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What is the quantum state transmitted to Bob in BB84 protocol?

Quantum Computing Asked on January 14, 2021

Sender (Alice) wants to exchange private encryption key with Bob using BB84 protocol. She generate a binary string (1010100010010111),and encode using Hadamard gate and identity gate (HHHIIHIHHIIIHIIH).

What is the quantum state she transmits to reciever (Bob)?

One Answer

Application of Hadamard gates changes states $|0rangle$ and $|1rangle$ followingly:

  • $mathrm{H}|0rangle = frac{1}{sqrt{2}}(|0rangle + |1rangle)$
  • $mathrm{H}|1rangle = frac{1}{sqrt{2}}(|0rangle - |1rangle)$

Identity operator does not change the state in any way, i.e. $mathrm{I}|0rangle = |0rangle$ and $mathrm{I}|1rangle = |1rangle$.

Hence if gates $mathrm{HHHII}dots$ are applied on input string of qubits $|10101dotsrangle$, the result is state $|-+-01dotsrangle$.

Answered by Martin Vesely on January 14, 2021

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