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 :-) $\endgroup$ – Adam Zalcman Feb 23 at 1:43
  • $\begingroup$ Regarding the first part of your question, does this post answer it? $\endgroup$ – Adam Zalcman Feb 23 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 Feb 23 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). $\endgroup$ – Adam Zalcman Feb 23 at 2:02
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    $\begingroup$ @lambda.: Please mention from which exercise this is taken. (page number of the book) $\endgroup$ – user27286 Feb 23 at 2:36

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 $P_0^2 = P_0$.

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    $\begingroup$ 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 Feb 23 at 2:47
  • $\begingroup$ You're welcome! I'm glad it was helpful :-) $\endgroup$ – Adam Zalcman Feb 23 at 2:49

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