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 1


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 $k,\ell$.)

  • $\begingroup$ 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 $X$?) $\endgroup$
    – glS
    Commented Jul 19, 2020 at 16:51
  • $\begingroup$ 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. $\endgroup$
    – glS
    Commented Jul 19, 2020 at 17:07
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jul 19, 2020 at 17:25

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