[cryptography] quantum key distribution

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BB84 is indeed a special case of B92, as B92 generalizes the concepts of BB84 with additional specific states. However, BB84 and B92 should not be considered special cases of the EPR Protocol, which fundamentally relies on entangled states, unlike the non-entangled states used in BB84 and B92. The EPR Protocol's flexibility stems from its use of entanglement, distinguishing it from the other two protocols. There is a wealth of resources available for comparing BB84, but fewer exist for B92 and the EPR Protocol. Understanding these distinctions is crucial for grasping the nuances of quantum key distribution.
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I'm reading up on quantum key distribution schemes, mostly: BB84, B92 and the EPR Protocol, and I'm having some problems relating them.

Is it correct to say that BB84 is a special case of B92(with extra specific states)? Is it also safe to say that both BB84 and B92 are special cases of EPR Protocol(with its entanglement being the most flexible possible)?

Anyone happen to know of a nice comparison? Theres a nice selection online for BB84, not much for B92 or the EPR Protocol unfortunately.

Thanks
 
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in advance!Yes, it is correct to say that BB84 is a special case of B92, since B92 is a more general version of BB84. However, it is not necessarily safe to say that both BB84 and B92 are special cases of the EPR Protocol, since the EPR Protocol is a different type of quantum key distribution protocol. The EPR Protocol uses entangled states whereas BB84 and B92 use non-entangled states. So while the EPR Protocol may be more flexible, it is not necessarily a special case of BB84 or B92.
 

Thread 'Two equivalent statements of time reversal symmetric Hamiltonian'

Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?
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