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Suppose that $A$ and $B$ are self-adjoint bounded linear operators on a Hilbert space and $\lambda \in \mathbb{C}$. It turns out that if $\lambda \notin \{-1, 1\}$ then $AB=\lambda BA \implies AB = BA = 0$.

Does anyone know of any applications of this result?

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In the physics context, with $A$ and $B$ creation operators of two identical particles, the fact that only $AB=+BA$ and $AB=-BA$ are nontrivially allowed implies that the particles must be either bosons (even under exchange) or fermions (odd under exchange).

$\require{enclose} \enclose{horizontalstrike}{\style{font-family:inherit;}{\scriptsize\text{Creation operators are not self-adjoint, but still if $AB=\lambda BA$ then $AB=\lambda^2 AB$ hence either $\lambda\in\{-1,1\}$ or $AB=0$.}}}$

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    $\begingroup$ For any $q\in (-1,1)$ one can construct operators $A$ and $B$ such that $AB = qBA$, so your last statement is not correct. In the self-adjoint case the condition $AB=\lambda BA$ indeed implies that also $BA = \lambda AB$, hence $AB = \lambda^{2} AB$, but that's not the case in general. $\endgroup$
    – Mateusz Wasilewski
    Commented May 25, 2020 at 13:13
  • $\begingroup$ Sooo ... one can't define creation operators for anyons? $\endgroup$
    – Michael Engelhardt
    Commented May 25, 2020 at 13:43
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    $\begingroup$ local creation operators are fermionic or bosonic, when you exchange the order only the initial and final position need to be specified, so their statistics is governed by the permutation group; for the exchange of anyons the path by which they are exchanged matters, governed by the braid group. $\endgroup$
    – Carlo Beenakker
    Commented May 25, 2020 at 14:17
  • $\begingroup$ Certainly, but it seems to me that your argument as stated would therefore need some qualification as to what types of creation operators $A$ and $B$ you're considering - roughly speaking, "local", as you say. If the operators involve some Wilson lines connecting them to some reference point, one gets into trouble. Evidently, the OP's observation extends beyond self-adjoint operators, but there are restrictions - it would be interesting to understand those. $\endgroup$
    – Michael Engelhardt
    Commented May 25, 2020 at 15:58

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