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