# Why does $\sum_n \langle n|M_m\rho M_m^\dagger|n\rangle$ simplify to $\langle \psi|M_m^\dagger M_m|\psi\rangle$?

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.

• @glS.: Sometimes good titles just do not come to mind. It was not intentional. Feb 25 at 10:04

## 1 Answer

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

• 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. Feb 25 at 3:03
• Yes. That is right. Feb 25 at 3:18