I'm following these notes https://www.tcm.phy.cam.ac.uk/~sea31/tiqit_complete_notes.pdf where in Section 4.6, the erasure channel is said to have the following Kraus operators. Similar descriptions are found in other notes too.

$$M_{0}=\left( \begin{array}{ccc}{\sqrt{1-p}} & {0} & {0} \\ {0} & {\sqrt{1-p}} & {0} \\ {0} & {0} & {0}\end{array}\right) M_{1}=\left( \begin{array}{ccc}{0} & {0} & {\sqrt{p}} \\ {0} & {0} & {0} \\ {0} & {0} & {0}\end{array}\right) M_{2}=\left( \begin{array}{ccc}{0} & {0} & {0} \\ {0} & {0} & {\sqrt{p}} \\ {0} & {0} & {0}\end{array}\right)$$

I don't see how this works since $\sum_i M^\dagger_i M_i \neq I$. One instead gets $$\left( \begin{array}{ccc}{1-p} & {0} & {0} \\ {0} & {1-p} & {0} \\ {0} & {0} & {2p}\end{array}\right)$$

What am I missing?

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    $\begingroup$ Can you also link other notes where you found this representation? $\endgroup$
    – exp ikx
    Apr 27, 2019 at 18:49
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    $\begingroup$ @expikx arxiv.org/pdf/1106.1445.pdf (Section 4.7.6) also has this representation $\endgroup$ Apr 27, 2019 at 18:51
  • $\begingroup$ My answer could be possibly wrong. Please consider Norbert's answer. $\endgroup$
    – exp ikx
    Apr 28, 2019 at 15:25
  • $\begingroup$ @user1936752 Wilde's notes you link in the comments do it right. There it is stated: "The output alphabet contains one more symbol than the input alphabet, namely, the erasure symbol e." -- The other notes you link, on the other hand, seem to have several shortcomings. $\endgroup$ May 5, 2019 at 20:41

1 Answer 1


The Kraus operators for the channel are incorrect. The erasure channel acts on a qubit and outputs a qutrit. In the (uncommon) convention $$ \mathcal E(\rho) = \sum M_i^\dagger \rho M_i $$ used in the paper, the correct matrices $M_i$ therefore need to have size $2\times 3$. They are exactly formed by the first two rows of the matrices above.

Then, you can indeed verify that $\sum_i M_i M_i^\dagger=I$, which in the convention above corresponds to a trace-preserving map.

  • $\begingroup$ My answer could be possibly wrong since I did not consider output being a qutrit. But I have certain doubts. Usually, the completeness relation for Kraus operators state $M_i^{\dagger}M_i = I$, but which is not the case here. I understand we get identity by considering the form you have mentioned. Also, I thought it could be non-unital because in the papers I have read, they usually classify erasure channels apart from unital channels. Apologies for confusion. $\endgroup$
    – exp ikx
    Apr 28, 2019 at 15:21
  • $\begingroup$ However, please tell me why can't trace-decreasing maps make no sense. There are trace-decreasing CP maps. $\endgroup$
    – exp ikx
    Apr 28, 2019 at 15:30
  • $\begingroup$ @expikx The erasure channel can be realized physically (deterministally). Thus, it is trace preserving. BTW, why do you put comments regarding your post under mine? Note that the position of the dagger depends where it is put in the def. of the channel, which is non-standard here. $\endgroup$ Apr 28, 2019 at 16:02

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