In the theory of quantum communication, it is commonly accepted that entanglement cannot increase the capacity of a classical communication channel as described by Shannon, that is, for an independent and identically distributed (i.i.d.) protocol. However, the capacity of an individual classical communication channel, which represents the amount of information you can transmit with a single use, can be increased if the sender and receiver share entanglement. Among the many papers exploring this concept, one of the more recent ones is an article by R. Prevedel, Y. Lu, W. Matthews, R. Kaltenbaek, and K.J. Resch entitled “Entanglement-assisted classical communication over a noisy channel”.[citation needed] In this paper, it is shown that a particular noisy classical channel has a higher success rate using a quantum strategy than using only a classical one. Even though the success rate boost may only be just under 7%, the paper still demonstrates the superiority of this approach. Furthermore, the publication even provides a physical experiment that can be used to implement their particular protocol, with experimental success rates very close to the theoretical ones and convincingly superior to the predicted theoretical results for a classical strategy.

Defining the classical channel

The entire sequence can be defined in the simple following way. A transmitter sends classical bit Q to a receiver over noisy classical communication channel N. First, the input is encoded locally, resulting in a classical bit X, then it is sent over N, resulting in a classical bit Y, and finally the receiver receives bit Y and decodes it into the classical bit Z. This is an extremely simple represention a purely classical information channel protocol.

Characterization of the channel N

To further characterize this simple classical protocol example to match the one described in the studied paper, 2 bits are introduced (B1 and B2), which outputs a trit (either 0,1, or 2) and a bit – T and B. There are three equi-probable outputs: (1, B1), (2, B2), or (P, parity of B1 and B2).

Maximizing the classical strategy

Our first task is to maximize this protocol for a one-time use of the channel. This will represent the best classical strategy. In order to do so, note that the channel takes as input two classical bits, but that we are looking for a strategy to optimize the transmission of a single classical bit. Thus, we will create code words in order to cut the input to a single bit – we will do so by choosing two out of the 4 possible inputs and mapping them to one of two binary code words (for example, by mapping (0,0) to 0 and (0,1) to 1). This will lead to one of the 6 possible outputs being cut as it will be impossible to reach, and we will be left with 5 possible outputs for 2 possible inputs. This can be more clearly seen in the graph mapping the inputs to the outputs below. A quick reminder on what we are looking for – we want to maximize the success rate of a particular strategy. Success can be defined as correctly guessing the input cbit based on the output cbit. Thus, we want to find a strategy to maximize the expected amount of correct input guesses based on all the possible outputs. In our particular case, in order to do so we should first note that 4 out of the 5 possible outputs have only a single associated input, and so we are left with only a single output that is ambiguous as it is associated with both the binary code words. Calculating the expected amount of correct guesses, we have that for each input, 2 out of 3 outputs are unambiguously associated to a given input, while the last one has one chance out of two of belonging to the other input. The total probability of success is thus 2/3 + ½ * 1/3 = 5/6. And so the best classical strategy will lead to a success rate of 5/6.

Simple entanglement-assisted classical protocol

Now consider the use of entanglement. First, we assume that Alice and Bob share an entangled qubit P. Now instead of sending an encoded cbit Q through our noisy channel N as in the classical case, Alice will use the input cbit Q to determine a measurement with which she will measure her half of the entangled qubit P. The measurement result, a classical bit, will be sent through N to Bob as in the classical case, only now Bob will not decode Y to Z, but rather use Y to select a measurement for measuring his half of the entangled qubit P, which will in turn result in the cbit Z. So basically, the idea is that instead of simply encoding and decoding a classical input bit, Alice and Bob will now use their classical bits to determine a measurement basis for encoding and decoding their halves of the entangled qubit P. This will yield a simple entanglement-assisted noisy classical protocol that can be visually represented by the following schema:

Maximizing the quantum strategy

CHSH-inequality game

Our goal now is to find a strategy that maximizes the success rate of this entanglement-assisted protocol. In order to do so, we will make use of the CHSH-inequality game. In this game, Alice and Bob are given input cbits A and B, respectively. They then perform some set of operations on their input cbit, and return cbits S and T, respectively. In order to win the game, the equation “S + T (mod 2) = AB” must hold. Note that Alice and Bob play as a team, and so they can choose a strategy before the game starts in order to maximize their success rate. We will now demonstrate that using a quantum strategy, Alice and Bob can get a higher success rate than using the best possible classical strategy.

Classical strategy

Let’s start with the classical strategy – looking at the table below which maps all possible inputs for A and B along with all the possible outputs for S and T, you will find that on average, Alice and Bob have a 50% chance of winning. This means that if they do not choose any strategy at all, they are as likely to win as they are to lose. What you might also notice is that the first and last rows have a ¾ success rate, while the middle two have only a ¼ success rate. Noting that the winning rows are (S, T) = { (0, 0), (1, 1) }, we can let Alice and Bob choose as strategy to agree that their output cbits will be equal – that is, we set S = T. This will yield the best possible classical strategy with a success rate of 75%.

Quantum strategy

Now let’s try to do better using a quantum strategy. First, we will let Alice and Bob share an entangled qubit BELL PHI MINUS IMAGE GOES HERE Now we perform the following measurements: One important detail to note here is that A and B here are used as the measurement bases for the shared maximally entangled qubit – if A is 0, Alice will measure her half of the maximally entangled qubit P in the basis X0, if A is 1 she will measure P in the basis X1. Bob’s situation is identical except that his input is B and his two bases are Y0 and Y2. The end result is that if Alice and Bob use this particular quantum strategy, they will obtain a success rate of cos^2(PI / 8), which is about 85%. This represents a 10% boost over the best possible classical strategy, and so we can see how a quantum strategy is superior to the best classical one for this particular game. It should also be noted that the CHSH-inequality itself is “| E(0,0) - E(0,1) + E(1,0) + E(1,1) | <= 2”, where the E(x,y) functions represent quantum correlations for all the possible output pairs. The inequality holds for the classical strategy, but is violated for the quantum case.

Optimal quantum strategy

Going back to the protocol, we concluded that the best classical strategy resulted in a 5/6 success rate, and we hoped to do better using a quantum approach. Based on the results of the CHSH-inequality game, we let Alice and Bob share a maximally entangled state, and define two measurement bases for Alice and Bob that will maximally violate the CHSH-inequality described above:

We should make note on what variables from the CHSH-inequality game we define in our current protocol – A and B will now become Q and V, where Q is the original message that Alice wants to transmit to Bob and V is chosen by Bob after he receives the output from the channel N. The outputs S and T will now become A (alpha) and B (beta), where A is the result of Alice’s measurement of her half of the maximally entangled qubit P in the basis determined by her input Q, and B is the measurement result of Bob’s half of the entangled qubit P using the basis selected by V. We also note that the two bits (B1, B2) that Alice sends through N are (Q, A). Here is a schema of the above description:

Note though that in the schema, lower-case letters are used and the final output Z is replaced by q-hat.

Now comes the crucial part – we know how the noisy classical channel N works, and we have a general idea of how we will be using our maximally entangled qubit P. All that is left to determine is how does Bob faithfully retrieve the initial cbit Q that Alice wants to transmit. We know that we have 3 equiprobable outputs for the channel N, and we also know that given that we have chosen a particular set of bases when measuring our entangled qubit P, the CHSH-inequality game equation holds with probability w = cos^2(PI / 8) = . Having this information, let’s see how we can determine the original message Alice wants to transmit using the information retrieved by Bob:

Let’s analyze this table row by row. For the first row, T = 1 and so B = Q. This means that Bob receives the original cbit message as-is — thus, he is done and there is no further decoding to be done. For the second row, T = 2 and so B = A. We thus need to take advantage of the equation “A + B (mod 2) = QV” to figure out a way to retrieve Q. We have A and we can obtain B by choosing V and measuring our maximally entangled qubit P. By choosing V = 1 and measuring P, our equation becomes “A + B (mod 2) = Q”, which leads to Q being easily obtained by taking the parity between A and B, both of which are known at this point. For the last row, T = P and we only know the parity between Q and A. By choosing V = 0, our equation becomes “A + B (mod 2) = 0”. Thus, we know that A and B are equal. Using this fact, we can obtain Q by applying the addition “Q + A + A (mod 2) = Q”. We already have as input the value “Q + A (mod 2)”, so all we need to do is modulo add to it the value of B, which as we’ve seen is equal to the value of A, resulting in the value of Q we’ve been looking for.

As we can see, for T = 1 Bob can retrieve Alice’s original message trivially, while for T = {2, P} Bob can deduct the message as long as the equation “A + B (mod 2) = QV” holds. Given that this equation is taken directly from the fact that we are running a CHSH-inequality game inside our protocol for T = {2,P}, we know that this equality holds 2/3 of the time with probability w = cos^2(PI / 8) =

And so, the success rate for the quantum strategy described is superior to the best possible classical strategy.

Practical applications

One might wonder what can be the practical use of such a result, as it is not an i.i.d. protocol and so it cannot be used for reliable communications. Indeed, when using an i.i.d. protocol the success rate approaches 100% vanishes as the number of channels grows large. Thus, the vast majority of the states you obtain are typical states from the typical subspace, and a typical projection may be used to slice out any states that are non-typical while preserving the quantum information of typical states. This effectively means that you can implement such a protocol as a reliable form of communication. This is not the case for one-shot protocols whose success rate is below 100% such as the one described in this paper.

However, one interesting parallel to make here is with the network communication protocols currently used between computers – specifically the TCP and UDP protocols. The TCP protocol is used all over the world to reliably transmit large amounts of information essentially error-free. It is not far-fetched to imagine that some i.i.d. quantum communication protocol will be used to do the same for some future quantum networks sometime in the not so distant future as it shares a lot of its characteristics, notably the reliability. The UDP protocol is also used in modern computer networks, though maybe not as often but for more specific tasks that require speed more than anything else. One such example is online video games, and particularly first-person shooters. In these fast-paced applications, what the sender and receiver are most worried about is the speed at which the data packets reach their destination rather than the reliability of the transmission, assuming a certain acceptable threshold – it doesn’t matter if a few packets are dropped, as long as certain threshold ratio of them reach their destination. The software game engine takes into account that some packets may be lost, and so there is an extra layer of error-correcting code built inside to make sure that the user’s experience is smooth. Thus, one-shot protocols such as the one studied in this paper can be seen as being similar to UDP communication.

Physical experiment

Finally, we should note that there has been a physical experiment that implements the described protocol. The experiment obtained a success rate of 0.89, which is quite close to the theoretical value of 0.90. This not only shows that the protocol is physically feasible, which is already impressive considering how far behind experimental quantum physics is behind the quantum theory, but that its results are also quite good, and at the very least above the theoretical maximum of classical communication alone, bringing additional proof to the superiority of quantum protocols over purely classical ones.


In summary, as we can see from the comparison of the classical and quantum results for the particular classical noisy channel, a quantum strategy is superior to a classical one when using this particular one-shot protocol.


Quantum information science
Quantum communication
Quantum algorithms
Quantum complexity theory
Quantum computing models
Decoherence prevention
Quantum optics
Ultracold atoms
This article uses material from the Wikipedia article One-Shot Entanglement-Enhanced Classical Communication, that was deleted or is being discussed for deletion, which is released under the Creative Commons Attribution-ShareAlike 3.0 Unported License.
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