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Hi Eric, I’ve been getting myself a bit confused with the 2014 exam, Q3.iii; could you please tell me the best way to approach it?

Thanks

This question is about the “completely dephasing map,” \[ \Lambda(\rho) = \sum_y | y \rangle\langle y | \rho | y \rangle\langle y |. \](1) In part (ii) of the question, you’re asked to prove it is a CPTP map. Then part (iii) asks you to find an expression for the von Neumann entropy of a state \(\sigma =\Lambda(\rho)\). We want to find the von Neumann entropy of a state, which we know is just the Shannon entropy of the eigenvalues of the state. So we just need to find the eigenvalues of \(\Lambda(\rho)\), using the expression (1). Try that, and ask again if you want more.

*Hint:* no extensive calculations are needed! If you don’t see it, it might help to write \(\Lambda(\rho)\) in a matrix representation.

How to prove superadditivity of Holevo capacity? It is given as Exercise 2 in notes 18: \[\chi^*(\Lambda_1 \otimes \Lambda_2) \geq \chi^*(\Lambda_1) + \chi^*(\Lambda_2).\]

The trick is that since the Holevo capacity is defined as a maximum over all ensembles, in particular it’s at least as large as its value on any tensor product ensemble.

P.S. for more on the additivity of the Holevo capacity, Section 20.5 of Mark Wilde’s book has an interesting discussion.

Could you please help me with an exercise given in the lectures (Notes 13, eq 1.18): show that there exist a set of unitary operators \(\{U_j\}_j\) and a probability distribution \(\{p_j\}_j\) such that \[ \rho_A \otimes \sigma_B = \sum\limits_j p_jU_j\rho_{AB}U_j^\dagger, \](1) where \(\sigma_B = I/d\) is a completely mixed state.

The exercise in (1.18) is a bit tricky— I wouldn’t know how to do it without having seen it before. The unitaries needed are described in Mark Wilde’s book. In Section 3.7.2 (p. 110), he defines qudit generalizations of the Pauli \(X\) and \(Z\) operators, which he calls the Heisenberg-Weyl operators. Then in Exercise 4.7.6 on p. 176 of the same book, he has an exercise to show that these operators “do the job” by averaging out to the completely mixed state (see equation (4.349) in the book). Note you have a doubly-indexed family of unitaries \(\{ X(i) Z(j) \}_{i,j}\), but that could always be reindexed to have only one index, like in the previous question. Actually, there is one more subtlety: in Exercise 4.7.6 the operators only act on one system; here you have two systems, but only want to map the \(B\) system to the completely mixed state. So you would use \(I\otimes X(i) Z(j)\) instead.