# Given a channel $\Phi(X)=\sum_k c_k(X)\sigma_k$, are there always $F_k\ge0$ such that $\Phi(X)=\sum_k \operatorname{tr}(F_k X)\sigma_k$?

Fix a finite number of states $$\sigma_k$$, and consider a channel of the form $$\Phi(X)=\sum_k c_{k}(X)\sigma_k.$$

For $$\Phi$$ to be linear and trace-preserving we must have: $$c_k(X+X') = c_k(X) + c_k(X'), \qquad \sum_k c_k(X)=1.$$ In other words, the coefficients must be linear functionals $$c_k\in\mathrm{Lin}(\mathcal X)^*$$ for all $$k$$.

Does this imply that there must be some positive operators $$F_k\ge0$$ such that $$c_k(X)=\operatorname{Tr}(F_k X)$$ for all $$k$$ (which in turn would imply $$\sum_k F_k=I$$ and thus that $$\{F_k\}_k$$ is a POVM)? What's a good way to show this?

## 1 Answer

The answer is no.

To this end, pick a linearly independent set $$\{\sigma_k\}$$ which spans the full matrix space (over $$\mathbb C)$$, that is, a basis. (This is always possible, as the positive operators span the hermitian ones over $$\mathbb R$$.)

Then pick a dual basis $$\sigma'_\ell$$ such that $$\mathrm{tr}[\sigma'_\ell \sigma_k]=\delta_{k\ell}\ .$$

Then, $$\Phi(X) = \sum_k \mathrm{tr}[\sigma'_k X]\,\sigma_k$$ is the identity channel, which cannot be written as a POVM $$F_k\ge0$$ followed by a preparation of $$\sigma_k$$ (as that channel would be entanglement breaking).

(Note that this shows that the dual basis $$\sigma'_\ell$$ has non-positive elements. This is not surprising, since otherwise the scalar product $$\mathrm{tr}[\sigma'_\ell\sigma_k]\ge0$$ for all $$k,\ell$$.)

• by "identity channel" you mean $\Phi(X)=X$? But if there are such states $\sigma_k\ge0$ s.t. $\Phi(X)=X=\sum_k c_k(X)\sigma_k$ for all $X$, then $c_k(X)=\operatorname{Tr}(\sigma_k X)$ and thus $\sum_k\sigma_k=I$ if $\Phi$ is trace-preserving (assuming these are an orthonormal basis... but if they are not, are we ensured that they can be used to decompose any $X$?) – glS Jul 19 '20 at 16:51
• ah, I think I got it. There is a (non-orthogonal) basis of states $\sigma_k$ such that we can write $X=c_k(X)\sigma_k$ for all $X$. However, it is not true that $c_k(X)=\operatorname{Tr}(\sigma_k X)$ because the basis is not made of orthogonal operators (and now I understand why you used the notion of dual basis here). So I guess the hypothesis is true only as long as restrict $\sigma_k$ to be orthogonal operators. – glS Jul 19 '20 at 17:07
• @glS The hypothesis is true if and only if the channel is entanglement breaking. I would have to look up myself whether an entanglement breaking channel can always be written with $\sigma_k$ an orthogonormal basis. On the spot, I don't see why. Just because this is not the case in my example does not mean it is required. – Norbert Schuch Jul 19 '20 at 17:25