Hadmardo vartai (angl. Hadamard gate ) – tai kvantiniai vartai naudojami kvantiniame kompiuteryje , kilę iš Hadamardo transformacijos, skirti kubitą pervesti į superpozicijos būseną. Hadamardo kvantiniai vartai neturių klasikinių vartų analogų. Hadamardo vartai gali operuoti tik ant vieno kubito, be to kiekvienam kubitui reikia savo atskirų Hadamardo vartų. Kubitas, kurio būsena yra |1>, pereidamas per Hadamardo vartus persikelia į būseną:
H | 1 ⟩ = 1 2 | 0 ⟩ − 1 2 | 1 ⟩ = | − ⟩ {\displaystyle H|1\rangle ={\frac {1}{\sqrt {2}}}|0\rangle -{\frac {1}{\sqrt {2}}}|1\rangle =|-\rangle } O |0> pereidamas per Hadamardo vartus pereina į superpozicijos būseną, kuri užrašoma šitaip:
H | 0 ⟩ = 1 2 | 0 ⟩ + 1 2 | 1 ⟩ = | + ⟩ {\displaystyle H|0\rangle ={\frac {1}{\sqrt {2}}}|0\rangle +{\frac {1}{\sqrt {2}}}|1\rangle =|+\rangle } Šiose būsenose kubitai pereidami dar kartą per Hadamardo vartus grįžta į pradinę būseną:
H ( 1 2 | 0 ⟩ − 1 2 | 1 ⟩ ) = 1 2 ( | 0 ⟩ + | 1 ⟩ ) − 1 2 ( | 0 ⟩ − | 1 ⟩ ) = | 1 ⟩ {\displaystyle H({\frac {1}{\sqrt {2}}}|0\rangle -{\frac {1}{\sqrt {2}}}|1\rangle )={\frac {1}{2}}(|0\rangle +|1\rangle )-{\frac {1}{2}}(|0\rangle -|1\rangle )=|1\rangle } H ( 1 2 | 0 ⟩ + 1 2 | 1 ⟩ ) = 1 2 1 2 ( | 0 ⟩ + | 1 ⟩ ) + 1 2 ( 1 2 | 0 ⟩ − 1 2 | 1 ⟩ ) = | 0 ⟩ {\displaystyle H({\frac {1}{\sqrt {2}}}|0\rangle +{\frac {1}{\sqrt {2}}}|1\rangle )={\frac {1}{\sqrt {2}}}{\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )+{\frac {1}{\sqrt {2}}}({\frac {1}{\sqrt {2}}}|0\rangle -{\frac {1}{\sqrt {2}}}|1\rangle )=|0\rangle }
Kartais vietoje 0 rašomas 1, o vietoje 1 rašomas -1:
| 0 ⟩ = | 1 ⟩ {\displaystyle |0\rangle =|1\rangle } | 1 ⟩ = | − 1 ⟩ {\displaystyle |1\rangle =|-1\rangle } Mes taip ir žymėsime. Pirmas (pvz.: 1) nuo viršaus kubitas rašomas pirmu, o antras (pvz:. -1) nuo viršaus antru:
| 1 ⟩ | − 1 ⟩ = | 1 , − 1 ⟩ {\displaystyle |1\rangle |-1\rangle =|1,-1\rangle } Kai 2 kubitai yra perėję Hadamardo vartus, tai jie turės 4 reikšmes:
1 2 ( | 1 ⟩ + | − 1 ⟩ ) 1 2 ( | 1 ⟩ − | − 1 ⟩ ) = 1 2 ( | 1 , 1 ⟩ + | − 1 , 1 ⟩ − | 1 , − 1 ⟩ − | − 1 , − 1 ⟩ ) {\displaystyle {\frac {1}{\sqrt {2}}}(|1\rangle +|-1\rangle ){\frac {1}{\sqrt {2}}}(|1\rangle -|-1\rangle )={\frac {1}{2}}(|1,1\rangle +|-1,1\rangle -|1,-1\rangle -|-1,-1\rangle )} Jei iš paskos šiuos 2 kubitus seka dar vienas kubitas H|-1>, tai bendras kubitų užrašymas atrodys taip:
1 2 ( | 1 , − 1 ⟩ + | − 1 , 1 ⟩ − | 1 , − 1 ⟩ − | − 1 , − 1 ⟩ ) 1 2 ( | 1 ⟩ − | − 1 ⟩ ) = {\displaystyle {\frac {1}{2}}(|1,-1\rangle +|-1,1\rangle -|1,-1\rangle -|-1,-1\rangle ){\frac {1}{\sqrt {2}}}(|1\rangle -|-1\rangle )=} = 1 2 3 ( | 1 , − 1 , 1 ⟩ + | − 1 , 1 , 1 ⟩ − | 1 , − 1 , 1 ⟩ − | − 1 , − 1 , 1 ⟩ − | 1 , − 1 , − 1 ⟩ − | − 1 , 1 , − 1 ⟩ + | 1 , − 1 , − 1 ⟩ + {\displaystyle ={\frac {1}{\sqrt {2^{3}}}}(|1,-1,1\rangle +|-1,1,1\rangle -|1,-1,1\rangle -|-1,-1,1\rangle -|1,-1,-1\rangle -|-1,1,-1\rangle +|1,-1,-1\rangle +} + | − 1 , − 1 , − 1 ⟩ ) {\displaystyle +|-1,-1,-1\rangle )}
⟨ ϕ | ( c 1 | ψ 1 ⟩ + c 2 | ψ 2 ⟩ ) = c 1 ⟨ ϕ | ψ 1 ⟩ + c 2 ⟨ ϕ | ψ 2 ⟩ . {\displaystyle \langle \phi |\;{\bigg (}c_{1}|\psi _{1}\rangle +c_{2}|\psi _{2}\rangle {\bigg )}=c_{1}\langle \phi |\psi _{1}\rangle +c_{2}\langle \phi |\psi _{2}\rangle .}
( | ψ ⟩ ⟨ ϕ | ) | v ⟩ = | ψ ⟩ ⟨ ϕ | v ⟩ . {\displaystyle (|\psi \rangle \langle \phi |)|v\rangle =|\psi \rangle \langle \phi |v\rangle .}
H | ψ ⟩ = 1 2 ( X | ψ ⟩ + Z | ψ ⟩ ) {\displaystyle H|\psi \rangle ={\frac {1}{\sqrt {2}}}(X|\psi \rangle +Z|\psi \rangle )} | ψ ⟩ = 1 2 ( | 0 ⟩ + | 1 ⟩ ) {\displaystyle |\psi \rangle ={\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )} H | ψ ⟩ = H 1 2 ( | 0 ⟩ + | 1 ⟩ ) = | 0 ⟩ {\displaystyle H|\psi \rangle =H{\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )=|0\rangle } H | ψ ⟩ = 1 2 ( X 1 2 ( | 0 ⟩ + | 1 ⟩ ) + Z 1 2 ( | 0 ⟩ + | 1 ⟩ ) ) = {\displaystyle H|\psi \rangle ={\frac {1}{\sqrt {2}}}(X{\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )+Z{\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle ))=} = 1 2 ( ( 1 2 ( | 0 ⟩ + | 1 ⟩ ) + 1 2 ( | 0 ⟩ − | 1 ⟩ ) ) = | 0 ⟩ {\displaystyle ={\frac {1}{\sqrt {2}}}(({\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle )+{\frac {1}{\sqrt {2}}}(|0\rangle -|1\rangle ))=|0\rangle } Kur X yra kvantiniai NOT vartai , o Z yra fazės vartai .
| ψ ⟩ = a | 00 ⟩ + b | 10 ⟩ + c | 01 ⟩ + d | 11 ⟩ {\displaystyle |\psi \rangle =a|00\rangle +b|10\rangle +c|01\rangle +d|11\rangle } tai
⟨ ψ | ψ ⟩ = a 2 + b 2 + c 2 + d 2 = 1 {\displaystyle \langle \psi |\psi \rangle =a^{2}+b^{2}+c^{2}+d^{2}=1} Pavyzdžiui,
| ψ ⟩ = 1 2 ( | 0 ⟩ + | 1 ⟩ ) 1 2 ( | 0 ⟩ − | 1 ⟩ ) = 1 2 ( | 00 ⟩ + | 10 ⟩ − | 01 ⟩ − | 11 ⟩ ) {\displaystyle |\psi \rangle ={\frac {1}{\sqrt {2}}}(|0\rangle +|1\rangle ){\frac {1}{\sqrt {2}}}(|0\rangle -|1\rangle )={\frac {1}{2}}(|00\rangle +|10\rangle -|01\rangle -|11\rangle )} Tada,
⟨ ψ | ψ ⟩ = 4 ⋅ ( 1 2 ) 2 = 1. {\displaystyle \langle \psi |\psi \rangle =4\cdot ({\frac {1}{2}})^{2}=1.} ⟨ ψ | x ⟩ = 1 2 n , {\displaystyle \langle \psi |x\rangle ={\frac {1}{\sqrt {2^{n}}}},} kur n, kubitų skaičius. Pavyzdžiui, jei n=4, tai:
⟨ ψ | x ⟩ = 1 2 4 = 1 16 = 1 4 {\displaystyle \langle \psi |x\rangle ={\frac {1}{\sqrt {2^{4}}}}={\frac {1}{\sqrt {16}}}={\frac {1}{4}}} O jei n=2, tai:
⟨ ψ | x ⟩ = 1 2 2 = 1 4 = 1 2 {\displaystyle \langle \psi |x\rangle ={\frac {1}{\sqrt {2^{2}}}}={\frac {1}{\sqrt {4}}}={\frac {1}{2}}} ⟨ x | x ⟩ = 1 {\displaystyle \langle x|x\rangle =1} ⟨ x | y ⟩ = ⟨ y | x ⟩ = 0 {\displaystyle \langle x|y\rangle =\langle y|x\rangle =0}
Hadamardo vartai naudojami:
Kiti kvantiniai vartai