1
Fork 0
mirror of https://github.com/Steffo99/appunti-magistrali.git synced 2024-11-28 21:04:19 +00:00
appunti-steffo/7 - Introduction to quantum information processing/5 - Cose strane/costruire un Hardy state.md

100 lines
2.1 KiB
Markdown
Raw Normal View History

2024-07-05 17:22:13 +00:00
Si vuole creare un [[Hardy state]] su due [[qbit]] nello stato neutro applicandovi due [[gate quantistico universale|gate quantistici universali]].
2024-06-04 06:16:05 +00:00
2024-07-05 17:22:13 +00:00
## Obiettivo
2024-06-04 06:16:05 +00:00
2024-07-05 17:22:13 +00:00
Si vogliono quindi trovare i valori di $\mathbf{T}$ e $\mathbf{U}$ per cui:
$$
\def \kzero {{\color{darkgreen} 3}}
\def \kone {{\color{forestgreen} 1}}
\def \ktwo {{\color{limegreen} 1}}
\def \kthree {{\color{lightgreen} -1}}
\large
{\color{mediumpurple} \mathbf{T}}
{\color{mediumorchid} \mathbf{U}}
\ket{00}
=
\frac{
\kzero \cdot \ket{00} +
\kone \cdot \ket{01} +
\ktwo \cdot \ket{10} +
\kthree \cdot \ket{11}
}{\sqrt{12}}
$$
Ovvero:
$$
{\color{mediumpurple} \mathbf{T}}
\times
{\color{mediumorchid} \mathbf{U}}
\times
{
\begin{bmatrix}
1\\
0\\
0\\
0
\end{bmatrix}
}
=
\frac{1}{\sqrt{12}}
\cdot
{
\begin{bmatrix}
\kzero\\
\kone\\
\ktwo\\
\kthree
\end{bmatrix}
}
$$
## Separazione e raccolta nell'[[Hardy state]]
Ricordando che è possibile separare i [[qbit]]:
$$
\def \noteA {{\color{orangered} \Leftarrow}}
\def \noteB {{\color{dodgerblue} \Rightarrow}}
2024-06-04 06:16:05 +00:00
2024-07-05 17:22:13 +00:00
\displaylines{
\ket{00} = \ket{0}_\noteA \otimes \ket{0}_\noteB \\
\ket{01} = \ket{0}_\noteA \otimes \ket{1}_\noteB \\
\ket{10} = \ket{1}_\noteA \otimes \ket{0}_\noteB \\
\ket{11} = \ket{1}_\noteA \otimes \ket{1}_\noteB
}
$$
Possiamo separare i [[qbit]] dell'[[Hardy state]] in:
$$
\frac{1}{\sqrt{12}}
\cdot
\left\{
\begin{matrix}
\kzero & \cdot & (\ket{0}_\noteA \otimes \ket{0}_\noteB) \\
& + \\
\kone & \cdot & (\ket{0}_\noteA \otimes \ket{1}_\noteB) \\
& + \\
\ktwo & \cdot & (\ket{1}_\noteA \otimes \ket{0}_\noteB) \\
& + \\
\kthree & \cdot & (\ket{1}_\noteA \otimes \ket{1}_\noteB)
\end{matrix}
\right\}
$$
Poi, possiamo raccogliere lo stato di uno dei due [[qbit]], per esempio $\noteB$, ottenendo:
$$
\frac{1}{\sqrt{12}}
\cdot
\left\{
\begin{matrix}
(\ \kzero \cdot \ket{0}_\noteA + \ktwo \cdot \ket{1}_\noteA\ ) & \otimes & \ket{0}_\noteB \\
& + \\
(\ \kone \cdot \ket{0}_\noteA + \kthree \cdot \ket{1}_\noteA\ ) & \otimes & \ket{1}_\noteB
\end{matrix}
\right\}
$$
2024-06-04 06:16:05 +00:00
2024-07-05 17:22:13 +00:00
## Determinare gli elementi di ${\color{mediumorchid}\mathbf{U}}$
2024-06-04 06:16:05 +00:00
2024-07-05 17:22:13 +00:00
==TODO==