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I am currently working through a set of lecture notes on operator theory.

For self - adjoint operators, I just showed that, if $B_1$ and $B_2 \in \mathcal{B}(\mathcal{H},)$ are self - adjoint then $B_1B_2$ is self - adjoint if and only if $B_1$ and $B_2$ commute. (Here, $\mathcal{H}$ denotes a Hilbert Space, and $\mathcal{B}(\mathcal{H})$ stands for the space of bounded operators defined on $\mathcal{H}$.

Now I am wondering - are there actually self - adjoint operators that do not commute?

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Did you try 2x2 matrices? –  Did Apr 15 '12 at 22:48
    
AAh :) thanks for that hint! –  harlekin Apr 15 '12 at 22:57
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You could add your own answer and then accept it. –  Nate Eldredge Apr 16 '12 at 1:23
    
Let $B_1= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ and $B_1= \begin{pmatrix} 0&i\\ -i&0 \end{pmatrix}$ then $ \left[B_1,B_2\right]_-=2i\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\neq 0 $ –  draks ... Apr 16 '12 at 7:13

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Example given by draks in the comments: the matrices $B_1= \begin{pmatrix} 0&1\\ 1&0 \end{pmatrix}$ and $B_1= \begin{pmatrix} 0&i\\ -i&0 \end{pmatrix}$ do not commute: $$\left[B_1,B_2\right]_-=2i\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\neq 0$$

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