#  Superintegrable systems with spin and second-order (pseudo)tensor integrals
of motion

İsmet Yurduşen1 , O. Oğulcan Tuncer1 , and Pavel Winternitz2,3

1Department of Mathematics, Hacettepe University,
06800 Beytepe, Ankara, Turkey

2Centre de Recherches Mathématiques, Université de Montréal,
CP 6128, Succ. Centre-Ville, Montréal, Quebec H3C 3J7, Canada

3Département de Mathématiques et de Statistique, Université de Montréal,
CP 6128, Succ. Centre-Ville, Montréal, Quebec H3C 3J7, Canada E-mail address:
<EMAIL_ADDRESS>address<EMAIL_ADDRESS>address<EMAIL_ADDRESS>
###### Abstract

We investigate a quantum non-relativistic system describing the interaction of
two particles with spin $\frac{1}{2}$ and spin 0, respectively. Assuming that
the Hamiltonian is rotationally invariant and parity conserving we identify
all such systems which allow additional (pseudo)tensor integrals of motion
that are second order matrix polynomials in the momenta. Previously we found
all the (pseudo)scalar and (axial)vector integrals of motion. No non-obvious
tensor integrals exist. However, nontrivial pseudo-tensor integrals do exist.
Together with our earlier results we give a complete list of such
superintegrable Hamiltonian systems allowing second-order integrals of motion.

PACS numbers: 02.30.Ik, 03.65.-w, 11.30.-j, 25.80.Dj

##  1 Introduction

The purpose of this article is to report on a research program which
investigates the superintegrability properties of systems involving particles
with spin. This time the contribution is given by analyzing the second-order
(pseudo)tensor integrals of motion. In this program we consider two non-
relativistic particles with spin $s=1/2$ and $s=0$, respectively, which can be
interpreted e.g. as a nucleon–pion interaction or an electron–$\alpha$
particle one. An earlier article [1] was devoted to the investigation of the
second-order (pseudo)scalar and (axial)vector integrals of motion for those
Hamiltonian systems in a real three-dimensional Euclidean space. Hence, this
work completes the full classification problem for those systems with second-
order integrals of motion. The other articles in this program were devoted to
the investigation of the same problem with first-order integrals of motion in
$E_{2}$ and $E_{3}$ [2, 3, 4] and second-order integrals of motion in $E_{2}$
[5].

For a Hamiltonian system with $n$ degrees of freedom, integrability (in the
Liouville sense, of course) corresponds to existing of $n$ independent
commuting integrals of motion (including the Hamiltonian). If more than $n$
independent integrals exist, the system is called superintegrable.

The best known superintegrable systems are: Kepler, or Coulomb system
($V_{0}=\frac{\alpha}{r}$) and the Harmonic oscillator ($V_{0}=\omega r^{2}$).
These are characterized by the fact that all finite classical trajectories in
these systems are periodic. Indeed, due to Bertrand’s theorem [6] these are
the only two spherically symmetric potentials in which all bounded
trajectories are closed. Releasing the restriction of spherical symmetry for
the potential can lead to many new possibilities e.g. the anisotropic harmonic
oscillator with rational ratio of frequencies [7].

A systematic search for the properties of superintegrable systems was started
quite some time ago [8, 9, 10]. Originally the approach concentrated on the
natural Hamiltonians of type

$H=-\frac{1}{2}\Delta+V(\vec{r}\,)\,,$ (1.1)

with integrals of motion that are second-order polynomials in the momenta and
directly related with the multiseparability in two- and three-dimensional
Euclidean spaces [8, 11, 12, 13]. This relationship between integrability and
separability breaks down in other cases. For example, for natural Hamiltonians
(1.1), the existence of third-order integrals does not lead to the separation
of variables [14, 15, 16]. Furthermore, if we consider velocity dependent
potentials

$H=-\frac{1}{2}\Delta+V(\vec{r}\,)+(\vec{A},\vec{p})\,,$ (1.2)

then quadratic integrability no longer implies the separation of variables
[17, 18].

Second-order superintegrability has also been studied in 2- and 3-dimensional
spaces of constant and nonconstant curvature [19, 20, 21, 22],[23, 24, 25, 26,
27, 28, 29, 30] and also in $n$ dimensions [31, 32, 33].

After the discovery of infinite families of classical and quantum systems with
integrals of arbitrary order [34], the direction of the research has been
shifted to higher-order integrability/superintegrability [35, 36, 37, 38, 39,
40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50]. Recently extended reviews have
been published describing the current status of the subject [48, 49].

More recently, exotic and standard potentials appearing in classical and
quantum superintegrable systems have been studied both in Cartesian and polar
coordinates [51, 52, 53, 54, 55, 56, 57, 58, 59, 60].

Previous studies of superintegrable systems with spin include [61, 62, 63, 64,
65, 66, 67, 68, 69].

The outline of the paper is as follows. In the next section, we introduce the
formulation of the problem. Then, in section 3, we give the most general
second-order tensor integrals of motion, obtain the determining equations and
solve them in order to find all the second-order tensor integrals of motion.
In section 4, we repeat the similar analysis for pseudo-tensor integrals of
motion. Finally, in the last section we give a complete list of second-order
integrals of motion and conclude about our results.

##  2 Formulation of the problem

We consider the following Hamiltonian

$H=-\frac{\hbar^{2}}{2}\Delta+V_{0}(r)+V_{1}(r)\,(\vec{\sigma},\vec{L})\,,$
(2.1)

in real three-dimensional Euclidean space $E_{3}$. Here $H$ is a matrix
operator acting on a two-component spinor and we decompose it in terms of the
$2\times 2$ identity matrix $I$ and Pauli matrices (we drop the matrix $I$
whenever this does not cause confusion). We assume that the scalar potential
$V_{0}(r)$ and the spin-orbital one $V_{1}(r)$ depend on the (scalar) distance
$r$ only. The same system was already considered in [1, 4]. In [1] the search
for superintegrable systems was restricted to second-order (pseudo)scalar and
(axial)vector integrals of motion and in [4] the search was restricted to
first-order integrals.

Here we will investigate the second-order (pseudo)tensor integrals of motion.
These are second-order matrix polynomials in the momenta. Our notations are
the same as in [4], i.e.

$p_{k}=-i\hbar\partial_{x_{k}}\,,\qquad
L_{k}=-i\hbar\epsilon_{klm}x_{l}\partial_{x_{m}}\,,$ (2.2)

are the linear and angular momentum respectively and

$\sigma_{1}=\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\,,\qquad\sigma_{2}=\left(\begin{array}[]{cc}0&-i\\\
i&0\end{array}\right)\,,\qquad\sigma_{3}=\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right)\,,$ (2.3)

are the Pauli matrices.

The system (2.1) is integrable by construction. Since $V_{0}(r)$ and
$V_{1}(r)$ depend only on the distance $r$ and $(\vec{\sigma},\vec{L})$ is a
scalar, the Hamiltonian $H$ commutes with the total angular momentum
$\vec{\mathcal{J}}$ which is defined by

$\vec{\mathcal{J}}=\vec{L}+\dfrac{\hbar}{2}\vec{\sigma}.$ (2.4)

To obtain nontrivial results, we impose from the beginning that the spin-
orbital interaction be present ($V_{1}\neq 0$). We also recall a result from
[4], namely for

$V_{1}(r)=\frac{\hbar}{r^{2}}\,,\qquad V_{0}(r)\quad\text{arbitrary}\,,$ (2.5)

the Hamiltonian (2.1) allows 2 first-order axial vector integrals of motion.
They are $\vec{\mathcal{J}}$ and

$\vec{S}=-\frac{\hbar}{2}\vec{\sigma}+\hbar\frac{\vec{x}}{r^{2}}(\vec{x},\vec{\sigma})\,.$
(2.6)

For

$V_{1}(r)=\frac{\hbar}{r^{2}}\,,\qquad V_{0}(r)=\frac{\hbar^{2}}{r^{2}}\,,$
(2.7)

it allows 2 first-order axial vector integrals and 1 first-order vector
integral. They are $\vec{\mathcal{J}}$, $\vec{S}$ and

$\vec{\Pi}=\vec{p}-\frac{\hbar}{r^{2}}(\vec{x}\wedge\vec{\sigma})\,.$ (2.8)

These two systems are first-order superintegrable and the term
$V_{1}(r)=\frac{\hbar}{r^{2}}$ can be induced from a Hamiltonian with
$V_{1}(r)=0$ by a gauge transformation [4]. For the gauge induced potentials
given in (2.5), the integrals of motion are either just the gauge transforms
of the terms obtained from $\vec{L}$, $\vec{\sigma}$, or else leave in the
enveloping algebra of a direct sum of the algebra $o(3)$ with itself
${o}(3)\oplus{o}(3)=\\{\vec{\mathcal{J}}-\vec{S}\\}\oplus\\{\vec{S}\\}$. For
the gauge induced potentials given in (2.7), such integrals of motion are
either the gauge transforms of the terms obtained from $\vec{L}$, $\vec{p}$,
$\vec{\sigma}$, or else leave in the enveloping algebra of a direct sum of the
Euclidean Lie algebra $e(3)$ with the algebra $o(3)$:
$e(3)\oplus{o}(3)=\\{\vec{\mathcal{J}}-\vec{S},\vec{\Pi}\\}\oplus\\{\vec{S}\\}$.
(For details; see [4].) Hence, in the subsequent analysis we exclude the cases
when the spin-orbital potential is a gauged induced one. However, we will
present those pseudo(tensor) integrals of motion corresponding to gauge
induced potentials in the appendix in order to explicitly verify that they
reduce to the well-known results in the absence of the spin-orbit interaction.

In sections 3 and 4, we search for second-order (pseudo)tensor integrals
commuting with the Hamiltonian (2.1). Such commutations yield the so-called
determining equations which will then be solved to find the integrals of
motion. The Hamiltonian (2.1) is invariant under rotations and reflections in
$E_{3}$ which, however, can transform the integrals of motion into new
invariants. Thus, instead of considering the solution of the whole set of
determining equations, it is reasonable to simplify the problem by classifying
the integrals of motion into irreducible $O(3)$ multiplets.

At our disposal are two vectors $\vec{x}$ and $\vec{p}$ and one pseudo-vector
$\vec{\sigma}$. The integrals we are considering can involve at most second-
order powers of $\vec{p}$ and first-order powers of $\vec{\sigma}$, but
arbitrary powers of $\vec{x}$.

We shall construct symmetric two-component tensors and pseudo-tensors in the
space:

$\\{\\{\vec{x}\\}^{n}\times\vec{p}\times\vec{\sigma}\\}.$ (2.9)

The quantities $\vec{x}$, $\vec{p}$ and $\vec{\sigma}$ allow us to define six
independent ‘directions’ in the direct product of the Euclidean space and the
spin one, namely

$\\{\vec{x},\vec{p},\vec{L}=\vec{x}\wedge\vec{p},\vec{\sigma},\vec{\sigma}\wedge\vec{x},\vec{\sigma}\wedge\vec{p}\\},$

and any $O(3)$ tensor can be expressed in terms of these. The positive integer
$n$ in (2.9) is arbitrary and any scalar in $\vec{x}$ space will be written as
$f(r)$, where $f$ is an arbitrary function of $r=\sqrt{x^{2}+y^{2}+z^{2}}$.
Since $\vec{p}$ figures at most quadratically and $\vec{\sigma}$ at most
linearly, we can form 28 two-component symmetric tensors and 26 symmetric
pseudo-tensors. In the following sections we will separately and explicitly
give the form of these integrals of motion.

##  3 Tensor integrals of motion

Two index tensors are expressed as follows:

$\displaystyle T_{1}^{ij}=x^{i}x^{j},\quad
T_{2}^{ij}=(\vec{x},\vec{p})x^{i}x^{j},\quad
T_{3}^{ij}=(\vec{\sigma},\vec{L})x^{i}x^{j},\quad
T_{4}^{ij}=x^{i}p^{j}+p^{i}x^{j},$ $\displaystyle
T_{5}^{ij}=x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j},\quad
T_{6}^{ij}=(\vec{x},\vec{p})(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}),\quad
T_{7}^{ij}=L^{i}\sigma^{j}+\sigma^{i}L^{j},$ $\displaystyle
T_{8}^{ij}=p^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}p^{j},\quad
T_{9}^{ij}={\vec{p}}^{\>2}x^{i}x^{j},\quad
T_{10}^{ij}=\vec{L}^{2}x^{i}x^{j},\quad
T_{11}^{ij}=(\vec{x},\vec{p})(\vec{\sigma},\vec{L})x^{i}x^{j},$ $\displaystyle
T_{12}^{ij}=L^{i}L^{j}+L^{j}L^{i},\quad
T_{13}^{ij}=(\vec{\sigma},\vec{L})(x^{i}p^{j}+p^{i}x^{j}),\quad
T_{14}^{ij}=\vec{p}^{\>2}(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}),$
$\displaystyle
T_{15}^{ij}=\vec{L}^{2}(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}),\quad
T_{16}^{ij}=(\vec{x},\vec{p})(x^{i}(\vec{p}\wedge\vec{\sigma})^{j}+(\vec{p}\wedge\vec{\sigma})^{i}x^{j}),\quad
T_{17}^{ij}=p^{i}p^{j},$ $\displaystyle
T_{18}^{ij}=(\vec{x},\vec{p})(p^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}p^{j}),\quad
T_{19}^{ij}=p^{i}(\vec{p}\wedge\vec{\sigma})^{j}+(\vec{p}\wedge\vec{\sigma})^{i}p^{j},$
$\displaystyle
T_{20}^{ij}=x^{i}(\vec{p}\wedge\vec{\sigma})^{j}+(\vec{p}\wedge\vec{\sigma})^{i}x^{j},\quad
T_{21}^{ij}=(\vec{x},\vec{p})(L^{i}\sigma^{j}+\sigma^{i}L^{j}),\quad
T_{22}^{ij}=(\vec{x},\vec{p})(x^{i}p^{j}+p^{i}x^{j}),$ $\displaystyle
T_{23}^{ij}=(\vec{x},\vec{p})^{2}x^{i}x^{j},\quad
T_{24}^{ij}=(\vec{x},\vec{p})^{2}(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}),\quad
T_{25}^{ij}=(\vec{\sigma},\vec{x})(x^{i}L^{j}+L^{i}x^{j}),$ $\displaystyle
T_{26}^{ij}=(\vec{x},\vec{p})(\vec{\sigma},\vec{x})(x^{i}L^{j}+L^{i}x^{j}),\quad
T_{27}^{ij}=(\vec{\sigma},\vec{x})(p^{i}L^{j}+L^{i}p^{j}),\quad
T_{28}^{ij}=(\vec{\sigma},\vec{p})(x^{i}L^{j}+L^{i}x^{j}).$ (3.1)

Each of the quantities in (3) can be multiplied by a scalar $f(r)$ without
changing its properties under rotations or reflections.

It can be shown that we have three linear relations

$\displaystyle T_{20}^{ij}=-T_{7}^{ij}+T_{8}^{ij},\qquad
T_{21}^{ij}=-T_{16}^{ij}+T_{18}^{ij},\qquad
T_{28}^{ij}=T_{14}^{ij}-T_{16}^{ij}+T_{13}^{ij}\,,$ (3.2)

and even though all the above tensors but $T_{20}^{ij},T_{21}^{ij}$ and
$T_{28}^{ij}$ are linearly independent, higher-order polynomial relations
between them exist. Indeed, we have the following syzygies:

$\displaystyle
T_{22}^{ij}=\frac{1}{2}(\vec{r}^{\>2}T_{17}^{ij}+T_{12}^{ij}+T_{9}^{ij}),$
$\displaystyle T_{23}^{ij}=\vec{r}^{\>2}T_{9}^{ij}-T_{10}^{ij},$
$\displaystyle T_{24}^{ij}=\vec{r}^{\>2}T_{14}^{ij}-T_{15}^{ij},$
$\displaystyle T_{25}^{ij}=-\vec{r}^{\>2}T_{7}^{ij}+T_{6}^{ij}+T_{3}^{ij},$
$\displaystyle T_{26}^{ij}=-\vec{r}^{\>2}T_{16}^{ij}+T_{24}^{ij}+T_{11}^{ij},$
$\displaystyle T_{27}^{ij}=-\vec{r}^{\>2}T_{19}^{ij}+T_{18}^{ij}+T_{13}^{ij}.$
(3.3)

We use relations (3.2) and (3.3) to remove the left-hand sides of these
equations from the analysis completely.

###  3.1 The commutativity condition and determining equations

In this subsection, we take the linear combinations of all the two-component
tensors and then fully symmetrize them in order to obtain the determining
equations from the commutativity condition.

Let us take the linear combination of the independent tensors given in (3) as

$X_{T}^{ij}=\sum\limits_{k=1}^{19}f_{k}(r)T_{k}^{ij},$

which can be fully symmetrized as decribed in [1]. In the commutativity
relation $[H,X_{T}^{ij}]=0$ it is enough to consider only two indices, say
$i=1,j=2$ since the others then necessarily commute due to the rotations. In
the analysis we consider the full symmetric form of $X_{T}^{12}$ which,
however, is rather long to be presented here. From the requirement
$[H,X_{T}^{ij}]=0$, we obtain the determining equations.

The determining equations obtained by equating the coefficients of third-order
terms to zero in the commutativity equation give us

$\displaystyle f_{10}=0,\quad f_{9}=0,\quad f_{12}=c_{1},\quad
f_{17}=c_{2},\quad f_{11}=-f_{15},\vskip 6.0pt plus 2.0pt minus 2.0pt$ (3.4)
$\displaystyle f_{16}=-f_{14},\quad f_{19}=-r^{2}f_{18}+c_{4},\quad
f_{13}=r^{2}f_{15}+f_{14}+f_{18},\quad f_{14}=c_{3},\vskip 6.0pt plus 2.0pt
minus 2.0pt$ (3.5) $\displaystyle 2rf_{15}V_{1}+\hbar f_{15}^{\prime}=0,$
(3.6) $\displaystyle 2c_{3}V_{1}-\hbar(f_{15}+rf_{15}^{\prime})=0,$ (3.7)
$\displaystyle 2r(c_{3}+r^{2}f_{15}-f_{18})V_{1}-\hbar f_{18}^{\prime}=0,$
(3.8) $\displaystyle\hbar f_{18}+2c_{4}V_{1}+\hbar(c_{3}+rf_{18}^{\prime})=0,$
(3.9)

where $c_{i}\,(i=1,2,3,4)$ are integration constants.

Introducing the relations given in (3.4) and (3.5) into the determining
equations obtained by equating the coefficients of the second-order terms to
zero in the commutativity equation, we get

$\displaystyle f_{3}=f_{6}+c_{5},\quad f_{7}=r^{2}f_{6}+c_{6},$ (3.10)
$\displaystyle rf_{15}V_{1}-f_{2}^{\prime}=0,$ (3.11) $\displaystyle
4rf_{2}+r(c_{3}+2r^{2}f_{15}+f_{18})V_{1}+f_{4}^{\prime}-2c_{4}V_{1}^{\prime}=0,$
(3.12) $\displaystyle 2c_{5}\hbar
r+2rf_{6}(-\hbar+r^{2}V_{1})-\hbar(f_{8}^{\prime}+2c_{2}V_{1}^{\prime})=0,$
(3.13) $\displaystyle 2rf_{6}V_{1}+{\hbar}f_{6}^{\prime}=0,$ (3.14)
$\displaystyle-3\hbar rf_{6}-2r(c_{6}-c_{1}\hbar+f_{8})V_{1}-\hbar
f_{8}^{\prime}=0,$ (3.15) $\displaystyle
2rf_{2}+r(-3c_{3}-2r^{2}f_{15}+3f_{18})V_{1}+f_{4}^{\prime}=0,$ (3.16)
$\displaystyle
f_{4}+(3c_{4}+c_{3}r^{2}+r^{4}f_{15}-2r^{2}f_{18})V_{1}+c_{4}rV_{1}^{\prime}=0,$
(3.17) $\displaystyle 2c_{5}\hbar r+\hbar
f_{6}+2r(c_{6}-c_{1}\hbar+r^{2}f_{6}+f_{8})V_{1}-2c_{2}\hbar V_{1}^{\prime}=0$
(3.18) $\displaystyle-2c_{6}V_{1}+\hbar(f_{6}+2c_{1}V_{1}+rf_{6}^{\prime})=0,$
(3.19) $\displaystyle
f_{8}-2c_{2}V_{1}+r(3rf_{6}+r^{2}f_{6}^{\prime}+f_{8}^{\prime})=0,$ (3.20)

where, again, $c_{5}$ and $c_{6}$ are integration constants.

Now, introducing the relations given in (3.4), (3.5) and (3.10) into the
determining equations obtained by equating the coefficients of the first-order
terms to zero in the commutativity equation yields

$\displaystyle
r(2c_{5}-f_{6})V_{1}-f_{1}^{\prime}+(r^{2}f_{6}+f_{8})V_{1}^{\prime}=0,$
(3.21) $\displaystyle 12\hbar rf_{15}V_{1}+8\hbar^{2}f_{15}^{\prime}-2\hbar
f_{18}^{\prime}V_{1}+4f_{5}^{\prime}-4c_{3}\hbar V_{1}^{\prime}-4\hbar
f_{18}V_{1}^{\prime}+4f_{4}V_{1}^{\prime}+4rf_{2}(-2V_{1}+rV_{1}^{\prime})$
$\displaystyle\qquad\qquad\qquad+\hbar^{2}rf_{15}^{\prime\prime}=0,$ (3.22)
$\displaystyle-4rf_{5}+6\hbar rf_{18}V_{1}+4rf_{4}V_{1}-2\hbar
rf_{15}(\hbar-5r^{2}V_{1})+8\hbar^{2}r^{2}f_{15}^{\prime}+10\hbar^{2}f_{18}^{\prime}-2\hbar
r^{2}f_{18}^{\prime}V_{1}$
$\displaystyle\qquad\qquad\qquad+4c_{4}V_{0}^{\prime}+4c_{4}\hbar
V_{1}^{\prime}+\hbar^{2}r^{3}f_{15}^{\prime\prime}+2\hbar^{2}rf_{18}^{\prime\prime}=0.$
(3.23)

Finally, introducing the relations given in (3.4), (3.5) and (3.10) into the
determining equations obtained by equating the coefficients of the zeroth-
order terms to zero in the commutativity equation, we obtain

$\displaystyle-2c_{2}V_{0}^{\prime}+r\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}-r\big{(}6f_{1}^{\prime}+3V_{1}(-2c_{5}r+3rf_{6}+r^{2}f_{6}^{\prime}+f_{8}^{\prime})+rf_{1}^{\prime\prime}\big{)}+2c_{2}V_{0}^{\prime\prime}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}=0,$
(3.24) $\displaystyle 12r^{2}(-rf_{5}V_{1}+3\hbar
f_{2}^{\prime})+4r^{2}(r^{2}f_{2}+f_{4})V_{0}^{\prime}+\hbar^{2}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}-4f_{4}^{\prime}+r\big{(}4f_{4}^{\prime\prime}+r(13rf_{2}^{\prime\prime}+r^{2}f_{2}^{(3)}+f_{4}^{(3)})\big{)}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}=0,$
(3.25) $\displaystyle 10\hbar
r^{3}f_{2}V_{1}-2r^{3}f_{5}V_{1}+\hbar\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}-2c_{4}V_{0}^{\prime}+r\big{(}r(2V_{1}(r^{2}f_{2}^{\prime}+f_{4}^{\prime})-6f_{5}^{\prime}+(c_{3}+f_{18})V_{0}^{\prime}-rf_{5}^{\prime\prime})$
$\displaystyle\qquad\qquad\qquad+2c_{4}V_{0}^{\prime\prime}\big{)}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}=0,$
(3.26) $\displaystyle
24\hbar^{2}r^{2}f_{6}^{\prime}+2r^{2}V_{1}\big{(}4rf_{1}+\hbar(-2c_{5}r+3rf_{6}+r^{2}f_{6}^{\prime}+f_{8}^{\prime})\big{)}+4r^{2}(r^{2}f_{6}+f_{8})V_{0}^{\prime}$
$\displaystyle\qquad\qquad\qquad+\hbar^{2}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}-4f_{8}^{\prime}+r\big{(}4f_{8}^{\prime\prime}+r(11rf_{6}^{\prime\prime}+r^{2}f_{6}^{(3)}+f_{8}^{(3)})\big{)}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}=0.$
(3.27)

We notice that there are four types of determining equations:

  1. 1.

Those that are independent of the potentials $V_{1}$ and $V_{0}$, namely
(3.4), (3.5) and (3.10). We input these relations into the other determining
equations. This eliminates some of the equations involving the potential and
we have presented only the remaining ones.

  2. 2.

Equations involving only $V_{1}$ but not its derivative (3.6)-(3.9), (3.11),
(3.14)-(3.16), (3.19) and (3.20).

  3. 3.

Those involving $V_{1}$ and $V^{\prime}_{1}$ which are (3.12), (3.13), (3.17),
(3.21) and (3.22).

  4. 4.

Those involving $V_{0}$, $V_{1}$ and their derivatives.

###  3.2 Solutions of the determining equations

In this subsection, we find the solutions of the determining equations coming
from the tensors and then obtain the tensor integrals of motion. We present a
complete classification by following the above four classes of determining
equations sequentially. As indicated before, the first type of determining
equations had already been introduced into the others to eliminate or simplify
them. So, we start the analysis with ten determining equations of second type.
These equations can be viewed as linear algebraic equations for $V_{1}$. They
must all be satisfied simultaneously. Thus they must all be multiples of just
one equation. Its solution determines $V_{1}$ uniquely (though there may be
several parametric versions of it depending on the coefficients $f_{i}$,
$c_{a}$).

Let us first consider (3.6). Assuming $f_{15}\neq 0$, we obtain

$V_{1}=-\dfrac{\hbar f_{15}^{\prime}}{2rf_{15}}.$ (3.28)

This equation can be introduced into (3.7) to give

$f_{15}=\dfrac{-c_{3}+\epsilon\sqrt{{c_{3}}^{2}+\gamma_{1}r^{2}}}{r^{2}},$
(3.29)

where $c_{3}$ and $\gamma_{1}$ are real constants. First, it would be better
to see what happens when $c_{3}=0$. The reason for this will become clear
shortly. In this case, we immediately find

$f_{15}=\dfrac{\alpha}{r},$ (3.30)

where $\alpha$ is an integration constant. Introducing this into (3.6), we
find

$2\alpha V_{1}-\dfrac{\alpha\hbar}{r^{2}}=0,$ (3.31)

which gives us two options, either $\alpha=0$ or
$V_{1}=\dfrac{\hbar}{2r^{2}}$. But, the first one yields $f_{15}=0$ which is
undesirable due to the assumption. Then, let us take
$V_{1}=\dfrac{\hbar}{2r^{2}}$. If this potential is considered in (3.8) and
(3.9), one can easily get $c_{3}=0$ and $\alpha=0$, which again yields
$f_{15}=0$. So, we need to take $c_{3}\neq 0$ which lead to (3.29) to be
rewritten as

$f_{15}=-\dfrac{c_{3}(1+\epsilon\sqrt{1+\beta r^{2}})}{r^{2}},$ (3.32)

where $\epsilon^{2}=1$ and $\beta$ is a real constant. From (3.28) and (3.32)
we get the following potential:

$V_{1}=\dfrac{\hbar}{2r^{2}}\left(1+\dfrac{\epsilon}{\sqrt{1+\beta
r^{2}}}\right).$ (3.33)

It is obvious that $V_{1}=\dfrac{\hbar}{r^{2}}$ and
$V_{1}=\dfrac{\hbar}{2r^{2}}$ are special solutions for
$(\epsilon,\beta)=(1,0)$ and $(1,\infty)$, respectively. The first special
potential is a gauge induced one and had already been thoroughly considered in
[4]. Hence, we exclude the analysis of this option which means that we will
assume $\beta\neq 0$.

By considering the above relations (3.32) and (3.33) in the equations (3.8)
and (3.9), one can obtain

$f_{18}=\dfrac{-c_{3}r^{2}(1+\beta r^{2})+c_{4}(1+\epsilon\sqrt{1+\beta
r^{2}})}{r^{2}}.$ (3.34)

Substituting this back into (3.9) yields $\beta=0$ or $c_{3}=0$. The case
$\beta=0$ gives either $V_{1}=0$ or $V_{1}=\dfrac{\hbar}{r^{2}}$ both of which
are excluded. On the other hand, from (3.32) the case $c_{3}=0$ means that
$f_{15}=0$ which contradicts with the assumption $f_{15}\neq 0$.

The above discussion states that $f_{15}$ must vanish. So, (3.7) directly
yields $c_{3}=0$. Keeping in mind these two facts, let us continue the
analysis by considering (3.8). Assume that $f_{18}\neq 0$. Then, one can
easily get

$V_{1}=-\dfrac{\hbar f_{18}^{\prime}}{2rf_{18}},$ (3.35)

and introducing this into (3.9) we obtain

$f_{18}=\dfrac{c_{4}+\epsilon\sqrt{{c_{4}}^{2}+\gamma_{2}r^{2}}}{r^{2}},$
(3.36)

where $c_{4}$ and $\gamma_{2}$ are real constants. Similar to the above, it
can be easily checked that the case $c_{4}=0$ will eventually cause
$f_{18}=0$. Hence, we take $c_{4}\neq 0$ which allows us to rewrite (3.36) as

$f_{18}=\dfrac{c_{4}(1+\epsilon\sqrt{1+\beta
r^{2}})}{r^{2}},\quad\epsilon^{2}=1.$ (3.37)

Then, from the last two relations (3.33) and (3.34) we again have the
potential

$V_{1}=\dfrac{\hbar}{2r^{2}}\left(1+\dfrac{\epsilon}{\sqrt{1+\beta
r^{2}}}\right).$ (3.38)

Then, by introducing these into (3.14), (3.15), (3.19) and (3.20) one can find

$\displaystyle f_{6}=\dfrac{(c_{1}\hbar-c_{6})(1+\epsilon\sqrt{1+\beta
r^{2}})}{r^{2}},$ (3.39) $\displaystyle f_{8}=(c_{6}-c_{1}\hbar)(1+\beta
r^{2}+\epsilon\sqrt{1+\beta r^{2}})-\dfrac{\hbar c_{2}(1+\epsilon\sqrt{1+\beta
r^{2}})}{r^{2}}.$ (3.40)

Substituting these relations into (3.15) gives $\beta(c_{6}-c_{1}\hbar)=0$.
Hence, we directly have $c_{6}=c_{1}\hbar$ since the case $\beta=0$ is
excluded.

The remaining two determining equations of second type, namely (3.11) and
(3.16), give

$f_{2}=d_{1},\quad
f_{4}=-d_{1}r^{2}+\dfrac{3c_{4}\hbar(1+\epsilon\sqrt{1+\beta
r^{2}})}{2r^{2}}+d_{2},$ (3.41)

where $d_{1}$ and $d_{2}$ are real constants. Then, we see that for the
relations (3.37), (3.39), (3.41) and the potential given in (3.38) all of the
determining equations of second type are satisfied. Now, we need to continue
the analysis with the determining equations of third type. However, one can
easily check that introducing these relations into (3.12) immediately yields
either $(d_{1},\beta)=(0,0)$ or $(d_{1},c_{4})=(0,0)$. We already know that
the first case is excluded. The second case gives us a contradiction. This
means that we do not need to examine the case $f_{18}\neq 0$ further.

Hence, $f_{18}$ must vanish too, which immediately gives us $c_{4}=0$. We
continue the analysis with the relations $f_{15}=0$, $c_{3}=0$, $f_{18}=0$ and
$c_{4}=0$. By introducing these relations into (3.11) and (3.16) we find

$f_{2}=d_{1},\quad f_{4}=-d_{1}r^{2}+d_{2},$ (3.42)

where $d_{1}$ and $d_{2}$ are integration constants.

Assuming $f_{6}\neq 0$ in (3.14) we obtain

$V_{1}=-\dfrac{\hbar f_{6}^{\prime}}{2rf_{6}},$ (3.43)

and introducing this into (3.19) yields

$f_{6}=\dfrac{-(c_{6}-c_{1}\hbar)+\epsilon\sqrt{(c_{6}-c_{1}\hbar)^{2}+\gamma_{3}r^{2}}}{r^{2}},$
(3.44)

where $c_{1}$, $c_{6}$ and $\gamma_{3}$ are real constants. From the remaining
determining equations, one can see that the special case when
$c_{6}=c_{1}\hbar$ will give rise to $f_{6}=0$. Therefore, we need to assume
$c_{6}\neq c_{1}\hbar$ which gives

$f_{6}=\dfrac{(c_{1}\hbar-c_{6})(1+\epsilon\sqrt{1+\beta
r^{2}})}{r^{2}},\quad\epsilon^{2}=1.$ (3.45)

Then, the last two relations once again give the potential

$V_{1}=\dfrac{\hbar}{2r^{2}}\left(1+\dfrac{\epsilon}{\sqrt{1+\beta
r^{2}}}\right).$ (3.46)

One can easily check that substituting these relations into (3.15) and (3.19)
yields $(c_{6}-c_{1}\hbar)\beta=0$. Thus, we have $c_{6}=c_{1}\hbar$ due to
the fact that $\beta\neq 0$. However, this contradicts with the assumption
$f_{6}\neq 0$. Therefore, we need to take $f_{6}=0$ which directly yields
$c_{6}=c_{1}\hbar$ by (3.19).

The above discussion implies that two determining equations of second type
remain:

$2rf_{8}V_{1}+\hbar f_{8}^{\prime}=0,\quad
f_{8}+rf_{8}^{\prime}-2c_{2}V_{1}=0.$ (3.47)

Let us first assume $f_{8}\neq 0$. In this case, it is rather easy to see that
$c_{2}$ cannot vanish for this case. So, similar steps to the previous cases
yield the relation

$f_{8}=-\dfrac{c_{2}\hbar(1+\epsilon\sqrt{1+\beta
r^{2}})}{r^{2}},\quad\epsilon^{2}=1,$ (3.48)

and the potential

$V_{1}=\dfrac{\hbar}{2r^{2}}\left(1+\dfrac{\epsilon}{\sqrt{1+\beta
r^{2}}}\right).$ (3.49)

So, all of the determining equations of second type are satisfied. But, by
introducing these relations into (3.13) we see that either
$(c_{5},\beta)=(0,0)$ or $(c_{5},c_{2})=(0,0)$. Again, the first case is
excluded and the second one causes a contradiction.

Eventually, by solving the determining equations of second type we just find
the following relations for an arbitrary spin-orbital potential:

$f_{15}=0,\quad f_{18}=0,\quad f_{6}=0,\quad f_{8}=0,\quad c_{3}=0,\quad
c_{4}=0,\quad c_{6}=c_{1}\hbar,\quad c_{2}=0.$ (3.50)

Let us examine the determining equations of third type. Introducing the
relations in (3.50) into the determining equations (3.12), (3.13), (3.17),
(3.21) and (3.22), one can easily obtain

$f_{2}=0,\quad f_{4}=0,\quad f_{1}=k_{1},\quad c_{5}=0,$ (3.51)

where $k_{1}$ is an integration constant. By these relations, all of the
determining equations of third type are satisfied. Considering (3.50) and
(3.51) in the remaining equations, one can see that $f_{1}=0$ and those
equations are satisfied for arbitrary scalar potential $V_{0}$. Hence, we have
just two functions $f_{12}$ and $f_{7}$ that do not vanish.

Finally, all the determining equations are satisfied for arbitrary potentials
$V_{0}=V_{0}(r)$ and $V_{1}=V_{1}(r)$. Having left with just one arbitrary
constant $c_{1}$, we have the following trivial integral of motion

$\mathcal{T}_{1}^{\,ij}=\hbar(L^{i}\sigma^{j}+\sigma^{i}L^{j})+L^{i}L^{j}+L^{j}L^{i}=\\{\mathcal{J}_{i},\mathcal{J}_{j}\\}-\delta_{ij}\dfrac{\hbar}{2},$
(3.52)

where $\\{\cdot,\cdot\\}$ stands for the anti-commutator.

##  4 Pseudo-tensor integrals of motion

Two index pseudo-tensors are expressed as follows:

$\displaystyle Y_{1}^{ij}=x^{i}\sigma^{j}+\sigma^{i}x^{j},\quad
Y_{2}^{ij}=(\vec{x},\vec{p})(x^{i}\sigma^{j}+\sigma^{i}x^{j}),\quad
Y_{3}^{ij}=p^{i}\sigma^{j}+\sigma^{i}p^{j},\quad
Y_{4}^{ij}=(\vec{\sigma},\vec{x})x^{i}x^{j},$ $\displaystyle
Y_{5}^{ij}=(\vec{\sigma},\vec{p})x^{i}x^{j},\quad
Y_{6}^{ij}=(\vec{x},\vec{p})(\vec{\sigma},\vec{x})x^{i}x^{j},\quad
Y_{7}^{ij}=(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j}),$ $\displaystyle
Y_{8}^{ij}=x^{i}L^{j}+L^{i}x^{j},\quad
Y_{9}^{ij}={\vec{p}}^{\>2}(x^{i}\sigma^{j}+\sigma^{i}x^{j}),\quad
Y_{10}^{ij}=\vec{L}^{2}(x^{i}\sigma^{j}+\sigma^{i}x^{j}),$ $\displaystyle
Y_{11}^{ij}=(\vec{x},\vec{p})(p^{i}\sigma^{j}+\sigma^{i}p^{j}),\quad
Y_{12}^{ij}=(\vec{x},\vec{p})((\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}),$
$\displaystyle
Y_{13}^{ij}=(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j},\quad
Y_{14}^{ij}=(\vec{\sigma},\vec{x})\vec{L}^{\>2}x^{i}x^{j},\quad
Y_{15}^{ij}=(\vec{x},\vec{p})(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j}),$
$\displaystyle Y_{16}^{ij}=(\vec{\sigma},\vec{x})p^{i}p^{j},\quad
Y_{17}^{ij}=(\vec{x},\vec{p})(x^{i}L^{j}+L^{i}x^{j}),\quad
Y_{18}^{ij}=(\vec{\sigma},\vec{L})(x^{i}L^{j}+L^{i}x^{j}),$ $\displaystyle
Y_{19}^{ij}=p^{i}L^{j}+L^{i}p^{j},\quad
Y_{20}^{ij}=(\vec{\sigma},\vec{p})(x^{i}p^{j}+p^{i}x^{j}),\quad
Y_{21}^{ij}=(\vec{x},\vec{p})(L^{i}\sigma^{j}+\sigma^{i}L^{j}),$
$\displaystyle Y_{22}^{ij}=(\vec{\sigma},\vec{x})(L^{i}L^{j}+L^{j}L^{i}),\quad
Y_{23}^{ij}=(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j},\quad
Y_{24}^{ij}=(\vec{x},\vec{p})(\vec{\sigma},\vec{p})x^{i}x^{j},$ $\displaystyle
Y_{25}^{ij}=(\vec{x},\vec{p})^{2}(x^{i}\sigma^{j}+\sigma^{i}x^{j}),\quad
Y_{26}^{ij}=(\vec{x},\vec{p})^{2}(\vec{\sigma},\vec{x})x^{i}x^{j}.$ (4.1)

Each of the quantities in (4) can be multiplied by a scalar $f(r)$ without
changing its properties under rotations or reflections.

It can be shown that we have one linear relation

$Y_{20}^{ij}=Y_{9}^{ij}+Y_{16}^{ij}-Y_{11}^{ij}-Y_{13}^{ij},$ (4.2)

and even though all the above pseudo-tensors except $Y_{20}^{ij}$ are linearly
independent, higher-order polynomial relations between them exist. Indeed, we
have the following syzygies:

$\displaystyle Y_{21}^{ij}=-\vec{r}^{\>2}Y_{13}^{ij}+Y_{12}^{ij}+Y_{18}^{ij},$
$\displaystyle
Y_{22}^{ij}=-\vec{r}^{\>2}Y_{16}^{ij}+2Y_{15}^{ij}-Y_{21}^{ij},$
$\displaystyle
Y_{23}^{ij}=-\vec{r}^{\>2}Y_{3}^{ij}+Y_{2}^{ij}-Y_{5}^{ij}+Y_{7}^{ij},$
$\displaystyle
Y_{24}^{ij}=-\vec{r}^{\>2}(Y_{13}^{ij}+Y_{11}^{ij}-Y_{9}^{ij})+Y_{18}^{ij}-Y_{10}^{ij}+Y_{15}^{ij}-Y_{21}^{ij},$
$\displaystyle Y_{25}^{ij}=\vec{r}^{\>2}Y_{9}^{ij}-Y_{10}^{ij},$
$\displaystyle Y_{26}^{ij}=\vec{r}^{\>2}Y_{22}^{ij}-Y_{14}^{ij}.$ (4.3)

We use relations (4.2) and (4.3) to remove the left-hand sides of these
equations from the analysis completely.

###  4.1 The commutativity condition and determining equations

Here, we take the linear combinations of all the two-component pseudo-tensors
and then fully symmetrize them in order to obtain the determining equations
from the commutativity condition.

Let us take the linear combination of the independent pseudo-tensors given in
(4) as

$X_{Y}^{ij}=\sum\limits_{k=1}^{19}f_{k}(r)Y_{k}^{ij},$

which can be fully symmetrized as decribed in [1]. Similar to the tensors it
is sufficient to consider the full symmetric form of $X_{Y}^{12}$. From the
requirement $[H,X_{Y}^{ij}]=0$, we obtain the determining equations.

The determining equations obtained by equating the coefficients of third-order
terms to zero in the commutativity equation, are given as follows:

$\displaystyle f_{17}=0,\quad f_{19}=c_{1},\quad f_{10}=f_{12}-f_{18},\quad
f_{9}=-f_{11},\vskip 6.0pt plus 2.0pt minus 2.0pt$ (4.4) $\displaystyle
f_{13}=f_{11}-r^{2}f_{12}+c_{2},\quad f_{11}=r^{2}f_{15}+c_{3},\quad
f_{16}=-r^{2}f_{15}+\dfrac{c_{4}}{r},\vskip 6.0pt plus 2.0pt minus 2.0pt$
(4.5) $\displaystyle 2rf_{14}V_{1}+\hbar f_{14}^{\prime}=0,$ (4.6)
$\displaystyle\hbar f_{12}+2c_{2}V_{1}+\hbar rf_{12}^{\prime}=0,$ (4.7)
$\displaystyle\hbar
rf_{14}+2rf_{18}V_{1}+\hbar(f_{12}^{\prime}+r^{2}f_{14}^{\prime})=0,$ (4.8)
$\displaystyle 2rf_{12}V_{1}+\hbar f_{18}^{\prime}=0,$ (4.9) $\displaystyle
2\hbar(c_{4}-r^{3}(f_{15}+f_{18}))-4r^{2}(c_{4}(c_{2}+c_{3})r-r^{3}f_{12})V_{1}=0,$
(4.10)
$\displaystyle\hbar(f_{12}+2r^{2}f_{14}+f_{15}-f_{18})-2(c_{3}+r^{2}(-f_{12}+f_{15}+f_{18}))V_{1}=0,$
(4.11)
$\displaystyle-2c_{3}V_{1}+f_{12}(\hbar+4r^{2}V_{1})+\hbar(f_{15}-f_{18}+r(f_{12}^{\prime}+f_{15}^{\prime}+f_{18}^{\prime}))=0,$
(4.12)

where $c_{i}\,(i=1,2,3,4)$ are integration constants. Introducing the
relations given in (4.4) and (4.5) into the determining equations obtained by
equating the coefficients of the second-order terms to zero in the
commutativity equation, we get

$\displaystyle f_{2}^{\prime}+c_{1}V_{1}^{\prime}=0,$ (4.13) $\displaystyle
f_{2}-c_{1}V_{1}+rf_{2}^{\prime}=0,$ (4.14) $\displaystyle
rf_{2}+c_{1}rV_{1}+f_{3}^{\prime}=0,$ (4.15) $\displaystyle\hbar
f_{7}+2(-c_{1}\hbar+f_{3})V_{1}=0,$ (4.16) $\displaystyle 2\hbar f_{5}+2\hbar
r^{2}f_{6}+\hbar f_{7}+2c_{1}\hbar V_{1}+2r^{2}f_{2}V_{1}-2f_{3}V_{1}+\hbar
rf_{7}^{\prime}=0,$ (4.17) $\displaystyle-2\hbar
rf_{6}-2r(f_{2}+f_{7})V_{1}-\hbar f_{7}^{\prime}+c_{1}\hbar V_{1}^{\prime}=0,$
(4.18) $\displaystyle 2rf_{6}V_{1}+{\hbar}f_{6}^{\prime}=0,$ (4.19)
$\displaystyle 4\hbar
rf_{6}+2r(f_{2}+2f_{5}+f_{7})V_{1}+\hbar(-f_{2}^{\prime}+f_{7}^{\prime})=0,$
(4.20)
$\displaystyle\hbar(f_{2}+2f_{5}+f_{7})+(c_{1}\hbar-2(f_{3}+r^{2}f_{7}))V_{1}=0,$
(4.21) $\displaystyle
2r(-f_{2}+f_{7})V_{1}+2rf_{6}(3\hbar-2r^{2}V_{1})+\hbar(f_{2}^{\prime}+2f_{5}^{\prime}+f_{7}^{\prime})=0,$
(4.22) $\displaystyle 4\hbar
rf_{6}+2r(f_{2}+2f_{5}+f_{7})V_{1}+\hbar(f_{7}^{\prime}+c_{1}V_{1}^{\prime})=0,$
(4.23) $\displaystyle 2\hbar
rf_{6}-2rf_{2}V_{1}+\hbar(f_{5}^{\prime}+r^{2}f_{6}^{\prime})=0,$ (4.24)
$\displaystyle
rf_{8}+(2c_{4}+3(c_{2}+c_{3})r+r^{3}(-3f_{12}+f_{15}+f_{18}))V_{1}+c_{2}r^{2}V_{1}^{\prime}=0,$
(4.25) $\displaystyle 2\hbar
rf_{6}+2r(f_{2}+f_{7})V_{1}+\hbar(-2f_{2}^{\prime}+f_{7}^{\prime})-3c_{1}\hbar
V_{1}^{\prime}=0,$ (4.26) $\displaystyle
2r(f_{2}+2f_{5}-f_{7})V_{1}+4rf_{6}(-\hbar+r^{2}V_{1})-\hbar(2f_{2}^{\prime}+2f_{5}^{\prime}+f_{7}^{\prime})+c_{1}\hbar
V_{1}^{\prime}=0.$ (4.27)

In like manner, introducing the relations given in (4.4) and (4.5) into the
determining equations obtained by equating the coefficients of the first-order
terms to zero in the commutativity equation yields

$\displaystyle
4r(r(2f_{5}-f_{7})V_{1}+c_{1}V_{0}^{\prime})+2r(-c_{5}+2c_{6}r^{2}+2f_{3})V_{1}^{\prime}=0,$
(4.28) $\displaystyle 6c_{4}\hbar^{3}-4\hbar
r^{5}f_{4}+8\hbar^{2}r^{5}f_{14}(\hbar+2r^{2}V_{1})$
$\displaystyle\quad+r^{2}\bigg{(}2V_{1}\Big{(}-3c_{4}\hbar^{2}+r^{3}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}-2f_{1}+\hbar\big{(}2f_{8}+\hbar(f_{12}+5(f_{15}+f_{18})-rf_{15}^{\prime})\big{)}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}\Big{)}$
$\displaystyle\quad+\hbar
r(12\hbar^{2}rf_{12}^{\prime}+20\hbar^{2}r^{3}f_{14}^{\prime}+14\hbar^{2}rf_{15}^{\prime}+4c_{4}V_{0}^{\prime}+2c_{2}rV_{0}^{\prime}+2c_{3}rV_{0}^{\prime}+2c_{4}\hbar
V_{1}^{\prime}+3c_{2}\hbar rV_{1}^{\prime}$ $\displaystyle\quad-c_{3}\hbar
rV_{1}^{\prime}-2\hbar
r^{3}f_{15}V_{1}^{\prime}+2\hbar^{2}r^{2}f_{12}^{\prime\prime}+2\hbar^{2}r^{4}f_{14}^{\prime\prime}+2\hbar^{2}r^{2}f_{15}^{\prime\prime})\bigg{)}=0,$
(4.29) $\displaystyle 4c_{4}\hbar^{2}+4r^{3}f_{1}-8\hbar
r^{5}f_{14}(\hbar-r^{2}V_{1})+r^{2}\Big{(}2\big{(}\hbar(-4c_{4}+c_{2}r-c_{3}r)+r^{3}(5\hbar
f_{12}+\hbar f_{18}+2f_{8})\big{)}V_{1}$ $\displaystyle\quad+2\hbar
rf_{15}(-2\hbar+5r^{2}V_{1})+r^{2}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}4c_{2}V_{0}^{\prime}+\hbar\big{(}4c_{2}V_{1}^{\prime}+\hbar(6f_{12}^{\prime}+6f_{15}^{\prime}+6f_{18}^{\prime}$
$\displaystyle\quad+r(f_{12}^{\prime\prime}+f_{15}^{\prime\prime}+f_{18}^{\prime\prime}))\big{)}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}\Big{)}=0,$
(4.30) $\displaystyle 2(-2c_{4}\hbar+r^{3}(13\hbar f_{12}+7\hbar f_{15}-7\hbar
f_{18}+2f_{8}))V_{1}-40\hbar
r^{3}f_{14}(\hbar-r^{2}V_{1})+r\big{(}-4rf_{1}^{\prime}-6\hbar^{2}rf_{12}^{\prime}$
$\displaystyle\quad+10\hbar^{2}rf_{15}^{\prime}+18\hbar^{2}rf_{18}^{\prime}+4c_{2}rV_{0}^{\prime}+8c_{4}\hbar
V_{1}^{\prime}+4\hbar
r(c_{2}+c_{3})V_{1}^{\prime}+\hbar^{2}r^{2}(-f_{12}^{\prime\prime}+f_{15}^{\prime\prime}+3f_{18}^{\prime\prime})\big{)}=0,$
(4.31)
$\displaystyle-12c_{4}\hbar^{3}+8\hbar^{2}r^{5}f_{14}(\hbar-15r^{2}V_{1})+16r^{5}f_{4}(\hbar-r^{2}V_{1})+r^{2}(V_{1}(8c_{4}\hbar^{2}+2r^{3}(-4f_{1}+\hbar(2f_{8}$
$\displaystyle\quad+\hbar(-27f_{12}-13f_{15}+17f_{18}+2rf_{15}^{\prime}))))+\hbar
r(4rf_{1}^{\prime}+30\hbar^{2}rf_{12}^{\prime}-80\hbar^{2}r^{3}f_{14}^{\prime}-6\hbar^{2}rf_{15}^{\prime}$
$\displaystyle\quad-42\hbar^{2}rf_{18}^{\prime}-8c_{4}V_{0}^{\prime}+4c_{3}rV_{0}^{\prime}+8r^{3}f_{15}V_{0}^{\prime}-4c_{4}\hbar
V_{1}^{\prime}-6c_{2}\hbar rV_{1}^{\prime}-14c_{3}\hbar rV_{1}^{\prime}+4\hbar
r^{3}f_{15}V_{1}^{\prime}$
$\displaystyle\quad+5\hbar^{2}r^{2}f_{12}^{\prime\prime}-8\hbar^{2}r^{4}f_{14}^{\prime\prime}+15\hbar^{2}r^{2}f_{15}^{\prime\prime}-7\hbar^{2}r^{2}f_{18}^{\prime\prime}+8c_{4}\hbar
rV_{1}^{\prime\prime}+2\hbar^{2}r^{3}f_{15}^{(3)}))=0.$ (4.32)

Finally, introducing the relations given in (4.4) and (4.5) into the
determining equations obtained by equating the coefficients of the zeroth-
order terms to zero in the commutativity equation, we obtain

$\displaystyle
8r^{4}f_{4}V_{1}-\hbar(4c_{4}V_{0}^{\prime}+r^{3}(8f_{4}^{\prime}-4f_{15}V_{0}^{\prime}+rf_{4}^{\prime\prime})-2c_{4}rV_{0}^{\prime\prime})=0,$
(4.33) $\displaystyle
2r^{2}f_{1}V_{1}+4r^{2}f_{4}(-\hbar+r^{2}V_{1})-\hbar(4rf_{1}^{\prime}+(-2c_{4}-c_{2}r+c_{3}r+2r^{3}f_{15})V_{0}^{\prime}+r^{2}f_{1}^{\prime\prime})=0,$
(4.34) $\displaystyle
2\hbar(-3\hbar+4r^{2}V_{1})f_{5}^{\prime}+56\hbar^{2}r^{2}f_{6}^{\prime}+8\hbar
r^{4}V_{1}f_{6}^{\prime}-6\hbar^{2}f_{7}^{\prime}+8\hbar
r^{2}V_{1}f_{7}^{\prime}+4r^{2}V_{0}^{\prime}(f_{5}+f_{7})$
$\displaystyle\quad+4r^{3}f_{6}(12\hbar
V_{1}+rV_{0}^{\prime})+6\hbar^{2}rf_{5}^{\prime\prime}+16\hbar^{2}r^{3}f_{6}^{\prime\prime}+6\hbar^{2}rf_{7}^{\prime\prime}+\hbar^{2}r^{2}(f_{5}^{(3)}+r^{2}f_{6}^{(3)}+f_{7}^{(3)})=0,$
(4.35) $\displaystyle 2c_{5}\hbar^{2}+24\hbar
r^{4}f_{6}(\hbar-r^{2}V_{1})+r\big{(}-2\hbar^{2}f_{3}^{\prime}-2\hbar
rV_{1}\Big{(}-c_{5}+4c_{6}r^{2}+r\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}f_{3}^{\prime}+r\big{(}2f_{5}+f_{7}$
$\displaystyle\quad+2r(f_{5}^{\prime}+r^{2}f_{6}^{\prime}+f_{7}^{\prime})\big{)}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}\Big{)}+r(12\hbar^{2}rf_{5}^{\prime}+4\hbar^{2}r^{3}f_{6}^{\prime}+8\hbar^{2}rf_{7}^{\prime}-2c_{5}rV_{0}^{\prime}+4c_{6}r^{3}V_{0}^{\prime}+4rf_{3}V_{0}^{\prime}$
$\displaystyle\quad+2\hbar^{2}f_{3}^{\prime\prime}+2\hbar^{2}r^{2}f_{5}^{\prime\prime}+\hbar^{2}r^{2}f_{7}^{\prime\prime}+\hbar^{2}rf_{3}^{(3)})\big{)}=0.$
(4.36)

Similar to the situation with tensors, we classify the above determining
equations into four types:

  1. 1.

Those that are independent of the potentials $V_{1}$ and $V_{0}$, namely (4.4)
and (4.5). These relations are substituted into the other determining
equations. This eliminates some of the equations involving the potential and
we have presented only the remaining ones.

  2. 2.

Those involving only $V_{1}$ but not its derivative (4.6)-(4.12),
(4.14)-(4.17), (4.19)-(4.22), (4.24).

  3. 3.

Those involving $V_{1}$ and $V^{\prime}_{1}$ which are (4.13), (4.18), (4.23),
(4.25)-(4.27).

  4. 4.

Those involving $V_{0}$, $V_{1}$ and their derivatives (4.28)-(4.36).

###  4.2 Solutions of the determining equations

The aim of this subsection is to present the solutions of the determining
equations given in section 4.1 and then reveal the pseudo-tensor integrals of
motion. Similar to the tensors, we shall analyze the determining equations by
following the above four classes and again the determining equations of first
type had already been substituted to the others to eliminate or simplify them.
Hence, we start with the determining equations of second type. But first, one
can see that integrating (4.13) allows us to consider it as a determining
equation of second type:

$f_{2}+c_{1}V_{1}+c_{5}=0,$ (4.37)

where $c_{1}$ and $c_{5}$ are real constants. This equation will simplify the
analysis.

Adding (4.37) to (4.13) yields

$rf_{2}^{\prime}+2f_{2}+c_{5}=0,$

which can be integrated to give

$f_{2}=-\dfrac{c_{5}}{2}+\dfrac{c_{6}}{r^{2}},$ (4.38)

where $c_{6}$ is an integration constant. Introducing (4.38) into (4.14), we
obtain

$c_{1}V_{1}+\dfrac{c_{6}}{r^{2}}+\dfrac{c_{5}}{2}=0.$ (4.39)

To continue analyzing, it is better to consider the following cases depending
upon the constant $c_{1}$: Case 1. $c_{1}\neq 0$. In this case, we immediately
get a potential as

$V_{1}=-\dfrac{c_{6}}{c_{1}r^{2}}-\dfrac{c_{5}}{2c_{1}}.$ (4.40)

Introducing this equation into (4.15) and integrating, we get

$f_{3}=\dfrac{c_{5}r^{2}}{2}+c_{7},$ (4.41)

where $c_{7}$ is an integration constant. Considering (4.40) and (4.41) in
equation (4.16), we obtain

$f_{7}=\dfrac{(2c_{7}-2c_{1}\hbar+c_{5}r^{2})(2c_{6}+c_{5}r^{2})}{c_{1}\hbar
r^{2}}.$ (4.42)

On the other hand, if we introduce (4.37), (4.40)-(4.42) into the equation
(4.17), we find

$f_{6}=\dfrac{4{c_{6}}^{2}-4c_{5}c_{6}r^{2}+c_{5}r^{2}(-4c_{7}+4c_{1}\hbar-5c_{5}r^{2})-4c_{1}\hbar
r^{2}f_{5}}{4c_{1}\hbar r^{4}},$ (4.43)

which can be considered in (4.24) to deduce

$c_{5}=0.$ (4.44)

If we introduce (4.38), (4.40), (4.42), (4.43) and (4.44) into (4.20), we
obtain

$\dfrac{(c_{6}+c_{1}\hbar)(-2c_{6}c_{7}+3c_{1}c_{6}\hbar-2c_{1}\hbar
r^{2}f_{5})}{{c_{1}}^{2}\hbar r^{3}}=0.$ (4.45)

Obviously, we encounter two different cases here. However, the case
$c_{6}=-c_{1}\hbar$ gives the potential $V_{1}=\frac{\hbar}{r^{2}}$ which can
be induced by a gauge transformation and had been considered thoroughly in
[4]. So, we take

$f_{5}=\dfrac{c_{6}(-2c_{7}+3c_{1}\hbar)}{2c_{1}\hbar r^{2}},$ (4.46)

which together with (4.43) can be considered in (4.19) to give

$\dfrac{c_{6}(c_{6}+2c_{1}\hbar)(-2c_{6}-2c_{7}+3c_{1}\hbar)}{{c_{1}}^{2}\hbar
r^{5}}=0.$ (4.47)

Then we have three subcases. The first one is the case $c_{6}=0$. However, it
leads to $V_{1}=0$. Therefore, we analyze the following two subcases.

  * Subcase 1. $c_{6}=-2c_{1}\hbar$.

From (4.40) and (4.44), we immediately obtain

$V_{1}=\dfrac{2\hbar}{r^{2}}.$ (4.48)

Then, from (4.21) it is easy to show that

$c_{7}=\dfrac{3c_{1}\hbar}{2}.$ (4.49)

By considering the above spin-orbital potential, let us solve all the
determining equations of second type. Integrating (4.6) and (4.7), we find

$f_{14}=\dfrac{k_{1}}{r^{4}},\quad f_{12}=\dfrac{4c_{2}+rk_{2}}{r^{2}},$
(4.50)

where $k_{1}$ and $k_{2}$ are integration constants. Then, from (4.8) we get

$f_{18}=\dfrac{8c_{2}+3k_{1}+k_{2}r}{4r^{2}}.$ (4.51)

Introducing (4.50) and (4.51) into (4.9) yields

$k_{1}=8c_{2},\quad k_{2}=0.$ (4.52)

By substituting (4.50), (4.51) and (4.52) into the equations (4.10), (4.11)
and (4.12), one can conclude that

$f_{15}=-\dfrac{4(c_{2}+c_{3})}{3r^{2}},\quad c_{3}=2c_{2},\quad c_{4}=0.$
(4.53)

Hence, all the determining equations of second type are satisfied. Thus, we
continue the analysis with the determining equations of third type.

Substituting (4.48), (4.50), (4.51) and (4.53) into (4.25) gives us

$f_{8}=\dfrac{2c_{2}\hbar}{r^{2}},$ (4.54)

which guarantees that the determining equations of third type are also
satisfied. Introducing the above relations into the remaining determining
equations yields

$V_{0}=\dfrac{3\hbar^{2}}{r^{2}},\quad
f_{1}=-\dfrac{9c_{2}\hbar^{2}}{r^{2}},\quad
f_{4}=\dfrac{24c_{2}\hbar^{2}}{r^{4}},$ (4.55)

where a redundant additive constant in $V_{0}$ is omitted.

Hence, all determining equations are satisfied. For this case, we have two
arbitrary constants $c_{1}$ and $c_{2}$. The two integrals of motion are:

$\displaystyle\mathcal{Y}_{1}^{\,ij}=$ $\displaystyle
p^{i}L^{j}+L^{i}p^{j}-\dfrac{2\hbar}{r^{2}}\big{(}\mathrm{i}\hbar+(\vec{x},\vec{p})\big{)}(x^{i}\sigma^{j}+\sigma^{i}x^{j})+\dfrac{3\hbar}{2}(p^{i}\sigma^{j}+\sigma^{i}p^{j})$
$\displaystyle+\dfrac{4\hbar}{r^{4}}\big{(}3\mathrm{i}\hbar+2(\vec{x},\vec{p})\big{)}(\vec{\sigma},\vec{x})x^{i}x^{j}-\dfrac{2\hbar}{r^{2}}(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j}),$
(4.56) $\displaystyle\mathcal{Y}_{2}^{\,ij}=$
$\displaystyle-2\big{(}\dfrac{2\hbar^{2}}{r^{2}}-\dfrac{2\mathrm{i}\hbar}{r^{2}}(\vec{x},\vec{p})-\vec{p}^{\>2}+\dfrac{2}{r^{2}}\vec{L}^{\>2}\big{)}(x^{i}\sigma^{j}+\sigma^{i}x^{j})-\big{(}4\mathrm{i}\hbar+2(\vec{x},\vec{p})\big{)}(p^{i}\sigma^{j}+\sigma^{i}p^{j})$
$\displaystyle-\dfrac{8}{r^{2}}\big{(}\dfrac{\mathrm{i}\hbar}{r^{2}}(\vec{x},\vec{p})(\vec{\sigma},\vec{x})-\dfrac{2\hbar^{2}}{r^{2}}(\vec{\sigma},\vec{x})+\mathrm{i}\hbar(\vec{\sigma},\vec{p})-\dfrac{2}{r^{2}}(\vec{\sigma},\vec{x})\vec{L}^{\>2}\big{)}x^{i}x^{j}+8(\vec{\sigma},\vec{x})p^{i}p^{j}$
$\displaystyle+\dfrac{4}{r^{2}}(\vec{x},\vec{p})\big{(}(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}\big{)}-5\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}$
$\displaystyle-\dfrac{4}{r^{2}}(\vec{x},\vec{p})(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})+\dfrac{2}{r^{2}}\big{(}\hbar+4(\vec{\sigma},\vec{L})\big{)}(x^{i}L^{j}+L^{i}x^{j}).$
(4.57)

######  Remark 4.1.

Note that only the off-diagonal elements of the above pseudo-tensors commute
with the Hamiltonian. Observe that they are not traceless and none of the
diagonal elements are integrals of motion. However, we know that traces of
such integrals of motion appear separately as (pseudo)scalars. Indeed, if the
diagonal elements are redefined as
$\mathcal{Y}^{ii}-\frac{1}{3}\,tr(\mathcal{Y}^{ij})$, then these new elements
commute with the Hamiltonian as well. For example, the trace of
$\mathcal{Y}_{1}^{\,ij}$ is the pseudo-scalar $3(\vec{\sigma},\vec{p})$ and so
$\mathcal{Y}_{1}^{\,ii}-(\vec{\sigma},\vec{p})$ commutes with the Hamiltonian.
Note that in the rest of the paper only one pseudo-tensor integral of motion
$\mathcal{Y}_{6}^{\,ij}$ will be traceless.

* Subcase 2. $-2c_{6}-2c_{7}+3c_{1}\hbar=0$. By using the relation $c_{7}=-c_{6}+\dfrac{3c_{1}\hbar}{2}$ in (4.21), one can directly obtain

$\dfrac{{c_{6}}^{2}}{{c_{1}}^{2}}-\hbar^{2}=0,$

which gives either $c_{6}=-c_{1}\hbar$ or $c_{6}=c_{1}\hbar$. However, the
first one is excluded due to the same reason mentioned before. Then, we
continue to analyze by taking $c_{6}=c_{1}\hbar$, which in the vicinity of
(4.40) gives

$V_{1}=-\dfrac{\hbar}{r^{2}}.$ (4.58)

By following very similar steps to Subcase 1, one can obtain the relations
given as:

$f_{14}=0,\quad f_{12}=-\dfrac{2c_{2}}{r^{2}},\quad
f_{18}=\dfrac{2c_{2}}{r^{2}},\quad c_{3}=-2c_{2},\quad c_{4}=0,\quad
f_{15}=0.$ (4.59)

With these relations all the determining equations of second type are
satisfied. On the other hand, introducing the relations (4.58) and (4.59) into
(4.25) we get

$f_{8}=\dfrac{3c_{2}\hbar}{r^{2}}.$

Then, the determining equations of third type are also satisfied. Finally,
considering the above relations in the remaining determining equations give us

$V_{0}=\alpha,\quad f_{1}=-\dfrac{3c_{2}\hbar^{2}}{2r^{2}},\quad f_{4}=0,$

where $\alpha$ is a real constant. Therefore, all the determining equations
are satisfied for the spin-orbital potential $V_{1}=-\frac{\hbar}{r^{2}}$ and
the scalar one $V_{0}=\alpha$. Having left with two arbitrary constants
$c_{1}$ and $c_{2}$, there exist two integrals of motion which read:

$\displaystyle\mathcal{Y}_{3}^{\,ij}=$ $\displaystyle
p^{i}L^{j}+L^{i}p^{j}+\dfrac{\hbar}{r^{2}}\big{(}\mathrm{i}\hbar+(\vec{x},\vec{p})\big{)}(x^{i}\sigma^{j}+\sigma^{i}x^{j})+\dfrac{\hbar}{2}(p^{i}\sigma^{j}+\sigma^{i}p^{j})$
$\displaystyle-\dfrac{\hbar}{r^{2}}(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})+\dfrac{2\hbar}{r^{2}}(\vec{\sigma},\vec{p})x^{i}x^{j},\vskip
6.0pt plus 2.0pt minus 2.0pt$ (4.60) $\displaystyle\mathcal{Y}_{4}^{\,ij}=$
$\displaystyle(\dfrac{3\hbar^{2}}{2r^{2}}-\dfrac{2\mathrm{i}\hbar}{r^{2}}(\vec{x},\vec{p})+2\vec{p}^{\>2}-\dfrac{4}{r^{2}}\vec{L}^{\>2})(x^{i}\sigma^{j}+\sigma^{i}x^{j})+2(\mathrm{i}\hbar-(\vec{x},\vec{p}))(p^{i}\sigma^{j}+\sigma^{i}p^{j})$
$\displaystyle+\dfrac{2\mathrm{i}\hbar}{r^{2}}\big{(}2(\vec{\sigma},\vec{p})x^{i}x^{j}-(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})\big{)}-\dfrac{2}{r^{2}}(\vec{x},\vec{p})\big{(}(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}\big{)}$
$\displaystyle+\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}+\dfrac{1}{r^{2}}(2(\vec{\sigma},\vec{L})+3\hbar)(x^{i}L^{j}+L^{i}x^{j})\,.$
(4.61)

Case 2. $c_{1}=0$, $V_{1}$ unspecified. In this case, (4.14) together with
(4.38) yield

$c_{5}=0,\quad c_{6}=0.$ (4.62)

Introducing these into (4.15) and using (4.38) give

$f_{3}=c_{8},\quad c_{8}\in\mathbb{R}.$ (4.63)

By considering these relations in (4.24) and then integrating, we find

$f_{5}=-r^{2}f_{6}+c_{9},$ (4.64)

where $c_{9}$ is an integration constant. Introducing this equation into the
sum of (4.16) and (4.17), we obtain

$f_{7}=-c_{9}+\dfrac{c_{10}}{r^{2}},\quad c_{10}\in\mathbb{R},$ (4.65)

which can be substituted back into (4.16) to obtain

$2c_{8}V_{1}+\dfrac{\hbar c_{10}}{r^{2}}-\hbar c_{9}=0.$ (4.66)

If $c_{8}\neq 0$, we have

$V_{1}=\dfrac{\hbar(c_{9}r^{2}-c_{10})}{2c_{8}r^{2}}.$ (4.67)

Introducing (4.62), (4.64), (4.65) and (4.67) into (4.20) and (4.22) and after
making some computations we obtain

$c_{9}=0,\quad f_{6}=0,\quad c_{10}(2c_{8}+c_{10})=0\,,$

which implies the existence of two subcases, namely $c_{10}=0$ or
$c_{10}=-2c_{8}$ to consider. However, the former yields $V_{1}=0$ and the
latter gives the gauge induced potential $V_{1}=\frac{\hbar}{r^{2}}$.

Thus, we take $c_{8}=0$ which immediately yields $c_{9}=0$ and $c_{10}=0$.
Then, by (4.20) we find $f_{6}=0$. We continue the analysis by considering the
determining equation (4.7) which now reads

$\hbar f_{12}+2c_{2}V_{1}+\hbar rf_{12}^{\prime}=0.$ (4.68)

Let us examine the following subcases based upon the constant $c_{2}$.

  * Subcase 1. $c_{2}=0$.

This case directly yields

$f_{12}=\dfrac{c_{11}}{r},$ (4.69)

where $c_{11}$ is a real constant. One can show after a routine computation
that the case $c_{11}=0$ gives either $V_{1}=0$ or
$V_{1}=\frac{\hbar}{r^{2}}$. Hence, we assume $c_{11}\neq 0$. Then, from (4.9)
we obtain

$V_{1}=-\dfrac{\hbar f_{18}^{\prime}}{2c_{11}}.$ (4.70)

Introducing these into (4.12), (4.10) and (4.8), respectively, we find

$\displaystyle f_{18}=\dfrac{c_{12}-rc_{11}f_{15}}{c_{3}-rc_{11}},\quad
f_{15}=-\dfrac{c_{3}c_{11}(c_{4}+rc_{13})-r(c_{4}({c_{11}^{2}}+c_{12})+r{c_{11}^{2}}c_{13})}{r^{2}c_{11}(c_{4}+r(c_{3}-rc_{11}))},$
(4.71) $\displaystyle
f_{14}=-\dfrac{(c_{3}c_{11}-r(c_{11}^{2}+c_{12})-c_{11}c_{13})(2c_{4}c_{11}+r(c_{3}c_{11}-rc_{11}^{2}+rc_{12}+c_{11}c_{13}))}{2r^{2}c_{11}(c_{4}+r(c_{3}-rc_{11}))^{2}}+\dfrac{c_{14}}{r},$
(4.72)

where $c_{i}$ ($i=12,13,14$) are integration constants. If the relations
(4.69)-(4.72) are introduced into the remaining determining equations of
second type, the following relations are obtained:

$c_{14}=0,\quad(3c_{11}^{2}+c_{12})(c_{11}^{2}+c_{12})(c_{11}^{2}-c_{12})=0.$
(4.73)

Therefore, we have the following three possibilities: I.
$c_{12}=-3c_{11}^{2}$. This immediately gives us the potential

$V_{1}=\dfrac{3\hbar}{2r^{2}}.$

Then, we obtain

$c_{13}=0,\quad c_{3}=0,\quad c_{4}=0.$ (4.74)

Hence, the determining equations of second type are satisfied. By introducing
these relations into the remaining equations, we get

$f_{8}=0,\quad f_{1}=-\dfrac{6\hbar^{2}c_{11}}{r},\quad
f_{4}=\dfrac{15\hbar^{2}c_{11}}{r^{3}}.$

Now, all the determining equations are satisfied for any $V_{0}=V_{0}(r)$.
Having just one arbitrary constant $c_{11}$ we have the following integral of
motion:

$\displaystyle\mathcal{Y}_{5}^{\,ij}=$
$\displaystyle-r\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}+\dfrac{(\vec{x},\vec{p})}{r}\big{(}(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}\big{)}$
$\displaystyle+\dfrac{3(\vec{\sigma},\vec{L})}{r}(x^{i}L^{j}+L^{i}x^{j})-\mathrm{i}\hbar
r(p^{i}\sigma^{j}+\sigma^{i}p^{j})+\dfrac{\mathrm{i}\hbar}{r}(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})$
(4.75)
$\displaystyle+\dfrac{2}{r}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}\dfrac{(\vec{\sigma},\vec{x})}{r^{2}}\big{(}4\vec{L}^{\>2}+\hbar^{2}\big{)}-\mathrm{i}\hbar(\vec{\sigma},\vec{p})\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}x^{i}x^{j}+\dfrac{1}{r}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}-2\vec{L}^{\>2}+\mathrm{i}\hbar(\vec{x},\vec{p})-\dfrac{3\hbar^{2}}{2}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}(x^{i}\sigma^{j}+\sigma^{i}x^{j}).$

II. $c_{12}=-c_{11}^{2}$. In this case, we directly find

$V_{1}=\dfrac{\hbar}{2r^{2}}.$

Similar to the above case, introducing these into the remaining determining
equations yields

$\displaystyle c_{13}=0,\quad c_{3}=0,\quad c_{4}=0,\quad f_{8}=\dfrac{\hbar
c_{11}}{r},\quad f_{1}=0,\quad f_{4}=0.$

So, all the determining equations are satisfied for arbitrary
$V_{0}=V_{0}(r)$. We have just one arbitrary constant $c_{11}$ which yields
the following integral of motion:

$\displaystyle\mathcal{Y}_{6}^{\,ij}=$
$\displaystyle-r\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}+\dfrac{(\vec{x},\vec{p})}{r}\big{(}(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}\big{)}$
$\displaystyle+\dfrac{1}{r}\big{(}(\vec{\sigma},\vec{L})+\hbar\big{)}(x^{i}L^{j}+L^{i}x^{j})-\mathrm{i}\hbar
r(p^{i}\sigma^{j}+\sigma^{i}p^{j})+\dfrac{\mathrm{i}\hbar}{r}(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})$
$\displaystyle-\dfrac{2\mathrm{i}\hbar}{r}(\vec{\sigma},\vec{p})x^{i}x^{j}+\dfrac{\hbar}{r}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}\mathrm{i}(\vec{x},\vec{p})-\dfrac{\hbar}{2}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}(x^{i}\sigma^{j}+\sigma^{i}x^{j}).$
(4.76)

However, the above pseudo-tensor integral of motion can be written as

$\mathcal{Y}_{6}^{\,ij}=\\{X_{V}^{i},\mathcal{J}_{j}\\}+\delta_{ij}\dfrac{\hbar^{2}X_{P}}{2},$

where

$X_{P}=\dfrac{(\vec{\sigma},\vec{x})}{r}$

is a first-order pseudo-scalar integral of motion and

$\vec{X}_{V}=\\{\vec{\mathcal{J}},X_{P}\\}$

is a first-order vector integral of motion for the case
$V_{1}=\dfrac{\hbar}{2r^{2}}$ and $V_{0}=V_{0}(r)$, (For details see; [4, 1]).
Hence, we say that $\mathcal{Y}_{6}^{\,ij}$ is an obvious integral of motion.
III. $c_{12}=c_{11}^{2}$. This case yields the potential

$V_{1}=-\dfrac{\hbar}{2r^{2}}.$

By following similar steps to the above cases, we find the following
relations:

$\displaystyle c_{13}=0,\quad c_{3}=0,\quad c_{4}=0,\quad f_{8}=-\dfrac{2\hbar
c_{11}}{r},\quad f_{1}=0,\quad f_{4}=0.$

Again, all the determining equations are satisfied for arbitrary scalar
potential $V_{0}=V_{0}(r)$ and having just one arbitrary constant $c_{11}$, we
obtain the following integral of motion:

$\displaystyle\mathcal{Y}_{7}^{\,ij}=$
$\displaystyle-r\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}+\dfrac{(\vec{x},\vec{p})}{r}\big{(}(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}\big{)}$
$\displaystyle-\dfrac{1}{r}\big{(}(\vec{\sigma},\vec{L})+2\hbar\big{)}(x^{i}L^{j}+L^{i}x^{j})+\mathrm{i}\hbar\big{(}\dfrac{(\vec{\sigma},\vec{x})}{r}-r\big{)}(p^{i}\sigma^{j}+\sigma^{i}p^{j})$
$\displaystyle-\dfrac{2\mathrm{i}\hbar}{r}(\vec{\sigma},\vec{p})x^{i}x^{j}+\dfrac{1}{r}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}2\vec{L}^{\>2}+\mathrm{i}\hbar(\vec{x},\vec{p})-\dfrac{3\hbar^{2}}{2}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}(x^{i}\sigma^{j}+\sigma^{i}x^{j}).$
(4.77)

* Subcase 2. $c_{2}\neq 0$. With this choice, the equation (4.68) directly gives

$V_{1}=-\dfrac{\hbar(f_{12}+rf_{12}^{\prime})}{2c_{2}},$

and upon introducing this into (4.9), we get

$f_{18}=\dfrac{r^{2}f_{12}^{2}}{2c_{2}}+c_{15},$ (4.78)

where $c_{15}$ is an integration constant. Using (4.6), (4.8) and (4.12), we
obtain

$\displaystyle
f_{15}=\dfrac{2c_{2}(rc_{15}+c_{16})-2(c_{2}+c_{3})rf_{12}+r^{3}f_{12}^{2}}{2c_{2}r},$
(4.79) $\displaystyle
f_{14}=\dfrac{2c_{2}rc_{15}f_{12}+r^{3}f_{12}^{3}-2c_{2}(c_{2}-r^{2}c_{15})f_{12}^{\prime}+r^{4}f_{12}^{2}f_{12}^{\prime}}{2c_{2}r(c_{2}+r^{2}f_{12}+r^{3}f_{12}^{\prime})},$
(4.80)

where $c_{16}$ is a real constant. Note that the denominator of (4.80) is
assumed to be nonvanishing. Otherwise it leads to
$f_{12}=\dfrac{c_{2}}{r^{2}}+\dfrac{a_{1}}{r}$ for a constant $a_{1}$ and
gives the potential $V_{1}=\dfrac{\hbar}{2r^{2}}$. In order to satisfy the
other determining equations we need to set $c_{2}=0$ which of course
contradicts with the beginning assumption that $c_{2}\neq 0$.

By substituting (4.78) and (4.79) in the (4.10), we find

$f_{12}=\dfrac{c_{4}+(c_{2}+c_{3})r+\epsilon\sqrt{c_{4}^{2}+2c_{3}c_{4}r+r^{2}(c_{2}^{2}+c_{3}^{2}+2c_{2}(c_{3}-r(rc_{15}+c_{16}))+c_{17})}}{r^{3}},$
(4.81)

where $c_{17}$ is a real constant and $\epsilon^{2}=1$. After introducing
these relations into the remaining determining equations of second type, we
obtain the following relations:

$c_{4}=0,\quad c_{16}=0,\quad
c_{15}(c_{2}^{2}-c_{3}^{2})(2c_{2}c_{3}+c_{3}^{2}+c_{17})=0.$ (4.82)

Here, we have three cases to consider.

The case $c_{15}=0$ yields $c_{17}=-c_{2}^{2}-2c_{2}c_{3}-c_{3}^{2}$ by the
help of (4.11). This implies that $f_{12}=\dfrac{c_{2}+c_{3}}{r^{2}}$ and
$f_{14}=\dfrac{-c_{2}^{2}+c_{3}^{2}}{2c_{2}r^{4}}$. If we introduce these
relations into the other determining equations, we find $c_{3}=\pm c_{2}$.
Then, it is easy to see that the option $c_{3}=c_{2}$ gives
$V_{1}=\frac{\hbar}{r^{2}}$ while the other one $c_{3}=-c_{2}$ causes the
potential $V_{1}$ to be vanished.

The case $c_{3}^{2}-c_{2}^{2}=0$ gives $c_{15}=0$. Then, we are back in the
above case.

Finally, for the possibility $2c_{2}c_{3}+c_{3}^{2}+c_{17}=0$ we easily find
that either $c_{3}=0$ or $c_{3}=\pm 2c_{2}$. If $c_{3}=0$, then we get
$c_{15}=0$ which yields either $f_{12}=0$ or $f_{12}=\dfrac{2c_{2}}{r^{2}}$
both of which give the excluded potentials. If $c_{3}=2c_{2}$, then we obtain
$c_{15}=0$. This directly yields either $f_{12}=\dfrac{2c_{2}}{r^{2}}$ or
$f_{12}=\dfrac{4c_{2}}{r^{2}}$. However, we do not need to continue the
analysis here since the first case yields a known potential
$V_{1}=\dfrac{\hbar}{r^{2}}$ and the second one gives
$V_{1}=\dfrac{2\hbar}{r^{2}}$ which has already been investigated in Subcase 1
of Case 1. Similarly, the case $c_{3}=-2c_{2}$ yields either $V_{1}=0$ or
$V_{1}=-\dfrac{\hbar}{r^{2}}$ which has been analyzed in Subcase 2 of Case 1.

The above facts state that no new pseudo-tensor integrals of motion are found
in this case.

##  5 Conclusions

A classification of superintegrable systems preserving rotational invariance,
involving spin interaction and admitting second-order integrals of motion was
pursued in [1]. However, the results presented in [1] were restricted to
scalar, pseudo-scalar, vector and axial vector integrals of motion. This paper
serves as a continuation of [1] by considering integrals that are two index
tensors and pseudo-tensors. Hence, the classification of such superintegrable
systems is completed.

After a complete analysis, we found 6 different tensor integrals which can be
seen in (3.52), (A.1)–(A.5). However, the first one exists for all $V_{0}(r)$
and $V_{1}(r)$ whereas the others correspond to the gauge induced potential
$V_{1}=\frac{\hbar}{r^{2}}$. Among the tensor integrals for the gauge induced
potential, (A.1), (A.3) and (A.5) exist for any scalar potential $V_{0}(r)$.
However, the integrals (A.2) and (A.4) exist only for the scalar potential
$V_{0}(r)=\frac{\hbar^{2}}{r^{2}}+\alpha r^{2}$. Such a system can be viewed
as a deformation of the Harmonic oscillator ($V_{0}(r)=\alpha r^{2}$). This
naturally should bring to mind a well-known integral of motion, the quadrupole
tensor [7] (also known as the Fradkin tensor [70]) for the harmonic
oscillator. Notice that in the limit $\hbar\rightarrow 0$ the tensor integral
(A.2) gives the Fradkin tensor for the spinless case whereas (A.4) vanishes.
We conclude that no nontrivial second order tensor integrals of motion exist
for the system (2.1).

On the other hand, $12$ different pseudo-tensor integrals of motion have been
obtained, seven of which are nontrivial (see (4.56), (4.57), (4.60), (4.61),
(4.75)–(4.77)). However, it has been demonstrated that the integral given in
(4.76) can be obtained from a first-order pseudo-scalar integral and a first-
order vector integral, so it is an obvious integral of motion.

The other five integrals given in (A.6)–(A.10) correspond to the gauge induced
potential. Among these integrals, (A.6)–(A.9) do exist for
$V_{0}=\frac{\hbar^{2}}{r^{2}}$. There are no pseudo-tensor integrals of
motion for arbitrary scalar potential $V_{0}(r)$. In the case of
$V_{0}=\frac{\hbar^{2}}{r^{2}}-\frac{\alpha}{r}$, we have the pseudo-tensor
integral (A.10). This superintegrable system represents a deformation of the
Kepler–Coulomb one $V_{0}=\frac{\alpha}{r}$. It is easy to see that (A.10)
reduces to zero for the spinless case.

All the results presented in [1] and obtained in this paper are summed up in
Table 1 which represents the complete list of superintegrable potentials and
their integrals of motion for the current Hamiltonian system under
investigation. Column 2 gives the spin-orbital potentials arising from the
analysis. In column 3, the scalar potentials are given where $V_{0}(r)$ means
an arbitrary potential. The results obtained in [1] are listed in columns 4, 5
and 6. However, we leave blank some entries corresponding to the gauge induced
potential in these columns since this potential was excluded from the analysis
in [1]. As expected there may be some (pseudo)scalar or (axial)vector
integrals of motion for these cases but it is not necessary to present them
here. Finally, integrals of motion corresponding to superintegrable potentials
obtained in this paper are presented in columns 7 and 8. The real constants
$\alpha$ and $\beta$ are arbitrary and $\epsilon=\pm 1$. One can conclude from
the table that we have four new nontrivial superintegrable potentials given in
No. 3, 4, 5 and 6. The remaining one given in No. 2 has been previously found
in [1] for pseudo-scalars and vectors. We have shown that there is also an
obvious pseudo-tensor integral of motion in addition to those for this case.

A study of the algebras of the integrals of motion presented in this paper is
in progress and will be presented in a future article.

Table 1: Complete list of superintegrable potentials and their integrals of motion No | $V_{1}$ | $V_{0}$ | Pseudo-scalars | Vectors | Axial Vectors | Tensors | Pseudo-Tensors
---|---|---|---|---|---|---|---
1 | $\frac{\hbar}{r^{2}}$ | $V_{0}(r)$ |  |  | (2.6) | (A.1), (A.3), (A.5) | $-$
|  | $\frac{\hbar^{2}}{r^{2}}$ |  | (2.8) | (2.6) | (A.1), (A.3), (A.5) | (A.6)$-$(A.9)
|  | $\frac{\hbar^{2}}{r^{2}}+\alpha r^{2}$ |  |  |  | (A.1)$-$(A.5) | $-$
|  | $\frac{\hbar^{2}}{r^{2}}-\frac{\alpha}{r}$ |  |  |  | $-$ | (A.10)
2 | $\frac{\hbar}{2r^{2}}$ | $V_{0}(r)$ | (B.1), (B.2) | (B.10), (B.11) | $-$ | $-$ | (4.76)
|  | $\frac{3\hbar^{2}}{8r^{2}}-\frac{\alpha}{r}$ | (B.1), (B.2) | (B.8)$-$(B.11) | $-$ | $-$ | $-$
3 | $-\frac{\hbar}{2r^{2}}$ | $V_{0}(r)$ | $-$ | $-$ | $-$ | $-$ | (4.77)
|  | $-\frac{\hbar^{2}}{8r^{2}}+\alpha r^{2}$ | (B.3) | $-$ | $-$ | $-$ | $-$
4 | $\frac{3\hbar}{2r^{2}}$ | $V_{0}(r)$ | $-$ | $-$ | $-$ | $-$ | (4.75)
|  | $\frac{15\hbar^{2}}{8r^{2}}+\alpha r^{2}$ | (B.4) | $-$ | $-$ | $-$ | $-$
5 | $-\frac{\hbar}{r^{2}}$ | $\alpha$ | $-$ | $-$ | $-$ | $-$ | (4.60), (4.61)
|  | $\alpha r^{2}$ | $-$ | $-$ | (B.15) | $-$ | $-$
6 | $\frac{2\hbar}{r^{2}}$ | $\frac{3\hbar^{2}}{r^{2}}$ | $-$ | $-$ | $-$ | $-$ | (4.56), (4.57)
|  | $\frac{3\hbar^{2}}{r^{2}}+\alpha r^{2}$ | $-$ | $-$ | (B.16) | $-$ | $-$
7 | $\frac{\alpha}{2r^{2}}+\beta$ | $\frac{\alpha(\alpha+2\hbar)+4\beta r^{4}}{8r^{2}}$ | $-$ | (B), (B.13) | $-$ | $-$ | $-$
8 | $\frac{\hbar}{2r^{2}}\left(1+\frac{\epsilon}{\sqrt{1+\beta r^{2}}}\right)$ | $\frac{\hbar^{2}}{2r^{2}}\left(1+\frac{\epsilon}{\sqrt{1+\beta r^{2}}}\right)$ | (B.5), (B.6) | (B.14) | $-$ | $-$ | $-$
9 | $\frac{\hbar}{2r^{2}}\left(1+\frac{2\epsilon}{\sqrt{1+\beta r^{2}}}\right)$ | (5.21) | (B.7) | $-$ | $-$ | $-$ | $-$
10 | $\frac{\epsilon\hbar}{r^{2}\sqrt{1+\beta r^{2}}}$ | (5.22) | $-$ | $-$ | (B.17) | $-$ | $-$
11 | $\frac{\hbar}{r^{2}}\left(1+\frac{\epsilon}{\sqrt{1+\beta r^{2}}}\right)$ | (5.23) | $-$ | $-$ | (B.18) | $-$ | $-$

The scalars potentials that do not fit into the table are given below:

$V_{0}=\frac{\hbar^{2}}{8r^{2}\left(1+\beta
r^{2}\right)^{2}}\left(7+10r^{2}\beta+8\epsilon\left(1+\beta
r^{2}\right)^{3/2}\right)-\frac{\alpha}{4\beta(1+\beta\,r^{2})}\,,$ (5.21)
$\displaystyle V_{0}=\frac{\hbar^{2}}{8r^{2}\left(1+\beta
r^{2}\right)^{2}}\left(4+6\beta r^{2}-r^{4}\beta^{2}+4\epsilon\,\left(1+\beta
r^{2}\right)^{\frac{3}{2}}\right)+\frac{\alpha}{1+\beta r^{2}}\,,$ (5.22)
$\displaystyle V_{0}=\frac{3\hbar^{2}\Big{(}4+5\beta
r^{2}+4\epsilon\left(1+\beta
r^{2}\right)^{\frac{3}{2}}\Big{)}}{8r^{2}\left(1+\beta
r^{2}\right)^{2}}-\frac{\alpha}{2\beta\left(1+\beta r^{2}\right)}\,.$ (5.23)

##  6 Acknowledgments

İY thanks the Centre de Recherches Mathématiques, Université de Montréal,
where part of this work was done, for the kind hospitality.

This research program, systematic investigation of superintegrability with
spin, was initiated by Pavel Winternitz in 2005. Throughout the program he was
the coauthor of the most of the articles. This last work started in 2018 when
İY was visiting the Centre de Recherches Mathématiques, Université de Montréal
on a sabbatical leave. Pavel Winternitz was involved in almost every stage of
the article and was eager to see its completion. However, he passed away on
February 13, 2021. We dedicate this paper to his memory.

##  Appendix A Integrals of Motion for the Gauge Induced Potential

In the previous sections, it has been pointed out that we excluded the special
case $V_{1}=\frac{\hbar}{r^{2}}$ since it is a gauge induced potential.
Nevertheless, we present the (pseudo)tensor integrals of motion appeared in
such cases.

For the gauge induced potential $V_{1}=\frac{\hbar}{r^{2}}$, we have the
following five tensor integrals of motion:

$\displaystyle\allowdisplaybreaks\mathcal{T}_{2}^{\,ij}=$
$\displaystyle(L^{i}L^{j}+L^{j}L^{i})+2\hbar(L^{i}\sigma^{j}+\sigma^{i}L^{j})+\dfrac{2\hbar}{r^{2}}\big{(}2(\vec{\sigma},\vec{L})+\hbar\big{)}x^{i}x^{j}$
$\displaystyle-2\hbar\big{(}p^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}p^{j}\big{)}+\dfrac{2\hbar}{r^{2}}\big{(}(\vec{x},\vec{p})+\mathrm{i}\hbar\big{)}\big{(}x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}\big{)},$
(A.1) $\displaystyle\mathcal{T}_{3}^{\,ij}=$ $\displaystyle
2(p^{i}p^{j}+2\alpha\,x^{i}x^{j})-\dfrac{2\hbar}{r^{2}}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}\dfrac{\hbar}{r^{2}}x^{i}x^{j}+\big{(}p^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}p^{j}$
$\displaystyle+\dfrac{\mathrm{i}\hbar}{r^{2}}(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j})\big{)}\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}},$
(A.2) $\displaystyle\mathcal{T}_{4}^{\,ij}=$
$\displaystyle\mathrm{i}\hbar(L^{i}\sigma^{j}+\sigma^{i}L^{j})-\big{(}2(\vec{\sigma},\vec{L})+\hbar\big{)}(x^{i}p^{j}+p^{i}x^{j})-(\vec{x},\vec{p})(x^{i}(\vec{p}\wedge\vec{\sigma})^{j}+(\vec{p}\wedge\vec{\sigma})^{i}x^{j})$
$\displaystyle+\dfrac{2}{r^{2}}\big{(}\mathrm{i}\hbar^{2}+\hbar(\vec{x},\vec{p})+2\big{(}(\vec{x},\vec{p})+\mathrm{i}\hbar\big{)}(\vec{\sigma},\vec{L})\big{)}x^{i}x^{j}+r^{2}(p^{i}(\vec{p}\wedge\vec{\sigma})^{j}+(\vec{p}\wedge\vec{\sigma})^{i}p^{j})$
$\displaystyle-\big{(}(\vec{x},\vec{p})+2\,\mathrm{i}\hbar\big{)}(p^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}p^{j})$
$\displaystyle+\big{(}-\dfrac{2}{r^{2}}\vec{L}^{\>2}+\vec{p}^{\>2}-\dfrac{3\hbar^{2}}{2r^{2}}\big{)}(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}),$
(A.3) $\displaystyle\mathcal{T}_{5}^{\,ij}=$
$\displaystyle\dfrac{(3\hbar+2(\vec{\sigma},\vec{L}))}{r^{2}}\scalerel*[5pt]{\big{(}}{\ensurestackMath{\addstackgap[1.5pt]{\big{(}}}}\dfrac{2\,\mathrm{i}\hbar}{r^{2}}x^{i}x^{j}+(x^{i}p^{j}+p^{i}x^{j})\scalerel*[5pt]{\big{)}}{\ensurestackMath{\addstackgap[1.5pt]{\big{)}}}}-(p^{i}(\vec{p}\wedge\vec{\sigma})^{j}+(\vec{p}\wedge\vec{\sigma})^{i}p^{j})$
$\displaystyle+\dfrac{2}{r^{2}}\big{(}(\vec{x},\vec{p})+\mathrm{i}\hbar\big{)}(p^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}p^{j})$
$\displaystyle+\big{(}2\alpha-\dfrac{\hbar}{r^{4}}(3\hbar-2\,\mathrm{i}(\vec{x},\vec{p}))\big{)}(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}),$
(A.4) $\displaystyle\mathcal{T}_{6}^{\,ij}=$
$\displaystyle-(L^{i}\sigma^{j}+\sigma^{i}L^{j})-\dfrac{2}{r^{2}}\big{(}2(\vec{\sigma},\vec{L})+\hbar\big{)}x^{i}x^{j}+2(p^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}p^{j})$
$\displaystyle-\dfrac{2}{r^{2}}\big{(}(\vec{x},\vec{p})+\mathrm{i}\hbar\big{)}(x^{i}(\vec{x}\wedge\vec{\sigma})^{j}+(\vec{x}\wedge\vec{\sigma})^{i}x^{j}).$
(A.5)

We observe two scalar potentials for the gauge induced potential which are
$V_{0}=\frac{\hbar^{2}}{r^{2}}$ and $V_{0}=\frac{\hbar^{2}}{r^{2}}+\alpha
r^{2}$. We see that the integrals (A.1), (A.3) and (A.5) exist for any
$V_{0}(r)$ or $V_{0}=\frac{\hbar^{2}}{r^{2}}$. On the other hand, (A.2) and
(A.4) exist only for $V_{0}=\frac{\hbar^{2}}{r^{2}}+\alpha r^{2}$.

For the gauge induced potential $V_{1}=\frac{\hbar}{r^{2}}$, we have the
following pseudo-tensor integrals of motion:

$\displaystyle\mathcal{Y}_{8}^{\,ij}=$
$\displaystyle(p^{i}L^{j}+L^{i}p^{j})-\dfrac{\hbar}{r^{2}}\big{(}(\vec{x},\vec{p})+\mathrm{i}\hbar\big{)}(x^{i}\sigma^{j}+\sigma^{i}x^{j})+\dfrac{\hbar}{r^{2}}(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})$
$\displaystyle+\dfrac{2\hbar}{r^{2}}\big{(}(\vec{\sigma},\vec{p})+\dfrac{2\mathrm{i}\hbar}{r^{2}}(\vec{\sigma},\vec{x})\big{)}x^{i}x^{j}+\dfrac{\hbar}{2}(p^{i}\sigma^{j}+\sigma^{i}p^{j}),$
(A.6) $\displaystyle\mathcal{Y}_{9}^{\,ij}=$
$\displaystyle\dfrac{2\mathrm{i}\hbar}{r^{2}}\big{(}(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})-2(\vec{\sigma},\vec{p})x^{i}x^{j}-r^{2}(p^{i}\sigma^{j}+\sigma^{i}p^{j})\big{)}$
$\displaystyle-\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}+\dfrac{2}{r^{2}}(\vec{x},\vec{p})\big{(}(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}\big{)}$
$\displaystyle+\dfrac{\hbar}{r^{2}}\big{(}2\mathrm{i}(\vec{x},\vec{p})-\dfrac{\hbar}{2}\big{)}(x^{i}\sigma^{j}+\sigma^{i}x^{j})+\dfrac{1}{r^{2}}\big{(}2(\vec{\sigma},\vec{L})+3\hbar\big{)}(x^{i}L^{j}+L^{i}x^{j}),$
(A.7) $\displaystyle\mathcal{Y}_{10}^{\,ij}=$
$\displaystyle-\dfrac{\hbar}{r^{2}}(x^{i}L^{j}+L^{i}x^{j})-\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}-(\vec{x},\vec{p})(p^{i}\sigma^{j}+\sigma^{i}p^{j})$
$\displaystyle-\dfrac{2}{r^{2}}(\vec{\sigma},\vec{x})\big{(}(\vec{x},\vec{p})+\mathrm{i}\hbar\big{)}(x^{i}p^{j}+p^{i}x^{j})+\vec{p}^{\>2}(x^{i}\sigma^{j}+\sigma^{i}x^{j})$
$\displaystyle+4(\vec{\sigma},\vec{x})p^{i}p^{j}+\dfrac{4\hbar}{r^{4}}(\vec{\sigma},\vec{x})\big{(}\hbar-\mathrm{i}(\vec{x},\vec{p})\big{)}x^{i}x^{j},$
(A.8) $\displaystyle\mathcal{Y}_{11}^{\,ij}=$
$\displaystyle(p^{i}\sigma^{j}+\sigma^{i}p^{j})-\dfrac{2}{r^{2}}(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})-\dfrac{4\mathrm{i}\hbar}{r^{4}}(\vec{\sigma},\vec{x})x^{i}x^{j},$
(A.9) $\displaystyle\mathcal{Y}_{12}^{\,ij}=$
$\displaystyle\dfrac{2}{r^{2}}\bigg{(}-2\mathrm{i}\hbar\big{(}(\vec{\sigma},\vec{p})+\dfrac{1}{r^{2}}(\vec{x},\vec{p})(\vec{\sigma},\vec{x})\big{)}+\big{(}\dfrac{\alpha}{r}+\dfrac{\hbar^{2}}{r^{2}}\big{)}(\vec{\sigma},\vec{x})\bigg{)}x^{i}x^{j}$
$\displaystyle-\big{(}2\mathrm{i}\hbar+(\vec{x},\vec{p})\big{)}(p^{i}\sigma^{j}+\sigma^{i}p^{j})-\dfrac{2}{r^{2}}(\vec{x},\vec{p})(\vec{\sigma},\vec{x})(x^{i}p^{j}+p^{i}x^{j})+4(\vec{\sigma},\vec{x})p^{i}p^{j}$
$\displaystyle-2\big{(}(\vec{p}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{p}\wedge\vec{\sigma})^{j}\big{)}+\dfrac{2}{r^{2}}(\vec{x},\vec{p})\big{(}(\vec{x}\wedge\vec{\sigma})^{i}L^{j}+L^{i}(\vec{x}\wedge\vec{\sigma})^{j}\big{)}$
$\displaystyle+\big{(}\dfrac{2\mathrm{i}\hbar}{r^{2}}(\vec{x},\vec{p})+\vec{p}^{\>2}-(\dfrac{\alpha}{r}+\dfrac{3\hbar^{2}}{r^{2}})\big{)}(x^{i}\sigma^{j}+\sigma^{i}x^{j})+\dfrac{2}{r^{2}}\big{(}(\vec{\sigma},\vec{L})+\hbar\big{)}(x^{i}L^{j}+L^{i}x^{j})\,.$
(A.10)

The last pseudo-tensor integral of motion can also be represented as

$\displaystyle\mathcal{Y}_{12}^{\,ij}=$
$\displaystyle\mathcal{Y}_{9}^{\,ij}+\mathcal{Y}_{10}^{\,ij}+\dfrac{\alpha}{r}\big{(}\dfrac{2}{r^{2}}(\vec{\sigma},\vec{x})x^{i}x^{j}-(x^{i}\sigma^{j}+\sigma^{i}x^{j})\big{)}.$
(A.11)

There are two scalar potentials for this case that are
$V_{0}=\frac{\hbar^{2}}{r^{2}}$ and
$V_{0}=\frac{\hbar^{2}}{r^{2}}-\frac{\alpha}{r}$. The integrals (A.6)–(A.9)
exist for $V_{0}=\frac{\hbar^{2}}{r^{2}}$ whereas the other one (A.10) exists
for the potential $V_{0}=\frac{\hbar^{2}}{r^{2}}-\frac{\alpha}{r}$.

##  Appendix B Pseudo-scalar, Vector and Axial Vector Integrals of Motion

In this appendix, we remind the pseudo-scalar, vector and axial vector
integrals of motion presented in [1]. No nontrivial scalar integrals of motion
exist.

Pseudo-scalars:

$\displaystyle X_{P}^{1}=\frac{(\vec{\sigma},\vec{x})}{r}\,,$ (B.1)
$\displaystyle
X_{P}^{2}=-r(\vec{\sigma},\vec{p})+\frac{1}{r}\,(\vec{\sigma},\vec{x})\,(\vec{x},\vec{p})-\frac{i\hbar}{r}\,(\vec{\sigma},\vec{x})\,,$
(B.2) $\displaystyle
X_{P}^{3}=\frac{(\vec{\sigma},\vec{x})}{r}\,\left(\frac{3\hbar^{2}}{2\,r^{2}}+4\alpha\,r^{2}-2(\vec{p},\vec{p})\right)+\frac{4}{r}\,\big{(}(\vec{x},\vec{p})-i\hbar\big{)}\,(\vec{\sigma},\vec{p})\,,$
(B.3) $\displaystyle
X_{P}^{4}=\frac{(\vec{\sigma},\vec{x})}{r}\,\left(-\frac{5\hbar^{2}}{2\,r^{2}}+4\,\alpha\,r^{2}-\frac{20i\hbar}{r^{2}}\,(\vec{x},\vec{p})-2(\vec{p},\vec{p})+\frac{8}{r^{2}}\big{(}\vec{x},(\vec{x},\vec{p})\,\vec{p}\big{)}\right)-\frac{4}{r}\,\big{(}(\vec{x},\vec{p})-2i\hbar\big{)}\,(\vec{\sigma},\vec{p})\,,$
(B.4) $\displaystyle X_{P}^{5}=-\frac{1}{\beta}\sqrt{1+\beta
r^{2}}\,(\vec{\sigma},\vec{p})+\frac{(\vec{\sigma},\vec{x})}{-\epsilon+\sqrt{1+\beta
r^{2}}}\,\,\Big{(}(\vec{x},\vec{p})-i\hbar\Big{)}\,,$ (B.5) $\displaystyle
X_{P}^{6}=\frac{2}{r^{2}}\big{(}1+\epsilon\,\sqrt{1+\beta\,r^{2}}\big{)}\,(\vec{\sigma},\vec{x})\,\Big{(}\hbar^{2}+3i\hbar\,(\vec{x},\vec{p})-\big{(}\vec{x},(\vec{x},\vec{p})\,\vec{p}\big{)}\Big{)}$
$\displaystyle\qquad\quad-2i\hbar\Big{(}2+\epsilon\,\sqrt{1+\beta\,r^{2}}\Big{)}\,(\vec{\sigma},\vec{p})+2\,(\vec{x},\vec{p})\,(\vec{\sigma},\vec{p})+2\epsilon\,\sqrt{1+\beta\,r^{2}}\,(\vec{\sigma},\vec{x})\,(\vec{p},\vec{p})\,,$
(B.6) $\displaystyle
X_{P}^{7}=\frac{2\,r^{4}\,\alpha\,(1+\beta\,r^{2})-\hbar^{2}(1+2\beta\,r^{2})\,\Big{(}1+4\epsilon\sqrt{1+\beta\,r^{2}}+4r^{2}\beta\,\big{(}1+\epsilon\,\sqrt{1+\beta\,r^{2}}\big{)}\Big{)}}{2\,r^{3}\,(1+\beta\,r^{2})^{2}}\,(\vec{\sigma},\vec{x})$
$\displaystyle\qquad\quad-\frac{2i\hbar}{r^{3}}\left(5+3\beta\,r^{2}-\frac{\epsilon}{\sqrt{1+\beta\,r^{2}}}+6\,\epsilon\,\sqrt{1+\beta\,r^{2}}\right)\,(\vec{\sigma},\vec{x})\,(\vec{x},\vec{p})$
$\displaystyle\qquad\quad+\frac{2i\hbar}{r}\left(1+\beta\,r^{2}-\frac{\epsilon}{\sqrt{1+\beta\,r^{2}}}+4\,\epsilon\,\sqrt{1+\beta\,r^{2}}\right)\,(\vec{\sigma},\vec{p})$
$\displaystyle\qquad\quad-\frac{2}{r}\,(1+\beta\,r^{2})\,(\vec{\sigma},\vec{x})\,(\vec{p},\vec{p})+\frac{2}{r^{3}}\,\Big{(}2+\beta\,r^{2}+2\,\epsilon\,\sqrt{1+\beta\,r^{2}}\Big{)}\,(\vec{\sigma},\vec{x})\,\big{(}\vec{x},(\vec{x},\vec{p})\vec{p}\big{)}$
$\displaystyle\qquad\quad-\frac{4\epsilon}{r}\,\sqrt{1+\beta\,r^{2}}\,(\vec{x},\vec{p})\,(\vec{\sigma},\vec{p})\,.$
(B.7)

Vectors:

$\displaystyle\vec{X}_{V}^{1}=-\big{(}\vec{\sigma},\vec{L}\big{)}\,\vec{p}+\frac{3\hbar}{2}\vec{p}-\frac{\hbar\vec{x}}{2r^{2}}(\vec{x},\vec{p})+\frac{i\hbar}{2}(\vec{\sigma}\wedge\vec{p})+i\hbar^{2}\frac{\vec{x}}{2r^{2}}+\frac{4\alpha
r-\hbar^{2}}{4r^{2}}(\vec{x}\wedge\vec{\sigma})\,,$ (B.8)
$\displaystyle\vec{X}_{V}^{2}=2\vec{x}\vec{p}^{\,\,2}-2(\vec{x},\vec{p})\,\vec{p}-\frac{\hbar\vec{x}}{r^{2}}\big{(}\vec{\sigma},\vec{L}\big{)}+2i\hbar\vec{p}-\hbar(\vec{\sigma}\wedge\vec{p})+i\hbar^{2}\frac{(\vec{x}\wedge\vec{\sigma})}{2r^{2}}+\frac{\vec{x}}{r^{2}}(\hbar^{2}-2\alpha
r)\,,$ (B.9)
$\displaystyle\vec{X}_{V}^{3}=\frac{(\vec{x},\vec{\sigma})}{r}\vec{L}+\frac{\hbar}{2r}\big{(}\vec{x}-i(\vec{x}\wedge\vec{\sigma})\big{)}\,,$
(B.10)
$\displaystyle\vec{X}_{V}^{4}=r\vec{L}(\vec{\sigma},\vec{p})+\frac{\hbar
r}{2}\vec{p}-\frac{i\hbar
r}{2}(\vec{\sigma}\wedge\vec{p})+\vec{X}_{V}^{3}\,\Big{(}i\hbar-(\vec{x},\vec{p})\Big{)}\,,$
(B.11)
$\displaystyle\vec{X}_{V}^{5}=\vec{x}\,\Big{(}2\vec{p}^{\,2}-\left(\frac{2\beta
r^{2}-\alpha}{r^{2}}\right)\Big{(}\big{(}\vec{\sigma},\vec{L}\big{)}+\hbar\Big{)}\Big{)}+2\big{(}i\hbar-(\vec{x},\vec{p})\big{)}\,\vec{p}-\hbar(\vec{\sigma}\wedge\vec{p})$
$\displaystyle\qquad\,\,+i\hbar\left(\frac{2\beta
r^{2}-\alpha}{2r^{2}}\right)\,(\vec{\sigma}\wedge\vec{x})\,,$ (B.12)
$\displaystyle\vec{X}_{V}^{6}=\Big{(}\frac{1}{2}(2\hbar+\alpha-2\beta
r^{2})+\big{(}\vec{\sigma},\vec{L}\big{)}\Big{)}\,\vec{p}+\frac{\vec{x}}{2}\left(\frac{2\beta
r^{2}-\alpha}{r^{2}}\right)\Big{(}(\vec{x},\vec{p})-i\hbar\Big{)}+\frac{i\hbar}{2}(\vec{\sigma}\wedge\vec{p})$
$\displaystyle\qquad\quad-\frac{\hbar}{4}\left(\frac{2\beta
r^{2}-\alpha}{r^{2}}\right)(\vec{\sigma}\wedge\vec{x})\,,$ (B.13)
$\displaystyle\vec{X}_{V}^{7}=\\{X_{P}^{1},\vec{\mathcal{J}}\\}\,.$ (B.14)

Axial vectors:

$\displaystyle\vec{X}_{A}^{1}=-\big{(}2\alpha
r^{2}+\vec{p}^{\,\,2}\big{)}\vec{\sigma}+2(\vec{\sigma},\vec{p})\vec{p}$
$\displaystyle\qquad\quad+\frac{2}{r^{2}}\Bigg{(}\vec{x}\Big{(}(\vec{\sigma},\vec{x})\,\vec{p}^{\,\,2}+2i\hbar(\vec{\sigma},\vec{p})-2(\vec{x},\vec{p})(\vec{\sigma},\vec{p})\Big{)}+i\hbar\big{(}\vec{\sigma}(\vec{x},\vec{p})-(\vec{\sigma},\vec{x})\vec{p}\,\big{)}\Bigg{)}\,,$
(B.15) $\displaystyle\vec{X}_{A}^{2}=\big{(}3\vec{p}^{\,\,2}-2\alpha
r^{2}+\frac{4}{r^{2}}\left(i\hbar(\vec{x},\vec{p})-(\vec{x},(\vec{x},\vec{p})\vec{p})\right)\big{)}\vec{\sigma}-\frac{2}{r^{2}}\big{(}\hbar+(\vec{L},\vec{\sigma})\big{)}\vec{L}-2\left((\vec{\sigma},\vec{p})-\frac{3i\hbar(\vec{\sigma},\vec{x})}{r^{2}}\right)\vec{p}$
$\displaystyle\qquad\quad+\frac{2\vec{x}}{r^{4}}\bigg{(}3i\hbar
r^{2}(\vec{\sigma},\vec{p})-(\vec{\sigma},\vec{x})\big{(}3h^{2}-2r^{4}\alpha+2r^{2}\vec{p}^{\,2}+12i\hbar(\vec{x},\vec{p})\big{)}-4(\vec{x},(\vec{x},\vec{p})\vec{p})\bigg{)}\,,$
(B.16)
$\displaystyle\vec{X}_{A}^{3}=\left(2i\hbar\frac{(\vec{\sigma},\vec{x})}{r^{2}}Q_{+}-4\epsilon\sqrt{1+\beta
r^{2}}\,(\vec{\sigma},\vec{p})\right)\vec{p}-\frac{4}{r^{2}}\left(\hbar+2\alpha
r^{2}+\frac{\epsilon\hbar}{\sqrt{1+\beta
r^{2}}}+q(\vec{\sigma},\vec{L})\right)\,\vec{L}$
$\displaystyle\qquad\quad+\frac{\vec{\sigma}}{2r^{2}}\Bigg{(}Y+8i\hbar\left(1+\frac{\beta}{2}r^{2}\right)(\vec{x},\vec{p})-8q(\vec{x},(\vec{x},\vec{p})\vec{p})+4r^{2}\left(2q-1\right)\vec{p}^{\,\,2}\Bigg{)}$
$\displaystyle\qquad\quad+\frac{2\vec{x}}{r^{2}}\Bigg{(}\bigg{(}i\hbar
Q_{-}+4\epsilon\sqrt{1+\beta
r^{2}}(\vec{x},\vec{p})\bigg{)}(\vec{\sigma},\vec{p})-2\epsilon(\vec{\sigma},\vec{x})\sqrt{1+\beta
r^{2}}\vec{p}^{\,\,2}\Bigg{)}\,,$ (B.17)
$\displaystyle\vec{X}_{A}^{4}=\Bigg{(}\frac{2i\hbar^{2}(\vec{\sigma},\vec{x})}{r^{2}}\widetilde{Q}-4\epsilon\,\sqrt{1+\beta
r^{2}}\,(\vec{\sigma},\vec{p})\Bigg{)}\vec{p}-\frac{4}{r^{2}}\Bigg{(}\hbar+\frac{\epsilon\hbar}{\sqrt{1+\beta
r^{2}}}+q(\vec{\sigma},\vec{L})\Bigg{)}\vec{L}$
$\displaystyle\qquad\quad+\frac{2\vec{\sigma}}{r^{2}}\Bigg{(}2i\hbar
q(\vec{x},\vec{p})-\widetilde{Y}-2q(\vec{x},(\vec{x},\vec{p})\vec{p})+r^{2}\left(2q-1\right)\vec{p}^{\,\,2}\Bigg{)}+\frac{2\vec{x}}{r^{4}}\Bigg{(}(\vec{\sigma},\vec{x})\Big{(}4q(\vec{x},(\vec{x},\vec{p})\vec{p})$
$\displaystyle\qquad\quad-2r^{2}\left(2q-1-\epsilon\,\sqrt{1+\beta
r^{2}}\right)\vec{p}^{\,\,2}-Z-4i\hbar
W(\vec{x},\vec{p})\Big{)}+i\hbar\widetilde{Q}(\vec{\sigma},\vec{p})\Bigg{)}\,,$
(B.18)

where $Q_{\pm}$, $q$, $Y$, $\widetilde{Q}$, $\widetilde{Y}$, $Z$ and $W$ are
given by the following relations

$\displaystyle Q_{\pm}=1+\frac{\beta}{2}r^{2}\pm\epsilon\frac{3+4\beta
r^{2}}{\sqrt{1+\beta r^{2}}}\,,\qquad
q=1+\frac{\beta}{2}r^{2}+\epsilon\,\sqrt{1+\beta r^{2}}\,,$ $\displaystyle
Y=\frac{-4\hbar^{2}-8r^{2}\alpha-6\hbar^{2}r^{2}\beta-8r^{4}\alpha\beta+\hbar^{2}r^{4}\beta^{2}}{(1+\beta
r^{2})^{2}}-\frac{4\epsilon\hbar^{2}}{\sqrt{1+\beta r^{2}}}\,,$
$\displaystyle\widetilde{Q}=3+\frac{5\beta}{2}r^{2}+\epsilon\frac{3+4\beta
r^{2}}{\sqrt{1+\beta r^{2}}}\,,\qquad W=3+2\beta
r^{2}+\epsilon\,\frac{6+7\beta r^{2}}{2\sqrt{1+\beta r^{2}}}\,,$
$\displaystyle\widetilde{Y}=\frac{\hbar^{2}r^{4}(4\alpha-6\beta^{2})+\hbar^{2}r^{6}(3\beta^{3}+4\beta\alpha)-6\hbar^{2}\beta
r^{2}}{4(1+\beta r^{2})^{2}}+\frac{2\epsilon\hbar^{2}\beta
r^{2}}{\sqrt{1+\beta r^{2}}}\,,$ $\displaystyle Z=\frac{-4r^{4}\alpha(1+\beta
r^{2})+3\hbar^{2}\big{(}2+\beta r^{2}(7+6\beta r^{2})\big{)}}{2(1+\beta
r^{2})^{2}}+\epsilon\frac{3\hbar^{2}(1+2\beta r^{2})}{\sqrt{1+\beta
r^{2}}}\,.$

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