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

1

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)?

Ba. Taj

Posted 2020-02-28T16:12:57.247

Reputation: 338

1

Does the Description section of https://en.wikipedia.org/wiki/BB84 answer your question?

– Victory Omole – 2020-02-28T16:46:37.250

1Hi @ba. taj, this sounds like a homework problem. Can you talk about what you've tried so far? – Mohammad Athar – 2020-02-28T17:30:21.713

Answers

4

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

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

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

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

Martin Vesely

Posted 2020-02-28T16:12:57.247

Reputation: 7 763