Workings of Deutsch-Josza Alririthm : One Qubit
As an excersize, we look into the detail workings of the Deutsch-Josza algorithm in the example of one qubit to one qubit oracle. The four unitary matrices U_a, U_b, U_c, and U_d are given by
We have |
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So, the quantum phase oracle V_f for f=a, b, c, and d are given by |
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[ 1
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0
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0
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0 ]
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V_a =
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[ 0
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-1
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0
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0 ]
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[ 0
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0
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1
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0 ]
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[ 0
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‚O
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0
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-1 ]
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[ 1
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0
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0
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0 ]
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V_b =
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[ 0
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-1
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0
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0 ]
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[ 0
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0
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-1
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0 ]
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[ 0
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‚O
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0
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1 ]
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[ -1
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0
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0
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0 ]
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V_c =
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[ 0
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1
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0
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0 ]
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[ 0
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0
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1
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0 ]
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[ 0
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‚O
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0
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-1 ]
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[ -1
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0
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0
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0 ]
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V_b =
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[ 0
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1
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0
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0 ]
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[ 0
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0
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-1
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0 ]
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[ 0
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‚O
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0
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1 ]
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Considering the state
| \psi > = (H | 0 >) | 0 >, namely |
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[1]
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| \psi > =1/\sqrt2
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[0]
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[1]
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[0]
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we obtain
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< \psi | V_a | \psi > = 1 , < \psi | V_a | \psi > = 0
< \psi | V_c | \psi > = 0 , < \psi | V_d | \psi > = -1
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whose squares give P. This furnishes the proof of the assertion of the last section. Readers are encouraged to extend this to the sisteen oracles that appear in the two qubit case.
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