Classical Superintegrable Systems in a Magnetic Field

Classical Superintegrable Systems in a Magnetic Field
that Separate in Cartesian Coordinates

Antonella MARCHESIELLO † and Libor ŠNOBL ‡

A. Marchesiello and L. Šnobl

† Czech Technical University in Prague, Faculty of Information Technology,
† Department of Applied Mathematics, Thákurova 9, 160 00 Prague 6, Czech
Republic<EMAIL_ADDRESS>
‡ Czech Technical University in Prague, Faculty of Nuclear Sciences and
Physical Engineering,
‡ Department of Physics, Břehová 7, 115 19 Prague 1, Czech Republic
<EMAIL_ADDRESS>
Received November 05, 2019, in final form March 06, 2020; Published online
March 12, 2020

We consider superintegrability in classical mechanics in the presence of
magnetic fields. We focus on three-dimensional systems which are separable in
Cartesian coordinates. We construct all possible minimally and maximally
superintegrable systems in this class with additional integrals quadratic in
the momenta. Together with the results of our previous paper [J. Phys. A:
Math. Theor. 50 (2017), 245202, 24 pages], where one of the additional
integrals was by assumption linear, we conclude the classification of three-
dimensional quadratically minimally and maximally superintegrable systems
separable in Cartesian coordinates. We also describe two particular methods
for constructing superintegrable systems with higher-order integrals.

integrability; superintegrability; higher-order integrals; magnetic field

37J35; 78A25

##  1 Introduction

In this paper we investigate superintegrability of three-dimensional systems
that separate in Cartesian coordinates in the presence of a magnetic field. We
say that a mechanical system is superintegrable if it is Liouville integrable
and possesses additional independent integrals of motion. Depending on their
number we distinguish minimal superintegrability when only one additional
integral is present, and maximal superintegrability when the number of
additional integrals is the maximal possible, i.e., equal to the number of
degrees of freedom minus one. (In three spatial dimensions there is no other
possibility.)

The study of superintegrability with magnetic fields was initiated in [5] and
subsequently followed in both two spatial dimensions [3, 4, 24, 25] and three
spatial dimensions [2, 14, 15, 16, 17]; relativistic version of the problem
was recently considered too, cf. [11]. Separability of three-dimensional
systems with magnetic fields was considered in the papers [1, 27]. Particular
planar two-body systems, e.g., Coulomb, in perpendicular constant magnetic
field were also studied from the point of solvability and superintegrability,
see, e.g., [8, 28, 29, 30].

It turns out that the presence of magnetic field significantly increases the
complexity of both calculations and structure of these systems. E.g., contrary
to the case without magnetic field separability in orthogonal coordinates and
integrability with integrals at most quadratic in the momenta are no longer
equivalent, namely separability is stronger and implies the existence of at
least one integral linear in the momenta. Similarly, the explicit construction
of superintegrable systems and their classification become much harder when
magnetic fields are present.

In the present paper we attempt to approach the problem from a different
viewpoint. We exploit the fact that in certain situations the three-
dimensional system can be rewritten as effectively a two-dimensional one
without magnetic field, thus generalizing the principal idea of [15]. In other
cases we show that the existence of a quadratic integral necessarily implies
the existence of an integral in a particular simpler form, which makes our
calculations tractable. When the results of the present paper and [14, 16] are
viewed together, they provide an exhaustive list of three-dimensional
quadratically minimally and maximally superintegrable systems with magnetic
fields separable in Cartesian coordinates.

We shall investigate the superintegrability of the system defined on the phase
space $\mathbb{R}^{6}$, with the canonical coordinates $(\vec{x},\vec{p})$, by

$\displaystyle
H(\vec{x},\vec{p})=\frac{1}{2}\big{(}\big{(}p_{1}^{A}\big{)}^{2}+\big{(}p_{2}^{A}\big{)}^{2}+\big{(}p_{3}^{A}\big{)}^{2}\big{)}+W(\vec{x}),$
(1.1)

where $W(\vec{x})$ denotes the so called electrostatic or effective potential,
$p_{j}^{A}$ are the covariant expressions for the momenta

$\displaystyle p_{j}^{A}=p_{j}+A_{j}(\vec{x}),\qquad j=1,2,3,$ (1.2)

and $A_{j}(\vec{x})$ are the components of the vector potential. The magnetic
field $\vec{B}(\vec{x})$ is related to $\vec{A}(\vec{x})$ through

$\displaystyle\vec{B}(\vec{x})=\nabla\times\vec{A}(\vec{x}).$

Newtonian equations of motion and thus also the physical dynamics are gauge
invariant, i.e., depend only on $B(\vec{x})$ and $\nabla W(\vec{x})$. However,
in the Hamiltonian formulation gauge transformations can be seen as canonical
transformations (cf. [12, Problem 11.25]), namely they alter the Hamiltonian,
the corresponding Hamilton’s equations of motion and the Hamilton–Jacobi
equation in a prescribed way. Separation of variables in the Hamilton–Jacobi
equation is related to a specific choice of the coordinate system and is not
preserved under canonical transformations – on the contrary, one looks for a
suitable canonical transformation such that the system becomes separable after
it. Since we are interested in systems that separate in Cartesian coordinates,
we find it preferable to work in a suitably chosen fixed gauge adapted to the
separation.

Furthermore, we will sometimes use canonical transformations to reduce to
cyclic coordinates corresponding to integrals. Also in this perspective, it is
helpful to fix an appropriate gauge. However, the final results, in particular
the superintegrable systems found shall be given in the gauge covariant form,
so to express them in the most general way.

In gauge dependent form the Hamiltonian (1.1) reads

$\displaystyle
H(\vec{x},\vec{p})=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}+A_{1}(\vec{x})p_{1}+A_{2}(\vec{x})p_{2}+A_{3}(\vec{x})p_{3}+V(\vec{x}),$
(1.3)

where the gauge dependent “scalar” potential $V(\vec{x})$, i.e., the momentum-
free term in (1.3), is related to the gauge invariant electrostatic potential
$W(\vec{x})$ via

$\displaystyle
V(\vec{x})=W(\vec{x})+\frac{1}{2}\big{|}\vec{A}(\vec{x})\big{|}^{2}.$

There are only two cases in which the system (1.3) separates in Cartesian
coordinates [1, 27], up to a canonical permutation of the variables. Let us
write them in both gauge dependent and gauge covariant form:

Case I

$\displaystyle
V(\vec{x})=V_{1}(x_{1})+V_{2}(x_{2}),\qquad\vec{A}(\vec{x})=(0,0,u_{1}(x_{2})-u_{2}(x_{1})),$
(1.4)

therefore

$\displaystyle\vec{B}(\vec{x})=(u_{1}^{\prime}(x_{2}),u_{2}^{\prime}(x_{1}),0),\qquad
W(\vec{x})=V_{1}(x_{1})+V_{2}(x_{2})-\frac{1}{2}(u_{1}(x_{2})-u_{2}(x_{1}))^{2}.$
(1.5)

Case II

$\displaystyle
V(\vec{x})=V_{1}(x_{1}),\qquad\vec{A}(\vec{x})=(0,u_{3}(x_{1}),-u_{2}(x_{1})),$
(1.6)

thus

$\displaystyle\vec{B}(\vec{x})=(0,u_{2}^{\prime}(x_{1}),u_{3}^{\prime}(x_{1})),\qquad
W(\vec{x})=V_{1}(x_{1})-\frac{1}{2}\big{(}u_{3}(x_{1})^{2}+u_{2}(x_{1})^{2}\big{)}.$
(1.7)

In these two cases the system admits two Cartesian-type integrals, related to
the separation of variables:

$\displaystyle
X_{1}=\big{(}p_{1}^{A}\big{)}^{2}-2(u_{2}(x_{1})(p_{3}^{A}-u_{1}(x_{2})+u_{2}(x_{1}))-V_{1}(x_{1}))=p_{1}^{2}-2(u_{2}(x_{1})p_{3}-V_{1}(x_{1})),$
$\displaystyle
X_{2}=\big{(}p_{2}^{A}\big{)}^{2}+2(u_{1}(x_{2})(p_{3}^{A}-u_{1}(x_{2})+u_{2}(x_{1}))+V_{2}(x_{2}))=p_{2}^{2}+2(u_{1}(x_{2})p_{3}+V_{2}(x_{2}))\\!\\!\\!\\!$
(1.8)

for (1.4) and

$\displaystyle X_{1}=p_{2}^{A}-u_{3}(x_{1})=p_{2},\qquad
X_{2}=p_{3}^{A}-u_{2}(x_{1})=p_{3}$ (1.9)

for (1.6).

######  Remark.

$X_{0}=p_{3}^{A}-u_{1}(x_{2})+u_{2}(x_{1})=p_{3}$ is another integral of
(1.5), though dependent on the Hamiltonian and (1.8).

Minimal superintegrability due to the existence of another first-order
integral has been studied in [14, 16]. Here we investigate the conditions for
the existence of an additional integral of order at least two for the systems
(1.5), (1.7). We give an exhaustive list of systems for which an additional
second-order integral exists, and are able to answer the question on the
existence of higher-order integrals in special cases.

Sections 2 and 3 present two propositions for finding out whether certain
classes of systems are superintegrable by reducing to a two-dimensional (2D)
problem without magnetic field. In this way we also construct families of
systems with higher-order integrals. Next, in Section 4 we address the problem
of second-order superintegrability. The determining equations for second-order
integrals are given in gauge covariant form, together with their compatibility
conditions. In Section 5 we give a necessary condition for second-order
superintegrability, which is used in Sections 6 and 7 to simplify the
structure of the integral for the classes (1.5), (1.7), respectively. With
these simplifications at hand, the determining equations for the integral can
be solved. In Section 9.1 we list the superintegrable systems so found; their
explicit derivation is rather technical and tedious and we review it in
Appendices A, B and C. The special case in which the magnetic field is
constant and the functions $V_{j}$ in (1.5) and (1.7) are at most quadratic
polynomials is studied in Section 8. Finally, in Section 9.2 we discuss the
approaches to construction of higher-order integrals.

##  2 Minimal superintegrability for Case I
when all the integrals commute with one linear momentum

Let us consider the natural Hamiltonian systems on the phase space
$(x_{1},x_{2},p_{1},p_{2})$, for $\kappa$ $\in\mathbb{R}$, $\kappa\neq 0$

$\displaystyle\mathcal{H}^{\kappa}_{0}(x_{1},x_{2},p_{1},p_{2})=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}\big{)}+\kappa(u_{1}(x_{2})-u_{2}(x_{1}))+V_{1}(x_{1})+V_{2}(x_{2}).$
(2.1)

For the sake of clarity let us refer here to the Hamiltonian of Case I as to
$\mathcal{H}$. Since $p_{3}$ is an integral of motion for (1.4), by setting
$p_{3}=\kappa$,
$\mathcal{H}_{0}^{\kappa}=\mathcal{H}(x,y,z,p_{1},p_{2},\kappa)-\frac{1}{2}\kappa^{2}$.
Both systems have a pair of second-order integrals corresponding to
separation: $X_{j}$ as in (1.8) for $\mathcal{H}$ and clearly

$\displaystyle\mathcal{I}_{j}^{\kappa}=X_{j}(x_{1},x_{2},p_{1},p_{2},\kappa),\qquad
j=1,2$

for (2.1).

If (1.6) possesses any additional integral $X_{3}$ independent of the variable
$x_{3}$, then $\mathcal{I}_{3}^{\kappa}(x_{1},x_{2},p_{1},\allowbreak
p_{2})=X_{3}(x_{1},x_{2},p_{1},p_{2},\kappa)$ would be an integral for (2.1).
And vice versa, any additional integral $\mathcal{I}_{3}^{\kappa}$ of (2.1),
would correspond to an integral $X_{3}$ of (1.6), obtained by simply replacing
$\kappa$ by $p_{3}$, i.e.,
$X_{3}(x_{1},x_{2},x_{3},p_{1},p_{2},p_{3})=\mathcal{I}_{3}^{p_{3}}(x_{1},x_{2},p_{1},p_{2})$.
Indeed,

$\displaystyle\\{\mathcal{H},X_{3}\\}=\sum_{i=1}^{2}\left(\frac{\partial\mathcal{H}_{0}^{p_{3}}}{\partial{x_{i}}}\frac{\partial\mathcal{I}_{3}^{p_{3}}}{\partial{p_{i}}}-\frac{\partial\mathcal{I}_{3}^{p_{3}}}{\partial{x_{i}}}\frac{\partial\mathcal{H}_{0}^{p_{3}}}{\partial{p_{i}}}\right)+\frac{1}{2}\big{\\{}p_{3}^{2},X_{3}\big{\\}}=0,$

where $\\{\,,\,\\}$ is the Poisson bracket on the phase space
$\mathbb{R}^{6}$. The right hand side of the equality is zero since both
$\mathcal{H}$ and $X_{3}$ do not depend on $x_{3}$ and
$\mathcal{I}_{3}^{p_{3}}$ is an integral of $\mathcal{H}_{0}^{p_{3}}$. Thus,
we arrive at the following immediate conclusion

######  Proposition 2.1.

Let us consider the Hamiltonian system defined by (1.1) on the phase space
$(x_{1},x_{2},x_{3},p_{1},p_{2},p_{3})$ with magnetic field and effective
potential as in (1.5). Such system admits an additional independent integral
$I_{3}$ such that $\\{I_{3},p_{3}\\}=0$ if and only if (2.1) is
superintegrable on the phase space $(x_{1},x_{2},p_{1},p_{2})$.

Therefore all the systems of the form (1.5) that are minimally
superintegrable, with an additional integral independent of Cartesian
coordinate, can be deduced from 2D natural superintegrable systems of the form
(2.1). And vice versa, every superintegrable system in two degrees of freedom
can be extended to a minimally superintegrable system in three degrees of
freedom with magnetic field. Superintegrable systems of the form (2.1) have
been widely studied. In particular they have been completely classified for
integrals up to third order [22]. Concerning higher-order integrals, many
examples are known, including the harmonic oscillator and the caged oscillator
[10, 26], and a wide class of so called exotic potentials [6, 7, 20].

###  2.1 Example: extension of 2D second-order superintegrable systems

Table 1 contains all three-dimensional systems that can be proven to be (at
least) minimally quadratically superintegrable by applying Proposition 2.1 to
2D superintegrable systems that separate in Cartesian coordinates and have
integrals at most quadratic. The list of 2D systems is taken from [22], from
which we consider only the systems on real phase space. To obtain the most
general family of systems (and recalling that the Hamiltonian must depend
linearly on $\kappa$), we renamed all the parameters as
$c_{j}=a_{j}\kappa+b_{j}$, $a_{j}$ not all vanishing, then set $p_{3}=\kappa$
and applied Proposition 2.1. The third integral, leading to
superintegrability, can then be found from the integral $\mathcal{I}_{3}$ of
the 2D system, by substituting $c_{j}=a_{j}p_{3}+b_{j}$. Since the dependence
on the constants $c_{j}$ is linear, the order of the so obtained integral
remains quadratic.

Table 1: 3D (at least) minimally quadratically superintegrable extensions of 2D quadratically superintegrable systems that separate in Cartesian coordinates. For the reader’s convenience, we give the Hamiltonian expressed in the gauge choice (1.4), but also the functions $u_{j}$ and $V_{j}$ that allow to find the magnetic field $\vec{B}$ and potential $W$ as in the more general gauge invariant form (1.5). In the integrals, $L_{3}$ denotes angular momentum on the plane, $L_{3}=x_{1}p_{2}-x_{2}p_{1}$.  | 2D system and its third integral | 3D system
---|---|---
$\mathcal{E}_{1}$: | $\begin{array}[]{c}\mathcal{H}_{0}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}\big{)}+c_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+\frac{c_{2}}{x_{1}^{2}}+\frac{c_{3}}{x_{2}^{2}}\\\ \mathcal{I}_{3}=L_{3}^{2}+2\big{(}c_{2}\frac{x_{2}^{2}}{x_{1}^{2}}+c_{3}\frac{x_{1}^{2}}{x_{2}^{2}}\big{)}\end{array}$ |  $\begin{array}[]{c}\mathcal{H}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}+\big{(}a_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+\frac{a_{2}}{x_{1}^{2}}+\frac{a_{3}}{x_{2}^{2}}\big{)}p_{3}+b_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+\frac{b_{2}}{x_{1}^{2}}+\frac{b_{3}}{x_{2}^{2}}\tsep{2pt}\\\ u_{1}(x_{2})=a_{1}x_{2}^{2}+\frac{a_{3}}{x_{2}^{2}},\qquad u_{2}(x_{1})=-a_{1}x_{1}^{2}-\frac{a_{2}}{x_{1}^{2}}\\\ V_{1}(x_{1})=b_{1}x_{1}^{2}+\frac{b_{2}}{x_{1}^{2}},\qquad V_{2}(x_{2})=b_{1}x_{2}^{2}+\frac{b_{3}}{x_{2}^{2}}\end{array}$ 2pt
$\mathcal{E}_{2}$: | $\begin{array}[]{c}\mathcal{H}_{0}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}\big{)}+c_{1}\big{(}4x_{1}^{2}+x_{2}^{2}\big{)}+c_{2}x_{1}+\frac{c_{3}}{x_{2}^{2}}\\\ \mathcal{I}_{3}=p_{2}L_{3}-x_{2}^{2}\big{(}2c_{1}x_{1}+\frac{c_{2}}{2}\big{)}+2c_{3}\frac{x_{1}}{x_{2}^{2}}\end{array}$ |  $\begin{array}[]{c}\mathcal{H}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}+\big{(}a_{1}\big{(}4x_{1}^{2}+x_{2}^{2}\big{)}+a_{2}x_{1}+\frac{a_{3}}{x_{2}^{2}}\big{)}p_{3}+b_{1}\big{(}4x_{1}^{2}+x_{2}^{2}\big{)}+b_{2}x_{1}+\frac{b_{3}}{x_{2}^{2}}\tsep{2pt}\\\ u_{1}(x_{2})=a_{1}x_{2}^{2}+\frac{a_{3}}{x_{2}^{2}},\qquad u_{2}(x_{1})=-4a_{1}x_{1}^{2}-a_{2}x_{1}\\\ V_{1}(x_{1})=4b_{1}x_{1}^{2}+b_{2}x_{1},\qquad V_{2}(x_{2})=b_{1}x_{2}^{2}+\frac{b_{3}}{x_{2}^{2}}\end{array}$ 2pt
$\mathcal{E}_{3}$: | $\begin{array}[]{c}\mathcal{H}_{0}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}\big{)}+c_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+c_{2}x_{1}+c_{3}x_{2}\\\ \mathcal{I}_{3}=p_{1}p_{2}+2c_{1}x_{1}x_{2}+c_{2}x_{2}+c_{3}x_{1}\end{array}$ |  $\begin{array}[]{c}\mathcal{H}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}+\big{(}a_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+a_{2}x_{1}+a_{3}x_{2}\big{)}p_{3}+b_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+b_{2}x_{1}+b_{3}x_{2}\tsep{2pt}\\\ u_{1}(x_{2})=a_{1}x_{2}^{2}+a_{3}x_{2},\qquad u_{2}(x_{1})=-a_{1}x_{1}^{2}-a_{2}x_{1}\\\ V_{1}(x_{1})=b_{1}x_{1}^{2}+b_{2}x_{1},\qquad V_{2}(x_{2})=b_{1}x_{2}^{2}+b_{3}x_{2}\end{array}$ 2pt

###  2.2 Example: a family of higher-order superintegrable systems
from the 2D caged oscillator

Let us consider the two-dimensional caged anisotropic oscillator

$\displaystyle\mathcal{H}_{0}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}\big{)}+\omega\big{(}\ell^{2}x_{1}^{2}+m^{2}x_{2}^{2}\big{)}+\frac{\alpha}{x_{1}^{2}}+\frac{\beta}{x_{2}^{2}}$
(2.2)

for $\omega\in\mathbb{R}\setminus\\{0\\}$, $\ell$, $m$ nonvanishing integers
and $\alpha,\beta\in\mathbb{R}$. The system is well known to be
superintegrable if $\frac{\ell}{m}$ rational [10, 26]. A first straightforward
extension to a 3D superintegrable system is given by

$\displaystyle\mathcal{H}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}+\big{(}\ell^{2}x_{1}^{2}+m^{2}x_{2}^{2}\big{)}p_{3}+\frac{\alpha}{x_{1}^{2}}+\frac{\beta}{x_{2}^{2}},$
(2.3)

that can be transformed into (2.2) by simply reducing $p_{3}=\omega$.

A more general extension can be constructed as in the previous example. Let us
set

$\displaystyle\ell^{2}=\ell_{1}\kappa+\ell_{2},\qquad\alpha=\alpha_{1}\kappa+\alpha_{2},\qquad
m^{2}=m_{1}\kappa+m_{2},\qquad\beta=\beta_{1}\kappa+\beta_{2}.$ (2.4)

The system (2.2) can then be seen as the 2D reduction of

$\displaystyle\mathcal{H}=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}+\left(\omega\big{(}\ell_{1}x_{1}^{2}+m_{1}x_{2}^{2}\big{)}+\frac{\alpha_{1}}{x_{1}^{2}}+\frac{\beta_{1}}{x_{2}^{2}}\right)p_{3}$
$\displaystyle\hphantom{\mathcal{H}=}{}+\omega\big{(}\ell_{2}x_{1}^{2}+m_{2}x_{2}^{2}\big{)}+\frac{\alpha_{2}}{x_{1}^{2}}+\frac{\beta_{2}}{x_{2}^{2}},$
(2.5)

by substituting $p_{3}=\kappa$. We obtain in this way the three-dimensional
integrable system (2.5) that becomes superintegrable when the frequency ratio
of (2.2) (where (2.4) has to be taken into account) is a rational number,
i.e., when

$\displaystyle\frac{\ell_{1}p_{3}+\ell_{2}}{m_{1}p_{3}+m_{2}}=\frac{\ell^{2}}{m^{2}},\qquad\frac{\ell}{m}\in\mathbb{Q},$
(2.6)

for every possible value of the phase space variable $p_{3}$. Equivalently,
(2.6) can be written as

$\displaystyle\big{(}m^{2}\ell_{1}-\ell^{2}m_{1}\big{)}p_{3}+m^{2}\ell_{2}-m_{2}\ell^{2}=0,\qquad\frac{\ell}{m}\in\mathbb{Q}.$

The above equation contains a polynomial in $p_{3}$ that must be identically
zero. This is possible only when the coefficient of each power of $p_{3}$
vanishes. Namely, when

$\displaystyle\frac{\ell_{1}}{m_{1}}=\frac{\ell_{2}}{m_{2}}=\frac{\ell^{2}}{m^{2}},\qquad\frac{\ell}{m}\in\mathbb{Q}.$
(2.7)

Thus, the family of systems (2.5) is superintegrable if and only its
parameters satisfy (2.7) (and in that case also (2.2) is superintegrable). For
$\ell_{j}=m_{j}=0$ for some $j$ (not both $j=1,2$), the previous condition
reduces to

$\displaystyle\frac{\ell_{j}}{m_{j}}=\frac{\ell^{2}}{m^{2}},\qquad\frac{\ell}{m}\in\mathbb{Q}.$

For $\alpha_{1}=\beta_{1}=\ell_{2}=m_{2}=0$ we have the simpler system (2.3).
The case $\alpha_{j}=\beta_{j}=0$, $\ell_{j}=m_{j}=\pm 1$, $j=1,2$ was studied
in [14] and it is shown there to be quadratically minimally superintegrable,
with the fourth independent integral (besides the two Cartesian ones)
inherited from the 2D caged oscillator, of first order. In the more general
case (2.5), the order of the fourth integral can be arbitrarily high,
depending on the value of $\frac{\ell}{m}$. Notice that all the systems in
Table 1 are contained in the family (2.5), except the systems
$\mathcal{E}_{2}$ and $\mathcal{E}_{3}$ for the special case $a_{1}=b_{1}=0$
(i.e., $c_{1}=0$), in which the linear terms in the space variables cannot be
eliminated by translation, due to the absence of quadratic terms.

##  3 Maximal superintegrable class canonically conjugated
to natural 2D systems

Let us consider the system whose magnetic field and effective potential read

$\displaystyle\vec{B}(\vec{x})=(0,\gamma,0),\qquad\gamma\in\mathbb{R}\setminus\\{0\\}$
(3.1)

and

$\displaystyle W(\vec{x})=V(x_{2}),$ (3.2)

respectively. This system can be written in the form (1.4), with the gauge
chosen as

$\displaystyle\vec{A}(\vec{x})=(0,0,-\gamma x_{1}).$

Its Hamiltonian reads

$\displaystyle H=\frac{1}{2}\big{(}p_{1}^{2}+p_{2}^{2}+p_{3}^{2}\big{)}-\gamma
x_{1}p_{3}+\frac{\gamma^{2}}{2}x_{1}^{2}+V(x_{2}).$ (3.3)

Actually by a different choice of the gauge and a canonical permutation of the
variables $x_{1}$ and $x_{2}$ we see that the system belongs also to Case II.
The Hamiltonian (3.3) admits three independent first-order integrals [14]

$\displaystyle I_{1}=p_{1}-\gamma x_{3},\qquad I_{2}=p_{3},\qquad
I_{3}=2l_{2}+\gamma\big{(}x_{1}^{2}-x_{3}^{2}\big{)}.$ (3.4)

Out of them, we can construct two Cartesian-type integrals,

$\displaystyle X_{1}=I_{1}^{2}+\gamma I_{3},\qquad
X_{2}=2H-I_{1}^{2}-I_{2}^{2}-\gamma X_{3}.$ (3.5)

The system can be reduced to two degrees of freedom through the following
canonical transformation

$\displaystyle x_{1}=X+\frac{P_{3}}{\gamma},\qquad x_{2}=Y,\qquad
x_{3}=Z+\frac{1}{\gamma}P_{1},\qquad p_{j}=P_{j},\qquad j=1,2,3,$ (3.6)

with the second type generating function

$\displaystyle
G(\vec{x},\vec{P})=\left(x_{1}-\frac{1}{\gamma}P_{3}\right)P_{1}+x_{2}P_{2}+x_{3}P_{3}.$

The Hamiltonian in the new coordinates reads

$\displaystyle\mathcal{K}(\vec{X},\vec{P})=\frac{1}{2}\big{(}P_{1}^{2}+P_{2}^{2}\big{)}+\frac{1}{2}\gamma^{2}X^{2}+V(Y),$
(3.7)

i.e., it is effectively in two degrees of freedom and without magnetic field.
This system (3.7) has two cyclic coordinates in the full phase space
$(\vec{X},\vec{P})$, namely $Z$ and $P_{3}$, that are therefore both
integrals. Expressed in the original variables, these integrals correspond to
$p_{3}$ and $\frac{I_{1}}{\gamma}$ as in (3.4). Moreover (3.7) separates in
the Cartesian coordinates $(X,Y)$, and the corresponding Cartesian-type
integrals, $\mathcal{I}_{1}$, $\mathcal{I}_{2}$, once written in the original
coordinates, provide (3.5). Thus we have

######  Proposition 3.1.

The system with the magnetic field (3.1) and potential (3.2) is maximally
superintegrable if and only if (3.7), seen as a system in two degrees of
freedom on the phase space $(X,Y,P_{1},P_{2})$ has one additional integral of
motion, besides $\mathcal{I}_{1}$, $\mathcal{I}_{2}$, and independent of them.

Therefore the problem of maximal superintegrability of (3.3) has been reduced
to the two-dimensional problem of superintegrability of (3.7). In particular,
all the potentials $V(Y)$ that make (3.7) superintegrable give (by simply
replacing $Y=x_{2}$) the effective potentials that render (3.3)
superintegrable.

The cases

$\displaystyle V(Y)=\frac{c}{Y^{2}}+\frac{\gamma^{2}Y^{2}}{8},$ (3.8)

and

$\displaystyle V(Y)=\frac{\gamma^{2}}{2}Y^{2},$ (3.9)

that correspond to 3D superintegrable systems with additional second-order
integral have already been found in [14] with a different approach.

All the potentials $V(Y)$ that lead to second and third-order
superintegrability in 2D have been classified [22]. If we focus on second-
order integrals, they are listed in Table 1. The systems that can be obtained
from it, after applying the transformation (3.6) and are still quadratically
superintegrable are given by (3.8), and

$\displaystyle V(Y)=\frac{\gamma^{2}}{2}Y^{2}+cY,$

that, since $\gamma\neq 0$, can be reduced to (3.9) by translation in $Y$.

However, higher-order superintegrable systems can be generated, e.g., from

$\displaystyle V(Y)=\frac{c}{Y^{2}}+\frac{\gamma^{2}Y^{2}}{2},\qquad c\geq 0.$
(3.10)

The additional integral of (3.7) is second order and reads (see Table 1)

$\displaystyle\mathcal{X}_{4}=\mathcal{L}_{3}^{2}+2c\frac{X^{2}}{Y^{2}}.$

Here $\mathcal{L}_{3}$ denotes the third component of the angular momentum
with respect to the coordinates $(X,Y,Z,P_{1},P_{2},P_{3})$. Inverting the
transformation (3.6), it gives the fourth-order integral

$\displaystyle
X_{4}=\frac{1}{\gamma^{2}}\left(\big{(}p_{2}^{A}p_{3}^{A}+\gamma
p_{1}^{A}x_{2}\big{)}^{2}+2c\frac{\big{(}p_{3}^{A}\big{)}^{2}}{x_{2}^{2}}\right).$

Actually, by polynomial combinations with the other integrals, it can be
reduced to the third order one

$\displaystyle X_{5}=2\gamma
p_{2}^{A}p_{3}^{A}l_{3}^{A}+\gamma^{2}\Big{(}x_{1}^{2}\big{(}p_{2}^{A}\big{)}^{2}+x_{2}^{2}\Big{(}\big{(}p_{3}^{A}\big{)}^{2}-\big{(}p_{1}^{A}\big{)}^{2}\Big{)}\Big{)}+2\gamma\frac{x_{1}}{x_{2}^{2}}\big{(}\gamma^{2}x_{2}^{4}+2c\big{)}p_{3}^{A}$
$\displaystyle\hphantom{X_{5}=}{}+\gamma^{2}\frac{x_{1}^{2}}{x_{2}^{2}}\big{(}\gamma^{2}x_{2}^{4}+2c\big{)},$

that cannot be further reduced to lower order by using any of the integrals
(3.4) nor (3.5).

A more general 3D infinite family of maximally superintegrable system,
including the previous cases (3.8) and (3.10) and the one found in [15],
corresponds to the caged oscillator

$\displaystyle
V(Y)=\frac{c}{Y^{2}}+\frac{m^{2}}{\ell^{2}}\gamma^{2}Y^{2},\qquad\ell,m\in\mathbb{N}$
(3.11)

If we compare it with (2.5), we see that for $\gamma^{2}=\omega l_{2}^{2}$,
$\alpha_{2}=0$, $\beta_{2}=c$ and $m_{2}$ satisfying (2.7) the two obtained 3D
families would have the same scalar potential. However, the magnetic fields
differ, rendering (3.11) maximally superintegrable, while (2.5) – as far as we
can see – is only minimally superintegrable.

##  4 Second-order integrals

Any second-order integral of motion we can write

$\displaystyle
X=\sum_{j=1}^{3}h_{j}(\vec{x})p_{j}^{A}p_{j}^{A}+\sum_{j,k,l=1}^{3}\frac{1}{2}|\epsilon_{jkl}|n_{j}(\vec{x})p_{k}^{A}p_{l}^{A}+\sum_{j=1}^{3}s_{j}(\vec{x})p_{j}^{A}+m(\vec{x}),$
(4.1)

where $\epsilon_{jkl}$ is the completely antisymmetric tensor with
$\epsilon_{123}=1$.

The condition that the Poisson bracket

$\displaystyle\\{a(\vec{x},\vec{p}),b(\vec{x},\vec{p})\\}=\sum_{j=1}^{3}\left(\frac{\partial
a}{\partial{x_{j}}}\frac{\partial b}{\partial{p_{j}}}-\frac{\partial
b}{\partial{x_{j}}}\frac{\partial a}{\partial{p_{j}}}\right)$

of the integral (4.1) with the Hamiltonian (1.1) vanishes

$\displaystyle\\{H,X\\}=0$

seen as a polynomial in the momenta leads to the determining equations for the
unknown functions $h_{j}$, $n_{j}$, $s_{j}$, $j=1,2,3$ and $m$ in the
integral. Order by order (from the third to the zeroth) they read (cf. [16]):

$\displaystyle\begin{aligned}
&\partial_{x_{1}}h_{1}=0,\qquad&&\partial_{x_{2}}h_{1}=-\partial_{x_{1}}n_{3},\qquad&&\partial_{x_{3}}h_{1}=-\partial_{x_{1}}n_{2},&\\\
&\partial_{x_{1}}h_{2}=-\partial_{x_{2}}n_{3},\qquad&&\partial_{x_{2}}h_{2}=0,\qquad&&\partial_{x_{3}}h_{2}=-\partial_{x_{2}}n_{1},&\\\
&\partial_{x_{1}}h_{3}=-\partial_{x_{3}}n_{2},\qquad&&\partial_{x_{2}}h_{3}=-\partial_{x_{3}}n_{1},\qquad&&\partial_{x_{3}}h_{3}=0,\\\
&\nabla\cdot\vec{n}=0,&&&&&\end{aligned}$ (4.2)
$\displaystyle\partial_{x_{1}}s_{1}=n_{2}B_{2}-n_{3}B_{3},$
$\displaystyle\partial_{x_{2}}s_{2}=n_{3}B_{3}-n_{1}B_{1},$
$\displaystyle\partial_{x_{3}}s_{3}=n_{1}B_{1}-n_{2}B_{2},$
$\displaystyle\partial_{x_{2}}s_{1}+\partial_{x_{1}}s_{2}=n_{1}B_{2}-n_{2}B_{1}+2(h_{1}-h_{2})B_{3},$
(4.3)
$\displaystyle\partial_{x_{3}}s_{1}+\partial_{x_{1}}s_{3}=n_{3}B_{1}-n_{1}B_{3}+2(h_{3}-h_{1})B_{2},$
$\displaystyle\partial_{x_{2}}s_{3}+\partial_{x_{3}}s_{2}=n_{2}B_{3}-n_{3}B_{2}+2(h_{2}-h_{3})B_{1},$
$\displaystyle\partial_{x_{1}}m=2h_{1}\partial_{x_{1}}W+n_{3}\partial_{x_{2}}W+n_{2}\partial_{x_{3}}W+s_{3}B_{2}-s_{2}B_{3},$
$\displaystyle\partial_{x_{2}}m=n_{3}\partial_{x_{1}}W+2h_{2}\partial_{x_{2}}W+n_{1}\partial_{x_{3}}W+s_{1}B_{3}-s_{3}B_{1},$
(4.4)
$\displaystyle\partial_{x_{3}}m=n_{2}\partial_{x_{1}}W+n_{1}\partial_{x_{2}}W+2h_{3}\partial_{x_{3}}W+s_{2}B_{1}-s_{1}B_{2},$
$\displaystyle\vec{s}\cdot\nabla W=0.$ (4.5)

The equations (4.2) prescribe that the functions $h_{j}$, $n_{j}$ are such
that the highest-order terms in the integral (4.1) are linear combinations of
products of the generators $p_{1}$, $p_{2}$, $p_{3}$, $l_{1}$, $l_{2}$,
$l_{3}$ of the Euclidean group, where
$l_{j}=\sum\limits_{k,l}\epsilon_{jkl}x_{k}p_{l}$ [16]. Explicitly, in terms
of the expressions (1.2), we have

$\displaystyle X=\sum_{i,j\colon i\leq
j}\alpha_{ij}l_{i}^{A}l_{j}^{A}+\sum_{i,j}\beta_{ij}p_{i}^{A}l_{j}^{A}+\sum_{i,j:\;i\leq
j}\gamma_{ij}p_{i}^{A}p_{j}^{A}+\sum_{j=1}^{3}s_{j}(\vec{x})p_{j}^{A}+m(\vec{x}),$
(4.6)

where $l_{j}^{A}=\sum\limits_{k,l}\epsilon_{jkl}x_{k}p_{l}^{A}$. By
subtracting the Hamiltonian and the two Cartesian integrals we can a priori
set $\gamma_{11}=\gamma_{22}=\gamma_{33}=0$. There are compatibility
conditions on equations (4.3), consequence of the following conditions on the
derivatives of the functions $s_{j}$, namely,

$\displaystyle\partial^{2}_{x_{2}}\partial_{x_{1}}s_{1}+\partial^{2}_{x_{1}}\partial_{x_{2}}s_{2}=\partial_{x_{1}}\partial_{x_{2}}(\partial_{x_{2}}s_{1}+\partial_{x_{1}}s_{2}),$
$\displaystyle\partial^{2}_{x_{3}}\partial_{x_{1}}s_{1}+\partial^{2}_{x_{1}}\partial_{x_{3}}s_{3}=\partial_{x_{1}}\partial_{x_{3}}(\partial_{x_{3}}s_{1}+\partial_{x_{1}}s_{3}),$
$\displaystyle\partial^{2}_{x_{3}}\partial_{x_{2}}s_{2}+\partial^{2}_{x_{2}}\partial_{x_{3}}s_{3}=\partial_{x_{2}}\partial_{x_{3}}(\partial_{x_{3}}s_{2}+\partial_{x_{2}}s_{3}),$
$\displaystyle\partial_{x_{1}}\partial_{x_{3}}(\partial_{x_{2}}s_{1}+\partial_{x_{1}}s_{2})=2\partial_{x_{2}}\partial_{x_{3}}(\partial_{x_{1}}s_{1})-\partial_{x_{1}}\partial_{x_{2}}(\partial_{x_{3}}s_{1}+\partial_{x_{1}}s_{3})+\partial^{2}_{x_{1}}(\partial_{x_{3}}s_{2}+\partial_{x_{2}}s_{3}),$
$\displaystyle\partial_{x_{2}}\partial_{x_{3}}(\partial_{x_{2}}s_{1}+\partial_{x_{1}}s_{2})=2\partial_{x_{1}}\partial_{x_{3}}(\partial_{x_{2}}s_{2})-\partial_{x_{1}}\partial_{x_{2}}(\partial_{x_{3}}s_{2}+\partial_{x_{2}}s_{3})+\partial^{2}_{x_{2}}(\partial_{x_{3}}s_{1}+\partial_{x_{1}}s_{3}),$
$\displaystyle\partial_{x_{2}}\partial_{x_{3}}(\partial_{x_{3}}s_{1}+\partial_{x_{1}}s_{3})=2\partial_{x_{1}}\partial_{x_{2}}(\partial_{x_{3}}s_{3})-\partial_{x_{1}}\partial_{x_{3}}(\partial_{x_{3}}s_{2}+\partial_{x_{2}}s_{3})+\partial^{2}_{x_{3}}(\partial_{x_{2}}s_{1}+\partial_{x_{1}}s_{2}).\\!\\!\\!\\!$
(4.7)

These translate into compatibility conditions on the magnetic field and the
constants in the coefficients of the second-order terms. Further compatibility
constraints come from (4.4), consequence of

$\displaystyle\partial_{x_{i}}\partial_{x_{j}}m=\partial_{x_{j}}\partial_{x_{i}}m,\qquad
i,j=1,2,3,\quad i\neq j.$ (4.8)

##  5 A necessary condition for second-order superintegrability

Both classes of systems that separate in Cartesian coordinates have at least
one first-order integral and it is always possible to choose a gauge so that
such integral reads as one of the linear momenta. To fix the ideas, let us
work in such a gauge choice and assume that the constant momentum is $p_{3}$.
If a second-order integral $X$ exists, then $K_{1}=\\{X,p_{3}\\}$ is still an
integral at most of second order or a constant. Since the highest-order terms
in $X$ are as in (4.6), they can be at most quadratic in $x_{3}$. This means
that if $K_{1}$ is quadratic in the momenta, its second-order terms are at
most linear in $x_{3}$, since $K_{1}=\\{X,p_{3}\\}=\frac{\partial X}{\partial
x_{3}}$. Thus, $K_{2}=\\{K_{1},p_{3}\\}$ can be, as above, either an integral
at most quadratic or a constant. If $K_{2}$ is again quadratic,
$K_{3}=\\{K_{2},p_{3}\\}$ can be now at most linear in the momenta, since the
highest-order terms in $K_{2}$ do not depend on $x_{3}$. Therefore, we can
conclude that if a second-order independent integral $X$ exists, then
necessarily there must exist a second-order integral (which could be $X$
itself) such that $\\{X,p_{3}\\}$ is at most linear in the momenta. In
general, for a conserved momentum $p_{j}$, the result is the same, it is
enough to replace $x_{3}$ by $x_{j}$ in the argument above. Thus, we obtain
the following

######  Proposition 5.1.

Let the system defined by $H$ as in (1.1) separate in Cartesian coordinates
and have a quadratic integral $\mathcal{I}$ independent of the Cartesian
integrals. Then there exists a second-order integral $X$, not necessarily
different from $\mathcal{I}$, such that $\\{X,p_{j}\\}$ is a polynomial
expression in the momenta of at most first order, for some $j$.

Thus, to answer the question on the existence of an additional second-order
integral for the class of systems we are considering here, we can start by
answering the simpler question on the existence of the necessary integral $X$
that satisfies the above property. This is done in the following Sections 6
and 7 and Appendices A, B, C.

Since we found that the special case in which the magnetic field is constant
and the functions $V_{j}$ are second-order polynomials in the respective
variables appears several times in the computation therein, we discuss it at
once in the separate Section 8.

##  6 Quadratic superintegrability in Case I

We start with the class of systems in (1.5). To fix the ideas, let us choose a
gauge as in (1.4) and assume that there exists a quadratic independent
integral $\mathcal{I}$. Thus, by Proposition 5.1 there exist another quadratic
integral $X$ such that $\\{X,p_{3}\\}$ is at most first order as a polynomial
in the momenta. Here we consider only the case in which the two Cartesian-type
integrals do not reduce to first-order integrals. In case one of them does,
then the system is at the intersection of Case I and Case II (up to a
permutation of indices) and it is treated at once in Section 7. Moreover, we
assume there does not exist a linear integral, other than $p_{3}$. If it
exists, the corresponding systems can be found in [14], where there is a
complete study of quadratically superintegrable systems with Cartesian
integrals and one independent first-order integral. We can have several cases:

  * (i)

$\\{X,p_{3}\\}$ is at most linear and not vanishing. Thus the only possibility
of finding something new is in assuming that $\\{X,p_{3}\\}$ is a dependent
integral or a constant (we excluded the case there is an independent first-
order integral). We therefore look for a quadratic integral $X$ such that

$\displaystyle\partial_{x_{3}}X=\\{X,p_{3}\\}=c_{1}p_{3}+c_{0},\qquad
c_{j}\in\mathbb{R},$ (6.1)

and $c_{j}$ not both vanishing, $j=0,1$.

  * (ii)

$\\{X,p_{3}\\}=0$ and there exist no quadratic integral independent of the
Cartesian integrals and commuting with $p_{3}$. Then $X$ is trivial, in the
sense that it depends on the Cartesian integrals and $p_{3}$. However, to have
a quadratic superintegrable system, a quadratic integral $\mathcal{I}$ as in
Proposition 5.1 must exist. Without loss of generality, we can assume
$X=\\{\mathcal{I},p_{3}\\}$ with

$\displaystyle\\{\mathcal{I},p_{3}\\}=a_{0}p_{3}^{2}+a_{1}X_{1}+a_{2}X_{2}+c_{1}p_{3}+c_{0},$
(6.2)

where $X_{1}$ and $X_{2}$ are as in (1.8),
$a_{0},a_{1},a_{2},c_{0},c_{1}\in\mathbb{R}$, not all $a_{j}$ vanishing,
otherwise we are in the previous point i).

  * (iii)

$\\{X,p_{3}\\}=0$ and $X$ is independent of the Cartesian integrals. Since $X$
commutes with $p_{3}$, it satisfies the assumptions of Proposition 2.1. Thus,
the corresponding systems can be found in Table 1. If an additional quadratic
independent integral exists, then its Poisson bracket with $p_{3}$ cannot
vanish. This is a consequence of the fact that the 2D system (2.1) cannot have
more than $3$ independent integrals. However, as in the previous point, there
could exist a quadratic independent integral $\mathcal{I}$ such that
$\\{\mathcal{I},p_{3}\\}$ depends on the others, namely

$\displaystyle\\{\mathcal{I},p_{3}\\}=a_{0}p_{3}^{2}+a_{1}X_{1}+a_{2}X_{2}+a_{3}X_{3}+c_{1}p_{3}+c_{0},$
(6.3)

where $a_{0},a_{1},a_{2},a_{3},c_{0},c_{1}\in\mathbb{R}$ and not all $a_{j}$
are vanishing (otherwise we are in case (i)), $X_{1}$, $X_{2}$ as in (1.8) and
$X_{3}=X$.

Let us investigate the possibilities for $X_{3}$ in (6.3). Its highest-order
terms should come from a Poisson bracket of the quadratic terms of
$\mathcal{I}$ with $p_{3}$, i.e., their derivatives with respect to $x_{3}$.
Moreover, by assumption $X_{3}$ does not depend on $x_{3}$. Thus, its second-
order terms can arise only by taking derivatives of a second-order polynomial
that contains terms of the form $p_{i}\cdot l_{j}$, $i=1,2,3$, $j=1,2$. By
computing their Poisson bracket with $p_{3}$, we see that the only outcome
(for an integral $X_{3}$ independent of $X_{1}$ and $X_{2}$) is in terms of
the type $p_{i}p_{\ell}$, $i\neq\ell$. Looking at the integrals of the of 2D
systems in Table 1, and the dependent integrals obtained by their Poisson
bracket with the Cartesian integrals, we see that the only possibility is
(8.4) below.

Now that we outlined all the possibilities, we need to solve the determining
equations (4.3)–(4.5), for the different cases. For this, it is necessary to
work in gauge covariant setting.

The conditions (6.1), (6.2) and (6.3) can be written together as (we can now
set $a_{3}=0$):

$\displaystyle\partial_{x_{3}}X=a_{0}\big{(}p_{3}^{A}-u_{1}(x_{2})+u_{2}(x_{1})\big{)}^{2}+a_{1}X_{1}+a_{2}X_{2}+c_{1}\big{(}p_{3}^{A}-u_{1}(x_{2})+u_{2}(x_{1})\big{)}+c_{0},\\!\\!\\!$
(6.4)

where with an abuse in the notation we denoted $\mathcal{I}$ as $X$ (the
unknown independent integral we are looking for), with
$a_{j},c_{j}\in\mathbb{R}$ and not all vanishing. For $a_{j}=0$, $j=0,1,2$ we
are in case (i).

Equation (6.4) implies the following values for the second-order terms of $X$
as in (4.6):

$\displaystyle\alpha_{11}=\alpha_{22}=\alpha_{12}=\alpha_{13}=\alpha_{23}=\beta_{31}=\beta_{32}=0,\qquad\beta_{11}=\beta_{22},$
$\displaystyle a_{0}=0,\qquad a_{1}=\beta_{12},\qquad a_{2}=-\beta_{21}.$
(6.5)

Moreover, since $\vec{p}\cdot\vec{L}=0$ we can set $\beta_{22}=0$ (and
consequently also $\beta_{11}=0$).

Concerning the lower-order terms, by integrating the right-hand side of (6.4),
we obtain the following restriction on the structure of $X$:

$\displaystyle s_{j}=S_{j}(x_{1},x_{2}),\qquad j=1,2,$ $\displaystyle
s_{3}=S_{3}(x_{1},x_{2})-(2\beta_{12}u_{2}(x_{2})+2\beta_{21}u_{1}(x_{2})-c_{1})x_{3},$
$\displaystyle
m_{3}=c_{0}x_{3}+(u_{1}(x_{2})-u_{2}(x_{1}))\left((2\beta_{12}u_{2}(x_{1})+2\beta_{21}u_{1}(x_{2})-c_{1})x_{3}\right)$
$\displaystyle\hphantom{m_{3}=}{}+(2\beta_{12}V_{1}(x_{1})-2\beta_{21}V_{2}(x_{2}))x_{3}+M(x_{1},x_{2}).$
(6.6)

With this simplifications at hand, we can solve equations (4.3)–(4.5).

Let as assume that $a_{1}$ and $a_{2}$ in (6.4) are not both zero; e.g., let
it be $a_{1}\neq 0$. Then we can shift both the potential $V_{1}(x_{1})$ and
the third component of the vector potential by a constant, thus absorbing the
constants $c_{0}$ and $c_{1}$. Similarly, if $a_{2}\neq 0$ we could use
$X_{2}$. Therefore, by (6.5), we see that if either $\beta_{12}\neq 0$ or
$\beta_{21}\neq 0$, we can proceed in the solution of (4.3)–(4.5) as if
$c_{1}=c_{0}=0$. We obtain that no new superintegrable system can be found in
this case. The details of the computation are in Appendix B.

For $\beta_{12}=\beta_{21}=0$ we find it convenient to start from (4.3), in
which the third equation simplifies to

$\displaystyle(\beta_{33}x_{1}+\gamma_{23})u_{1}^{\prime}(x_{2})+(\beta_{33}x_{2}-\gamma_{13})u_{2}^{\prime}(x_{1})-c_{1}=0.$
(6.7)

The above equation could be trivially satisfied for some of the functions
$u_{j}$ or not. This determines a major splitting in the computation. For the
details see Appendix A, the resulting list of systems is given in the
conclusions, Section 9.1.

##  7 Quadratic superintegrability in Case II

For the class of systems (1.7) we can choose a gauge so that there are two
mutually orthogonal conserved linear momenta. Let us assume that they are
$p_{2}$ and $p_{3}$ as in (1.9). As above, we assume there exists an
independent quadratic integral. Thus, by Proposition 5.1 we can have two
possibilities:

  * (i)

there exists a quadratic integral $X$ such that
$\\{X,p_{2}\\}=\\{X,p_{3}\\}=0$. Then $X$ is an integral of the reduced system
obtained from (1.7) by setting the conserved momenta to constants, i.e.,
function of the 1-dimensional Hamiltonian. Thus, it is dependent on the
Hamiltonian and the conserved momenta. The only hope to find something
interesting is to look for a quadratic integral $\mathcal{I}$ such that
$\\{\mathcal{I},p_{j}\\}=X$ for some $j$.

  * (ii)

There exists a quadratic integral $X$ such that $\\{X,p_{j}\\}$ is linear and
not vanishing for at least one $p_{j}$, $j=2,3$. Without loss of generality we
can assume that $\\{X,p_{3}\\}\neq 0$, otherwise we permute the coordinates
$x_{2}$ and $x_{3}$.

Let us set $j=3$ in both cases and with an abuse of notation let us rename
$\mathcal{I}$ in case (i) as $X$. Thus, we look for a quadratic integral $X$
such that

$\displaystyle\\{X,p_{3}\\}=2a_{0}\big{(}H-X_{1}^{2}-X_{2}^{2}\big{)}+a_{1}X_{1}^{2}+a_{2}X_{2}^{2}+a_{3}X_{1}X_{2}+c_{0}+c_{1}X_{1}+c_{2}X_{2},$
(7.1)

$X_{1}$, $X_{2}$ as in (1.9). For $a_{j}=0$, $j=1,\dots,4$, we have case (ii).

Equation (7.1) implies the following conditions on the coefficients of the
higher-order terms of the integral, expressed as in (4.6) (again, we use the
condition $\vec{p}\cdot\vec{L}=0$)

$\displaystyle\alpha_{11}=\alpha_{22}=\alpha_{12}=\alpha_{13}=\alpha_{23}=\beta_{11}=\beta_{22}=\beta_{32}=0,$
$\displaystyle a_{0}=\beta_{12},\qquad a_{1}=-\beta_{21},\qquad a_{2}=0,\qquad
a_{3}=-\beta_{31}.$ (7.2)

Moreover, by subtracting $X_{1}X_{2}$ from $X$, we can set $\gamma_{23}=0$.

Still as a consequence of (7.1), we have further conditions on the
coefficients of the lower-order terms

$\displaystyle s_{1}=S_{1}(x_{1},x_{2}),\qquad
s_{2}=S_{2}(x_{1},x_{2})+(2(\beta_{12}+\beta_{21})u_{3}(x_{1})-\beta_{31}u_{2}(x_{1})+c_{1})x_{3},$
$\displaystyle
s_{3}=S_{3}(x_{1},x_{2})+z(\beta_{31}u_{3}(x_{1})-2\beta_{12}u_{2}(x_{1})+c_{2})$

and

$\displaystyle
m=M(x_{1},x_{2})-\big{(}(2\beta_{12}+\beta_{21})u_{3}(x_{1})^{2}-\beta_{31}u_{2}(x_{1})u_{3}(x_{1})$
$\displaystyle\hphantom{m=}{}+c_{1}u_{3}(x_{1})-c_{2}u_{2}(x_{1})+2\beta_{12}u_{2}(x_{1})^{2}-2\beta_{12}V_{1}(x_{1})-c_{0}\big{)}x_{3}.$

With these simplifications at hand, we are able to solve the determining
equations (4.2)–(4.5).

Let us perform the substitution

$\displaystyle u_{j}(x_{1})=U_{j}^{\prime}(x_{1}),\qquad j=2,3.$ (7.3)

Since $u_{j}$ are defined in (1.5) up to addition of arbitrary constants and
$U_{j}$ is defined as in (7.3), in the following we can set to zero all the
coefficient of first and zero-order powers of $x_{1}$ in the solutions for
$U_{j}$.

From (4.3) we find

$\displaystyle
S_{1}(x_{1},x_{2})=s_{1}(x_{2})+\beta_{12}U_{2}(x_{1})+(\beta_{13}-2\alpha_{33}x_{2})U_{3}(x_{1})-(\beta_{12}x_{1}+\beta_{33}x_{2}-\gamma_{13})U_{2}^{\prime}(x_{1})$
$\displaystyle\hphantom{S_{1}(x_{1},x_{2})=}{}+(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+\beta_{23}x_{2}-\gamma_{12})U_{3}^{\prime}(x_{1}),$
$\displaystyle
S_{2}(x_{1},x_{2})=s_{2}(x_{1})-\left(\alpha_{33}x_{1}x_{2}^{2}-\beta_{13}x_{1}x_{2}+\frac{1}{2}\beta_{23}x_{2}^{2}-\gamma_{12}x_{2}\right)U_{3}^{\prime\prime}(x_{1}),$
$\displaystyle
S_{3}(x_{1},x_{2})=s_{3}(x_{1})+c_{1}x_{2}+\beta_{31}x_{2}U_{2}^{\prime}(x_{1})-2(\beta_{12}+\beta_{21})x_{2}U_{3}^{\prime}(x_{1})$
$\displaystyle\hphantom{S_{3}(x_{1},x_{2})=}{}+\left(\alpha_{33}x_{1}x_{2}^{2}-\beta_{13}x_{1}x_{2}+\frac{1}{2}\beta_{23}x_{2}^{2}-\gamma_{12}x_{2}\right)U_{2}^{\prime\prime}(x_{1})$
$\displaystyle\hphantom{S_{3}(x_{1},x_{2})=}{}-\left(\beta_{12}x_{1}x_{2}+\frac{1}{2}\beta_{33}x_{2}^{2}-\gamma_{13}x_{2}\right)U_{3}^{\prime\prime}(x_{1}),$

where $U_{j}$ and $s_{\ell}$ must satisfy the third, fourth and fifth equation
of (4.3). Let us continue by considering the third of these equations, namely

$\displaystyle
U_{2}^{\prime\prime}(x_{1})(\beta_{12}x_{1}+\beta_{33}x_{2}-\gamma_{13})+2\beta_{12}U_{2}^{\prime}(x_{1})-\beta_{31}U_{3}^{\prime}(x_{1})-c_{2}=0,$
(7.4)

together with the compatibility conditions (4.7). The first one is trivially
satisfied, while the remaining five read

$\displaystyle\beta_{33}U_{2}^{\prime\prime\prime}(x_{1})=0,$
$\displaystyle(2\alpha_{33}x_{1}+\beta_{23})U_{2}^{\prime\prime\prime}(x_{1})+6\alpha_{33}U_{2}^{\prime\prime}(x_{1})-\beta_{33}U_{3}^{\prime\prime\prime}(x_{1})=0,$
$\displaystyle(\beta_{12}x_{1}+\beta_{33}x_{2}-\gamma_{13})U_{2}^{(4)}(x_{1})+4\beta_{12}U_{2}^{\prime\prime\prime}(x_{1})-\beta_{31}U_{3}^{\prime\prime\prime}(x_{1})=0,$
$\displaystyle-(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+\beta_{23}x_{2}-\gamma_{12})U_{3}^{(4)}(x_{1})+4(\beta_{13}-2\alpha_{33}x_{2})U_{3}^{\prime\prime\prime}(x_{1})-\beta_{21}U_{2}^{\prime\prime\prime}(x_{1})=0,$
$\displaystyle(8\alpha_{33}x_{2}-4\beta_{13}-\beta_{31})U_{2}^{\prime\prime\prime}(x_{1})-(4\beta_{12}+\beta_{21})U_{3}^{\prime\prime\prime}(x_{1})$
(7.5)
$\displaystyle\qquad{}+(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+\beta_{23}x_{2}-\gamma_{12})U_{2}^{(4)}(x_{1})-(\beta_{12}x_{1}+\beta_{33}x_{2}-\gamma_{13})U_{3}^{(4)}(x_{1})=0.$

We can have different subcases according to whether the equations (7.4), (7.5)
are trivially satisfied for some of the functions $U_{j}$ or not. This
determines a major splitting in the computation. The details are given in
Appendix C.

##  8 Constant magnetic field
and second-order polynomial potentials

Let us consider the particular case in which the magnetic field is constant

$\displaystyle\vec{B}(\vec{x})=(a_{1},a_{2},0)$ (8.1)

and in (1.5) we have

$\displaystyle V_{1}(x)=v_{11}x_{1}+v_{12}x_{1}^{2},\qquad
V_{2}(x_{2})=v_{21}x_{2}+v_{22}x_{2}^{2},\qquad u_{1}=a_{1}x_{2},\qquad
u_{2}=-a_{2}x_{1}.$

This system appears in various branches of calculation in the appendices; thus
we find it practical to discuss it separately here.

Notice that since the magnetic field is constant, by rotation around
$x_{3}$-axis we could reduce it to the case in which it is aligned with one of
the Cartesian axis. However, the system would no longer separate in the
corresponding rotated Cartesian coordinates, therefore we prefer not to
perform such a rotation.

Let us also point out that if $V_{1}(x_{1})=0$, for constant magnetic field a
rotation around $x_{2}$ brings the system (1.5) into (1.7). Thus, what we will
deduce in the following for $V_{1}=0$ applies also for (1.7).

###  8.1 $\boldsymbol{v_{12}}$ and $\boldsymbol{v_{22}}$ both not vanishing

Let us assume $v_{12}$ and $v_{22}$ are both not vanishing. Then by the
translation of the coordinate system we can set $v_{11}=v_{21}=0$ without loss
of generality.

Then, similarly to Section 3, we can reduce to a natural Hamiltonian system
through canonical transformations. Namely, let us take as the generating
function

$\displaystyle
G=\left(x_{1}-\frac{a_{2}P_{3}}{2v_{12}}\right)P_{1}+\left(\frac{a_{1}P_{3}}{2v_{22}}+x_{2}\right)P_{2}+x_{3}P_{3},$
(8.2)

so that $p_{j}=P_{j}$, $j=1,2,3$ and

$\displaystyle x_{1}=X+\frac{a_{2}P_{3}}{2v_{12}},\qquad
x_{2}=Y-\frac{a_{1}P_{3}}{2v_{22}},\qquad
x_{3}=Z+\frac{a_{2}v_{22}P_{1}-a_{1}v_{12}P_{2}}{2v_{12}v_{22}}.$

After the transformation, with gauge chosen as in (1.4), the Hamiltonian reads

$\displaystyle
H=\frac{1}{2}\left(P_{1}^{2}+P_{2}^{2}+\left(1-\frac{a_{1}^{2}}{2v_{22}}-\frac{a_{2}^{2}}{2v_{12}}\right)P_{3}^{2}\right)+v_{12}X^{2}+v_{22}Y^{2}.$

If $\frac{a_{1}^{2}}{2v_{22}}+\frac{a_{2}^{2}}{2v_{12}}\neq 1$ we can, by a
canonical transformation

$\displaystyle P_{3}=\frac{1}{\lambda}\tilde{P}_{3},\qquad
Z=\lambda\tilde{Z},\qquad\lambda^{2}=\left|1-\frac{a_{1}^{2}}{2v_{22}}-\frac{a_{2}^{2}}{2v_{12}}\right|$

scale the $P_{3}^{2}$ term to have the Hamiltonian of the form

$\displaystyle
H=\frac{1}{2}\big{(}P_{1}^{2}+P_{2}^{2}\pm\tilde{P}_{3}^{2}\big{)}+v_{12}X^{2}+v_{22}Y^{2}.$

The system can therefore be reduced to a system determined by a two-
dimensional, possibly inverted, anisotropic harmonic oscillator and free
motion along the $Z$-direction. The original 3D system is minimally
superintegrable if and only if the corresponding 2D oscillator is
superintegrable as a system in the $(X,Y,P_{1},P_{2})$ space. If
$v_{12}=v_{22}$ we have a special case of the system $\mathcal{E}_{3}$ in
Table 1. If $\frac{v_{12}}{v_{22}}\in\mathbb{Q}$, $\frac{v_{12}}{v_{22}}\neq
1$ we have a higher-order integral when expressed in the variables $x_{j}$,
$p_{j}$.

If $\frac{a_{1}^{2}}{2v_{22}}+\frac{a_{2}^{2}}{2v_{12}}=1$, the coordinate
$P_{3}$ becomes cyclic and its conjugated variable $Z$ is an independent
constant of motion. In this case the system (1.5) becomes at least minimally
superintegrable. It is maximally superintegrable if and only if its reduction
on the $(X,Y,P_{1},P_{2})$ space is superintegrable. Indeed, we have reduced
to the system (3.7) for $V(Y)=v_{22}Y^{2}$ and $2v_{12}=\gamma^{2}=a_{2}^{2}$.
Its maximally superintegrable exception is included in the family of systems
(3.11).

###  8.2 $\boldsymbol{v_{22}=0}$ and $\boldsymbol{v_{12}}$ not vanishing

In this case by translation in $x_{1}$ we can still set $v_{11}=0$. Then by a
canonical transformation such that

$\displaystyle x_{1}=X+\frac{a_{2}}{2v_{12}}P_{3},\qquad x_{2}=Y,\qquad
x_{3}=Z+\frac{a_{2}}{2v_{12}}P_{1},\qquad p_{j}=P_{j},\qquad j=1,2,3,$

we obtain the system

$\displaystyle
H=\frac{1}{2}\big{(}P_{1}^{2}+P_{2}^{2}+P_{3}^{2}\big{)}+v_{12}X^{2}+a_{1}P_{3}Y+v_{21}Y.$

If $a_{1}=0$ we have reduced to a natural system. By reducing the integral
$P_{3}$ we have a 2D system that, to our knowledge, is not superintegrable.

If $a_{1}\neq 0$ we have reduced to the case with magnetic field aligned along
one axis. The effective potential of the so obtained system reads

$\displaystyle W=v_{12}X^{2}+v_{21}Y-\big{(}a_{1}^{2}Y^{2}\big{)}/2.$ (8.3)

Thus, by the translation

$\displaystyle Y\rightarrow Y+\frac{v_{21}}{a_{1}^{2}}$

we can eliminate the linear term from the effective potential. A shift of the
vector potential by a constant, i.e.,

$\displaystyle A_{3}\rightarrow A_{3}-\frac{v_{21}}{a_{1}},$

gives

$\displaystyle
H=\frac{1}{2}\big{(}P_{1}^{2}+P_{2}^{2}+P_{3}^{2}\big{)}+v_{12}X^{2}+a_{1}P_{3}Y.$

Thus, without loss of generality, we can set $v_{21}=0$. By plugging (8.3) and
(8.1) with these simplifications into the determining equations for a second-
order integral as in (4.6), we find that they have no solution.

###  8.3 $\boldsymbol{v_{12}=v_{22}=0}$

We have a subcase of the system $\mathcal{E}_{3}$ in Table 1. Thus, the system
admits a second-order integral. With the gauge chosen as in (1.4), that
integral reads

$\displaystyle
X_{3}=p_{1}p_{2}+(v_{11}-a_{2}p_{3})x_{2}+(v_{21}+a_{1}p_{3})x_{1}.$

Equivalently, in gauge covariant form, we have

$\displaystyle
X_{3}=p_{1}^{A}p_{2}^{A}+(a_{1}x_{1}-a_{2}x_{2})p_{3}^{A}-\big{(}a_{1}^{2}+a_{2}^{2}\big{)}x_{1}x_{2}+a_{1}a_{2}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+v_{11}x_{2}+v_{21}x_{1},$

corresponding to the fact that the system actually separates in any rotated
system of Cartesian coordinates, since the Hamiltonian is linear in the space
variables. Without altering the structure of the Cartesian-type integrals, we
can therefore by rotation align the magnetic field along one Cartesian axis,
let us say the $x_{2}$-axis. Thus, without lost of generality, let us assume
$a_{1}=0$. The determining equations for an additional second-order integral
can be solved. We find for $v_{11}=0$ one maximally superintegrable system:

$\displaystyle\vec{B}(\vec{x})=(0,a_{2},0),\qquad
W(\vec{x})=v_{21}x_{2}-\frac{1}{2}a_{2}^{2}x_{1}^{2},$ (8.4)

with the integral

$\displaystyle
X_{4}=3p_{3}^{A}l_{1}^{A}-p_{1}^{A}l_{3}^{A}-\frac{3v_{21}}{a_{2}}l_{2}^{A}+a_{2}x_{1}x_{2}p_{3}^{A}+3a_{2}x_{1}l_{1}^{A}+v_{21}x_{1}^{2}+a_{2}^{2}x_{1}^{2}x_{2}$
$\displaystyle\hphantom{X_{4}}{}=3p_{3}l_{1}-p_{1}l_{3}-\frac{3}{a_{2}}\big{(}3v_{21}l_{2}+2a_{2}^{2}x_{1}x_{2}p_{3}+2a_{2}v_{21}x_{1}^{2}\big{)}.$

##  9 Conclusions

Let us summarize our results. We have provided an exhaustive determination of
quadratically superintegrable systems which separate in the Cartesian
coordinates with magnetic field. In addition, we have found classes of systems
minimally and maximally superintegrable with higher-order integrals. We list
them below for reader’s convenience.

###  9.1 Superintegrable systems with second-order integrals

We have constructed an exhaustive list of quadratically superintegrable
systems with nonvanishing magnetic field which separate in Cartesian
coordinates. Under the assumption that there is no independent first-order
integral other than the Cartesian ones (in that case we refer the reader to
our previous work [14, 16]) we have found 8 classes of minimally
superintegrable systems, among which one contains a quadratically maximally
superintegrable subclass, cf. (8.4).

For brevity, we write here the magnetic field, the electrostatic potential and
the leading order terms in the integral(s) together with the reference to the
equation in which the system was introduced. We refer the reader to the
relevant formulas therein encoding the complete information about the
integral(s).

Case I,

i.e., the magnetic field and potential are of the form (1.5) and the Cartesian
integrals as in (1.8). The superintegrable systems read

(a)

$\displaystyle\vec{B}(\vec{x})=\big{(}a{\mathrm{e}}^{bx_{2}},c,0\big{)},\qquad
W(\vec{x})=a\left(w+\frac{c}{b}x_{1}\right){\mathrm{e}}^{bx_{2}}-\frac{a^{2}}{2b^{2}}{\mathrm{e}}^{2bx_{2}},$

$X_{3}=p_{1}^{A}p_{3}^{A}+\cdots$, cf. (A.11).

(b)

$\displaystyle\vec{B}(\vec{x})=2\left(a_{1}x_{2}-\frac{a_{3}}{x_{2}^{3}},-a_{1}x_{1}+\frac{a_{2}}{x_{1}^{3}},0\right),$
$\displaystyle
W(\vec{x})=-\frac{1}{2}a_{1}^{2}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}^{2}-\frac{a_{2}^{2}}{2x_{1}^{4}}-\frac{a_{3}^{2}}{2x_{2}^{4}}-a_{1}\left(a_{2}\frac{x_{2}^{2}}{x_{1}^{2}}+a_{3}\frac{x_{1}^{2}}{x_{2}^{2}}\right)$
$\displaystyle\hphantom{W(\vec{x})=}{}-\frac{a_{2}a_{3}}{x_{2}^{2}x_{1}^{2}}+\frac{b_{3}}{x_{2}^{2}}+b_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+\frac{b_{2}}{x_{1}^{2}},$

$X_{3}=\big{(}l_{3}^{A}\big{)}^{2}+\cdots$, cf. Table 1.

(c)

$\displaystyle\vec{B}(\vec{x})=\left(2\left(a_{1}x_{2}-\frac{a_{3}}{x_{2}^{3}}\right),-8a_{1}x_{1}-a_{2},0\right),$
$\displaystyle
W(\vec{x})=-\frac{a_{1}^{2}}{2}\big{(}4x_{1}^{2}+x_{2}^{2}\big{)}^{2}-\frac{a_{2}^{2}}{2}x_{1}^{2}-\frac{a_{3}^{2}}{2x_{2}^{4}}-a_{2}a_{3}\frac{x_{1}}{x_{2}^{2}}$
$\displaystyle\hphantom{W(\vec{x})=}{}-a_{1}a_{2}x_{1}\big{(}4x_{1}^{2}+x_{2}^{2}\big{)}-4a_{1}a_{3}\frac{x_{1}^{2}}{x_{2}^{2}}+\frac{b_{3}}{x_{2}^{2}}+b_{1}\big{(}4x_{1}^{2}+x_{2}^{2}\big{)}+b_{2}x_{1},$

$X_{3}=p_{2}^{A}l_{3}^{A}+\cdots$, cf. Table 1.

(d)

$\displaystyle\vec{B}(\vec{x})=(2a_{1}x_{2}+a_{3},-2a_{1}x_{1}-a_{2},0),$
$\displaystyle
W(\vec{x})=-\frac{a_{1}^{2}}{2}\left(x_{1}^{2}+x_{2}^{2}\right)^{2}-\frac{a_{3}^{2}}{2}x_{2}^{2}-\frac{a_{2}^{2}}{2}x_{1}^{2}-a_{2}a_{1}x_{1}\left(x_{1}^{2}+x_{2}^{2}\right)$
$\displaystyle\hphantom{W(\vec{x})=}{}-a_{2}a_{3}x_{1}x_{2}-a_{1}a_{3}x_{2}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+b_{1}\big{(}x_{1}^{2}+x_{2}^{2}\big{)}+b_{2}x_{1}+b_{3}x_{2},$

$X_{3}=p_{1}^{A}p_{2}^{A}+\cdots$, cf. Table 1. When $a_{1}=a_{3}=0$ and
$b_{1}=b_{2}=0$ the system becomes maximally superintegrable, with the
additional integral of the form
$X_{4}=p_{1}^{A}l_{3}^{A}-3p_{3}^{A}l_{1}^{A}+\cdots$, cf. (8.4).

Case II,

i.e., the magnetic field and potential are of the form (1.7) and the Cartesian
integrals as in (1.9). The superintegrable systems read

(a)

$\displaystyle\vec{B}(\vec{x})=\big{(}0,a{\rm e}^{bx_{1}},0\big{)},\qquad
W(\vec{x})=wx_{1}+ce^{bx_{1}}-\frac{1}{2}\frac{a^{2}}{b^{2}}{\rm
e}^{2bx_{1}},$

$X_{3}=p_{1}^{A}p_{2}^{A}-bp_{3}^{A}l_{1}^{A}+\cdots$, cf. (C.7),

(b)

$\displaystyle\vec{B}(\vec{x})=\big{(}0,a(b-2)x_{1}^{b-3},0\big{)},\qquad
W(\vec{x})=-\frac{a^{2}x_{1}^{2(b-2)}}{2}+a(b-2)cx_{1}^{b-2}+\frac{w}{x_{1}^{2}},$

$X_{3}=p_{1}^{A}l_{3}^{A}-bp_{3}^{A}l_{1}^{A}+\cdots$, cf. (C.9),

(c)

$\displaystyle\vec{B}(\vec{x})=\left(0,\frac{a}{x_{1}},0\right),\qquad
W(\vec{x})=-\frac{1}{2}a^{2}\left(\ln|x_{1}|\right)^{2}+b\ln|x_{1}|+\frac{w}{x_{1}^{2}},$

$X_{3}=2p_{1}^{A}l_{3}^{A}-p_{3}^{A}l_{1}^{A}+\cdots$, cf. (C.10),

(d)

$\displaystyle\vec{B}(\vec{x})=\left(0,0,\frac{a}{x_{1}^{3}}\right),\qquad
W(\vec{x})=-\frac{ab\ln|x_{1}|}{x_{1}^{2}}-\frac{a^{2}}{8x_{1}^{4}}+\frac{w}{x_{1}^{2}},$

$X_{3}=p_{1}^{A}l_{2}^{A}+\cdots$, cf. (C.18).

Our approach also demonstrates that any quadratically maximally
superintegrable system with magnetic field which separates in Cartesian
coordinates would necessarily appear at the intersection of the presented
classes. Given the different structure of the magnetic field in each of the
cases we find only few potential candidates. One is the intersection of Case
I.d and Case II.b which, as we already observed, leads to the maximally
superintegrable system (8.4). Another is the system Case I.a which for $c=0$
reduces to the system Case II.a (upon interchange of the $x_{1}$ and $x_{2}$
coordinates and momenta). However, the integral $X_{3}$ of Case I.a when $c=0$
becomes a function of the two first-order Cartesian integrals, i.e., it is not
independent anymore. Last but not least, the systems Case I.b, Case I.c, Case
II.b and Case II.d (after a permutation of coordinates) overlap for
$a_{1}=a_{2}=0$ (Case I.b/c) and $b=0$ (Case II.b/d) but the integrals again
turn our to be dependent.

Thus we conclude that no other quadratically maximally superintegrable systems
separating in Cartesian coordinates other than (8.4) and the ones found in
[14] exist.

###  9.2 Superintegrable systems with higher-order integrals

Above we have provided a complete answer to the problem of quadratic
superintegrability for the considered classes of systems (1.5) and (1.7). As
we have seen, maximal superintegrability via at most quadratic integrals is
very rare in the presence of magnetic field, as opposed to numerous purely
scalar maximally superintegrable systems discussed, e.g., in [9, 13]. Thus one
should consider also the possible existence of higher-order integrals.
However, these are computationally very difficult to find. In this paper we
have presented two propositions, namely Propositions 2.1 and 3.1 which can be
used to construct three-dimensional maximally superintegrable systems with
magnetic field out of two-dimensional scalar ones. In particular, Proposition
3.1 states that a system with

$\displaystyle\vec{B}(\vec{x})=(0,\gamma,0),\qquad\gamma\neq 0,\qquad
W(\vec{x})=V(x_{2}),$

is maximally superintegrable if and only if the two-dimensional system with
the Hamiltonian

$\displaystyle\mathcal{K}(\vec{X},\vec{P})=\frac{1}{2}\big{(}P_{1}^{2}+P_{2}^{2}\big{)}+\frac{1}{2}\gamma^{2}X^{2}+V(Y)$

is superintegrable (where $V$ is the same function of a single variable).

Using Proposition 3.1 we have arrived at an explicit example of maximally
superintegrable system with

$\displaystyle\vec{B}(\vec{x})=(0,\gamma,0),\qquad
W(\vec{x})=\frac{c}{x_{2}^{2}}+\frac{m^{2}}{\ell^{2}}\gamma^{2}x_{2}^{2},\qquad\ell,m\in\mathbb{N},\qquad
c\in\mathbb{R},$

cf. (3.11), with three first-order integrals (3.4) and an additional integral
coming from the integral of two-dimensional caged oscillator through the
change of variables (3.6).

Similarly, Proposition 2.1 led us to minimally superintegrable systems (2.5)

$\displaystyle\vec{B}(\vec{x})=2\left(\omega
m_{1}x_{2}-\frac{\beta_{1}}{x_{2}^{3}},-\omega\ell_{1}x_{1}+\frac{\alpha_{1}}{x_{1}^{3}},0\right),$
$\displaystyle
W(\vec{x})=-\frac{\omega^{2}}{2}\big{(}\ell_{1}x_{1}^{2}+m_{1}x_{2}^{2}\big{)}^{2}+\omega\left(\ell_{2}x_{1}^{2}+m_{2}x_{2}^{2}-\alpha_{1}m_{1}\frac{x_{2}^{2}}{x_{1}^{2}}-\beta_{1}\ell_{1}\frac{x_{1}^{2}}{x_{2}^{2}}\right)$
$\displaystyle\hphantom{\vec{B}(\vec{x})=}{}+\frac{\alpha_{2}}{x_{1}^{2}}+\frac{\beta_{2}}{x_{2}^{2}}-\frac{1}{2}\left(\frac{\alpha_{1}}{x_{1}^{2}}+\frac{\beta_{1}}{x_{2}^{2}}\right)^{2},\qquad\frac{l_{1}}{m_{1}}=\frac{l_{2}}{m_{2}}=\frac{l^{2}}{m^{2}},\qquad
l,m\in\mathbb{Z}.$

Systems I.b and I.c of Section 9.1 with $a_{1}\neq 0$ are special subcases of
it when the integral $X_{3}$ becomes second order one.

Another class of minimally superintegrable systems with

$\displaystyle\vec{B}(\vec{x})=(a_{1},a_{2},0),\qquad
W(\vec{x})=v_{12}x_{1}^{2}+v_{22}x_{2}^{2}-\frac{1}{2}\left(a_{2}x_{1}+a_{1}x_{2}\right)^{2},$
$\displaystyle\frac{v_{12}}{v_{22}}\in\mathbb{Q},\qquad v_{12},v_{22}\neq 0$

can be constructed out of anisotropic harmonic oscillator in two dimensions
through the canonical transformation (8.2).

Of course, more efficient and widely applicable tools for construction of
higher-order superintegrable systems are needed. Given the recent rapid
progress on a similar problem for scalar potentials [7, 18, 19, 21, 23] (see
also references in [22]) we hope that in foreseeable future we will be able to
report on further development also in the case with magnetic field.

##  Appendix A Solution of the determining equations
in Case I for $\boldsymbol{\beta_{12}=\beta_{21}=0}$

We consider here $\beta_{12}=\beta_{21}=0$. Thus $c_{0}$ and $c_{1}$ cannot
both vanish, since by assumption the right-hand side of (6.4) is not
identically zero and, by (6.5), $a_{j}=0$, $j=1,2$. There are several
subcases, according to whether equation (6.7) is trivially satisfied or not.
For the reader’s convenience, let us type it here again

$\displaystyle(\beta_{33}x_{1}+\gamma_{23})u_{1}^{\prime}(x_{2})+(\beta_{33}x_{2}-\gamma_{13})u_{2}^{\prime}(x_{1})-c_{1}=0.$
(A.1)

We can have

  * $(a)$

the above equation is not trivially satisfied for $u_{2}$ nor for $u_{1}$,
thus $\beta_{33}\neq 0$ or $\beta_{33}=0$ and both $\gamma_{13}$,
$\gamma_{23}$ not vanishing;

  * $(b)$

equation (A.1) is trivially satisfied for $u_{2}$ but not for $u_{1}$, thus
$\beta_{33}=\gamma_{13}=0$ and $\gamma_{23}\neq 0$;

  * $(c)$

equation (A.1) is trivially satisfied for $u_{1}$ but not for $u_{2}$, thus
$\beta_{33}=\gamma_{23}=0$ and $\gamma_{13}\neq 0$;

  * $(d)$

equation (A.1) is trivially satisfied for both $u_{j}$, thus
$c_{1}=\beta_{33}=\gamma_{23}=\gamma_{13}=0$.

Notice that Case $(b)$ can be reduced to Case $(c)$, and viceversa, by a
canonical exchange of $p_{1}$ with $p_{2}$. Thus both cases are recovered by
Appendix A.3. Case $(a)$ is splitted for convenience into Appendices A.1 and
A.2, while Case $(d)$ is treated in Appendix A.4.

###  A.1 $\boldsymbol{\beta_{33}\neq 0}$

In this case, by translation in $x_{1}$ and $x_{2}$, we can set
$\gamma_{13}=\gamma_{23}=0$.

By taking second derivatives of (A.1) with respect to $x_{1}$ and $x_{2}$ and
by equating them to zero, we find that

$\displaystyle u_{1}(x_{2})=\frac{a_{12}}{2}x_{2}^{2}+a_{11}x_{2},\qquad
u_{2}(x_{1})=-\frac{a_{12}}{2}x_{1}^{2}+a_{21}x_{1},\qquad
a_{ij}\in\mathbb{R}.$ (A.2)

By imposing then that (A.2) solves (A.1), we obtain a polynomial expression in
$x_{1}$ and $x_{2}$ which must vanish

$\displaystyle c_{1}-(a_{11}x_{1}-a_{21}x_{2})\beta_{33}=0.$

We conclude $a_{11}=a_{21}=c_{1}=0$.

Solving the remaining equations in (4.3) we find the functions $S_{j}$. Next
we look at the first-order equations (4.4), and in particular at the third
one, for $m_{x_{3}}$. We proceed as above to solve it for $V_{j}$. By
considering its second-order derivatives with respect to $x_{1}$ and $x_{2}$
and by equating them to zero, we find that also the functions $V_{j}$ must be
second-order polynomials. Then we plug such solution back into the equation
for $m_{x_{3}}$ and find a polynomial in $x_{1}$ and $x_{2}$ that must vanish.
This implies $c_{0}=0$. Thus $\\{X_{3},p_{3}\\}=0$. Nothing new can be found
in this case.

###  A.2 $\boldsymbol{\beta_{33}=0}$, $\boldsymbol{\gamma_{13}\neq 0}$,
$\boldsymbol{\gamma_{23}\neq 0}$

Proceeding as above we find from (A.1) that

$\displaystyle c_{1}=\left(a_{1}\gamma_{23}-a_{2}\gamma_{13}\right)$

and

$\displaystyle u_{1}(x_{2})=a_{1}x_{2},\qquad u_{2}(x_{1})=a_{2}x_{1},\qquad
a_{ij}\in\mathbb{R},$ (A.3)

where $|a_{1}|^{2}+|a_{2}|^{2}\neq 0$. The solution (A.3) implies that the
magnetic field is constant. By substituting (A.3) into the compatibility
conditions (4.7) for the magnetic field, they imply $\alpha_{33}=0$. The
second-order equations (4.3) can now be solved and give

$\displaystyle
S_{1}(x_{1},x_{2})=a_{2}\gamma_{13}x_{1}+(S-a_{1}\gamma_{13}a_{2}\gamma_{23})x_{2}+s_{11},\qquad
S_{2}(x_{1},x_{2})=-Sx_{1}-a_{1}\gamma_{23}x_{2}+s_{21},$ $\displaystyle
S_{3}(x_{1},x_{2})=\frac{1}{2}\big{(}a_{1}\beta_{13}x_{1}^{2}+a_{2}\beta_{23}x_{2}^{2}\big{)}-a_{2}\gamma_{1}x_{2}+(a_{2}\beta_{13}x_{2}+a_{1}\gamma_{1})x_{1},\qquad
S,s_{ij}\in\mathbb{R},$

together with the condition

$\displaystyle\beta_{13}a_{2}=\beta_{23}a_{1}.$

We now look at the first-order equations (4.4), starting with the equation for
$m_{x_{3}}$. Its first-order derivatives with respect to $x_{1}$ and $x_{2}$
imply that each $V_{j}$ is a second-order polynomial in its variable, $j=1,2$.
This case is discussed in Section 8.

###  A.3 $\boldsymbol{\beta_{33}=\gamma_{23}=0}$, $\boldsymbol{\gamma_{13}\neq
0}$

In this case equation (A.1) is trivially satisfied for $u_{1}$ and as an
equation for $u_{2}$ implies

$\displaystyle u_{2}(x_{1})=-\frac{c_{1}x_{1}}{\gamma_{13}}.$ (A.4)

By solving the second-order equations (4.3) for $s_{j}$, $j=1,2,3$, we find

$\displaystyle
S_{1}(x_{1},x_{2})=-c_{1}x_{1}-Sx_{2}-\gamma_{13}u_{1}(x_{2})+s_{11},\qquad
S_{2}(x_{1},x_{2})=Sx_{1}+s_{21},$ $\displaystyle
S_{3}(x_{1},x_{2})=\frac{c_{1}x_{2}\big{(}2\alpha_{33}x_{1}x_{2}-2\beta_{13}x_{1}-\frac{1}{2}\beta_{23}x_{2}+\gamma_{12}\big{)}}{\gamma_{13}}$
$\displaystyle\hphantom{S_{3}(x_{1},x_{2})=}{}+x_{1}\left(x_{1}\left(-\alpha_{33}x_{2}+\frac{1}{2}\beta_{13}\right)-\beta_{23}x_{2}+\gamma_{12}\right)u_{1}^{\prime}(x_{2}),$

where $S,s_{ij}\in\mathbb{R}$ and $u_{1}$ has to solve the remaining
compatibility conditions (4.7) (some are already satisfied by the conditions
imposed on the constant $\alpha_{ij}$, $\beta_{ij}$, $\gamma_{ij}$ and (A.4)):

$\displaystyle\frac{6c_{1}\alpha_{33}}{\gamma_{13}}-\beta_{23}x_{2}u_{1}^{(3)}(x_{2})+\gamma_{12}u_{1}^{(3)}(x_{2})-4\beta_{23}u_{1}^{\prime\prime}(x_{2})=0,$
(A.5)
$\displaystyle(2\alpha_{33}x_{2}-\beta_{13})u_{1}^{\prime\prime}(x_{2})+6\alpha_{33}u_{1}^{\prime}(x_{2})=0.$
(A.6)

The second derivative with respect to $x_{1}$ of the third first-order
equation in (4.4) implies

$\displaystyle V_{1}(x_{1})=v_{11}x_{1}+v_{12}x_{1}^{2},\qquad
v_{ij}\in\mathbb{R}.$ (A.7)

To proceed, we must solve the equations (A.5)–(A.6) for $u_{1}$. There can be
several subcases.

####  A.3.1 $\boldsymbol{\alpha_{33}=\beta_{13}=\beta_{23}=\gamma_{12}=0}$

In this case (A.5)–(A.6) are trivially satisfied. Let us consider the second
set of compatibility conditions (4.8) coming from the first-order equations,
which simplify to

$\displaystyle\frac{{c_{1}}^{2}}{\gamma_{13}}-2v_{12}\gamma_{13}-Su_{1}^{\prime}(x_{2})=0,$
(A.8) $\displaystyle
c_{1}\left(\frac{S}{\gamma_{13}}+u_{1}^{\prime}(x_{2})\right)-s_{21}u_{1}^{\prime\prime}(x_{2})=0.$
(A.9)

Let us first assume $u_{1}^{\prime\prime}\neq 0$. If $c_{1}=0$, then
$S=v_{12}=s_{21}=0$ and $u_{2}(x_{1})=0$. Equation (4.5) implies that then
$u_{1}$ is a constant, in contradiction with the assumption
$u_{1}^{\prime\prime}\neq 0$. Thus, let us consider the case
$u_{1}^{\prime\prime}$ and $c_{1}$ both not zero. From equation (A.9) we
obtain $S=0$ and

$\displaystyle u_{1}(x_{2})=\frac{a_{1}s_{21}}{c_{1}}{\rm
e}^{\frac{c_{1}x_{2}}{s_{21}}},\qquad a_{1}\in\mathbb{R}$

together with

$\displaystyle v_{12}=\frac{c_{1}^{2}}{2\gamma_{13}^{2}}$

that we substitute into (A.7).

The solution of the compatibility constraints (A.8)–(A.9) assures that a
solution of (4.4) for $m$ exists. Thus, let us look at the zero-order equation
(4.5), which in this case reduces to

$\displaystyle c_{1}v_{11}x_{1}-\frac{a_{1}s_{21}{\rm
e}^{\frac{c_{1}x_{2}}{s_{21}}}\big{(}4v_{11}\gamma_{13}^{2}+c_{1}s_{11}\big{)}}{c_{1}\gamma_{13}}+s_{11}v_{11}+s_{21}V_{2}^{\prime}(x_{2})=0.$
(A.10)

Therefore $v_{11}=0$ and (A.7) simplifies to

$\displaystyle V_{1}(x_{1})=\frac{c_{1}^{2}x_{1}^{2}}{2\gamma_{13}^{2}}.$

Let us notice that if $s_{21}=0$ we have constant magnetic field along the
$x_{2}$ direction and arbitrary $V_{2}$. Thus, we obtain the class of systems
already studied in Section 3. Otherwise, if $s_{21}$ is not zero, by solving
(A.10) we find

$\displaystyle V_{2}(x_{2})=\frac{a_{1}s_{11}s_{21}{\rm
e}^{\frac{c_{1}x_{2}}{s_{21}}}}{c_{1}\gamma_{13}}.$

Thus, we arrive at the system determined by

$\displaystyle\vec{B}(\vec{x})=\big{(}a{\mathrm{e}}^{bx_{2}},c,0\big{)},\qquad
W(\vec{x})=a\left(w+\frac{c}{b}x_{1}\right){\mathrm{e}}^{bx_{2}}-\frac{a^{2}}{2b^{2}}{\mathrm{e}}^{2bx_{2}},$
(A.11)

where we relabelled the integration constants as $a_{1}=a$,
$c_{1}=-\gamma_{13}c$, $s_{11}=\gamma_{13}bw$,
$s_{21}=-\frac{c\gamma_{13}}{b}$, with $a,b,c,w\in\mathbb{R}$ such that $a$,
$b$, and $c$ are not vanishing. The system (A.11) is (at least) minimally
superintegrable, with the integral $X_{3}$ as in (4.6), where all the
coefficients of the second-order terms are zero except $\gamma_{13}$, and
$S_{j}$ and $m_{3}$ of equation (6.6) are given by

$\displaystyle
S_{1}(x_{1},x_{2})=\gamma_{13}\left(wb-{\mathrm{e}}^{bx_{2}}\frac{a}{b}+cx_{1}\right),\qquad
S_{2}(x_{1},x_{2})=-\gamma_{13}\frac{c}{b},\qquad S_{3}(x_{1},x_{2})=0,$
$\displaystyle
m_{3}(\vec{x})=\gamma_{13}c\left({\mathrm{e}}^{bx_{2}}\frac{a}{b}-cx_{1}-wb\right)x_{3}.$

It remains to be considered the case in which $u_{1}^{\prime\prime}=0$,
therefore $u_{1}(x_{2})=a_{1}x_{2}$, the magnetic field is constant. If
$c_{1}=0$, from the zero-order equation (4.5) we get that either $V_{2}$ is a
second-order polynomial (not of interest here since $V_{1}$ is already a
second-order polynomial, too), or $S=s_{21}=v_{11}=0$ and $v_{12}=0$ from
(A.8). This gives $V_{1}(x_{1})=0$. From the first-order equations (4.4) we
find $c_{0}=0$. Nothing new here.

If $c_{1}$ is not zero, the compatibility condition (A.8), (A.9) together with
the zero-order equation (4.5) imply that either $V_{1}$ and $V_{2}$ are both
second-order polynomials or $V_{1}=0$ and the magnetic field is constant and
aligned along the $x_{2}$-axis. Both these cases are of no interest in this
section since they are considered elsewhere, in Section 8 and Appendix C.

####  A.3.2 $\boldsymbol{\alpha_{33}=0}$, $\boldsymbol{\beta_{13}\neq 0}$

By translation in $x_{1}$ we can assume $\gamma_{12}=0$.

The compatibility condition (A.5) reduces to

$\displaystyle-\beta_{23}\big{(}x_{2}u_{1}^{(3)}(x_{2})+4u_{1}^{\prime\prime}(x_{2})\big{)}=0,$
$\displaystyle\beta_{13}u_{1}^{\prime\prime}(x_{2})=0.$ (A.12)

Since $\beta_{13}$ is not zero, (A.12) gives

$\displaystyle u_{1}(x_{2})=a_{1}x_{2},\qquad a_{1}\in\mathbb{R}.$

The compatibility conditions (4.8) for $m_{3}$ imply that $V_{2}$ is a second-
order polynomial. This case is of no interest here and it is studied in
Section 8.

####  A.3.3 $\boldsymbol{\alpha_{33}=0}$, $\boldsymbol{\beta_{13}=0}$,
$\boldsymbol{\beta_{23}\neq 0}$

Still, by translation in $x_{1}$, we can set $\gamma_{12}=0$. Equations (4.3)
imply

$\displaystyle u_{1}(x_{2})=\frac{a_{2}}{6x_{2}^{2}}.$

In order to have nonvanishing magnetic field and nonvanishing Poisson bracket
in (6.1), we find from equations (4.4) and (4.5) that we must have $S=0$,
$a_{2}=0$ and $c_{0}=\frac{c_{1}s_{11}}{\gamma_{13}}$. Looking for a solution
of (4.5) for $V_{2}$ not second or lower-order polynomial we find

$\displaystyle\vec{B}(\vec{x})=\left(0,-\frac{c_{1}}{\gamma_{13}},0\right),\qquad
V_{1}(x_{1})=\frac{c_{1}^{2}x_{1}^{2}}{2\gamma_{13}^{2}},\qquad
V_{2}(x_{2})=\frac{c_{1}^{2}x_{2}^{2}}{8\gamma_{13}^{2}}-\frac{v_{21}}{2x_{2}^{2}},$

which is the already known system (3.8).

####  A.3.4 $\boldsymbol{\alpha_{33}=\beta_{13}=\beta_{23}=0,$$\gamma_{12}\neq
0}$

The compatibilities (A.5)–(A.6) reduce to the sole equation

$\displaystyle\gamma_{12}u_{1}^{(3)}(x_{2})=0,$

therefore

$\displaystyle u_{1}(x_{2})=a_{1}x_{2}+a_{2}x_{2}^{2},\qquad
a_{j}\in\mathbb{R}.$

The compatibility conditions (4.8), together with the zero-order equation,
imply that either the magnetic field is constant and $V_{2}$ is a second-order
polynomial, or $V_{1}(x_{1})=u_{2}(x_{1})=0$ and $X_{2}$ reduces to a first-
order integral. Both these cases are of no interest here.

####  A.3.5 $\boldsymbol{\alpha_{33}\neq 0}$

By translation in $x_{1}$ and $x_{2}$ we can set $\beta_{13}=\beta_{23}=0$.
Thus, (A.6) gives

$\displaystyle u_{1}(x_{2})=\frac{a}{2x_{2}^{2}},\qquad a\in\mathbb{R}.$

By plugging this solution into (4.3) we obtain that it must be $c_{1}=0$ and
therefore by (A.5) $\gamma_{12}=0$. By looking for a solution of the equations
(4.8) and (4.4) we find that also $c_{0}=0$, i.e., $\\{X,p_{3}\\}=0$, and this
case is of no interest here.

###  A.4 $\boldsymbol{\beta_{33}=\gamma_{23}=\gamma_{13}=c_{1}=0}$

In this case equation (A.1) is trivially satisfied for both $u_{j}$. The
remaining second-order equations give for $S_{j}$:

$\displaystyle S_{1}(x_{1},x_{2})=Sx_{2}+s_{11},\qquad
S_{2}(x_{1},x_{2})=-Sx_{1}+s_{21},$ $\displaystyle
S_{3}(x_{1},x_{2})=-\frac{x_{1}}{2}(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+2\beta_{23}x_{2}-2\gamma_{12})u_{1}^{\prime}(x_{2})$
$\displaystyle\hphantom{S_{3}(x_{1},x_{2})=}{}-2x_{2}(\alpha_{33}x_{2}-\beta_{13})u_{2}(x_{1})+S_{31}(x_{2}),\qquad
S,s_{ij}\in\mathbb{R},$

where $S_{31}(x_{2})$ must solve

$\displaystyle\left(\alpha_{33}x_{1}^{2}x_{2}-\frac{\beta_{13}}{2}x_{1}^{2}+\beta_{23}x_{1}x_{2}-\gamma_{12}x_{1}\right)u_{1}^{\prime\prime}(x_{2})+3x_{1}(\alpha_{33}x_{1}+\beta_{23})u_{1}^{\prime}(x_{2})$
$\displaystyle\qquad{}-((\beta_{13}-2\alpha_{33}x_{2})x_{1}-\beta_{23}x_{2}+\gamma_{12})u_{2}^{\prime}(x_{1})$
$\displaystyle\qquad{}+2(2\alpha_{33}x_{2}-\beta_{13})u_{2}(x_{1})-S_{31}^{\prime}(x_{2})=0$

and $u_{j}$ satisfy the compatibilities (4.7), that in this case reduce to

$\displaystyle(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+\beta_{23}x_{2}-\gamma_{12})u_{1}^{\prime\prime\prime}(x_{2})-4(2\alpha_{33}x_{1}+\beta_{23})u_{1}^{\prime\prime}(x_{2})$
$\displaystyle\qquad{}+(2\alpha_{33}x_{1}+\beta_{23})u_{2}^{\prime\prime}(x_{1})+6\alpha_{33}u_{2}^{\prime}(x_{1})=0,$
(A.13) $\displaystyle
6\alpha_{33}u_{1}^{\prime}(x_{2})+(2\alpha_{33}x_{2}-\beta_{13})(u_{1}^{\prime}(x_{2})+8u_{2}^{\prime}(x_{1}))$
$\displaystyle\qquad{}-(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+\beta_{23}x_{2}-\gamma_{12})u_{2}^{\prime\prime\prime}(x_{1})=0.$
(A.14)

As in the above Appendix A.3, to solve (A.13)–(A.14) we have to distinguish
several subcases.

####  A.4.1 $\boldsymbol{\alpha_{33}\neq 0}$

By translation in $x_{1}$ and $x_{2}$ we can eliminate $\beta_{13}$ and
$\beta_{23}$. By taking the third-order derivatives
$\partial^{2}_{x_{1}}\partial_{x_{2}}$, $\partial^{2}_{x_{2}}\partial_{x_{1}}$
of (A.13) and (A.14), respectively we get

$\displaystyle
u_{2}(x_{1})=a_{22}x_{1}^{2}-\frac{a_{23}}{x_{1}^{2}}+a_{21}x_{1},\qquad
u_{1}(x_{2})=a_{12}x_{2}^{2}-\frac{a_{13}}{x_{2}^{2}}+a_{11}x_{2},\qquad
a_{ij}\in\mathbb{R}.$

By plugging the above solution into the third condition in (4.4) we see that,
as a polynomial in $x_{1}$ and $x_{2}$, it can be vanishing only if $c_{0}=0$.
Since also $c_{1}=0$, there is nothing new here.

####  A.4.2 $\boldsymbol{\alpha_{33}=0}$, $\boldsymbol{\beta_{13}\neq 0}$

By translation in $x_{1}$ we can assume $\gamma_{12}=0$. By considering the
first-order derivative of (A.13) with respect to $x_{2}$ and of (A.14) with
respect to $x_{1}$, we find

$\displaystyle u_{1}(x_{2})=a_{11}x_{2}+a_{12}x_{2}^{2}+a_{13}x_{2}^{3},\qquad
u_{2}(x_{1})=a_{21}x_{1}+a_{22}x_{1}^{2}+\frac{a_{23}}{x_{1}^{2}},\qquad
a_{ij}\in\mathbb{R}.$ (A.15)

We plug (A.15) into (A.13)–(A.14) and we obtain the conditions
$a_{23}=a_{13}=0$ and

$\displaystyle a_{22}\beta_{23}=0.$

If $\beta_{23}\neq 0$, then $a_{22}=0$. The equation for $m_{x_{3}}$ in (4.4),
together with (4.8), implies that $V_{j}$ are second-order polynomials and
$a_{12}=0$, i.e., the magnetic field is constant. This case is of no interest
here, cf. Section 8.

If $\beta_{23}=0$ we again consider the third equation in (4.4), together with
(4.8) and their first order and second order mixed derivatives with respect to
$x_{1}$ and $x_{2}$. If $a_{22}$ is not vanishing, we can find a solution for
both $V_{j}$ which is not a second-order polynomial (or lower) but only
assuming that $c_{0}=0$. Therefore nothing new arises in this case.

If $a_{22}=0$, we can find a solution for $V_{j}$ only if the magnetic field
is vanishing, or $c_{0}=0$, or the solution is equivalent to the system (3.8).
Anyway, not of interest here (we recall that in Appendix A.4 we have
$c_{1}=0$).

####  A.4.3 $\boldsymbol{\alpha_{33}=\beta_{13}=0}$,
$\boldsymbol{\beta_{23}\neq 0}$

We can eliminate $\gamma_{12}$ by translation in $x_{2}$, thus setting
$\gamma_{12}=0$.

The compatibility conditions (A.13), (A.14) imply that

$\displaystyle u_{1}(x_{2})=a_{12}x_{2}^{2}+\frac{a_{11}}{x_{2}^{2}},\qquad
u_{2}(x_{1})=-4a_{12}x_{1}^{2}+a_{21}x_{1},\qquad a_{ij}\in\mathbb{R}.$

By imposing that also the remaining equations (4.3), (4.4) and (4.5) are
satisfied we find that either $c_{0}=0$ which implies $\\{X,p_{3}\\}=0$ in
contradiction with our assumption, or we find $a_{11}=a_{12}=0$ and

$\displaystyle V_{1}(x_{1})=\frac{1}{2}a_{21}^{2}x_{1}^{2},\qquad
V_{2}(x_{2})=\frac{a_{21}^{2}}{8}x_{2}^{2}-\frac{v_{21}}{2x_{2}^{2}},$

i.e., a system equivalent to (3.8).

####  A.4.4 $\boldsymbol{\alpha_{33}=\beta_{13}=\beta_{23}=0}$,
$\boldsymbol{\gamma_{12}\neq 0}$

The compatibility conditions (A.13), (A.14) imply that

$\displaystyle u_{1}(x_{2})=a_{12}x_{2}^{2}+a_{11}x_{2},\qquad
u_{2}(x_{1})=-a_{12}x_{1}^{2}+a_{21}x_{1},\qquad a_{ij}\in\mathbb{R}.$

By imposing that also the remaining equations are satisfied we find that
necessarily $a_{12}=0$ and

$\displaystyle
V_{1}(x_{1})=\frac{1}{2}\big{(}a_{11}^{2}+a_{21}^{2}\big{)}x_{1}^{2}+v_{12}x_{1},\qquad
V_{2}(x_{2})=\frac{1}{2}\big{(}a_{11}^{2}+a_{21}^{2}\big{)}x_{2}^{2}+\frac{a_{21}v_{12}}{a_{11}}x_{2}.$

Thus this case was already studied in Section 8.

##  Appendix B Solution of the determining equations
in Case I for $\boldsymbol{\beta_{12}}$, $\boldsymbol{\beta_{21}}$ not both
vanishing

Here we distinguish between the cases

  * $(a)$

$\beta_{12}$ and $\beta_{21}$ both not vanishing;

  * $(b)$

$\beta_{12}\neq 0$ and $\beta_{21}=0$. Notice that, by a canonical permutation
of the variables, this case is equivalent to $\beta_{21}\neq 0$ and
$\beta_{12}=0$.

Case $(a)$ is treated in the following Appendix B.1. Case $(b)$ follows in
Appendix B.2.

###  B.1 $\boldsymbol{\beta_{12}\neq 0}$, $\boldsymbol{\beta_{21}\neq 0}$

Since $\beta_{12}$ and $\beta_{21}$ are both nonvanishing, by translation in
$x_{1}$ and $x_{2}$ we can set $\gamma_{13}=\gamma_{23}=0$.

Let us start from the compatibility conditions (4.7), that read

$\displaystyle(\beta_{33}x_{1}+\beta_{21}x_{2})u_{1}^{\prime\prime\prime}(x_{2})+4\beta_{21}u_{1}^{\prime\prime}(x_{2})=0,$
$\displaystyle\beta_{33}(u_{1}^{\prime\prime}(x_{2})+u_{2}^{\prime\prime}(x_{1}))=0,$
$\displaystyle(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+\beta_{23}x_{2}-\gamma_{12})u_{1}^{\prime\prime\prime}(x_{2})+4(2\alpha_{33}x_{1}+\beta_{23})u_{1}^{\prime\prime}(x_{2})$
$\displaystyle\qquad{}+(2\alpha_{33}x_{1}+\beta_{23})u_{2}^{\prime\prime}(x_{1})+6\alpha_{33}u_{2}^{\prime}(x_{1})=0,$
$\displaystyle(\beta_{12}x_{1}+\beta_{33}x_{2})u_{2}^{\prime\prime\prime}(x_{1})+4\beta_{12}u_{2}^{\prime\prime}(x_{1})=0,$
$\displaystyle\beta_{21}u_{2}^{\prime\prime}(x_{1})+\beta_{12}u_{1}^{\prime\prime}(x_{2})=0,$
$\displaystyle(2\alpha_{33}x_{1}x_{2}-\beta_{13}x_{1}+\beta_{23}x_{2}-\gamma_{12})u_{2}^{\prime\prime\prime}(x_{1})-4(\beta_{13}-2\alpha_{33}x_{2})u_{2}^{\prime\prime}(x_{1})$
$\displaystyle\qquad{}-(\beta_{13}-2\alpha_{33}x_{2})u_{1}^{\prime\prime}(x_{2})+6\alpha_{33}u_{1}^{\prime}(x_{2})=0.$
(B.1)

The above equations could be trivially satisfied for $u_{j}$ or not, depending
on the constants $\alpha_{ij}$, $\beta_{ij}$, $\gamma_{ij}$. This determines a
splitting in the computation.

####  B.1.1 $\boldsymbol{\beta_{33}\neq 0}$

The second equation in (B.1) implies

$\displaystyle u_{1}(x_{2})=\frac{a_{12}}{2}x_{2}^{2}+a_{11}x_{2},\qquad
u_{2}(x_{1})=-\frac{a_{12}}{2}x_{1}^{2}+a_{21}x_{1},\qquad
a_{ij}\in\mathbb{R}.$

By imposing that the above solution satisfies the remaining compatibilities
(B.1) and that the magnetic field does not vanish, we find that
$a_{12}=\alpha_{33}=0$, therefore the magnetic field is constant. The third
second-order equation simplifies to

$\displaystyle(\beta_{33}a_{11}+3\beta_{12}a_{21})x_{1}+(3\beta_{21}a_{11}+\beta_{33}a_{21})x_{2}=c_{1},$

which must hold for all values of $x_{1}$ and $x_{2}$. Without loss of
generality, we can assume that one component of the magnetic fields is not
vanishing, e.g., $a_{11}$. Thus the above equation implies

$\displaystyle\beta_{33}=-3\frac{\beta_{12}a_{21}}{a_{11}},\qquad\beta_{21}=\frac{a_{21}^{2}}{a_{11}^{2}}\beta_{12},\qquad
c_{1}=0.$ (B.2)

With this assumption the first and second-order equations (4.3) can be solved
for $S_{1}$ and $S_{2}$. We plug the so found solution into the third equation
in (4.4) and take its third-order derivatives
$\partial^{2}_{x_{1}}\partial_{x_{2}}$ and
$\partial^{2}_{x_{2}}\partial_{x_{1}}$. In this way we obtain the condition

$\displaystyle\beta_{12}a_{21}V_{j}^{\prime\prime\prime}(x_{j})=0,\qquad
j=1,2.$

Since the magnetic field is already constant, we do not consider here
solutions for $V_{j}$ in the form of at most quadratic polynomials. Thus
necessarily $a_{21}=0$. However, from (B.2) we have $\beta_{21}=0$, which
violates our assumption for this subcase.

####  B.1.2 $\boldsymbol{\beta_{33}=0}$

Equations (B.1) can be solved for nonvanishing magnetic field only if
$\alpha_{33}=0$. In this case we find the solution

$\displaystyle u_{1}(x_{2})=a_{1}x_{2},\qquad u_{2}(x_{1})=a_{2}x_{1},\qquad
a_{j}\in\mathbb{R},$

corresponding to constant magnetic field. The third second-order equation then
reduces to

$\displaystyle a_{1}\beta_{21}x_{2}+a_{2}\beta_{12}x_{1}=c_{1},$

which must hold for all $x_{1}$, $x_{2}$. Since $\beta_{12}$ and $\beta_{21}$
are assumed to be not vanishing in this section, we conclude that there is no
solution for nonvanishing magnetic field.

###  B.2 $\boldsymbol{\beta_{12}\neq 0}$, $\boldsymbol{\beta_{21}=0}$

Again, we start by the compatibilities (B.1), now simplified by the condition
$\beta_{21}=0$. We can proceed as above, considering first the case
$\beta_{33}\neq 0$. Then, by translation in $x_{1}$ and $x_{2}$ we can assume
$\gamma_{23}=\gamma_{13}=0$. We proceed as in Appendix B.1.1 with the
simplification $\beta_{21}=0$. Equation (B.2) implies that the magnetic field
has to vanish.

Therefore, we continue in the following by assuming $\beta_{33}=0$. Since
$\beta_{12}\neq 0$, by translation in $x_{1}$ we can still set
$\gamma_{13}=0$. The computation then splits into two major subcases,
according to whether $\gamma_{23}=0$ or not.

####  B.2.1 $\boldsymbol{\gamma_{23}\neq 0}$

The third second-order equation simplifies to

$\displaystyle
2\beta_{12}u_{2}(x_{1})+\gamma_{23}u_{1}^{\prime}(x_{2})+\beta_{12}x_{1}u_{2}^{\prime}(x_{1})=c_{1}.$
(B.3)

By solving for $u_{j}$ we find

$\displaystyle u_{1}(x_{2})=a_{1}x_{2},\qquad
u_{2}(x_{1})=\frac{a_{2}}{2x_{1}^{2}}+\frac{c_{1}-a_{1}\gamma_{23}}{2\beta_{12}},\qquad
a_{j}\in\mathbb{R}.$

The remaining second-order equations can be solved for $S_{j}$ only under the
condition
$a_{1}\beta_{23}=a_{2}\beta_{23}=a_{2}\gamma_{12}=a_{1}\alpha_{33}=0$. We find

$\displaystyle
S_{1}(x_{1},x_{2})=-s_{21}x_{2}-\frac{a_{2}\beta_{12}}{x_{1}}+s_{12},$
$\displaystyle
S_{2}(x_{1},x_{2})=\frac{a_{1}\beta_{12}}{2}x_{1}^{2}-a_{1}\gamma_{23}x_{2}+s_{21}x_{1}+\frac{a_{2}\gamma_{23}}{2x_{1}^{2}}+s_{22},$
$\displaystyle
S_{3}(x_{1},x_{2})=\frac{\beta_{13}a_{1}}{2}x_{1}^{2}+\gamma_{12}a_{1}x_{1}-\alpha_{33}a_{2}\frac{x_{2}^{2}}{x_{1}^{2}}+\beta_{13}a_{2}\frac{x_{2}}{x_{1}^{2}}.$

Equations (4.4) imply the following constraint on our integration constants:
$a_{2}s_{21}=0$. The compatibility conditions (4.8) can be solved for $V_{j}$
and together with (4.5) imply $a_{2}=0$. In order to have nonvanishing
magnetic field we thus must have $\alpha_{33}=\beta_{23}=0$. Equation (4.5)
further implies $s_{21}=s_{12}=0$ together with

$\displaystyle
V_{1}(x_{1})=\frac{1}{8}a_{1}^{2}x_{1}^{2}-\frac{v_{1}}{2x_{1}^{2}},$
$\displaystyle
V_{2}(x_{2})=\frac{1}{2}a_{1}^{2}x_{2}^{2}+\big{(}a_{1}^{2}\gamma_{23}^{2}-a_{1}c_{1}\gamma_{23}-2c_{0}\beta_{12}\big{)}\frac{x_{2}}{2\gamma_{23}\beta_{12}}+\frac{c_{0}x_{2}}{\gamma_{23}},$

i.e., we arrived at a system equivalent to (3.8) (after translation in
$x_{2}$).

####  B.2.2 $\boldsymbol{\gamma_{23}=0}$

Equation (B.3) does not contain $u_{1}$ anymore, while for $u_{2}$ implies

$\displaystyle
u_{2}(x_{1})=\frac{a_{2}}{x_{1}^{2}}+\frac{c_{1}}{2\beta_{12}},\qquad
a_{2}\in\mathbb{R}.$

From the last but one equation in (B.1) we see that

$\displaystyle u_{1}(x_{2})=a_{1}x_{2},\qquad a_{1}\in\mathbb{R},$

while the remaining equations (B.1) read

$\displaystyle
a_{2}\beta_{23}=0,\qquad\alpha_{33}a_{1}x_{1}^{5}-4a_{2}(\beta_{23}x_{2}-\gamma_{12})=0,$
(B.4)

i.e., $a_{2}\beta_{23}=\alpha_{33}a_{1}=a_{2}\gamma_{12}=0$.

The second-order equations (4.3) can be solved for $S_{1}$ and $S_{2}$. We
find

$\displaystyle
S_{1}(x_{1},x_{2})=Sx_{2}-\frac{2a_{2}\beta_{12}}{x_{1}}+s_{11},\qquad
S_{2}(x_{1},x_{2})=-Sx_{1}+\frac{1}{2}a_{1}\beta_{12}x_{1}^{2}+s_{21},\qquad
s_{ij}\in\mathbb{R}.$

The remaining second-order equations for $S_{3}$ imply that $\beta_{23}=0$,
otherwise they would imply that the magnetic field must vanish.

To proceed, we need to solve (B.4). We have to distinguish between the cases
$a_{2}=0$ and $a_{2}\neq 0$.

Let us start by assuming $a_{2}=0$. The compatibility conditions (4.8)
together with the zero-order equation (4.5) can be solved only if
$\alpha_{33}=\gamma_{12}=c_{1}=0$ and $S=s_{11}=s_{21}=0$. However, the
resulting system is equivalent to the already known maximally superintegrable
system (3.8), up to a permutation of the canonical variables.

If $a_{2}\neq 0$, then from (B.4) we have $\gamma_{12}=0$ and
$\alpha_{33}a_{1}=0$. Under these two conditions, also the remaining second-
order equations (4.3) can be solved for $S_{3}$. We find

$\displaystyle
S_{3}(x_{1},x_{2})=\frac{1}{2}a_{1}\beta_{13}x_{1}^{2}+2a_{2}\left(\beta_{13}\frac{x_{2}}{x_{1}^{2}}-\alpha_{33}\frac{x_{2}^{2}}{x_{1}^{2}}\right).$

The compatibility conditions for the first-order equations (4.8) can be solved
for $V_{1}$. They admit a solution for $V_{2}$ only if
$\alpha_{33}=\beta_{13}=0$ (and in this way also the remaining (B.4) is
satisfied). After solving for $V_{1}$ and imposing the previous condition, we
see that they are satisfied for any $V_{2}$. To find $V_{2}$ we look at the
zero-order equation, in which we impose all the conditions we obtained till
now and the solutions found for $V_{1}$ and $S_{j}$. We obtain in this way a
polynomial in $x_{1}$ (whose some coefficients contains equations for $V_{2}$)
that must vanish. By collecting the different powers of $x_{1}$ and impose
that they are all equal to zero, we arrive at the condition
$a_{2}\beta_{12}=0$, which cannot be satisfied in the case we are considering
here. Thus, no new system can be found.

##  Appendix C Solution of the determining equations for Case II

Let us start by some preliminary considerations. By taking second-order
derivatives with respect to $x_{2}$ of (4.4) and (4.5) we obtain the
conditions

$\displaystyle
s_{1}^{\prime\prime}(x_{2})U_{2}^{\prime\prime}(x_{1})=0\qquad\text{and}\qquad
s_{1}^{\prime\prime}(x_{2})W^{\prime}(x_{1})=0.$

If $s_{1}^{\prime\prime}\neq 0$, we therefore have
$U_{2}^{\prime\prime}=W^{\prime}=0$. As a consequence, new solutions can arise
here only for $U_{3}^{\prime\prime\prime}\neq 0$ (i.e., for nonconstant
magnetic field). If so, (7.5) imply $\beta_{33}=\beta_{31}=0$. However then
the compatibility conditions (4.8) cannot be solved for
$s_{1}^{\prime\prime}\neq 0$.

Thus, necessarily $s_{1}^{\prime\prime}=0$, which gives

$\displaystyle s_{1}(x_{2})=s_{11}+s_{12}x_{2}.$ (C.1)

Let us proceed by looking at (7.5) in case $\beta_{33}\neq 0$. Then
$U_{2}^{\prime\prime\prime}=0$ and the third equation in (7.5) can be solved
for $U_{3}$. Then the remaining equations imply that the magnetic field and
the potential $W$ are constant, which is not of interest here. Thus,
necessarily $\beta_{33}=0$ and (C.1) holds.

###  C.1 $\boldsymbol{\alpha_{33}\neq 0}$

By translation in both $x_{1}$ and $x_{2}$ we can set
$\beta_{13}=\beta_{23}=0$. The conditions (7.5) and (4.3) can be solved. We
find

$\displaystyle
U_{2}(x_{1})=\frac{a_{1}}{2x_{1}},\qquad\\!U_{3}(x_{1})=\frac{a_{2}}{2x_{1}},\qquad\\!s_{3}(x_{1})=0,\qquad\\!s_{2}(x_{1})=-s_{12}x_{1},\qquad\\!a_{1},a_{2},s_{12}\in\mathbb{R},$

together with $c_{2}=\beta_{21}=\beta_{31}=\gamma_{13}=\gamma_{12}=0$. This
give the solution for $S_{j}$, $j=1,2,3$. The compatibility conditions (4.8)
can then be solved for $W$ and by imposing also (4.5) give

$\displaystyle W(x_{1})=-\frac{a}{8x_{1}^{4}}+\frac{w}{x_{1}^{2}},\qquad
a,w\in\mathbb{R},$ (C.2)

together with $s_{11}=s_{12}=\beta_{12}=a_{2}=0$, where $a\equiv a_{1}$. The
magnetic field is

$\displaystyle\vec{B}(\vec{x})=\big{(}0,ax_{1}^{-3},0\big{)},\qquad
a\in\mathbb{R}.$ (C.3)

The first-order equations (4.8) give

$\displaystyle
M(x_{1},x_{2})=\alpha_{33}\left(\frac{2wx_{2}^{2}}{x_{1}^{2}}-\frac{a^{2}x_{2}^{2}}{2x_{1}^{4}}\right)$

with the condition $c_{0}=c_{1}=0$. This system is a special case of the
systems in Table 1, namely it can be expressed in the forms both
$\mathcal{E}_{1}$ and $\mathcal{E}_{2}$. The integral constructed here is
included in the ones coming from $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$,
i.e., the system is only minimally quadratically superintegrable.

From now on we continue our search by assuming $\alpha_{33}=0$.

###  C.2 $\boldsymbol{\alpha_{33}=0}$, $\boldsymbol{\beta_{23}\neq 0}$

By translation in $x_{2}$ we can set $\gamma_{12}=0$. Then the second and
fourth equation in (7.5) imply

$\displaystyle U_{2}(x_{1})=a_{11}x_{1}^{2},\qquad
U_{3}(x_{1})=a_{21}x_{1}^{2}+a_{22}x_{1}^{3},\qquad a_{ij}\in\mathbb{R}.$

From the compatibility conditions (4.8) and the zero-order equation (4.5) we
easily see that the only possibility is polynomial potential at most quadratic
and $a_{22}=0$ (and $a_{21}a_{11}=0$), i.e., constant magnetic field. Thus,
this is not of interest here.

###  C.3 $\boldsymbol{\alpha_{33}=\beta_{23}=0}$, $\boldsymbol{\beta_{31}\neq
0}$

Equation (7.4) reads

$\displaystyle(\beta_{12}x_{1}-\gamma_{13})U_{2}^{\prime\prime}(x_{1})+2\beta_{12}U_{2}^{\prime}(x_{1})-\beta_{31}U_{3}^{\prime}(x_{1})-c_{2}=0,$

which, integrated by $x_{1}$ and neglecting integration constants (which do
not affect the magnetic field), gives

$\displaystyle
U_{3}(x_{1})=\frac{1}{\beta_{31}}\big{(}(\beta_{12}x_{1}-\gamma_{13})U_{2}^{\prime}(x_{1})+\beta_{12}U_{2}(x_{1})+c_{2}x_{1}\big{)}.$
(C.4)

Conditions (4.8) together with (4.4) and (4.5) seen as a polynomial in $x_{2}$
imply that unless $s_{12}=0$ the magnetic field vanishes. Since this is of no
interest here, we continue in the following by assuming $s_{12}=0$. From the
second-order equations (4.3), considered as polynomials in $x_{2}$, we see
that necessarily $s_{3}(x_{1})=s_{2}(x_{1})=0$. We shall distinguish several
subcases.

####  C.3.1 $\boldsymbol{\beta_{12}=\gamma_{13}=0}$

Let us start by the easiest case in which $\beta_{12}=\gamma_{13}=0$.

By (C.4) we have here $U_{3}^{\prime\prime}=0$. Thus $U_{2}^{\prime\prime}$
cannot vanish, otherwise we have no magnetic field. From equations (4.3) we
see that necessarily $\beta_{21}=0$. At this point, all (4.3) are solved,
except for the condition

$\displaystyle(\beta_{13}x_{1}+\gamma_{12})U_{2}^{\prime\prime\prime}(x_{1})+(3\beta_{13}+\beta_{31})U_{2}^{\prime\prime}(x_{1})=0.$
(C.5)

The zero-order equation reads

$\displaystyle\left(s_{11}+\frac{c_{2}\gamma_{12}}{\beta_{31}}\right)W^{\prime}(x_{1})=0.$

For $s_{11}\neq-\frac{c_{2}\gamma_{12}}{\beta_{31}}$ the above condition
implies vanishing potential $W(x_{1})$. From (4.4) we see that then also
$U_{2}^{\prime\prime}$ is constant. Thus, this solution is of no interest
here.

Thus, let us set $s_{11}=-\frac{c_{2}\gamma_{12}}{\beta_{31}}$. In this case
the third equation in (4.4) is solved for
$c_{0}=-\frac{c_{1}c_{2}}{\beta_{31}}$. The remaining first-order equations
can be solved for $M$ and give

$\displaystyle
M(x_{1},x_{2})=(\beta_{13}x_{1}+\gamma_{12})x_{2}W^{\prime}(x_{1})$

for $W(x_{1})$ satisfying

$\displaystyle\big{(}W^{\prime\prime}(x_{1})+U_{2}^{\prime\prime}(x_{1})^{2}\big{)}(\beta_{13}x_{1}+\gamma_{12})+(c_{1}-\beta_{31}U_{2}^{\prime}(x_{1}))U_{2}^{\prime\prime}(x_{1})+3\beta_{13}W^{\prime}(x_{1})=0.$
(C.6)

Recall that $U_{2}$ needs to satisfy (C.5), whose solution depends on the
constants involved there. We can have

(a) $\boldsymbol{\beta_{13}=0}$. If $\gamma_{12}\neq 0$, (C.5) is solved by

$\displaystyle U_{2}(x_{1})=\frac{a}{b^{2}}{\rm e}^{bx_{1}},\qquad
a,b\in\mathbb{R},\qquad b=-\frac{\beta_{31}}{\gamma_{12}}\in\mathbb{R},$

giving a new superintegrable system

$\displaystyle\vec{B}(\vec{x})=\big{(}0,a{\rm e}^{bx_{1}},0\big{)},\qquad
W(\vec{x})=wx_{1}+ce^{bx_{1}}-\frac{1}{2}\frac{a^{2}}{b^{2}}{\rm
e}^{2bx_{1}},\qquad a,w\in\mathbb{R},$ (C.7)

where $c=-\frac{ac_{1}\gamma_{12}}{\beta^{2}_{31}}\in\mathbb{R}$.

For $\gamma_{12}=0$, equation (C.5) reduces to
$\beta_{31}U_{2}^{\prime\prime}(x_{1})=0$, implying
$U_{2}^{\prime\prime}(x_{1})=0$ and therefore no magnetic field.

(b) $\boldsymbol{\beta_{31}\neq-\beta_{13}\neq\frac{1}{2}\beta_{31}}$,
$\boldsymbol{\beta_{13}\neq 0}$. By translation in $x_{1}$ we can set
$\gamma_{12}=0$. After the substitution $\beta_{31}=-b\beta_{13}$, the
solution of (C.5) in this case reads

$\displaystyle U_{2}(x_{1})=\frac{ax_{1}^{b}}{b-1},\qquad
a\in\mathbb{R},\qquad b\neq 0,1,2.$ (C.8)

By solving (C.6) for $W$, we arrive at a new system determined by

$\displaystyle\vec{B}(\vec{x})=\big{(}0,a(b-2)x_{1}^{b-3},0\big{)},\qquad
W(\vec{x})=-\frac{a^{2}x_{1}^{2(b-2)}}{2}+a(b-2)cx_{1}^{b-2}+\frac{w}{x_{1}^{2}},$
(C.9)

where $a,b,w,c=\frac{c_{1}}{b(2-b)\beta_{13}}\in\mathbb{R}$, $b\neq 0,1,2$.

Notice that in the limit $b\rightarrow 0$ we reduce (C.9) to the system
determined by (C.2) and (C.3), that therefore can be seen as a special case of
(C.9) (which is already known to have two second-order integrals, with
highest-order terms $\big{(}l_{3}^{A}\big{)}^{2}$ and $p_{1}^{A}l_{3}^{A}$,
that are however dependent once also the Cartesian integrals (1.9) and the
Hamiltonian are taken into account).

(c) $\boldsymbol{\beta_{13}=-\frac{1}{2}\beta_{31}}$. Since in this case
$\beta_{13}=-\frac{1}{2}\beta_{31}\neq 0$, by translation in $x_{1}$ we can
still set $\gamma_{12}=0$. Equation (C.5) has solution

$\displaystyle U_{2}(x_{1})=ax_{1}\ln|x_{1}|,\qquad a\in\mathbb{R},$

that gives the new system

$\displaystyle\vec{B}(\vec{x})=\big{(}0,ax_{1}^{-1},0\big{)},\qquad
W(\vec{x})=-\frac{1}{2}a^{2}(\ln|x_{1}|)^{2}+b\ln|x_{1}|+\frac{w}{x_{1}^{2}},\qquad
a,w\in\mathbb{R},$ (C.10)

where $b=\frac{ac_{1}}{\beta_{31}}-a^{2}$.

(d) $\boldsymbol{\beta_{13}=-\beta_{31}}$. As above, we can still set
$\gamma_{12}=0$ by translation. Equation (C.5) admits the solution

$\displaystyle U_{2}(x_{1})=-a\ln|x_{1}|,\qquad a\in\mathbb{R}.$ (C.11)

By denoting $c=-\frac{c_{1}a}{\beta_{13}}$, from (C.6) and (C.11) we obtain

$\displaystyle\vec{B}(\vec{x})=\big{(}0,ax_{1}^{-2},0\big{)},\qquad
W(\vec{x})=\frac{c}{x_{1}}+\frac{w}{x_{1}^{2}},\qquad a,c,w\in\mathbb{R}.$
(C.12)

Thus, though (C.8) is singular for $b=1$ (and we obtain a different solution
(C.11) for (C.5) if $b=1$), indeed such a singularity does not appear in the
solution for $\vec{B}$ and $W$ (it is lost by differentiating) and the system
(C.9) becomes (C.12) for $b=1$.

####  C.3.2 $\boldsymbol{\gamma_{13}\neq 0}$, $\boldsymbol{\beta_{12}=0}$

In this case it is convenient to start by looking at the zeroth-order equation
(4.5), that simplifies to

$\displaystyle\big{(}\gamma_{13}\big{(}(\beta_{13}x_{1}+\gamma_{12})U_{2}^{\prime\prime}(x_{1})-(\beta_{13}-\beta_{31})\big{)}U_{2}^{\prime}(x_{1})+s_{11}\beta_{31}+c_{2}\gamma_{12}\big{)}W^{\prime}(x_{1})=0.$
(C.13)

For $W^{\prime}\neq 0$, we solve the above equation for $U_{2}$. Once we plug
the so found solution into the remaining equations we see that there can be
several subcases according to different values of the constants involved.
However, in most cases, the solution for $U_{2}$ is polynomial and at most of
first order. By (C.4) also $U_{3}$ results into a polynomial of order at most
one. Thus in these cases we have vanishing magnetic field. The only exception
is the solution

$\displaystyle
U_{2}\left(x_{1}\right)=\frac{2\gamma_{12}a^{2}\big{(}b_{2}^{2}+\frac{1}{2}\big{)}{e}^{b_{1}x_{1}}-\big{(}(2b_{1}b_{2}s_{11}+c_{1})a+w_{2}b_{1}^{2}\gamma_{12}\big{)}b_{1}x_{1}}{b_{1}^{2}\gamma_{12}a\big{(}2b_{2}^{2}+1\big{)}},$
$\displaystyle\vec{B}(\vec{x})=(0,1,b_{2})\cdot a{\textrm{e}}^{b_{1}x_{1}},$
$\displaystyle
W(\vec{x})=-\frac{\big{(}b_{2}^{2}+1\big{)}}{2b_{1}^{2}}a^{2}{\textrm{e}}^{2b_{1}x_{1}}+w_{2}{\textrm{e}}^{b_{1}x_{1}}+x_{1}w_{1},$
$\displaystyle
M(x_{1},x_{2})=-\frac{\gamma_{12}}{b_{1}}\big{(}a^{2}\big{(}b_{2}^{2}+1\big{)}{\textrm{e}}^{2b_{1}x_{1}}-w_{2}b_{1}^{2}{\textrm{e}}^{b_{1}x_{1}}-w_{1}b_{1}\big{)}x_{2},$
(C.14)

where $s_{12}=0$, $\beta_{13}=0$, $\beta_{21}=b_{1}b_{2}\gamma_{12}$,
$\beta_{31}=-b_{1}\gamma_{12}$, $\gamma_{13}=b_{2}\gamma_{12}$,
$c_{0}=\gamma_{12}b_{2}w_{1}+\frac{b_{1}^{3}\gamma_{12}w_{2}^{2}b_{2}^{3}}{a^{2}(2b_{2}^{2}+1)^{2}}-\frac{b_{1}b_{2}(2b_{1}b_{2}s_{11}+c_{1})w_{2}}{a(2b_{2}^{2}+1)^{2}}+\frac{(b_{1}s_{11}-b_{2}c_{1})(b_{1}b_{2}s_{11}+b_{2}^{2}c_{1}+c_{1})}{b_{1}\gamma_{12}(2b_{2}^{2}+1)^{2}}$
and
$c_{2}=\frac{ab_{1}s_{11}-w_{2}b_{1}^{2}b_{2}\gamma_{12}-ab_{2}c_{1}}{a\left(2b_{2}^{2}+1\right)}$.
By rotation of the coordinate frame around the $x_{1}$-axis, which preserves
the existence of the Cartesian integrals $p_{2}$ and $p_{3}$, the system
(C.14) can be brought to the form (C.7).

For $W^{\prime}=0$ the equation (C.13) does not give any condition on $U_{2}$,
that is therefore constrained only by the remaining second and first-order
equations (4.3), (4.4). However, also in this case we always arrive to a
solution for $U_{2}$ at most linear in $x_{1}$, except for
$U_{2}=\frac{1}{b^{2}}(a_{1}\sin(bx_{1})+a_{2}\cos(bx_{1}))$ that leads to the
superintegrable system already found in [16], whose magnetic field has
components

$\displaystyle B_{1}=0,\qquad B_{2}=a_{1}\sin(bx_{1})+a_{2}\cos(bx_{1}),\qquad
B_{3}=a_{2}\sin(bx_{1})-a_{1}\cos(bx_{1}),$

where $a_{1},a_{2}\in\mathbb{R}$ and $b=\frac{\beta_{31}}{\gamma_{13}}$ (which
must for this case be also equal to $b=\frac{\beta_{21}}{\gamma_{12}}$).

####  C.3.3 $\boldsymbol{\beta_{12}\neq 0}$

By translation in $x_{1}$, let us set $\gamma_{13}=0$. We can proceed as above
and, after a long and tedious computation, arrive at two solutions

  * •

if $W$ is not identically zero, we have

$\displaystyle
U_{2}(x_{1})=\frac{1}{b_{1}-1}\left(ax_{1}^{b_{1}-1}+\frac{b_{1}s_{11}}{b_{2}\beta_{13}}\right),$
$\displaystyle\vec{B}(\vec{x})=(0,1,b_{2})\cdot a(b_{1}-2)x_{1}^{b_{1}-3},$
$\displaystyle
W(\vec{x})=-\frac{a^{2}}{2}\big{(}b_{2}^{2}+1\big{)}x_{1}^{2(b_{1}-2)}+w_{2}x_{1}^{b_{1}-2}+\frac{w_{1}}{x_{1}^{2}},$
$\displaystyle
M(x_{1},x_{2})=\beta_{13}x_{2}\left((b_{1}-2)\big{(}w_{2}x_{1}^{b_{1}}-a^{2}\big{(}b_{2}^{2}+1\big{)}x_{1}^{2(b_{1}-2)}\big{)}-2\frac{w_{1}}{x_{1}^{2}}\right),$
(C.15)

where
$c_{0}=-\frac{c_{2}}{a}w_{2}-\frac{(b_{1}-1)b_{2}c_{2}^{2}}{b_{1}^{2}\beta_{13}}$,
$c_{1}=-2(b_{1}-1)\frac{b_{2}}{b_{1}}c_{2}-\frac{b_{1}}{a}\beta_{13}w_{2}$,
$s_{12}=0$, $\beta_{21}=b_{1}b_{2}\beta_{13}$, $\beta_{12}=-b_{2}\beta_{13}$,
$\beta_{31}=-b_{1}\beta_{13}$ and $\gamma_{12}=0$. By rotation of the
coordinate frame around the $x_{1}$-axis, which preserves the existence of the
Cartesian integrals $p_{2}$ and $p_{3}$, the system (C.15) can be brought to
the form (C.9).

  * •

For $W(\vec{x})=0$ we find

$\displaystyle
U_{2}(x_{1})=-a\ln|x_{1}|,\qquad\vec{B}(\vec{x})=(0,1,b)\cdot\frac{a}{x_{1}^{2}},\qquad
W(\vec{x})=0,\qquad M(x_{1},x_{2})=0,$

where $c_{0}=0$, $c_{1}=0$, $s_{11}=-ab\beta_{31}$, $\beta_{13}=-\beta_{31}$,
$\beta_{21}=-b\beta_{31}$ and $\gamma_{12}=0$. However, this solution can be
obtained as a limiting case of (C.15) with $b_{1}=1$, $b_{2}=b$, $w_{2}=0$ and
$w_{1}=\frac{a^{2}(b_{2}^{2}+1)}{2}$.

###  C.4 $\boldsymbol{\alpha_{33}=\beta_{23}=\beta_{31}=0}$

Equation (7.4) now reads

$\displaystyle(\beta_{12}x_{1}-\gamma_{13})U_{2}^{\prime\prime}(x_{1})+2\beta_{12}U_{2}^{\prime}(x_{1})-c_{2}=0.$
(C.16)

Thus, it does not imply any relationship between $U_{2}$ and $U_{3}$. It only
gives a condition on $U_{2}$ that could be trivially satisfied or not,
according to the values of the constants involved. By the same argument used
in Appendix C.3, we see that also in this case $s_{12}=0$ and
$s_{3}(x_{1})=s_{2}(x_{1})=0$.

####  C.4.1 $\boldsymbol{\beta_{21}\neq 0}$

Let us look at the situation in which $\beta_{21}\neq 0$ and recall that we
are looking for an integral $X$ of the form (7.1). Its Poisson bracket with
$X_{1}=p_{2}$, i.e., $\\{X,p_{2}\\}$, is again an integral, by assumption
expressible in terms of the known integrals, similarly to equation (7.1).
However, these two expressions can be interchanged under the permutation of
$x_{2}$ and $x_{3}$; thus we can reduce the considered problem into an already
discussed one. We have that

$\displaystyle(p_{1},p_{2},p_{3},l_{1},l_{2},l_{3})\rightarrow(p_{1},p_{3},p_{2},-l_{1},-l_{3},-l_{2}),$

while the system (1.7) transforms into

$\displaystyle W(x_{1})\rightarrow
W(x_{1}),\qquad(u_{2},u_{3})\rightarrow(-u_{3},u_{2}).$ (C.17)

Therefore, all the conditions imposed in this section (including
$\beta_{21}\neq 0$) on the coefficients of the second-order terms in the
integral change into $\alpha_{22}=\beta_{32}=\beta_{21}=\beta_{22}=0$,
$\beta_{31}\neq 0$ and (7.2) turns into $a_{0}=-\beta_{13}$, $a_{1}=0$,
$a_{2}=\beta_{31}$, $a_{3}=\beta_{21}$ and

$\displaystyle\alpha_{11}=\alpha_{12}=\alpha_{13}=\alpha_{23}=\alpha_{33}=\beta_{11}=\beta_{33}=\beta_{23}=0.$

Thus we recover all the conditions imposed on our parameters in Appendix C.3.
By (C.17) any system belonging into this section would transform into some
system already found in Appendix C.3. Therefore in the following we can assume
that $\beta_{21}=0$ without lost of generality.

We distinguish several subcases depending on whether (C.16) is trivially
satisfied or not.

####  C.4.2 $\boldsymbol{\beta_{21}=0}$, $\boldsymbol{\beta_{12}\neq 0}$

By translation in $x_{1}$ we can set $\gamma_{13}=0$. Equation (C.16) gives

$\displaystyle
U_{2}(x_{1})=\frac{a_{1}}{2x_{1}}+\frac{c_{2}}{2\beta_{12}}x_{1},\qquad
a_{1}\in\mathbb{R}.$

We look at the remaining second-order equations (4.3), that read

$\displaystyle(\beta_{13}x_{1}+\gamma_{12})U_{3}^{\prime\prime\prime}(x_{1})+3\beta_{13}U_{3}^{\prime\prime}(x_{1})=0,\qquad\beta_{12}x_{1}^{5}U_{3}^{\prime\prime\prime}(x_{1})+3\beta_{12}x_{1}^{4}U_{3}^{\prime\prime}(x_{1})-3\gamma_{12}a_{1}=0.$

If $\beta_{13}=0$ and $\gamma_{12}\neq 0$ they have a solution only if
$a_{1}=0$ and $U_{3}=0$, i.e., the magnetic field vanishes.

If $\beta_{13}=\gamma_{12}=0$, the above equations admit solution for

$\displaystyle U_{3}(x_{1})=\frac{a_{2}}{x_{1}}.$

The zeroth-order equation (4.5) has solution for $W(x_{1})=0$. If
$\beta_{12}\neq 0$ the remaining equations (4.4) lead to the solution

$\displaystyle U_{2}(x_{1})=0,\qquad
U_{3}(x_{1})=\frac{a}{2x_{1}},\qquad\vec{B}(\vec{x})=\left(0,0,\frac{a}{x_{1}^{3}}\right),$
$\displaystyle
W(\vec{x})=-\frac{ab\ln|x_{1}|}{x_{1}^{2}}-\frac{a^{2}}{8x_{1}^{4}}+\frac{w}{x_{1}^{2}},$
(C.18)

where $a_{1}=0$, $c_{0}=0$, $c_{1}=2\beta_{12}b$, $c_{2}=0$, $s_{11}=0$,
$\beta_{13}=0$, $\gamma_{12}=0$ and $M(x_{1},x_{2})=0$. This system is new
when $b\neq 0$, otherwise it can be turned into a subcase of (C.9) by
permutation of $x_{2}$ and $x_{3}$.

If $\beta_{13}\neq 0$ we can set $\gamma_{12}=0$. We have a solution for
$U_{3}$ given by

$\displaystyle U_{3}(x_{1})=\frac{4a_{2}}{\beta_{13}x_{1}},\qquad
b\in\mathbb{R}.$

Equations (4.4) and (4.5) can be solved for $W$ and $M$ under the conditions
$s_{11}=0$ and $a_{2}=-\frac{a_{1}\beta_{12}}{8}$. We find

$\displaystyle
M(x_{1},x_{2})=\beta_{13}\big{(}b^{2}+1\big{)}a^{2}\frac{x_{2}}{2x_{1}^{4}}-\beta_{13}c\frac{2\ln|x_{1}|-1}{x_{1}^{2}}-2\beta_{13}w\frac{x_{2}}{x_{1}^{2}}$

and

$\displaystyle
W(x_{1})=-\frac{\big{(}b^{2}+1\big{)}a^{2}}{8x_{1}^{4}}+\frac{c\ln|x_{1}|+w}{x_{1}^{2}},\qquad
w\in\mathbb{R},$ (C.19)

where $b=-\frac{\beta_{12}}{\beta_{13}}\neq 0$, $a=a_{1}$,
$c=\frac{ac_{1}}{2\beta_{13}}$ and by using the physically irrelevant shift of
the potential by an additive constant we set $c_{0}=c_{2}=0$.

The magnetic field reads

$\displaystyle\vec{B}(\vec{x})=(0,1,b)\cdot ax_{1}^{-3},\qquad
a,b\in\mathbb{R},\qquad b\neq 0.$ (C.20)

However, the reference frame can be rotated around $x_{1}$ without affecting
the fact that $p_{2}$ and $p_{3}$ are integrals. Such a rotation brings the
magnetic field (C.20) and the potential (C.19) and the integral into the form
(C.18).

####  C.4.3 $\boldsymbol{\beta_{21}=\beta_{12}=0}$,
$\boldsymbol{\gamma_{13}\neq 0}$

From equation (C.16) we have

$\displaystyle U_{2}(x_{1})=-\frac{c_{2}x_{1}^{2}}{2\gamma_{13}}.$

From the second and zeroth-order equations we get

$\displaystyle U_{3}(x_{1})=ax_{1}^{2},\qquad a\in\mathbb{R}.$

The first-order equations can be solved for $W$ and give a polynomial solution
at most quadratic. Since the magnetic field is constant, nothing of interest
can be found here.

####  C.4.4 $\boldsymbol{\beta_{21}=\beta_{12}=\gamma_{13}=0}$

Equation (C.16) implies $c_{2}=0$ and it is trivially satisfied for $U_{2}$.
The remaining second-order equations read

$\displaystyle\beta_{13}\big{(}3U_{j}^{\prime\prime}(x_{1})+x_{1}U_{j}^{\prime\prime\prime}(x_{1})\big{)}+\gamma_{12}U_{j}^{\prime\prime\prime}(x_{1})=0,\qquad
j=2,3.$

If $\beta_{13}=0$ we easily conclude that the only possibility is constant
magnetic field and vanishing potential. For $\beta_{13}\neq 0$ we can set
$\gamma_{12}=0$ by translation in $x_{1}$. The above equations imply

$\displaystyle U_{j}(x_{1})=\frac{a_{j}}{2x_{1}},\qquad j=2,3.$

From the first-order equations, seen as polynomials in $x_{2}$, we get the
conditions

$\displaystyle\frac{a_{2}^{2}}{x_{1}^{5}}+\frac{a_{3}^{2}}{x_{1}^{5}}-2W^{\prime}(x_{1})=0,\qquad
a_{2}(2a_{3}\beta_{13}+3s_{11}x_{1})+c_{1}a_{3}x_{1}^{2}=0,$

together with two differential equations for $M$. From the second equation
above we see that we have two possibilities for nonvanishing magnetic field:
$a_{2}=c_{1}=0$ or $a_{3}=s_{11}=0$. In the first case, when imposing also the
remaining zero-order equation (4.5) we get that necessarily $a_{3}=0$, i.e.,
the magnetic field vanishes. In the second case we have the solution

$\displaystyle
W(x_{1})=\frac{c\ln|x_{1}|+w}{x_{1}^{2}}-\frac{a^{2}}{8x_{1}^{4}},\qquad
a,c,w\in\mathbb{R},$

with $c=\frac{ac_{1}}{2\beta_{13}}$, $a=a_{2}$. The magnetic field reads

$\displaystyle\vec{B}(\vec{x})=\big{(}0,ax_{1}^{-3},0\big{)},\qquad
a\in\mathbb{R}.$

By solving the remaining equations for $M$, we find

$\displaystyle
M(x_{1},x_{2})=a^{2}\beta_{13}\frac{x_{2}}{2x_{1}^{4}}-2\beta_{13}c\ln|x_{1}|\frac{x_{2}}{x_{1}^{2}}+\beta_{13}(c-2w)\frac{x_{2}}{x_{1}^{2}}.$

However, this system is just a special case of (C.19), (C.20) for $b=0$.

### Acknowledgments

This paper was supported by the Czech Science Foundation (Grant Agency of the
Czech Republic), project 17-11805S. This paper is dedicated to our son Flavio
born just after the submission of the original manuscript.

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