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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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 Fall '13
 BruceMacLennan

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