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I think this might be trivial but I have a question about the commutator. I want $A$ to be an operator (might be non-linear) on a Hilbert space (e.g. $L^2$). Is $$ [A,-A] =0$$ always true? I think this is wrong, my attempt: By definition $$[A, -A] = A(-A)- (-AA) = A(-A) +A^2$$ and $A(-A) \neq -A^2$ might be the case because $A$ might be non-linear. Is that correct or did I miss something there?

Ottavio
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kade
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  • If the operator is not linear you have first to define the addition of operator so in order for the commutation makes sense – amilton moreira Nov 27 '20 at 13:09
  • @amiltonmoreira I guess the OP takes $$(A+B)(u) := A(u) + B(u)$$ even if $A,B$ are non-linear. –  Nov 27 '20 at 13:10
  • $-1$ is just a number/scalar. The question either doesn't make sense or could be reformulated more generally for an arbitrary number: $[A, cA]$. – Roger V. Nov 27 '20 at 13:14
  • @kade In this case you have to define what do you mean by $-A$ – amilton moreira Nov 27 '20 at 13:16
  • If we define $(cA)(u) = c \cdot A(u)$, then we do not care whether $A$ is a linear function or not; we would have $[A,-A] = \textbf{0}$ always. –  Nov 27 '20 at 13:18
  • @kade What I mean if the minus in the commutator is the same minus in the operator – amilton moreira Nov 27 '20 at 13:20
  • Ok yes I think I get the problem about the necessity of a definition. So let's define it in the way @AndreasMastronikolis did. But how is $[A,-A]=0$ then? Let's take some $u$, then $$A,-A=(A(-A) - (-AA))(u)=A(-A)(u) - (-A A)(u) = A(-A(u)) - (-A(A(u))) = A(-A(u)) + A(A(u))$$ but how can we then conclude that this is $0$? – kade Nov 27 '20 at 13:27
  • Ok but isn't that linearity of the operator that $A(-w) = -A(w)$? I mean, take $\mathbb{R}$ as the Hilbert space, $A(x):= x^2$ for $x \in \mathbb{R}$. Then $$A(-A(x)) = A(-x^2) = (-x^2)^2 = x^4 \neq -x^4 = -A(x^2) = -A(A(x))$$ – kade Nov 27 '20 at 13:35
  • @kade Yes, I think you are right. –  Nov 27 '20 at 13:36
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    A more semantic comment: I would never use the word operator on a Hilbert space to mean anything else than an endomorphism (maybe a homomorphism between Hilbert spaces, but I prefer an operator to have same domain and codomain).

    So something that is not structure preserving should always be called a function.

     – Richard Jensen Nov 27 '20 at 14:20
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    @RichardJensen I kind of agree with that but the convention in my classes is always that an operator is an arbitrary function between a vector space and itself – kade Nov 27 '20 at 14:27
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    @kade Right, that clears that up :) – Richard Jensen Nov 27 '20 at 14:27

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Your commutator vanishes only if $A(-f) = -A(f)$. For a counter example, consider $A(f) = \text{abs} f = |f|$: the two terms in the commutator,

\begin{align} \text{abs}( -\text{abs}(f)) &= +\text{abs}(f) \\ -\text{abs}( \text{abs}(f)) &= -\text{abs}(f) \end{align}

are not identical, so the commutator does not vanish.

rob
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