Physical interpretation of operator-sum measurement of state after measuring

Question

On page 362 of Nielsen & Chuang, in the attached reference, why do we have

$$|e_k\rangle \langle e_k|U(P \oplus |e_0\rangle \langle e_0|)U^{\dagger} |e_k\rangle \langle e_k|$$

instead of

$$|e_k\rangle \langle e_k|U(P \oplus |e_0\rangle \langle e_0|)U^{\dagger} (|e_k\rangle \langle e_k|)^{\dagger} \ \ ?$$

They are trying there to give a physical interpretation of the operator sum representation.

I thought for getting the state after measurement we must do $M_m * $ operator $M_m$ normalized?

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Answer 1

Recall that the projector onto a pure state is self-adjoint${}^{(1)}$, i.e. $(|e_k\rangle\langle e_k|)^\dagger=|e_k\rangle\langle e_k|$ so the two expressions from your post are the same.

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${}^{(1)}$ This comes from the identity $(|\psi\rangle\langle\phi|)^\dagger=|\phi\rangle\langle\psi|$ for the Hermitian adjoint which holds due to \begin{align*} \big\langle x,\color{red}{(|\psi\rangle\langle\phi|)^\dagger} y\big\rangle\overset{\text{Def.}}=\big\langle (|\psi\rangle\langle\phi|)x,y\big\rangle&=\big\langle \langle\phi,x\rangle\psi,y\big\rangle\\ &=\langle x,\phi\rangle\langle\psi,y\rangle\\ &=\big\langle x,\langle\psi,y\rangle\phi\big\rangle=\big\langle x,\color{red}{(|\phi\rangle\langle\psi|)}y\big\rangle\,. \end{align*}