1
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

2
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .