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?


2 Answers 2


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.


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