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1

Let's start by considering specific density matrices $\rho=|i\rangle\langle i|$. This immediately tells you that $$ \langle i|\sum_kE_k^\dagger E_k|i\rangle=1, $$ and hence all diagonal elements of $\sum_kE_k^\dagger E_k$ are 1. Next, consider a more general $\rho$, which we divide into diagonal and off-diagonal components, $$ \rho=\rho_d+\rho_o. $$ We ...


2

Suppose that $$ \mathrm{tr}\left(\sum_k E_k\rho E_k^\dagger\right) = \mathrm{tr}(\rho) $$ for all $\rho$. Then $$ \mathrm{tr}\left(\sum_k E_k^\dagger E_k\rho\right) = \mathrm{tr}(I\rho) $$ for all $\rho$. The last equation can be rewritten in terms of Hilbert-Schmidt inner product as $$ \left\langle \sum_k E_k^\dagger E_k,\rho\right\rangle_{HS} = \left\...


2

For every matrix $A=\begin{pmatrix}a & x-iy\\x+iy & b\end{pmatrix}$(hermitian here) with real number $a,b,x,y$. And $A$ satisfy $Tr(A\rho)$ and $Tr(\rho)=1$, let's consider two by two matrix for example, we can choose $\rho=|0\rangle\langle 0|$ to make sure $Tr(A\rho)=I$. Then $A_{11}$ must be 1. For the same reason we can get $A_{22}=1$. Now $A=\...


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