Understanding Continuous Variable QKD

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Dopplershift
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So, I am doing my undergraduate research project in Quantum Cryptography, and I have some confusion in a few areas, especially in the topic of continuous variable quantum key distribution.

From what I understand,
Discrete Variable - Single photon. That is, for example, the BB84 protocol. Bob is measuring 1 photon at a time.
Continuous Variable- Alice encodes key bits (is it more than 1 bit of the key?) in the quadrature X and P.

I understand that is Bob measures in the right quadrature in the BB-84 protocol, then he'll get the right corresponding bit. However, if he chooses the wrong quadrature, he'll get the right bit 50% of the time.

But in CV-QKD, if Bob chooses the right quadrature either X or P, then he'll get the right information, but what happens if he chooses the wrong quadrature?

One more question, in continuous-variable QKD, what is the difference between coherent states and squeeze states. I understand that squeeze states either increase or decrease the certainty in the quadrature X and P because of the uncertainty principle. Is coherent states still used in DV-QKD (similar to weak pulses)?

Thanks for your help!
 
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In Continuous Variable (CV) QKD, if Bob chooses the wrong quadrature then he will not get the right information. Instead, he will get random data that has no meaning. This is why it is important for Bob and Alice to agree on which quadrature to use in advance. Coherent states are used in both Discrete Variable (DV) and Continuous Variable (CV) QKD. In DV-QKD, coherent states are usually used in the form of weak pulses. In CV-QKD, coherent states can be used to encode information in either X or P quadrature. Squeezed states are also used in CV-QKD, but unlike coherent states they can reduce the uncertainty of either the X or P quadrature. This allows for increased security and better performance in noisy channels.