\chapter{Coupled Supersymmetric Quantum Mechanics and Associated Integral Operators} \label{chapter-coupledsusy}



In Chapter 3 generalized Fourier transforms that emerged from an axiomatic framework were investigated. These were identified as the unitary transforms that diagonalized certain singular Laplacian operators $\Delta_n = -\frac{d}{dx}\frac{1}{x^{2n-2}}\frac{d}{dx}$. The motivation for the present work came from the question whether these transforms are related to a type of oscillator Hamiltonian---or Hamiltonian-like operator. This is indeed the case. If $a_n = \frac{1}{\sqrt{2}} \left(\frac{1}{x^{n-1}} \frac{d}{dx} + x^n\right)$, then $\mathcal H = a_n^* a_n$ has the set of eigenvalues $\{2kn, 2kn+2n-1\}_{k=0}^\infty$ and the corresponding eigenfunctions are complete in $L^2(\R)$.

While supersymmetric quantum mechanics has a somewhat nice \emph{algebraic} structure, the algebraic structure is not strong enough to give meaningful information about the analytic structure of the underlying Hamiltonians. In this chapter, we develop a family of supersymmetric systems which have a rich analytic structure and also a richer algebraic structure than traditional supersymmetric quantum mechanics. We also develop analogues of the Fourier transform for these systems and present further generalizations thereof.


  
\section{Coupled Supersymmetry and Its Lie Algebra}



In Chapter 3, it was shown that the QMHO has two separate factorizations: $\mathcal{H}_{\text{HO}} = a^*a + \frac{1}{2}$ and $\mathcal{H}_{\text{HO}} = aa^*-\frac{1}{2}$. Defining then the SUSY Hamiltonian $\mathcal{H}_1 = a^*a$, its SUSY partner Hamiltonian is $\mathcal{H}_2 = aa^*$. From the two factorizations of the QMHO, it is clear that $\mathcal{H}_1$ and $\mathcal{H}_2$ each have two distinct factorizations, even though these two factorizations are intimately connected.

That $\mathcal{H}_2 = \mathcal{H}_1+1$ is a restatement of the commutation relation for $a$ and $a^*$---which is equivalent to the canonical commutation relation. The canonical commutation relation can be a very rigid structure \cite[p.274]{reed2} and as such cannot be a point of generalization. Note that the canonical commutation relations do not necessarily guarantee that $[a^*,a] = -I$ is solved by $a = x+ip$, see \cite[p.275]{reed2}. The Schr\"odinger representation is not strong enough to guarantee uniqueness, however the exponentiated (group) representation is. As the canonical commutation relations can be very rigid, we instead use the property that the QMHO and its partner Hamiltonian both have two distinct factorizations to develop our theory and this leads into our first definition.

\begin{definition}
Let $a,b$ be closed, densely defined operators on a Hilbert space $\mathfrak{H}$, $a^*$ and $b^*$ be their adjoints, and $\gamma,\delta\in\R$ with $\gamma<\delta$. Furthermore, suppose that $\dom a = \dom b$, $\dom a^* = \dom b^*$, $\ran a\subseteq \dom a^*$ (and vice versa), and similarly for $b$ and $b^*$. The ordered quadruplet $\{a,b,\gamma,\delta\}$ defines a \emph{coupled supersymmetric system} (\emph{coupled SUSY system}) if it satisfies
%
\begin{align}
    a^*a &= b^*b + \gamma, \label{eq:susy_1} \\
    aa^* &= bb^* + \delta. \label{eq:susy_2}
\end{align}

\noindent The operators $a^*a$, $aa^*$, $b^*b$ and $bb^*$ will be referred to as Hamiltonians throughout the chapter.
\end{definition}

It is easily seen that the coupled SUSY conditions are equivalent to the system of equations for commutators $[a^*,a] = [b^*,b]+\gamma-\delta$ and anticommutators $\{a^*,a\} = \{b^*,b\} + \gamma+\delta$. Note that \eqref{eq:susy_1} and \eqref{eq:susy_2} do not imply each other, as evidenced by taking $a = \frac{1}{\sqrt{2}}\left(\frac{d}{dx}+x\right)$ and $b = \frac{1}{\sqrt{2}} e^{ix}\left(\frac{d}{dx}+x\right)$. In this case, \eqref{eq:susy_1} holds whereas \eqref{eq:susy_2} does not. Placing the exponential on the right in the definition of $b$ shows that the reverse implication does not hold either.

The condition that $\gamma \neq \delta$ will play a crucial role in what follows. We assumed without loss of generality that $\gamma < \delta$; otherwise, if $\gamma > \delta$, we may simply switch the roles of $a$ and $a^*$ as well as $b$ and $b^*$ and the conditions for Definition 1 would still hold. 

\begin{example}
Definition 1 includes the QMHO by letting $b^* = \frac{1}{\sqrt{2}} \left(\frac{d}{dx}+x\right) = a$, $-\gamma = 1 = \delta$. There exists an infinite family of examples satisfying the coupled SUSY equations. A straightforward calculation shows that, for $n\in\N$, the operators
%
\begin{align}
    a_n &= \frac{1}{\sqrt{2}} \left(\frac{1}{x^{n-1}} \frac{d}{dx} + x^n\right) \\
    b_n &= \frac{1}{\sqrt{2}} \left(-\frac{1}{x^{n-1}} \frac{d}{dx} + x^n\right)
\end{align}

\noindent taken with their adjoints on $L^2(\R)$ also define a coupled SUSY system when restricted to an appropriate subspace where $\gamma = -1$ and $\delta = 2n-1$. This family of examples is closely related to standard SUSY with the anharmonic superpotential $W_1(x) = x^{2n-1}$ where the charge operator $\frac{d}{dx}+x^{2n-1}$ has been multiplied on the left by $\frac{1}{x^{n-1}}$. This relationship has connections to canonical transformations and mutually unbiased bases \cite{kouri3}.
\end{example}

In SUSY, one considers broken and unbroken (or exact) systems. SUSY is unbroken if at least one of $\mathcal{Q}_1$ and $\mathcal{Q}_1^*$ in the factorization $\mathcal{H}_1 = \mathcal{Q}_1^*\mathcal{Q}_1$ annihilates a state and is broken if neither annihilate a state \cite{cooper}. Unbroken SUSY is the primary focus in the study of basic supersymmetric quantum mechanics and is the focus of much of the remainder of this chapter. To this end, we make the following definition.

\begin{definition}
Let $\{a,b,\gamma,\delta\}$ form a coupled SUSY system. We say that it is an \emph{unbroken} coupled SUSY system if $a$ annihilates a state and $b^*$ annihilates a state. Otherwise we say that it is a \emph{broken} coupled SUSY system.
\end{definition}

\begin{remark}
If $\{a,b,\gamma,\delta\}$ form an unbroken coupled SUSY system, it cannot be the case that $a^*$ and $b$ annihilate states since otherwise $\delta < \gamma$, contradicting the assumption that $\gamma < \delta$. If however \eqref{eq:susy_1} and \eqref{eq:susy_2} hold but $a^*$ and $b$ annihilate states, we may reverse the roles of $a$ and $a^*$ (and $b$ and $b^*$) to obtain an unbroken coupled SUSY system.
\end{remark}

Coupled SUSY, as the name suggests, automatically comes with a SUSY structure by noting that $\mathcal{H}_1 = a^*a$ and $\mathcal{H}_2 = aa^*$ come equipped with the usual SUSY ladder structure. In our first theorem, we show that a ladder structure exists for coupled SUSY.

\begin{thm}
 If $\{a,b,\gamma,\delta\}$ form a coupled SUSY system, then $a^*b$ and $b^*a$ act as ladder operators for $a^*a$ (and $b^*b$) while $ab^*$ and $ba^*$ act as ladder operators for $aa^*$ (and $bb^*$). Moreover, the triples $\{a^*a - \frac{\gamma}{2}, a^*b, b^*a\}$ and $\{aa^*-\frac{\delta}{2},ab^*,ba^*\}$ generate Lie algebras isomorphic to $\mathfrak{su}(1,1)$.
\end{thm}

\begin{proof}
To prove this, we proceed in much the same way as in the standard QMHO by considering the commutator of $a^*a$ with $a^*b$, $a^*a$ with $b^*a$, and $a^*b$ with $b^*a$. The other cases with $aa^*$, $ab^*$ and $ba^*$ follow the same logic and so they are omitted for the sake of brevity.
%
\begin{align*}
    [a^*a,a^*b] &= a^*aa^*b - a^*b a^*a \\
    &= a^*aa^*b - a^*b(b^*b+\gamma) \\
    &= a^*(aa^*-bb^*)b - \gamma a^*b \\
    &= (\delta-\gamma)a^*b.
\end{align*}

Similar reasoning shows that $[a^*a,b^*a] = -(\delta-\gamma)b^*a$. Since $\gamma<\delta$, $a^*b$ is a raising operator for $a^*a$ and $b^*a$ is a lowering operator for $a^*a$. To show that these generate a Lie algebra, we  inspect the commutator of $a^*b$ and $b^*a$:
%
\begin{align*}
    [a^*b,b^*a] &= a^*bb^*a - b^*aa^*b \\
    &= a^*(aa^*-\delta)a - b^*(bb^*+\delta)b \\
    &= (a^*a)^2 - (b^*b)^2 - \delta(a^*a + b^*b) \\
    &= 2(\gamma-\delta)\left(a^*a - \frac{\gamma}{2}\right).
\end{align*}

Thus, the triple generates a Lie algebra as it is closed under commutation. After adding a multiple of the identity to $a^* a$ and rescaling,
%
\begin{equation}
   \mathcal K_+ = \frac{1}{\delta-\gamma} a^* b, \quad \mathcal K_- = \frac{1}{\delta-\gamma} b^* a, \quad \mathcal K_0 = \frac{1}{\delta-\gamma} \left(a^* a - \frac \gamma 2 \right)
\end{equation}

\noindent are seen to verify
%
\begin{equation}
    [\mathcal{K}_0, \mathcal{K}_{\pm}] = \pm \mathcal{K}_{\pm}, \quad [\mathcal{K}_+, \mathcal{K}_-] = -2\mathcal{K}_0
\end{equation}

\noindent the commutation relations of $\mathfrak{su}(1,1)$ \cite{perelomov}, hence the Lie algebra associated to coupled SUSY is found to be isomorphic to the $\mathfrak{su}(1,1)$ Lie algebra.
\end{proof}

We note that the discrete series of $\mathfrak{su}(1,1)$ representations correspond to a choice of $\frac{\gamma}{\delta-\gamma} \in \mathbb Z$, $\frac{\gamma}{\delta-\gamma} \le -2$ \cite{perelomov}.

As with any mathematical structure, it is of interest to %inspect whether there are edge cases to consider, or if the structure tells the whole story. Particularly, a natural question to 
ask is if there are systems for which the commutation relations in Theorem 1 hold that do not correspond to a coupled SUSY system---be it a broken or unbroken coupled SUSY. In the next theorem, we prove that this is not the case under the modest assumption that $\ker a^* = \{0\} = \ker b$.

\begin{thm}
If $a,a^*,b,b^*$ are operators satisfying $\ker a^* = \{0\} = \ker b$ and
%
\begin{align}
    [a^*a,a^*b] &= \lambda a^*b, \\
    [bb^*,ba^*] &= \lambda' ba^*, \\
    [a^*b,b^*a] &= \mu a^*a + \nu, \\
    [ba^*,ab^*] &= \mu'aa^* + \nu',
\end{align}

\noindent where $\lambda,\lambda',\nu,\nu'\in\R$, $\lambda \ne 0$, then $b^*b = \alpha a^*a + \gamma$ and $bb^* = \beta aa^* + \delta$ for some $\alpha,\beta,\gamma,\delta\in\R$.
\end{thm}

\begin{proof}
Suppose that $a^*a = b^*b + S$ and $aa^* = bb^* + T$ for some as-of-yet undetermined operators $S$ and $T$. Inspecting commutation relations, we have
%
\begin{align*}
    [a^*a,a^*b] &= a^*(aa^*b - ba^*a) \\
    &= a^*((bb^*+T)b - b(b^*b + S)) \\
    &= a^*(Tb-bS).
\end{align*}

\noindent Likewise, it follows that
%
\begin{equation*}
    [bb^*,ba^*] = b(a^*T-Sa^*).
\end{equation*}

\noindent Equating the above with the ladder operator relations, it can be seen that
%
\begin{align*}
    [a^*a,a^*b] = \lambda a^*b &= a^*(Tb-bS) \\
    [bb^*,ba^*] = \lambda' ba^* &= b(a^*T-Sa^*).
\end{align*}

\noindent Since $a^*$ and $b$ have trivial kernel, we must have that $Tb-bS = \lambda b$ and similarly $a^*T-Sa^* = \lambda'a^*$. We may use these to prove our result.
%
\begin{align*}
    [ba^*, ab^*] &= b(b^*b + S)b^* - a(a^*a-S)a^* \\
    &= (bb^*)^2 + bSb^* - (aa^*)^2 + aSa^* \\
    &= (aa^* - T)^2 + bSb^* - (aa^*)^2 + aSa^*\\
    &= -aa^*T - Taa^* + T^2 + bSb^* + aSa^* \\
    &= -aa^*T-T(bb^*+T) + T^2 + bSb^* + aSa^* \\
    &= -a(a^*T-Sa^*) - (Tb-bS)b^* \\
    &= -\lambda'aa^* - \lambda bb^*
\end{align*}

\noindent Making use of our relations above, it follows that
%
\begin{equation*}
    [ba^*,ab^*] = \mu' aa^*+\nu' = -\lambda'aa^* - \lambda bb^*.
\end{equation*}

\noindent Therefore
%
\begin{equation}
    bb^* = -\frac{\lambda'+\mu'}{\lambda}aa^* - \frac{\nu'}{\lambda}
\end{equation}

By repeating these steps with $[a^*b,b^*a]$ and using the above relations $Tb-bS = \lambda b$ and $a^*T-Sa^* = \lambda'a^*$, $[a^* b, b^* a]= - \lambda b^* b - \lambda' a^* a = \mu a^* a + \nu$ the claimed affine relationship between $a^*a$ and $b^*b$ holds as well.
\end{proof}

Taking $\lambda=\lambda'$ in the previous theorem gives exactly that $a,b$ form a coupled SUSY system as $\alpha = 1 = \beta$ in this case.

It is natural to ask if a coupled SUSY system is unique, i.e., given $a$, $\gamma$, $\delta$ if $b$ is unique or if there are many possible choices for $b$. We prove a uniqueness result in the next theorem.

\begin{thm}
Suppose that $\{a,b,\gamma,\delta\}$ and $\{a,c,\gamma,\delta\}$ define coupled SUSY systems. Then $b = U (c^* c)^{1/2} $ and $c = V (b^* b)^{1/2}$ for some partial isometries $U,V$.
\end{thm}

\begin{proof}
 Since we have that
 %
 \begin{align}
     b^*b + \gamma &= a^*a = c^*c + \gamma, \\
     bb^* + \delta &= aa^* = cc^* + \delta,
 \end{align}
 It follows that $b^*b = c^*c$ and $bb^* = cc^*$. The equivalences are warranted as the operators are defined on the same subspaces. From the polar decomposition for closed operators \cite{reed1}, the first equality guarantees that $b = U(c^* c)^{1/2}$ for some partial isometry $U$. Switching roles gives $c = V (b^* b)^{1/2}$ for some partial isometry $V$.
\end{proof}



\section{An Energy Ladder Structure for Coupled SUSY}



With ladder operators established for coupled SUSY systems, it is natural to inquire about the eigenvalues of the coupled SUSY Hamiltonians $a^*a$, $aa^*$, $b^*b$, and $bb^*$. As in standard SUSY, $a^*a$ and $aa^*$ share the same eigenvalues---up to a possible eigenvalue of $0$. Likewise, $b^*b$ and $bb^*$ share the same eigenvalues---up to a possible eigenvalue of $0$. Moreover, the spectra of $a^*a$ and $b^*b$ are related by a shift of $\gamma$ since $a^*a = b^*b + \gamma$. Thus it is sufficient to study one of $a^*a$ and $aa^*$ to fully understand the eigenvalues of any of the Hamiltonians in a coupled SUSY system.

\begin{thm}
If $\{a,b,\gamma,\delta\}$ form an unbroken coupled SUSY system, then the eigenvalues of $a^*a$ are given by $m(\delta-\gamma)$ and $m(\delta-\gamma)+\delta$ where $m\in\N_0$.
\end{thm}

\begin{proof}
We first note that if $\psi\in\dom{a}$ is an eigenfunction of $a^*a$, then $a^*b\psi\in\dom{a}$; particularly, it is normalizable. To see this, note that $\langle a^*b\psi,a^*b\psi\rangle = \langle \psi,b^*aa^*b\psi \rangle = \langle \psi,b^*(bb^*+\delta)b\psi\rangle < \infty$ since $\psi$ is also an eigenstate of $b^*b$. An analogous result holds for $aa^*$.

Since $a^*b$ is a raising operator for $a^*a$ and $0$ is an eigenvalue of $a^*a$, $m(\delta-\gamma)$ is an eigenvalue for $a^*a$. 
Moreover, $\delta$ is an eigenvalue for $aa^*$ since $0$ is an eigenvalue of $bb^*$. $ba^*$ is a raising operator for $aa^*$, so $\delta+m(\delta-\gamma)$ is an eigenvalue for $aa^*$. Since $a^*a$ and $aa^*$ share eigenvalues---up to $0$---$m(\delta-\gamma)+\delta$ is an eigenvalue of $a^*a$.

These are indeed \emph{all} of the eigenvalues for $a^*a$. If $\lambda$ is an eigenvalue of $a^*a$, then so is $\lambda-(\delta-\gamma)$ since $b^*a$ is a lowering operator for $a^*a$. If an eigenvalue $\lambda$ (corresponding to the eigenfunction $\psi$) existed between $0$ and $\delta-\gamma$, then $\lambda-(\delta-\gamma) < 0$ would be an eigenvalue of $a^*a$ (corresponding to the eigenfunction $b^*a\psi$) which contradicts the positivity of $a^*a$. Similar reasoning shows that no eigenvalue can exist between $m(\delta-\gamma)$ and $m(\delta-\gamma)+\delta$, which proves the theorem.
\end{proof}

A consequence of this is that there exist no bounded operator representations for a coupled SUSY system since the eigenvalues are unbounded. Particularly, no finite-dimensional (matricial) representations exist.

With the ladder structure for $a^*a$ (and $aa^*$) via $a^*b$ and $b^*a$ (and $ba^*$ and $ab^*$),  a richer ladder structure exists than the standard SUSY or QMHO structure. We already know that $a$ and $a^*$ transfer between the sectors generated by $a^*a$ and $aa^*$ so we wish to explore the structure that lies beyond this.

In general, there need not be only one state that is annihilated by $a$ (or more generally two states annihilated by $b^*a$). For instance, it could be that
%
\begin{equation}
    a = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} \frac{d}{dx}+x & 0 \\ 0 & \frac{d}{dx}+x \end{array}\right),
\end{equation}
which annihilates the states $(\exp(-x^2/2),0)^{\operatorname{T}}$ and $(0,\exp(-x^2/2))^{\operatorname{T}}$. As such we define the following notation.

\begin{definition}
Let $\{a,b,\gamma,\delta\}$ be an unbroken coupled SUSY system. Let $\psi_{i,0}$, $i\in I$ for some finite or at most countable index set $I$, be an orthonormal family of (normalized) state vectors annihilated by $a$ and $\phi_{j,0}$, $j\in J$ for some finite or at most countable index set $J$, be an orthonormal family of state vectors annihilated by $b^*a$ but not annihilated by $a$. Define then $\psi_{i,m} = (a^*b)^m\psi_{i,0}/\|(a^*b)^m\psi_{i,0}\|$ and $\phi_{j,m} = (a^*b)^m \phi_{j,0}/\|(a^*b)^m \phi_{j,0}\|$.
\end{definition}

In the case of the harmonic oscillator, the coupled SUSY system collapses because $b=a^*$ and the kernel of $a$ is spanned by
the Gaussian $h_0$, while a vector annihilated by $b^* a = a^2$ but not by $a$ is necessarily a multiple of the first excited state $h_1$.
Note that in general the states annihilated by $b^*a$ but not by $a$ are in one-to-one correspondence with the states annihilated by $b^*$ via the usual SUSY ladder structure.

Consider the following pairs of operators $a$ and $b$ which have the following (infinite) matrix representations in an orthonormal basis:

\begin{align}
    a &= \begin{pmatrix} 
     {\bf 0}_{n\times m} & \sqrt{\delta}{\bf 1}_{n\times n} & & \\[1.5em]
     & \ddots & \sqrt{\delta-\gamma}{\bf 1}_{m\times m} &  \\
     & & \ddots & \ddots
    \end{pmatrix}, \\
    b &= \begin{pmatrix}
    {\bf 0}_{n\times m} & & \\
    \sqrt{-\gamma}{\bf 1}_{m\times m} & \ddots & \\
    & \sqrt{\delta-\gamma}{\bf 1}_{n\times n} & \\
    & \ddots & \ddots
    \end{pmatrix},
\end{align}

\noindent where ${\bf 0}_{n_1\times n_2}$ is the $n_1\times n_2$ zero matrix, ${\bf 1}_{n_1\times n_2}$ is the $n_1\times n_2$ ones matrix, and all other elements are taken to be zero. Then taking their adjoints, we have that $a^*a = b^*b + \gamma$ and $aa^* = bb^*+\delta$, however $\dim \ker a \neq \dim \ker b^*$ unless $m=n$, thus $|I|\neq |J|$ in general. Moreover, this shows that one can \emph{build} coupled SUSY systems without much difficulty.

\begin{thm}
Let $\{a,b,\gamma,\delta\}$ define an unbroken coupled SUSY system. For any $i\in I$, $j\in J$, and $m\in\N_0$, $\psi_{i,m}$ and $\phi_{j,m}$ are eigenfunctions of $a^*a$ and any normalized eigenfunction of $a^*a$ is of the form $\sum_i \lambda_i \psi_{i,m}$ or $\sum_j \eta_j \phi_{j,m}$ for some finite collection $\lambda_i,\eta_j\in\C$.
\end{thm}

\begin{proof}
It is clear by earlier arguments that $\psi_{i,m},\phi_{j,m}$ are eigenfunctions of $a^*a$. Conversely, every eigenfunction is linear combination of the $\psi_{i,m}$ or $\phi_{j,m}$ for fixed $m$. To prove this, we proceed by induction on the eigenvalue of a given eigenfunction $\zeta$.

Without loss of generality, we consider the case that $a^*a\zeta = m(\delta-\gamma)\zeta$. The case of $a^*a\zeta = (m(\delta-\gamma)+\delta)\zeta$ proceeds similarly. For $m=0$ this is trivial since $\ker a = \operatorname{span}\{\psi_{i,0}:i\in I\}$ and $\ker a^* = \{0\}$. Assume that $a^*a\zeta = m(\delta-\gamma)\zeta$ implies that $\zeta = \sum_i \lambda_i \psi_{i,m}$ for some finite collection $\lambda_i\in\C$. Suppose then that $a^*a\wt{\zeta} = (m+1)(\delta-\gamma)\wt\zeta$. We wish to show that $\wt\zeta = \sum_i\wt{\lambda_i}\psi_{i,m+1}$ for some $\wt{\lambda_i}\in\C$.

Applying $b^*a$ to $\wt\zeta$ yields a state with eigenvalue $m(\delta-\gamma)$ and so $b^*a\zeta = \sum_i\lambda_i\psi_{i,m}$ for some $\lambda_i\in\C$ by the inductive hypothesis. Applying then $a^*b$ we have that
%
\begin{align*}
    a^*b\sum \lambda_i \psi_{i,m} &= a^*bb^*a\wt\zeta \\
    &= a^*(aa^*-\delta)a\wt\zeta \\
    &= ((a^*a)^2-\delta a^*a)\wt\zeta \\
    &= ((m+1)^2(\delta-\gamma)^2-(m+1)\delta(\delta-\gamma))\wt\zeta.
\end{align*}

\noindent Because $\gamma \le 0$ and $m\ge 0$, $(\delta-\gamma)(m+1)-\delta$ is never zero. Thus $\wt\zeta = \sum_i \wt{\lambda_i}\psi_{i,m+1}$ for some $\wt{\lambda_i}$ as claimed.
\end{proof}

As noted above, there is a correspondence between eigenfunctions of $a^*a$ and $aa^*$ via $a$ and $a^*$ in typical SUSY fashion. Thus we can write the eigenfunctions of $aa^*$ as $\wt{\psi}_{i,m} = a\psi_{i,m}/\|a\psi_{i,m}\|$ (where $m\neq 0$ since $a$ annihilates $\psi_{i,0}$) and $\wt{\phi}_{j,m} = a\phi_{j,m}/\|a\phi_{j,m}\|$. As noted above, the states $\wt{\phi}_{j,0}$ are annihilated by $b^*$. The following figures summarize the coupled SUSY ladder structure.


\begin{figure}[ht!]
\centering
\begin{tikzpicture}
\node (a) at (1.5,0) {};
\node (b) at (3.5,0) {};
\draw (a) to node [above] {$a^*a$} (b);

\node (c) at (7.5,0) {};
\node (d) at (9.5,0) {};
\draw (c) to node [above] {$aa^*$} (d);

\draw (1,-5)--(4,-5);
\draw (1,-4)--(4,-4);
\draw (1,-3.5)--(4,-3.5);
\draw (1,-2.5)--(4,-2.5);
\draw (1,-2)--(4,-2);
\draw (1,-1)--(4,-1);
\draw (1,-0.5)--(4,-0.5);

\draw (7,-4)--(10,-4);
\draw (7,-3.5)--(10,-3.5);
\draw (7,-2.5)--(10,-2.5);
\draw (7,-2)--(10,-2);
\draw (7,-1)--(10,-1);
\draw (7,-0.5)--(10,-0.5);

\node (1a) at (4,-4) {};
\node (1b) at (7,-4) {};

\node (2a) at (4,-2.5) {};
\node (2b) at (7,-1) {};

\node at (0.5,-5) {$\psi_{i,0}$};
\node at (0.5,-4) {$\phi_{j,0}$};
\node at (0.5,-3.5) {$\psi_{i,1}$};
\node at (0.5,-2.5) {$\phi_{j,1}$};
\node at (0.5,-2) {$\psi_{i,2}$};
\node at (0.5,-1) {$\phi_{j,2}$};
\node at (0.5,-0.5) {$\psi_{i,3}$};

\node at (10.5,-4) {$\wt\phi_{j,0}$};
\node at (10.5,-3.5) {$\wt\psi_{i,1}$};
\node at (10.5,-2.5) {$\wt\phi_{j,1}$};
\node at (10.5,-2) {$\wt\psi_{i,2}$};
\node at (10.5,-1) {$\wt\phi_{j,2}$};
\node at (10.5,-0.5) {$\wt\psi_{i,3}$};

\draw[->] (1a) to [bend right=30] node[below] {$a$} (1b);
\draw[->] (1b) to [bend right=30] node[below] {$a^*$} (1a);

\draw[->] (2a) to [bend right=30] node[below] {$b$} (2b);
\draw[->] (2b) to [bend right=30] node[below] {$b^*$} (2a);
\end{tikzpicture}
\caption{The actions of $a$, $b$, $a^*$, and $b^*$ in a coupled SUSY system.}
\vspace{.5em}
\begin{tikzpicture}
\node (a) at (1.5,0) {};
\node (b) at (3.5,0) {};
\draw (a) to node [above] {$a^*a$} (b);

\node (c) at (7.5,0) {};
\node (d) at (9.5,0) {};
\draw (c) to node [above] {$aa^*$} (d);

\draw (1,-2)--(4,-2);
\draw (1,-1)--(4,-1);
\draw (1,-0.5)--(4,-0.5);

\draw (7,-1)--(10,-1);
\draw (7,-0.5)--(10,-0.5);

\node (1a) at (4,-0.5) {};
\node (1b) at (7,-0.5) {};

\node (2a) at (4,-2) {};
\node (2b) at (7,-0.5) {};

\node (3a) at (0.1,-2) {};
\node (3b) at (0.1,-0.5) {};

\node at (0.5,-2) {$\psi_{i,0}$};
\node at (0.5,-1) {$\phi_{j,0}$};
\node at (0.5,-0.5) {$\psi_{i,1}$};

\node at (10.5,-1) {$\wt\phi_{j,0}$};
\node at (10.5,-0.5) {$\wt\psi_{i,1}$};

\draw[->] (1b) to [bend right=30] node[below] {$a^*$} (1a);

\draw[->] (2a) to [bend right=30] node[below] {$b$} (2b);

\draw[->] (3a) to [bend left=30] node[left] {$a^*b$} (3b);
\end{tikzpicture}
\caption{The raising operator structure for the first sector in a coupled SUSY system.}
\vspace{.5em}
\begin{tikzpicture}
\node (a) at (1.5,0) {};
\node (b) at (3.5,0) {};
\draw (a) to node [above] {$a^*a$} (b);

\node (c) at (7.5,0) {};
\node (d) at (9.5,0) {};
\draw (c) to node [above] {$aa^*$} (d);

\draw (1,-2)--(4,-2);
\draw (1,-1)--(4,-1);
\draw (1,-0.5)--(4,-0.5);

\draw (7,-1)--(10,-1);
\draw (7,-0.5)--(10,-0.5);

\node (1a) at (4,-0.5) {};
\node (1b) at (7,-0.5) {};

\node (2a) at (4,-2) {};
\node (2b) at (7,-0.5) {};

\node (3a) at (0.1,-2) {};
\node (3b) at (0.1,-0.5) {};

\node at (0.5,-2) {$\psi_{i,0}$};
\node at (0.5,-1) {$\phi_{j,0}$};
\node at (0.5,-0.5) {$\psi_{i,1}$};

\node at (10.5,-1) {$\wt\phi_{j,0}$};
\node at (10.5,-0.5) {$\wt\psi_{i,1}$};

\draw[<-] (1b) to [bend right=30] node[below] {$a$} (1a);

\draw[<-] (2a) to [bend right=30] node[below] {$b^*$} (2b);

\draw[<-] (3a) to [bend left=30] node[left] {$b^*a$} (3b);
\end{tikzpicture}
\caption{The lowering operator structure for the first sector in a coupled SUSY system.}
\end{figure}

\begin{example}
 Returning to the family in Example 1, a simple inductive proof shows that, for a fixed $n\in\N$, the functions
 %
 \begin{align}
  \psi_{2m}^{(n)}(x) &= e^{\frac{x^{2n}}{2n}} \left(\frac{d}{dx}\frac{1}{x^{2n-2}}\frac{d}{dx}\right)^m e^{-\frac{x^{2n}}{n}} \\
  \psi_{2m+1}^{(n)}(x) &= e^{\frac{x^{2n}}{2n}} \left(\frac{d}{dx}\frac{1}{x^{2n-2}}\frac{d}{dx}\right)^m\left(x^{2n-1} e^{-\frac{x^{2n}}{n}}\right).
 \end{align}

 \noindent are an orthogonal collection of eigenfunctions of $a_n^*a_n$ satisfying $\psi_{m+2}^{(n)} = \lambda_{m,n} a_n^*b_n \psi_m^{(n)}$ for some $\lambda_{m,n}$. It is not hard to see that these are polynomials multiplying $e^{-\frac{x^{2n}}{2n}}$ and analysis similar to that in Chapter 3 shows that they form a basis for $L^2(\R)$. In the case of $n=1$, these become the usual Hermite functions up to normalization. Similar relations hold for the eigenfunctions of $a_n a_n^*$.
\end{example}




\section{Coupled SUSY and Other Oscillator Systems}



Traditionally the QMHO is associated to the 1D Heisenberg-Weyl Lie algebra as this is  the Lie algebra which corresponds to the canonical commutation relations which is reflected in the algebra generated by the ladder operators. This is not the only Lie algebra which may be associated to the QMHO. There are two other treatments of the QMHO: Schwinger's ``spinification" of the two-particle QMHO and the $\mathfrak{su}(1,1)$ treatment of the QMHO. Coupled SUSY is to some degree a unification of the two treatments.

In Schwinger's ``spinification" of the QMHO \cite{schwinger}, one considers two independent oscillators and defines the operators
%
\begin{equation}
    \mathcal{Q}_{\nu} = \sum_{i,j=1}^2\frac{1}{2} a_i^* (\sigma_{\nu})^{ij} a_j,
\end{equation}
where for $\nu = 1,2,3$, $\sigma_{\nu}$ is the $\nu$th Pauli matrix. Explicitly, $\mathcal{Q}_1 = \frac{1}{2} (a_1^*a_2+a_1a_2^*)$, $\mathcal{Q}_2 = -\frac{i}{2}(a_1^*a_2-a_1 a_2^*)$, and $\mathcal{Q}_3 = \frac{1}{2}(a_1^*a_1 - a_2^*a_2)$. The operators $\mathcal{Q}_{\nu}$ form the $\mathfrak{su}(2)$ Lie algebra, for instance
%
\begin{align*}
    [\mathcal{Q}_1, \mathcal{Q}_2] &= -\frac{i}{4} [a_1^* a_2 + a_1 a_2^*, a_1^* a_2 - a_1 a_2^*] \\
    &= \frac{i}{2} [a_1^* a_2, a_1 a_2^*] \\
    &= \frac{i}{2} (a_1^*a_1 a_2 a_2^* - a_1 a_1^* a_2^* a_2) \\
    &= \frac{i}{2}(a_1^* a_1 - a_2^* a_2) \\
    &= i \mathcal{Q}_3.
\end{align*}

The finite dimensional representations in this system have a fixed quantum number which corresponds to the energy difference between the two oscillators and the individual spin states for a fixed energy correspond to the different configurations within each energy level. The Lie algebra $\mathfrak{su}(2)$ is equipped with its own ladder operators. In this case, defining $\mathcal{Q}_{\pm} = \mathcal{Q}_1 \pm i\mathcal{Q}_2$, we have
%
\begin{equation}
    [\mathcal{Q}_3, \mathcal{Q}_{\pm}] = \pm \mathcal{Q}_{\pm}.
\end{equation}

\noindent A simple computation shows that $\mathcal{Q}_+ = a_1^*a_2$ and $\mathcal{Q}_- = a_2^*a_1$. These are analogous to the 
(quadratic) ladder operators for a coupled SUSY system.

Similar to Schwinger's $\mathfrak{su}(2)$ ``spinification" of the QMHO is an $\mathfrak{su}(1,1)$ representation of the QMHO \cite{perelomov}. If $a,a^*$ represent the usual QMHO ladder operators, then letting $\mathcal{K}_0 = a^*a+\frac{1}{2}$, $\mathcal{K}_+ = (a^*)^2$, and $\mathcal{K}_- = a^2$, we have that
%
\begin{equation}
    [\mathcal{K}_0, \mathcal{K}_{\pm}] = \pm \mathcal{K}_{\pm}, \quad [\mathcal{K}_+, \mathcal{K}_-] = -2\mathcal{K}_0,
\end{equation}

\noindent which is exactly the $\mathfrak{su}(1,1)$ Lie algebra.

Coupled SUSY has elements of both of the above approaches to the QMHO. The ladder operators in all three cases are second order; in coupled SUSY, the ladder operators are a combination of the operators coming from two separate Hamiltonians; and coupled SUSY retains the $\mathfrak{su}(1,1)$ structure implicitly built into the QMHO.



\section{Coherent States for Coupled SUSY Systems}



In traditional SUSY, coherent states have been developed, however the coherent states do not mimic those of the QMHO even though they do exploit the SUSY structure \cite{bagarello,kouri2,david}. Particularly, QMHO coherent states are often taken to be eigenfunctions of the lowering operator; SUSY coherent states do not enjoy this property or a property similar to it. Part of the reason for this is the lack of a true ladder structure in SUSY. As seen above, coupled SUSY has a rich ladder structure that has elements of both SUSY and the QMHO. This suggests that the coherent states for a coupled SUSY system should behave somewhat analogously to that of the QMHO while retaining a SUSY flavor.

In general, there are several kinds of coherent states, depending on the context in which one is interested. Coherent states can be seen to be uncertainty minimizers, eigenstates of a lowering operator, specific infinite series of basis functions, or generalized displacements of a cyclic vector. In the case of the QMHO, these are all the same, however in general this is not the case. For $\mathfrak{su}(1,1)$, the various forms of coherent states have been studied extensively \cite{brif,gerry,perelomov}. We elect to use the displacement operator definition for coherent states as it uses the Lie algebraic properties of $\mathfrak{su}(1,1)$ and ties well into the coupled SUSY formalism as there are natural cyclic vectors in $\psi_{i,0}$, $\phi_{j,0}$, $\wt\psi_{i,1}$, and $\wt\phi_{j,0}$.

If one has $\mathfrak{su}(1,1)$ operators $\mathcal{K}_0, \mathcal{K}_{\pm}$ where
%
\begin{equation}
    [\mathcal{K}_0,\mathcal{K}_{\pm}] = \pm \mathcal{K}_{\pm}, \quad [\mathcal{K}_+,\mathcal{K}_-] = -2\mathcal{K}_0, \label{eq:su(1,1)}
\end{equation}

\noindent then the operator $\mathcal{D}(z) = \exp(z\mathcal{K}_+-\bar{z}\mathcal{K}_-)$ defines an $\mathfrak{su}(1,1)$ displacement operator \cite[p. 74]{perelomov}. Suppose that $\mathcal{K}_0\psi_m = (m+k)\psi_m$ for a basis of states $\psi_m$, where $\psi_{m+1} = \mathcal{K}_+\psi_m/\|\mathcal{K}_+\psi_m\|$. The coherent state generated by $\mathcal{D}(z)$ can be written as
%
\begin{equation}
    |z;k\rangle = \mathcal{D}(z)\psi_0 = (1-|z|^2)^k \sum_{m=0}^{\infty} \left(\frac{\Gamma(m+2k)}{m!\Gamma(2k)}\right)^{\frac{1}{2}} z^m \psi_m.
\end{equation}

Suppose that $\{a,b,\gamma,\delta\}$ defines a coupled SUSY system. Define then the following operators
%
\begin{align}
    \mathcal{K}_0 = \frac{1}{\delta-\gamma}\left(a^*a-\frac{\gamma}{2}\right), \quad \mathcal{K}_+ = \frac{1}{\delta-\gamma} a^*b, \quad \mathcal{K}_- = \frac{1}{\delta-\gamma} b^*a, \\
    \wt{\mathcal{K}_0} = \frac{1}{\delta-\gamma}\left(aa^*-\frac{\delta}{2}\right), \quad \wt{\mathcal{K}_+} = \frac{1}{\delta-\gamma} ba^*, \quad \wt{\mathcal{K}_-} = \frac{1}{\delta-\gamma} ab^*.
\end{align}

\noindent A straightforward calculation shows that the relations in \eqref{eq:su(1,1)} hold. Since we have four families of cyclic vectors (indexed by $i$ and $j$), we have four sets of coherent states (indexed by $i$ and $j$). A simple computation shows that
%
\begin{align}
    \mathcal{K}_0 \psi_{i,0} &= -\frac{\gamma}{2(\delta-\gamma)} \psi_{i,0} \\
    \mathcal{K}_0 \phi_{j,0} &= \left(\frac{\delta}{2(\delta-\gamma)}+\frac{1}{2}\right) \phi_{j,0} \\
    \wt{\mathcal{K}_0} \wt\psi_{i,1} &= \left(-\frac{\gamma}{2(\delta-\gamma)}+\frac{1}{2}\right) \wt\psi_{i,1} \\
    \wt{\mathcal{K}_0} \wt\phi_{j,0} &= \frac{\delta}{2(\delta-\gamma)}\wt\phi_{j,0},
\end{align}

\noindent from which we get the following coherent states:
%
\begin{align}
    \left|z; -\frac{\gamma}{2(\delta-\gamma)}\right\rangle_i &= (1-|z|^2)^{-\frac{\gamma}{2(\delta-\gamma)}} \sum_{m=0}^{\infty} \left(\frac{\Gamma\left(m-\frac{\gamma}{\delta-\gamma}\right)}{m!\,\Gamma\left(-\frac{\gamma}{\delta-\gamma}\right)}\right)^{\frac{1}{2}} z^m \psi_{i,m} \label{eq:coherent_state_1} \\
    \left|z;\frac{\delta}{2(\delta-\gamma)}+\frac{1}{2}\right\rangle_j &= (1-|z|^2)^{\frac{\delta}{2(\delta-\gamma)}+\frac{1}{2}} \sum_{m=0}^{\infty} \left(\frac{\Gamma\left(m+\frac{\delta}{\delta-\gamma}+1\right)}{m!\, \Gamma\left(\frac{\delta}{\delta-\gamma}+1\right)}\right)^{\frac{1}{2}} z^m \phi_{j,m} \label{eq:coherent_state_2}\\
    \left|z;-\frac{\gamma}{2(\delta-\gamma)}+\frac{1}{2}\right\rangle_i &= (1-|z|^2)^{-\frac{\gamma}{2(\delta-\gamma)}+\frac{1}{2}} \sum_{m=1}^{\infty} \left(\frac{\Gamma\left(m-\frac{\gamma}{\delta-\gamma}+1\right)}{m!\, \Gamma\left(-\frac{\gamma}{\delta-\gamma}+1\right)}\right)^{\frac{1}{2}} z^m\wt\psi_{i,m} \label{eq:coherent_state_3} \\
    \left|z; \frac{\delta}{2(\delta-\gamma)}\right\rangle_j &= (1-|z|^2)^{\frac{\delta}{2(\delta-\gamma)}} \sum_{m=0}^{\infty} \left(\frac{\Gamma\left(m+\frac{\delta}{\delta-\gamma}\right)}{m!\,\Gamma\left(\frac{\delta}{\delta-\gamma}\right)}\right)^{\frac{1}{2}} z^m \wt\phi_{j,m} \label{eq:coherent_state_4}
\end{align}

In the case of the QMHO, the coherent states are eigenstates of the lowering operator. Since coupled SUSY behaves so similarly to the QMHO, it is natural to ask what happens to the coupled SUSY coherent states under an application of the lowering operators. The lowering operators are composed of $a$ and $b^*$, so we want to investigate the action of $a$ and $b^*$ on the coherent states. To this end, we have the following lemma.

\begin{lemma}
Let $\psi_{i,m}$, $\phi_{j,m}$, $\wt\phi_{i,m}$, and $\wt\phi_{j,m}$ be as above. Then
%
\begin{align}
    a\psi_{i,m} &= \sqrt{m(\delta-\gamma)} \wt{\psi}_{i,m} \label{eq:lowering_1} \\
    a\phi_{j,m} &= \sqrt{(\delta-\gamma)\left(m+\frac{\delta}{\delta-\gamma}\right)} \wt{\phi}_{j,m} \\
    b^*\wt{\psi}_{i,m} &= \sqrt{(\delta-\gamma)\left(m-\frac{\delta}{\delta-\gamma}\right)} \psi_{i,m-1} \\
    b^*\wt{\phi}_{j,m} &= \sqrt{m(\delta-\gamma)} \phi_{j,m-1} \label{eq:lowering_4}
\end{align}
\end{lemma}

\begin{proof}
We only prove the first as the others follow in exactly the same manner. We assume without loss of generality that the proportionality is pure real as a global phase does not change the underlying mathematics. By definition, $a\psi_{i,m} = \lambda \wt\psi_{i,m}$, so we need only to solve for $\lambda$.
%
\begin{align*}
    |\lambda|^2 &= \langle a\psi_{i,m}, a\psi_{i,m}\rangle \\
    &= \langle a^*a \psi_{i,m},\psi_{i,m}\rangle \\
    &= m(\delta-\gamma).
\end{align*}

\noindent The last equality follows since $\psi_{i,m}$ is an eigenfunction of $a^*a$ with eigenvalue $m(\delta-\gamma)$. Thus $\lambda = \sqrt{m(\delta-\gamma)}$ as desired.
\end{proof}

Because our ladder operators can only relate $\psi_{i,m}$ with $\wt\psi_{i,m'}$ and $\phi_{j,m}$ with $\wt\phi_{j,m'}$ (and vice versa), it is clear that we can only relate \eqref{eq:coherent_state_1} with \eqref{eq:coherent_state_3} and \eqref{eq:coherent_state_2} with \eqref{eq:coherent_state_4} (and vice versa). Making use of relations \eqref{eq:lowering_1}-\eqref{eq:lowering_4}, we have that
%
\begin{align}
    a\left|z;-\frac{\gamma}{2(\delta-\gamma)}\right\rangle_i &= \sqrt{-\gamma} \frac{z}{\sqrt{1-|z|^2}} \left|z;-\frac{\gamma}{2(\delta-\gamma)}+\frac{1}{2}\right\rangle_i, \\
    b^*\left|z;\frac{\delta}{2(\delta-\gamma)}\right\rangle_j &= \sqrt{\delta} z\sqrt{1-|z|^2} \left|z;-\frac{\delta}{2(\delta-\gamma)}+\frac{1}{2}\right\rangle_j.
\end{align}

If we were to apply $b^*$ to the first or $a$ to the second, we would not retain a multiple of the original state as the relations \eqref{eq:lowering_1}-\eqref{eq:lowering_4} indicate. Thus, while the coherent states are not eigenstates of the lowering operators $b^*a$ or $ab^*$, applying half of one of the lowering operators can convert one coherent state into another---up to a multiplicative factor (since the operators are not unitary). This appears to be a new structure in SUSY and QMHO coherent states.



\section{Uncertainty Principles for Coupled SUSY Systems}



The canonical uncertainty principle in quantum mechanics is the Heisenberg uncertainty principle which is an uncertainty principle between the position operator $x$ and the momentum operator $p$. The Heisenberg uncertainty principle says that, in natural units, the standard deviation in 
position and momentum is bounded below by
%
$
    \sigma_x \sigma_p \ge \frac{1}{2}.
$

The minimizer of the uncertainty principle is the Gaussian (and translations and modulations thereof). This is easily proved via Cauchy-Schwarz techniques \cite{grochenig}. Since $x$ and $p$ can be written as linear combinations of the QMHO ladder operators (i.e., $x = \frac{1}{\sqrt{2}}(a+a^*)$ and $p = \frac{i}{\sqrt{2}}(a-a^*)$),  we expect to realize uncertainty principles for coupled SUSY systems in a similar way via their ladder operators.

\begin{definition}
Let $\{a,b,\gamma,\delta\}$ define a coupled SUSY system. We define the following analogues of the traditional position and momentum operators for the separate sectors:
%
\begin{align}
    \mathcal{L} &= - \frac{1}{2}(a^*b+b^*a), \quad \mathcal{A} = \frac{i}{2}(a^*b - b^*a), \\
    \widetilde{\mathcal{L}} &= -\frac{1}{2}(ba^*+ab^*), \quad \widetilde{\mathcal{A}} = \frac{i}{2}(ba^* - ab^*).
\end{align}
\end{definition}

In the case of the family of examples in Example 1, $\mathcal{L} = -\frac{1}{2} \frac{d}{dx}\frac{1}{x^{2n-2}}\frac{d}{dx} - \frac{1}{2} x^{2n}$ resembles a Lagrangian and is exactly a Lagrangian operator $-\frac{1}{2} \frac{d^2}{dx^2} -\frac{1}{2} x^2$ when $n=1$, whereas $\mathcal{A} = \frac{1}{2}\{x,p\}$ is precisely the dilation operator.

\begin{thm}
Let $\{a,b,\gamma,\delta\}$ be an unbroken coupled SUSY system, $\mathcal{L}$ and $\mathcal{A}$ be as above, and, also as above, $\ker a = \{\psi_{i,0}:i\in I\}$ for some index set $I$. An uncertainty principle holds for $\mathcal{L}$ and $\mathcal{A}$ and the minimizers are the states $\psi_{i,0}$.
\end{thm}

\begin{proof}
 Let $\psi$ be a normalized wavefunction. Note that Robertson's uncertainty relation \cite[p.53]{galindo} gives us that
 %
 \begin{align*}
     (\sigma_{\mathcal{L}}\sigma_{\mathcal{A}})_{\psi} & \ge \frac{1}{2}|\langle\psi|[\mathcal{L},\mathcal{A}]|\psi\rangle| \\
     &= \frac{1}{4}|\langle\psi|[a^*b,b^*a]-[b^*a, a^*b]|\psi\rangle| \\
     &= \frac{\delta-\gamma}{4}|2\langle\psi|a^*a|\psi\rangle - \gamma|.
 \end{align*}
 
 \noindent Because $\gamma<0$, $a$ has annihilating states $\psi_{i,0}$, and $a^*a$ is self-adjoint, the lower bound given by these states is
 %
 \begin{equation}
     \sigma_{\mathcal{L}}\sigma_{\mathcal{P}} \ge \frac{1}{4}(\delta-\gamma)|\gamma|.
 \end{equation}
 
 This does not guarantee that this lower bound is indeed attained. For the states $\psi_{i,0}$, we have
 %
 \begin{align*}
     \langle \mathcal{L}\rangle &= \langle \phi_{i,0}| \mathcal{L}|\psi_{i,0}\rangle \\
     &= \frac{1}{2}\langle \phi_{i,0} |a^*b+b^*a|\psi_{i,0}\rangle \\
     &= 0.
 \end{align*}
 
 \noindent Similarly, $\langle \mathcal{A}\rangle = 0$. Evaluating $\langle \mathcal{L}^2\rangle$, making use of the fact that $a$ annihilates the states $\psi_{i,0}$, and employing the coupled SUSY structure, it follows that
 %
 \begin{align*}
     \langle\mathcal{L}^2\rangle &= \frac{1}{4}\langle\psi_{i,0}|(b^*a + a^*b)^2|\psi_{i,0}\rangle \\
     &= \frac{1}{4}\langle\psi_{i,0}|b^*aa^*b|\psi_{i,0}\rangle \\
     &= \frac{1}{4}\langle\psi_{i,0}|b^*(bb^*+\delta)b|\psi_{i,0}\rangle \\
     &= \frac{1}{4}\langle\psi_{i,0}|(a^*a-\gamma)^2+\delta(a^*a-\gamma)|\psi_{i,0}\rangle \\
     &= \frac{1}{4}(\delta-\gamma)|\gamma|.
 \end{align*}
 
 \noindent An identical result holds for $\langle\mathcal{A}^2\rangle$, thus we have that
 %
 \begin{align*}
     \sigma_{\mathcal{L}}\sigma_{\mathcal{A}} &= \sqrt{\langle\mathcal{L}^2\rangle - \langle\mathcal{L}\rangle^2}\sqrt{\langle\mathcal{A}^2\rangle - \langle\mathcal{A}\rangle^2} \\
     &= \frac{1}{4}(\delta-\gamma)|\gamma|.
 \end{align*}
 
 The states $\psi_{i,0}$ are indeed the \emph{only} minimizers. If $a\psi \neq 0$, then $\langle\psi|a^*a|\psi\rangle > 0$. Because $\gamma \le 0$, it follows that the uncertainty product is strictly greater than $\frac{1}{4}(\delta-\gamma)|\gamma|$.
\end{proof}

\begin{thm}
 Let $\{a,b,\gamma,\delta\}$ define an unbroken coupled SUSY system, $\widetilde{\mathcal{L}}$ and $\widetilde{\mathcal{A}}$ be as above, and, also as above, $\ker b^* = \{\wt\phi_{j,0}:j\in J\}$ for some index set $J$. An uncertainty principle holds for $\widetilde{\mathcal{L}}$ and $\widetilde{\mathcal{A}}$ and the minimizers are the states $\wt\phi_{j,0}$.
\end{thm}

The proof of this theorem is nearly identical to that of the previous one so we omit it, however the uncertainty principle is now given by
%
\begin{equation}
    \sigma_{\widetilde{\mathcal{L}}}\sigma_{\widetilde{\mathcal{A}}} \ge \frac{1}{4}(\delta-\gamma)\delta.
\end{equation}

Often in SUSY one treats the first and second sectors simultaneously in a matrix formulation by defining the operators $\mathcal H$, $\mathcal Q$, and $\mathcal Q^*$ which act on the direct sum of the two sectors as follows:
%
\begin{align}
    \mathbb{H} = \left(\begin{array}{cc} \mathcal{H}_1 & 0 \\ 0 & \mathcal{H}_2\end{array}\right), \quad \mathbb{Q} = \left(\begin{array}{cc} 0 & 0 \\ \mathcal{Q}_1 & 0 \end{array}\right), \quad \mathbb{Q}^* = \left(\begin{array}{cc} 0 & \mathcal{Q}_1^* \\ 0 & 0 \end{array}\right).
\end{align}

The logic being that the joint Hamiltonian should act on the two subspaces separately by their own Hamiltonians and is therefore diagonal, and the joint charge operators should be off-diagonal because $\mathcal{Q}_1$ and $\mathcal{Q}_1^*$ transfer between the two sectors.

This allows us to define a tertiary set of first order position and momentum operators. Previously, the analysis was relegated to second order position and momentum operators because the operators $\mathcal{L}$, $\mathcal{A}$, $\wt{\mathcal{L}}$, and $\wt{\mathcal{A}}$ acted within a sector; however by combining the two sectors into one framework, we allow ourselves the ability to drop down to first order operators, analogous to the usual QMHO case.

\begin{definition}
Again, let $\{a,b,\gamma,\delta\}$ define a coupled SUSY system. We define the operators $\mathcal{X}$ and $\mathcal{P}$ on the direct sum of the two sectors as follows:
%
\begin{equation}
    \mathbb{X} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc} 0 & a^*+b^* \\ a+b & 0 \end{array}\right), \quad \mathbb{P} = -\frac{i}{\sqrt{2}} \left(\begin{array}{cc} 0 & a^*-b^* \\ -a+b & 0 \end{array}\right).
\end{equation}
\end{definition}

For the infinite family of operators in Example 1, $a_n+b_n$ yields $\sqrt{2} x^n$ and $a_n-b_n$ yields $\frac{\sqrt{2}}{x^{n-1}}\frac{d}{dx}$. Hence $a_n+b_n$ extracts the coordinate-like object corresponding to the coupled SUSY system, whereas $a_n-b_n$ extracts the derivative-like object corresponding to the coupled SUSY system. Therefore, $\mathbb{X}$ plays the role of a generalized notion of position and $\mathbb{P}$  plays the role of a generalized notion of momentum \cite{kouri2}.

\begin{thm}
Let $\{a,b,\gamma,\delta\}$ define an unbroken coupled SUSY system and $\mathbb{X}$ and $\mathbb{P}$ be as above, then the following uncertainty principle holds for $\mathbb{X}$ and $\mathbb{P}$:
%
\begin{equation}
    \sigma_{\mathbb{X}}\sigma_{\mathbb{P}} \ge \frac{1}{2}\min\{|\gamma|,\delta\}.
\end{equation}
\end{thm}

\begin{proof}
Let $\Psi = (\psi_1, \psi_2)^{\operatorname{T}}$ be the state in which we are evaluating the expectation, then $1 = \|\psi_1\|^2 + \|\psi_2\|^2$. Again making use of Robertson's inequality, we have that
%
\begin{align}
    (\sigma_{\mathbb{X}}\sigma_{\mathbb{P}})_{\Psi} & \ge \frac{1}{2} |\langle\Psi| [\mathbb{X}, \mathbb{P}]|\Psi\rangle| \\
    &= \frac{1}{4} \left|\left\langle\Psi\left|\left[\left(\begin{array}{cc} 0 & a^*+b^* \\ a+b & 0 \end{array}\right),\left(\begin{array}{cc} 0 & a^*-b^* \\ -a+b & 0 \end{array}\right)\right]\right|\Psi\right\rangle\right| \\
    &= \frac{1}{4} \left|\left\langle\Psi\left|\left(\begin{array}{cc} -2\gamma & 0 \\ 0 & 2\delta \end{array}\right)\right|\Psi\right\rangle\right| \\
    &= \frac{1}{2} (|\gamma|\|\psi_1\|^2 + \delta\|\psi_2\|^2).
\end{align}

\noindent Since $\|\psi_2\|^2 = 1-\|\psi_1\|^2$, the above is a convex combination of $|\gamma|$ and $\delta$, so indeed we have that
%
\begin{equation}
    \sigma_{\mathbb{X}}\sigma_{\mathbb{P}} \ge \frac{1}{2}\min\{|\gamma|,\delta\}.
\end{equation}

We now show that the value of $\frac{1}{2}|\gamma|$ is attainable. Let $\Psi = (\psi_{i,0},0)^{\operatorname{T}}$, where $\psi_{i,0}$ is as above. The case of $\frac{1}{2}\delta$ proceeds similarly by taking $\Psi = (0,\wt\phi_{j,0})^{\operatorname{T}}$, where $\wt\phi_{j,0}$ is also as above. For this choice of $\Psi$, it follows that
%
\begin{align}
    \langle\Psi|\mathbb{X}|\Psi\rangle &= \frac{1}{\sqrt{2}}(\langle\psi_{i,0} |a^* + b^*|0\rangle + \langle 0| a+b|\psi_{i,0}\rangle) = 0, \\
    \langle\Psi|\mathbb{P}|\Psi\rangle &= -\frac{i}{\sqrt{2}}(\langle\psi_{i,0} |a^*-b^*|0\rangle + \langle 0|-a+b|\psi_{i,0}\rangle) = 0.
\end{align}

\noindent Computing $\mathbb{X}^2$ and $\mathbb{P}^2$ yields
%
\begin{align}
    \mathbb{X}^2 &= \frac{1}{2}\left(\begin{array}{cc} (a^*+b^*)(a+b) & 0 \\ 0 & (a+b)(a^*+b^*)\end{array}\right) \\
    &= \frac{1}{2}\left(\begin{array}{cc} a^*a + a^*b + b^*a + b^*b & 0 \\ 0 & aa^* + ab^* + ba^* + bb^*\end{array}\right) \\
    \mathbb{P}^2 &= \frac{1}{2}\left(\begin{array}{cc} (a^*-b^*)(a-b) & 0 \\ 0 & (a-b)(a^*-b^*)\end{array}\right) \\
    &= \frac{1}{2} \left(\begin{array}{cc} a^*a - a^*b - b^* a + b^*b & 0 \\ 0 & aa^* - ab^* - ba^* + bb^* \end{array}\right)
\end{align}

Inspecting the diagonal terms, it is clear that this uncertainty principle is quite different from that of $\mathcal{L}$ and $\mathcal{A}$ (and from that of $\widetilde{\mathcal{L}}$ and $\widetilde{\mathcal{A}}$). We are only concerned with the upper left elements since we are considering states of the form $\Psi = (\psi_{i,0}, 0)^{\operatorname{T}}$. Noting that $a$ annihilates $\psi_{i,0}$ and using $b^*b = a^*a - \gamma$, it follows that
%
\begin{equation}
    \langle \Psi|\mathbb{X}^2|\Psi\rangle = -\frac{1}{2}\gamma = \langle\Psi|\mathbb{P}^2|\Psi\rangle.
\end{equation}

\noindent Since $\gamma \le 0$, the result follows.
\end{proof}

\begin{remark}
The above uncertainty principles agree exactly with the traditional Heisenberg uncertainty principle in the case of the QMHO since $\gamma = -1$ and $\delta = 1$, giving an uncertainty bound of $\frac{1}{2}$ for each with the minimizers being Gaussians. For $n>1$, the uncertainty product $\sigma_{\mathcal{X}}\sigma_{\mathcal{P}}$ has a lower bound of $\frac{1}{2}$, just as in the Heisenberg uncertainty principle with minimizers $\exp(-x^{2n}/2n)$, but the uncertainty products $\sigma_{\mathcal{L}}\sigma_{\mathcal{A}}$ and $\sigma_{\wt{\mathcal{L}}}\sigma_{\wt{\mathcal{A}}}$ are for $n>1$ bounded by a larger constant  since $\delta-\gamma = 2n$.
\end{remark}




\section{A Family of Coupled SUSY Integral Transforms}



The normalized eigenfunctions of $a^*a$ were denoted by $\psi_{i,m}$ and $\phi_{j,m}$ and the normalized eigenfunctions for $aa^*$ were denoted by $\widetilde{\psi}_{i,m}$ and $\widetilde{\phi}_{j,m}$ for some $i\in I$, $j\in J$ and $m$. It is not clear that the coupled SUSY algebra is sufficient in determining the completeness of the eigenfunctions of $a^*a$ or $aa^*$. To this end, define the following Hilbert subspaces for fixed $i$, $j$:
%
\begin{align}
    \mathfrak{H}_i^+ &:= \overline{\operatorname{span}}\{\psi_{i,m}:m\in\N_0\}, \\
    \mathfrak{H}_j^- &:= \overline{\operatorname{span}}\{\phi_{j,m}:m\in\N_0\}, \\
    \wt{\mathfrak{H}}_i^+ &:= \overline{\operatorname{span}} \{\wt{\psi}_{i,m}:m\in\N\}, \\
    \wt{\mathfrak{H}}_j^- &:= \overline{\operatorname{span}} \{\wt{\phi}_{j,m}:m\in\N_0\}
\end{align}

Assume that the parent Hilbert space is realized as an $L^2(\Omega)$ space for some measure space $\Omega$. Define the $L^2$ distributions $c_i$, $s_j$, $\wt{c_i}$, and $\wt{s_j}$ by
%
\begin{align}
    c_i(x,y) &= \sum_{m=0}^{\infty} (-1)^m \psi_{i,m}(x) \psi_{i,m}(y), \\
    s_j(x,y) &= -\sum_{m=0}^{\infty} (-1)^m \phi_{j,m}(x) \phi_{j,m}(y), \\
    \wt{c}_i(x,y) &= \sum_{m=1}^{\infty} (-1)^m \wt{\psi}_{i,m}(x) \wt{\psi}_{i,m}(y), \\
    \wt{s}_j(x,y) &= -\sum_{m=0}^{\infty} (-1)^m \wt{\phi}_{j,m}(x) \wt{\phi}_{j,m}(y).
\end{align}

\noindent The functions $c_i$, etc., can be thought of as elements of a rigged Hilbert space \cite{antoine}. Particularly, taking $\mathfrak{X}_i^+\subseteq L^2(\Omega)$ to be $\operatorname{span}\{\psi_{i,m}: m\in\N_0\}$, then the linear maps $y\mapsto\langle f, c_i(\cdot, y)\rangle$ are bounded. Thus $c_i$, etc., define linear functionals on $\mathfrak{X}_i^+$, etc., that is they can be realized as elements of $(\mathfrak{X}_i^+)^*$, etc., and can be considered $L^2$ ``distributions.''

The above equalities are to be understood in the weak sense, e.g., for sufficiently nice $f\in\mathfrak{H}_i^+$, we have
%
\begin{equation}
    \langle f,c_i(\cdot,y)\rangle := \sum_{m=0}^{\infty} (-1)^m \psi_{i,m}(y) \langle f,\psi_{i,m}\rangle.
\end{equation}

\noindent To this end, define the integral operators $\Phi_i^+$, $\Phi_j^-$, $\wt{\Phi}_i^+$, and $\wt{\Phi}_j^-$ by
%
\begin{align}
    \Phi_i^+f_1(y) &:= \langle f_1, c_i(\cdot, y)\rangle, \\
    \Phi_j^-f_2(y) &:= \langle f_2, s_j(\cdot, y)\rangle, \\
    \wt{\Phi}_i^+ f_3(y) &:= \langle f_3, \wt{c}_i(\cdot, y)\rangle, \\
    \wt{\Phi}_j^- f_4(y) &:= \langle f_4, \wt{s}_j(\cdot, y)\rangle.
\end{align}

The integral operators are well-defined on the bases generated by the eigenfunctions of the coupled SUSY Hamiltonians and have eigenvalues $\pm 1$. It is a straightforward calculation to show that the coupled SUSY eigenfunctions are also eigenfunctions of the integral operators (in the appropriate ways). As such, the integral operators extend to unitaries on their respective Hilbert spaces.

The integral operators are intertwined and skew-intertwined via the coupled SUSY charge operators, that is
%
\begin{align}
    a \Phi_i^+ &= \wt{\Phi}_i^+ a & \Phi_i^+ a^* &= a^* \wt{\Phi}_i^+ \\
    a \Phi_j^- &= \wt{\Phi}_j^- a & \Phi_j^- a^* &= a^* \wt{\Phi}_j^- \\
    b \Phi_i^+ &= -\wt{\Phi}_i^+ b & \Phi_i^+ b^* &= -b^* \wt{\Phi}_i^+ \\
    b \Phi_j^- &= -\wt{\Phi}_j^- b & \Phi_j^- b^* &= -b^* \wt{\Phi}_j^-
\end{align}

This is easily proved by considering basis elements and making use of the fact that the charge operators are, by assumption, closed, e.g.,
%
\begin{align}
    a\Phi_i^+ \psi_{i,m'}(y) &= a \sum_{m=0}^{\infty} (-1)^m \psi_{i,m} \langle \psi_{i,m'}, \psi_{i,m}\rangle \\
    &= a \sum_{m=0}^{\infty} (-1)^m \delta_{m,m'} \psi_{i,m} \\
    &= (-1)^{m'} a \psi_{i,m'}
\end{align}

\noindent Noting that $a\psi_{i,m}= \lambda_m \wt{\psi}_{i,m}$ for some scalar $\lambda_m$ by definition, we have that
%
\begin{align}
    \wt{\Phi}_i^+ a\psi_{i,m'}(y) &= \langle a\psi_{i,m'},\wt{c}_i(\cdot, y)\rangle \\
    &= \sum_{m=1}^{\infty} (-1)^m \wt{\psi}_{i,m} \langle a\psi_{i,m'}, \wt{\psi}_{i,m}\rangle \\
    &= \sum_{m=1}^{\infty} (-1)^m \lambda_m \wt{\psi}_{i,m} \langle \wt{\psi}_{i,m'}, \wt{\psi}_{i,m}\rangle \\
    &= \sum_{m=1}^{\infty} (-1)^m \lambda_m \wt{\psi}_{i,m}\delta_{m,m'} \\
    &= (-1)^{m'} a \psi_{i,m'}.
\end{align}

\noindent The other cases proceed similarly. 

These integral operators generalize the usual Fourier sine and cosine transforms. In the harmonic oscillator case, these integral operators are exactly the Fourier sine and cosine transforms. Furthermore, the $\Phi_n$ and $\widetilde{\Phi}_n$ transforms are subsumed in this as well by letting $a = \frac{1}{x^{n-1}}\frac{d}{dx} + x^n$ and $b = -\frac{1}{x^{n-1}}\frac{d}{dx}$.



\section{The Coupled SUSY Algebra}



As noted in Chapter 4, the matrix formulation of SUSY carries with it an algebra structure. In coupled SUSY, there is not just one $\Q$ and $\Q^*$ like in traditional SUSY, rather, but two of each: one corresponding to $a$ (or $a^*$) and one corresponding to $b$ (or $b^*$). Define then
%
\begin{alignat}{2}
    & \Q_a = \left(\begin{array}{cc} 0 & 0 \\ a & 0 \end{array}\right),  && \Q_b = \left(\begin{array}{cc} 0 & 0 \\ b & 0 \end{array}\right), \\
    & \Q_a^* = \left(\begin{array}{cc} 0 & a^* \\ 0 & 0 \end{array}\right), && \Q_b^* = \left(\begin{array}{cc} 0 & b^* \\ 0 & 0 \end{array}\right), \\
    & \H_a = \left(\begin{array}{cc} a^*a & 0 \\ 0 & aa^* \end{array}\right), \qquad && \H_b = \left(\begin{array}{cc} b^*b & 0 \\ 0 & bb^*\end{array}\right)
\end{alignat}

The usual SUSY superalgebra is present as expected but there is a richer structure for coupled SUSY. Define the (lowering and raising) operators $\L$ and $\L^*$ by
%
\begin{equation}
    \L = \left(\begin{array}{cc} b^*a & 0 \\ 0 & ab^*\end{array}\right), \quad \L^* = \left(\begin{array}{cc} a^*b & 0 \\ 0 & ba^* \end{array}\right).
\end{equation}

\noindent We have then the following commutation and anti-commutation relations:
%
\begin{alignat}{3}
 & [\H_a, \H_a] = 0, && [\H_b, \H_b] = 0, \qquad\qquad\quad && [\H_a, \H_b] = 0, \\
 & [\H_a, \Q_a] = 0, && [\H_a, \Q_a^*] = 0,  &&\\
 & [\H_b, \Q_a] = -(\delta-\gamma) \Q_a, \quad && [\H_b, \Q_a^*] = (\delta-\gamma) \Q_a^*, && \\
 & [\H_a, \Q_b] = (\delta-\gamma) \Q_b, && [\H_a, \Q_b^*] = -(\delta-\gamma) \Q_b^*,  &&\\
 & [\H_b, \Q_b] = 0, && [\H_b, \Q_b^*] = 0, && \\
 & [\H_a, \L] = -(\delta-\gamma)\L, && [\H_a, \L^*] = (\delta-\gamma) \L^* && \\
 & [\H_b, \L] = -(\delta-\gamma)\L, && [\H_b, \L^*] = (\delta-\gamma) \L^* && \\[5pt]
 & \{\Q_a, \Q_a\} = 0, && \{\Q_a^*, \Q_a^*\} = 0, && \{\Q_a, \Q_a^*\} = \H_a, \\
 & \{\Q_b, \Q_b\} = 0, && \{\Q_b^*, \Q_b^*\} = 0, && \{\Q_b, \Q_b^*\} = \H_b, \\
 & \{\Q_a, \Q_b\} = 0, && \{\Q_a^*, \Q_b^*\} = 0, && \\
 & \{\Q_a, \Q_b^*\} = \L, && \{\Q_a^*, \Q_b\} = \L^*, && \\
 & [\Q_a, \L] = 0, && [\Q_a^*, \L] = -(\delta-\gamma) \Q_b^*, && \\
 & [\Q_a, \L^*] = (\delta-\gamma) \Q_b, && [\Q_a^*, \L^*] = 0, && \\
 & [\Q_b, \L] = -(\delta-\gamma) \Q_a, && [\Q_b^*, \L] = 0, && \\
 & [\Q_b, \L^*] = 0, && [\Q_b^*, \L^*] = (\delta-\gamma) \Q_a^*, && \\[5pt]
 & [\L^*, \L] = -(\delta-\gamma) \H_a - (\delta-\gamma) \H_b.
\end{alignat}

The vector space $\X$ generated by these operators is graded. To this end define the following vector spaces: $\X_{\H} = \operatorname{span}\{\H_a,\H_b\}$, $\X_a = \operatorname{span}\{\Q_a,\Q_a^*\}$, $\X_b = \operatorname{span}\{\Q_b, \Q_b^*\}$, and $\X_{\pm} = \operatorname{span}\{\L, \L^*\}$, then clearly $\X = \X_{\H}\oplus \X_a \oplus \X_b \oplus \X_{\pm}$ as vector spaces.

Let $[\![\cdot,\cdot]\!]$ denote the Lie (super)bracket on $\X$ where it reduces to a commutator or anti-commutator when appropriate. The above relations can be summarized neatly via the summands for $\X$ in the following ``Cayley table", where the ``product" denotes the target space:
%
\begin{center}
 \begin{tabular}{c|cccc}
    [\![$\cdot$,$\cdot$]\!] & $\X_{\H}$  & $\X_a$     & $\X_b$     & $\X_{\pm}$ \\
    \hline
    $\X_{\H}$  & $\X_{\H}$  & $\X_a$     & $\X_b$     & $\X_{\pm}$ \\
    $\X_a$     & $\X_a$     & $\X_{\H}$  & $\X_{\pm}$ & $\X_b$     \\
    $\X_b$     & $\X_b$     & $\X_{\pm}$ & $\X_{\H}$  & $\X_a$     \\
    $\X_{\pm}$ & $\X_{\pm}$ & $\X_b$     & $\X_a$     & $\X_{\H}$  
 \end{tabular}
\end{center}

The Cayley table mimics that of the usual Klein four-group $\Z_2\times\Z_2$, and so the grading on this algebra is a $\Z_2\times\Z_2$-grading. Klein graded algebras are not entirely new \cite{tsartsaflis,noyvert,boboc,rittenberg1,rittenberg2,aizawa}. If $(0,0)$ denotes $\X_{\H}$, $(1,0)$ denotes $\X_a$, $(0,1)$ denotes $\X_b$, and $(1,1)$ denotes $\X_{\pm}$, then the Lie (super)bracket on $\X$ can be written in the following compact form:
%
\begin{equation}
    [\![x,y]\!] = -(-1)^{(i_0+j_0)(i_1+j_1)} [\![y,x]\!],
\end{equation}

\noindent where $x\in (i_0, j_0)$ and $y\in (i_1, j_1)$. In the case that $i_0 = 0 = j_0$ or $i_1 = 0 = j_1$, then the Lie (super)bracket is a commutator, likewise if $i_0 = 1 = j_0$ or $i_1 = 1 = j_1$, otherwise the Lie (super)bracket is an anti-commutator. This perfectly captures the above commutation and anti-commutation relations. The Jacobi identity for this $\Z_2\times\Z_2$-graded algebra is
%
\begin{equation}
    [\![x,[\![y,z]\!]]\!] = [\![[\![x,y]\!],z]\!] + (-1)^{(i_0+j_0)(i_1+j_1)} [\![y,[\![x,z]\!]]\!],
\end{equation}

\noindent where $x\in(i_0,j_0)$ and $y\in(i_1,j_1)$.

Define now the operators $\mathcal{F}^+$ and $\mathcal{F}^-$ for fixed $i$ and $j$ by
%
\begin{equation}
    \mathcal{F}^+ = \begin{pmatrix} \Phi_i^+ & 0 \\ 0 & \wt{\Phi}_i^+\end{pmatrix},\qquad \mathcal{F}^- = \begin{pmatrix} \Phi_j^- & 0 \\ 0 & \wt{\Phi}_j^-\end{pmatrix}.
\end{equation}

\noindent The intertwining/skew-intertwining relations yield the following relations
%
\begin{align}
    \Q_a \mathcal{F}^{\pm} &= \mathcal{F}^{\pm} \Q_a & \Q_a^* \mathcal{F}^{\pm} &= \mathcal{F}^{\pm} \Q_a^* \\
    \Q_b \mathcal{F}^{\pm} &= -\mathcal{F}^{\pm} \Q_b & \Q_b^* \mathcal{F}^{\pm} &= -\mathcal{F}^{\pm} \Q_b^* \\
    \L \mathcal{F}^{\pm} &= \mathcal{F}^{\pm} \L & \L^* \mathcal{F}^{\pm} &= \mathcal{F}^{\pm} \L^* \\
    \H_a \mathcal{F}^{\pm} &= \mathcal{F}^{\pm} \H_a & \H_b \mathcal{F}^{\pm} &= \mathcal{F}^{\pm} \H_b
\end{align}

\noindent For instance,
%
\begin{align*}
    \Q_a \mathcal{F}^+ &= \begin{pmatrix} 0 & 0 \\ a & 0 \end{pmatrix} \begin{pmatrix} \Phi_i^+ & 0 \\ 0 & \wt{\Phi}_i^+\end{pmatrix} \\
    &= \begin{pmatrix} 0 & 0 \\ a\Phi_i^+ & 0 \end{pmatrix} \\
    &= \begin{pmatrix} 0 & 0 \\ \wt{\Phi}_i^+ a & 0 \end{pmatrix} \\
    &= \begin{pmatrix} \Phi_i^+ & 0 \\ 0 & \wt{\Phi}_i^+ \end{pmatrix} \begin{pmatrix} 0 & 0 \\ a & 0 \end{pmatrix} \\
    &= \mathcal{F}^+ \Q_a.
\end{align*}

\noindent The other cases follow similarly.




\section{Future Work: A Generalization of Coupled SUSY}



In the theory of coupled SUSY, two ladders existed and the ladder operators alternated between the two structures. Indeed, there are only two true sectors if one generalizes the algorithm for SUSY in the following way: instead of finding a function $W$ such that $\frac{d}{dx} + W(x)$ annihilates the lowest energy state, one finds an operator $a$ such that $a$ annihilates the state. In this language, it is clear that there are only two true sectors in coupled SUSY since the lowest energy state for $aa^*$ was seen to be $\widetilde{\varphi}_{j,0}$ and it was assumed to be annihilated by the operator $b^*$. Developing the Hamiltonians $bb^*$ and $b^*b$, one sees that the Hamiltonians that arise in this fashion are directly related to the Hamiltonians $aa^*$ and $a^*a$. Thus there are two sectors which are repeated \emph{ad infinitum} and so one may choose to view coupled SUSY as a supersymmetric theory with only two true sectors, much like the quantum mechanical harmonic oscillator having only one sector and its supersymmetric partner sectors being carbon copies thereof.

It is natural to ask what, if any, generalization of coupled SUSY there is. The most natural question to ask is if there is a supersymmetric theory with $n$ sectors. Indeed, there is, and it may be established quite simply. We make a minor change in notation from coupled SUSY.

\begin{definition}
 Let $a_1,\ldots, a_n$ be closed, densely defined operators on a Hilbert space $\mathfrak{H}$, $a_1^*, \ldots, a_n^*$ be their adjoints, and $\gamma_1,\ldots,\gamma_n\in\R$ with $\gamma_1+\cdots+\gamma_n \neq 0$. Furthermore, suppose that $\dom a_1^* = \dom a_2$, $\dom a_2^* = \dom a_3$, and so on, $\ran a_n^* \subseteq \dom a_{n-1}^*$, $\ran a_{n-1}^* \subseteq \dom a_{n-2}^*$ and so on. We say that these generate an \emph{$n$th order coupled SUSY system} if
 %
 \begin{align}
  a_1 a_1^* &= a_2^*a_2 + \gamma_1 \\
  a_2 a_2^* &= a_3^*a_3 + \gamma_2 \\
  &\quad \vdots \\
  a_n a_n^* &= a_1^*a_1 + \gamma_n.
 \end{align}
\end{definition}

Notice that the ordering is not the same as it is in the original realization of coupled SUSY. In coupled SUSY, $a^*a = b^*b + \gamma$ and $aa^* = bb^* + \delta$, whereas in this new notation, it would be replaced with $aa^* = b^*b + \gamma$ and $bb^* = a^*a + \delta$. This is not a significant change as it corresponds to interchanging the roles of $a$ and $a^*$ and replacing $\delta$ with $-\delta$. The reason for this change of notation is that it allows for the results to be very simply stated. We have the following theorem.

\begin{theorem}
 Let $a_1,\ldots,a_n$ define an $n$th order generalized coupled SUSY system, then the $n$th order operators $a_n\cdots a_2 a_1$ and $a_1^* a_2^* \cdots a_n^*$ are ladder operators for $a_1^* a_1$. Cyclic permutations of these operators act as ladder operators for the other Hamiltonians $a_j^*a_j$. The operators $a_1a_2\cdots a_n$ and $a_n^*\cdots a_2^*a_1^*$ are ladder operators for $a_1a_1^*$. Cyclic permutations of these operators act as ladder operators for the other Hamiltonians $a_ja_j^*$.
\end{theorem}

\begin{proof}
 Instead of computing the commutator, we find an expression for $a_1^*a_1a_1^* a_2^*\cdots a_n^*$ which will be closely related to the commutator.
 
 \begin{align*}
 a_1^*a_1a_1^* a_2^*\cdots a_n^* &= a_1^*a_1 a_1^*a_2^*\cdots a_n^* \\
 &= a_1^*(a_2^*a_2 + \gamma_1)a_2^*\cdots a_n^* \\
 &= \gamma_1 a_1^*a_2^*\cdots a_n^* + a_1^*a_2^*a_2a_2^*\cdots a_n^* \\
 &= \gamma_1 a_1^*a_2^*\cdots a_n^* + a_1^*a_2^*(a_3^*a_3+\gamma_2)\cdots a_n^* \\
 &= (\gamma_1+\gamma_2) a_1^*a_2^*\cdots a_n^* + a_1^*a_2^*a_3^*a_3\cdots a_n^* \\
 & \,\,\,\vdots \\
 &= (\gamma_1+\gamma_2+\cdots+\gamma_{n-1})a_1^*a_2^*\cdots a_n^* + a_1^*a_2^*\cdots a_n^*a_na_n^* \\
 &= (\gamma_1+\gamma_2+\cdots+\gamma_{n-1})a_1^*a_2^*\cdots a_n^* + a_1^*a_2^*\cdots a_n^*(a_1^*a_1+\gamma_n) \\
 &= (\gamma_1+\gamma_2+\cdots+\gamma_n) a_1^*a_2^*\cdots a_n^* + a_1^*a_2^*\cdots a_n^* a_1^* a_1
\end{align*}

\noindent Subtracting $a_1^*a_2^*\cdots a_n^* a_1^* a_1$ gives 
%
\begin{equation}
 [a_1^*a_1,a_1^* a_2^*\cdots a_n^*] = (\gamma_1+\cdots+\gamma_n) a_1^*a_2^*\cdots a_n^*.
\end{equation}

\noindent A similar analysis shows that $a_n\cdots a_2 a_1$ is also a ladder operator for $a_1^*a_1$.
\end{proof}

The operators $a_1,a_2,\ldots,a_n$ and $a_1^*, a_2^*,\ldots,a_n^*$ act as transfer operators between the different sectors, and the sectors are cyclicly related via these operators. In general, it is not clear what Lie algebra lies behind this generalization of coupled SUSY as the next result suggests.
%
\begin{theorem}
 The commutator of $a_n\cdots a_2 a_1$ and $a_1^* a_2^*\cdots a_n^*$ is an $(n-1)$st degree polynomial in $a_1^*a_1$.
\end{theorem}

\begin{proof}
 We have that
 %
 \begin{align*}
  a_n \cdots a_2 a_1 a_1^* a_2^* \cdots a_n^* &= a_n \cdots a_2 (a_2^*a_2 + \gamma_1)a_2^* \cdots a_n^* \\
  &= a_n \cdots (a_2 a_2^*)^2\cdots a_n^* + \gamma_1 a_n \cdots a_2 a_2^* \cdots a_n^* \\
  &= a_n \cdots (a_3^* a_3 + \gamma_2)^2 \cdots a_n^* + \gamma_1 a_n \cdots (a_3^* a_3 +\gamma_2)\cdots a_n^*
 \end{align*}

 A simple induction focusing on the first term yields a term of the form $(a_n a_n^*)^{n-1}$ which can be readily recognized as being related to $(a_1^* a_1)^{n-1}$ and so the claim is shown.
\end{proof}

These results generalize those obtained in the case of traditional coupled SUSY. It is of interest to explore the Lie structure for $n$th order coupled SUSY and what the graded structure is.












