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I am studying quantum cryptography and I have a very basic question. Suppose A and B share a secret key k, where k=0 or 1. A wants to send one qubit to B. What A does is, if k=1, she 'flips' the qubit (i.e. applies the Pauli X matrix), otherwise she leaves the qubit as it is. She sends this to B. Now B, knowing the value of k, either flips it or leaves it as it is.

I am wondering why this protocol is insecure. How can an eavesdropper E cause problems?

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I don't think they can –  namehere Dec 3 '12 at 10:15
How is this different from a classical protocol in which the parties flip a classical bit depending on the value of $k$? What is the goal of the protocol? –  Juan Miguel Arrazola Dec 3 '12 at 23:15

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I don't know what the point of your question is, so I have two answers.

Suppose the qubit is in the $\sigma_x$ basis. In other words, its state is $$| \pm \rangle = \frac{1}{\sqrt{2}} \left(|0\rangle \pm |1\rangle\right).$$ Then we have $\sigma_x | + \rangle = | + \rangle$ and $\sigma_x | - \rangle = - | -\rangle$, so Eve could measure the qubit in the $\sigma_x$ basis and obtain its state without disturbing it. You could fix this by applying one of the three Pauli matrices or the identity matrix, each with probability $1/4$.

The other possible question I think you might be asking is: why use quantum key distribution (e.g. BB84) when this variant of the one-time pad (aka the Vernam cypher) works fine. The reason is that BB84 supplies a secret key (one-tine pad) to Alice and Bob, and secret keys can be very difficult to distribute while ensuring that they have not been compromised.

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I think you mean $\sigma_x$ instead of $\sigma_z$? –  Juan Miguel Arrazola Dec 3 '12 at 23:08
@Juan: (slapping myself on the forehead) yes, thanks. I'll fix it. –  Peter Shor Dec 3 '12 at 23:14

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