2
$\begingroup$

In the book Quantum computation and quantum information, it says to evaluate $tr(A|\psi\rangle\langle\psi|)$ using Gram-Schmidt procedure to extend $|\psi\rangle$ to an orthonormal basis $|i\rangle$ which includes $|\psi\rangle$ as the first element. Then:

$$tr(A|\psi\rangle\langle\psi|)=\sum_i\langle i|A|\psi\rangle\langle\psi|i\rangle\tag{2.60}$$ $$=\langle\psi|A|\psi\rangle\tag{2.61}$$

I understood that equation 2.61 that uses the special basis $|i\rangle$ described. But in equation 2.60, how $\sum_i\langle i|M|i\rangle$ is correlated to $tr(M)$ ? Can you help me with a more detailed description of it ?

$\endgroup$
2
  • 3
    $\begingroup$ The equation in bold is just the definition of a trace computed in the computational basis $\{|i\rangle \}$ where $i=1\dots d$ and $d$ is the dimension of the system. You can pick any basis since trace is basis-independent $\endgroup$
    – forky40
    Feb 1 at 18:30
  • 3
    $\begingroup$ Note that by definition: $Tr(M) = \sum_i M_{ii} $ and note that $ \langle i| M | i\rangle = M_{ii} $. $\endgroup$
    – KAJ226
    Feb 1 at 18:52
2
$\begingroup$

Thanks for the comments, so $tr(M)$ is exactly $\sum_i\langle i|M|i\rangle$ as pointed. I just discovered the same question answered with a proof in the physics stackexchange: https://physics.stackexchange.com/a/104155/273977

$\endgroup$

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.