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I am trying to better understand the meaning of Holevo information $\chi$. Suppose Alice starts with data encoded on qubits in the $\{0,1\}$ basis. She takes one of these qubits, originally in pure state $\lvert0\rangle$; randomly chooses from a uniform distribution of SU(2) transformations with which to encrypt the qubit; sends the qubit to Bob.

I have the formula $\chi=S(\sum_j p_j \rho_j)-\sum_j p_j S(\rho_j)$, where $\rho_j$ are all the encrypted states Alice chooses from, with corresponding probabilities $p_j$. The first term is clearly equal to $1$ (the entropy of a completely mixed qubit state); the second term is zero, since Alice always sends pure states (albeit unknown ones); so $\chi=1$.

Have I done this correctly? It makes sense to me, but it's not the way I see others handling the calculation.

edit: note that Bob does not know Alice's key. See comment for clarification.

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  • $\begingroup$ The part 'randomly encrypts..' is a bit unclear. Do you mean encode? If you use tge Holevo quantity like this, then badically you are enconding a continous alphabet. So this beeing one makes sense. But if you say you only have 0 as the alphabet, then it the information is 0. Can you be more explicit whar information is encoded by alice? $\endgroup$ – lalala Mar 17 '18 at 23:00
  • $\begingroup$ Without getting too much into my application, let me restate: Alice sends 0 or 1 to Bob, but in a random basis. Let's say Eve intercepts the qubit, but we're guaranteed that she doesn't know the basis. What is the Holevo bound on Eve's information about Alice's plaintext? So I do mean in "encrypt" in the sense that the key is meant to be secret. $\endgroup$ – ostrichCamel Mar 17 '18 at 23:33
  • $\begingroup$ In that case it is not true. The alphabet she receives is actually for each 0 or 1 the fully mixed state. So the Holevo quantity is 0 for this setup. $\endgroup$ – lalala Mar 18 '18 at 8:25
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One should consider the states associated with each of the plaintext messages Alice could send, in this case 0 and 1. For both of these cases, the state that reaches Bob is effectively mixed: even though it is actually pure, it is indistinguishable from a mixed state (e.g., one qubit of an entangled pair).

I found Preskill's notes helpful: http://www.theory.caltech.edu/~preskill/ph219/chap10_6A.pdf

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  • $\begingroup$ Could the down-voters please say what's wrong with this answer? $\endgroup$ – ostrichCamel Mar 18 '18 at 1:45
  • $\begingroup$ I didnt vote down, but the answer is not true. If you send Bob tge encoding (which is the whole point about encryption) then bob gets a different Holevo quantity. The answer is true if you replace bob with eve. $\endgroup$ – lalala Mar 18 '18 at 14:52
  • $\begingroup$ I understand the question may have been unclear, but it's the same thing whether we talk about Bob or Eve. I didn't say Bob has the key (in my application he does not). $\endgroup$ – ostrichCamel Mar 21 '18 at 19:16

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