# Erasure channel Kraus operators

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?

• Can you also link other notes where you found this representation? – Abhay Hegde Apr 27 '19 at 18:49
• @expikx arxiv.org/pdf/1106.1445.pdf (Section 4.7.6) also has this representation – user1936752 Apr 27 '19 at 18:51
• My answer could be possibly wrong. Please consider Norbert's answer. – Abhay Hegde Apr 28 '19 at 15:25
• @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. – Norbert Schuch May 5 '19 at 20:41

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.
• 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. – Abhay Hegde Apr 28 '19 at 15:21