Algebraic approach to the Dunkl–Coulomb problem and Dunkl oscillator in arbitrary dimensions

Abstract

The Dunkl–Coulomb and the Dunkl oscillator models in arbitrary space-dimensions are introduced. These models are shown to be maximally superintegrable and exactly solvable. The energy spectrum and the wave functions of both systems are obtained using different realizations of the Lie algebra $\text{ so }(1,2).$ The N-dimensional Dunkl oscillator admits separation of variables in both Cartesian and polar coordinates and the corresponding separated solutions are expressed in terms of the generalized Hermite, Laguerre and Gegenbauer polynomials. The N-dimensional Dunkl–Coulomb Hamiltonian admits separation of variables in polar coordinates and the separated wave functions are expressed in terms of the generalized Laguerre and Gegenbauer polynomials. The symmetry operators generalizing the Runge–Lenz vector operator are given. Together with the Dunkl angular momentum operators and reflection operators they generate the symmetry algebra of the N-dimensional Dunkl–Coulomb Hamiltonian which is a deformation of $\text{ so }(N+1)$ by reflections for bound states and is a deformation of $\text{ so }(N,1)$ by reflections for positive energy states. The symmetry operators of the N-dimensional Dunkl oscillator are obtained by the Schwinger construction and generate its invariance algebra which is a deformation of ${\textit{su}}(N)$ by reflections.

Appendices

Appendix: Proof of Theorem 4.1

First evaluate

$$\begin{aligned} a_{j,k,l,m}= & {} \left[ A_{j}^{+}A_{k}^{-}, A_{l}^{+}A_{m}^{-}\right] .\\ b_{j,k,l,m}= & {} \left[ A_{j}^{+}A_{k}^{-}, A_{m}^{+}A_{l}^{-}\right] .\\ c_{j,k,l,m}= & {} \left[ A_{k}^{+}A_{j}^{-}, A_{l}^{+}A_{m}^{-}\right] .\\ d_{j,k,l,m}= & {} \left[ A_{k}^{+}A_{j}^{-}, A_{m}^{+}A_{l}^{-}\right] . \end{aligned}$$

Using the commutation relations given in Definition 4.1

$$\begin{aligned} a_{j,k,l,m}= & {} A_{j}^{+}\left[ A_{k}^{-}, A_{l}^{+}\right] A_{m}^{-}+A_{l}^{+}\left[ A_{j}^{+}, A_{m}^{-}\right] A_{k}^{-}.\\= & {} \omega \left( 1+2\mu _{k}R_{k}\right) \delta _{l}^{k}A_{j}^{+}A_{m}^{-}-\omega \left( 1+2\mu _{j}R_{j}\right) \delta _{j}^{m}A_{l}^{+}A_{k}^{-}. \end{aligned}$$

Similarly

$$\begin{aligned} b_{j,k,l,m}= & {} \omega \left( 1+2\mu _{k}R_{k}\right) \delta _{k}^{m}A_{j}^{+}A_{l}^{-}-\omega \left( 1+2\mu _{j}R_{j}\right) \delta _{j}^{l}A_{m}^{+}A_{k}^{-}.\\ c_{j,k,l,m}= & {} \omega \left( 1+2\mu _{j}R_{j}\right) \delta _{j}^{l}A_{k}^{+}A_{m}^{-}-\omega \left( 1+2\mu _{k}R_{k}\right) \delta _{k}^{m}A_{l}^{+}A_{j}^{-}.\\ d_{j,k,l,m}= & {} \omega \left( 1+2\mu _{j}R_{j}\right) \delta _{j}^{m}A_{k}^{+}A_{l}^{-}-\omega \left( 1+2\mu _{k}R_{k}\right) \delta _{k}^{l}A_{m}^{+}A_{j}^{-}. \end{aligned}$$

Therefore

$$\begin{aligned} \left[ C_{j,k}, C_{l,m}\right]= & {} a_{j,k,l,m}+b_{j,k,l,m}+c_{j,k,l,m}+d_{j,k,l,m}\\= & {} \omega (1+2\mu _{j}R_{j})\delta _{j}^{l}D_{k,m}\\&+\omega (1+2\mu _{k}R_{k})\delta _{k}^{m}D_{j,l}+\omega (1+2\mu _{k}R_{k})\delta _{l}^{k}D_{j,m}\\&+\omega (1+2\mu _{j}R_{j})\delta _{m}^{j}D_{k,l}.\\ \left[ C_{j,k},D_{l,m}\right]= & {} a_{j,k,l,m}-b_{j,k,l,m}+c_{j,k,l,m}-d_{j,k,l,m}\\= & {} \omega (1+2\mu _{j}R_{j})\delta _{j}^{l}C_{k,m}\\&-\omega (1+2\mu _{k}R_{k})\delta _{k}^{m}C_{j,l}+\omega (1+2\mu _{k}R_{k})\delta _{l}^{k}C_{j,m}\\&-\omega (1+2\mu _{j}R_{j})\delta _{m}^{j}C_{k,l}.\\ \left[ D_{j,k},D_{l,m}\right]= & {} a_{j,k,l,m}-b_{j,k,l,m}-c_{j,k,l,m}+d_{j,k,l,m}\\= & {} -\omega (1+2\mu _{j}R_{j})\delta _{j}^{l}D_{k,m}\\&-\omega (1+2\mu _{k}R_{k})\delta _{k}^{m}D_{j,l}+\omega (1+2\mu _{k}R_{k})\delta _{l}^{k}D_{j,m}\\&+\omega (1+2\mu _{j}R_{j})\delta _{m}^{j}D_{k,l}. \end{aligned}$$

In similar fashion evaluate

$$\begin{aligned} a^{1}_{j,k,l}= & {} \left[ A_{j}^{+}A_{k}^{-}, A_{l}^{+}A_{l}^{-}\right] .\\ b^{1}_{j,k,l}= & {} \left[ A_{j}^{+}A_{k}^{-}, A_{l}^{-}A_{l}^{+}\right] .\\ c^{1}_{j,k,l}= & {} \left[ A_{k}^{+}A_{j}^{-}, A_{l}^{+}A_{l}^{-}\right] .\\ d^{1}_{j,k,l}= & {} \left[ A_{k}^{+}A_{j}^{-}, A_{l}^{-}A_{l}^{+}\right] . \end{aligned}$$

Using the commutation relations given in Definition 4.1

$$\begin{aligned} a^{1}_{j,k,l}= & {} A_{j}^{+}\left[ A_{k}^{-}, A_{l}^{+}\right] A_{l}^{-}+A_{l}^{+}\left[ A_{j}^{+}, A_{l}^{-}\right] A_{k}^{-}.\\= & {} \omega \left( 1+2\mu _{k}R_{k}\right) \delta _{l}^{k}A_{j}^{+}A_{l}^{-}-\omega \left( 1-2\mu _{j}R_{j}\right) \delta _{j}^{l}A_{l}^{+}A_{k}^{-}. \end{aligned}$$

Again

$$\begin{aligned} b^{1}_{j,k,l}= & {} -\omega \left( 1+2\mu _{j}R_{j}\right) \delta _{j}^{l}A_{k}^{-}A_{l}^{+}+\omega \left( 1-2\mu _{k}R_{k}\right) \delta _{l}^{k}A_{l}^{-}A_{j}^{+}.\\ c^{1}_{j,k,l}= & {} \omega \left( 1+2\mu _{j}R_{j}\right) \delta _{j}^{l}A_{k}^{+}A_{l}^{-}-\omega \left( 1-2\mu _{k}R_{k}\right) \delta _{k}^{l}A_{l}^{+}A_{j}^{-}.\\ d^{1}_{j,k,l}= & {} -\omega \left( 1+2\mu _{k}R_{k}\right) \delta _{k}^{l}A_{j}^{-}A_{l}^{+}+\omega \left( 1-2\mu _{j}R_{j}\right) \delta _{j}^{l}A_{l}^{-}A_{k}^{+}. \end{aligned}$$

Therefore

$$\begin{aligned} \left[ C_{j,k}, E_{l}\right]= & {} (1/2)(a^{1}_{j,k,l}+b^{1}_{j,k,l}+c^{1}_{j,k,l}+d^{1}_{j,k,l})=\left\{ \begin{array}{ll} 0; &{} j\ne k\ne l. \\ -\omega D_{j,k}; &{} j\ne k, l=j. \\ \omega D_{j,k}; &{} j\ne k, l=k. \end{array} \right. \\ \left[ D_{j,k}, E_{l}\right]= & {} (1/2)(a^{1}_{j,k,l}+b^{1}_{j,k,l}-c^{1}_{j,k,l}-d^{1}_{j,k,l})=\left\{ \begin{array}{ll} 0; &{} j\ne k\ne l. \\ -\omega C_{j,k}; &{} j\ne k, l=j. \\ \omega C_{j,k}; &{} j\ne k, l=k. \end{array} \right. \end{aligned}$$

From which we deduce

$$\begin{aligned} \left[ B, C_{j,k}\right]= & {} \left[ B, D_{j,k}\right] =0, \ 1\le j<k\le N. \end{aligned}$$

The other commutation relations can be proved in similar fashion.

Appendix: Proof of Proposition 5.3

Here we derive the relation (5.10), which is a quadratic relation involving the total Dunkl angular momentum ${\mathbf {J}}^{2},$ the Dunkl Coulomb Hamiltonian ${\mathcal {H}}$ and the square ${\mathbf {A}}^{2}$ of the Dunkl Runge–Lenz operators.

We use the definition (5.3) for ${\mathbf {A}}_{j}:$

$$\begin{aligned} {\mathbf {A}}_{j}^{2}-\frac{\alpha ^{2}{\mathbf {x}}_{j}^{2}}{r^{2}}=a_{j}-b_{j}-ic_{j}-id_{j}-\alpha e_{j}-\alpha f_{j} , \end{aligned}$$

where

$$\begin{aligned} a_{j}= & {} \left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}\right) ^{2},\\ b_{j}= & {} \left( {\mathbf {p}}_{j}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) ^{2},\\ c_{j}= & {} \left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}\right) {\mathbf {p}}_{j}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ,\\ d_{j}= & {} {\mathbf {p}}_{j}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}\right) ,\\ e_{j}= & {} \left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}-i{\mathbf {p}}_{j}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \frac{{\mathbf {x}}_{j}}{r},\\ f_{j}= & {} \frac{{\mathbf {x}}_{j}}{r}\left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}-i{\mathbf {p}}_{j}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) . \end{aligned}$$

We need the following Lemma

Lemma B.1

The following identities hold:

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}a_{j}= & {} {\mathbf {p}}^{2}{\mathbf {J}}^{2}-2i\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}{\mathbf {p}}_{k}\displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}{\mathbf {J}}_{k,j}. \end{aligned}$$

(B.1)

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}b_{j}= & {} {\mathbf {p}}^{2}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ^{2}\nonumber \\&-2\left( \displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}^{2}\mu _{j}{\mathbf {R}}_{j}\right) \left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

(B.2)

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}c_{j}= & {} 2i{\mathbf {p}}^{2}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ^{2}\nonumber \\&-2i\left( \displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}^{2}\mu _{j}{\mathbf {R}}_{j}\right) \left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

(B.3)

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}d_{j}= & {} -2\displaystyle \sum _{j=1}^{N}\mu _{j}{\mathbf {R}}_{j}{\mathbf {p}}_{j}\displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}. \end{aligned}$$

(B.4)

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}e_{j}= & {} \left( {\mathbf {J}}^{2}+i({\mathbf {p}}\cdot {\mathbf {x}} )\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) r^{-1}. \end{aligned}$$

(B.5)

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}f_{j}= & {} r^{-1}\left( {\mathbf {J}}^{2}-i({\mathbf {x}}\cdot {\mathbf {p}})\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) . \end{aligned}$$

(B.6)

Proof

The proof of (B.2) is obvious.

Derivation of (B.1): From the commutation relation

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k}, {\mathbf {p}}_{l}\right]= & {} i(1+2\mu _{l}{\mathbf {R}}_{l})\delta _{j}^{l}{\mathbf {p}}_{k}-i(1+2\mu _{l}{\mathbf {R}}_{l})\delta _{k}^{l}{\mathbf {p}}_{j}, \end{aligned}$$

(B.7)

we deduce

$$\begin{aligned} a_{j}= & {} \displaystyle \sum _{k,l=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,l} =\displaystyle \sum _{k,l=1}^{N}{\mathbf {p}}_{k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}-i\displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {p}}_{j}{\mathbf {J}}_{j,k} -2i\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}{\mathbf {p}}_{k}{\mathbf {p}}_{j}{\mathbf {J}}_{k,j}. \end{aligned}$$

Then

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}a_{j}= & {} \displaystyle \sum _{j,k,l=1}^{N}{\mathbf {p}}_{k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}-i\displaystyle \sum _{j,k=1}^{N}{\mathbf {p}}_{k}{\mathbf {p}}_{j}{\mathbf {J}}_{j,k} -2i\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}{\mathbf {p}}_{k}\displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}{\mathbf {J}}_{k,j}. \end{aligned}$$

(B.8)

Since ${\mathbf {J}}_{j,k}=-{\mathbf {J}}_{k,j},$ we deduce

$$\begin{aligned} \displaystyle \sum _{j,k=1}^{N}{\mathbf {p}}_{k}{\mathbf {p}}_{j}{\mathbf {J}}_{j,k}=0. \end{aligned}$$

(B.9)

Now note that

$$\begin{aligned} \displaystyle \sum _{j,k,l=1}^{N}{\mathbf {p}}_{k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}= & {} \displaystyle \sum _{s=1}^{N}{\mathbf {p}}_{s}^{2}\left( \displaystyle \sum _{\overset{j=1}{j\ne s}}^{N}{\mathbf {J}}_{j,s}^{2}\right) +\displaystyle \sum _{k\ne l,j\ne k,j\ne l}{\mathbf {p}}_{k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l} =\displaystyle \sum _{s=1}^{N}{\mathbf {p}}_{s}^{2}\left( \displaystyle \sum _{\overset{j=1}{j\ne s}}^{N}{\mathbf {J}}_{j,s}^{2}\right) \\&+\displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}\left( \displaystyle \sum _{j<l,j\ne k, l\ne k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}+ \displaystyle \sum _{j>l,j\ne k, l\ne k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}\right) \\= & {} \displaystyle \sum _{s=1}^{N}{\mathbf {p}}_{s}^{2}\left( \displaystyle \sum _{\overset{j=1}{j\ne s}}^{N}{\mathbf {J}}_{j,s}^{2}\right) +\displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}\left( \displaystyle \sum _{j<l,j\ne k, l\ne k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}+ {\mathbf {p}}_{j}{\mathbf {J}}_{k,l}{\mathbf {J}}_{j,l}\right) . \end{aligned}$$

From (5.9):

$$\begin{aligned} {\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}+{\mathbf {p}}_{j}{\mathbf {J}}_{k,l}{\mathbf {J}}_{j,l}={\mathbf {p}}_{k}{\mathbf {J}}_{j,l}^{2}. \end{aligned}$$

Then

$$\begin{aligned} \displaystyle \sum _{j,k,l=1}^{N}{\mathbf {p}}_{k}{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}{\mathbf {J}}_{j,l}= & {} {\mathbf {p}}^{2}{\mathbf {J}}^{2} \end{aligned}$$

(B.10)

Inserting (B.9) and (B.10) into (B.8), we obtain the final result.

Derivation of (B.3): From the commutation relation (2.20) we deduce:

$$\begin{aligned} {\mathbf {p}}_{k}{\mathbf {J}}_{j,k}={\mathbf {p}}_{k}^{2}{\mathbf {x}}_{j}-{\mathbf {x}}_{k}{\mathbf {p}}_{k}{\mathbf {p}}_{j}+i\left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}_{j}; \ j\ne k. \end{aligned}$$

Then

$$\begin{aligned} \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}={\mathbf {p}}^{2}{\mathbf {x}}_{j}-({\mathbf {x}}\cdot {\mathbf {p}}){\mathbf {p}}_{j}+2i{\mathbf {p}}_{j} +i\left( N-1+2\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}-2\mu _{j}{\mathbf {R}}_{j}\right) {\mathbf {p}}_{j}. \end{aligned}$$

Using (3.11), we obtain

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}\left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}\right) {\mathbf {p}}_{j}= & {} 2i{\mathbf {p}}^{2}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) -2i\left( \displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}^{2}\mu _{j}{\mathbf {R}}_{j}\right) , \end{aligned}$$

which gives the desired result.

Derivation of (B.4): From the commutation relation (2.19) we deduce:

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}d_{j}= & {} -2\displaystyle \sum _{j=1}^{N}\mu _{j}{\mathbf {R}}_{j}{\mathbf {p}}_{j}\displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}\nonumber \\&+ \left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}\left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}\right) . \end{aligned}$$

Using

$$\begin{aligned} {\mathbf {p}}_{k}{\mathbf {J}}_{j,k}={\mathbf {x}}_{j}{\mathbf {p}}_{k}^{2}-{\mathbf {p}}_{j}{\mathbf {p}}_{k}{\mathbf {x}}_{k}; \ j\ne k, \end{aligned}$$

we deduce

$$\begin{aligned} \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}= & {} {\mathbf {x}}_{j}{\mathbf {p}}^{2}+{\mathbf {p}}_{j}^{2}{\mathbf {x}}_{j}-{\mathbf {x}}_{j}{\mathbf {p}}_{j}^{2}- {\mathbf {p}}_{j}\left( {\mathbf {p}}\cdot {\mathbf {x}}\right) =-2i{\mathbf {p}}_{j}+{\mathbf {x}}_{j}{\mathbf {p}}^{2}-{\mathbf {p}}_{j}\left( {\mathbf {p}}\cdot {\mathbf {x}}\right) . \end{aligned}$$

Then

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}\left( \displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}\right)= & {} -2i{\mathbf {p}}^{2}+({\mathbf {p}}\cdot {\mathbf {x}}){\mathbf {p}}^{2}-{\mathbf {p}}^{2}\left( {\mathbf {p}}\cdot {\mathbf {x}}\right) =0. \end{aligned}$$

The last equality follows from

$$\begin{aligned} \left[ ({\mathbf {p}}\cdot {\mathbf {x}}),{\mathbf {p}}^{2}\right]= & {} 2i{\mathbf {p}}^{2}. \end{aligned}$$

Derivation of (B.5): We observe that:

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}e_{j}= & {} \left( \displaystyle \sum _{j=1}^{N}\displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}{\mathbf {x}}_{j}\right) r^{-1} -i\displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {x}}_{j}r^{-1}. \end{aligned}$$

From the commutation relation (2.19) we obtain:

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}{\mathbf {p}}_{j}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {x}}_{j}= & {} ({\mathbf {p}}\cdot {\mathbf {x}} )\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \\&-2\displaystyle \sum _{j=1}^{N}\mu _{j}{\mathbf {R}}_{j}{\mathbf {p}}_{j}{\mathbf {x}}_{j}. \end{aligned}$$

Using (B.7)

$$\begin{aligned} {\mathbf {p}}_{k}{\mathbf {J}}_{j,k}{\mathbf {x}}_{j}= & {} {\mathbf {J}}_{j,k}{\mathbf {x}}_{j}{\mathbf {p}}_{k}+i\left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}_{j}{\mathbf {x}}_{j}; \ j \ne k. \end{aligned}$$

Then

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}\displaystyle \sum _{k=1}^{N}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}{\mathbf {x}}_{j}= & {} {\mathbf {J}}^{2}+2i({\mathbf {p}}\cdot {\mathbf {x}} )\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) -2i\displaystyle \sum _{j=1}^{N}\mu _{j}{\mathbf {R}}_{j}{\mathbf {p}}_{j}{\mathbf {x}}_{j}. \end{aligned}$$

Here we have used the fact that:

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}\displaystyle \sum _{\overset{k=1}{k\ne j}}^{N}{\mathbf {J}}_{j,k}{\mathbf {x}}_{j}{\mathbf {p}}_{k}= & {} \displaystyle \sum _{j=1}^{N}\left( \displaystyle \sum _{j<k}{\mathbf {J}}_{j,k}{\mathbf {x}}_{j}{\mathbf {p}}_{k}+\displaystyle \sum _{j>k}{\mathbf {J}}_{j,k}{\mathbf {x}}_{j}{\mathbf {p}}_{k}\right) = \displaystyle \sum _{j=1}^{N}\displaystyle \sum _{j<k}{\mathbf {J}}_{j,k}^{2}={\mathbf {J}}^{2}. \end{aligned}$$

Collect these results to obtain the desired result.

Derivation of (B.6): This follows from the fact that:

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}\displaystyle \sum _{k=1}^{N}{\mathbf {x}}_{j}{\mathbf {p}}_{k}{\mathbf {J}}_{j,k}= & {} {\mathbf {J}}^{2}. \end{aligned}$$

$\square$

To prove Proposition 5.3, we use the previous lemma to write

$$\begin{aligned} {\mathbf {A}}^{2}-\alpha ^{2}= & {} {\mathbf {p}}^{2}{\mathbf {J}}^{2}+{\mathbf {p}}^{2}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ^{2}\\&-2\alpha r^{-1}{\mathbf {J}}^{2}-i\alpha \left( ({\mathbf {p}}\cdot {\mathbf {x}})r^{-1}-r^{-1}({\mathbf {x}}\cdot {\mathbf {p}})\right) \left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \\= & {} 2{\mathcal {H}}\left( {\mathbf {J}}^{2}+\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ^{2}\right) . \end{aligned}$$

Here we have used:

$$\begin{aligned} ({\mathbf {p}}\cdot {\mathbf {x}})r^{-1}-r^{-1}({\mathbf {x}}\cdot {\mathbf {p}})= & {} -2i r^{-1}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

Appendix: Proof of Proposition 5.4

First observe that we can write

$$\begin{aligned} {\mathbf {B}}_{j}^{\alpha }{\mathbf {B}}_{j}^{\beta }= & {} \frac{1}{4}\left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) \left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) \\&-\frac{\beta }{4}\left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) {\mathbf {x}}_{j}-\frac{\alpha }{4}x_{j}^{2}{\mathbf {p}}^{2}+ \frac{\alpha }{2}{\mathbf {x}}_{j}{\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \\&+\frac{\alpha \beta }{4}x_{j}^{2}-\frac{1}{2}a_{j}-\frac{1}{2}b_{j}+c_{j}+\frac{\beta }{2}d_{j}, \end{aligned}$$

where

$$\begin{aligned} a_{j}= & {} \left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) ,\\ b_{j}= & {} {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) ,\\ c_{j}= & {} {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) ,\\ d_{j}= & {} {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) x_{j}. \end{aligned}$$

Using (3.9), we get

$$\begin{aligned} \left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) \left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right)= & {} {\mathbf {x}}_{j}\left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}-2i{\mathbf {p}}_{j}\right) {\mathbf {p}}^{2},\\ \left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) {\mathbf {x}}_{j}= & {} {\mathbf {x}}_{j}\left( {\mathbf {p}}^{2}{\mathbf {x}}_{j}\right) ={\mathbf {x}}_{j}\left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}-2i{\mathbf {p}}_{j}\right) \end{aligned}$$

then

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}\left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) \left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right)= & {} r^{2}{\mathbf {p}}^{4}-2i\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}^{2},\end{aligned}$$

(C.1)

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}\left( {\mathbf {x}}_{j}{\mathbf {p}}^{2}\right) {\mathbf {x}}_{j}= & {} r^{2}{\mathbf {p}}^{2}-2i\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) . \end{aligned}$$

(C.2)

By means of (3.11),

$$\begin{aligned} a_{j}= & {} \left( {\mathbf {x}}_{j}{\mathbf {p}}_{j}\right) \left( {\mathbf {p}}^{2}({\mathbf {x}}\cdot {\mathbf {p}})-i{\mathbf {p}}^{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \\= & {} \left( {\mathbf {x}}_{j}{\mathbf {p}}_{j}\right) \left( \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}^{2}-2i{\mathbf {p}}^{2}-i{\mathbf {p}}^{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) , \end{aligned}$$

then

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}a_{j}= & {} \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ^{2}{\mathbf {p}}^{2}-2i\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}^{2}-i\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}^{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) .\nonumber \\ \end{aligned}$$

(C.3)

A similar argument gives

$$\begin{aligned} b_{j}= & {} \left( \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) -i\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \left( {\mathbf {x}}_{j}{\mathbf {p}}_{j}-i\left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) \right) {\mathbf {p}}^{2}, \end{aligned}$$

then

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}b_{j}= & {} \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ^{2}{\mathbf {p}}^{2}-i\left( \frac{3N-1}{2}+3\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}^{2}\nonumber \\&-\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( N+2\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}^{2}. \end{aligned}$$

(C.4)

Rewrite $c_{j}$ as

$$\begin{aligned} c_{j}= & {} I_{1}-iI_{2}-iI_{3}-I_{4}, \end{aligned}$$

where

$$\begin{aligned} I_{1}= & {} {\mathbf {p}}_{j}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}_{j}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ,\\ I_{2}= & {} {\mathbf {p}}_{j}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ,\\ I_{3}= & {} {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}_{j}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ,\\ I_{4}= & {} {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

Using (2.19) and (3.13) , we obtain

$$\begin{aligned} I_{1}= & {} \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}_{j}^{2}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) -i(1+2\mu _{j}{\mathbf {R}}_{j}){\mathbf {p}}_{j}^{2}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ,\\ I_{2}= & {} \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}_{j}^{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) -i(1+2\mu _{j}{\mathbf {R}}_{j}){\mathbf {p}}_{j}^{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ,\\ I_{3}= & {} \left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}-2\mu _{j}{\mathbf {R}}_{j}\right) {\mathbf {p}}_{j}^{2}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ,\\ I_{4}= & {} \left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}-2\mu _{j}{\mathbf {R}}_{j}\right) {\mathbf {p}}_{j}^{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

Then

$$\begin{aligned}&\displaystyle \sum _{j=1}^{N}c_{j}=\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ^{2}{\mathbf {p}}^{2}-i\left( N+2\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}^{2}\\&\quad -\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( \frac{N+1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}^{2} \end{aligned}$$

In similar fashion, we can prove

$$\begin{aligned} d_{j}= & {} \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {x}}_{j}{\mathbf {p}}_{j}-i\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) \left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) \\&-i\left( {\mathbf {x}}_{j}{\mathbf {p}}_{j}-i\left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) \right) \left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

This implies

$$\begin{aligned} \displaystyle \sum _{j=1}^{N}d_{j}= & {} \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ^{2}-i\left( \frac{3N-1}{2}+3\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) \nonumber \\&-\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( N+2\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

(C.5)

After simplification we get

$$\begin{aligned} {\mathbf {B}}^{\alpha }.{\mathbf {B}}^{\beta }= & {} \frac{1}{4}\ r^{2}{\mathbf {p}}^{4}-\frac{1}{4} \ (\alpha +\beta )r^{2}{\mathbf {p}}^{2}+ \frac{1}{4} \ \alpha \beta r^{2}-\frac{i}{2}\ \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) {\mathbf {p}}^{2}+ \frac{1}{2} \ (\alpha +\beta )\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ^{2}\\&-\frac{3i\beta }{2}\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) -\frac{1}{2}\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}^{2}\\&\quad -\frac{i\alpha }{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) \\&\quad -\frac{\beta }{2}\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( N+2\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

Now using (3.10)

$$\begin{aligned} r^{2}{\mathbf {p}}^{4}= & {} r\left( r{\mathbf {p}}^{2}\right) {\mathbf {p}}^{2}=r\left( {\mathbf {p}}^{2}r+2ir^{-1}\left( \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) -i\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \right) {\mathbf {p}}^{2}\\= & {} \left( r{\mathbf {p}}^{2}\right) ^{2}+2i\left( \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) -i\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) {\mathbf {p}}^{2} \end{aligned}$$

and

$$\begin{aligned} \beta r^{2}{\mathbf {p}}^{2}= & {} \beta \left( r{\mathbf {p}}^{2}\right) r+2i\beta \left( \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) -i\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \end{aligned}$$

Hence

$$\begin{aligned} {\mathbf {B}}^{\alpha }.{\mathbf {B}}^{\beta }= & {} \frac{1}{4}\left( r{\mathbf {p}}^{2}-\alpha r\right) \left( r{\mathbf {p}}^{2}-\beta r\right) +\frac{\alpha +\beta }{2}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ^{2}\\&-\frac{i\alpha }{2}\left( \frac{N-3}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) -\frac{i\beta }{2}\left( \frac{3N-1}{2}+3\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) \\&-\frac{\beta }{2}\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \left( N+1+2\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

Finally, substituting

$$\begin{aligned} \left( {\mathbf {x}}\cdot {\mathbf {p}}\right)= & {} {\mathbf {T}}_{2}+i\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \\ \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) ^{2}= & {} {\mathbf {T}}_{2}^{2}+2i\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {T}}_{2}-\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ^{2} \end{aligned}$$

gives

$$\begin{aligned} {\mathbf {B}}^{\alpha }.{\mathbf {B}}^{\beta }= & {} \frac{1}{4}\left( r{\mathbf {p}}^{2}-\alpha r\right) \left( r{\mathbf {p}}^{2}-\beta r\right) +\frac{\alpha +\beta }{2}\left( {\mathbf {T}}_{2}^{2}-\left( \frac{N-1}{2}+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right) \\&+i(\alpha -\beta )\left( \frac{N+1}{4}+\frac{1}{2}\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {T}}_{2}. \end{aligned}$$

Appendix: Proof of Theorem 5.1

Let us derive the commutation relation between ${\mathbf {J}}_{j,k}$ with ${\mathbf {A}}_{l}.$ We use the representation (5.3) for ${\mathbf {A}}_{l},$ then

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{l}\right]= & {} \displaystyle \sum _{m=1}^{N}\left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{m}{\mathbf {J}}_{l,m}\right] -i \left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{l}\left( (N-1)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right] \\&-\alpha \left[ {\mathbf {J}}_{j,k},{\mathbf {x}}_{l}r^{-1}\right] . \end{aligned}$$

By the definition of ${\mathbf {J}}_{j,k}$, (2.20) and (5.18), we see that:

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {x}}_{l}r^{-1}\right]= & {} \left\{ \begin{array}{ll} 0, &{} j\ne k \ne l. \\ i(1+2\mu _{j}{\mathbf {R}}_{j}){\mathbf {x}}_{k}r^{-1}, &{} j \ne k, l=j. \\ -i(1+2\mu _{k}{\mathbf {R}}_{k}){\mathbf {x}}_{j}r^{-1}, &{} j \ne k, l=k. \end{array} \right. \end{aligned}$$

Using (5.15), (B.7) and the fact that $[a,bc]=[a,b]c+b[a,c],$ we deduce

$$\begin{aligned} \displaystyle \sum _{m=1}^{N}\left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{m}{\mathbf {J}}_{l,m}\right]= & {} 2i\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) \left( {\mathbf {p}}_{j}{\mathbf {J}}_{k,l}+{\mathbf {p}}_{k}{\mathbf {J}}_{l,j}\right) \\+ & {} i\delta _{l}^{j}\displaystyle \sum _{m=1}^{N}{\mathbf {p}}_{m}\left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) {\mathbf {J}}_{k,m}-i\delta _{l}^{k}\displaystyle \sum _{m=1}^{N}{\mathbf {p}}_{m}\left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {J}}_{j,m}. \end{aligned}$$

In similar fashion, we can prove

$$\begin{aligned}&\left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{l}\left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \right] \\&\quad = i\left( (1+2\mu _{l}{\mathbf {R}}_{l})\delta _{j}^{l}{\mathbf {p}}_{k}-(1+2\mu _{l}{\mathbf {R}}_{l})\delta _{k}^{l}{\mathbf {p}}_{j}\right) \left( (N-1)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) \\&\qquad -2{\mathbf {p}}_{l}\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {J}}_{j,k}. \end{aligned}$$

Collect these results and use the fact that

$$\begin{aligned} {\mathbf {p}}_{j}{\mathbf {J}}_{k,l}+{\mathbf {p}}_{l}{\mathbf {J}}_{j,k}+{\mathbf {p}}_{k}{\mathbf {J}}_{l,j}=0; \ j\ne k\ne l, \end{aligned}$$

we obtain

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{l}\right]= & {} \left\{ \begin{array}{ll} 0, &{} j\ne k\ne l. \\ i\left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) {\mathbf {A}}_{k}, &{} j\ne k, l=j. \\ -i\left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {A}}_{j}, &{} j\ne k, l=k. \end{array} \right. \end{aligned}$$

(D.1)

Let us derive the commutation relation between ${\mathbf {A}}_{j}$ with ${\mathbf {A}}_{k}.$ We use the representation (5.2) for ${\mathbf {A}}_{j},$ then

$$\begin{aligned} \left[ {\mathbf {A}}_{j},{\mathbf {A}}_{k}\right]= & {} \frac{1}{4}\left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2}, {\mathbf {x}}_{k}{\mathbf {p}}^{2}\right] +\left[ {\mathbf {x}}_{j}{\mathcal {H}}, {\mathbf {x}}_{k}{\mathcal {H}}\right] +a_{j,k}-\frac{1}{2} \left( b_{j,k}-b_{k,j}\right) \nonumber \\&+\frac{1}{2} \left( c_{j,k}-c_{k,j}\right) - \left( d_{j,k}-d_{k,j}\right) , \end{aligned}$$

(D.2)

where

$$\begin{aligned} a_{j,k}= & {} \left[ {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right) , {\mathbf {p}}_{k}\left( ({\mathbf {x}}\cdot {\mathbf {p}}) \right. \right. \\&\quad \left. \left. -i\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right) \right] \\ b_{j,k}= & {} \left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2}, {\mathbf {p}}_{k}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right) \right] \\ c_{j,k}= & {} \left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2},{\mathbf {x}}_{k}{\mathcal {H}} \right] \\ d_{j,k}= & {} \left[ {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right) ,{\mathbf {x}}_{k}{\mathcal {H}}\right] . \end{aligned}$$

Using (3.9) and the fact that $[a,bc]=[a,b]c+b[a,c],$ we deduce

$$\begin{aligned} \left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2}, {\mathbf {x}}_{k}{\mathbf {p}}^{2}\right] =-2i \ {\mathbf {J}}_{j,k} \ {\mathbf {p}}^{2}. \end{aligned}$$

(D.3)

In similar fashion, we can prove

$$\begin{aligned} \left[ {\mathbf {x}}_{j}{\mathcal {H}}, {\mathbf {x}}_{k}{\mathcal {H}}\right] =-i \ {\mathbf {J}}_{j,k} {\mathcal {H}}. \end{aligned}$$

(D.4)

It is easy to see that

$$\begin{aligned} a_{j,k}= & {} \alpha _{j,k}-i(\beta _{j,k}-\beta _{k,j})-\gamma _{j,k}, \end{aligned}$$

where

$$\begin{aligned} \alpha _{j,k}= & {} \left[ {\mathbf {p}}_{j}({\mathbf {x}}\cdot {\mathbf {p}}),{\mathbf {p}}_{k}({\mathbf {x}}\cdot {\mathbf {p}})\right] .\\ \beta _{j,k}= & {} \left[ {\mathbf {p}}_{j}({\mathbf {x}}\cdot {\mathbf {p}}),{\mathbf {p}}_{k}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right] .\\ \gamma _{j,k}= & {} \left[ {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) ,{\mathbf {p}}_{k}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right] . \end{aligned}$$

Using (3.13)

$$\begin{aligned} \alpha _{j,k}= & {} {\mathbf {p}}_{j}\left[ ({\mathbf {x}}\cdot {\mathbf {p}}),{\mathbf {p}}_{k}\right] ({\mathbf {x}}\cdot {\mathbf {p}})+ {\mathbf {p}}_{k}\left[ {\mathbf {p}}_{j},({\mathbf {x}}\cdot {\mathbf {p}})\right] ({\mathbf {x}}\cdot {\mathbf {p}})\\= & {} 2i\left( \mu _{k}{\mathbf {p}}_{j}{\mathbf {R}}_{k}{\mathbf {p}}_{k}-\mu _{j}{\mathbf {p}}_{k}{\mathbf {R}}_{j}{\mathbf {p}}_{j}\right) ({\mathbf {x}}\cdot {\mathbf {p}}).\\ \beta _{j,k}= & {} \left( i{\mathbf {p}}_{j}{\mathbf {p}}_{k}+2i\mu _{k}{\mathbf {p}}_{j}{\mathbf {R}}_{k}{\mathbf {p}}_{k}\right) \left( (N-3)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) -2\mu _{j}{\mathbf {p}}_{k}{\mathbf {R}}_{j}{\mathbf {p}}_{j}({\mathbf {x}}\cdot {\mathbf {p}}).\\ \gamma _{j,k}= & {} 2\left( \mu _{k}{\mathbf {p}}_{j}{\mathbf {R}}_{k}{\mathbf {p}}_{k}-\mu _{j}{\mathbf {p}}_{k}{\mathbf {R}}_{j}{\mathbf {p}}_{j}\right) \left( (N-3)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

Then

$$\begin{aligned} a_{j,k}= & {} 0. \end{aligned}$$

(D.5)

It is easy to see that

$$\begin{aligned} b_{j,k}= & {} \left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2}, {\mathbf {p}}_{k}({\mathbf {x}}\cdot {\mathbf {p}})\right] -i\left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2},{\mathbf {p}}_{k}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right] . \end{aligned}$$

Using (2.20), (3.11) and (3.12)

$$\begin{aligned} \left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2}, {\mathbf {p}}_{k}({\mathbf {x}}\cdot {\mathbf {p}})\right]= & {} {\mathbf {x}}_{j}{\mathbf {p}}_{k}\left[ {\mathbf {p}}^{2},({\mathbf {x}}\cdot {\mathbf {p}})\right] + {\mathbf {p}}_{k}\left[ {\mathbf {x}}_{j},({\mathbf {x}}\cdot {\mathbf {p}})\right] {\mathbf {p}}^{2}+\left[ {\mathbf {x}}_{j},{\mathbf {p}}_{k}\right] ({\mathbf {x}}\cdot {\mathbf {p}}){\mathbf {p}}^{2}\\= & {} -i{\mathbf {x}}_{j}{\mathbf {p}}_{k}{\mathbf {p}}^{2}+2i\mu _{j}{\mathbf {p}}_{k}{\mathbf {x}}_{j}{\mathbf {R}}_{j}{\mathbf {p}}^{2}+ \left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) \delta _{j}^{k}{\mathbf {p}}^{2}\\&+i\left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) \delta _{j}^{k}({\mathbf {x}}\cdot {\mathbf {p}}){\mathbf {p}}^{2}. \end{aligned}$$

In similar fashion, we can prove

$$\begin{aligned}&\left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2},{\mathbf {p}}_{k}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots + \mu _{N}{\mathbf {R}}_{N}\right) \right] \\&\quad =2\mu _{j}{\mathbf {p}}_{k}{\mathbf {x}}_{j}{\mathbf {R}}_{j}{\mathbf {p}}^{2} +i\left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) \delta _{j}^{k}\left( (N-3)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {p}}^{2} \end{aligned}$$

Then

$$\begin{aligned} b_{j,k}-b_{k,j}= & {} -i \ {\mathbf {J}}_{j,k}\ {\mathbf {p}}^{2}. \end{aligned}$$

(D.6)

Using (3.9), we obtain

$$\begin{aligned} c_{j,k}= & {} {\mathbf {x}}_{j}{\mathbf {x}}_{k}\left[ {\mathbf {p}}^{2},{\mathcal {H}}\right] + {\mathbf {x}}_{j}\left[ {\mathbf {p}}^{2},{\mathbf {x}}_{k}\right] {\mathcal {H}}+\frac{{\mathbf {x}}_{k}}{2} \left[ {\mathbf {x}}_{j},{\mathbf {p}}^{2}\right] {\mathbf {p}}^{2}\\= & {} {\mathbf {x}}_{j}{\mathbf {x}}_{k}\left[ {\mathbf {p}}^{2},{\mathcal {H}}\right] -2i{\mathbf {x}}_{j}{\mathbf {p}}_{k}{\mathcal {H}}+i{\mathbf {x}}_{k}{\mathbf {p}}_{j}{\mathbf {p}}^{2}. \end{aligned}$$

Then

$$\begin{aligned} c_{j,k}-c_{k,j}= & {} -i \ {\mathbf {J}}_{j,k}\ \left( 2{\mathcal {H}}+{\mathbf {p}}^{2}\right) . \end{aligned}$$

(D.7)

It is easy to see that

$$\begin{aligned} d_{j,k}= & {} \left[ {\mathbf {p}}_{j}({\mathbf {x}}\cdot {\mathbf {p}}), {\mathbf {x}}_{k}{\mathcal {H}}\right] -i\left[ {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) ,{\mathbf {x}}_{k}{\mathcal {H}}\right] . \end{aligned}$$

Using (3.11), (3.14), (2.21) and (2.20), we deduce

$$\begin{aligned} \left[ {\mathbf {p}}_{j}({\mathbf {x}}\cdot {\mathbf {p}}), {\mathbf {x}}_{k}{\mathcal {H}}\right]= & {} {\mathbf {p}}_{j}{\mathbf {x}}_{k}\left[ ({\mathbf {x}}\cdot {\mathbf {p}}),{\mathcal {H}}\right] +{\mathbf {p}}_{j}\left[ ({\mathbf {x}}\cdot {\mathbf {p}}),{\mathbf {x}}_{k}\right] {\mathcal {H}}+{\mathbf {x}}_{k}\left[ {\mathbf {p}}_{j},{\mathcal {H}}\right] ({\mathbf {x}}\cdot {\mathbf {p}})\\&+ \left[ {\mathbf {p}}_{j},{\mathbf {x}}_{k}\right] {\mathcal {H}}({\mathbf {x}}\cdot {\mathbf {p}})\\= & {} \frac{i}{2}\left( {\mathbf {x}}_{k}{\mathbf {p}}_{j}-i\left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) \delta _{j}^{k}\right) {\mathbf {p}}^{2}-2i\mu _{k}{\mathbf {p}}_{j}{\mathbf {x}}_{k} {\mathbf {R}}_{k}{\mathcal {H}}\\&-i\alpha {\mathbf {x}}_{k}{\mathbf {x}}_{j}r^{-3}({\mathbf {x}}\cdot {\mathbf {p}})-i\left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) \delta _{j}^{k}{\mathcal {H}}({\mathbf {x}}\cdot {\mathbf {p}}). \end{aligned}$$

In similar fashion, we can prove

$$\begin{aligned}&\left[ {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) ,{\mathbf {x}}_{k}{\mathcal {H}}\right] = -2\mu _{k}{\mathbf {p}}_{j}{\mathbf {x}}_{k}{\mathbf {R}}_{k}{\mathcal {H}}\\&-i\left( \left( 1+2\mu _{j}{\mathbf {R}}_{j}\right) \delta _{j}^{k}{\mathcal {H}}+\alpha {\mathbf {x}}_{k}{\mathbf {x}}_{j}r^{-3}\right) \left( (N-3)/2+\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right) . \end{aligned}$$

Then

$$\begin{aligned} d_{j,k}-d_{k,j}= & {} -\frac{i}{2} \ {\mathbf {J}}_{j,k}\ {\mathbf {p}}^{2}. \end{aligned}$$

(D.8)

Finally, inserting (D.3) to (D.8) into (D.2), we obtain

$$\begin{aligned} \left[ {\mathbf {A}}_{j},{\mathbf {A}}_{k}\right]= & {} -2i{\mathcal {H}} \ {\mathbf {J}}_{j,k}. \end{aligned}$$

(D.9)

Investigation of Casimir operator: We first observe that the operator ${\mathbf {C}}_{1}$ defined by (5.24) commutes with the reflection operators ${\mathbf {R}}_{j}, \ j=1,\dots ,N.$ From the definition of the ${\mathbf {A}}^{2}$, (D.1) and (D.9) it can be seen that

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {A}}^{2}\right]= & {} \displaystyle \sum _{l=1}^{N}\left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{l}^{2}\right] = \left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{j}^{2}\right] +\left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{k}^{2}\right] \\= & {} \left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{j}\right] {\mathbf {A}}_{j}+{\mathbf {A}}_{j}\left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{j}\right] +\left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{k}\right] {\mathbf {A}}_{k}+{\mathbf {A}}_{k}\left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{k}\right] \\= & {} -2i\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) \left[ {\mathbf {A}}_{j},{\mathbf {A}}_{k}\right] \\= & {} -4\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) {\mathcal {H}} \ {\mathbf {J}}_{j,k}. \end{aligned}$$

Since ${\mathcal {H}}$ commutes with ${\mathbf {R}}_{j}$ and ${\mathbf {J}}_{j,k},$ then

$$\begin{aligned} -4\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) {\mathcal {H}} \ {\mathbf {J}}_{j,k}= & {} \left[ {\mathbf {J}}_{j,k},2{\mathcal {H}}\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) \right] \\= & {} \left[ {\mathbf {J}}_{j,k},2{\mathcal {H}}\left( \mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right] \end{aligned}$$

and therefore

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {A}}^{2}-2{\mathcal {H}}\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}\right]= & {} 0. \end{aligned}$$

(D.10)

Using (D.1) we can write

$$\begin{aligned} \left[ {\mathbf {J}}^{2},{\mathbf {A}}_{l}\right]= & {} \displaystyle \sum _{j<k}\left[ {\mathbf {J}}_{j,k}^{2},{\mathbf {A}}_{l}\right] = \displaystyle \sum _{j<k}{\mathbf {J}}_{j,k}\left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{l}\right] +\left[ {\mathbf {J}}_{j,k},{\mathbf {A}}_{l}\right] {\mathbf {J}}_{j,k}\\= & {} \displaystyle \sum _{k=l+1}^{N}{\mathbf {J}}_{l,k}\left[ {\mathbf {J}}_{l,k},{\mathbf {A}}_{l}\right] +\left[ {\mathbf {J}}_{l,k},{\mathbf {A}}_{l}\right] {\mathbf {J}}_{l,k} +\displaystyle \sum _{k=1}^{l-1}{\mathbf {J}}_{k,l}\left[ {\mathbf {J}}_{k,l},{\mathbf {A}}_{l}\right] +\left[ {\mathbf {J}}_{k,l},{\mathbf {A}}_{l}\right] {\mathbf {J}}_{k,l}. \end{aligned}$$

By means of (D.1), we see that

$$\begin{aligned} {\mathbf {J}}_{l,k}\left[ {\mathbf {J}}_{l,k},{\mathbf {A}}_{l}\right] +\left[ {\mathbf {J}}_{l,k},{\mathbf {A}}_{l}\right] {\mathbf {J}}_{l,k}= & {} i\left\{ {\mathbf {J}}_{l,k},{\mathbf {A}}_{k}\right\} +2i\mu _{l}{\mathbf {R}}_{l}\left[ {\mathbf {A}}_{k},{\mathbf {J}}_{l,k}\right] .\\ {\mathbf {J}}_{k,l}\left[ {\mathbf {J}}_{k,l},{\mathbf {A}}_{l}\right] +\left[ {\mathbf {J}}_{k,l},{\mathbf {A}}_{l}\right] {\mathbf {J}}_{k,l}= & {} i\left\{ {\mathbf {J}}_{l,k},{\mathbf {A}}_{k}\right\} +2i\mu _{l}{\mathbf {R}}_{l}\left[ {\mathbf {A}}_{k},{\mathbf {J}}_{l,k}\right] . \end{aligned}$$

Then

$$\begin{aligned} \left[ {\mathbf {J}}^{2},{\mathbf {A}}_{l}\right]= & {} \displaystyle \sum _{k=1, k\ne l}^{N}i\left\{ {\mathbf {J}}_{l,k},{\mathbf {A}}_{k}\right\} +2i\mu _{l}{\mathbf {R}}_{l}\displaystyle \sum _{k=1, k\ne l}^{N}\left[ {\mathbf {A}}_{k},{\mathbf {J}}_{l,k}\right] \\= & {} \displaystyle \sum _{k=1, k\ne l}^{N}i\left\{ {\mathbf {J}}_{l,k},{\mathbf {A}}_{k}\right\} +2i\mu _{l}{\mathbf {R}}_{l}\displaystyle \sum _{k=1, k\ne l}^{N}i\left( 1+2\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {A}}_{l}. \end{aligned}$$

Therefore

$$\begin{aligned} \left[ {\mathbf {J}}^{2},{\mathbf {A}}_{l}\right]= & {} \displaystyle \sum _{k=1, k\ne l}^{N}i\left\{ {\mathbf {J}}_{l,k},{\mathbf {A}}_{k}\right\} -2(N-1)\mu _{l}{\mathbf {R}}_{l}{\mathbf {A}}_{l}-4\mu _{l}{\mathbf {R}}_{l}\left( \displaystyle \sum _{k=1, k\ne l}^{N}\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {A}}_{l}. \end{aligned}$$

(D.11)

Using (D.9), we can prove

$$\begin{aligned} \left[ {\mathbf {A}}_{l},{\mathbf {A}}^{2}\right]= & {} \displaystyle \sum _{k=1}^{N}\left[ {\mathbf {A}}_{l},{\mathbf {A}}_{k}^{2}\right] = \displaystyle \sum _{k=1}^{N}\left[ {\mathbf {A}}_{l},{\mathbf {A}}_{k}\right] {\mathbf {A}}_{k}+{\mathbf {A}}_{k}\left[ {\mathbf {A}}_{l},{\mathbf {A}}_{k}\right] \nonumber \\= & {} -2i{\mathcal {H}}\displaystyle \sum _{k=1, k\ne l}^{N}\left\{ {\mathbf {J}}_{l,k},{\mathbf {A}}_{k}\right\} . \end{aligned}$$

(D.12)

From (D.11) and (D.12) we deduce

$$\begin{aligned} \left[ {\mathbf {A}}^{2}-2{\mathcal {H}}\left( {\mathbf {J}}^{2}+(N-1)\mu _{l}{\mathbf {R}}_{l}+2\mu _{l}{\mathbf {R}}_{l}\displaystyle \sum _{k=1, k\ne l}^{N}\mu _{k}{\mathbf {R}}_{k}\right) ,{\mathbf {A}}_{l}\right]= & {} 0. \end{aligned}$$

(D.13)

Since $\left[ \mu _{k}{\mathbf {R}}_{k},{\mathbf {A}}_{l}\right] =0; \ k\ne l,$ we get

$$\begin{aligned} \displaystyle \sum _{k=1, k\ne l}\left[ \mu _{k}{\mathbf {R}}_{k},{\mathbf {A}}_{l}\right]= & {} \displaystyle \sum _{j<k, j\ne l, j\ne k}\left[ \mu _{j}\mu _{k}{\mathbf {R}}_{j}{\mathbf {R}}_{k},{\mathbf {A}}_{l}\right] =0. \end{aligned}$$

(D.14)

Multiply (D.14) by $2{\mathcal {H}}$ and add (D.13) to that equation, we get

$$\begin{aligned} \left[ {\mathbf {C}}_{1},{\mathbf {A}}_{l}\right] =0, \ l=1,\dots ,N, \end{aligned}$$

(D.15)

where

$$\begin{aligned} {\mathbf {C}}_{1}={\mathbf {A}}^{2}-2{\mathcal {H}}\left( {\mathbf {J}}^{2}+(N-1)\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}+2\displaystyle \sum _{j<k}\mu _{j}\mu _{k} {\mathbf {R}}_{j}{\mathbf {R}}_{k}\right) . \end{aligned}$$

(D.16)

Now using (D.10), Proposition 5.5 and writing

$$\begin{aligned} {\mathbf {C}}_{1}={\mathbf {A}}^{2}-2{\mathcal {H}}\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}-2{\mathcal {H}}{\mathbf {C}} \end{aligned}$$

we deduce

$$\begin{aligned} \left[ {\mathbf {C}}_{1},{\mathbf {J}}_{j,k}\right] =0, \ j<k. \end{aligned}$$

(D.17)

Appendix: Proof of Theorem 5.2

It suffices to show (4) and (6). Let us derive the commutation relation between ${\mathbf {J}}_{j,k}$ with ${\mathbf {U}}_{l}.$ We use the representation (5.25) for ${\mathbf {U}}_{l},$ then

$$\begin{aligned} \sqrt{2|{\mathcal {E}}|}\left[ {\mathbf {J}}_{j,k},{\mathbf {U}}_{l}\right]= & {} (1/2)\left[ {\mathbf {J}}_{j,k},{\mathbf {x}}_{l}{\mathbf {p}}^{2}\right] - \left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{l}({\mathbf {x}}\cdot {\mathbf {p}})\right] +i((N-3)/2)\left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{l}\right] \\&+i\left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{l}(\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N})\right] +{\mathcal {E}}\left[ {\mathbf {J}}_{j,k},{\mathbf {x}}_{l}\right] . \end{aligned}$$

Using (5.17) and (2.20)

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k}, {\mathbf {x}}_{l}{\mathbf {p}}^{2}\right]= & {} \left[ {\mathbf {J}}_{j,k}, {\mathbf {x}}_{l}\right] {\mathbf {p}}^{2}+{\mathbf {x}}_{l}\left[ {\mathbf {J}}_{j,k}, {\mathbf {p}}^{2}\right] \\ {}= & {} i{\mathbf {x}}_{k}(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{l}{\mathbf {p}}^{2} -i{\mathbf {x}}_{j}(1+2\mu _{k}{\mathbf {R}}_{k})\delta _{l}^{k}{\mathbf {p}}^{2}. \end{aligned}$$

By (2.20), (3.12) and (3.13)

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k}, {\mathbf {p}}_{l}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) \right]= & {} \left[ {\mathbf {J}}_{j,k}, {\mathbf {p}}_{l}\right] \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) +{\mathbf {p}}_{l}\left[ {\mathbf {J}}_{j,k}, \left( {\mathbf {x}}\cdot {\mathbf {p}}\right) \right] \\= & {} i(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{l}{\mathbf {p}}_{k}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) - i(1+2\mu _{k}{\mathbf {R}}_{k})\delta _{k}^{l}{\mathbf {p}}_{j}\left( {\mathbf {x}}\cdot {\mathbf {p}}\right) \\&- 2i{\mathbf {p}}_{l}\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {J}}_{j,k}. \end{aligned}$$

By (2.20),

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k}, {\mathbf {p}}_{l}\right]= & {} i(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{l}{\mathbf {p}}_{k}- i(1+2\mu _{k}{\mathbf {R}}_{k})\delta _{k}^{l}{\mathbf {p}}_{j}.\\ \left[ {\mathbf {J}}_{j,k}, {\mathbf {x}}_{l}\right]= & {} i{\mathbf {x}}_{k}(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{l} -i{\mathbf {x}}_{j}(1+2\mu _{k}{\mathbf {R}}_{k})\delta _{l}^{k}. \end{aligned}$$

Using (5.13) and (5.14)

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {p}}_{l}(\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N})\right]= & {} \left( i(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{l}{\mathbf {p}}_{k}- i(1+2\mu _{k}{\mathbf {R}}_{k})\delta _{k}^{l}{\mathbf {p}}_{j}\right) \\&\displaystyle \sum _{k=1}^{N}\mu _{k}{\mathbf {R}}_{k}-2{\mathbf {p}}_{l}\left( \mu _{j}{\mathbf {R}}_{j}+\mu _{k}{\mathbf {R}}_{k}\right) {\mathbf {J}}_{j,k}. \end{aligned}$$

Then

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {U}}_{l}\right]= & {} \left\{ \begin{array}{lr} 0, &{} j\ne k \ne l. \\ i(1+2\mu _{j}{\mathbf {R}}_{j}) {\mathbf {U}}_{k}, &{} j\ne k, j=l. \\ -i(1+2\mu _{k}{\mathbf {R}}_{k}) {\mathbf {U}}_{j}, &{} j\ne k, l=k. \end{array} \right. \end{aligned}$$

(6) We proceed as in the proof of (5.23). We use the representation (5.25) for ${\mathbf {U}}_{j},$ we get

$$\begin{aligned} 2|{\mathcal {E}}| \ \left[ {\mathbf {U}}_{j},{\mathbf {U}}_{k}\right] =({\mathcal {E}}/2)(a_{j,k}-a_{k,j})-{\mathcal {E}}(b_{j,k}-b_{k,j}), \end{aligned}$$

where

$$\begin{aligned} a_{j,k}= & {} \left[ {\mathbf {x}}_{j}{\mathbf {p}}^{2}, {\mathbf {x}}_{k}\right] ,\\ b_{j,k}= & {} \left[ {\mathbf {p}}_{j}\left( ({\mathbf {x}}\cdot {\mathbf {p}})-i\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \right) ,{\mathbf {x}}_{k}\right] . \end{aligned}$$

Using (3.9)

$$\begin{aligned} a_{j,k}= & {} {\mathbf {x}}_{j}\left[ {\mathbf {p}}^{2}, {\mathbf {x}}_{k}\right] =-2i{\mathbf {x}}_{j}{\mathbf {p}}_{k}. \end{aligned}$$

From (2.20) and (3.12)

$$\begin{aligned} b_{j,k}= & {} \left[ {\mathbf {p}}_{j}({\mathbf {x}}\cdot {\mathbf {p}}),{\mathbf {x}}_{k}\right] -i\left[ {\mathbf {p}}_{j}\left( \frac{N-3}{2}+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) ,{\mathbf {x}}_{k}\right] .\\= & {} {\mathbf {p}}_{j}\left[ ({\mathbf {x}}\cdot {\mathbf {p}}), {\mathbf {x}}_{k}\right] +\left[ {\mathbf {p}}_{j}, {\mathbf {x}}_{k}\right] ({\mathbf {x}}\cdot {\mathbf {p}})-i{\mathbf {p}}_{j}\left[ ((N-3)/2 +\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}),{\mathbf {x}}_{k}\right] \\- & {} i\left[ {\mathbf {p}}_{j}, {\mathbf {x}}_{k}\right] ((N-3)/2 +\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N})\\= & {} -i{\mathbf {p}}_{j}{\mathbf {x}}_{k}-i(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{k}({\mathbf {x}}\cdot {\mathbf {p}})-(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{k}((N-3)/2 +\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N})\\= & {} -i{\mathbf {x}}_{k}{\mathbf {p}}_{j}-(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{k}\\&\quad -i(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{k}({\mathbf {x}}\cdot {\mathbf {p}})-(1+2\mu _{j}{\mathbf {R}}_{j})\delta _{j}^{k}((N-3)/2 +\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}). \end{aligned}$$

Then

$$\begin{aligned} b_{j,k}-b_{k,j}=i{\mathbf {x}}_{j}{\mathbf {p}}_{k}-i{\mathbf {x}}_{k}{\mathbf {p}}_{j}, \end{aligned}$$

and

$$\begin{aligned} \left[ {\mathbf {U}}_{j},{\mathbf {U}}_{k}\right] =-i\frac{{\mathcal {E}}}{|{\mathcal {E}}|}({\mathbf {x}}_{j}{\mathbf {p}}_{k}-{\mathbf {x}}_{k}{\mathbf {p}}_{j})=-i\sigma \ {\mathbf {J}}_{j,k}. \end{aligned}$$

Appendix: Proof of Theorem 5.3

It suffices to show (5.30), (5.31) and

$$\begin{aligned} \left[ {\mathbf {T}}_{3}, {\mathbf {V}}_{j}\right]= & {} 0, \ j=1,\dots ,N.\\ \left[ {\mathbf {T}}_{3}, {\mathbf {J}}_{j,k}\right]= & {} 0, \ 1\le j<k\le N. \end{aligned}$$

(5.30) follows immediately from (5.26). From (3.20) and (3.21) we have the identity

$$\begin{aligned} {\mathbf {J}}^{2}+{\mathbf {T}}_{1}^{2}+{\mathbf {T}}_{2}^{2}={\mathbf {T}}_{3}^{2}-\left( \frac{N-1}{2}+\displaystyle \sum _{j=1}^{N}\mu _{j}{\mathbf {R}}_{j}\right) ^{2}+\left( \frac{N-1}{2}+\displaystyle \sum _{j=1}^{N}\mu _{j}{\mathbf {R}}_{j}\right) . \end{aligned}$$

Substituting $\alpha =\beta =1$ into Eq. (5.11), we obtain

$$\begin{aligned} {\mathbf {V}}^{2}={\mathbf {T}}_{1}^{2}+{\mathbf {T}}_{2}^{2}-\left( \frac{N-1}{2}+\displaystyle \sum _{j=1}^{N}\mu _{j}{\mathbf {R}}_{j}\right) . \end{aligned}$$

By inserting the previous relations into (5.30) we obtain

$$\begin{aligned} {\mathbf {C}}= & {} {\mathbf {T}}_{3}^{2}-\left( \left( \frac{N-1}{2}\right) ^{2}+\displaystyle \sum _{j=1}^{N}\mu _{j}^{2}\right) . \end{aligned}$$

From the definition of ${\mathbf {J}}_{j,k}, \ 1\le j<k\le N,$

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {T}}_{3}\right]= & {} \left[ {\mathbf {X}}_{j}{\mathbf {P}}_{k}-{\mathbf {X}}_{k}{\mathbf {P}}_{j}, {\mathbf {T}}_{3}\right] \\= & {} {\mathbf {X}}_{j}\left[ {\mathbf {P}}_{k}, {\mathbf {T}}_{3}\right] +\left[ {\mathbf {X}}_{j}, {\mathbf {T}}_{3}\right] {\mathbf {P}}_{k} -{\mathbf {X}}_{k}\left[ {\mathbf {P}}_{j}, {\mathbf {T}}_{3}\right] -\left[ {\mathbf {X}}_{k}, {\mathbf {T}}_{3}\right] {\mathbf {P}}_{j}. \end{aligned}$$

Using (2.21), (3.9) and the fact that $[a,bc]=[a,b]c+b[a,c],$ then

$$\begin{aligned} \left[ {\mathbf {X}}_{j}, {\mathbf {T}}_{3}\right]= & {} (1/2)\left[ {\mathbf {X}}_{j},\ R {\mathbf {P}}^{2}+R\right] =(1/2)R\left[ {\mathbf {X}}_{j}, {\mathbf {P}}^{2}\right] =iR{\mathbf {P}}_{j}.\\ \left[ {\mathbf {P}}_{k}, {\mathbf {T}}_{3}\right]= & {} (1/2)\left[ {\mathbf {P}}_{k},\ R {\mathbf {P}}^{2}+R\right] =(1/2)\left[ {\mathbf {P}}_{k}, R\right] \left( 1+{\mathbf {P}}^{2}\right) \\= & {} -(i/2)R^{-1}{\mathbf {X}}_{k}\left( 1+{\mathbf {P}}^{2}\right) . \end{aligned}$$

Collect these results to obtain

$$\begin{aligned} \left[ {\mathbf {J}}_{j,k},{\mathbf {T}}_{3}\right] =0, \ 1\le j<k\le N. \end{aligned}$$

From the definitions (3.8) and (5.29),

$$\begin{aligned} \left[ {\mathbf {V}}_{j},{\mathbf {T}}_{3}\right] =\frac{1}{4}\left[ {\mathbf {X}}_{j}{\mathbf {P}}^{2}, R\right] +\frac{1}{4}\left[ {\mathbf {X}}_{j}{\mathbf {P}}^{2}, R{\mathbf {P}}^{2}\right] -\frac{1}{4}\left[ {\mathbf {X}}_{j}, R{\mathbf {P}}^{2}\right] -\frac{a_{1}}{2}-\frac{a_{2}}{2}, \end{aligned}$$

where

$$\begin{aligned} a_{1}= & {} \left[ {\mathbf {P}}_{j}\Bigr ({\mathbf {X}}\cdot {\mathbf {P}}-i\left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \Bigl ),R\right] ,\\ a_{2}= & {} \left[ {\mathbf {P}}_{j}\Bigr ({\mathbf {X}}\cdot {\mathbf {P}}-i\left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \Bigl ),R{\mathbf {P}}^{2}\right] . \end{aligned}$$

Using (3.9) and (3.10)

$$\begin{aligned} \left[ {\mathbf {X}}_{j}{\mathbf {P}}^{2}, R\right]= & {} {\mathbf {X}}_{j}\left[ {\mathbf {P}}^{2}, R\right] \\= & {} -2i R^{-1}{\mathbf {X}}_{j}\Bigr ({\mathbf {X}}\cdot {\mathbf {P}}-i\left( (N-1)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \Bigl ),\\ \left[ {\mathbf {X}}_{j}{\mathbf {P}}^{2}, R{\mathbf {P}}^{2}\right]= & {} \left( {\mathbf {X}}_{j}{\mathbf {P}}^{2}R-R{\mathbf {P}}^{2}{\mathbf {X}}_{j}\right) {\mathbf {P}}^{2}={\mathbf {X}}_{j}\left[ {\mathbf {P}}^{2}, R\right] {\mathbf {P}}^{2}+R\left[ {\mathbf {X}}_{j},{\mathbf {P}}^{2}\right] {\mathbf {P}}^{2},\\= & {} -2iR^{-1} {\mathbf {X}}_{j}\Bigr ({\mathbf {X}}\cdot {\mathbf {P}}-i\left( (N-1)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \Bigl ){\mathbf {P}}^{2}\\&+2iR{\mathbf {P}}_{j}{\mathbf {P}}^{2},\\ \left[ {\mathbf {X}}_{j}, R{\mathbf {P}}^{2}\right]= & {} R\left[ {\mathbf {X}}_{j}, {\mathbf {P}}^{2}\right] =2iR{\mathbf {P}}_{j}. \end{aligned}$$

From (2.21) and (3.14)

$$\begin{aligned} a_{1}= & {} \left[ {\mathbf {P}}_{j}\left( {\mathbf {X}}\cdot {\mathbf {P}}\right) , R\right] -i\left[ {\mathbf {P}}_{j} \left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) , R\right] \\= & {} {\mathbf {P}}_{j}\left[ \left( {\mathbf {X}}\cdot {\mathbf {P}}\right) ,R\right] +\left[ {\mathbf {P}}_{j}, R \right] \left( {\mathbf {X}}\cdot {\mathbf {P}}\right) \\&-i\left[ {\mathbf {P}}_{j}, R \right] \left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \\= & {} -iR{\mathbf {P}}_{j}-iR^{-1}{\mathbf {X}}_{j}\left( {\mathbf {X}}\cdot {\mathbf {P}}\right) -R^{-1}{\mathbf {X}}_{j}\left( (N-1)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) . \end{aligned}$$

It is easy to see that

$$\begin{aligned} a_{2}= & {} \left[ {\mathbf {P}}_{j}\left( {\mathbf {X}}\cdot {\mathbf {P}}\right) , R{\mathbf {P}}^{2}\right] -i\left[ {\mathbf {P}}_{j}\left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) , R{\mathbf {P}}^{2}\right] . \end{aligned}$$

From (2.21), (3.11) and (3.14)

$$\begin{aligned} \left[ {\mathbf {P}}_{j}\left( {\mathbf {X}}\cdot {\mathbf {P}}\right) , R{\mathbf {P}}^{2}\right]= & {} {\mathbf {P}}_{j}R\left[ \left( {\mathbf {X}}\cdot {\mathbf {P}}\right) ,{\mathbf {P}}^{2}\right] +{\mathbf {P}}_{j}\left[ \left( {\mathbf {X}}\cdot {\mathbf {P}}\right) , R\right] {\mathbf {P}}^{2}+\left[ {\mathbf {P}}_{j},R\right] {\mathbf {P}}^{2}\left( {\mathbf {X}}\cdot {\mathbf {P}}\right) \\= & {} iR{\mathbf {P}}_{j}{\mathbf {P}}^{2}-R^{-1}{\mathbf {X}}_{j}{\mathbf {P}}^{2}-iR^{-1}{\mathbf {X}}_{j}\left( {\mathbf {X}}\cdot {\mathbf {P}}\right) {\mathbf {P}}^{2}, \end{aligned}$$

and

$$\begin{aligned}&\left[ {\mathbf {P}}_{j}\left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) , R{\mathbf {P}}^{2}\right] \\&\quad =\left[ {\mathbf {P}}_{j},R\right] {\mathbf {P}}^{2}\left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) \\&\quad =-iR^{-1}{\mathbf {X}}_{j}{\mathbf {P}}^{2}\left( (N-3)/2+\mu _{1}{\mathbf {R}}_{1}+\dots +\mu _{N}{\mathbf {R}}_{N}\right) . \end{aligned}$$

Collecting these results

$$\begin{aligned} \left[ {\mathbf {V}}_{j},{\mathbf {T}}_{3}\right] =0, \ j=1,\dots ,N. \end{aligned}$$