Unformatted text preview: Uf |x, i = ( )f (x) |xi| i, where ( )n is an
abbreviation for ( 1)n .
¶8. From the deﬁnition of | i and Uf , Uf |x, i = |xi p2 (|f (x)i |¬f (x)i). 1
¶9. Since f (x) 2 2, p2 (|f (x)i |¬f (x)i) = | i if f (x) = 0, and it =
if f (x) = 1.
This establishes the claim. |i ¶10. Function application: Since Uf |x, y i = |x, y f (x)i, you can see
p ( ) f (x ) | x , i .
| 2 i = Uf | 1 i =
x 22 n
¶11. The top n lines contain a superposition of the 2n simultaneous evaluations of f . To see how we can make use of this information, let’s
consider their state in more detail.
¶12. For a single bit you can show (exercise!):
p ( )xz |z i.
H | xi =
(This is just another way of writing H |0i =
p ( | 0i
¶13. Therefore, for the n bits: 1
H ⌦ n | x 1 , x2 , . . . , x n i = p
2n X z1 ,...,zn 22 1
p ( | 0i
2 + |1i) and H |1i = ( )x1 z1 +···+xn zn |z1 , z2 , . . . , zn i 1X
( )x·z |zi,
2 z 22 n (III.22) where x · z is the bitwise inner product. (It doesn’t matter if you do
addition or since only the parity of the resul...
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- Fall '13
- Quantum computer, Qubit, Theoretical Computer Science, Quantum information science