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The question

Consider the following linear, trace preserving quantum map over $d \times d$ quantum states $\sigma$,

$$\Phi(\sigma) = \Pi_R \sigma \Pi_R + (1 - \text{Tr}(\Pi_R \sigma)) \frac{\Pi_R}{R}$$

for $\Pi_R$ a rank $R$ projector. This is a simplified version of the quantum channel used in Schumacher's compression protocol (see Eq. 18.9 on page 525 of Wilde's "From Classical to Quantum Shannon Theory" v8). Thus, this map must actually be a quantum channel which is completely positive and trace preserving (CPTP). For my pedagogy, I wanted to actually show complete positivity explicitly.

As Norbert Schuch points out, it is clearly also CP by simple physical interpretation. I've now verified this (below). But I would also like to prove it from the Choi matrix construction. In attempting to do so on a qudit example, I ended up showing (incorrectly) that the Choi matrix has negative eigenvalues. What did I do wrong?

Showing map is CPTP by composition of CPTP maps

Since the composition of CPTP maps is also CPTP, we can prove $\Phi$ is CPTP by constructing it from known CPTP maps. In short, we can build $\Phi$ up from (a) a quantum instrument (see Section 4.6.8 in Wilde), (b) a measurement, and (c) the replacement channel.

For (a), we consider two completely positive trace-non increasing maps $\Lambda_1(\rho) = \Pi_R \rho \Pi_R$ and $\Lambda_2(\rho) = (I - \Pi_R) \rho (I - \Pi_R)$ (just two POVM elements). Then the quantum instrument is given by $\Lambda(\rho) = \Lambda_1(\rho) \otimes |0\rangle \langle 0 | + \Lambda_2(\rho) \otimes |1 \rangle \langle 1|$. For (b), we simply measure the ancilla system which returns $0$ with probability $p_0 = \text{Tr}[\Pi_R \rho]$ and $1$ with probability $p_1 = \text{Tr}[(I - \Pi_R) \rho]$. If the ancilla reads $0$, we do nothing to the system, and if it reads $1$, we apply the replacement channel to $\Pi_R / R$. This leads to a final state $(p_0 \Pi_R \rho \Pi_R) / p_0 + p_1 (\Pi_R / R)$ which is the desired channel $\Phi$.

Incorrect Choi matrix calculation for a specific qudit example

The corresponding Choi matrix is given by

$$\chi = \sum_{i=0,j=0}^{d-1} |i\rangle \langle j| \otimes \Phi(|i\rangle \langle j|)$$

which we can calculate readily since

$$ \Phi(|i \rangle \langle j |) = \begin{cases} |i \rangle \langle j| + (1 - \delta_{i,j}) (\Pi_R / R) & \Pi_R |i \rangle = |i \rangle \ \& \ \Pi_R |j \rangle = |j \rangle \\ \Pi_R / R & \text{otherwise}. \end{cases} $$

As a special case, consider $d = 3, R = 2, \Pi_2 = |0\rangle\langle0| + |1 \rangle \langle 1|$. In this case, the Choi (block) matrix is of the form

$$ \chi_{(d=3,R=2)} = \begin{pmatrix} |0 \rangle \langle 0 | & |0 \rangle \langle 1 | + \Pi_2 / 2 & \Pi_2 / 2 \\ %----- |1 \rangle \langle 0 | + \Pi_2 / 2 & |1 \rangle \langle 1 | & \Pi_2 / 2 \\ \Pi_2 / 2 & \Pi_2 / 2 & \Pi_2 / 2 \end{pmatrix} $$ with spectrum $\{2.32288, 1.20711, -0.5, 0.5, -0.322876, -0.207107, 0., 0., 0.\}$ which contains several negative values and is hence not positive semi-definite. This suggests $\Phi$ is not CP in general which contradicts the earlier calculation. What is wrong with my Choi matrix calculation?

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2 Answers 2

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Your map is not CPTP. In fact, it is not even linear, so it is unclear what the Choi state even means.

The correct map would be $$\Phi(\sigma) = \Pi_R \sigma \Pi_R + (\mathrm{tr}(\sigma) - \mathrm{tr}(\Pi_R \sigma)) \frac{\Pi_R}{R}$$ or, by thinking of a two-outcome POVM $\{\Pi_R,I-\Pi_R\}$, $$\Phi(\sigma) = \Pi_R \sigma \Pi_R + \mathrm{tr}((I-\Pi_R) \sigma) \frac{\Pi_R}{R}\ .$$ (This also matches what Wilde writes.)


Note that this map will no longer have the $\Pi_2/2$ on the off-diagonal in the Choi state (which in its current form indeed does not match the intuition for what the channel is supposed to do: Why would it output $\Pi_R/2$ on an off-diagonal input element?).

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    $\begingroup$ In your defense, implicitly assuming a properly normalized input is a popular mistake. $\endgroup$ Dec 6, 2023 at 14:07
  • $\begingroup$ Thanks! This implicit assumption was driving me crazy, and now I know to look out for it in my future debugging. $\endgroup$
    – Nic
    Dec 6, 2023 at 18:34
  • $\begingroup$ @Nic You'd think so, but I was also overlooking it at first, even though I already fell for this mistake a few times. It was really that the $\Pi/2$ on the off-diagonal in the Choi state felt fishy which made me realize. $\endgroup$ Dec 6, 2023 at 19:02
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When you show that the map is CPTP, you write $p_1 = tr[ (1-\Pi) \rho ]$, but in the definition of your map $\Phi$, you write $1 - tr[\Pi \rho]$. The former is correct. Both expressions are equal for states $\rho$, but $\Phi$ must be a linear map that can be applied to any operator. With your definition, $\Phi$ is not linear.

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  • $\begingroup$ Argh, I didn't see the edit in the other answer before replying $\endgroup$
    – Noiralef
    Dec 6, 2023 at 14:36
  • $\begingroup$ I appreciate this answer because it helped me understand my mistake very clearly. I've accepted Norbert's answer simply because it was first and makes explicit reference to the Choi matrix which matches the title better. Sorry! The edit confusion is partly my fault for editing the question! $\endgroup$
    – Nic
    Dec 6, 2023 at 18:32

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