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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.

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    $\begingroup$ If $\vert \tilde{\varphi}_k\rangle$ are basis states, the result follows directly as those are orthogonal. $\endgroup$
    – nippon
    Commented Aug 26, 2020 at 14:18
  • $\begingroup$ @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? $\endgroup$
    – mbuchberger1967
    Commented Aug 26, 2020 at 14:25

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Let $U$ be a unitary matrix. By definition $U^\dagger U=I$. So, if I take an off-diagonal element, this corresponds to ($j\neq k$) $$ (U^\dagger U)_{j,k}=I_{j,k}=0, $$ and of course $$ (U^\dagger U)_{j,k}=\sum_iu_{ji}^\star u_{ik}. $$

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  • $\begingroup$ Thanks and I understand your explanation as you look on one unitary U, and the indices refer to elements of the matrix U. But in my case $u_{ij}$ are i*j different unitary matrices. How does your explanation take that into account? Sorry if this is a stupid question, I'm a software dev, not a mathematican. $\endgroup$
    – mbuchberger1967
    Commented Aug 26, 2020 at 14:35
  • $\begingroup$ yes, that's right. $\endgroup$
    – DaftWullie
    Commented Aug 26, 2020 at 14:36
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    $\begingroup$ No, $u_{ij}$ are not different matrices. There is one unitary matrix with elements $u_{i,j}$. $\endgroup$
    – DaftWullie
    Commented Aug 26, 2020 at 14:42
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    $\begingroup$ It's not written as clearly as it might be. It might be clearer to say "for some set of coefficients $u_{ij}$ comprising a unitary matrix". This is stated more clearly right at the end of the proof of Theorem 2.6 where it talks about finding a unitary matrix $w$ and then refers to the components in the following formula. $\endgroup$
    – DaftWullie
    Commented Aug 29, 2020 at 13:03
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    $\begingroup$ Yes, they mean the Hermitian conjugate. Indeed, that is what is written in my version of Nielsen & Chuang. Again, it's a bit of an abuse of notation. What they're really meaning, of course is not to take the Hermitian conjugate of a number, but to take the hermitian conjugate of the underlying matric and then just extract the relevant coefficient. $\endgroup$
    – DaftWullie
    Commented Sep 16, 2020 at 6:36

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