Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

How we could define the classical communication in the quantum teleportation protocol?

I mean, classical communication means to send a classical signal. But what happens if we are in an unclear situation in which we send a signal that we cannot classify as neither classical neither non-classical, because of lack of criteria?

Conceptually speaking, how critical is this classical in the protocol?

share|improve this question
add comment

1 Answer

Classical here refers to the protocol being deterministic. In other words, if we ignore noise, you can abstract the classical side of the protocol as '0 goes in, 0 comes out, 1 goes in, 1 comes out' and ignore the physical layer.

That said, noise is a very real phenomenon, and means that no channel is truly perfect, nor truly deterministic. However, there is a large body of knowledge (classical communications theory, error correcting codes, etc.) that is very well tested (e.g. various layers of the internet) to compensate for this side of the problem and ensure, with arbitrarily high probability, that the two classical bits make it through unharmed.

On the other hand, if you wanted to use a single photon source and on/off keying[1] to send the classical bits, you could in theory do that. A single photon state is entirely non-classical (so you have a 'non-classical signal' if you squint in just the right way), but the communications protocol is deterministic; that's the important part.

[1] That is, break time into a series of slots. Send a single photon to denote a '1' and don't send one to denote a '0' in a given slot. There are large technical difficulties associated with this currently, such as the non-availability of high quantum efficiency single photon sources, but we're talking about theory here.

share|improve this answer
Classical does not mean deterministic. The whole quantum teleportation protocol is deterministic, but that sure does not make it classical. The classicality of the communication means that either zero or one is sent, but not their general superposition. And that has nothing to do with the communication being deterministic or probabilistic. –  Ondřej Černotík Nov 11 '12 at 21:32
I will try to go deeper in the question thanks to these comments. When you are sending 0 or 1 from Alice to Bob, in practice Alice is sending an electromagnetic field (used in standard communication), which is converted to a number by Bob using universal protocols (the one in our computers for instance). In this picture, what is the classical feature? The fact that we are sending a classical electromagnetic field? If yes, what if we send something that is not clear if it is classical or not classical? –  Bob Nov 12 '12 at 0:13
@Bob You can always use quantum systems to send classical information -- you can, e.g., use polarization of a single photon. The classicality comes from the fact that you always send either zero (e.g. horizontal polarization) or one (vertical polarization) but NOT their superposition (i.e., any other linear, circular or elliptical polarization). –  Ondřej Černotík Nov 12 '12 at 8:47
@Ondřej Černotík The fact that a photon is a vertical or horizontal polarization it depends strongly from which basis you are measuring: $|H\rangle=\frac{|H\rangle+|V\rangle}{2}+\frac{|H\rangle-|V\rangle}{2}$, and changing basis you get $\frac{|phi\rangle+|\phi^\bot}{\sqrt{2}}$. So what are you telling depends strongly from an universal protocol that Alice and Bob knows, e.g., in your case, in which basis you have to measure to retrieve the information that Alice wants to send to Bob. –  Bob Nov 12 '12 at 10:37
@Bob Of course that depends on convention. But if Alice and Bob agreee upon which polarization represents zero and which represents one, in classical communication they never send a superposition of these two, though they can use single photons to transmit the information. –  Ondřej Černotík Nov 12 '12 at 10:43
show 4 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.