I am currently reading Nielsen & Chuang, and one of the questions asks to calculate a density matrix with the following mixed state,

$$ \frac{1}{9}\begin{bmatrix} 5 & 1 & −i \\ 1 & 2 & -2i \\ i & 2i & 2 \end{bmatrix}$$

and retrieving the probability of measuring

$$ |0\rangle = \begin{pmatrix} 1 \\ 0\\ 0\\ \end{pmatrix}$$

How would I do this?

  • $\begingroup$ Welcome to QCSE! Please ask one focused question per post. It's no problem to submit multiple posts :-) $\endgroup$ Commented Feb 23, 2021 at 1:43
  • $\begingroup$ Regarding the first part of your question, does this post answer it? $\endgroup$ Commented Feb 23, 2021 at 1:53
  • $\begingroup$ @AdamZalcman Hi, sorry I didnt realise, my apologies. I did try a few times but I cant seem to get the results for either, also shall I submit a new post with these questions seperately or is it okay to keep this post as it is and remember for future? $\endgroup$
    – lambda
    Commented Feb 23, 2021 at 1:58
  • $\begingroup$ You an edit this post so it contains only one question and submit the other as a separate post. That said, the first part of this question (density matrix of $|0\rangle$) is really a duplicate of the question I linked to above (just with different numbers substituted into the same formula). $\endgroup$ Commented Feb 23, 2021 at 2:02
  • 1
    $\begingroup$ @lambda.: Please mention from which exercise this is taken. (page number of the book) $\endgroup$
    – user27286
    Commented Feb 23, 2021 at 2:36

1 Answer 1


Let $\rho$ denote the density matrix specified in the question. The probability of obtaining the result corresponding to $|0\rangle$ when measuring $\rho$ in the computational basis is

$$ p(0|\rho) = \mathrm{tr}(P_0\rho) = \mathrm{tr}(|0\rangle\langle 0|\rho) = \langle 0|\rho|0 \rangle = \frac{5}{9} $$

where $P_0=|0\rangle\langle 0|$ is the projector onto the subspace spanned by $|0\rangle$. See equation $(2.159)$ on page 102 in Nielsen & Chuang where $M_m = P_0$ and note that $P_0=P_0^\dagger$ and $P_0^2 = P_0$.

  • 2
    $\begingroup$ Thank you, i've had a recap through that section and along with your explanation it makes perfect sense, I appreciate all the help and the warm welcome to QCSE :) $\endgroup$
    – lambda
    Commented Feb 23, 2021 at 2:47
  • $\begingroup$ You're welcome! I'm glad it was helpful :-) $\endgroup$ Commented Feb 23, 2021 at 2:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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