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?

  • 2
    $\begingroup$ Did you try 2x2 matrices? $\endgroup$ – Did Apr 15 '12 at 22:48
  • $\begingroup$ AAh :) thanks for that hint! $\endgroup$ – harlekin Apr 15 '12 at 22:57
  • $\begingroup$ 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 $ $\endgroup$ – draks ... Apr 16 '12 at 7:13

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: $$\left[B_1,B_2\right]_-=2i\begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}\neq 0$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.