0
$\begingroup$

I want to prove that if I have a density matrix of the form: $$ \begin{pmatrix} p_{++}& p_{+-}\\ p_{-+}&p_{--} \end{pmatrix} $$

then $|p_{+-}|^2 = |p_{-+}|^2 \le p_{++}p_{--}$. (This was stated here). However, I don't know where to start. I appreciate if you could point the way to prove this. Thanks

$\endgroup$
2
$\begingroup$

Since density matrix is hermitian, $p_{+-}=p_{-+}^*$. As the eigenvalues are necessarily non-negative: $$ \hbox{Det}\begin{pmatrix} p_{++}& p_{+-}\\ p_{-+}&p_{--} \end{pmatrix}= p_{++}p_{--}-\vert p_{+-}\vert^2 \ge 0 $$ and the result follows.

$\endgroup$

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.