d, and not the 3d Kepler/hydrogen problem considered here!

By suitably normalizing the LRL vector to ${\mathbf D}\equiv {\mathbf A}/\sqrt{-2H}$ and unravelling to chiral components, A = L + D , B = L - D , you can show the A s close into an su(2) commuting with the su(2) of the B s. So, then, a total so(4). This, now untwisted, so(4) is a bona-fide symmetry and entirely canonical, and a subgroup of sp(6). That is, for the six parameters a, b, $$\delta {\mathbf q} = \{ {\mathbf q} , {\mathbf a\cdot A}\},\qquad \delta {\mathbf p} = \{ {\mathbf p} , {\mathbf a\cdot A}\} \\ \delta' {\mathbf q} = \{ {\mathbf q} , {\mathbf b\cdot B}\},\qquad \delta '{\mathbf p} = \{ {\mathbf p} , {\mathbf b\cdot B}\} $$ preserve the Poisson brackets, as you may (should) demonstrate. (For example, as per your request in the comment, the first order variation of $\{ q^i,p^j \}=\delta^{ij}$ vanishes, $$ \{ \delta q^i,p^j \} + \{ q^i,\delta p^j \} = \{ \{ q^i, {\mathbf a\cdot A}\} \},p^j \} + \{ q^i, \{ p^j,{\mathbf a\cdot A}\} \} \} \\ =-\{ \delta^{ij} , {\mathbf a\cdot A}\} - \{ \{ {\mathbf a\cdot A}\},p^j \}, q^i \} + \{ q^i, \{ p^j,{\mathbf a\cdot A}\} \} \}=0 $$ by the Jacobi identity, and likewise for all other PBs and generators.)

It hardly matters, since Pauli's algebraic maneuver, upon quantization, works just fine, anyway. One never really had to think about canonical transformations, and how motion is a canonical transformation with time as the infinitesimal parameter, and the whole shebang...

• In conclusion, your connection to the issues of the Physics Forums question appears completely spurious.