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

2026-07-30 15:04:27 +02:00
parent 5aafe8898d
commit 0763b687f5
3 changed files with 8 additions and 2 deletions
@@ -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 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]]
+