Why does a quantum operation being trace-preserving imply that $\sum_k E_k^\dagger E_k=I$?

Question

I am reading Nielsen Chuang Chapter 8. They say that if a quantum operation is trace-preserving, then \begin{equation} Tr\left(\sum_k E_k^{\dagger}E_k \rho\right) = 1 \end{equation} which I understand. They however then claim that as this is true for all $\rho$, then \begin{equation} \sum_k E_k^{\dagger}E_k = I. \end{equation} This seems reasonable, but I cannot clearly write down the proof. Further, they say if it is non-trace preserving then $\sum_k E_k^{\dagger}E_k \leq I$, which is something I do not understand. How are we saying one matrix is smaller than the other. Do they mean that $\sum_k E_k^{\dagger}E_k = cI$, where $c \leq 1$ ? However then can someone please sketch the proof?

Answer 1

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

$$ \left\langle \sum_k E_k^\dagger E_k,\rho\right\rangle_{HS} = \left\langle I,\rho\right\rangle_{HS}\\ \left\langle I-\sum_k E_k^\dagger E_k,\rho\right\rangle_{HS} = 0 $$

for all $\rho$. Now, for any inner product $\langle .,.\rangle$, the only vector orthogonal to all other vectors is the zero vector. Therefore,

$$ I-\sum_k E_k^\dagger E_k = 0 $$

and so

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

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The notation $A\le B$ means that $B-A$ is positive semi-definite.

Answer 2

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 already know that $\text{Tr}(\sum_kE_k^\dagger E_k\rho_d)=1$. This must mean that $\text{Tr}(\sum_kE_k^\dagger E_k\rho_o)=0$. This must also be true for any linear combinations of $\rho_o$. So, again, consider two specific instances of $\rho$: $$ \rho=(|i\rangle+|j\rangle)(\langle i|+\langle j|), \qquad \rho'==(|i\rangle+i|j\rangle)(\langle i|-i\langle j|). $$ By taking an appropriate linear combination, we can get the off-diagonal term $|i\rangle\langle j|$. By this token, we prove that every off-diagonal element of $\sum_kE_k^\dagger E_k$ is 0. Thus, $\sum_kE_k^\dagger E_k=I$.

As for the meaning of $$ \sum_kE_k^\dagger E_k\leq I, $$ you should interpret this as the eigenvalues of $\sum_kE_k^\dagger E_k$ all being less than or equal to 1.

Answer 3

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 $A=\begin{pmatrix}1 & x-iy\\x+iy & 1\end{pmatrix}$. Furthermore, we can choose $\rho=\begin{pmatrix}1 & 1 \\ 1&0\end{pmatrix}$, then to satisfy $Tr(A\rho)=I$, we have $x=0$. And by choosing $\rho=\begin{pmatrix}1 & -i\\i & 0\end{pmatrix}$, we can make sure $y = 0$.

The higher-dimensional condition can be similarly generalized.