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Holevo's theorem says that no more than n bits can be stored (and retrieved) in n qubits. Indeed, allowing error can't improve this either -- the probability of retrieving the correct information is no better than that which could be transmitted in the same number of bits and guessing at the rest.

Superdense coding is one way around this bound: if the receiver shares n maximally entangled qubits with the sender, the sender can manipulate them such that when she gives the receiver her n qubits the receiver can obtain 2n bits of information. Perhaps this is not surprising, though, since he has to measure 2n qubits to get the data.

Is this the limit of quantum information capacity? That is, say sender and receiver share a large number N of entangled qubits and (after judicious manipulation and selection) the sender gives n of them to the receiver. Can more than 2n bits be transmitted in this way?

It would seem that the answer is "no", but I'd like a double-check. I'm very much a beginner, just working through Michael Nielsen's tutorials and Scott Aaronson's book. This question is similar to another question here but my question is different and not answered there.

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up vote 1 down vote accepted

No, only $2n$ bits can be transferred. The maximum capacity of superdense coding is actually known explicitly, and it is given by $\log_2(d) - S(A|B) = \log_2(d) - S(AB) + S(B)$. Here $d$ is the dimension of the system, in case of two level systems $d = 2$. This means that conditional entropy tells you by how much the standard classical capacity of $\log_2(d)$ is either attenuated or increased. It can be increased only because of the peculiar situation that the quantum conditional entropy can be negative.

We know that $\log_2(d) - S(A|B) = S(\rho^{AB} || 1/d \otimes \rho^B)$, where $S(\rho || \sigma)$ is the relative entropy. This quantity satisfies $0 \leq S(\rho^{AB} || 1/d \otimes \rho^B) \leq 2 \log_2(d)$ in case that both $A$ and $B$ systems have equal dimension. If they do not then replace $d = \min(d_A, d_B)$.

So for $n$ entangled qubits you will have $d = 2^n$ and therefore you can transmit at most $2 \log_2(2^n) = 2n$ bits of information.

If Alice posseses $N$ entangled qubits (I will imagine this means $N/2$ pairs) initially and sends $n$ of them to Bob, them $d = \min(d_A, d_B) = \min(2^{N-n}, 2^n) = 2^n$. This implies that the capacity is $2n$ bits. So having $N$ entangled qubits by itself does not offer any advantage over classical coding unless their counterparts are already with the receiver.

For further reading see for instance, or Nielsen & Chuang. If you want a more general picture you can also read about this in the introductory chapters of my thesis

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Since I'm new to quantum information theory, would you define your terms? In particular, what are A, B, S, and $\rho$? I notice that you never used my N, so would you verify that having N > n entangled qubits does not allow more than 2n bits to be received? (This seems to follow from the no-communication theorem, but I'd like confirmation.) – Charles Jun 13 '13 at 0:12
Notation: $\rho$ is the quantum state in question, $A$ refers to Alice's part of the state, $B$ to Bob's. So $\rho^A = Tr_B(\rho^{AB})$, ie. the state that Alice sees, having no access to Bob's part of the state. $S$ is entropy, $S(\rho) = -Tr[\rho \log_2(\rho)]$, $S(A|B)$ is the conditional entropy, ie $S(\rho^{AB}) - S(\rho^B)$ and $S(\rho || \sigma) = Tr[\rho(\log_2(\rho) - \log_2(\sigma))]$ is the relative entropy of states $\rho$ and $\sigma$. For $N$ I updated the answer as there is not enough space in this comment. – SMeznaric Jun 13 '13 at 18:05
Thank you! I will try to read the papers now. – Charles Jun 13 '13 at 18:38

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