# When is the Choi matrix of a channel pure?

For a quantum channel $$\mathcal{E}$$, the Choi state is defined by the action of the channel on one half of an unnormalized maximally entangled state as below:

$$J(\mathcal{E}) = (\mathcal{E}\otimes I)\sum_{ij}\vert i\rangle\langle j\vert\otimes \vert i\rangle\langle j\vert$$

For isometric channels, the Choi state is also a pure state. What about the converse statement? Does the Choi state being pure give us any information about the properties of the channel?

It works the other way around too. A pure state is rank $$1$$, and any channel with more than one Kraus operator will give a higher-rank Choi matrix, which can be easily seen from the definition.

You can also work it out on a different condition for pure states:

For any pure state $$\rho$$ we have $$\rho^{2} = \rho.$$

$$(J(\mathcal{E}))^{2} = \sum_{ij}\sum_{kl} \mathcal{E}(|i\rangle \langle j|)\mathcal{E}(|k\rangle \langle l|) \otimes |i\rangle \langle j|k\rangle\langle l| = \sum_{ijl}\mathcal{E}(|i\rangle \langle j|)\mathcal{E}(|j\rangle \langle l|) \otimes |i\rangle \langle l|$$

so if $$J(\mathcal{E})$$ is pure then:

$$\sum_{ij}\mathcal{E}(|i\rangle \langle j|) \otimes |i\rangle \langle j| = \sum_{ijl}\mathcal{E}(|i\rangle \langle j|)\mathcal{E}(|j\rangle \langle l|) \otimes |i\rangle \langle l|$$ which, when relabeling $$j <-> l$$ on the right hand side, leads to: $$\sum_{ij}\mathcal{E}(|i\rangle \langle j|) \otimes |i\rangle \langle j| = \sum_{l}\sum_{ij}\mathcal{E}(|i\rangle \langle l|)\mathcal{E}(|l\rangle \langle j|) \otimes |i\rangle \langle j|$$ Since all different $$|i\rangle\langle j |$$ are orthogonal, this needs to hold term-by-term:

$$\mathcal{E}(|i\rangle \langle j|) = \sum_{l}\mathcal{E}(|i\rangle \langle l|)\mathcal{E}(|l\rangle \langle j|).$$

Writing $$\mathcal{E}$$ in it's Kraus decomposition $$\{A_{k}\}$$ sheds some extra light:

$$\sum_{k} A_{k}|i\rangle \langle j | A_{k}^{\dagger} = \sum_{l} \sum_{k'}\sum_{k''} A_{k'}|i\rangle \langle l | A_{k'}^{\dagger} A_{k''}|l\rangle \langle j | A_{k''}^{\dagger}$$

noting that $$\sum_{l} \langle l| A^{\dagger}_{k'} A_{k''}|l\rangle = \mathrm{tr}[A^{\dagger}_{k'} A_{k''}]$$, we get:

$$\sum_{k} A_{k}|i\rangle \langle j | A_{k}^{\dagger} = \sum_{k'}\sum_{k''}\mathrm{tr}[A^{\dagger}_{k'} A_{k''}] A_{k'}|i\rangle \langle j | A_{k''}^{\dagger}$$

and taking the trace and using its cyclic property on either side we get: $$\sum_{k'k''}\delta_{k'k''} \langle j | A_{k''}^{\dagger}A_{k'}|i\rangle = \sum_{k'}\sum_{k''}\mathrm{tr}[A^{\dagger}_{k'} A_{k''}] \langle j | A_{k''}^{\dagger}A_{k'}|i\rangle$$

Importantly, this works for every $$|i\rangle, | j \rangle$$, so the above equation can only hold if $$\delta_{k'k''} = \mathrm{tr}[A^{\dagger}_{k'} A_{k''}]$$, which is evidently only true if the Kraus operators are orthogonal and of unit length. But then they are unitary, which means there is only a single Kraus operator, necessarily unitary.

Let $$\Phi$$ have Kraus representation $$\Phi(X)=\sum_a A_a X A_a^\dagger$$. Its Choi then has the form $$J(\Phi) = \sum_a \mathrm{vec}(A_a)\mathrm{vec}(A_a)^\dagger$$. This works both ways: if the Choi has this form, then the $$A_a$$ are Kraus operators for $$\Phi$$ (note that the normalization condition on the Kraus operators translates into the condition $$\mathrm{Tr}_1(J(\Phi))=I$$ for the Choi).

Therefore, if the Choi has rank one, i.e. $$J(\Phi)=\mathrm{vec}(A)\mathrm{vec}(A)^\dagger$$ for some linear operator $$A$$, then $$\Phi(X)=AXA^\dagger$$. This implies that $$\Phi$$ is an isometric channel.