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I have this assignment:

Show that super-dense coding protocol with the key in the state $\frac{|00⟩⟨00|+|11⟩⟨11|}{2}$ is equivalent (in a sense of transmission rate and security) with classical one-time pad.

I understand that super-dense coding protocol is a coding protocol, not an encryption protocol, so what exactly is the phrase "...super-dense coding protocol with the key..." supposed to mean? How is a key, usually used in encryption protocols, related to this coding protocol? Or has the word key a different meaning here?

Thank you for your answer.

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I recommend to read this: marozols.wordpress.com/2012/05/23/… –  Piotr Ćwikliński Dec 5 '13 at 15:22
    
You are correct that superdense coding doesn't relate directly to encryption, so I think the wording of this as "key" is confusing and misleading. In principle what you're transferring with superdense coding is just a sequence of bits. I think it was probably asked this way because one of the main contexts in which this might be useful quantum encryption in which two parties are exchanging encryption keys over a quantum channel. –  Iguananaut Dec 6 '13 at 16:03
    
I also recommend to read the post suggested by Piotr! ;) –  Māris Ozols Dec 6 '13 at 16:24
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1 Answer 1

I will assume the author means that each party has one qubit of the entangled state

$$\rho=\frac {|00\rangle \langle 00|+|11\rangle\langle11|}{2}$$

Which is interesting because $\rho$ isn't a pure state.

Anyway this state is a "key"; in the sense that without it parties should not make any sense of the data being transmitted.


I am not going any further into the details, because of the "homework" tag and the rules we have.

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