# Nielsen & Chuang Theorem 2.6 Proof

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.

• If $\vert \tilde{\varphi}_k\rangle$ are basis states, the result follows directly as those are orthogonal. Aug 26 '20 at 14:18
• @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? Aug 26 '20 at 14:25

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 $$(U^\dagger U)_{j,k}=\sum_iu_{ji}^\star u_{ik}.$$
• 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. Aug 26 '20 at 14:35
• No, $u_{ij}$ are not different matrices. There is one unitary matrix with elements $u_{i,j}$. Aug 26 '20 at 14:42
• 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. Aug 29 '20 at 13:03