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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?

user1936752
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1 Answers1

6

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.

Norbert Schuch
  • 22,105