Is the convex hull of product channels strictly contained in the set of all channels

Question

For two finite dimensional Hilbert spaces $\mathcal{H},\mathcal{H'}$ let $T(\mathcal{H},\mathcal{H'})$ be the set of all CPTP maps (quantum channels) $L(\mathcal{H})\to L(\mathcal{H}')$.

Let $\mathcal{H},\mathcal{H_1},\mathcal{H_2}$ be finite dimensional Hilbert spaces, I am wondering whether the convex hull of $T(\mathcal{H_1},\mathcal{H})\otimes T(\mathcal{H_2},\mathcal{H})$ is strictly contained in $T(\mathcal{H_1}\otimes\mathcal{H_2},\mathcal{H}\otimes\mathcal{H})$ (it seems that it does but a proof eludes me).

If so, is there a way to express this set, or an as small as possible set containing this set, as the solution set of a linear matrix inequality?

Answer 1

Yes, this set is strictly contained in the other.

Note that by the main body of your question you are asking whether or not the convex hull of product channels (not separable) is strictly contained in the set of all channels. Separable channels are defined as those for which their Kraus operators have a tensor product form.

To see a counterexample we can look at Example 6.21 from [1] you can find a free version on the author's website. Define a channel $\Phi$ by the action, $$ \Phi(|i\rangle \langle j| \otimes |k \rangle \langle l|) = \begin{cases} |i \rangle \langle i | \otimes |i \rangle \langle i | \qquad &\text{if } i=j \text{ and } k=l \\ 0 &\text{otherwise} \end{cases} $$

Now let us suppose we could write $\Phi = \sum_a p(a) \Phi_a \otimes \Psi_a$ (i.e. a convex combination of product channels). Well then we have for any normalized state $\rho$ $$ \begin{aligned} \Phi(|0\rangle \langle 0 | \otimes \rho) &= |0 \rangle \langle 0|\otimes |0\rangle \langle 0 | \\ &= \sum_a p(a) \Phi_a(|0\rangle\langle 0|) \otimes \Psi_a(\rho) \end{aligned} $$ and then taking the partial trace over the first subsystem we may conclude that $$ \sum_a p(a) \Psi_a(\rho) = |0\rangle \langle 0|. $$ However, if we repeat these steps with the input state $|1 \rangle \langle 1| \otimes \rho$ we will instead conclude that $\sum_a p(a) \Psi_a(\rho) = |1\rangle \langle 1 |$. Thus we run into a contradiction and our assumption that our channel can be written as a convex combination of product channels is false.

[1]: Watrous, John, The theory of quantum information, Cambridge: Cambridge University Press (ISBN 978-1-107-18056-7/hbk; 978-1-316-84814-2/ebook). viii, 590 p. (2018). ZBL1393.81004.