1
$\begingroup$

Suppose 3 operators $A$, $B$, $C$ are Hermitian operators. Assume $A$ has a non-degenerate spectrum, and $AB$ and $AC$ are also Hermitian. Show that $$[B,C] = 0$$

From the conditions $A$, $B$, $C$, $AB$ and $AC$ are Hermitian operators, one can derive that $$[A,B]=[A,C]=0$$ How can one proceed to show that $[B,C]=0$?

$\endgroup$
  • 1
    $\begingroup$ Note that $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$. Think: how does that look if $A, B$ and $AB$ are Hermitian? what does that tell us? $\endgroup$ – yu-v Feb 3 at 17:11
  • $\begingroup$ I know that gives us $[A,B]=0$, same argument gives $[A,C]=0$, but how do we relate this to [B,C]? $\endgroup$ – LY3000 Feb 3 at 17:16
  • $\begingroup$ oh sorry, I didn't read the question carefully enough. You can show that $[A, BC]=0$. This, together with the fact that $A$ is non degenerate, should allow you to show that $BC$ is also hermitian. $\endgroup$ – yu-v Feb 3 at 17:30
  • $\begingroup$ This is where I get unsure, I tried to write $A=\sum_i a_i|a_i><a_i|$, with $a_i \neq a_j$. But I'm not sure how to proceed $\endgroup$ – LY3000 Feb 3 at 17:46
  • $\begingroup$ Great, so eigenvectors are orthogonal, $A=\sum_i a_i ~|i\rangle\langle i|$, $B=\sum_i b_i ~|i\rangle\langle i|$, and $C=\sum_i c_i ~|i\rangle\langle i|$. $\endgroup$ – Cosmas Zachos Feb 3 at 22:25
4
$\begingroup$

You are nearly there. If $A$ commutes with $B$ it means that they can be diagonalized simultaneously. Now use the fact the the eigenvalues of $A$ are non-degenerate. This means that also $B$ is diagonal in the same basis. Repeat with $C$ and you are done.

$\endgroup$

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.