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I have trouble understanding this proof in Nielsen & Chuang, specifically the identity in $(10.20)$, which reads $$ U_k^\dagger P_k F_l \sqrt{\rho} = U_k^\dagger P_k^\dagger P_k^\dagger F_l P \sqrt{\rho}.$$

By playing around with the definitions, the only way how I would get the rightmost $P$ to appear would be by assuming $\sqrt{\rho} = P \sqrt{\rho}$. In this answer it is stated that "$P$ is the projector onto the code space and thus does nothing to $\rho$". But I don't get this statement.

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Recall from the sentence immediately preceding $(10.20)$ that $\rho$ is a state in the code subspace $C$ and from the statement of theorem $10.1$ that $P$ is a projector onto $C$. We will show that $\sqrt{\rho}=P\sqrt{\rho}$.

Let $|i_L\rangle$ be an orthonormal basis of the code subspace. Then $$ P=\sum_i|i_L\rangle\langle i_L|.\tag1 $$ Therefore, for any $|j_L\rangle$, we have$^1$ $$ P|j_L\rangle=\sum_i|i_L\rangle\langle i_L|j_L\rangle=|j_L\rangle.\tag2 $$ Now, if $\rho$ has eigendecomposition $$ \rho=\sum_kp_k|\psi_k\rangle\langle\psi_k|\tag3 $$ then $$ \sqrt{\rho}=\sum_k\sqrt{p_k}|\psi_k\rangle\langle\psi_k|.\tag4 $$ But $\rho$ is in the code subspace, so $|\psi_k\rangle=\sum_i\alpha_{ki}|i_L\rangle$ for some $\alpha_{ki}\in\mathbb{C}$. Substituting into $(4)$, we get $$ \sqrt{\rho}=\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}|i_L\rangle\langle j_L|.\tag5 $$ Finally, calculate $$ \begin{align} P\sqrt{\rho}&=P\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}|i_L\rangle\langle j_L|\tag6\\ &=\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}P|i_L\rangle\langle j_L|\tag7\\ &=\sum_{ijk}\sqrt{p_k}\alpha_{ki}\overline{\alpha_{kj}}|i_L\rangle\langle j_L|\tag8\\ &=\sqrt{\rho}\tag9 \end{align} $$ where we used $(5)$, linearity of $P$, $(2)$, and finally $(5)$ again.


$^1$ Equivalently, one may take $(2)$ together with the requirement that $P$ vanishes on vectors orthogonal to the space spanned by $|i_L\rangle$ as the definition of the projector.

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  • $\begingroup$ Very helpful, thanks! $\endgroup$
    – qntdni
    Dec 1, 2023 at 16:21
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$\rho$ is a state in the code. $P$ is the projector onto the codespace. That means (by definition) that for any state $|\psi\rangle$ in the code, $P|\psi\rangle=|\psi\rangle$.

$\rho$ is one such state (except that it's possibly mixed), so it must satisfy $P\rho=\rho$. Thus, if $\rho$ appears somewhere in an equation, you can arbitrarily choose to replace it with $P\rho$ if you want. Furthermore, if $\rho$ is in the code, all its eigenvectors are in the code. Hence $\sqrt{\rho}$, which has the same eigenvectors, must also be in the code.

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  • $\begingroup$ Thanks, but shouldn't it be $P \rho P = \rho$? $\endgroup$
    – qntdni
    Dec 1, 2023 at 16:25
  • $\begingroup$ That is also true. Think about a pure state $P|\psi\rangle=|\psi\rangle$. So, if I had $\rho=|\psi\rangle\langle\psi|$, then it is true that $\rho=P\rho=\rho P=P\rho P$ $\endgroup$
    – DaftWullie
    Dec 1, 2023 at 16:31

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