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I have two operators, $A$ and $B$. I want to figure out their commutator, $[A,\,B]$. The commutator $\left[A,\,B^2\right]=C$ is known. Equivalently, I want to compute $\left[A,\,\sqrt{D}\right]$ given a known $[A,\,D]=C$.

The best I have been able to do is to take the identity $$\left[A,\,B^\beta\right] = \left[A,\,B\right]B^{\beta-1} + B \left[A,\, B^{\beta-1}\right]$$ to get that $$C = [A,\,B]B + B[A,\,B].$$ This can be solved for the desired commutator in two ways: \begin{align} [A,\,B] & = CB^+ - B[A,\,B]B^+,\ \mathrm{and} \\ & = B^+ C - B^+ [A,\, B] B, \end{align} where $B^+$ is the a pseudo-inverse of $B$ (e.g. the Moor-Penrose inverse). Either of those equations can be applied recursively to get the formulae \begin{align} [A,\,B] & = \sum_{j=0}^\infty (-1)^j B^j C B^+ \left(B^j\right)^+,\ \mathrm{and} \\ & = \sum_{j=0}^\infty (-1)^j \left(B^j\right)^+ B^+ C B^j. \end{align}

Is this the best that can be done? If so, how can I tell if those infinite series are convergent? Can more be said if $B$ is a positive semi-definite Hermitian operator (i.e. all eigenvalues are $\ge 0$, but $0$ is definitely one of the eigenvalues)?

In the exact use-case, $A=a(\mathbf{k})$, $B = \sqrt{a^\dagger(\mathbf{k})\, a(\mathbf{k})}$, and $[a(\mathbf{k}),\, a^\dagger(\mathbf{k}')\,a(\mathbf{k}')] = a(\mathbf{k})\,\delta(\mathbf{k}-\mathbf{k}')$. There are a few more identities (like $[a(\mathbf{k}),\, a^\dagger(\mathbf{k}')] = \delta(\mathbf{k}-\mathbf{k}')$) and that the eigenvalues of $a^\dagger(\mathbf{k})\,a(\mathbf{k})$ are discrete, starting at $0$ and going up to infinity. That is, they're Bosonic creation and annihilation operators in quantum field theory.

Sean Lake
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  • So, you only know $B$ and $[A,B^2]$, and want to find $[A,B]$? What precisely is known? – Mark Schultz-Wu Nov 28 '17 at 22:48
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    What if $B^2$ = 0 and $B$ is not central? (Which is to say, I don't think you're going to get much without some involvement with or at least assumption on $A$.) – anomaly Nov 28 '17 at 22:50
  • What does "$B$ is not central" mean? – Sean Lake Nov 29 '17 at 03:17
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    @SeanLake: $B$ is called central if it commutes with every other element: $XB=BX$ for all $X$. Or equivalently, $[X, B] = 0$ for all $X$. An example for the situation anomaly describes is $B = \pmatrix{0&1\0&0}$ in the matrix algebra $M_2(\mathbb{R})$. This $B$ is not central since, for example, $[ \pmatrix{1&0\0&-1}, B] = 2B$. But $B^2 = 0$ and hence $[A, B^2] = 0$ for all matrices $A$, even though $[A, B]$ is sometimes 0 and sometimes not. – Torsten Schoeneberg Nov 29 '17 at 03:57
  • (In my answer, $B$ is not central but $B^2$ is, which suffices for an example where $[A,B^2] = 0$ for every $A$, while $[A, B]$ can be many different things for different $A$.) – Torsten Schoeneberg Nov 29 '17 at 04:04

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I don't know if this translates to your setting, but if you look at the (Hamilton) quaternions and let $A, B$ be pure quaternions, then $B^2 \in \mathbb{R}$ and hence $[A, B^2] = 0$. But $[A, B]$ could be any pure quaternion (or at least all those $[A,B]$ span the full space of pure quaternions, I have not checked the stronger statement.) So there is not much to infer about it.

Torsten Schoeneberg
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