## 7.9 The mathematics of “can” and “cannot”

So what is state-channel duality good for? To start with, it can be used to asses whether or not a given map \mathcal{B}(\mathcal{H})\to\mathcal{B}(\mathcal{H}') can actually be implemented, i.e. if it is a physically realisable CPTP map. Indeed, all we have to do is to check if the corresponding Choi matrix is a density matrix. Let’s look at a simple example.

Consider the map \mathcal{E}\colon |i\rangle\langle j| \longmapsto p|j\rangle\langle i|+(1-p)\delta_{ij} \frac{1}{2}|i\rangle\langle j| where 0\leqslant p\leqslant 1 is some fixed parameter. This map acts on a density operator \rho via \rho \longmapsto \rho' = p\rho^T +(1-p) \frac{1}{2}\mathbf{1} where \rho^T is the transpose of \rho.

But is this map a quantum channel? That is, does it represent a physical process that can be implemented in a lab?

The expression for \mathcal{E}, i.e. the convex sum, can be interpreted as follows: take the input state \rho and either (i) apply the transpose, or (ii) replace it with the maximally mixed state (with probabilities p and 1-p, respectively). This is fine, except that the transpose operation is not completely positive, and, as such, is not physically admissible — it cannot be implemented. But does it mean that the map \mathcal{E} itself cannot be implemented? Not necessarily!

In fact, the answer depends on the value of p. The case p=0 corresponds to just replacing the input with the maximally mixed state, which is something that can be easily implemented. However, as p increases from 0 to 1, at some critical point the map switches from completely positive to merely positive. In order to find this critical value of p, we first calculate the \mathcal{E} (|i\rangle\langle j|), as follows: \begin{aligned} |0\rangle\langle 0| &= \begin{bmatrix}1&0\\0&0\end{bmatrix} \longmapsto \begin{bmatrix}\frac{1+p}{2}&0\\0&\frac{1-p}{2}\end{bmatrix} \\|0\rangle\langle 1| &= \begin{bmatrix}0&1\\0&0\end{bmatrix} \longmapsto \begin{bmatrix}0&0\\p&0\end{bmatrix}, \\|1\rangle\langle 0| &= \begin{bmatrix}0&0\\1&0\end{bmatrix} \longmapsto \begin{bmatrix}0&p\\0&0\end{bmatrix} \\|1\rangle\langle 1| &= \begin{bmatrix}0&0\\0&1\end{bmatrix} \longmapsto \begin{bmatrix}\frac{1-p}{2}&0\\0&\frac{1+p}{2}\end{bmatrix}, \end{aligned} We can then write down the Choi matrix: \widetilde{\mathcal{E}} = \frac12 \begin{bmatrix} \mathcal{E}(|0\rangle\langle 0|) & \mathcal{E}(|0\rangle\langle 1|) \\\mathcal{E}(|1\rangle\langle 0|) & \mathcal{E}(|1\rangle\langle 1|) \end{bmatrix} = \frac{1}{2} \left[ \begin{array}{cc|cc} \frac{1+p}{2} & 0 & 0 & 0 \\0 & \frac{1-p}{2} & p & 0 \\\hline 0 & p & \frac{1-p}{2} & 0 \\0 & 0 & 0 & \frac{1+p}{2} \end{array} \right] which lets us apply state-channel duality: \mathcal{E} is completely positive (and hence physically realisable) if and only if \widetilde{\mathcal{E}}\geqslant 0; the latter is true only when p\leqslant\frac13 (note that the eigenvalues of \widetilde{\mathcal{E}} are \frac{1}{4}(1+p) and \frac{1}{4}(1-3p)).