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Lecture Notes 3.D

# 1 9 since f x 2 2 p2 f xi f xi i if f x

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Unformatted text preview: Uf |x, i = ( )f (x) |xi| i, where ( )n is an abbreviation for ( 1)n . 1 ¶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 that: X1 p ( ) f (x ) | x , i . | 2 i = Uf | 1 i = 2n 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!): X1 p ( )xz |z i. H | xi = 2 z 22 (This is just another way of writing H |0i = 1 p ( | 0i |1i).) 2 ¶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 =p ( )x·z |zi, n 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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