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

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)?

• Does the Description section of en.wikipedia.org/wiki/BB84 answer your question? – Victory Omole Feb 28 at 16:46
• Hi @ba. taj, this sounds like a homework problem. Can you talk about what you've tried so far? – Mohammad Athar Feb 28 at 17:30

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