In Nielsen's book when proving "Unitary freedom in the ensemble for density matrices"(Theorem 2.6):

$$\text{Suppose }|\widetilde{\psi_i}\rangle = \sum\limits_{j}u_{ij} |\widetilde{\phi_j}\rangle$$ Then in Equation 2.168: $$ \sum_i |\widetilde{\psi_i}\rangle \langle\widetilde{\psi_i}| = \sum_{ijk} u_{ij} u_{ik}^{*}|\widetilde{\phi_j}\rangle \langle\widetilde{\phi_k}|$$

In equation 2.168 adjoint of the tilded psi has now the element in the unitary matrix u being ik conjugated($u_{ik}^*$). Now I understand that the column index after the adjoint will not be the same due to the transpose(hence k instead of j), what I don't understand is why the row index (i) is unchanged. I know it's probably something simple that I am missing, but I would appreciate your help.


The proof begins with let $|\psi_i\rangle = \sum_j u_{ij} |\varphi_j\rangle$ where $U = (u_{ij})_{ij}$ is some unitary matrix. But now, $$ \begin{aligned} |\psi_i\rangle \langle \psi_i| &= \left(\sum_j u_{ij} |\varphi_j\rangle \right)\left(\sum_k u_{ik} |\varphi_k\rangle\right)^{\dagger} \\ &= \left(\sum_j u_{ij} |\varphi_j\rangle \right)\sum_k u_{ik}^* \langle\varphi_k| \\ &= \sum_{jk} u_{ij} u_{ik}^* |\varphi_j\rangle \langle\varphi_k|. \end{aligned} $$

On the first line $\dagger$ denotes the hermitian conjugate (adjoint operator); on the second line we used that $\dagger$ is conjugate-linear (here $u_{ik}^*$ is the complex conjugate of the complex number $u_{ik}$) and on the last line we just rearranged the sums and moved the complex numbers to the front.

  • $\begingroup$ Thank you! I didn't notice that I needed to add a new index 'k' for multiplying with the other summation. But I guess it's only logical :) $\endgroup$ – Omar Hossam Ahmed Mar 10 at 14:28

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