2
$\begingroup$

Suppose Alice prepares $\rho_x$ with probabilities $p_x$ and sends it to Bob. I would say this is the same thing as "Alice prepares $\rho = \sum_x p_x \rho_x$ and sends it to Bob", but Preskill's lecture and Nielsen & Chuang seems to disagree.

I will be more precise.

If I want to compress the n-fold tensor product of $\rho$ I can do it down to $n \ S(\rho)$ qubits ($S$ is the Von Neumann entropy), this is the Schumacher theorem as stated in Nielsen & Chuang, page 544.

If I want to compress the ensemble $\epsilon = \{p_x,\rho_x\}$ Schumacher-like, I can do it down to $n \ \chi(\epsilon)$ qubits, as stated in http://www.theory.caltech.edu/people/preskill/ph229/notes/chap5.pdf, page 31. (Maybe if Preskill gave a definition for "ensemble fidelity" the issue would be easier to tackle...)

Note that $n \ \chi(\epsilon)$ is ensemble-dependent and in general different from the Von-Neumann entropy. But this seems to suggest that different ways of looking at a density matrix allows different ways of encoding it: If I think I am sending a $\rho_x$ with probabilities $p_x$ I must use the $\chi$ and hence a certain number of qubits, if I send $\rho = \sum_x p_x \rho_x$ I must use the entropy, and hence another number of qubits...

This bother me because I was intimately convinced that the ensemble description is not physical: Classical and quantum probabilities in density matrices,

$\endgroup$
2
$\begingroup$

There are two types of information being talked about, the information of a density matrix of a random ensemble (which you seem to understand) and the information of a sequence of states (which seems to be the source of confusion).

The key idea is that Alice prepares a sequence made out of $N$ states $\rho_j \in \{\rho_x\}$, and we want to know the sequence $\{\rho_1,\rho_2,\dots,\rho_N\}$. You can compute the average $\rho=\sum p_x\rho_x$, but this only gives (asymptotically) the average of the sequence i.e. $$\rho\approx\left<\rho_j\right>_j,$$ which is not the same as the information contained in the sequence itself.

As a (classical) example a book contains a sequence of letters. The information is containted in the particular order of the letters, whereas the "average symbol state" will only tell you e.g. what percentage of characters are the letter "s" etc. which is not the information you are trying to get out of reading a book.

$\endgroup$
  • $\begingroup$ Thanks for your answer, let me rephrase it with my own words and tell me if it is corrected: The idea is that Bob knows exactly what Alice is sending. They want to devise a scheme to compress the sequence of states such that when "averaged" over the p(x) they have a high fidelity. In this situation sending \rho = sum p(x) rho(x) with probability 1 is different than sending rho(x) with probability p(x). It is not like if Bob has to understand the sequence sent either, in such a situation the state he had had to use would be rho = sum p(x) rho(x). $\endgroup$ – giulio bullsaver May 29 '14 at 8:35
  • $\begingroup$ @giuliobullsaver Alice & Bob are communicating so both know the characteristics of the channel and have agreed on an encoding scheme. Therefore Bob knows the set of possible states $\{\rho_x\}$ & associated probabilities $p_x$. W/o compression Alice sends a string of states $S=\{\rho_1,\rho_2,\dots\}$ & that's what Bob gets. W/compression Alice takes $S$ compresses and sends a smaller string $S'$. Once received, Bob tries to decompress $S'$ to get $S$ & a necessary condition for success is that there is enough information available to encode the full string (i.e. at least $n\chi$ qbits). $\endgroup$ – Punk_Physicist May 30 '14 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.