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

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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)?

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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

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 $$|-+-01\dots\rangle$$.