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I was trying to derive the formula for $p(m)$ in exercise 8.2 on page 357 in Nielsen & Chuang. But I am wondering what rule I can apply to simplify this

$$\mathrm{tr}(\mathcal{E}_m(\rho) )= \sum_n \langle n | M_m \rho M_m^{\dagger} | n\rangle = \sum_n \langle n | M_m | \psi\rangle \langle \psi | M_m^{\dagger} | n\rangle $$

to $\langle\psi|M_m^{\dagger}M_m|\psi\rangle$?

Because after this I can’t find any idea to boil down to this.

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  • $\begingroup$ @glS.: Sometimes good titles just do not come to mind. It was not intentional. $\endgroup$
    – user27286
    Feb 25, 2021 at 10:04

1 Answer 1

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This is because:

$$ \sum_n \langle n | M_m | \psi\rangle \langle \psi | M_m^{\dagger} | n\rangle = \sum_n \langle \psi | M_m^{\dagger} | n\rangle \langle n | M_m | \psi\rangle = \langle \psi | M_m^{\dagger} I M_m | \psi\rangle = \langle \psi | M_m^{\dagger} M_m | \psi\rangle $$

note that $\sum_n|n\rangle\langle n| = I $

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  • $\begingroup$ As it is scalar we can order them anyway right? Is this the main idea? First time it could be ordered because of that...and then in the second case as well..two scalars multiplication would be same from both side. Please confirm. $\endgroup$
    – user27286
    Feb 25, 2021 at 3:03
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    $\begingroup$ Yes. That is right. $\endgroup$
    – KAJ226
    Feb 25, 2021 at 3:18

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