Use of maximally entangled states in dense coding and quantum teleportation

Question

In studying about the procedures of superdense coding and quantum teleportation, I have seen that maximally entangled states are used in both cases to show that how entanglement can be used as resource. Why are arbitrary entangled states (not necessarily maximally entangled) not used? What is the speciality of maximally entangled states that they are specifically used in the discussions?

Answer 1

Maximally entangled states just work better for these two applications:

• for superdense coding, you need an orthonormal basis which can be mapped between each other using unitaries that are local to just one qubit. If you used $\alpha|00\rangle+\beta|11\rangle$ (you don't lose generality here, due to the Schmidt decomposition), then applying $X$ to one qubit produces an orthogonal state, but unless $|\alpha|=1/\sqrt{2}$, there is no other unitary that produces a different orthogonal state. That means your procedure one transmits 1 bit, not 2.

• for teleportation, the important thing is to be able to correct for getting all the possible different measurement results. For any entangled state, you could probably make a scheme that would work for specific measurement outcomes (at worst, I can probabilistically convert the state into a maximally entangled state). But it has to work for all outcomes. Again, the key is essentially that the Bell basis consists of 4 orthonormal states that can be mapped to each other by local Pauli rotations. We can propagate the Paulis through the circuit and see how to correct them. If the state isn't maximally entangled, we don't have that property.

Answer 2

Others have given a more heuristic explanation. I will try to give a more mathematical one. In a general dense coding protocol between a sender, Alice, and a receiver, Bob, sharing a quantum state $\rho_{AB}$, we concern ourselves with mutual information between Alice's encoding (associated with a random variable $X$) and Bob's decoding (associated with a random variable $Y$), given by $I(X;Y)$. When you maximise mutual information over all decoding measurements by Bob, you get accessible information, $I_{acc}$, associated with the encoded ensemble. We want to maximise accessible information over all encodings. It is well-known that accessible information is upper bounded by the Holevo bound. It has been shown that this bound can be asymptotically saturated, therefore, the maximum potential of a dense coding protocol would mean maximising the Holevo quantity over all encodings. The Holevo quantity looks like \begin{equation} I(X;Y) \le I_{acc}(\mathcal{E}) \le \chi(\mathcal{E}) = S(\rho_{\mathcal{E}}) - \sum_x p_X(x)\rho_x \end{equation} where the encoded ensemble looks like \begin{equation} \mathcal{E} = \{p_X(x), (U_A^x \otimes\mathbb{I})\rho_{AB}({U_A^x}^{\dagger} \otimes\mathbb{I})\} \equiv \{p_X(x), \rho_x \} \end{equation} where $\rho_{\mathcal{E}} = \sum_xp_X(x)\rho_x$ is the density matrix of the encoded ensemble and $p_X(x)$ is the probability with which Alice applies unitary $U_A^x$ to her part of the system before she sends it to Bob.

It has been shown in paper that if one maximises $\chi(\mathcal{E})$ over all possible encodings, then we get: \begin{equation} \max_{\{p_X(x), U_A^x\}_x} \chi(\mathcal{E}) = \chi_{DC} = \log(d_A) + S(\rho_B) - S(\rho_{AB}) \end{equation}

Clearly, the above quantity maximises for a maximally entangled state where it becomes $\log(d_Ad_B)$ ($2\log{d_A}$ for $d_A=d_B$). In fact, it is an if and only if condition.

Because dense coding and quantum teleportation are equivalent protocols (this can be rigorously shown), that is, one takes you from quantum to classical, and the other is from classical to quantum, you can make similar arguments for quantum teleportation.