The mathematics of “can” and “cannot”
So what is channel-state 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 physically implemented, i.e. if it is a 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}\mathbf{1}
where 0\leqslant p\leqslant 1 is some fixed parameter.
This map acts on a density operator \rho via
\rho
\longmapsto 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?
We can interpret the convex-sum expression
\mathcal{E}(|i\rangle\langle j|)
= p|j\rangle\langle i|+(1-p)\delta_{ij} \frac{1}{2}|i\rangle\langle j|
as follows: take the input state \rho and either (i) apply the transpose, with probability p, or (ii) replace it with the maximally mixed state, with probability 1-p.
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 this 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 \mathcal{E} (|i\rangle\langle j|) for i,j\in\{0,1\} as follows:
\begin{aligned}
|0\rangle\langle 0|
&= \begin{bmatrix}1&0\\0&0\end{bmatrix}
\overset{\mathcal{E}}{\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}
\overset{\mathcal{E}}{\longmapsto}
\begin{bmatrix}0&0\\p&0\end{bmatrix},
\\|1\rangle\langle 0|
&= \begin{bmatrix}0&0\\1&0\end{bmatrix}
\overset{\mathcal{E}}{\longmapsto}
\begin{bmatrix}0&p\\0&0\end{bmatrix}
\\|1\rangle\langle 1|
&= \begin{bmatrix}0&0\\0&1\end{bmatrix}
\overset{\mathcal{E}}{\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}}
= \frac{1}{2}
\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 channel-state duality: \mathcal{E} is completely positive (and hence physically realisable) if and only if \widetilde{\mathcal{E}}\geqslant 0, and 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)).