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	The Resource Diagrams: Proofs

The proofs   The C diagram   The Q diagram   The E diagram

In the following we are going to give an elementary and self-contained proof to show that the resource diagrams are correct as depicted.

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The Q diagram

For your convenience we start with an enlarged reproduction of the diagram:

To begin with, we are going to focus on the teleportation point, confirming that at least two classical bits (henceforth, for simplicity: c-bits) are needed to send one qubit, regardless of the amount of entanglement at hand. The proof is by contradiction, showing that if less than two c-bits were sufficient, we could use a combination of dense coding and entanglement-assisted teleportation to transmit messages superluminally, violating causality.

Suppose that Alice (A) and Bob (B) jointly run a superdense coding machine, and thus can reliably encode two c-bits into one qubit, with the help of one bit of entanglement (henceforth: e-bit) they share. Their colleagues, Charlie (C) and Dave (D), have a teleportation facility, which allows Charlie to reliably send to Dave one qubit using two c-bits and one shared e-bit.
Assume that Alice wants to transmit a classical message to Bob using their dense coding machine. She encodes two c-bits at a time into one qubit. But instead of sending the qubit directly, she passes it on to Charlie, who teleports the signal to Dave, who, on his part, then sends the signal to Bob. Of course, for sending longer messages this protocol will have to be repeated a number of times.
The following picture sketches the setup. The dense coding machine jointly run by Alice and Bob is represented by the blue boxes labelled A and B, respectively. The green boxes dubbed C and D stand for the teleportation facility operated by Charlie and Dave. In the diagram, each C-bit is represented by an arrow with solid line, whereas wavy lines and dotted lines symbolize qubits and e-bits, respectively.

Now suppose that Charlie and Dave could reliably encode one qubit into less than two c-bits, if need be with the help of some additional entanglement. In this case, if Dave were impatient and decided to guess the message Charlie sends to him instead of waiting for it to arrive, his chance of guessing correctly would be greater than one fourth. He could then proceed in the protocol and pass his guess on to Bob, which implies that classical information could be transmitted superluminally from Alice to Bob, clearly violating causality. This shows that no less than two c-bits are necessary to reliably send one qubit.

From this it easily follows that for c = 0, at least one qubit is required for one-qubit-transmission, no matter how much additional entanglement is brought into play. For suppose that less than one qubit was sufficient. Then, by quantum teleportation, one qubit could also be sent with less than two c-bits and some extra amount of entanglement, which we know is impossible.
We thus see that for the noiseless qubit channel, the so-called entanglement-assisted quantum capacity , Q_E, i.e., the maximum asymptotic rate of reliable qubit transmission with the help of unlimited prior entanglement between sender and receiver, is precisely one bit, i.e., Q_E = 1.

This result can be used to confirm that the teleportation point is optimal also with regard to the amount of entanglement, in the sense that for q = 0, at least one e-bit is needed to generate one qubit: From the E diagram we see that one e-bit can be substituted completely by one qubit. So if we needed less than one e-bit to send one qubit, it could be replaced completely by less than one qubit, implying Q_E > 1, a contradiction.
Based on the same requirement, q + e >= 1, we see that no point on the ocherous wedge-shaped face allows to send more than one qubit.

To conclude the proof it remains to show that the straw yellow face of the Q diagram is not vaulted: Suppose we were given a point with coordinates (c,q,e),

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The C diagram

The proof we are going to give is very close in spirit to the one for the Q diagram. Using teleportation and dense coding, it may be shown that the entanglement-assisted quantum capacity , Q_E, is precisely half the entanglement-assisted classical capacity , C_E. From the Q section we know that Q_E = 1 for the ideal qubit channel; thus C_E = 2, implying that, for c = 0, one c-bit cannot be sent with less than half a qubit, regardless of the amount of entanglement at hand. This shows that dense coding is optimal regarding the amount of qubits used, and proves that the edge running upwards from the dense coding point is correct as drawn.

In the remainder of the proof we are going to make use of the fact that one c-bit cannot be sent with less than one qubit, which is a direct consequence of the Holevo bound (see, for instance: M. A. Nielsen, I. L. Chuang, Quantum Computation and Quantum Information , Cambridge University Press, Cambridge 2000, p. 531), and also manifests itself in the observation that the quantum channel capacity is no bigger than the classical channel capacity.

This said, we are now going to show that dense coding is optimal also with regard to the amount of entanglement, in the sense that, for c = 0 and q = 0.5, at least half an e-bit is necessary to generate one c-bit: Assume that e' < 0.5 e-bits would do the job. From the E diagram we see that one e-bit can be substituted by one q-bit. Therefore, we could attain one c-bit with q + e' < 1 qubits and no classical communication, violating the Holevo bound. Combining our results, we may infer that on the c = 0 plane no point other than the ones depicted can be used to attain one c-bit.

Next, we are going to focus on the yellow triangle: Since c-bits and e-bits can be substituted completely by an equal amount of qubits, the Holevo bound requires that

But what about the points that, while satisfying c + q + e >=1, lie between the straw yellow face and the planes c = 0 and q = 0, respectively? May one of these be used to send more than one c-bit? To see that this is not the case, note that these points have coordinates in the range

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The E diagram

This diagram looks simpler than the others, and correspondingly simple is the proof. That classical communication is useless for the generation of entanglement follows from the very definition of separability. It therefore remains to show that no point on the ocherous face can be used to generate more than one e-bit. But this, as above, follows from the requirement e + q >= 1.

Questions and comments

Last modified: 06 Feb 2003

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