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November 26, 2014

November 26, 2014

Posted by **s** on Friday, April 19, 2013 at 5:15am.

If we run for m+1 additional steps (i.e. total of 2m+1 steps from the initial state), what is the resulting superposition? Note that you can describe the superposition by specifying two numbers, αy and αx for x≠y. You may use K to denote the total number of elements. (K should be uppercase.) HINT: You may want to review Problems 2, 3, and 5 of Assignment 6.

Now if we run for another m steps, what is the resulting superposition?

What about after yet another m+1 steps?

- Quantum Physics -
**Salmon Kahn**, Monday, October 14, 2013 at 10:39pmThis is fairly simple. You just need to remember that Grover's algorithm is cyclical

You start out with an equal superposition for both the y x answers

0 steps: ay = 1/sqrt(K), ax = 1/sqrt(k)

then after m steps, you solve it

@m steps: ay - 1, ax = o

Since you started from 0 instead of 1, it takes M+1 steps to get it back to the original position, but it's inverted

@2m+1 steps: ay = -1/sqrt(K), ax = -1/sqrt(k)

Then it takes another m steps to find the answer (but inverted)

@3m+1 steps: ay = -1, ax = 0

and then m+1 steps later, you're back to the initial position.

@4m+2 steps: ay = 1/sqrt(K), ax = 1/sqrt(k)

and the cycle repeats.

@5m+2 steps: ay = 1, ax = 0

@6m+3 steps: ay = -1/sqrt(K), ax = -1/sqrt(k)

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