\chapter{Supersymmetric Quantum Mechanics} \label{chapter-susy}



\section{The Quantum Mechanical Harmonic Oscillator}



In the language of quantum mechanics, observables such as energy, momentum, and position are realized as densely-defined operators on a Hilbert space $\mathfrak{H}$ \cite[p.26]{dirac} which have specific representations in a given coordinate system. Energy is replaced with the Hamiltonian operator $\mathcal{H}$, momentum is replaced with the momentum operator $p$, and position is replaced with the position operator $x$. In spatial coordinates, the momentum operator is represented by $p = -i\hbar\nabla$, where $\nabla$ is the gradient operator, and the position operator is represented by the multiplication operator $\mathcal{M}_x$, often simply denoted by $x$. Two important operations in quantum mechanics are the commutator and anticommutator. Given (possibly unbounded) densely-defined operators $A,B$ on a Hilbert space $\mathfrak{H}$, their commutator and anticommutator are respectively given by
%
\begin{equation}
 [A,B] = AB-BA, \qquad \{A,B\} = AB+BA.
\end{equation}

\noindent The existence of commutators is nontrivial in the case of unbounded operators \cite[p.275]{reed1}. In what follows, the existence of each is guaranteed however. In general, if both operators share an operator core, formal manipulation is justified. Unless stated otherwise or noted, this is assumed.

In this language, the energy equation for the classical harmonic oscillator in one variable becomes the operator
%
\begin{equation*}
 \mathcal{H}_{\text{HO}} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}kx^2,
\end{equation*}

\noindent where $\hbar$ is Planck's reduced constant, $m$ is the mass of the oscillator, and $k$ is the spring constant. A natural question to ask about any operator is what, if any, eigenvalues and eigenfunctions (or eigenvectors) it may have. To this end, we wish to solve the equation $\mathcal{H}_{\text{HO}}\psi = \lambda \psi$. This results in the following second order linear differential equation after a suitable change of variables
%
\begin{equation}
 -\frac{1}{2}\psi''(x) + \frac{1}{2}x^2\psi(x) = \lambda \psi(x).
\end{equation}

In the theory of differential equations, it is common to try series solution techniques for linear differential equations. In this one case, one may expand $\psi$ as a series and find a recursion relation for its coefficients \cite[p.251]{strauss}. By requiring an $L^2$-normalizable solution, we get a constraint on the choices of $\lambda$ and immediately determine the eigenvalues. This however is not a very enlightening technique and completely neglects the rich algebraic structure lurking behind the quantum-mechanical harmonic oscillator.

Perhaps the most elegant technique for solving this differential equation is via a factorization approach \cite[p.136]{dirac}. The operator $-\frac{1}{2} \frac{d^2}{dx^2} + \frac{1}{2}x^2$ resembles a difference of squares from elementary algebra and so it is natural to try to factor it with the same technique. The natural factorization might be
%
\begin{equation}
 -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 \stackrel{?}{=} \frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx}+x\right).
\end{equation}

\noindent However expanding the right hand side yields
%
\begin{equation*}
 \frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx} + x\right) = -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 + \frac{1}{2}\left(-\frac{d}{dx} x + x\frac{d}{dx}\right).
\end{equation*}

The term $-\frac{d}{dx} x + x\frac{d}{dx}$ can be readily simplified by noting that these are operators and must act on Hilbert space functions. Assuming $f\in L^2(\R)$ and $f'$ exists almost everywhere, then
%
\begin{equation*}
 -\frac{d}{dx}(xf) + x\frac{df}{dx} = -f(x) - xf'(x) + xf'(x) = -f.
\end{equation*}

\noindent Thus we may really view $-\frac{d}{dx}x + x\frac{d}{dx}$ as the operator $-I$ on such functions, and so our previous relation becomes
%
\begin{equation}
 \frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx} + x\right) = -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 - \frac{1}{2}I.
\end{equation}

The operators $\mathcal{H}_1 = -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2$ and $\mathcal{H}_2 = \frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx} + x\right)$ will share many of the same properties, e.g. share eigenfunctions, and moreover their spectra will only be different by an overall shift of $\frac{1}{2}$ because,
%
\begin{align*}
 \sigma(\mathcal{H}_1) &= \{\lambda\in\C: \mathcal{H}_1 - \lambda I \text{ not invertible}\} \\
 &= \left\{\lambda\in\C: \mathcal{H}_2 + \frac{1}{2} I -\lambda I \text{ not invertible}\right\} \\
 &= \left\{\lambda\in\C: \mathcal{H}_2 - \left(\lambda-\frac{1}{2}\right)I \text{ not invertible}\right\} \\
 &= \left\{\lambda+\frac{1}{2}\in\C:\mathcal{H}_2 - \lambda I \text{ not invertible}\right\} \\
 &= \frac{1}{2} + \sigma(\mathcal{H}_2).
\end{align*}

The factored Hamiltonian $\mathcal{H}_2$ is of a slightly simpler form, so we prefer to consider it from this point on. Many properties of $\mathcal{H}_2$ can be translated into properties of $\mathcal{H}_1$ without much difficulty as they only differ by an overall additive constant.

Define $\mathcal{H}$ then to be $\frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx} + x\right)$ and define the operator $a$ by
%
\begin{equation}
 a = \frac{1}{\sqrt{2}}\left(\frac{d}{dx} + x\right).
\end{equation}

\noindent $a$ is defined on those $f\in L^2(\R)$ such that $f'\in L^2(\R)$ and $xf\in L^2(\R)$. If $f,g\in L^2(\R)\cap AC(\R)$ such that $f',g'\in L^2(\R)$ and $xf,xg\in L^2(\R)$, then the following equalities hold and are well-defined
%
\begin{align*}
 \int_{-\infty}^{\infty} \left(\frac{d}{dx}+x\right)f(x)\overline{g(x)}\,dx &= \int_{-\infty}^{\infty} (f'(x) + xf(x))\overline{g(x)}\,dx \\
 &= f(x)\overline{g(x)}\bigg|_{-\infty}^{\infty} + \int_{-\infty}^{\infty} f(x)\overline{(-g'(x)+xg(x))}\,dx \\
 &= \int_{-\infty}^{\infty}f(x)\overline{\left(-\frac{d}{dx}+x\right)g(x)}\,dx.
\end{align*}

\noindent Because $f\overline{g}\in AC(\R)\cap L^1(\R)$, $fg$ vanishes at infinity, justifying the final equality.

The operator $\frac{1}{\sqrt{2}}\left(-\frac{d}{dx}+x\right)$ can be realized as the adjoint of $a$, and is denoted as $a^*$. Thus the Hamiltonian $\mathcal{H}$ may be realized as being equivalent to
%
\begin{equation}
 \mathcal{H} = a^*a.
\end{equation}

In our analysis, there was an implicit choice made when ``factoring'' the Hamiltonian $-\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2$. The factorization chosen was $\frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx} + x\right)$, however there is no reason that an alternative factorization could not be made, particularly one may choose to instead factor the Hamiltonian as $\frac{1}{2}\left(\frac{d}{dx} + x\right)\left(-\frac{d}{dx} + x\right)$. Notice that $-\frac{d}{dx}$ now appears in the second term rather than the first. Expanding this product gives
%
\begin{align*}
 \frac{1}{2}\left(\frac{d}{dx} + x\right)\left(-\frac{d}{dx} + x\right) &= -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 + \frac{1}{2}\left(\frac{d}{dx}x - x\frac{d}{dx}\right) \\
 &= -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 + \frac{1}{2}I.
\end{align*}

\noindent Thus $\mathcal{H} = \frac{1}{2}\left(\frac{d}{dx} + x\right)\left(-\frac{d}{dx} + x\right) - I$ which is equivalent to $\mathcal{H} = aa^* - I$. Equating the two expressions for $\mathcal{H}$, it follows that
%
\begin{equation}
 a^*a = aa^* - I.
\end{equation}

\noindent In commutator language, this becomes $[a^*,a] = -I$.

This commutation relation achieves both a rich algebraic structure and a rich analytic structure. This commutation relation shows that the operators $a$, $a^*$ and $I$ generate a closed Lie algebra. This Lie algebra is the Heisenberg-Weyl Lie algebra \cite[p.8]{perelomov}. Many algebraic structures underlying operators do not have associated analytic structures, cf. supersymmetry and supersymmetric quantum mechanics. This makes the harmonic oscillator unique as we will show that it also has a nice analytic structure.

Note first that the function $\psi_0(x) = \exp(-x^2/2)\in L^2(\R)$ is annihilated by $a$, that is to say that $a\psi_0 = 0$:
%
\begin{equation}
 \left(\frac{d}{dx} + x\right)e^{-\frac{x^2}{2}} = -x e^{-\frac{x^2}{2}} + x e^{-\frac{x^2}{2}} = 0.
\end{equation}

\noindent Trivially then $a^*a \psi_0 = 0$ since $a\psi_0 = 0$. Thus $a^*a$ has $\psi_0$ as an eigenfunction with eigenvalue 0. The function $\exp(x^2/2)$ is annihilated by $a^*$ however it is not in $L^2(\R)$, so while $aa^*\exp(x^2/2) = 0$ and so $a^*a\exp(x^2/2) = -\exp(x^2/2)$, it is not an eigenfunction as we are only interested in $L^2(\R)$ eigenfunctions, not generalized eigenfunctions.

The commutation relation $[a^*,a] = -I$ yields two further commutation relations which will serve as the basis for the rest of the analysis of the quantum-mechanical harmonic oscillator. Applying $a^*$ to $a^*a$ on the left gives
%
\begin{equation*}
 a^*aa^* = a^*(a^*a+I) = a^*a^*a + a^*,
\end{equation*}

\noindent which may be realized as a commutation relation:
%
\begin{equation}
 [a^*a,a^*] = a^*.
\end{equation}

\noindent A similar approach shows that $[a^*a,a] = -a$. This leads into our next definition.

\begin{definition}
 Let $0\neq \lambda\in\C$ and $\mathcal{N},\mathcal{X}$ be densely-defined operators on a Banach space $\X$ be such that $[\mathcal{N},\mathcal{X}] = \lambda \mathcal{X}$, $\ran \mathcal{X} \subseteq \dom \mathcal{N}$ and $\ran \mathcal{N}\subseteq \dom\mathcal{X}$. We say that $\mathcal{X}$ is a \emph{ladder operator} for $\mathcal{N}$. If $\lambda \in\R$ and $\lambda > 0$, we say that $\mathcal{X}$ is a \emph{raising operator} for $\mathcal{N}$. Likewise, if $\lambda\in \R$ and $\lambda < 0$, we say that $\mathcal{X}$ is a \emph{lowering operator} for $\mathcal{N}$.
\end{definition}

This is fundamentally a restatement of the notion of roots of Lie algebras cast in the language of unbounded operators \cite[p.176]{hall}. Such commutation relations are important as they encode a lot of important information regarding the analytic structure of operators. This is discussed in the next theorem.
%
\begin{theorem}
 Let $0\neq \lambda\in\C$ and $\mathcal{N},\mathcal{X}$ be densely-defined operators on a Banach space $\X$ be such that $[\mathcal{N},\mathcal{X}] = \lambda \mathcal{X}$, $\ran \mathcal{X} \subseteq \dom \mathcal{N}$ and $\ran \mathcal{N}\subseteq \dom\mathcal{X}$. If $\psi$ is an eigenvector for $\mathcal{N}$ with eigenvalue $\mu$, then $\mathcal{X}\psi$ is also an eigenvector for $\mathcal{N}$ provided that $\mathcal{X}\psi \neq 0$.
\end{theorem}

\begin{proof}
 We have that
%
\begin{equation}
 \mathcal{N}\mathcal{X}\psi = (\mathcal{X}\mathcal{N} + \lambda \mathcal{X})\psi = (\mu+\lambda)\mathcal{X}\psi.
\end{equation}

\noindent Provided that $\mathcal{X}\psi \neq 0$, $\mathcal{X}\psi$ is also an eigenfunction of $\mathcal{N}$ with eigenvalue $\mu+\lambda$.
\end{proof}

From the previous argument, $a$ and $a^*$ are ladder operators for $a^*a$, particularly $a$ is a lowering operator and $a^*$ is a raising operator for $a^*a$. This allows us to exactly solve for the eigenvalues for $a^*a$. Since $0$ is an eigenvalue of $a^*a$, it follows that $m$ is an eigenvalue for $a^*a$ for any $m\in\N_0$.

Suppose then that $\lambda$ is an eigenvalue of $a^*a$ corresponding to eigenfunction $\psi$ such that $m < \lambda < m+1$ for some $m\in\N_0$. $a^m\psi$ is then an eigenfunction of $a^*a$ with eigenvalue $0 < \lambda-m < 1$. Applying $a$ once more yields another eigenfunction $a^{m+1}\psi$ which has eigenvalue $\lambda-m-1 < 0$. However,
%
\begin{align*}
 0 & \le \langle a^{m+2}\psi, a^{m+2} \psi\rangle \\
 &= \langle a^*a (a^{m+1} \psi), a^{m+1} \psi\rangle \\
 &= (\lambda-m-1)\langle a^{m+1}\psi, a^{m+1} \psi\rangle \\
 & < 0.
\end{align*}

\noindent Thus, there cannot be an eigenfunction with eigenvalue $\lambda$ such that $m < \lambda < m+1$. A not-too-dissimilar argument shows that the only eigenfunctions are exactly those of the form $(a^*)^m\psi_0$. We define then $\psi_m$ by
%
\begin{equation}
 \psi_m(x) = \frac{1}{2^{m/2}} \left(-\frac{d}{dx}+x\right)^m\psi_0(x).
\end{equation}

\noindent A straightforward calculation shows that
%
\begin{equation}
 \langle \psi_{m+1},\psi_{m+1}\rangle = \sqrt{m}\langle \psi_m,\psi_m\rangle
\end{equation}

\noindent so that $\psi_m\in L^2(\R)$ for all $m$ since $\psi_0\in L^2(\R)$.

It is a well-known fact that the entire spectrum of the quantum-mechanical harmonic oscillator is indeed comprised only of its eigenvalues \cite[p.131]{galindo}, that is to say that there is no approximate point spectrum or residual spectrum. This is a direct consequence of the fact that the functions $\{\psi_m:m\in\N_0\}$ form a basis for $L^2(\R)$.



\section{Supersymmetric Quantum Mechanics}



With the broad success of the factorization approach to the quantum-mechanical harmonic oscillator, physicists and mathematicians applied similar techniques to more general systems. In the quantum-mechanical harmonic oscillator, both $a^*a$ and $aa^*$ played a crucial role in the general theory. Let $\mathcal{H}_1$ be a one dimensional, time-independent Hamiltonian with potential $V$ such that $\mathcal{H}_1$ is positive and has an eigenvalue of $0$, then in the coordinate representation and neglecting physical constants, $\mathcal{H}_1$ may be written as
%
\begin{equation}
 \mathcal{H}_1 = -\frac{1}{2}\frac{d^2}{dx^2} + V(x).
\end{equation}

Provided that one is studying a system with finitely many negative energy states, the condition of positivity is not a significant assumption, as the operator $\mathcal{H}_1+\lambda I$ will be positive for sufficiently large $\lambda$. Similarly, assuming that $0$ is an eigenvalue is not a significant assumption by taking $\lambda$ to be the negative of the lowest energy eigenvalue.

In the case of the quantum-mechanical harmonic oscillator, $V(x) = \frac{1}{2}x^2$. We define the operator $\mathcal{Q}$ and its (formal) adjoint $\mathcal{Q}^*$ on $L^2(\R)$ by
%
\begin{equation}
 \mathcal{Q} = \frac{1}{\sqrt{2}}\left(\frac{d}{dx} + W(x)\right), \quad \mathcal{Q}^* = \frac{1}{\sqrt{2}}\left(-\frac{d}{dx} + W(x)\right),
\end{equation}

\noindent where it is further assumed that $\mathcal{H}_1 = \mathcal{Q}^*\mathcal{Q}$, that is to say that
%
\begin{equation}
 V(x) = \frac{1}{2} W(x)^2 - \frac{1}{2} W'(x).
\end{equation}

Here it is assumed that $W$ is continuously differentiable. If $W$ is also polynomially bounded from above and bounded from below, then the harmonic oscillator eigenfunctions are naturally in the domain of the operators $\mathcal{Q}$ and $\mathcal{Q}^*$ so that both are densely-defined. The image of the harmonic oscillator eigenfunctions will also be in the domain of $\mathcal{Q}^*$ and vice versa. Therefore if we restrict the domain of $\mathcal{Q}$ to the set of functions $\psi$ such that $\mathcal{Q}\psi$ is in $L^2(\R)$ and that $\mathcal{Q}^*\mathcal{Q}\psi$ is well-defined as a differential operator acting on $\psi$ and in $L^2(\R)$, our formal manipulations to follow are well-defined.

This is a Ricatti equation and is quite difficult to solve for general potentials $V$ due to its nonlinear nature. If instead, the nodeless eigenfunction $\psi_0$ corresponding to eigenvalue $0$ is known, we may define $W$ by
%
\begin{equation}
 W(x) = -\frac{d}{dx}\log \psi_0(x).
\end{equation}

\noindent Direct computation shows that such a choice of $W$ solves the Ricatti equation and that $\psi_0$ may be represented by
%
\begin{equation}
 \psi_0(x) = A\exp\left(-\int_0^x W(x')\,dx'\right).
\end{equation}

A simple application of Cauchy-Schwarz shows that if $\exp(-\int_0^x W(x')\,dx')$ is in $L^2(\R)$, then $\exp(\int_0^x W(x')\,dx')$ cannot possibly be in $L^2(\R)$. This is in tune with what was seen in the case of the quantum-mechanical harmonic oscillator.

Since $aa^*$ was also critical in the analysis of the quantum-mechanical harmonic oscillator, it is natural to inspect the nature of $\mathcal{Q}\mathcal{Q}^*$ as well. The essence of supersymmetric quantum mechanics is the analysis of the operators $\mathcal{H}_1$ and $\mathcal{H}_2$ and the relations between them. Defining $\mathcal{H}_2 = \mathcal{Q}\mathcal{Q}^*$ yields the operator
%
\begin{equation}
 \mathcal{H}_2 = \mathcal{Q}\mathcal{Q}^* = -\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2} W(x)^2 + \frac{1}{2}W'(x).
\end{equation}

Note that we need not duplicate the argument above for $\mathcal{Q}^*$ since $\mathcal{Q}$ and $\mathcal{Q}^*$ are of effectively the same form and so if $\mathcal{Q}^*\mathcal{Q}\psi$ is well-defined, so will be $\mathcal{Q}\mathcal{Q}^*\psi$. Note that both operators are densely-defined.

In the quantum-mechanical harmonic oscillator, $[a^*,a] = -I$; assuming a form of $a = \frac{d}{dx} + W(x)$, direct computation shows that $[a^*,a] = -I$ if and only if $W(x) = x+\alpha$ for some $\alpha\in\R$. Thus, in general, supersymmetric quantum mechanics should not be expected to have as rich an analytic structure as the quantum-mechanical harmonic oscillator. Indeed, many quantum-mechanical systems do not have nice algebraic relations between their eigenvalues, and so a true ladder structure will not exist involving the operators $\mathcal{H}_1$, $\mathcal{H}_2$, $\mathcal{Q}$, and $\mathcal{Q}^*$.

Instead, a different structure exists for supersymmetric quantum mechanics. It is straightforward that $\mathcal{Q}\mathcal{H}_1 = \mathcal{H}_2\mathcal{Q}$ and $\mathcal{H}_1 \mathcal{Q}^* = \mathcal{Q}^* \mathcal{H}_2$. These intertwining relations lead to the well-known intertwining between the eigenvalues and eigenfunctions of $\mathcal{H}_1$ and $\mathcal{H}_2$.

\begin{theorem}
 Let $\psi$ be an eigenfunction of $\mathcal{H}_1$ with eigenvalue $\lambda\neq 0$, then $\mathcal{Q}\psi$ is an eigenfunction of $\mathcal{H}_2$ with eigenfunction $\lambda$.
\end{theorem}

\begin{proof}
 First note that $\langle \mathcal{Q}\psi, \mathcal{Q}\psi\rangle = \langle \mathcal{Q}^*\mathcal{Q}\psi,\psi\rangle = \lambda\langle\psi,\psi\rangle<\infty$ so $Q\psi$ is a permissible eigenfunction of $\mathcal{H}_2$ since it is in $L^2(\R)$ and $\mathcal{Q}\psi \neq 0$. Moreover, $\mathcal{H}_2\mathcal{Q}\psi = \mathcal{Q}\mathcal{H}_1\psi = \lambda \mathcal{Q}\psi$. Hence $\mathcal{Q}\psi$ is an eigenfunction of $\mathcal{H}_2$ with eigenvalue of $\lambda$. The assumption that $\lambda\neq 0$ is critical as it prevents $\mathcal{Q}\psi = 0$.
\end{proof}

A similar argument shows that if $\varphi$ is an eigenfunction of $\mathcal{H}_2$ with eigenvalue $\lambda$, then $\mathcal{Q}\varphi$ is an eigenfunction of $\mathcal{H}_1$ with eigenvalue $\lambda$. The case of $\mathcal{Q}^*\psi = 0$ is of no consequence since if $\exp\left(-\int_0^x W(x')\,dx'\right)$ is in $L^2(\R)$, $\exp\left(\int_0^x W(x')\,dx'\right)$ cannot be in $L^2(\R)$.

The supersymmetric operators $\mathcal{H}_1$, $\mathcal{H}_2$, $\mathcal{Q}$, and $\mathcal{Q}^*$ can be combined into one framework via a matricial representation on $L^2(\R)\oplus L^2(\R)$. Define matrix operators $\mathbb{H}$, $\mathbb{Q}$ and $\mathbb{Q}^*$ by
%
\begin{equation}
 \mathbb{H} = \begin{pmatrix}\mathcal{H}_1 & 0 \\ 0 & \mathcal{H}_2\end{pmatrix}, \quad \mathbb{Q} = \begin{pmatrix} 0 & 0 \\ \mathcal{Q} & 0 \end{pmatrix}, \quad \mathbb{Q}^* = \begin{pmatrix} 0 & \mathcal{Q}^* \\ 0 & 0 \end{pmatrix}.
\end{equation}

\noindent These operators enjoy an algebraic structure:
%
\begin{align}
 [\mathbb{H},\mathbb{Q}] &= 0, &[\mathbb{H}, \mathbb{Q}^*] &= 0 \\
 \{\mathbb{Q},\mathbb{Q}^*\} &= \mathbb{H}, & \{\mathbb{Q},\mathbb{Q}\} &= 0 = \{\mathbb{Q}^*, \mathbb{Q}^*\}.
\end{align}

\noindent This is a $\Z_2$-graded Lie superalgebra \cite{nijenhuis,corwin} where the grading is generated by the subspaces $\X_1 = \operatorname{span}\mathbb{H}$ and $\X_2 = \operatorname{span}\{\mathbb{Q}, \mathbb{Q}^*\}$.

Diagrammatically, the intertwining for $\mathcal{H}_1$ and $\mathcal{H}_2$ can be easily summarized. Let $\{\psi_m:m\in\N_0\}$ and $\{\varphi_m:m\in\N_0\}$ denote the eigenfunctions for $\mathcal{H}_1$ and $\mathcal{H}_2$, respectively. In the language of supersymmetry, the eigenstates of $\mathcal{H}_1$ and $\mathcal{H}_2$ form what are called \emph{sectors}. Figure \ref{fig:2} shows the structure relating the two sectors.
%
\begin{figure}[ht!]
 \centering
 \begin{tikzpicture}
  \node (a) at (1.5,0) {};
  \node (b) at (3.5,0) {};
  \draw (a) to node [above] {$\mathcal{H}$} (b);
  
  \node (c) at (5.5,0) {};
  \node (d) at (7.5,0) {};
  \draw (c) to node [above] {$\widetilde{\mathcal{H}}$} (d);
  
  \draw[black,fill=black] (2.5,-.75) circle (.2ex);
  \draw[black,fill=black] (2.5,-1) circle (.2ex);
  \draw[black,fill=black] (2.5,-1.25) circle (.2ex);
  
  \draw (1,-2)--(4,-2);
  \draw (1,-3)--(4,-3);
  \draw (1,-4)--(4,-4);
  
  \node at (0,-2) {$\psi_2$};
  \node at (0,-3) {$\psi_1$};
  \node at (0,-4) {$\psi_0$};
  
  \node (e) at (3.5,-2) {};
  \node (f) at (3.5,-3) {};
  
  \draw[black,fill=black] (6.5,-.75) circle (.2ex);
  \draw[black,fill=black] (6.5,-1) circle (.2ex);
  \draw[black,fill=black] (6.5,-1.25) circle (.2ex);
  
  \draw (5,-2)--(8,-2);
  \draw (5,-3)--(8,-3);
  \draw[dashed] (5,-4)--(8,-4);
  
  \node at (9,-2) {$\varphi_1$};
  \node at (9,-3) {$\varphi_0$};
  \node at (9,-4) {No state};
  
  \node (g) at (5.5,-2) {};
  \node (h) at (5.5,-3) {};
  
  \draw[->] (e) to [bend left=60] node[above] {$Q$} (g);
  \draw[<-] (f) to [bend right=60] node[above] {$Q^*$} (h);
 \end{tikzpicture}
 
 \caption{The ladder structure for supersymmetric quantum mechanics.} \label{fig:2}
\end{figure}

As seen above, for supersymmetric quantum mechanics, the ``ladder'' structure is quite different from the quantum-mechanical harmonic oscillator insofar that the charge operators only allow one to move left and right between sectors, rather than up and down within either sector. In supersymmetric quantum mechanics, one often defines a hierarchy of Hamiltonians by repeating the above process with $\varphi_0$ as the new ground state. This creates a series of Hamiltonians which are intertwined via the charge operators, however again one may only move left and right between sectors, thus it is not a true ladder structure. In the next chapter, a new structure is proposed to remedy this which melds the graded structure of supersymmetry (and supersymmetric quantum mechanics) with the analytic structure of the harmonic oscillator. Particularly, knowing the ground states for each sector, one may generate all higher excited states.



