Suppose we ran m steps of Grover's algorithm on some function f (which has one marked element y) and the resulting superposition was exactly |y⟩.
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.
αy :
αx for x≠y:
Now if we run for another m steps, what is the resulting superposition?
αy :
αx for x≠y:
What about after yet another m+1 steps?
αy :
αx for x≠y:
Stucklike you :(, anyone please
I have done 6A:
(1/2 5/2)
(5/2 1/2)
6C and 6D: 0
others pleasee
6B is
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))
6B: exp(-3it/sqrt(2)|+> and exp(2it/sqrt(2)|->
Problem 5 plz
Thank you guys!!
4b= 4
someone for the first??plzz
Any kind of help is well received ;)
3C=0
1st is part c):
-z-h-
-x-h-