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Math Help - Help with Wikipedia Shor's Algorithm Derivation

  1. #1
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    Help with Wikipedia Shor's Algorithm Derivation

    Looking at Shor's algorithm - Wikipedia, the free encyclopedia, I'm finding it hard to understand part of step four of the period-finding subroutine derivation. Why does w^(x0 * y) not contribute to the probability amplitude? It must be equal to unity but I don't understand why?
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  2. #2
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    Re: Help with Wikipedia Shor's Algorithm Derivation

    i find that hard to read (i know very little about quantum mechanics or computing) but it seems to me what is happening is this:

    \left|\sum_b \omega^{(x_0 + rb)y}\right| = \left|\sum_b \omega^{x_0y}\omega^{rby}\right|

    and since x0y is independent of b:

    = |\omega^{x_0y}|\left|\sum_b \omega^{rby}\right| = \left|\sum_b \omega^{rby}\right|

    since ω lies on the unit circle, and thus any power of it has magnitude 1 (whereas the sum of such unit vectors may not lie on the unit circle).
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  3. #3
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    Re: Help with Wikipedia Shor's Algorithm Derivation

    Thanks a lot - I definitely should have seen this!
    Quote Originally Posted by Deveno View Post
    i find that hard to read (i know very little about quantum mechanics or computing) but it seems to me what is happening is this: \left|\sum_b \omega^{(x_0 + rb)y}\right| = \left|\sum_b \omega^{x_0y}\omega^{rby}\right|and since x0y is independent of b: = |\omega^{x_0y}|\left|\sum_b \omega^{rby}\right| = \left|\sum_b \omega^{rby}\right|since ω lies on the unit circle, and thus any power of it has magnitude 1 (whereas the sum of such unit vectors may not lie on the unit circle).
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