2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 :) $\endgroup$ Mar 10 at 14:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.