In my reference, Page 360, Operator-sum representation, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, it is given that
\begin{align}
1&=tr\Big(\mathcal{E}(\rho)\Big)\\
&=tr\Big(\sum_k E_k\rho E_k^\dagger\Big)\\
&=tr\Big(\sum_k E_k^\dagger E_k\rho\Big)
\end{align}
since this is true for all $\rho$ we must have $\sum_k E_k^\dagger E_k=I$
where $\rho$ is positive semidefinite such that $tr(\rho)=1$ and $E_k=(I\otimes\langle e_k|)U(I\otimes|e_o\rangle)$ where $U$ is unitary and $\{|e_k\rangle\}$ is an orthonormal basis.
The original problem statement is that,
Let $|e_k\rangle$ be an orthonormal basis for the state space of the environment and $\rho_{env}=|e_0\rangle\langle e_0|$ be the initial state of the environment. Then
\begin{align}
\mathcal{E}(\rho)=tr_{env}\Big[U(\rho\otimes\rho_{env})U^\dagger\Big]=\sum_k(I\otimes\langle e_k|)(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger)(I\otimes|e_k\rangle)
\end{align}
Now,
\begin{align}
\rho\otimes|e_0\rangle\langle e_0|&=(\rho\otimes I)(I\otimes|e_0\rangle)(I\otimes\langle e_0|)=((\rho I)\otimes(I|e_0\rangle))(I\otimes\langle e_0|)\\
&=((I\rho)\otimes(|e_0\rangle.1))(I\otimes\langle e_0|)=(I\otimes|e_0\rangle)(\rho\otimes 1)(I\otimes \langle e_0|)\\
&=(I\otimes|e_0\rangle)\rho(I\otimes \langle e_0|)
\end{align}
Substituting into the equation,
\begin{align}
\mathcal{E}(\rho)&=\sum_k I\otimes\langle e_k|(U(\rho\otimes|e_0\rangle\langle e_0|)U^\dagger)I\otimes|e_k\rangle\\
&=\sum_k \color{blue}{(I\otimes\langle e_k|)U(I\otimes|e_0\rangle)}\rho\color{blue}{(I\otimes \langle e_0|)U^\dagger(I\otimes|e_k\rangle)}\\
&=\sum_k E_k\rho E_k^\dagger
\end{align}
where $E_k=(I\otimes\langle e_k|)U(I\otimes|e_o\rangle)$.
It is required that $tr(\mathcal{E}(\rho))=1$ as $tr(\rho)=1$ since the eigenvalues constitute a probability distribution and must be added up to 1.
Since $tr(AB)=tr(BA)$ we have $tr(\mathcal{E}(\rho))=tr(\sum_k E_k\rho E_k^\dagger)=tr(\sum_k E_k^\dagger E_k\rho)=1$ which is true for all $\rho$.
How does this imply that $\sum_k E_k^\dagger E_k=I$ ?
Best Answer
You have $\mathrm{tr}[ X \rho] =1$ for all $\rho$ where $X$ and $\rho$ are both positive semidefinite matrices.
Now as $X$ is PSD we know by the spectral theorem that $$ X= U D U^\dagger = \sum_i \lambda_i |u_i\rangle \langle u_i| $$ where $\lambda_i$ are the nonnegative eigenvalues of $X$ and $|u_i\rangle$ are the corresponding orthonormal eigenvectors. Note that $X = I$ if we can show that each $\lambda_i =1$. But this follows from the condition above by taking $\rho = |u_i\rangle \langle u_i|$ we have $$ \mathrm{tr}[ X |u_i\rangle \langle u_i|]=1 \implies \lambda_i = 1 $$ Repeating for each eigenvector we see that $X = U I U^\dagger = U U^\dagger = I$.