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a more general form of type of Fourier transform). Fourier sampling is easy
on a quantum computer, but appears to be diﬃcult to carry out on a classical 4.4. PHASE STATE 41 computer. In what follows, we will explore some of the power of Fourier
sampling. 4.4 Phase State We will now see how to set up an interesting state for fourier sampling. Given
a classical circuit for computing a boolean function f : {0, 1}n → {0, 1}, this
procedure due to Deutsch and Jozsa, shows how to transform it into a quantum
circuit that produces the quantum state φ = 1/2n/2 x (−1)f (x) x.
The quantum algorithm to carry out this task uses two quantum registers,
the ﬁrst consisting of n qubits, and the second consisting of a single qubit.
• Start with the registers in the state 0n 0
• Compute the Fourier transform on the ﬁrst register to get
0.
• Compute f to get x  x f ( x) . • Apply a conditional phase based on f (x) to get
• Uncompute f to get 4.5 x (−1) f ( x)  x ⊗  0 . x (−1) x∈ { 0 ,1 } n x⊗ f ( x)  x  f ( x ) . Extracting n bits with 2 evaluations of
Boolean Function Suppose we are given a black box (or an obfuscated classical circuit) that
computes the function function fs : {0, 1}n → {1, −1}, where f (x) = s · x.
s · x denotes the dot product s1 x1 + · · · + sn xn mod 2. The challenge is to use
this black box to eﬃciently determine s.
It is easy to see how to perform this task with n queries to the black
box: simply input in turn the n inputs x of Hamming weight 1. The outputs
of the black box are the bits of s. Since each query reveals only one bit of
information, it is clear that n queries are necessary.
Remarkably there is a quantum algorithm (the base case of the BernsteinVazirani algorithm) that requires only two (quantum) queries to the black
box:
• Use the black box to set up the phase state φ = 1/2n/2 x (−1) f (x) x. 42 CHAPTER 4. FOURIER SAMPLING & SIMON’S ALGORITHM
• Apply the Fourier transform H ⊗n and measure. The outcome of the
measurement is s. To that the outcome of the measurement is s, recall that H ⊗n s =
see
s·x x = φ. Since H ⊗n is its own inverse, it follows that
x (−1)
⊗n  φ =  s .
H
More generally, the transformation H ⊗n maps the standard basis s to
the fourier basis φs = 1/2n/2 x (−1)s·x x and viceversa.
We have shown that a quantum algorithm can be more eﬃcient than any
probabilistic algorithm in terms of the number of queries. One way to use this
diﬀerence in the number of queries in order to demonstrate a gap between
quantum and probabilistic algorithms is to make the queries very expensive.
Then the quantum algorithm would be n/2 times faster than any probabilistic
algorithm for the given task. But this does not help us in our goal, which is
to show that quantum computers violate the extended ChurchTuring thesi...
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 Fall '13

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