# Prove that $tr\Big(\sum_k E_k^\dagger E_k\rho\Big)=1$ for all $\rho$ implies $\sum_k E_k^\dagger E_k=I$

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 $$\sum_k E_k^\dagger E_k=I$$ ?

• If $\operatorname{tr}(X\rho)=1$ for all PSD matrices $\rho$ with trace $1$, then $\operatorname{tr}((X-I)\rho)=0$ for all PSD matrices $\rho$ (regardless of $\operatorname{tr}(\rho)$). In turn, $\operatorname{tr}((X-I)\rho)=0$ for all Hermitian matrices $\rho$, because every Hermitian matrix is the difference of two PSD matrices. Hence $\|X-I\|_F^2=\operatorname{tr}((X-I)^2)=0$ and $X=I$. Commented Sep 11, 2022 at 19:18

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 $$X = U I U^\dagger = U U^\dagger = I$$.
• A matrix $A$ is positive semidefinite iff $v^\dagger Av\geq 0$ for all $v\neq 0$. So, for $A=E_k^\dagger E_k$ we have $v^\dagger Av=v^\dagger E_k^\dagger E_kv=||E_kv||\geq 0\implies A=E_k^\dagger E_k$ is positive semidefinite. Commented Sep 11, 2022 at 19:53
• If $A$ and $B$ are positive semidefinite so is $A+B$.Proof : $v^\dagger Av\geq 0$ and $v^\dagger Bv\geq 0\implies v^\dagger (A+B)v\geq 0\implies A+B$ is positive semidefinite. Commented Sep 11, 2022 at 19:56
• Therefore, $X=\sum_kE_k^\dagger E_k$ is positive semidefinite. Then I think the required statement can be proven using your proof, right ? Commented Sep 11, 2022 at 20:36
Suppose $$\operatorname{Tr}(A^\dagger\rho)=\operatorname{Tr}(\rho)$$ for all matrices $$\rho$$. Checking this condition for $$\rho=|i\rangle\!\langle i|$$ tells you that $$A_{ii}=1$$. Checking it for $$\rho=|i\rangle\!\langle j|$$ you see that $$A_{ij}=0$$. Hence $$A=A^\dagger=I$$.