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

I am reading Nielsen Chuang Chapter 8. They say that if a quantum operation is trace-preserving, then $$$$Tr\left(\sum_k E_k^{\dagger}E_k \rho\right) = 1$$$$ which I understand. They however then claim that as this is true for all $$\rho$$, then $$$$\sum_k E_k^{\dagger}E_k = I.$$$$ This seems reasonable, but I cannot clearly write down the proof. Further, they say if it is non-trace preserving then $$\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?

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

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

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

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.

• Thank you for the answer. Can you also please comment on what the authors mean by $\sum_k E_k^{\dagger}E_k \leq I$ ? Commented Oct 20, 2021 at 6:44
• This linke might be helpful. Maybe someone else can describe this more reasonably. Commented Oct 20, 2021 at 6:50
• @alpha $\sum_k E_k^\dagger E_k \le I$ means $I - \sum_k E_k^\dagger E_k$ is positive semi-definite mathmatically. Commented Oct 20, 2021 at 7:01
• @alpha let me add that $A\leq I$ for a psd matrix $A$ in particular means that all eigenvalues of $A$ are less than 1. Commented Oct 20, 2021 at 7:08

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.