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vault backup: 2026-07-30 15:04:27

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Benjamin Neumann
2026-07-30 15:04:27 +02:00
parent 5aafe8898d
commit 0763b687f5
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.obsidian/workspace.json
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SS2026/Quantum Computing/Bernstein-Vazirani/Bernstein-Vazirani Algorithm.md
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@@ -12,4 +12,10 @@ We will look at a quantum algorithm that will find s with only one evaluation of
The top wire consists of n qubits in state 0. $\ket{0}^n = \ket{0} \otimes ... \otimes \ket{0}$ The top wire consists of n qubits in state 0. $\ket{0}^n = \ket{0} \otimes ... \otimes \ket{0}$
The bottom wire is in state 1. The bottom wire is in state 1.
Both wires together are in state $\ket{0^n1} = \ket{0}^n \$ Both wires together are in state $\ket{0^n1} = \ket{0}^n \otimes \ket{1}$
First, we apply the Hadamard gate on all qubits.
The resulting state is calculated as follows:
![[Pasted image 20260730145119.png]]