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I got a problem in understanding the proof of the Theorem 2.6 (Unitary freedom in the ensenble for density matrices), 2.168 and 2.169 in the Nielsen and Chuang book

**Equation 2.168**

Suppose $|{\tilde\psi_i}\rangle = \sum_j{u_{ij}|{\tilde\varphi_j}\rangle}$ for some unitary $u_{ij}$. Then $\sum_i{|{\tilde\psi_i}\rangle\langle\tilde\psi_i|} = \sum_{ijk}{u_{ij}u_{ik}^*|\tilde\varphi_j\rangle\langle\tilde\varphi_j|}$ (2.168)

I don't get this step. If I take $\langle\tilde\psi_i|=(|\tilde\psi_i\rangle)^\dagger=\sum_j(u_{ij}|\tilde\varphi_j\rangle)^\dagger=\sum_j{\langle\tilde\varphi_j|u_{ij}^\dagger}$ and substitute this in the outer product I receive $\sum_i{|{\tilde\psi_i}\rangle\langle\tilde\psi_i|} = \sum_{ijk}{u_{ij}|\tilde\varphi_j\rangle\langle\tilde\varphi_j|u_{ik}^\dagger}$

Can someone explain this to me please?

**Equation 2.169 -> 2.170**
$$\sum_{jk}{(\sum_i{u_{ki}^\dagger u_{ij})}|\tilde\varphi_j\rangle\langle\tilde\varphi_k|} = \sum_{jk}{\delta_{kj}|\tilde\varphi_j\rangle\langle\tilde\varphi_k|}$$
I can't understand why $(\sum_i{u_{ki}^\dagger u_{ij}}) = \delta_{kj}$.

I understand that $u_{ki}^\dagger u_{ij} = I$ for $k=j$, but why is it zero otherwise?

It would be so kind if someone could enlighten me.

1If $\vert \tilde{\varphi}_k\rangle$ are basis states, the result follows directly as those are orthogonal. – nippon – 2020-08-26T14:18:12.347

@nippon: you are refering to the 2nd question (2.169->2.170), aren't you? And you say $|\varphi_j\rangle \langle\varphi_k|$ is zero for j != k, because they are orthogonal, I agree. And $u_{ki}^\dagger u_{ij} = I$ for k=j, but for that to be true it must be $u_{ik}^\dagger u_{ij} = I$ (switch index ki -> ik) , right? – mbuchberger1967 – 2020-08-26T14:25:20.090