3
$\begingroup$

In Nielsen and Chuang's QCQI, I learned that the quantum tomography for n qubit can be described easily in math as we need to measure $Tr(\rho W_k),\forall k$ where $W_k\in\{I,\sigma_x,\sigma_y,\sigma_z\}^{\otimes n}$.

But what my understanding toward $Tr(\rho \hat{o})$ for some observable $\hat{o}$ is that, we measure with $\rho$ with measurement set from the spectrum decomposition of $\hat{o}$. E.g., the term $Tr(\rho\sigma_x)$ means we measure in this set of projective measurement: $|+\rangle\langle+|,|-\rangle\langle-|$ . But since $I$ have infinite spectrum decomposition(not unique because the multiplicity of the eigenvalues), so what does $Tr(\rho I)$ mean in the language of measurement?

$\endgroup$
4
  • 2
    $\begingroup$ why do you think the degeneracy of $I$ should be a problem here? $\endgroup$
    – glS
    Oct 8 '21 at 14:28
  • 1
    $\begingroup$ In addition to @gls 's comment, $I$ only has an infinite spectrum when you are considering infinite-dimensional Hilbert spaces. $\endgroup$
    – Condo
    Oct 8 '21 at 14:29
  • 6
    $\begingroup$ Generally, it means "don't measure that qubit". $\endgroup$
    – DaftWullie
    Oct 8 '21 at 14:30
  • 1
    $\begingroup$ Not having a unique eigenbasis also means that you can just pick your favorite one. Besides, @DaftWullie is right. Recall that the reduced state $\rho_B$ of a composite state $\rho_{AB}$ is uniquely defined by the equation $\mathrm{tr}(I \otimes X \rho_{AB}) = \mathrm{tr}(X\rho_B)$ for all operators $X$. $\endgroup$ Oct 10 '21 at 7:41

Your Answer

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

Browse other questions tagged or ask your own question.