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I was reading through the superdense coding protocol, that lets A convey two classical bits to B by sending one qubit (assuming B sends A a qubit beforehand). So B creates a 2-qubit state and sends the first qubit to A. A performs a transformation on this qubit and sends it back. Based on that, B can distinguish whether the two bits of A were 00, 01, 10 or 11.

The question is, why can this idea not be extended to convey more than 2 bits? Can A convey n bits to B this way? More specifically, can B create an n-qubit state and send the first qubit A? A can then apply a unitary transformation on this qubit and send it back to B, who can distinguish between the 2^n possibilities and figure out which n bits A had.

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Yeah I also don't see why you can't use single qubit operations on the first qubit to create 2^n orthogonal states, so long as your hilbert space has sufficiently large dimension... :( anyone? – user12863 Oct 7 '12 at 20:29

The problem is that you can't create $2^{n}$ orthogonal states on $n$ qubits using only operations on 1 qubit. The operations you can perform on the single qubit are $I,X,Z,XZ$ (or something of the same form). If you use those four you obtain four orthogonal states on $n$ qubits but any other operations will give states that are not orthogonal to these four. If the states are not orthogonal they are not reliably distinguishable. If you use states that are not distinguishable and then some strategy (such as maximum likelihood estimation) to try and guess the state you've received then you'll wind up with a noisy channel which has a capacity of less than or equal to $2n$ bits.

If you used $4$ qubits, sent $2$ and kept $2$ to act on and send later then you can send $4$ bits but obviously this is no better than using $2$ 2-qubit systems.

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Tthe operations on the single qubit could be more than 4. A has n bits. Based on the value of each bit, A could perform n different unitary transformations. Shouldn't this potentially give rise to 2^n orthogonal states for B? – user12752 Oct 4 '12 at 16:17
@braindead $2^{n}$ states, yes; $2^{n}$ orthogonal states, no. You can see this by decomposing our bipartite state using the Schmidt decomposition $|\psi_{tot}\rangle=\Sigma\lambda_{i}|\psi_{i}\rangle|e_{i}\rangle$. Restricting ourselves to unitaries on the second party (single qubit) we can only obtain $4$ mutually orthogonal states because the only things we can change are the Schmidt basis "$e$" and the phase relationship between the two terms. – Q-Anon Oct 8 '12 at 10:01
If you could produce $2^{n}$ orthogonal states this would allow you to signal because you would have to change the reduced density matrix of the $n-1$ qubits to obtain that many orthogonal states (since you will have as many states as it is possible to obtain on $n$ qubits) – Q-Anon Oct 8 '12 at 10:04

Not in the way that you suggest, but it is possible to transfer more with states that have higher dimension on both sides. Here is first the general theory regarding superdense coding. What you are looking at is a state $\rho$ on which Alice performs a local unitary transformation $U$ and sends her part of the state to Bob. You can show that the maximum amount that can be transferred through this way is given by the conditional entropy $\log_2(d_A) - S(A|B)$. Of course $\log_2(d_A)$ is just the maximum classical capacity so you can see that to beat classical you need a negative conditional entropy, a purely quantum phenomenon. This can be seen by using relative entropy. Since Bob knows the state $\rho^B$, he can gain at most $S(\rho^{AB} \lVert (1/d_A) \otimes \rho^B)$ when Alice sends him her part of the state, which is precisely equal to $\log_2(d_A) - S(A|B)$, where $d_A$ is the dimension of Alice's system (and $d_B$ of Bob's). For a detailed argument of this see Since $S(A|B) \geq -\log_2(\min(d_A,d_B))$, which is when the state is pure and maximally entangled, you can see that the maximum you can transfer this way is $\log_2(d_A \cdot \min(d_A,d_B))$. In the case that both Alice's and Bob's systems are equal in dimension, which is most of the time, you get $2 \log_2(d)$, where $d$ is the dimension. With $n$ qubits you have $d = 2^n$ so that $2n$ bits is maximum transmission rate with $n$ qubits.

So to answer your question. You have a n qubit state but only one qubit is sent to the other party. In this case, $S(A|B) \geq -\log_2(2) = -1$ so that maximum that can be sent is 2 bits. Intuitively, you can see this by noticing that what allows the additional capacity is the fact that the state looks like $|00> + |11>$ initially. If Bob has a higher dimension, they will share will have $|\phi_0>|0> + |\phi_1>|1>$ where $|\phi_0>, |\phi_1>$ now live in a higher dimensional Hilbert space. However, you still have at most two orthogonal states so that having higher dimension on one side does not allow a higher capacity than $|00> + |11>$.

You might also be interested that dense coding with multipartite states has also been considered in

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