The Quantum Mechanical Harmonic Oscillator
In my research, I've been working on a generalization of the quantum mechanical harmonic oscillator (QMHO). Much of the algebraic structure of the QMHO comes from the canonical commutation relation:
$$ [x,p] = i\hbar.\tag{1}$$
The time-independent QMHO is governed by the eigenvalue equation (in dimensionless units)
$$ \mathcal{H}\psi = -\frac{1}{2}\frac{d^2}{dx^2}\psi + \frac{1}{2}x^2\psi = E\psi. \tag{2}$$
The difference-of-squares-like expression suggests a factorization for the QMHO a la
$$ \frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx}+x\right).$$
However since these are now operators, it is not guaranteed that the product gives exactly $(2)$. Indeed, expanding the product, we instead get
$$-\frac{1}{2}\frac{d^2}{dx^2} + \frac{1}{2}x^2 - \frac{1}{2}.$$
This is not exactly $\mathcal{H}$, but is $\mathcal{H}-\frac{1}{2}$, that is to say that
$$\mathcal{H} = \frac{1}{2}\left(-\frac{d}{dx} + x\right)\left(\frac{d}{dx}+x\right) + \frac{1}{2}.\tag{3}$$
Defining $a = \frac{d}{dx} + x$, we see that the adjoint of $a$, denoted $a^{\dagger}$, which satisfies $\langle af,g\rangle = \langle f,a^{\dagger}g\rangle$, is given by $-\frac{d}{dx} + x$ (which can be verified by integration by parts). Noting that $[x,\frac{d}{dx}] = -1$ (which is directly related to $(1)$), we have the following commutation relation:
\begin{align} [a,a^{\dagger}] &= \left[\frac{d}{dx} + x,-\frac{d}{dx}+x\right] \\ &= \left[\frac{d}{dx},-\frac{d}{dx}\right] + \left[x,-\frac{d}{dx}\right] + \left[\frac{d}{dx},x\right] + [x,x] \\ &= 0 + 1 + 1 + 0 \\ &= 2. \end{align}
Therefore $a^{\dagger}a = aa^{\dagger} - 2$. Moreover we also have
\begin{align} [\mathcal{H},a^{\dagger}] &= \left[\frac{1}{2}a^{\dagger}a+\frac{1}{2},a^{\dagger}\right] \\ &= \frac{1}{2}(a^{\dagger}aa^{\dagger} - a^{\dagger}a^{\dagger}a) \\ &= \frac{1}{2} a^{\dagger}[a,a^{\dagger}] \\ &= a^{\dagger}.\end{align}
Thus $\mathcal{H}a^{\dagger} = a^{\dagger}\mathcal{H} + a^{\dagger}$. What this in turn gives is that if $\psi$ is an eigenfunction of $\mathcal{H}$, i.e. it solves $(2)$, with eigenvalue $E$, then $a^{\dagger}\psi$ also solves $(2)$, but it has an eigenvalue of $E+1$:
$$ \mathcal{H}a^{\dagger}\psi = (a^{\dagger}\mathcal{H} + a^{\dagger})\psi = a^{\dagger}(\mathcal{H}\psi) + a^{\dagger}\psi = a^{\dagger}(E\psi) + a^{\dagger}\psi = (E+1)a^{\dagger}\psi.$$
Analogously, $a\psi$ has an eigenvalue of $E-1$ (assuming $a\psi \neq 0$). Since $a^{\dagger}$ increases the eigenvalue, it is called a raising operator; similarly since $a$ decreases the eigenvalue, it is called a lowering operator. Collectively, they are called ladder operators. What we conclude is that the algebraic structure of the QMHO encodes deep information about the analytic nature of the QMHO.
The ladder structure is inherently an algebraic structure about the operators. The structure concerning the QMHO is at its most fundamental level the Heisenberg-Weyl Lie group/Lie algebra. Other Lie algebras have raising and lowering operators, such as $\mathfrak{su}(2)$.
A More General Setting
Without divulging the nature of my current project (which is as of posting unpublished), I stumbled upon the following result concerning pairs of operators $a$ and $b$ (and their adjoints):
Theorem If $a$ and $b$ are operators satisfying $a^{\dagger}a = b^{\dagger}b + \gamma$ and $aa^{\dagger} = bb^{\dagger}+\delta$ for $\gamma,\delta\in\mathbb{R}$, where $\gamma<\delta$, then the operator $a^{\dagger}b$ (resp. $b^{\dagger}a$) is a raising (resp. lowering) operator for $a^{\dagger}a$ (and $b^{\dagger}b$) and the operator $ba^{\dagger}$ (resp. $ab^{\dagger}$) is a raising (resp. lowering) operator for $aa^{\dagger}$ (and $bb^{\dagger}$).
Proof. Without loss of generality, we assume that $\gamma < \delta$ since otherwise we can just switch the roles of $a$ and $b$. Following the argument for the QMHO, we need only to show that $[a^{\dagger}a,a^{\dagger}b]$ is a multiple of $a^{\dagger}b$ and likewise for $ba^{\dagger}$ with $aa^{\dagger}$. To do that, we show a couple of simpler identities first.
\begin{align} aa^{\dagger}a &= ab^{\dagger}b + \gamma a \\ aa^{\dagger}a &= bb^{\dagger}a + \delta a \end{align}
Likewise we have
\begin{align} ba^{\dagger}a &= bb^{\dagger}b+\gamma b \\ aa^{\dagger}b &= bb^{\dagger}b + \delta b\end{align}
Subtracting these equations from each other gives the following "intertwining" relations:
\begin{align} bb^{\dagger}a - ab^{\dagger}b &= (\gamma-\delta)a \\ aa^{\dagger}b-ba^{\dagger}a &= (\gamma-\delta)b \end{align}
From this we get the identities
\begin{align} [b^{\dagger}b,b^{\dagger}a] &= b^{\dagger}(bb^{\dagger}a-ab^{\dagger}b) \\ &= (\gamma-\delta)b^{\dagger}a \\ [bb^{\dagger},ab^{\dagger}] &= (bb^{\dagger}a-ab^{\dagger}b)b^{\dagger} \\ &= (\gamma-\delta)ab^{\dagger} \\ [a^{\dagger}a,a^{\dagger}b] &= a^{\dagger}(aa^{\dagger}b-a^{\dagger}ba) \\ &= (\delta-\gamma) a^{\dagger}b \\ [aa^{\dagger},ba^{\dagger}] &= (aa^{\dagger}b-ba^{\dagger}a)a^{\dagger} \\ &= (\delta-\gamma) ba^{\dagger}\end{align}
\begin{align} aa^{\dagger}a &= ab^{\dagger}b + \gamma a \\ aa^{\dagger}a &= bb^{\dagger}a + \delta a \end{align}
Likewise we have
\begin{align} ba^{\dagger}a &= bb^{\dagger}b+\gamma b \\ aa^{\dagger}b &= bb^{\dagger}b + \delta b\end{align}
Subtracting these equations from each other gives the following "intertwining" relations:
\begin{align} bb^{\dagger}a - ab^{\dagger}b &= (\gamma-\delta)a \\ aa^{\dagger}b-ba^{\dagger}a &= (\gamma-\delta)b \end{align}
From this we get the identities
\begin{align} [b^{\dagger}b,b^{\dagger}a] &= b^{\dagger}(bb^{\dagger}a-ab^{\dagger}b) \\ &= (\gamma-\delta)b^{\dagger}a \\ [bb^{\dagger},ab^{\dagger}] &= (bb^{\dagger}a-ab^{\dagger}b)b^{\dagger} \\ &= (\gamma-\delta)ab^{\dagger} \\ [a^{\dagger}a,a^{\dagger}b] &= a^{\dagger}(aa^{\dagger}b-a^{\dagger}ba) \\ &= (\delta-\gamma) a^{\dagger}b \\ [aa^{\dagger},ba^{\dagger}] &= (aa^{\dagger}b-ba^{\dagger}a)a^{\dagger} \\ &= (\delta-\gamma) ba^{\dagger}\end{align}
Note the importance of the assumption that $\gamma\neq\delta$. Otherwise the mixed operators $a^{\dagger}b$, etc. would not resemble ladder operators in the truest sense. In the above commutation relations, $a^{\dagger}a$ can be replaced with $b^{\dagger}b$ and vice versa; likewise for the $aa^{\dagger}$ and $bb^{\dagger}$. $\Box$
The relations $a^{\dagger}a = b^{\dagger}b + \gamma$ and $aa^{\dagger} = bb^{\dagger} + \delta$ seem rather strange at first, but in physical terms, it means that the "Hamiltonians" $a^{\dagger}a$ and $b^{\dagger}b$ are really the same, but with an energy level shift of $\gamma$. They capture the same fundamental dynamics, but they have a different zero point. Likewise for $aa^{\dagger}$ and $bb^{\dagger}$. We have a further result:
Theorem The set $\{a^{\dagger}a-\frac{\gamma}{2},a^{\dagger}b,b^{\dagger}a\}$ forms a Lie algebra isomorphic to $\mathfrak{su}(1,1)$.
Proof. Based on the work above, we need only to consider $[a^{\dagger}b,b^{\dagger}a]$.
\begin{align} [a^{\dagger}b,b^{\dagger}a] &= a^{\dagger}bb^{\dagger}a - b^{\dagger}aa^{\dagger}b \\ &= a^{\dagger}(aa^{\dagger}-\delta)a - b^{\dagger}(bb^{\dagger}+\delta)b \\ &= (a^{\dagger}a)^2 - (b^{\dagger}b)^2 - \delta(a^{\dagger}a + b^{\dagger}b) \\ &= (a^{\dagger}a)^2 - (a^{\dagger}a - \gamma)^2 - \delta(a^{\dagger}a + a^{\dagger}a - \gamma) \\ &= 2(\gamma-\delta)a^{\dagger}a - \gamma(\gamma-\delta) \\ &= -2(\delta-\gamma)\left(a^{\dagger}a - \frac{\gamma}{2}\right) \end{align}
The $\mathfrak{su}(1,1)$ Lie algebra is generated by elements $K_3, K_+, K_-$ satisfying the commutation relations
$$ [K_3,K_{\pm}] = \pm K_{\pm}, \qquad [K_+,K_-] = -2K_3.$$
Letting $a^{\dagger}a-\frac{\gamma}{2} = (\delta-\gamma)K_3$, $a^{\dagger}b = \sqrt{\delta-\gamma}K_+$, and $b^{\dagger}a = \sqrt{\delta-\gamma}K_-$, our commutation relations exactly mimic those of $\mathfrak{su}(1,1)$. $\Box$
A similar analysis holds for $aa^{\dagger}$ and its ladder operators. The Lie algebra $\mathfrak{su}(1,1)$ does not have any finite dimensional unitary irreducible representations, however we can always make a finite dimensional representation of $\mathfrak{su}(2)$ into a finite dimensional representation of $\mathfrak{su}(1,1)$ by complexifying the operators in the proper way. The resulting representation will not be unitary. Finite dimensional representations correspond to matrices, so it is natural to ask if any pair of matrices satisfies the conditions for the theorem. We give an answer in the next result.
Theorem No pair of matrices $a$, $b$ can satisfy the conditions for the theorem.
Proof. Suppose such a pair of, say, $n\times n$ matrices existed. Taking the trace of $a^{\dagger}a$, we have
$$ \operatorname{tr}(a^{\dagger}a) = \operatorname{tr}(b^{\dagger}b+\gamma I) = \operatorname{tr}(b^{\dagger}b) + \operatorname{tr}(\gamma I) = \operatorname{tr}(b^{\dagger}b) + \gamma n.$$
However taking the trace of $aa^{\dagger}$ gives
$$ \operatorname{tr}(aa^{\dagger}) = \operatorname{tr}(bb^{\dagger} + \delta I) = \operatorname{tr}(bb^{\dagger}) + \operatorname{tr}(\delta I) = \operatorname{tr}(bb^{\dagger}) + \delta n.$$
Noting then that the trace is cyclic, we have $\operatorname{tr}(aa^{\dagger}) = \operatorname{tr}(a^{\dagger}a)$ and likewise $\operatorname{tr}(bb^{\dagger}) = \operatorname{tr}(b^{\dagger}b)$, which gives the following system of equations:
\begin{align} \operatorname{tr}(a^{\dagger}a) &= \operatorname{tr}(b^{\dagger}b) + \gamma n \\ \operatorname{tr}(a^{\dagger}a) &= \operatorname{tr}(b^{\dagger}b) + \delta n \end{align}
This system can only be solved if $\gamma = \delta$; however by supposition, $\gamma\neq\delta$ and so we must conclude that no such matrices exist. $\Box$
Thus every representation of the system must be infinite dimensional in nature. The above works in more generality. If one is working in an operator system with a tracial state, the above will show that again no such $a$ or $b$ exist.
Let $a,a^{\dagger}$ be the ladder operators for particle $A$ and $b,b^{\dagger}$ be the ladder operators for particle $B$. Form the composite system by tensoring, i.e. $a\otimes 1$ and $a^{\dagger}\otimes 1$ act only on the $A$ element and similarly for $1\otimes b$ and $1\otimes b^{\dagger}$. Define then the following operators
$$ J_+ = (a^{\dagger}\otimes 1)(1\otimes b),\qquad J_- = (1\otimes b^{\dagger})(a\otimes 1), \\ J_3 = (a^{\dagger}\otimes 1)(a\otimes 1) - (1\otimes b^{\dagger})(1\otimes b).$$
Then $J_+$ and $J_-$ act as ladder operators for $J_3$. The use of the letter $J$ suggests a connection to spin. Indeed, after a reparameterization, the above operators exactly satisfy the commutation relations for the spin operators, i.e. form the Lie algebra $\mathfrak{su}(2)$.
A seeming oversight of Schwinger's approach is that ultimately it is hinged on the QMHO and the first order commutation relations it obeys. These commutation relations are so deeply ingrained in the theory of the QMHO that it is difficult to see how to generalize. While my approach above is similar in effect, the idea was developed in a much different way. After some of my work is published, I will expand upon the ideas presented here and how I came to the result above.
I am unaware if this result has yet appeared anywhere in the literature. Given the odd nature of the set-up, I would not be surprised that it has not. In effect, what it is assuming is that two related Hamiltonians (via supersymmetric arguments) have two distinct factorizations which are governed by the same pairs of operators. (The distinctness comes from the fact that $\gamma\neq\delta$ - otherwise one may take $b = a$.)
In the case of the QMHO, one may take $b = a^{\dagger}$ and the above analysis gives second order ladder operators (which amount to $(a^{\dagger})^2$ and $a^2$) as opposed to the usual ladder operators.
Bonus: It is natural to ask if the condition $aa^{\dagger} = bb^{\dagger}+\delta$ is necessary or not in the theorem (taking $a^{\dagger}a = b^{\dagger}b + \gamma$ to be true) or if there is some weakening of this that will allow the theorem to remain true. As of posting this, I am unsure, however I am inclined to believe that it is indeed necessary.
Theorem The set $\{a^{\dagger}a-\frac{\gamma}{2},a^{\dagger}b,b^{\dagger}a\}$ forms a Lie algebra isomorphic to $\mathfrak{su}(1,1)$.
Proof. Based on the work above, we need only to consider $[a^{\dagger}b,b^{\dagger}a]$.
\begin{align} [a^{\dagger}b,b^{\dagger}a] &= a^{\dagger}bb^{\dagger}a - b^{\dagger}aa^{\dagger}b \\ &= a^{\dagger}(aa^{\dagger}-\delta)a - b^{\dagger}(bb^{\dagger}+\delta)b \\ &= (a^{\dagger}a)^2 - (b^{\dagger}b)^2 - \delta(a^{\dagger}a + b^{\dagger}b) \\ &= (a^{\dagger}a)^2 - (a^{\dagger}a - \gamma)^2 - \delta(a^{\dagger}a + a^{\dagger}a - \gamma) \\ &= 2(\gamma-\delta)a^{\dagger}a - \gamma(\gamma-\delta) \\ &= -2(\delta-\gamma)\left(a^{\dagger}a - \frac{\gamma}{2}\right) \end{align}
The $\mathfrak{su}(1,1)$ Lie algebra is generated by elements $K_3, K_+, K_-$ satisfying the commutation relations
$$ [K_3,K_{\pm}] = \pm K_{\pm}, \qquad [K_+,K_-] = -2K_3.$$
Letting $a^{\dagger}a-\frac{\gamma}{2} = (\delta-\gamma)K_3$, $a^{\dagger}b = \sqrt{\delta-\gamma}K_+$, and $b^{\dagger}a = \sqrt{\delta-\gamma}K_-$, our commutation relations exactly mimic those of $\mathfrak{su}(1,1)$. $\Box$
A similar analysis holds for $aa^{\dagger}$ and its ladder operators. The Lie algebra $\mathfrak{su}(1,1)$ does not have any finite dimensional unitary irreducible representations, however we can always make a finite dimensional representation of $\mathfrak{su}(2)$ into a finite dimensional representation of $\mathfrak{su}(1,1)$ by complexifying the operators in the proper way. The resulting representation will not be unitary. Finite dimensional representations correspond to matrices, so it is natural to ask if any pair of matrices satisfies the conditions for the theorem. We give an answer in the next result.
Theorem No pair of matrices $a$, $b$ can satisfy the conditions for the theorem.
Proof. Suppose such a pair of, say, $n\times n$ matrices existed. Taking the trace of $a^{\dagger}a$, we have
$$ \operatorname{tr}(a^{\dagger}a) = \operatorname{tr}(b^{\dagger}b+\gamma I) = \operatorname{tr}(b^{\dagger}b) + \operatorname{tr}(\gamma I) = \operatorname{tr}(b^{\dagger}b) + \gamma n.$$
However taking the trace of $aa^{\dagger}$ gives
$$ \operatorname{tr}(aa^{\dagger}) = \operatorname{tr}(bb^{\dagger} + \delta I) = \operatorname{tr}(bb^{\dagger}) + \operatorname{tr}(\delta I) = \operatorname{tr}(bb^{\dagger}) + \delta n.$$
Noting then that the trace is cyclic, we have $\operatorname{tr}(aa^{\dagger}) = \operatorname{tr}(a^{\dagger}a)$ and likewise $\operatorname{tr}(bb^{\dagger}) = \operatorname{tr}(b^{\dagger}b)$, which gives the following system of equations:
\begin{align} \operatorname{tr}(a^{\dagger}a) &= \operatorname{tr}(b^{\dagger}b) + \gamma n \\ \operatorname{tr}(a^{\dagger}a) &= \operatorname{tr}(b^{\dagger}b) + \delta n \end{align}
This system can only be solved if $\gamma = \delta$; however by supposition, $\gamma\neq\delta$ and so we must conclude that no such matrices exist. $\Box$
Thus every representation of the system must be infinite dimensional in nature. The above works in more generality. If one is working in an operator system with a tracial state, the above will show that again no such $a$ or $b$ exist.
Schwinger's "Spinification" of the QMHO
The above methodology is philosophically similar to that of Schwinger's "spinification" of the two-particle QMHO, but casts a much wider net than Schwinger's approach. Schwinger's approach to the two-particle QMHO is as follows:Let $a,a^{\dagger}$ be the ladder operators for particle $A$ and $b,b^{\dagger}$ be the ladder operators for particle $B$. Form the composite system by tensoring, i.e. $a\otimes 1$ and $a^{\dagger}\otimes 1$ act only on the $A$ element and similarly for $1\otimes b$ and $1\otimes b^{\dagger}$. Define then the following operators
$$ J_+ = (a^{\dagger}\otimes 1)(1\otimes b),\qquad J_- = (1\otimes b^{\dagger})(a\otimes 1), \\ J_3 = (a^{\dagger}\otimes 1)(a\otimes 1) - (1\otimes b^{\dagger})(1\otimes b).$$
Then $J_+$ and $J_-$ act as ladder operators for $J_3$. The use of the letter $J$ suggests a connection to spin. Indeed, after a reparameterization, the above operators exactly satisfy the commutation relations for the spin operators, i.e. form the Lie algebra $\mathfrak{su}(2)$.
A seeming oversight of Schwinger's approach is that ultimately it is hinged on the QMHO and the first order commutation relations it obeys. These commutation relations are so deeply ingrained in the theory of the QMHO that it is difficult to see how to generalize. While my approach above is similar in effect, the idea was developed in a much different way. After some of my work is published, I will expand upon the ideas presented here and how I came to the result above.
I am unaware if this result has yet appeared anywhere in the literature. Given the odd nature of the set-up, I would not be surprised that it has not. In effect, what it is assuming is that two related Hamiltonians (via supersymmetric arguments) have two distinct factorizations which are governed by the same pairs of operators. (The distinctness comes from the fact that $\gamma\neq\delta$ - otherwise one may take $b = a$.)
In the case of the QMHO, one may take $b = a^{\dagger}$ and the above analysis gives second order ladder operators (which amount to $(a^{\dagger})^2$ and $a^2$) as opposed to the usual ladder operators.
Bonus: It is natural to ask if the condition $aa^{\dagger} = bb^{\dagger}+\delta$ is necessary or not in the theorem (taking $a^{\dagger}a = b^{\dagger}b + \gamma$ to be true) or if there is some weakening of this that will allow the theorem to remain true. As of posting this, I am unsure, however I am inclined to believe that it is indeed necessary.
