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Method: Using O(m2 ) = O(log 2 M ) quantum operations perform the quantum FFT to obtain the superposition = M 1 j j j=0 Output: A random m-bit number j (that is, 0 j M 1), from the probability distribution P r[j] = | j |2 Quantum Fourier sampling is basically a quick way of getting a very rough idea about the output of the classical FFT, just detecting one of the larger components of the answer vector In fact, we don t even see the value of that component we only see its index How can we use such meager information In which applications of the FFT is just the index of the large components enough This is what we explore next
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Suppose that the input to the QFT, = ( 0 , 1 , , M 1 ), is such that i = j whenever i j mod k, where k is a particular integer that divides M That is, the array consists of M/k repetitions of some sequence ( 0 , 1 , , k 1 ) of length k Moreover, suppose that exactly one of the k numbers 0 , , k 1 is nonzero, say j Then we say that is periodic with period k and offset j Figure 105 Examples of periodic superpositions
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It turns out that if the input vector is periodic, we can use quantum Fourier sampling to compute its period! This is based on the following fact, proved in the next box: Suppose the input to quantum Fourier sampling is periodic with period k, for some k that divides M Then the output will be a multiple of M/k, and it is equally likely to be any of the k multiples of M/k Now a little thought tells us that by repeating the sampling a few times (repeatedly preparing the periodic superposition and doing Fourier sampling), and then taking the greatest common divisor of all the indices returned, we will with very high probability get the number M/k and from it the period k of the input!
The Fourier transform of a periodic vector
Suppose the vector = ( 0 , 1 , , M 1 ) is periodic with period k and with no offset (that is, the nonzero terms are 0 , k , 2k , ) Thus,
M/k 1
jk
We will show that its Fourier transform = ( 0 , 1 , , M 1 ) is also periodic, with period M/k and no offset Claim =
1 k k 1 jM j=0 k
Proof In the input vector, the coef cient is k/M if k divides , and is zero otherwise We can plug this into the formula for the jth coef cient of : 1 j = M
M 1 =0
k = M
M/k 1
jik
The summation is a geometric series, 1 + jk + 2jk + 3jk + , containing M/k terms and with ratio jk (recall that is a complex M th root of unity) There are two cases If the ratio is exactly 1, which happens if jk 0 mod M , then the sum of the series is simply the number of terms If the ratio isn t 1, we can apply the usual formula for geometric series to jk(M/k) Mj nd that the sum is 1 jk = 1 jk = 0 1 1 Therefore j is 1/ k if M divides jk, and is zero otherwise More generally, we can consider the original superposition to be periodic with period k, but with some offset l < k:
M/k 1 j=0
jk + l
Then, as before, the Fourier transform will have nonzero amplitudes precisely at multiples of M/k: Claim =
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