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Is commutator of two operators an operator? I searched google but still got no success! I'm very curious to know the answer to this!

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    $\begingroup$ What else would it be? $\endgroup$ – ACuriousMind Jun 7 '15 at 16:20
  • $\begingroup$ Is that something for me to solve! $\endgroup$ – Aman Jun 7 '15 at 16:22
  • $\begingroup$ @ACuriousMind to be pedantic, it could be even not an operator (i.e. defined only on the vector zero). $\endgroup$ – yuggib Jun 7 '15 at 16:23
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    $\begingroup$ @Aman if it is well defined on a reasonable (dense) set of vectors it satisfies indeed all the properties of a linear operator (trivial to check). Anyways it seems a question better suited for math SE (and shows no research effort in addition)... $\endgroup$ – yuggib Jun 7 '15 at 16:33
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    $\begingroup$ (Bounded) operators form an algebra (a vector space equipped with a product). Therefore the commutator of two operators is an operator, as it is a composition of the operations multiplication and addition. $\endgroup$ – Sebastian Riese Jun 7 '15 at 16:34
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Short answer: yes

A map $A: V \to V$ on a $\mathbb{K}$-vectorspace $V$ is linear if $\forall v_1, v_2 \in V$ and $\forall \lambda \in \mathbb{K}$ we have: $$A(\lambda v_1 + v_2)= \lambda A(v_1)+A(v_2)$$

If we have two such maps $A,B$ then $[A,B]=AB-BA$ is again linear because, well AB is linear:

$$(AB)(\lambda v_1+v_2)=A(B(\lambda v_1 + v_2))=A(\lambda B(v_1)+B(v_2))\\=\lambda A(B(v_1))+A(B(v_2))=\lambda AB(v_1)+AB(v_2)$$

(similarly BA is linear) and for any linear map $C$, $\alpha\, C$ is linear for $\alpha \in \mathbb{K}$, in this case $C=BA$ and $\alpha = -1$.

So $AB-BA$ can be written as the sum of two linear maps. The sum of two linear maps is also linear (if you have trouble seeing this orient yourself on how the statement that the composition of linear maps is again linear, and use $(A+B)(v) = A(v)+B(v)$).

This is pretty much whats is meant by the linear maps on a vector space being an algebra, you have an invertible and commutative addition operation between the maps, you can multiply constants and have a multiplicative operation (here the composition) between the maps that has an identity.

The long answer is that the commutator of two linear operators is again a linear operator on the space on which it is defined. Not every operator (and in infinte dimensional Hilbert spaces quite a lot of physically relevant operators) is defined on the entire vector space.

An example would be the differential operator $\partial_x$ on $L^2(\mathbb{R})$, not every element of $L^2(\mathbb{R})$ has a differentiable representative (not every function can be differentiated). If you look at the adjoint of $\partial_x$ you will find that it is defined on an even smaller subset.

But maybe these comments are going a bit too far afield.

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