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# Graded colour Lie superalgebras for solving Lévy-Leblond equations

Mitchell Ryan  School of Mathematics and Physics, University of Queensland,
St. Lucia, QLD 4072, Australia
ORCID: 0009-0006-2038-4410<EMAIL_ADDRESS>
###### Abstract.

The Lévy-Leblond equation with free potential admits a symmetry algebra that
is a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra (see
Aizawa–Kuznetsova–Tanaka–Toppan, 2016). We extend this result in two
directions by considering a time-independent version of the Lévy-Leblond
equation. First, we construct a $\mathbb{Z}_{2}^{3}$-graded colour Lie
superalgebra containing operators that leave the eigenspaces invariant and
demonstrate the utility of this algebra in constructing general solutions for
the free equation. Second, we find that the ladder operators for the harmonic
oscillator generate a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie
superalgebra and we use the operators from this algebra to compute the
spectrum. These results illustrate two points: the Lévy-Leblond equation
admits colour Lie superalgebras with gradings higher than
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and colour Lie superalgebras appear for
potentials besides the free potential.

###### Key words and phrases:

Color Lie (super)algebras, Graded Lie (super)algebras, Lévy-Leblond equation

###### 2020 Mathematics Subject Classification:

17B75, 17B70, 81Q05

##  1\. Introduction

The Lévy-Leblond equation is the non-relativistic limit of the Dirac equation
[1]. Dirac introduced his equation to describe a relativistic spin-$1/2$
particle and was the first quantum-mechanical equation to account for both
spin and special relativity [2]. To obtain his equation, Dirac took a ‘square
root’ of the Klein–Gordon equation (which describes a relativistic spin-$0$
particle) in order to replace the second-order time derivative with a first-
order derivative [2]. This square root required the introduction of gamma
matrices (which generate Clifford algebras) and naturally led to the
introduction of spin into the equation.

In a similar way, Lévy-Leblond [1] obtained his equation as a ‘square root’ of
the non-relativistic Schrödinger equation. Similar to the Dirac equation, this
square root required the introduction of gamma matrices and spin. Lévy-Leblond
showed that his equation is invariant under the Galilei group, predicts the
correct value for the magnetic moment of a spin-$1/2$ particle and can be
obtained as the non-relativistic limit of the Dirac equation [1].

Remarkably, it has since been shown that the Lévy-Leblond equation has
symmetry algebras which are $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour
Lie superalgebras [3, 4] . The authors of [3, 4] were motivated by the fact
that ungraded Schrödinger operators can give rise to $\mathbb{Z}_{2}$-graded
Lie superalgebras (see [5]); for example, the ladder operators for the quantum
harmonic oscillator generate a $\mathbb{Z}_{2}$-graded Lie superalgebra
$\mathfrak{osp}(1|2)$. With this motivation [3], it was investigated whether
replacing the ungraded Schrödinger operator with the $\mathbb{Z}_{2}$-graded
Lévy-Leblond operator would yield symmetry with a second
$\mathbb{Z}_{2}$-gradation, making the symmetry algebras
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded. Such
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded symmetry algebras did appear for
the free equation [3, 4] in the form of a colour Lie superalgebra. However,
this $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded symmetry disappeared when
considering the harmonic potential [3].

Colour Lie (super)algebras were first introduced in [6, 7] (however, see also
[8]) and generalise Lie superalgebras to grading by any abelian group. Since
their application to the Lévy-Leblond equation [3, 4], there has been an
increase in research activity surrounding applications of colour Lie
(super)algebras. For instance, colour Lie (super)algebras have been applied to
generalised quantum mechanics [9, 10, 11, 12, 13, 14, 15, 16, 17, 18] and
parastatistics [19, 20, 21, 22, 23, 24, 25, 26, 27, 28]. There also exists a
largely independent line of research focussing on graded (non-colour) Lie
algebras—see [29, 30] for applications of results for such graded (non-colour)
Lie algebras to the study of colour Lie (super)algebras.

Despite the recent activity surrounding colour Lie (super)algebras, the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded nature of the Lévy-Leblond
equation has received little attention since the first two papers [3, 4] (see
[31, 32] for recent work). This paper addresses two questions raised by the
work in [3, 4]:

  1. (1)

Do colour Lie superalgebras with gradings different from
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ appear?

  2. (2)

Do colour Lie superalgebras appear for potentials besides the free potential?

We demonstrate an affirmative answer to both questions. However, unlike [3,
4], we will not be focussed on symmetry algebras. Instead, we search for
algebras that aid in solving the time-independent equation.

The time independent equation is of the form
$H_{\textup{LL}}\ket{\psi}=\gamma_{+}E\ket{\psi}$. Here, $\gamma_{+}$ is a
gamma matrix, $E\in\mathbb{C}$, $\ket{\psi}$ is a state in the Hilbert space
and $H_{\textup{LL}}$ is a particular matrix differential operator which we
treat as if it were a Hamiltonian. Of particular note is the fact that
$\gamma_{+}$ is not invertible, and we are thus forced to leave $\gamma_{+}$
on the right-hand side of the equation. Regardless, we proceed with the
analysis as normal, and call $E$ a ‘$\gamma_{+}$-eigenvalue’.

For question 1, we examine a time-independent version of the $1+1$-dimensional
Lévy-Leblond equation with free potential. After solving a differential
equation, we find five linearly independent operators which leave the
$\gamma_{+}$-eigenspaces invariant: the identity, a gamma matrix multiplied by
the parity operator, the Schrödinger Hamiltonian, and two different ‘square
roots’ of the Schrödinger Hamiltonian. These five operators close to form a
$\mathbb{Z}_{2}^{3}$-graded colour Lie superalgebra $\mathfrak{D}$. We show
that this $\mathbb{Z}_{2}^{3}$-graded colour Lie superalgebra is fundamental
to the Lévy-Leblond equation: solutions to the Lévy-Leblond equation can be
expressed in terms of simultaneous $\gamma_{+}$-eigenstates of the two ‘square
roots’ of the Schrödinger Hamiltonian. As an illustration, we use this fact to
solve the Lévy-Leblond equation in this simple case.

To answer question 2, we take a slightly different approach from [3] and
introduce the harmonic potential in a different form. We then set up a
$\gamma_{+}$-eigenvalue equation and search for ladder operators. We find that
the ladder operators with the Lévy-Leblond Hamiltonian generate a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra. This
colour Lie superalgebra is the direct analogue of the $\mathfrak{osp}(1|2)$
superalgebra for the Schrödinger harmonic oscillator but, in the case of the
Lévy-Leblond equation, an extra $\mathbb{Z}_{2}$ gradation is required. We
then use the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie
superalgebra to compute the spectrum.

This paper is organised as follows: in Section 2 we introduce the Lévy-Leblond
equation and set up a time-independent version of the equation. In Section 3,
we construct the $\mathbb{Z}_{2}^{3}$-graded colour Lie superalgebra of
$\gamma_{+}$-eigenspace-preserving operators of the $(1+1)$-dimensional free
equation. In Section 4, we construct the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra generated
by the ladder operators of the $(1+1)$-dimensional Lévy-Leblond harmonic
oscillator. In Appendix A, we show how the gamma matrix relations used in this
paper can be obtained when considering the Lévy-Leblond equation as the non-
relativistic limit of the Dirac equation.

##  2\. The Lévy-Leblond equation and colour Lie superalgebras

###  2.1. Free equation

For the form of the free Lévy-Leblond equation, we will follow [3]. A Lévy-
Leblond equation is a first-order differential equation that is a square root
of the heat or Schrödinger equation in $(1+d)$-dimensions [3]. The goal of the
Lévy-Leblond equation is to reduce a second-order differential equation to a
first-order equation.

Let $t$ be the time coordinate, let $x_{j}$ be the $j$-th spacial coordinate
and take $\lambda\in\mathbb{C}$. Throughout this paper, we will use the
notation $\mathbf{x}=(x_{1},\ldots,x_{d})$ and
$\partial_{t}=\partialderivative{t}$,
$\partial_{j}=\partialderivative{x_{j}}$.The Lévy-Leblond operator (with free
potential) is

(1) $\Omega=\gamma_{+}\partial_{t}+\gamma_{-}\lambda+\gamma^{j}\partial_{j}$

where Einstein summation convention is used, and the coefficients
$\gamma_{+},\gamma_{-},\gamma^{j}\>$ are gamma matrices satisfying

(2) $\displaystyle\anticommutator{\gamma_{\pm}}{\gamma_{\pm}}$
$\displaystyle=0,$ $\displaystyle\anticommutator{\gamma_{+}}{\gamma_{-}}$
$\displaystyle=I,$ $\displaystyle\anticommutator{\gamma_{\pm}}{\gamma^{j}}$
$\displaystyle=0,$ $\displaystyle\anticommutator{\gamma^{j}}{\gamma^{k}}$
$\displaystyle=2\delta_{jk},$

(with $\delta_{jk}$ the Kronecker delta).

Using (2), the square of the Lévy-Leblond operator is

$\Omega^{2}=\lambda\partial_{t}+\Delta$

where $\Delta=\sum_{j}\partial_{j}^{2}$ is the Laplacian. If $\lambda$ is a
negative real number, then the partial differential equation induced by
$\Omega^{2}$ (i.e. $\Omega^{2}\Psi(t,\mathbf{x})=0$) becomes the heat
equation, and if $\lambda=i\beta$ for $\beta\in\mathbb{R}_{>0}$ (in
particular, $\beta=2m/\hbar$ where $m$ is mass and $\hbar$ is the reduced
Planck constant) then the partial differential equation induced by
$\Omega^{2}$ becomes the Schrödinger equation.

The gamma matrices in the Lévy-Leblond equation belong to some Clifford
algebra. The most general Clifford algebra required is
$C\ell_{2,d}(\mathbb{R})\otimes\mathbb{C}$. However, as in [3, Section 4], we
require that the gamma matrices for the $(1+1)$-dimensional equation also
satisfy the following relations

(3)
$(\gamma_{\pm})^{2}=0,\qquad(\gamma^{1})^{2}=I,\qquad\gamma_{\pm}\gamma_{\mp}=\frac{1}{2}(I\pm\gamma^{1}),\qquad\gamma^{1}\gamma_{\pm}=\pm\gamma_{\pm}=-\gamma_{\pm}\gamma^{1}.$

In this case, the gamma matrices can be chosen as elements of the Clifford
algebra $C\ell_{1,1}(\mathbb{R})\otimes\mathbb{C}$. More generally, we argue
in Appendix A that $C\ell_{1,d}(\mathbb{R})\otimes\mathbb{C}$ is the most
suitable Clifford algebra when $d$ is odd.

######  Remark 2.1.

We could work entirely with real matrices, and realise the imaginary unit as a
real matrix $J$ which commutes with all the gamma matrices and squares to the
negative identity matrix. This would allow us to realise the gamma matrices as
a representation of a _real_ Clifford algebra, and is the approach taken in
[3]. However, for simplicity, we will allow our gamma matrices to be over the
complex numbers.

To actually solve the Lévy-Leblond equation, we would need to choose a
specific matrix representation of the corresponding Clifford algebra (though
many of the properties that we will study are representation independent).
This makes the Lévy-Leblond equation a matrix differential equation. In
particular, if we realise relations (2) using $n\times n$-dimensional
matrices, the corresponding Hilbert space of the quantum system is
$L^{2}(\mathbb{R})\otimes\mathbb{C}^{n}$, whose elements we interpret as
$n$-component vectors of functions. This interpretation provides a natural
action of the Lévy-Leblond operator on the Hilbert space. For example, in
$(1+1)$-dimensions, such a representation (as used in [3]) is

(4) $\gamma^{1}=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix},\qquad\gamma_{+}=\begin{pmatrix}0&1\\\
0&0\end{pmatrix},\qquad\gamma_{-}=\begin{pmatrix}0&0\\\ 1&0\end{pmatrix}$

which acts on the Hilbert space

$L^{2}(\mathbb{R})\otimes\mathbb{C}^{2}=\left\\{\begin{pmatrix}f\\\
g\end{pmatrix}\mid f,g\in L^{2}(\mathbb{R})\right\\}$

(note that this representation satisfies (3)). If $\Psi(t,\mathbf{x})$ is a
solution to the free Lévy-Leblond equation (i.e.
$\Omega\Psi(t,\mathbf{x})=0$), then we also have
$\Omega^{2}\Psi(t,\mathbf{x})=0$, so each component of $\Psi(t,\mathbf{x})$
will be a solution to the Schrödinger equation.

###  2.2. Lévy-Leblond operator with potential

Up to this point, we have only been considering the free equation. There are
(at least) two ways to add a potential function to the Lévy-Leblond operator.

From a supersymmetry perspective, a natural choice for the Lévy-Leblond
operator with potential is the supercharge

(5)
$\Omega=\gamma_{+}\partial_{t}+\gamma_{-}\lambda+\gamma^{j}\partial_{j}+\gamma_{f}f(\mathbf{x})$

for some additional gamma matrix $\gamma_{f}$ that satisfies additional
anticommutation relations (omitted; see [3]). These anticommutation relations
for $\gamma_{f}$ ensure that the square of the Lévy-Leblond operator is

$\Omega^{2}=\lambda\partial_{t}+\Delta+(f(\mathbf{x}))^{2}+\gamma^{j}\gamma_{f}\partialderivative{f}{x_{j}}.$

In this case, the potential function is the square function
$(f(\mathbf{x}))^{2}$. This is the choice of Lévy-Leblond operator made in
[3].

From a gauge transformation perspective, the Lévy-Leblond operator can be
obtained by replacing $i\hbar\partial_{t}$ with
$i\hbar\partial_{t}-V(\mathbf{x})$ (where $V(\mathbf{x})$ is the potential
function). Working in natural units $\hbar=1$ and multiplying through by $i$,
the Lévy-Leblond operator then becomes

(6)
$\Omega=\gamma_{+}(i\partial_{t}-V(\mathbf{x}))-\gamma_{-}\beta+\gamma^{j}i\partial_{j}$

with square

$\Omega^{2}=-i\beta\partial_{t}-\Delta+\beta
V(\mathbf{x})-\gamma^{j}\gamma_{+}i\partialderivative{V}{x_{j}}.$

This was the choice of operator originally made by Lévy-Leblond [1] (in
combination with a magnetic vector potential which we have omitted).

When compared to the Schrödinger operator, the squares of both (5) and (6)
contain an additional term which depends on the derivative of a potential.
Thus, unlike the free equation, the components of a solution to the Lévy-
Leblond equation with non-vanishing potential are not necessarily a solution
to the Schrödinger equation.

In this paper, we will use the operator (6) with harmonic potential, since the
colour Lie superalgebra symmetries of (5) with harmonic potential have already
been studied in [3]. The operator (6) has the advantages that it does not
require the introduction of an additional gamma matrix and that we can work
directly with the potential function instead of a ‘square-root’ function $f$
with $(f(\mathbf{x}))^{2}=V(\mathbf{x})$. One disadvantage of (6) is that,
with a quadratic potential $V(x_{1})\propto{x_{1}}^{2}$, the square of (6)
contains an extra $\gamma^{1}\gamma_{+}x_{1}$ term, whereas the square of (5)
only has a extra $\gamma^{1}\gamma_{f}$ term that is independent of $x_{1}$.

###  2.3. Time-independent Lévy-Leblond equation

The differential equation induced by (6) is

(7)
$(\gamma_{-}\beta-\gamma^{j}i\partial_{j}+\gamma_{+}V(\mathbf{x}))\Psi(t,\mathbf{x})=\gamma_{+}i\partial_{t}\Psi(t,\mathbf{x}).$

We can choose to interpret
$\gamma_{-}\beta-\gamma^{j}i\partial_{j}+\gamma_{+}V(\mathbf{x})\eqqcolon
H_{\textup{LL}}$ as if it were a Hamiltonian and set up a time-independent
version of this equation:

(8)
$(\gamma_{-}\beta-\gamma^{j}i\partial_{j}+\gamma_{+}V(\mathbf{x}))\psi(\mathbf{x})=\gamma_{+}E\psi(\mathbf{x})$

for some $E\in\mathbb{C}$. A difficulty that we encounter with the Lévy-
Leblond equation (which we do not encounter while solving the Schrödinger
equation) is that $(\gamma_{+})^{2}=0$ and so $\gamma_{+}$ is not invertible.
This means that we are forced to leave $\gamma_{+}$ on the right-hand side of
(8).

Note that $H_{\textup{LL}}$ is not self-adjoint and its eigenvalues do not
correspond to the energy of the system. Regardless, we will call
$H_{\textup{LL}}$ the (Lévy-Leblond) Hamiltonian due to the role it plays in
our computations. This is a non-standard approach; we normally require
physical observables (such as energy) to be given by the spectra of self-
adjoint operators. Despite being non-standard, our approach is consistent with
the standard formulation of quantum mechanics: we will show (2.4) that the
energy levels $E$ of the time-independent equation (8) are contained in the
spectrum of the familiar self-adjoint Schrödinger Hamiltonian.

Let $\psi(\mathbf{x})$ be a solution to the time-independent equation (8). We
will call $E$ a $\gamma_{+}$-eigenvalue of $H_{\textup{LL}}$ with
$\gamma_{+}$-eigenstate $\psi(\mathbf{x})$ and use associated terminology
(such as $\gamma_{+}$-eigenspaces). If $\psi(\mathbf{x})$ is a solution to the
time-independent equation (8), then

$\Psi(t,\mathbf{x})=e^{-iEt}\psi(\mathbf{x})$

is a solution to the time-dependent equation (7).

######  Remark 2.2.

Instead of (8), we could consider the more general time-independent equation

(9)
$H_{\textup{LL}}\psi(\mathbf{x})=\gamma_{+}(E+\gamma_{+}E_{+}+\gamma_{-}E_{-}+\gamma^{j}E_{j})\psi(\mathbf{x}).$

If we choose a matrix representation for the gamma matrices, the corresponding
time-dependent solution is

$\Psi(t,\mathbf{x})=\exp(-i(E+\gamma_{+}E_{+}+\gamma_{-}E_{-}+\gamma^{j}E_{j})t)\psi(\mathbf{x})$

where this $\exp$ is a matrix exponential. For example, if we choose the
representation (4) then $\exp(-i\gamma_{-}E_{-}t)=I-i\gamma_{-}E_{-}t$.
However, the time-independent equation (8) is sufficient for our purposes.

If we apply $H_{\textup{LL}}$ to (8), then the differential equation becomes

(10) $\left(-\Delta+\beta
V(\mathbf{x})-\gamma^{j}\gamma_{+}i\partialderivative{V}{x_{j}}\right)\psi(\mathbf{x})=\beta
E\psi(\mathbf{x})$

and so $E$ is an eigenvalue of an operator that is similar to the Schrödinger
Hamiltonian:

(11)
$H_{\textup{SL}}=-(1/\beta)\Delta+V(\mathbf{x})-\gamma^{j}\gamma_{+}i\partialderivative{V}{x_{j}}/\beta.$

That is, any solution to the time-independent Lévy-Leblond equation (8) will
be a solution to the time-independent Schrödinger-like equation of (10).

In fact, the spectrum of $H_{\textup{SL}}$ is contained in the spectrum of the
familiar Schrödinger Hamiltonian $-(1/\beta)\Delta+V(x)$. We illustrate this
fact in the following example (and prove it more generally in 2.4).

######  Example 2.3.

Consider the representation (4) for the $(1+1)$-dimensional equation. In this
case, the Schrödinger-like equation (10) becomes

$H_{\textup{SL}}\begin{pmatrix}\psi_{1}(\mathbf{x})\\\
\psi_{2}(\mathbf{x})\end{pmatrix}=\begin{pmatrix}\left(-\frac{1}{\beta}(\partial_{1})^{2}+V(\mathbf{x})\right)\psi_{1}(\mathbf{x})-\frac{i}{\beta}V^{\prime}(\mathbf{x})\psi_{2}(\mathbf{x})\\\
\left(-\frac{1}{\beta}(\partial_{1})^{2}+V(\mathbf{x})\right)\psi_{2}(\mathbf{x})\end{pmatrix}=\begin{pmatrix}E\psi_{1}(\mathbf{x})\\\
E\psi_{2}(\mathbf{x})\end{pmatrix}.$

But the second component is just the familiar time-independent Schrödinger
equation. Therefore, $E$ is either an eigenvalue of the Schrödinger
Hamiltonian or $\psi_{2}=0$. But if $\psi_{2}=0$ then the first component
becomes the familiar Schrödinger equation—in either case, $E$ is an eigenvalue
of the Schrödinger Hamiltonian.

###  2.4. Spectrum formalism

In this paper, we will not concern ourselves with the full formality of
unbounded operators acting on a Hilbert space. However, given that we have
introduced a new concept of $\gamma_{+}$-eigenvalue, we will briefly explore
this concept more formally in this section.

Since $H_{\textup{LL}}$ is an unbounded operator, we should define the set of
all $\gamma_{+}$-eigenvalues $E$ similar to a spectrum: the
$\gamma_{+}$-spectrum of $H_{\textup{LL}}$ is the set of all $E\in\mathbb{C}$
such that $H_{\textup{LL}}-\gamma_{+}E$ has no everywhere-defined bounded
inverse. In general, a $\gamma_{+}$-eigenvector corresponding to a member of
the $\gamma_{+}$-spectrum may lie outside the Hilbert space.

Observe that $H_{\textup{LL}}-\gamma_{+}E$ has bounded inverse if and only if
$(H_{\textup{LL}}-\gamma_{+}E)^{2}=H_{\textup{LL}}^{2}-\beta
E=\beta(H_{\textup{SL}}-E)$ has bounded inverse (provided $H_{\textup{LL}}$ is
a closed operator), so the $\gamma_{+}$-spectrum of $H_{\textup{LL}}$ is
exactly the spectrum of $H_{\textup{SL}}$. In combination with the following
proposition (c.f. 2.3), we can conclude that the $\gamma_{+}$-spectrum of
$H_{\textup{LL}}$ is contained in the spectrum of the corresponding
Schrödinger Hamiltonian.

######  Proposition 2.4.

Assume that $H=-(1/\beta)\Delta+V(x)$ is self-adjoint and that
$H_{\textup{SL}}=H-\gamma^{j}\gamma_{+}(i/\beta)\partialderivative{V}{x_{j}}$
is a closed operator with the same domain as $H$. Then, the spectrum of
$H_{\textup{SL}}$ is contained in the spectrum of $H$.

###### Proof.

Denote $K=\ker\gamma_{+}$. Since $(\gamma_{+})^{2}=0$, we know that $K\neq 0$.
Let $K^{\perp}$ be the orthogonal complement of $K$ so that the Hilbert space
can be written as $K\oplus K^{\perp}$.

Consider an arbitrary state
$\ket{\psi}=\ket{\psi_{K}}\oplus\ket{\psi_{\perp}}\in K\oplus K^{\perp}$.
Applying $H_{\textup{SL}}-EI$ (for $E\in\mathbb{C}$ and $I$ the identity
operator), we find that

$\displaystyle(H_{\textup{SL}}-EI)\ket{\psi}$
$\displaystyle=(H-EI)\ket{\psi}-\gamma^{j}\frac{i}{\beta}\partialderivative{V}{x_{j}}\gamma_{+}\ket{\psi_{K}}+\gamma_{+}\gamma^{j}\frac{i}{\beta}\partialderivative{V}{x_{j}}\ket{\psi_{\perp}}$
$\displaystyle=(H-EI)\ket{\psi}+\gamma_{+}\gamma^{j}i\partialderivative{V}{x_{j}}\ket{\psi_{\perp}}.$

The gamma matrix $\gamma_{+}$ commutes with $H$, so
$\gamma_{+}(H\ket{\psi_{K}})=0$ and $H\ket{\psi_{K}}\in K$. Furthermore,
$H\ket{\psi_{\perp}}\in K^{\perp}$; indeed, we know that $H$ is self-adjoint,
so for any $\ket{\varphi}\in K$ we have
$\bra{\varphi}(H\ket{\psi_{\perp}})=(\bra{\psi_{\perp}}H\ket{\varphi})^{*}=0$
(because we have just shown that $H\ket{\varphi}\in K$). Finally, since
$(\gamma_{+})^{2}=0$ we have that
$\gamma_{+}\gamma^{j}(i/\beta)\partialderivative{V}{x_{j}}\in K$. Thus,

$(H_{\textup{SL}}-EI)\ket{\psi}=\left((H-EI)\ket{\psi_{K}}+\gamma_{+}\gamma^{j}\frac{i}{\beta}\partialderivative{V}{x_{j}}\ket{\psi_{\perp}}\right)\oplus(H-EI)\ket{\psi_{\perp}}\in
K\oplus K^{\perp}.$

Now, assume that $H-EI$ has a bounded inverse.

For injectivity, suppose $(H_{\textup{SL}}-EI)\ket{\psi}=0$. Then
$(H-EI)\ket{\psi_{\perp}}=0$ and hence $\ket{\psi_{\perp}}=0$. But since
$\ket{\psi_{\perp}}=0$, we have that $(H-EI)\ket{\psi_{K}}=0$ and hence
$\ket{\psi_{K}}=0$. Thus, $\ket{\psi}=0$ and $H_{\textup{SL}}-EI$ is
injective.

For surjectivity, take an arbitrary
$\ket{\varphi}=\ket{\varphi_{K}}\oplus\ket{\varphi_{\perp}}\in K\oplus
K^{\perp}$. By surjectivity of $H-EI$, we can choose $\ket{\psi_{\perp}}\in
K^{\perp}$ such that $(H-EI)\ket{\psi_{\perp}}=\ket{\varphi_{\perp}}$ and then
choose $\ket{\psi_{K}}\in K$ such that
$(H-EI)\ket{\psi_{K}}=\ket{\varphi_{K}}-\gamma_{+}\gamma^{j}(i/\beta)\partialderivative{V}{x_{j}}\ket{\psi_{\perp}}$.
That is, $(H_{\textup{SL}}-EI)\ket{\psi}=\ket{\varphi}$ and
$H_{\textup{SL}}-EI$ is surjective.

Finally, $H_{\textup{SL}}-EI$ is closed by assumption, and so its inverse is
bounded by the Closed Graph Theorem.

Taking the contrapositive, if $E$ is in the spectrum of $H_{\textup{SL}}$ then
$E$ is in the spectrum of $H$. ∎

If we take the representation in (4) with a free or harmonic potential, then
we can easily verify that $H$ is self-adjoint with the same domain as
$H_{\textup{SL}}$. Moreover, observe that

$\gamma_{+}\gamma^{j}\frac{i}{\beta}\partialderivative{V}{x_{j}}=-\gamma_{+}\frac{i}{\beta}V^{\prime}(x_{1})=\left(\gamma_{-}\frac{i}{\beta}(V^{\prime}(x_{1}))^{*}\right)^{\dagger}$

and so $H_{\textup{SL}}$ is closed in this representation, being the adjoint
of $H+\gamma_{-}(i/\beta)(V^{\prime}(x_{1}))^{*}$ (the adjoint of an operator
is always closed). Thus, we are justified in applying 2.4 to the operators
used in this paper.

###  2.5. Colour Lie superalgebras

The $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebras
generalise Lie superalgebras to grading by the group
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. Within this paper, every
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra appears as
a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded space of operators
$\mathfrak{g}=\mathfrak{g}_{00}\oplus\mathfrak{g}_{01}\oplus\mathfrak{g}_{10}\oplus\mathfrak{g}_{11}$
acting on a Hilbert space of states. This space $\mathfrak{g}$ is equipped
with a so-called colour bracket
$\cbrak{\cdot}{\cdot}\colon\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$
realised as

$\cbrak{x}{y}=xy-\varepsilon(\alpha,\beta)yx\qquad\text{for}\;x\in\mathfrak{g}_{\alpha},\,y\in\mathfrak{g}_{\beta}\;\text{and
satisfying}\;\cbrak{x}{y}\in\mathfrak{g}_{\alpha+\beta}$

with commutation factor
$\varepsilon(\alpha,\beta)=(-1)^{\alpha_{1}\beta_{1}+\alpha_{2}\beta_{2}}$
where
$\alpha=(\alpha_{1},\alpha_{2}),\,\beta=(\beta_{1},\beta_{2})\in\mathbb{Z}_{2}\times\mathbb{Z}_{2}$.
By replacing the commutation factor $\varepsilon$ with
$\varepsilon(\alpha,\beta)=(-1)^{\alpha_{1}\beta_{2}-\alpha_{2}\beta_{1}}$, we
obtain what is called a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie
algebra (as opposed to a _super_ algebra; however, despite the names, colour
Lie (super)algebras are neither Lie algebras nor Lie superalgebras). The
$\mathbb{Z}_{2}^{n}$-graded colour Lie superalebras are the natural
generalisation of $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie
superalgebras. The above definitions are sufficient for this paper; we direct
the interested reader to the more general definitions of colour Lie algebras
in [6, 7, 33].

In [3], the authors performed an exhaustive search of the first-order
differential operators which satisfied

$\commutator{\Omega}{Z}=\Phi_{Z}(\mathbf{x})\Omega\qquad\text{or}\qquad\anticommutator{\Omega}{Z}=\Phi_{Z}(\mathbf{x})\Omega$

where $\Omega$ is the Lévy-Leblond operator and $\Phi_{Z}$ is a matrix-valued
function of the space coordinates. By investigating the algebras generated by
these operators, the authors found a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra as an
extension of the Schrödinger symmetry algebra. However, it was proved that
there is no $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded extension of the
Schrödinger symmetry algebra for the Lévy-Leblond equation with harmonic
potential [3].

##  3\. Eigenspace-preserving operators of the free equation

Inspired by the symmetry algebras discovered in [3, 4], we wish to identify
the operators which leave the $\gamma_{+}$-eigenspaces invariant, and
determine if any $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie
(super)algebras appear. We will work in natural units, $\hbar=1$.

Recall from (8) that the time-independent $(1+1)$-dimensional Lévy-Leblond
equation (for the Schrödinger equation) with free potential is

(12)
$(\gamma_{-}\beta-\gamma^{1}i\partial_{1})\psi(x_{1})=\gamma_{+}E\psi(x_{1})$

where $t$ is the time coordinate and $x_{1}$ is the first (and only) spacial
coordinate. For ease of reading, we will omit the subscript and write $x$ as
the spacial coordinate.

By 2.4, $E$ is in the spectrum of the free Schrödinger Hamiltonian
$-(1/\beta)(\partial_{1})^{2}$. Since the free Schrödinger Hamiltonian is
proportional to the square of the self-adjoint momentum operator
$-i\partial_{1}$, we have that $E\in[0,\infty)$.

###  3.1. Eigenvalue conditions

Let $\ket{\psi}\equiv\psi(x)$ be a solution to the time-independent equation
(12). We wish to find an operator $A$ that leaves the $\gamma_{+}$-eigenspaces
invariant; that is,

(13)
$(\gamma_{-}\beta-\gamma^{1}i\partial_{1})A\ket{\psi}=\gamma_{+}EA\ket{\psi}$

for $\ket{\psi}$ a solution to the time-independent equation (12).

Note that we can rearrange equation (12) to solve for $\partial_{1}$:

(14) $\displaystyle\partial_{1}\ket{\psi}$
$\displaystyle=(\gamma^{1}\gamma_{+}iE-\gamma^{1}\gamma_{-}i\beta)\ket{\psi}$
$\displaystyle=(\gamma_{+}iE+\gamma_{-}i\beta)\ket{\psi}$

using relations (3). That is, the operator $\partial_{1}$ can be expressed in
terms of gamma matrices when acting on a $\gamma_{+}$-eigenspace. Therefore,
if we only consider the action on a $\gamma_{+}$-eigenspace, $A$ does not need
to contain any differential operators (this fact is unique to the
$(1+1)$-dimensional case). So the general form of $A$ (restricted to a
$\gamma_{+}$-eigenspace) is

(15) $A=c_{I}(x)I+c_{+}(x)\gamma_{+}+c_{-}(x)\gamma_{-}+c_{1}(x)\gamma^{1}$

for some complex-valued functions $c_{I},\,c_{+},\,c_{-},\,c_{1}$. We will
assume that these functions are differentiable.

We then substitute the general $A$ of (15) into (13), and use relations (3),
(14) and the product rule: $\partial_{1}f(x)=f(x)\partial_{1}+f^{\prime}(x)$.
In doing so, we obtain a coupled ODE:

(16) $\left\\{\begin{aligned} \derivative{c_{I}}{x}&=0\\\
\derivative{c_{+}}{x}&=-2iEc_{1}(x)\\\ \derivative{c_{-}}{x}&=2i\beta
c_{1}(x)\\\ \derivative{c_{1}}{x}&=-i\beta
c_{+}(x)+iEc_{-}(x).\end{aligned}\right.$

Obviously $c_{I}(x)\equiv c_{I}$ is a constant. Defining a new variable
$c(x)=-i\beta c_{+}(x)+iEc_{-}(x)$, yields the following coupled ODE:

$\left\\{\begin{aligned} \derivative{c}{x}&=-4E\beta c_{1}(x)\\\
\derivative{c_{1}}{x}&=c(x).\end{aligned}\right.$

which we can solve to find that

$\displaystyle c_{1}(x)$
$\displaystyle=ae^{i2\sqrt{E\beta}x}+be^{-i2\sqrt{E\beta}x}$ $\displaystyle
c(x)$
$\displaystyle=i2\sqrt{E\beta}(ae^{i2\sqrt{E\beta}x}-be^{-i2\sqrt{E\beta}x})$
$\displaystyle c_{+}(x)$
$\displaystyle=-a\sqrt{\tfrac{E}{\beta}}e^{i2\sqrt{E\beta}x}+b\sqrt{\tfrac{E}{\beta}}e^{-i2\sqrt{E\beta}x}+\frac{d}{\beta}$
$\displaystyle c_{-}(x)$
$\displaystyle=a\sqrt{\tfrac{\beta}{E}}e^{i2\sqrt{E\beta}x}-b\sqrt{\tfrac{\beta}{E}}e^{-i2\sqrt{E\beta}x}+\frac{d}{E}$

for some constants $a,\,b,\,d$ and assuming $E$ and $\beta$ are non-zero.
Substituting the expressions we have computed for
$c_{I},\,c_{+},\,c_{-},\,c_{1}$ back into the general form of $A$ in (15), we
find four linearly independent operators:

(17) $\displaystyle I$ (18) $\displaystyle\gamma_{+}iE+\gamma_{-}i\beta$ (19)
$\displaystyle-\gamma_{+}\sqrt{\frac{E}{\beta}}e^{i2\sqrt{E\beta}x}+\gamma_{-}\sqrt{\frac{\beta}{E}}e^{i2\sqrt{E\beta}x}+\gamma^{1}e^{i2\sqrt{E\beta}x}$
(20)
$\displaystyle\gamma_{+}\sqrt{\frac{E}{\beta}}e^{-i2\sqrt{E\beta}x}-\gamma_{-}\sqrt{\frac{\beta}{E}}e^{-i2\sqrt{E\beta}x}+\gamma^{1}e^{-i2\sqrt{E\beta}x}.$

We can easily verify that these operators leave the $\gamma_{+}$-eigenspaces
invariant.

For a specific choice of $E$, the operators 17, 18, 19 and 20 are only defined
on the corresponding $\gamma_{+}$-eigenspace. If we want operators defined
everywhere on (a dense subset of) the Hilbert space, then we can interpret 17,
18, 19 and 20 as eigenvalue conditions which the desired operators must
satisfy.

###  3.2. Finding operators

From (14) we immediately see that the operator $\partial_{1}$ has the same
action as (18) on the $\gamma_{+}$-eigenspace. We will multiply through by
$-i$ and instead instead consider the self-adjoint momentum operator
$-i\partial_{1}$. Since $-i\partial_{1}$ commutes with the Lévy-Leblond
Hamiltonian $H_{\textup{LL}}=\gamma_{-}\beta-\gamma^{1}i\partial_{1}$ and with
$\gamma_{+}E$, we can easily verify that $-i\partial_{1}$ does indeed leave
the $\gamma_{+}$-eigenspaces invariant. Surprisingly, (18) also gives rise to
another operator. If we multiply (18) by $-i\beta$ and use the fact that $E$
is an eigenvalue of the Schrödinger Hamiltonian, then we obtain the operator
$-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2}$. This operator does leave
the $\gamma_{+}$-eigenspaces invariant:

$\displaystyle
H_{\textup{LL}}(-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2})\ket{\psi}$
$\displaystyle=-(-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2})H_{\textup{LL}}\ket{\psi}+\anticommutator{H_{\textup{LL}}}{-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2}}\ket{\psi}$
$\displaystyle=-(-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2})\gamma_{+}E\ket{\psi}+I\beta^{2}\left(-\frac{1}{\beta}(\partial_{1})^{2}\right)\ket{\psi}$
$\displaystyle=\gamma_{+}E(-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2})\ket{\psi}-\anticommutator{-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2}}{\gamma_{+}E}\ket{\psi}+I\beta^{2}E\ket{\psi}$
$\displaystyle=\gamma_{+}E(-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2})\ket{\psi}-I\beta^{2}E\ket{\psi}+I\beta^{2}E\ket{\psi}$
$\displaystyle=\gamma_{+}E(-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2})\ket{\psi}.$

Now, we shall examine (20). We need to find an operator which has $\ket{\psi}$
as an eigenstate with eigenvalue involving $\sqrt{E}$. We would expect the
eigenvalues of $(\partial_{1})^{2}$ to be the squares of the eigenvalues of
$\partial_{1}$; that is, we expect the eigenvalues of $\partial_{1}$ to be
$\pm i\sqrt{E\beta}$. For now, we shall assume that $\ket{\psi}$ is an
eigenstate of $\partial_{1}$ with eigenvalue $i\sqrt{E\beta}$. With this
eigenvalue, we can find an operator corresponding to $e^{-2i\sqrt{E\beta}x}$:

$\displaystyle e^{-2i\sqrt{E\beta}x}\ket{\psi}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{(-2x)^{k}(i\sqrt{E\beta})^{k}\psi(x)}{k!}$
$\displaystyle\text{where}~{}\psi(x)\equiv\ket{\psi}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{(-2x)^{k}(\partial_{1})^{k}\psi(x)}{k!}$
$\displaystyle=\sum_{k=0}^{\infty}\frac{\psi^{(k)}(x)}{k!}(-x-x)^{k}.$

This final expression is the Taylor series for $\psi(-x)$ about the point $x$.
If we assume that $\psi$ is analytic everywhere, then
$e^{-2i\sqrt{E\beta}x}\ket{\psi}=\psi(-x)$. So, on an analytic function,
$e^{-2i\sqrt{E\beta}}$ acts the same as the parity operator $\mathscr{P}$,
defined by $\mathscr{P}(\chi(x))=\chi(-x)$ for all states $\chi$. Note that
$\partial_{1}\mathscr{P}=-\mathscr{P}\partial_{1}$ by the chain rule.

Assuming that $\ket{\psi}$ is an analytic function, from (20) we compute:

$\displaystyle\left(\gamma_{+}\sqrt{\frac{E}{\beta}}e^{-i2\sqrt{E\beta}x}-\gamma_{-}\sqrt{\frac{\beta}{E}}e^{-i2\sqrt{E\beta}x}+\gamma^{1}e^{-i2\sqrt{E\beta}x}\right)\ket{\psi}$
$\displaystyle=\left(\gamma_{+}\frac{iE}{i\sqrt{E\beta}}\mathscr{P}-\gamma_{-}\frac{i\beta}{i\sqrt{E\beta}}\mathscr{P}+\gamma^{1}\mathscr{P}\right)\ket{\psi}$
$\displaystyle=\mathscr{P}\left(\gamma^{1}\frac{1}{i\sqrt{E\beta}}\partial_{1}+\gamma^{1}\right)\ket{\psi}$
$\displaystyle=2\gamma^{1}\mathscr{P}\ket{\psi}$

using (14) and the fact that $\partial_{1}$ has eigenvalue $i\sqrt{E\beta}$.
Therefore, we guess that $\gamma^{1}\mathscr{P}$ is an operator which leaves
the $\gamma_{+}$-eigenspace invariant. And indeed, we can verify that this is
true for all eigenstates $\ket{\psi}$ (not just analytic ones):

$\displaystyle H_{\textup{LL}}(\gamma^{1}\mathscr{P}\ket{\psi})$
$\displaystyle=\gamma_{-}\beta\mathscr{P}\ket{\psi}-i\partial_{1}\mathscr{P}\ket{\psi}$
$\displaystyle=\gamma_{-}\beta\mathscr{P}\ket{\psi}+i\mathscr{P}\partial_{1}\ket{\psi}$
$\displaystyle=\gamma_{-}\beta\mathscr{P}\ket{\psi}+i\mathscr{P}(\gamma_{+}iE+\gamma_{-}i\beta)\ket{\psi}$
$\displaystyle=-\gamma_{+}E\mathscr{P}\ket{\psi}$
$\displaystyle=\gamma_{+}E(\gamma^{1}\mathscr{P}\ket{\psi}).$

Note that we did not use the assumption that the eigenvalue of $\partial_{1}$
is $i\sqrt{E\beta}$ in the above computation.

To find $\gamma^{1}\mathscr{P}$, we assumed that $\partial_{1}$ has eigenvalue
$i\sqrt{E\beta}$. If we instead assume that $\partial_{1}$ has eigenvalue
$-i\sqrt{E\beta}$, then the operator we obtain from (20) is $0$, since

$\displaystyle\left(\gamma_{+}\sqrt{\frac{E}{\beta}}e^{-i2\sqrt{E\beta}x}-\gamma_{-}\sqrt{\frac{\beta}{E}}e^{-i2\sqrt{E\beta}x}+\gamma^{1}e^{-i2\sqrt{E\beta}x}\right)\ket{\psi}$
$\displaystyle=e^{-i2\sqrt{E\beta}x}\left(\gamma_{1}\frac{1}{i\sqrt{E\beta}}\partial_{1}+\gamma^{1}\right)\ket{\psi}.$

The behaviour for (19) is similar: if $\partial_{1}$ has eigenvalue
$-i\sqrt{E\beta}$ then we again obtain the operator $\gamma^{1}\mathscr{P}$
and if $\partial_{1}$ has eigenvalue $i\sqrt{E\beta}$ then we obtain the zero
operator.

So far, we have identified three operators which leave the
$\gamma_{+}$-eigenspace invariant:

$\partial_{1},\qquad-\gamma_{+}(\partial_{1})^{2}+\gamma_{-}\beta^{2},\qquad\gamma^{1}\mathscr{P}.$

This is not an exhaustive list of such operators. Indeed, composing any number
of these three operators will yield further operators which leave the
$\gamma_{+}$-eigenspaces invariant. However, we will not consider any further
$\gamma_{+}$-eigenspace-preserving operators in this paper.

###  3.3. Grading the operators

We have found operators which leave the $\gamma_{+}$-eigenspaces invariant. We
now wish to determine if these operators close to form an algebra.

Let,

$H_{\textup{SL}}=-\frac{1}{\beta}(\partial_{1})^{2},\qquad\widehat{P}=-i\partial_{1},\qquad
D_{+}=-\gamma_{+}\frac{1}{\beta}(\partial_{1})^{2}+\gamma_{-}\beta,\qquad\mathscr{P}^{1}=\gamma^{1}\mathscr{P}.$

Computing the complete set of commutation and anticommutation relations we
find

(21) $\displaystyle\commutator*{\widehat{P}}{\widehat{P}}$ $\displaystyle=0$
$\displaystyle\anticommutator*{\widehat{P}}{\widehat{P}}$
$\displaystyle=2\beta H_{\textup{SL}}$
$\displaystyle\commutator*{D_{+}}{D_{+}}$ $\displaystyle=0$
$\displaystyle\anticommutator*{D_{+}}{D_{+}}$ $\displaystyle=2\beta
H_{\textup{SL}}$ $\displaystyle\commutator*{\mathscr{P}^{1}}{\mathscr{P}^{1}}$
$\displaystyle=0$
$\displaystyle\anticommutator*{\mathscr{P}^{1}}{\mathscr{P}^{1}}$
$\displaystyle=2I$ $\displaystyle\commutator*{\widehat{P}}{D_{+}}$
$\displaystyle=0$ $\displaystyle\anticommutator*{\widehat{P}}{D_{+}}$
$\displaystyle=2D_{+}\widehat{P}$
$\displaystyle\commutator*{\widehat{P}}{\mathscr{P}^{1}}$
$\displaystyle=-2\mathscr{P}^{1}\widehat{P}$
$\displaystyle\anticommutator*{\widehat{P}}{\mathscr{P}^{1}}$
$\displaystyle=0$ $\displaystyle\commutator*{D_{+}}{\mathscr{P}^{1}}$
$\displaystyle=-2\mathscr{P}^{1}D_{+}$
$\displaystyle\anticommutator*{D_{+}}{\mathscr{P}^{1}}$ $\displaystyle=0$

and $H_{\textup{SL}}$ commutes with the other three operators.

We can create colour Lie (super)algebras by assigning gradings to these
operators to choose which of the above (anti)commutation relations are
expressed.

######  Theorem 3.1.

Let $\mathfrak{A},\,\mathfrak{D}^{+},\,\mathfrak{D}^{1},\,\mathfrak{D}$ be
vector spaces. Define $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded sectors for
$\mathfrak{A},\,\mathfrak{D}^{+},\,\mathfrak{D}^{1}$ and
$\mathbb{Z}_{2}^{3}$-graded sectors for $\mathfrak{D}$ as follows:

$\displaystyle\mathfrak{A}_{00}$ $\displaystyle=0,$
$\displaystyle\mathfrak{A}_{01}$
$\displaystyle=\operatorname{span}\\{\widehat{P},D_{+}\\},$
$\displaystyle\mathfrak{A}_{10}$
$\displaystyle=\operatorname{span}\\{\mathscr{P}^{1}\\},$
$\displaystyle\mathfrak{A}_{11}$ $\displaystyle=0,$
$\displaystyle\mathfrak{D}^{+}_{00}$
$\displaystyle=\operatorname{span}\\{I,H_{\textup{SL}}\\},$
$\displaystyle\mathfrak{D}^{+}_{01}$
$\displaystyle=\operatorname{span}\\{\widehat{P}\\},$
$\displaystyle\mathfrak{D}^{+}_{10}$
$\displaystyle=\operatorname{span}\\{D_{+}\\},$
$\displaystyle\mathfrak{D}^{+}_{11}$
$\displaystyle=\operatorname{span}\\{\mathscr{P}^{1}\\},$
$\displaystyle\mathfrak{D}^{1}_{00}$
$\displaystyle=\operatorname{span}\\{I\\},$
$\displaystyle\mathfrak{D}^{1}_{01}$
$\displaystyle=\operatorname{span}\\{\mathscr{P}^{1}\\},$
$\displaystyle\mathfrak{D}^{1}_{10}$ $\displaystyle=0,$
$\displaystyle\mathfrak{D}^{1}_{11}$
$\displaystyle=\operatorname{span}\\{\widehat{P},D_{+}\\},$
$\displaystyle\mathfrak{D}_{000}$
$\displaystyle=\operatorname{span}\\{I,H_{\textup{SL}}\\},$
$\displaystyle\mathfrak{D}_{001}$
$\displaystyle=\operatorname{span}\\{\widehat{P}\\},$
$\displaystyle\mathfrak{D}_{010}$
$\displaystyle=\operatorname{span}\\{D_{+}\\},$
$\displaystyle\mathfrak{D}_{011}$ $\displaystyle=0,$
$\displaystyle\mathfrak{D}_{100}$ $\displaystyle=0,$
$\displaystyle\mathfrak{D}_{101}$ $\displaystyle=0,$
$\displaystyle\mathfrak{D}_{110}$ $\displaystyle=0,$
$\displaystyle\mathfrak{D}_{111}$
$\displaystyle=\operatorname{span}\\{\mathscr{P}^{1}\\}.$

We then define
$\mathfrak{A},\,\mathfrak{D}^{+},\,\mathfrak{D}^{1},\,\mathfrak{D}$ to be the
direct sum of their respective sectors. Define a bracket on each space by
$\cbrak{A}{B}=AB-\varepsilon(\alpha,\beta)BA$ for $A,\,B$ homogeneous of
degree $\alpha,\,\beta$ respectively. Here, $\varepsilon$ is the commutation
factor for the algebra

$\displaystyle\text{for}~{}\mathfrak{A}\colon$
$\displaystyle\varepsilon(\alpha_{1}\alpha_{2},\beta_{1}\beta_{2})$
$\displaystyle=(-1)^{\alpha_{1}\cdot\beta_{2}-\alpha_{2}\cdot\beta_{1}}$
$\displaystyle\text{for}~{}\mathfrak{D}^{+}~{}\text{and}~{}\mathfrak{D}^{1}\colon$
$\displaystyle\varepsilon(\alpha_{1}\alpha_{2},\beta_{1}\beta_{2})$
$\displaystyle=(-1)^{\alpha_{1}\cdot\beta_{1}+\alpha_{2}\cdot\beta_{2}}$
$\displaystyle\text{for}~{}\mathfrak{D}\colon$
$\displaystyle\varepsilon(\alpha_{1}\alpha_{2}\alpha_{3},\beta_{1}\beta_{2}\beta_{3})$
$\displaystyle=(-1)^{\alpha_{1}\cdot\beta_{1}+\alpha_{2}\cdot\beta_{2}+\alpha_{3}\cdot\beta_{3}}.$

Then $\mathfrak{D}^{+}$ closes to form a colour Lie algebra and
$\mathfrak{A}$, $\mathfrak{D}^{-}$, and $\mathfrak{D}$ close to form colour
Lie superalgebras.

###### Proof.

By examining (21), we verify that the colour bracket of the $\alpha$-sector
and the $\beta$-sector is a subspace of the $(\alpha+\beta)$-sector (for all
possible sectors). ∎

If we want to construct an algebra that contains only
$\widehat{P},\,D_{+},\,\mathscr{P}^{1}$, then we need to define a bracket
$\cbrak{\cdot}{\cdot}$ such that
$\cbrak*{\widehat{P}}{\widehat{P}}=\commutator*{\widehat{P}}{\widehat{P}}$ but
$\cbrak*{\widehat{P}}{\mathscr{P}^{1}}=\anticommutator*{\widehat{P}}{\mathscr{P}^{1}}$.
This is not possible with a Lie superalgebra, but is possible with a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie (super)algebra such as
$\mathfrak{A}$. However, $\mathfrak{A}$ is somewhat trivial because
$\cbrak{A}{B}=0$ for all $A,B\in\mathfrak{A}$.

The most interesting relations of (21) are
$\anticommutator{D_{+}}{D_{+}}=\beta H_{\textup{SL}}$ and
$\anticommutator{\mathscr{P}^{1}}{\mathscr{P}^{1}}=2I$. We would hope to
capture these relations in the colour Lie (super)algebra. Two options for a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded algebra which realise
$\anticommutator{D_{+}}{D_{+}}$ or
$\anticommutator{\mathscr{P}^{1}}{\mathscr{P}^{1}}$ are $\mathfrak{D}^{+}$ and
$\mathfrak{D}^{1}$ (respectively).

It is not possible to realise both $\anticommutator{D_{+}}{D_{+}}$ and
$\anticommutator{\mathscr{P}^{1}}{\mathscr{P}^{1}}$ simultaneously with a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-grading on at most five basis elements.
We can, however, do this with the $\mathbb{Z}_{2}^{3}$-graded algebra
$\mathfrak{D}$.

By considering only symmetry operators that are first- or second-order
differential operators, [3, 4] found a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra. But
observe that $\mathscr{P}^{1}$ is not a differential operator (though is a
symmetry operator). Thus, the appearance of a $\mathbb{Z}_{2}^{3}$-grade in
the above theorem suggests that a larger grading is required when considering
a broader class of symmetry operators. However, we are primarily concerned
with the utility of $\mathfrak{D}$ for finding solutions; a more in-depth
analysis for the symmetry operators of the free equation is beyond the scope
of this paper.

###  3.4. Relationship between the algebra and solutions

The $\mathbb{Z}_{2}^{3}$-graded algebra $\mathfrak{D}$ contains two operators,
$\widehat{P}$ and $D_{+}$, which both square to the Schrödinger Hamiltonian
$H_{\textup{SL}}$. Finding the eigenstates of either operator would then allow
us to solve the Schrödinger equation. We will prove that solving the Lévy-
Leblond equation is equivalent to finding _simultaneous_ eigenstates of these
operators. In a sense, this couples together these “square roots” of
$H_{\textup{SL}}$ and leads to the coupling of the components in the solutions
to the Lévy-Leblond equation.

######  Theorem 3.2.

Solving the time-independent Lévy-Leblond equation (12) with $E>0$ is
equivalent to finding the simultaneous eigenstates for $\widehat{P}$ and
$D_{+}$ with positive eigenvalues for both operators. Specifically,
$\ket{\psi}$ is a solution to the time independent Lévy-Leblond equation with
$E>0$ if and only if

$\ket{\psi}=a\ket{\varphi_{1}}+b\mathscr{P}^{1}\ket{\varphi_{2}}\qquad
a,b\in\mathbb{C}$

for some simultaneous eigenstates $\ket{\varphi_{1}},\,\ket{\varphi_{2}}$ of
$\widehat{P}$ and $D_{+}$ with all eigenvalues positive.

###### Proof.

For the forward direction, let $\ket{\psi}$ be a solution to the time-
independent Lévy-Leblond equation. Let
$\ket{\chi}=(1/\sqrt{E\beta})\widehat{P}\ket{\psi}$.

First, consider the case where $\ket{\chi}=c\ket{\psi}$ for some
$c\in\mathbb{C}$. Then $\ket{\psi}$ is an eigenstate for $\widehat{P}$. From
(14), we know that

$\widehat{P}\ket{\psi}=(\gamma_{+}E+\gamma_{-}\beta)\ket{\psi}=(-\gamma_{+}\frac{1}{\beta}(\partial_{1})^{2}+\gamma_{-}\beta)\ket{\psi}=D_{+}\ket{\psi}.$

Therefore, $\ket{\psi}$ is also an eigenstate for $D_{+}$. Since
$H_{\textup{SL}}=1/\beta(\widehat{P})^{2}$, we know that
$(\widehat{P})^{2}\ket{\psi}=E\beta\ket{\psi}$, and so $c^{2}=1$. If $c=1$
choose $\ket{\varphi_{1}}=\ket{\varphi_{2}}=\ket{\psi}$ and $a=1,\,b=0$. If
$c=-1$ choose $\ket{\varphi_{1}}=\ket{\varphi_{2}}=\mathscr{P}^{1}\ket{\psi}$
and $a=0,\,b=1$.

Now, consider the case $\ket{\chi}\neq c\ket{\psi}$ for any $c\in\mathbb{C}$.
Set $\ket{\varphi_{1}}=\ket{\psi}+\ket{\chi}$ and
$\ket{\varphi_{2}}=\mathscr{P}^{1}(\ket{\psi}-\ket{\chi})$. Since
$H_{\textup{SL}}=(1/\beta)\widehat{P}^{2}$, we know that
$(\widehat{P})^{2}\ket{\psi}=E\beta\ket{\psi}$, and so
$\widehat{P}\ket{\chi}=\sqrt{E\beta}\ket{\psi}$. Noting also that
$\widehat{P}\ket{\psi}=\sqrt{E\beta}\ket{\chi}$, it is easily verified that

$\widehat{P}\ket{\varphi_{1}}=\sqrt{E\beta}\ket{\varphi_{1}}\qquad\text{and}\qquad\widehat{P}\ket{\varphi_{2}}=\sqrt{E\beta}\ket{\varphi_{2}}.$

From (14), we know that

$\widehat{P}\ket{\psi}=(\gamma_{+}E+\gamma_{-}\beta)\ket{\psi}=\left(-\gamma_{+}\frac{1}{\beta}(\partial_{1})^{2}+\gamma_{-}\beta\right)\ket{\psi}=D_{+}\ket{\psi}.$

Therefore, $\ket{\chi}=(1/\sqrt{E\beta})D_{+}\ket{\psi}$ and a similar
argument shows that

$D_{+}\ket{\varphi_{1}}=\sqrt{E\beta}\ket{\varphi_{1}}\qquad\text{and}\qquad
D_{+}\ket{\varphi_{2}}=\sqrt{E\beta}\ket{\varphi_{2}}.$

That is, $\ket{\varphi_{1}}$ and $\ket{\varphi_{2}}$ are simultaneous
eigenstates for $\widehat{P}$ and $D_{+}$ with all eigenvalues positive.
Moreover, $\ket{\psi}=\ket{\varphi_{1}}+\mathscr{P}^{1}\ket{\varphi_{2}}$ so
we set $a=b=1$.

For the reverse direction, let $\ket{\varphi_{1}}$ and $\ket{\varphi_{2}}$ be
simultaneous eigenstates of $\widehat{P}$ and $D_{+}$ with only positive
eigenvalues. Since both operators square to $\beta H_{\textup{SL}}$ we have
that $\ket{\varphi_{1}}$ and $\ket{\varphi_{2}}$ are also eigenstates of
$H_{\textup{SL}}$. Let $E>0$ be the corresponding eigenvalue of
$H_{\textup{SL}}$. Then the eigenvalues for $\widehat{P}$ and $D_{+}$ must be
$\sqrt{E\beta}$ (note that we are assuming only positive eigenvalues). In
particular,

$-i\partial_{1}\ket{\varphi_{1}}=\sqrt{E\beta}\ket{\varphi_{1}}=\left(-\gamma_{+}\frac{1}{\beta}(\partial_{1})^{2}+\gamma_{-}\beta\right)\ket{\varphi_{1}}=(\gamma_{+}E+\gamma_{-}\beta)\ket{\varphi_{1}}.$

Rearranging, we get

$(\gamma_{-}\beta-\gamma^{1}i\partial_{1})\ket{\varphi_{1}}=\gamma_{+}E\ket{\varphi_{1}}$

so $\ket{\varphi_{1}}$ is a solution to the time-independent Lévy-Leblond
equation. Note that the eigenvalue of $\mathscr{P}^{1}\ket{\varphi_{2}}$ with
$\widehat{P}$ and $D_{+}$ is $-\sqrt{E\beta}$, so a similar argument shows
that $\mathscr{P}^{1}\ket{\varphi_{2}}$ is a solution to the Lévy-Leblond
equation. Since all the operators are linear, any linear combination of
$\ket{\varphi_{1}}$ and $\mathscr{P}^{1}\ket{\varphi_{2}}$ will also be a
solution to the Lévy-Leblond equation. ∎

3.2 further demonstrates that the operators of the
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded algebra $\mathfrak{D}$ are
fundamental to the free Lévy-Leblond equation.

###  3.5. Solving the free equation

To conclude this section, we will use 3.2 to solve the Lévy–Leblond equation.
This is mostly useful as an illustration; the $(1+1)$-dimensional Lévy-Leblond
equation is easy enough to solve without (directly) appealing to this theorem.
So far, our results have been representation independent (assuming relations
(3) hold). However, to find a solution we will choose the matrix
representation (4).

We are looking for simultaneous eigenstates of
$\widehat{P}=-i\partialderivative{x}$ and
$D_{+}=-\gamma_{+}(1/\beta)\partialderivative[2]{x}+\gamma_{-}\beta$ with only
positive eigenvalues. We will pre-emptively look for eigenstates with
eigenvalue $\sqrt{E\beta}$. We know that, up to scaling, the only
eigenfunction of the derivative operator with eigenvalue $i\sqrt{E\beta}$ is
the exponential function $\exp(i\sqrt{E\beta}x)$. Therefore, the eigenstates
for $\widehat{P}$ are of the form

$\begin{pmatrix}C_{1}e^{i\sqrt{E\beta}x}\\\ C_{2}e^{i\sqrt{E\beta}x}\\\
\end{pmatrix}$

for some constants $C_{1},\,C_{2}\in\mathbb{C}$. Now, we examine which of
these eigenstates are also eigenstates of $D_{+}$:

$\displaystyle D_{+}\begin{pmatrix}C_{1}e^{i\sqrt{E\beta}x}\\\
C_{2}e^{i\sqrt{E\beta}x}\\\ \end{pmatrix}$
$\displaystyle=\begin{pmatrix}C_{2}Ee^{i\sqrt{E\beta}x}\\\ C_{1}\beta
e^{i\sqrt{E\beta}x}\\\
\end{pmatrix}=\frac{C_{1}\beta}{C_{2}}\begin{pmatrix}\frac{(C_{2})^{2}E}{(C_{1})^{2}\beta}C_{1}e^{i\sqrt{E\beta}x}\\\
C_{2}e^{i\sqrt{E\beta}x}\\\ \end{pmatrix}$

so we must have $(C_{1})^{2}\beta=(C_{2})^{2}E$. Therefore, up to scaling, the
only simultaneous eigenstate with only positive eigenvalues is

$\begin{pmatrix}\sqrt{E}e^{i\sqrt{E\beta}x}\\\
\sqrt{\beta}e^{i\sqrt{E\beta}x}\end{pmatrix}.$

By 3.2, the general solution for $E>0$ is

$a\begin{pmatrix}\sqrt{E}e^{i\sqrt{E\beta}x}\\\
\sqrt{\beta}e^{i\sqrt{E\beta}x}\end{pmatrix}+b\gamma^{1}\mathscr{P}\begin{pmatrix}\sqrt{E}e^{i\sqrt{E\beta}x}\\\
\sqrt{\beta}e^{i\sqrt{E\beta}x}\end{pmatrix}=\begin{pmatrix}a\sqrt{E}e^{i\sqrt{E\beta}x}+b\sqrt{E}e^{-i\sqrt{E\beta}x}\\\
a\sqrt{\beta}e^{i\sqrt{E\beta}x}-b\sqrt{\beta}e^{-i\sqrt{E\beta}x}\end{pmatrix}$

for any $a,\,b\in\mathbb{C}$. Note that if we substitute $E=0$, then the above
is still an eigenvector. For every $k\in\mathbb{R}$, set

(22)
$\psi_{k}(x)=\frac{\beta^{1/4}}{\sqrt{2\pi(k^{2}+\beta)}}\begin{pmatrix}ke^{i\sqrt{\beta}kx}\\\
\sqrt{\beta}e^{i\sqrt{\beta}kx}\end{pmatrix}$

so that $\psi_{k}(x)$ and $\psi_{-k}(x)$ are the linearly independent
eigenvectors corresponding to $E=k^{2}$. (We can easily substitute the
eigenfunction $\psi_{k}(x)$ into (12) to verify that $\psi_{k}$ is indeed a
solution). Note that these eigenvectors are not square-integrable, so do not
live inside the Hilbert space $L^{2}(\mathbb{R})\otimes\mathbb{C}^{2}$.
Regardless, we find that

(23)
$\displaystyle\int_{-\infty}^{\infty}\differential{x}\psi_{j}^{\dagger}(x)\psi_{k}(x)$
$\displaystyle=\frac{jk+\beta}{\sqrt{(k^{2}+\beta)(j^{2}+\beta)}}\frac{\sqrt{\beta}}{2\pi}\int_{-\infty}^{\infty}\differential{x}e^{i\sqrt{\beta}(k-j)x}$
$\displaystyle=\frac{jk+\beta}{\sqrt{(k^{2}+\beta)(j^{2}+\beta)}}\delta(k-j)$
$\displaystyle=\delta(k-j)$

where $\delta$ is the Dirac delta function. Above, we used the fact that the
Fourier transform of $\delta$ is the constant function $1$.

Equation (23) tells us that the eigenstates $\psi_{k}$ are orthonormal. The
solutions to the time-dependent Lévy-Leblond equation (7) corresponding to
these eigenvectors are

$\Psi_{k}(t,x)=e^{-ik^{2}t}\psi_{k}(x)=\frac{\beta^{1/4}}{\sqrt{2\pi(k^{2}+\beta)}}\begin{pmatrix}ke^{i\sqrt{\beta}kx-
ik^{2}t}\\\ \sqrt{\beta}e^{i\sqrt{\beta}kx-ik^{2}t}\end{pmatrix}.$

We can find a more general solution to the time-dependent Lévy-Leblond
equation by taking a continuous linear combination:

$\Psi(t,x)=\int_{-\infty}^{\infty}\differential{k}f(k)\Psi_{k}(t,x)$

for some function $f\in L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$. Note that,
although the eigenstate solutions $\Psi_{k}(t,\cdot)$ are not in
$L^{2}(\mathbb{R})\otimes\mathbb{C}^{2}$ for any $t\in\mathbb{R}$, we can
choose $f$ so that the linear combination $\Psi(t,\cdot)$ will be in
$L^{2}(\mathbb{R})\otimes\mathbb{C}^{2}$ for each $t\in\mathbb{R}$.

We have just demonstrated how the operators of the colour Lie superalgebra
$\mathfrak{D}$ can be used to solve for the eigenstates (22) of the free Lévy-
Leblond equation. This demonstrates the fundamental nature of the
$\mathbb{Z}_{2}^{3}$-colour Lie superalgebra $\mathfrak{D}$ to this Lévy-
Leblond equation.

##  4\. Ladder operators of the harmonic oscillator

The $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded symmetry of the Lévy-Leblond
equation with harmonic potential has been investigated in [3], where the
constructed symmetry algebra did not have a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-grade (contrasting with the free
equation). Despite this, we construct a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded spectrum generating algebra for
the harmonic Lévy-Leblond equation by using a different form of the Lévy-
Levlond operator: from (8) (and assuming the relations (3) hold), the time
independent equation with potential $V(x)=(k/2)x^{2}$ is

(24)
$\left(\gamma_{-}\beta-\gamma^{1}i\partial_{1}+\gamma_{+}\frac{k}{2}x^{2}\right)\ket{\psi}=\gamma_{+}E\ket{\psi}$

where $k$ is the spring constant. Note that we are using natural units
$\hbar=1$.

The Lévy-Leblond Hamiltonian is

$H_{\textup{LL}}=\gamma_{-}\beta-\gamma^{1}i\partial_{1}+\gamma_{+}\frac{k}{2}x^{2}$

and the corresponding Schrödinger-like Hamiltonian is

$H_{\textup{SL}}=\frac{1}{\beta}(H_{\textup{LL}})^{2}=-\frac{1}{\beta}(\partial_{1})^{2}+\frac{k}{2}x^{2}-\gamma_{+}i\frac{k}{\beta}x.$

From 2.4, the spectrum of $H_{\textup{LL}}$ is contained in the spectrum of
the familiar harmonic Schrödinger Hamiltonian. Therefore, the spectrum of
$H_{\textup{LL}}$ is positive and discrete.

Consider a general first order differential operator

$A=(c_{I}(x)+c_{I1}(x)\partial_{1})I+(c_{+}(x)+c_{+1}(x)\partial_{1})\gamma_{+}+(c_{-}(x)+c_{-1}(x)\partial_{1})\gamma_{-}+(c_{1}(x)+c_{11}(x)\partial_{1})\gamma^{1}.$

Our strategy is to search for ladder operators, so we want

(25) $H_{\textup{LL}}A\ket{\psi}=\gamma_{+}(E+F)A\ket{\psi}$

for $E,F\in\mathbb{C}$ and for $\ket{\psi}$ a solution to the time-independent
equation (24).

If $\commutator{H_{\textup{LL}}-\gamma_{+}E}{A}=\gamma_{+}FA$ then $A$ will
satisfy (25). Indeed,

$(H_{\textup{LL}}-\gamma_{+}E)A\ket{\psi}=A(H_{\textup{LL}}-\gamma_{+}E)\ket{\psi}+\commutator{H_{\textup{LL}}-\gamma_{+}E}{A}\ket{\psi}=\gamma_{+}FA\ket{\psi}.$

Similarly, if $\anticommutator{H_{\textup{LL}}-\gamma_{+}E}{A}=\gamma_{+}FA$
then $A$ will also satisfy (25).

######  Remark 4.1.

Recall that, for the Schrödinger Hamiltonian $H$, if $\commutator{H}{A}=FA$
then $A$ is a Ladder operator. For the Lévy-Leblond Hamiltonian
$H_{\textup{LL}}$, we also need to consider the $\gamma_{+}E$ term because we
don’t necessarily have $\commutator{\gamma_{+}E}{A}=0$.

In the case that $\commutator{H_{\textup{LL}}-\gamma_{+}E}{A}=\gamma_{+}FA$,
we obtain the following ODE,

(26) $\left\\{\begin{aligned}
\derivative{c_{1}}{x}&=\frac{i\left(Fc_{-}{\left(x\right)}+kxc_{-1}{\left(x\right)}\right)}{2}\\\
\derivative{c_{11}}{x}&=\frac{iFc_{-1}{\left(x\right)}}{2}\\\
\derivative{c_{I}}{x}&=\frac{i\left(2Ec_{-}{\left(x\right)}+Fc_{-}{\left(x\right)}+2\beta
c_{+}{\left(x\right)}-kx^{2}c_{-}{\left(x\right)}-kxc_{-1}{\left(x\right)}\right)}{2}\\\
\derivative{c_{I1}}{x}&=i\left(Ec_{-1}{\left(x\right)}+\frac{Fc_{-1}{\left(x\right)}}{2}+\beta
c_{+1}{\left(x\right)}-\frac{kx^{2}c_{-1}{\left(x\right)}}{2}\right)\\\
\derivative{c_{+}}{x}&=i\left(-2Ec_{1}{\left(x\right)}-Fc_{1}{\left(x\right)}+Fc_{I}{\left(x\right)}+kx^{2}c_{1}{\left(x\right)}+kxc_{11}{\left(x\right)}+kxc_{I1}{\left(x\right)}\right)\\\
\derivative{c_{+1}}{x}&=-2iEc_{11}{\left(x\right)}-iFc_{11}{\left(x\right)}+iFc_{I1}{\left(x\right)}+ikx^{2}c_{11}{\left(x\right)}-2c_{+}{\left(x\right)}\\\
\derivative{c_{-}}{x}&=2i\beta c_{1}{\left(x\right)}\\\
\derivative{c_{-1}}{x}&=2i\beta
c_{11}{\left(x\right)}-2c_{-}{\left(x\right)}\\\ c_{+1}{\left(x\right)}&=0\\\
c_{-1}{\left(x\right)}&=0\end{aligned}\right.$

while in the case that
$\anticommutator{H_{\textup{LL}}-\gamma_{+}E}{A}=\gamma_{+}FA$, we obtain

(27) $\left\\{\begin{aligned}
\derivative{c_{1}}{x}&=\frac{i\left(2Ec_{-}{\left(x\right)}+Fc_{-}{\left(x\right)}-2\beta
c_{+}{\left(x\right)}-kx^{2}c_{-}{\left(x\right)}-kxc_{-1}{\left(x\right)}\right)}{2}\\\
\derivative{c_{11}}{x}&=iEc_{-1}{\left(x\right)}+\frac{iFc_{-1}{\left(x\right)}}{2}-i\beta
c_{+1}{\left(x\right)}-\frac{ikx^{2}c_{-1}{\left(x\right)}}{2}-2c_{1}{\left(x\right)}\\\
\derivative{c_{I}}{x}&=\frac{i\left(Fc_{-}{\left(x\right)}+kxc_{-1}{\left(x\right)}\right)}{2}\\\
\derivative{c_{I1}}{x}&=\frac{iFc_{-1}{\left(x\right)}}{2}-2c_{I}{\left(x\right)}\\\
\derivative{c_{+}}{x}&=i\left(2Ec_{I}{\left(x\right)}-Fc_{1}{\left(x\right)}+Fc_{I}{\left(x\right)}-kx^{2}c_{I}{\left(x\right)}-kxc_{11}{\left(x\right)}-kxc_{I1}{\left(x\right)}\right)\\\
\derivative{c_{+1}}{x}&=i\left(2Ec_{I1}{\left(x\right)}-Fc_{11}{\left(x\right)}+Fc_{I1}{\left(x\right)}-kx^{2}c_{I1}{\left(x\right)}\right)\\\
\derivative{c_{-}}{x}&=2i\beta c_{I}{\left(x\right)}\\\
\derivative{c_{-1}}{x}&=2i\beta c_{I1}{\left(x\right)}\\\
c_{11}{\left(x\right)}&=0\\\ c_{I1}{\left(x\right)}&=0\end{aligned}\right.$

Solving these two equations, we obtain the following operators and
(anti)commutation relations,

$\displaystyle H_{\textup{LL}}$
$\displaystyle=\gamma^{1}(-i\partial_{1})+\gamma_{-}\beta+\gamma_{+}\frac{k}{2}x^{2}$
$\displaystyle\commutator{H_{\textup{LL}}}{H_{\textup{LL}}}$ $\displaystyle=0$
$\displaystyle b$ $\displaystyle=\sqrt{\frac{\beta
k}{2}}x+\partial_{1}-\gamma_{+}i\sqrt{\frac{k}{2\beta}}$
$\displaystyle\commutator{H_{\textup{LL}}}{b}$
$\displaystyle=-\sqrt{\frac{2k}{\beta}}\gamma_{+}b$ $\displaystyle
b^{\dagger}$ $\displaystyle=\sqrt{\frac{\beta
k}{2}}x-\partial_{1}-\gamma_{+}i\sqrt{\frac{k}{2\beta}}$
$\displaystyle\commutator{H_{\textup{LL}}}{b^{\dagger}}$
$\displaystyle=\sqrt{\frac{2k}{\beta}}\gamma_{+}b^{\dagger}$ $\displaystyle c$
$\displaystyle=\gamma_{+}2i\partial_{1}+\gamma^{1}\beta=\beta
I-2\gamma_{+}H_{\textup{LL}}$
$\displaystyle\anticommutator{H_{\textup{LL}}}{c}$ $\displaystyle=0$

From the above relations, we obtain two additional operators $\gamma_{+}b$ and
$\gamma_{+}b^{\dagger}$. We can easily compute the anticommutation relations
for $\gamma_{+}b$ and $\gamma_{+}b^{\dagger}$ by using the derivation property

$\displaystyle\anticommutator{H_{\textup{LL}}}{\gamma_{+}b}$
$\displaystyle=\anticommutator{H_{\textup{LL}}}{\gamma_{+}}b-\gamma_{+}\commutator{H_{\textup{LL}}}{b}=\beta
b$ $\displaystyle\anticommutator{H_{\textup{LL}}}{\gamma_{+}b^{\dagger}}$
$\displaystyle=\anticommutator{H_{\textup{LL}}}{\gamma_{+}}b^{\dagger}-\gamma_{+}\commutator{H_{\textup{LL}}}{b^{\dagger}}=\beta
b^{\dagger}.$

Note that the operators $\gamma_{+}b$ and $\gamma_{+}b^{\dagger}$ are also
ladder operators for $H_{\textup{SL}}$ with
$\commutator{H_{\textup{SL}}}{\gamma_{+}b}=-\sqrt{(2k)/\beta}\gamma_{+}b$ and
$\commutator{H_{\textup{SL}}}{\gamma_{+}b^{\dagger}}=\sqrt{(2k)/\beta}\gamma_{+}b^{\dagger}$.
However, $\gamma_{+}b$ and $\gamma_{+}b^{\dagger}$ are _not_ ladder operators
for $H_{\textup{LL}}$. This emphasises a key distinction between the Lévy-
Leblond equation and the Schrödinger equation: some solutions of the
Schrödinger equation will not be solutions of the Lévy-Leblond equation.

The other (anti)commutation relations between these operators that are of
interest to us are

$\displaystyle\anticommutator{H_{\textup{LL}}}{H_{\textup{LL}}}=\anticommutator{b}{b^{\dagger}}$
$\displaystyle=2\beta H_{\textup{SL}}$
$\displaystyle\anticommutator{\gamma_{+}}{H_{\textup{LL}}}$
$\displaystyle=\beta I$
$\displaystyle\commutator{\gamma_{+}}{b}=\commutator{\gamma_{+}}{b^{\dagger}}=\anticommutator{\gamma_{+}}{c}$
$\displaystyle=0$
$\displaystyle\commutator{b}{c}=\commutator{b^{\dagger}}{c}=\anticommutator{c}{\gamma_{+}b}=\anticommutator{c}{\gamma_{+}b^{\dagger}}$
$\displaystyle=0$ $\displaystyle\anticommutator{c}{c}$
$\displaystyle=2\beta^{2}I$ $\displaystyle\anticommutator{b}{\gamma_{+}b}$
$\displaystyle=2\gamma_{+}b^{2}$
$\displaystyle\anticommutator{b^{\dagger}}{\gamma_{+}b^{\dagger}}$
$\displaystyle=2\gamma_{+}(b^{\dagger})^{2}$
$\displaystyle\anticommutator{b}{\gamma_{+}b^{\dagger}}=\anticommutator{b^{\dagger}}{\gamma_{+}b}$
$\displaystyle=2\gamma_{+}\beta H_{\textup{SL}}$
$\displaystyle\commutator{\gamma_{+}b}{\gamma_{+}b^{\dagger}}$
$\displaystyle=0$

as well as $\commutator{b}{b^{\dagger}}=\sqrt{2\beta k}I$.

If we want to include $\anticommutator{b}{b^{\dagger}}=2\beta
H_{\textup{SL}}$, then we are unable to realise the (anti)commutation
relations using only a $\mathbb{Z}_{2}$-graded Lie superalgebra, since we also
need $\commutator{H_{\textup{LL}}}{b}$,
$\commutator{H_{\textup{LL}}}{b^{\dagger}}$ but
$\anticommutator{H_{\textup{LL}}}{\gamma_{+}b}$,
$\anticommutator{H_{\textup{LL}}}{\gamma_{+}b^{\dagger}}$. These
(anti)commutation relations instead require (at least) a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour algebra. To this end,
define
$\mathfrak{L}=\mathfrak{L}_{00}\oplus\mathfrak{L}_{01}\oplus\mathfrak{L}_{10}\oplus\mathfrak{L}_{11}$
by

$\displaystyle\mathfrak{L}_{00}$
$\displaystyle=\operatorname{span}\\{I,H_{\textup{SL}},b^{2},(b^{\dagger})^{2}\\}$
$\displaystyle\mathfrak{L}_{01}$
$\displaystyle=\operatorname{span}\\{b,b^{\dagger}\\}$
$\displaystyle\mathfrak{L}_{10}$
$\displaystyle=\operatorname{span}\\{H_{\textup{LL}},c,\gamma_{+},\gamma_{+}H_{\textup{SL}},\gamma_{+}b^{2},\gamma_{+}(b^{\dagger})^{2}\\}$
$\displaystyle\mathfrak{L}_{11}$
$\displaystyle=\operatorname{span}\\{\gamma_{+}b,\gamma_{+}b^{\dagger}\\}.$

We can easily verify that $\mathfrak{L}$ closes to form a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra, and
realises the (anti)commutation relations listed above (besides, of course,
$\commutator{b}{b^{\dagger}}=\sqrt{2\beta k}I$).

However, if we instead include the relation for $\commutator{b}{b^{\dagger}}$
we can then realise the (anti)commutation relations as a
($\mathbb{Z}_{2}$-graded) Lie superalgebra with even sector spanned by
$H_{\textup{SL}},b,b^{\dagger},I$ and odd sector spanned by
$H_{\textup{LL}},\gamma_{+}b,\gamma_{+}b^{\dagger}$. To summarise: including
the relation for $\commutator{b}{b^{\dagger}}$ requires a
$\mathbb{Z}_{2}$-graded Lie superalgebra, whereas including the relation
$\anticommutator{b}{b^{\dagger}}$ requires a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra. This is
analogous to the situation for the harmonic Schrödinger equation: we can
realise the relation $\commutator{b}{b^{\dagger}}$ using an ungraded Lie
algebra, but we require a $\mathbb{Z}_{2}$-graded Lie superalgebra to realise
the relation $\anticommutator{b}{b^{\dagger}}$. Both of these relations are
useful for determining the $\gamma_{+}$-eigenvalues and
$\gamma_{+}$-eigenstates, as in the following theorem.

######  Theorem 4.2.

For the harmonic Lévy-Leblond Hamiltonian $H_{\textup{LL}}$, the
$\gamma_{+}$-eigenvalues are $\omega(n+1/2)$ where $\omega=\sqrt{2k/\beta}$ is
the angular frequency and $n$ is a non-negative integer. Moreover, the
corresponding eigenstates are

$(b^{\dagger})^{n}\ket{0}$

where $\ket{0}$ is a vacuum state satisfying $b\ket{0}=0$ and
$H_{\textup{LL}}\ket{0}=(1/2)\omega$.

###### Proof.

From 2.4, the $\gamma_{+}$-spectrum of $H_{\textup{LL}}$ is a subset of the
spectrum of the corresponding Schrödinger Hamiltonian. We wish to show that
every eigenvalue of $H_{\textup{SL}}$ is a $\gamma_{+}$-eigenvalue of
$H_{\textup{LL}}$.

We know that the eigenvalues of $H_{\textup{SL}}$ are positive and discrete.
Thus, there exists some eigenvector $\ket{0_{\textup{SL}}}$ for
$H_{\textup{SL}}$ with $b\ket{0_{\textup{SL}}}=0$ (that is,
$\ket{0_{\textup{SL}}}$ is a vacuum state for $H_{\textup{SL}}$). Then,

$\displaystyle H_{\textup{SL}}\ket{0_{\textup{SL}}}$
$\displaystyle=\frac{1}{2\beta}\anticommutator{b}{b^{\dagger}}\ket{0_{\textup{SL}}}$
$\displaystyle=\frac{1}{2\beta}(bb^{\dagger}+b^{\dagger}b)\ket{0_{\textup{SL}}}$
$\displaystyle=\frac{1}{2\beta}(2b^{\dagger}b+\commutator{b}{b^{\dagger}})\ket{0_{\textup{SL}}}$
$\displaystyle=\frac{1}{2}\sqrt{\frac{2k}{\beta}}\ket{0_{\textup{SL}}}$
$\displaystyle=\frac{1}{2}\omega\ket{0_{\textup{SL}}}$

That is, $(1/2)\omega$ will be an eigenvalue of $H_{\textup{SL}}$.

Now, consider the state $\ket{0}=(\beta I+c)\ket{0_{\textup{SL}}}$. We choose
this state because the operator $\beta I+c$ has useful properties. For one, it
is nearly idempotent:

$(\beta I+c)^{2}=\beta^{2}I+2\beta c+c^{2}=2\beta(\beta I+c)$

(recalling that $c^{2}=\beta^{2}I$). Moreover, by examining the
(anti)commutation relations, $\beta I+c$ commutes with $H_{\textup{SL}}$. We
have chosen to write this operator in terms of the elements of the colour Lie
superalgebra, but we can simplify this expression:

$\beta I+c=2\beta I-2\gamma_{+}H_{\textup{LL}}=2\beta
I+2H_{\textup{LL}}\gamma_{+}-2\anticommutator{\gamma_{+}}{H_{\textup{LL}}}=2H_{\textup{LL}}\gamma_{+}.$

Therefore,

$\displaystyle H_{\textup{LL}}\ket{0}$
$\displaystyle=H_{\textup{LL}}\frac{1}{2\beta}(\beta
I+c)^{2}\ket{0_{\textup{SL}}}$
$\displaystyle=\frac{1}{\beta}H_{\textup{LL}}H_{\textup{LL}}\gamma_{+}(\beta
I+c)\ket{0_{\textup{SL}}}$ $\displaystyle=H_{\textup{SL}}\gamma_{+}(\beta
I+c)\ket{0_{\textup{SL}}}$ $\displaystyle=\gamma_{+}(\beta
I+c)H_{\textup{SL}}\ket{0_{\textup{SL}}}$ $\displaystyle=\gamma_{+}(\beta
I+c)\frac{1}{2}\omega\ket{0_{\textup{SL}}}$
$\displaystyle=\gamma_{+}\frac{1}{2}\omega\ket{0}$

That is, provided $\ket{0}\neq 0$, we have shown that $\ket{0}$ is a vacuum
state for $H_{\textup{LL}}$ with $\gamma_{+}$-eigenvalue $(1/2)\omega$.

Now we show that we can choose $\ket{0_{\textup{SL}}}$ so that $\ket{0}\neq
0$. In the case that $\gamma_{+}\ket{0_{\textup{SL}}}=0$, instead choose
$\ket{0^{\prime}_{\textup{SL}}}=H_{\textup{LL}}\ket{0_{\textup{SL}}}$ to be
the vacuum state for $H_{\textup{SL}}$. Then,

$b\ket{0^{\prime}_{\textup{SL}}}=H_{\textup{LL}}b\ket{0_{\textup{SL}}}+\commutator{b}{H_{\textup{LL}}}\ket{0_{\textup{SL}}}=0$

so $\ket{0^{\prime}_{\textup{SL}}}$ is indeed a vacuum state. Moreover,

$\gamma_{+}\ket{0^{\prime}_{\textup{SL}}}=-H_{\textup{LL}}\gamma_{+}\ket{0_{\textup{SL}}}+\anticommutator{\gamma_{+}}{H_{\textup{LL}}}\ket{0_{\textup{SL}}}=\beta\ket{0_{\textup{SL}}}\neq
0.$

That is, without loss of generality, we can assume that
$\gamma_{+}\ket{0_{\textup{SL}}}\neq 0$. Since we are assuming that
$\gamma_{+}\ket{0_{\textup{SL}}}\neq 0$, we must also have
$\ket{0}=2H_{\textup{LL}}\gamma_{+}\ket{0_{\textup{SL}}}\neq 0$, for otherwise
$\gamma_{+}\ket{0_{\textup{SL}}}$ would be a $\gamma_{+}$-eigenvector for
$H_{\textup{LL}}$ with $\gamma_{+}$-eigenvalue $0$, but $0$ is not in the
$\gamma_{+}$-spectrum for $H_{\textup{LL}}$.

We have thus shown that $\ket{0}$ is an vacuum state with
$\gamma_{+}$-eigenvalue $(1/2)\omega$. Using the commutation relation
$\commutator{H_{\textup{LL}}}{b^{\dagger}}=\omega\gamma_{+}b^{\dagger}$, we
obtain the eigenstates $(b^{\dagger})^{n}\ket{0}$ with
$\gamma_{+}$-eigenvalues $\omega(n+1/2)$. Moreover, all eigenstates can be
obtained in this way: taking an arbitrary eigenstate, we can apply the
lowering operator $b$ until a vacuum state is reached and then use
$b^{\dagger}$ to raise back up to this arbitrary eigenstate. ∎

Of particular note is the appearance of the operators from the colour Lie
superalgebra $\mathfrak{L}$ in the above proof. This demonstrates the utility
of this colour Lie superalgebra in analysing the Lévy-Leblond equation.
However, unlike the colour Lie superalgebras of [3, 4], $\mathfrak{L}$
contains operators that are not symmetry operators. In summary, despite the
super Schrödinger symmetry algebra for the harmonic potential not admitting a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded extension [3], there does still
exist a (non-symmetry) algebra $\mathfrak{L}$ that is useful for computing the
spectrum.

##  5\. Conclusions

We have extended the results of [3, 4] by showing that higher gradings (in
particular, a $\mathbb{Z}_{2}^{3}$-grading) appear for the free Lévy-Leblond
equation and that the harmonic potential also admits a
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebra.

In [3, 4], the authors only searched for first- and second-order differential
operators. In this paper, we did not restrict ourselves to differential
operators, and searched for any operators which left the
$\gamma_{+}$-eigenspaces of the free Lévy-Leblond equation invariant. By
carefully examining eigenvalue constraints, we found operators that closed to
form the $\mathbb{Z}_{2}^{3}$-graded colour Lie superalgebra $\mathfrak{D}$.
Of particular note is the non-differential operator
$\gamma^{1}\mathscr{P}\in\mathfrak{D}$ where $\mathscr{P}$ is the parity
operator. This inclusion of a non-differential operator forced the grading to
be $\mathbb{Z}_{2}^{3}$, and it is likely that broader symmetry algebras will
require $\mathbb{Z}_{2}^{3}$ or even higher $\mathbb{Z}_{2}^{n}$-gradings
($n>3$). We then used $\mathfrak{D}$ to solve the Lévy-Leblond equation,
demonstrating the fundamental nature of the $\mathbb{Z}_{2}^{3}$-graded colour
Lie superalgebra $\mathfrak{D}$ to the Lévy-Leblond equation.

The super Schrödinger symmetry algebra for the harmonic Lévy-Leblond equation
in $(1+1)$-dimensions admits no $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded
extension [3]. Instead, we considered non-symmetry operators. In particular,
we discovered a $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie
superalgebra $\mathfrak{L}$ that is generated by a Hamiltonian-like operator
with the ladder operators. Despite $\mathfrak{L}$ containing non-symmetry
operators, the algebra has utility: we used the operators from $\mathfrak{L}$
to compute the spectrum of the harmonic Lévy-Leblond equation. To the author’s
knowledge, $\mathfrak{L}$ is the first colour Lie superalgebra that has been
found for a Lévy-Leblond equation with non-vanishing potential.

These results further highlight the utility of
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded colour Lie superalgebras in
examining the Lévy-Leblond equation, and potentially other physical systems.

### Acknowledgements

This work was partially supported by the Australian Research Council through
Discovery Project DP200101339. The author is especially grateful to Phillip
Isaac for proofreading the manuscript, and for his continuing support and
guidance.

##  Appendix A Non-relativistic limit of the Dirac equation

In this appendix, we show how the Lévy-Leblond equation (corresponding to the
Schrödinger equation) can be obtained as the non-relativistic limit of the
Dirac equation. This generalises the original argument of Lévy-Leblond [1] and
provides a realisation of the Lévy-Leblond gamma matrices in
$C\ell_{1,d}(\mathbb{R})\otimes\mathbb{C}$ when $d$ is odd.

Recall that the Dirac equation is

$(\widetilde{\gamma}^{0}(1/c)i\hbar\partial_{t}+\widetilde{\gamma}^{j}i\hbar\partial_{j}-mc)\Psi(t,\mathbf{x})=0$

and is a relativistic equation describing a free spin-$1/2$ particle (see e.g.
[34, Chapter 10]). The gamma matrices appearing in the Dirac equation satisfy
$\anticommutator{\widetilde{\gamma}^{j}}{\widetilde{\gamma}^{k}}=2\eta_{jk},\,(j,k=0,\ldots,d)$,
where $\eta_{jk}$ is a symmetric bilinear form with signature $(1,d)$:

$\eta_{jk}=\begin{cases}\delta_{jk}&\text{if}~{}j=0,\\\
-\delta_{jk}&\text{otherwise}.\end{cases}$

The gamma matrices for the Dirac equation are members of the complexified
Clifford algebra $C\ell_{1,d}(\mathbb{R})\otimes\mathbb{C}$ . From now on, we
will work in natural units, $c=\hbar=1$.

We can also write a time-independent version of the Dirac equation, by
replacing $i\partial_{t}$ with the total energy $\mathscr{E}$ ($mc^{2}+$
kinetic energy):

(28)
$(\widetilde{\gamma}^{0}\mathscr{E}+\widetilde{\gamma}^{j}i\partial_{j}-m)\Psi(\mathbf{x})=0.$

Let us recall the original limit argument of Lévy-Leblond [1]. For the moment,
we shall restrict ourselves to $(1+3)$-dimensions and choose the Dirac
representation of the gamma matrices in $4\times 4$ matrices:

$\widetilde{\gamma}^{0}=\begin{pmatrix}I&0\\\
0&-I\end{pmatrix},\qquad\widetilde{\gamma}^{j}=\begin{pmatrix}0&\sigma^{j}\\\
-\sigma^{j}&0\end{pmatrix}$

where $\sigma^{j}$ are the Pauli matrices and $I$ is the identity. In this
representation, the Dirac equation becomes a coupled differential equation

(29) $\left\\{\begin{aligned}
(\mathscr{E}-m)\varphi+\sigma^{j}i\partial_{j}\chi&=0\\\
(-\mathscr{E}-m)\chi-\sigma^{j}i\partial_{j}\varphi&=0\\\
\end{aligned}\right.$

where $\Psi=\begin{pmatrix}\varphi\\\ \chi\end{pmatrix}$ is a solution to the
Dirac equation. Lévy-Leblond discovered [1] that taking $\mathscr{E}=E+m$
(where $E$ is the non-relativistic kinetic energy), and assuming $E\ll m$, we
can approximate (29) by

(30) $\left\\{\begin{aligned} E\varphi+\sigma^{j}i\partial_{j}\chi&=0\\\
-2m\chi-\sigma^{j}i\partial_{j}\varphi&=0.\\\ \end{aligned}\right.$

If $\varphi$ and $\chi$ solve (30), then $\Psi=\begin{pmatrix}\chi\\\
\varphi\end{pmatrix}$ is a solution to the Lévy-Leblond equation, with

$\gamma_{+}=\begin{pmatrix}0&I\\\
0&0\end{pmatrix},\qquad\gamma_{-}=\begin{pmatrix}0&0\\\
I&0\end{pmatrix},\qquad\gamma^{j}=\begin{pmatrix}\sigma^{j}&0\\\
0&-\sigma^{j}\end{pmatrix}$

(notice that the components $\varphi$ and $\chi$ have swapped places when
compared to the solution of the Dirac equation). That is, the Lévy-Leblond
equation is the non-relativistic limit of the Dirac equation.

We can generalise the above argument of Lévy-Leblond for all odd space
dimensions in a representation independent way. Consider the time-independent
Dirac equation in $(1+d)$-dimensions, (28) and assume $d$ is odd. Let

$\widetilde{\gamma}^{\mathrm{chir}}=i^{(d+3)d/2}\prod_{j=0}^{d}\widetilde{\gamma}^{j}.$

Since the $\widetilde{\gamma}^{j}$ anticommute, we have that
$\widetilde{\gamma}^{j}\widetilde{\gamma}^{\mathrm{chir}}=(-1)^{d}\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{j},\,(j=0,\ldots,d)$.
Assuming that $d$ is odd, we have that
$\anticommutator{\widetilde{\gamma}^{j}}{\widetilde{\gamma}^{\mathrm{chir}}}=0$.
Additionally,

$\widetilde{\gamma}^{\mathrm{chir}}=i^{(d+3)d/2}(-1)^{d+(d-1)+\cdots+1}\prod_{j=0}^{d}\widetilde{\gamma}^{d-j}=i^{(d+3)d/2}(-1)^{(d+1)d/2}\prod_{j=0}^{d}\widetilde{\gamma}^{d-j}.$

Therefore,
$(\widetilde{\gamma}^{\mathrm{chir}})^{2}=(i^{2})^{(d+3)d/2}(-1)^{(d+1)d/2+d}I=I$.
In particular, $\widetilde{\gamma}^{\mathrm{chir}}$ is invertible so $\Psi$ is
a solution to the time-independent Dirac equation if and only if

(31)
$\widetilde{\gamma}^{\mathrm{chir}}(\widetilde{\gamma}^{0}\mathscr{E}+\widetilde{\gamma}^{j}i\partial_{j}-m)\Psi(\mathbf{x})=0.$

(The above multiplication by $\widetilde{\gamma}^{\mathrm{chir}}$ corresponds
to the swapping of components $\varphi$ and $\chi$ in Lévy-Leblond’s original
argument above.) Take the non-relativistic limit by setting $\mathscr{E}=E+m$
and assuming $E\ll m$. We can compute

$\displaystyle 0$
$\displaystyle=\widetilde{\gamma}^{\mathrm{chir}}(\widetilde{\gamma}^{0}\mathscr{E}+\widetilde{\gamma}^{j}i\partial_{j}-m)\Psi(\mathbf{x})$
$\displaystyle=(\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{0}(E+m)+\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{j}i\partial_{j}-\widetilde{\gamma}^{\mathrm{chir}}m)\Psi(\mathbf{x})$
$\displaystyle=\left(\frac{1}{2}(\widetilde{\gamma}^{\mathrm{chir}}+\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{0})E-\frac{1}{2}(\widetilde{\gamma}^{\mathrm{chir}}-\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{0})(E+2m)+\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{j}i\partial_{j}\right)\Psi(\mathbf{x})$
(32)
$\displaystyle\approx(\gamma_{+}E-\gamma_{-}2m+\gamma^{j}i\partial_{j})\Psi(\mathbf{x})$

where

(33)
$\gamma_{\pm}=\frac{1}{2}(\widetilde{\gamma}^{\mathrm{chir}}\pm\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{0}),\qquad\gamma^{j}=\widetilde{\gamma}^{\mathrm{chir}}\widetilde{\gamma}^{j},\,(j=1,\ldots,d).$

It is easily verified that $\gamma_{\pm},\,\gamma^{j}$ satisfy the anti-
commutation relations (2) by using the fact that
$\anticommutator{\widetilde{\gamma}^{j}}{\widetilde{\gamma}^{\mathrm{chir}}}=0$
(when $d$ is odd). We can now use the time independent Lévy-Leblond equation
(32) to derive a time-dependent Lévy-Leblond equation by replacing $E$ with
$i\partial_{t}$:

$(\gamma_{+}\partial_{t}+\gamma_{-}i2m+\gamma^{j}\partial_{j})\Psi(t,\mathbf{x})=0$

which has the desired form (1).

It is important to note that the realisation in (33) satisfies relations (such
as $\gamma_{+}\gamma^{j}+\gamma_{-}\gamma^{j}=\gamma^{j}$) that are not
satisfied by every representation of (2). In particular, when $d=1$, relations
(3) are satisfied. Despite being less general, the realisation in (33) has the
advantage that the gamma matrices can be defined in terms of a bilinear form
with signature $(1,d)$, representing $1$ time dimension and $d$ spacial
dimensions.

## References

  * [1] J.-M. Lévy-Leblond, Nonrelativistic particles and wave equations, Comm. Math. Phys. 6 (4) (1967) 286–311.  doi:10.1007/BF01646020.
  * [2] P. A. M. Dirac, The quantum theory of the electron, Proc. R. Soc. A 117 (778) (1928) 610–624.  doi:10.1098/rspa.1928.0023.
  * [3] N. Aizawa, Z. Kuznetsova, H. Tanaka, F. Toppan, $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded Lie symmetries of the Lévy-Leblond equations, PTEP. Prog. Theor. Exp. Phys. (12) (2016) 123A01, 26.  doi:10.1093/ptep/ptw176.
  * [4] N. Aizawa, Z. Kuznetsova, H. Tanaka, F. Toppan, Generalized supersymmetry and the lévy-leblond equation, in: S. Duarte, J.-P. Gazeau, S. Faci, T. Micklitz, R. Scherer, F. Toppan (Eds.), Physical and Mathematical Aspects of Symmetries, Springer International Publishing, Cham, 2017, pp. 79–84.
  * [5] F. Toppan, Symmetries of the Schrödinger equation and algebra/superalgebra duality, J. Phys. Conf. Ser. 597 (1) (2015) 012071.  doi:10.1088/1742-6596/597/1/012071.
URL https://dx.doi.org/10.1088/1742-6596/597/1/012071

  * [6] V. Rittenberg, D. Wyler, Generalized superalgebras, Nuclear Phys. B 139 (3) (1978) 189–202.  doi:10.1016/0550-3213(78)90186-4.
  * [7] V. Rittenberg, D. Wyler, Sequences of $Z_{2}\oplus Z_{2}$ graded Lie algebras and superalgebras, J. Math. Phys. 19 (10) (1978) 2193–2200.  doi:10.1063/1.523552.
  * [8] R. Ree, Generalized Lie elements, Canadian J. Math. 12 (1960) 493–502.  doi:10.4153/CJM-1960-044-x.
  * [9] A. J. Bruce, S. Duplij, Double-graded supersymmetric quantum mechanics, J. Math. Phys. 61 (6) (2020) 063503, 13.  doi:10.1063/1.5118302.
  * [10] N. Aizawa, Z. Kuznetsova, F. Toppan, $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded mechanics: the classical theory, Eur. Phys. J. C 80 (2020).  doi:10.1140/epjc/s10052-020-8242-x.
  * [11] N. Aizawa, Z. Kuznetsova, F. Toppan, $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded mechanics: The quantization, Nuclear Phys. B 967 (2021) 115426.  doi:10.1016/j.nuclphysb.2021.115426.
  * [12] N. Aizawa, K. Amakawa, S. Doi, $\mathcal{N}$-extension of double-graded supersymmetric and superconformal quantum mechanics, J. Phys. A 53 (6) (2020) 065205, 14.  doi:10.1088/1751-8121/ab661c.
  * [13] N. Aizawa, K. Amakawa, S. Doi, $\mathbb{Z}_{2}^{n}$-graded extensions of supersymmetric quantum mechanics via Clifford algebras, J. Math. Phys. 61 (5) (2020) 052105, 13.  doi:10.1063/1.5144325.
  * [14] S. Doi, N. Aizawa, $\mathbb{Z}_{2}^{3}$-graded extensions of Lie superalgebras and superconformal quantum mechanics, SIGMA Symmetry Integrability Geom. Methods Appl. 17 (2021) Paper No. 071, 14.  doi:10.3842/SIGMA.2021.071.
  * [15] N. Aizawa, K. Amakawa, S. Doi, Color algebraic extension of supersymmetric quantum mechanics, in: Quantum theory and symmetries, CRM Ser. Math. Phys., Springer, Cham, 2021, pp. 199–207.  doi:10.1007/978-3-030-55777-5\\_18.
  * [16] A. J. Bruce, Is the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded sine-Gordon equation integrable?, Nuclear Phys. B 971 (2021) Paper No. 115514, 10.  doi:10.1016/j.nuclphysb.2021.115514.
  * [17] C. Quesne, Minimal bosonization of double-graded quantum mechanics, Modern Phys. Lett. A 36 (33) (2021) Paper No. 2150238, 8.  doi:10.1142/S0217732321502382.
  * [18] N. Aizawa, R. Ito, Z. Kuznetsova, F. Toppan, New aspects of the $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded $1D$ superspace: induced strings and $2D$ relativistic models, Nuclear Phys. B 991 (2023) Paper No. 116202, 27.  doi:10.1016/j.nuclphysb.2023.116202.
URL https://doi.org/10.1016/j.nuclphysb.2023.116202

  * [19] W. Yang, S. Jing, A new kind of graded Lie algebra and parastatistical supersymmetry, Sci. China Ser. A 44 (9) (2001) 1167–1173.  doi:10.1007/BF02877435.
  * [20] S.-C. Jing, W.-M. Yang, P. Li, Graded Lie algebra generating of parastatistical algebraic relations, Commun. Theor. Phys. (Beijing) 36 (6) (2001) 647–650.  doi:10.1088/0253-6102/36/6/647.
  * [21] K. Kanakoglou, A. Herrera-Aguilar, Ladder operators, Fock-spaces, irreducibility and group gradings for the relative parabose set algebra, Int. J. Algebra 5 (9-12) (2011) 413–427.
  * [22] K. Kanakoglou, A. Herrera-Aguilar, Graded fock-like representations for a system of algebraically interacting paraparticles, J. Phys.: Conf. Ser. 287 (1) (2011) 012037.  doi:10.1088/1742-6596/287/1/012037.
  * [23] V. N. Tolstoy, Once more on parastatistics, Phys. Part. Nuclei Lett. 11 (2014) 933–937.  doi:10.1134/S1547477114070449.
  * [24] N. I. Stoilova, J. Van der Jeugt, The $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded Lie superalgebra $pso(2m+1|2n)$ and new parastatistics representations, J. Phys. A 51 (13) (2018) 135201, 17.  doi:10.1088/1751-8121/aaae9a.
  * [25] F. Toppan, $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded parastatistics in multiparticle quantum Hamiltonians, J. Phys. A 54 (11) (2021) Paper No. 115203, 35.  doi:10.1088/1751-8121/abe2f2.
  * [26] F. Toppan, Inequivalent quantizations from gradings and $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ parabosons, J. Phys. A 54 (35) (2021) Paper No. 355202, 21.  doi:10.1088/1751-8121/ac17a5.
  * [27] R. Zhang, Para-spaces, their differential analysis and an application to green’s quantisation (2023).  arXiv:2312.04250.
  * [28] N. I. Stoilova, J. V. d. Jeugt, Orthosymplectic $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$-graded Lie superalgebras and parastatistics, J. Phys. A 57 (9) (2024) Paper No. 095202, 13.
  * [29] Y. Bahturin, D. Pagon, Classifying simple color Lie superalgebras, in: Algebras, representations and applications, Vol. 483 of Contemp. Math., Amer. Math. Soc., Providence, RI, 2009, pp. 37–54.  doi:10.1090/conm/483/09433.
  * [30] M. Ryan, Refining the grading of irreducible lie colour algebra representations (2024).  arXiv:2403.02855.
  * [31] S. Bao, D. Constales, H. De Bie, T. Mertens, Solutions for the Lévy-Leblond or parabolic Dirac equation and its generalizations, J. Math. Phys. 61 (1) (2020) 011509, 12.  doi:10.1063/1.5135503.
  * [32] N. Faustino, On fractional semidiscrete Dirac operators of Lévy-Leblond type, Math. Nachr. 296 (7) (2023) 2758–2779.
  * [33] M. Scheunert, Generalized Lie algebras, J. Math. Phys. 20 (4) (1979) 712–720.  doi:10.1063/1.524113.
  * [34] P. Lounesto, Clifford algebras and spinors, 2nd Edition, Vol. 286 of London Mathematical Society Lecture Note Series, Cambridge University Press, Cambridge, 2001.  doi:10.1017/CBO9780511526022.