U ^ ( Q F T ) | 0 ⟩ = 1 N ∑ l = 0 N − 1 e 2 π i 0 l N | l ⟩ = 1 2 { | 0 ⟩ + | 1 ⟩ } {\displaystyle {\hat {U}}^{(QFT)}|0\rangle ={\frac {1}{\sqrt {N}}}\sum _{l=0}^{N-1}e^{\frac {2\pi i0l}{N}}|l\rangle ={\frac {1}{\sqrt {2}}}\{|0\rangle +|1\rangle \}}
U ^ ( Q F T ) | 1 ⟩ = 1 N ∑ l = 0 N − 1 e 2 π i 1 l N | l ⟩ = 1 2 { | 0 ⟩ − | 1 ⟩ } {\displaystyle {\hat {U}}^{(QFT)}|1\rangle ={\frac {1}{\sqrt {N}}}\sum _{l=0}^{N-1}e^{\frac {2\pi i1l}{N}}|l\rangle ={\frac {1}{\sqrt {2}}}\{|0\rangle -|1\rangle \}}
U ^ ( Q F T ) | 00 ⟩ = 1 4 ∑ l = 0 3 e 2 π i 0 l 4 | l ⟩ = 1 2 ( | 0 ⟩ + | 1 ⟩ + | 2 ⟩ + | 3 ⟩ ) {\displaystyle {\hat {U}}^{(QFT)}|00\rangle \ ={\frac {1}{\sqrt {4}}}\sum _{l=0}^{3}e^{\frac {2\pi i0l}{4}}|l\rangle ={\frac {1}{2}}(|0\rangle +|1\rangle +|2\rangle +|3\rangle )}
U ^ ( Q F T ) | 01 ⟩ = 1 4 ∑ l = 0 3 e 2 π i 1 l 4 | l ⟩ = 1 2 ( | 0 ⟩ + i | 1 ⟩ − | 2 ⟩ − i | 3 ⟩ ) {\displaystyle {\hat {U}}^{(QFT)}|01\rangle \ ={\frac {1}{\sqrt {4}}}\sum _{l=0}^{3}e^{\frac {2\pi i1l}{4}}|l\rangle ={\frac {1}{2}}(|0\rangle +i|1\rangle -|2\rangle -i|3\rangle )}
U ^ ( Q F T ) | 10 ⟩ = 1 4 ∑ l = 0 3 e 2 π i 2 l 4 | l ⟩ = 1 2 ( | 0 ⟩ − | 1 ⟩ + | 2 ⟩ − | 3 ⟩ ) {\displaystyle {\hat {U}}^{(QFT)}|10\rangle \ ={\frac {1}{\sqrt {4}}}\sum _{l=0}^{3}e^{\frac {2\pi i2l}{4}}|l\rangle ={\frac {1}{2}}(|0\rangle -|1\rangle +|2\rangle -|3\rangle )}
U ^ ( Q F T ) | 11 ⟩ = 1 4 ∑ l = 0 3 e 2 π i 3 l 4 | l ⟩ = 1 2 ( | 0 ⟩ − | 1 ⟩ + | 2 ⟩ − | 3 ⟩ ) {\displaystyle {\hat {U}}^{(QFT)}|11\rangle \ ={\frac {1}{\sqrt {4}}}\sum _{l=0}^{3}e^{\frac {2\pi i3l}{4}}|l\rangle ={\frac {1}{2}}(|0\rangle -|1\rangle +|2\rangle -|3\rangle )} || giống công thức trên, check lại sách
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U ^ ( Q F T ) | b n − 1 b n − 2 . . . b 0 ⟩ = 1 2 n { | 0 ⟩ + e 2 π i [ . b 0 ] | 1 ⟩ } { | 0 ⟩ + e 2 π i [ . b 1 b 0 ] | 1 ⟩ } { | 0 ⟩ + e 2 π i [ . b n − 1 b n − 2 . . . b 0 ] | 1 ⟩ } {\displaystyle {\hat {U}}^{(QFT)}|b_{n-1}b_{n-2}...b_{0}\rangle ={\frac {1}{\sqrt {2^{n}}}}\{|0\rangle \ +\ e^{2\pi i[.b_{0}]}\ |1\rangle \}\ \{|0\rangle \ +\ e^{2\pi i[.b_{1}b_{0}]}\ |1\rangle \}\ \{|0\rangle \ +\ e^{2\pi i[.b_{n-1}b_{n-2}...b_{0}]}\ |1\rangle \}} || công thức trong sách hơi ảo, xem lại
|| giống công thức trên, check lại sách
![{\displaystyle {\hat {U}}^{(QFT)}|b_{n-1}b_{n-2}...b_{0}\rangle ={\frac {1}{\sqrt {2^{n}}}}\{|0\rangle \ +\ e^{2\pi i[.b_{0}]}\ |1\rangle \}\ \{|0\rangle \ +\ e^{2\pi i[.b_{1}b_{0}]}\ |1\rangle \}\ \{|0\rangle \ +\ e^{2\pi i[.b_{n-1}b_{n-2}...b_{0}]}\ |1\rangle \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f9725296c502dea221bfd5b9cc748ddf05d500cf)
|| công thức trong sách hơi ảo, xem lại
