1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ 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 $X=I$. $\endgroup$
    – user1551
    Commented Sep 11, 2022 at 19:18

2 Answers 2

2
$\begingroup$

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 $X = U I U^\dagger = U U^\dagger = I$.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Sooraj S
    Commented Sep 11, 2022 at 19:53
  • $\begingroup$ 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. $\endgroup$
    – Sooraj S
    Commented Sep 11, 2022 at 19:56
  • $\begingroup$ Therefore, $X=\sum_kE_k^\dagger E_k$ is positive semidefinite. Then I think the required statement can be proven using your proof, right ? $\endgroup$
    – Sooraj S
    Commented Sep 11, 2022 at 20:36
  • $\begingroup$ Yes, I took these results for given. $\endgroup$
    – Rammus
    Commented Sep 11, 2022 at 21:01
0
$\begingroup$

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 $A=A^\dagger=I$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .