The following is some preamble for motivation which can be skipped; the quesiton will be posed at the end.

I recently started studying the creation and annihilation operators $\hat a^\dagger$ and $\hat a$ in a QM course I'm taking -- they've been introduced as satisfying $$\hat H=\hat a\hat a^\dagger\gamma+\zeta,$$ for $\gamma,\zeta\in\mathbb{C}$. In the course we were initially given a specific Hamiltonian to work with (single particle in an infinite square potential well), and we derived forms for $\hat a^\dagger$ and $\hat a$ based on this exact form of the Hamiltonian.

It occurred to me that we were effectively viewing the Hamiltonian as a multivariate polynomial over $\mathbb{C}$ with the observable operators as variables, for example $$\hat H=\frac{\hat p^2}{2m}+\hat V(\hat x)=\frac{1}{2m}\hat p^2+\frac{m\omega^2}{2}\hat x^2,$$ then using the Euclidean division algorithm to leverage the fact that polynomial rings are Euclidean domains to form a greatest common divisor of $\hat H$ and $\hat a^\dagger\hat a$, where $$\hat a=\sqrt{\frac{m\omega}{2\hbar}}\hat x+i\sqrt{\frac{1}{2\hbar m\omega}}\hat p.$$ In this case the greatest common divisor is $\gamma=\omega\hbar$ with a remainder of $\zeta=\frac{\omega\hbar}{2}$.

My question is this:

Is this process of viewing the Hamiltonian as a 'multivariate operator polynomial' of degree $\geq2$ in an Euclidean domain, where the variables are observable operators, physically signifigant somehow?

In particular, does the process of finding the greatest common divisor of $\hat H$ and $\hat a\hat a^\dagger$ in this setting have any established physical significance, where we view $\hat H$ as the total energy operator and $\hat a^\dagger,\hat a$ as the creation and annihilation operators?

Note that (like all algorithms) the Euclidean division algorithm is a finite recursion, so this question can somewhat more generally be viewed as a question about recursive relationships between the Hamiltonian of a system and the creation/annihilation operators for that system.

EDIT: Since the greatest common divisor (GCD) $\omega\hbar$ and the remainder $\frac{\omega\hbar}{2}$ have the units of energy in the specific case above, I suspect that the GCD might be related to quantization in some fashion.

SECOND EDIT: In response to a comment below, I'd like to clarify that it is completely fine to have a notion of a noncommutative polynomial, or in this case a lie-bracket commutative polynomial. This Hamiltonian polynomial would be such a structure, so $\hat x\hat p=\hat p\hat x+[\hat x,\hat p]=\hat p\hat x+i\hbar$ is still fine as an identity. In order for such a structure to be a Euclidean domain however, we must require that the overall polynomials commute with each other, which is fine since we can cancel any non-commutativity by adjoining the appropriate additional terms (this will determine the Euclidean domain subspace of the overall operator space in some sense).

ANOTHER EDIT: In response to another comment below, it is worth mentioning that multivariate polynomial rings are generally only unique factorization domains, not necessarily Euclidean domains (we can decompose multivariate polynomials into unique factors, but a unique GCD doesn't always exist between two elements). Despite this I believe it is possible to restrict to certain subsets of multivariate polynomial rings to obtain a Euclidean domain (roughly speaking look at the subsets connected by factors), and I suspect that $\hat a^\dagger\hat a$ and $\hat H$ will live together in such a subspace in general.

  • 1
    $\begingroup$ Any reason for the downvote? If I can clarify anything please let me know. $\endgroup$
    – Alec Rhea
    Nov 27, 2017 at 1:49
  • 9
    $\begingroup$ There is a powerful evil force on this forum that downvotes everything it doesn't like. And it hardly likes anything. It's practically organized crime that we can do nothing about. $\endgroup$
    – safesphere
    Nov 27, 2017 at 2:18
  • $\begingroup$ @safesphere Cest la vie -- thanks for the heads up. $\endgroup$
    – Alec Rhea
    Nov 27, 2017 at 3:23
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    $\begingroup$ I don’t think what you are suggesting makes much sense, but maybe I just don’t understand it completely. Please clarify the following: does your treatment take into account the noncommutativity of $x$ and $p$? It sounds like from your point of view, $xp$ and $px$ are the same thing, while in reality they are not. P.S. I didn’t downvote $\endgroup$ Nov 27, 2017 at 6:19
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/10804/2451 and links therein. $\endgroup$
    – Qmechanic
    Nov 27, 2017 at 9:55

1 Answer 1


Your question is actually closely related to a factorization property of some 2nd order ODEs. This was discussed in the thesis of T.E. Hull and published, with L. Infeld, as "Infeld, L., and T. E. Hull. The factorization method. Reviews of modern Physics 23.1 (1951): 21."

This factorization approach is closely related to supersymmetric quantum mechanics (see Cooper, Fred, Avinash Khare, and Uday Sukhatme. "Supersymmetry and quantum mechanics." Physics Reports 251.5-6 (1995): 267-385.)

Suppose we have a Hamiltonian $\hat H_-$ for which we already know the ground state solution $\vert{\psi_0}\rangle$ and the corresponding ground state energy $E_0$, i.e. we know (in the $x$ representation) that $\psi_0(x)$ is a solution to $$ \hat H\psi(x)=\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V_-(x)\right)\psi_0(x)=0\, , \qquad V_-(x)=V(x)-E_0 $$

Then one can show that Hamiltonian can be rewritten in the form $$ \hat H_-=\frac{\hbar^2}{2m}\left(\frac{d^2}{dx^2}+\frac{\psi_0^{''}(x)}{\psi_0(x)}\right)=\hat A^\dagger\hat A\, , $$ where $$ \hat A=\frac{\hbar}{\sqrt{2m}}\left(\frac{d}{dx}-\frac{\psi^{'}_0(x)}{\psi_0(x)}\right)\, ,\qquad \hat A^\dagger=\frac{\hbar}{\sqrt{2m}}\left(-\frac{d}{dx}-\frac{\psi^{'}_0(x)}{\psi_0(x)}\right) $$ and with $\hat A$ and $\hat A^\dagger$ satisfying $$ [\hat A,\hat A^\dagger]=\frac{2\hbar}{\sqrt{2m}}W'(x)\, ,\qquad W(x)=-\frac{\hbar}{\sqrt{2m}}\left(\frac{\psi^{'}_0(x)}{\psi_0(x)}\right)\, . $$ where $W(x)$ is the superpotential for the problem.

The (super)symmetry comes in by noting that \begin{align} \hat A\hat A^\dagger\equiv \hat H_+&=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V_+(x)\, ,\\ V_+(x)&=-V_-(x)+2W^2(x)\, , \\ V_{\pm}(x)&=W^2(x)\pm \frac{\hbar}{2m}W'(x) \end{align} $V_{\pm}$ are known as supersymmetric partners and support the same energy eigenvalues $E_{n}^{-}=E_{n}^+$ except that the lowest energy $E_{0}^-$, which does not have a counterpart in $V_+(x)$.

If $\psi_n^-(x)$ is an eigenfunction of $\hat H_-$ with eigenvalue $E_n^-$, then $\hat A\psi_n^-(x)$ is an eigenfunction of $\hat H_+$ with the same eigenvalue.

The simplest application of this is to the infinite well, with $E_0=\frac{\hbar^2\pi^2}{2ma^2}$ and $\psi_n(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{(n+1)\pi x}{a}\right)$. One can find the energies and wavefunctions of the ground and excited state for the potential $V_+(x)=\frac{\hbar^2 \pi^2}{2ma^2}\left(\frac{2}{\sin^2\left(\frac{\pi x}{a}\right)}-1\right)$.


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