What are examples of qubit channel with non-singular Choi state?

Question

Are there any obvious (ideally physically reasonable) examples of qubit channels that give rise to non-singular Choi states?

I've been exploring the Choi state of a variety of qubit channels but find that all of the obvious targets (specifically, the amplitude damping, depolarising and dephasing channels - if proofs are required, please ask in the comments and I will edit) have at least one vanishing eigenvalue. For a wider problem I'm working on at present, it would be convenient to have a non-singular Choi state.

In an attempt to be fully general for the moment, suppose $\mathcal{N}$ is a qubit quantum channel. It's Choi state is given by the action of it's trivial extension on the unnormalised maximally mixed state $|\Phi\rangle = |00\rangle + |11\rangle$:

$$\mathcal{J}(\mathcal{N}) = \sum_{i,j=1}^2\mathcal{N}(|i\rangle \langle j|)\otimes |i\rangle \langle j|.$$

Treat this is a $2\times2$ block matrix, itself made up of $2\times 2$ blocks:

$$\mathcal{J}(\mathcal{N}) = \begin{pmatrix} \mathcal{N}(|0\rangle \langle 0|)& \mathcal{N}(|0\rangle \langle 1|)\\ \mathcal{N}(|1\rangle \langle 0|)&\mathcal{N}(|1\rangle \langle 1|) \end{pmatrix}.$$

If either $\mathcal{N}(|0\rangle \langle 0|)$ or $\mathcal{N}(|1\rangle \langle 1|)$ is non-singular, we can take the determinant of this as either:

$$\text{det}(\mathcal{J}(\mathcal{N})) = \text{det} \biggl(\mathcal{N}(|0\rangle \langle 0|)\biggr)\text{det}\biggl(\mathcal{N}(|1\rangle \langle 1|)-\mathcal{N}(|1\rangle \langle 0|)\cdot [\mathcal{N}(|0\rangle \langle 0|)]^{-1}\cdot\mathcal{N}(|0\rangle \langle 1|)\biggr),$$

or

$$\text{det}(\mathcal{J}(\mathcal{N})) = \text{det}\biggl(\mathcal{N}(|0\rangle \langle 0|)-\mathcal{N}(|0\rangle \langle 1|)\cdot [\mathcal{N}(|1\rangle \langle 1|)]^{-1}\cdot\mathcal{N}(|1\rangle \langle 0|) \biggr)\text{det} \biggl(\mathcal{N}(|1\rangle \langle 1|)\biggr),$$

respectively. If both are singular, one can use the identity shown in this answer to exploit the potential invertibility of the other blocks. If all blocks are singular then I'm clearly looking at a singular Choi state anyway. Let us suppose for simplicity that our channel is such that $\mathcal{N}(|0\rangle \langle 0|)$ is invertible, then the Choi state is non-singular if and only if

$$\text{det}\biggl(\mathcal{N}(|1\rangle \langle 1|)-\mathcal{N}(|1\rangle \langle 0|)\cdot [\mathcal{N}(|0\rangle \langle 0|)]^{-1}\cdot\mathcal{N}(|0\rangle \langle 1|)\biggr)\not=0.$$

Since we're dealing with $2\times 2$ matrices, this can be written as the sum of the determinants of the two terms and the trace of a product of the terms (see this answer). If we assume further for the moment that a diagonal-sharing pair of the four blocks are non-singular (say, $\mathcal{N}(|0\rangle \langle 0|), \mathcal{N}(|1\rangle \langle 1|))$, then this condition reduces to:

$$ \text{Tr}\biggl(\mathcal{M}_{\mathcal{N}}\biggr) - \text{det}\biggl(\mathcal{M}_{\mathcal{N}}\biggr) \not= 1$$

where, $$\mathcal{M}_{\mathcal{N}} := [\mathcal{N}(|1\rangle \langle 1|)]^{-1}\cdot\mathcal{N}(|1\rangle \langle 0|)\cdot [\mathcal{N}(|0\rangle \langle 0|)]^{-1}\cdot\mathcal{N}(|0\rangle \langle 1|)$$

This conditions looks like it would be easy to satisfy, however this is proving not to be the case for the obvious channels. I'd like to find examples of qubit channels (or more general channels if that's easier) with non-singular Choi states so thanks for any input you may give!

Answer 1

Consider the completely depolarizing channel that enacts $$\mathcal{N}(\rho)=\sum_{i,j}\frac{1}{\sqrt{2}}|i\rangle\langle j|\rho|j\rangle\langle i|\frac{1}{\sqrt{2}}=\mathrm{Tr}(\rho)\frac{\mathbb{I}}{2}.$$ It is very important to include the factor $\mathrm{Tr}(\rho)$, because this formula holds even for "states" $\rho$ that do not have trace 1. Specifically, the depolarizing channel sends $$\mathcal{N}(|i\rangle\langle i|)=\frac{\mathbb{I}}{2}$$ but $$\mathcal{N}(|0\rangle\langle 1|)=\mathcal{N}(|1\rangle\langle 0|)=0\frac{\mathbb{I}}{2}=\begin{pmatrix}0&0\\0&0\end{pmatrix}.$$ Then your Choi matrix is indeed invertible, simply becoming $$\mathcal{J}=\rm{diag}(1,1,1,1)/2,$$ which is invertible. Or, look at your calculations and see that $\mathcal{M}_\mathcal{N}$ becomes the zero matrix and so its trace minus its determinant is 0, which indeed is $\neq 1$.

If one had erroniously applied the depolarizing channel to say "$\mathcal{N}(|1\rangle\langle 0|)=\frac{\mathbb{I}}{2}$" then the Choi matrix would have been singular; this is my guess for where the error is and is certainly an easy mistake to make (I have made it myself) but it could lead to nonphysical scenarios such as the trace increasing via depolarization.