Asked by qwerty
Let f:{0,1}->{0,1} with f(101)=1 and f(x)=0 for x ≠ 101 . In this question, we will run through two iterations of the algorithm.
(a) What is the superposition after the initialization step? Note that you can describe the superposition Σαx|x> by specifying two numbers α101 and αx for x ≠ 101 .
- α101=
- αx for x ≠ 101 =
(b) After the phase inversion in iteration 1?
- α101=
- αx for x ≠ 101 =
(c) After the inversion about mean in iteration 1?
- α101=
- αx for x ≠ 101 =
(d) After the phase inversion in iteration 2?
- α101=
- αx for x ≠ 101 =
(e) After the inversion about mean in iteration 2?
- α101=
- αx for x ≠ 101 =
(a) What is the superposition after the initialization step? Note that you can describe the superposition Σαx|x> by specifying two numbers α101 and αx for x ≠ 101 .
- α101=
- αx for x ≠ 101 =
(b) After the phase inversion in iteration 1?
- α101=
- αx for x ≠ 101 =
(c) After the inversion about mean in iteration 1?
- α101=
- αx for x ≠ 101 =
(d) After the phase inversion in iteration 2?
- α101=
- αx for x ≠ 101 =
(e) After the inversion about mean in iteration 2?
- α101=
- αx for x ≠ 101 =
Answers
Answered by
Anonymous
plz post other answers
Answered by
FLu
Anyone please?
Answered by
Anonymous
a) 1/sqrt(8), 1/sqrt(8)
b) -1/sqrt(8), 1/sqrt(8)
c) 5/(2*sqrt(8)), 1/(2*sqrt(8))
d) -5/(2*sqrt(8)), 1/(2*sqrt(8))
e) 22/(8*sqrt(8)), -1/(4*sqrt(8))
b) -1/sqrt(8), 1/sqrt(8)
c) 5/(2*sqrt(8)), 1/(2*sqrt(8))
d) -5/(2*sqrt(8)), 1/(2*sqrt(8))
e) 22/(8*sqrt(8)), -1/(4*sqrt(8))
Answered by
FLu
Thanks Anonymous!
Answered by
Anonymous
q5 pls
Answered by
Ar
PLease problem 5?
Answered by
Gyanno
problem 13 pls?
Answered by
Bel
12B pls???
Answered by
FLu
Click on FLu and you see the answers guys.
Finally, anyone for Problem 5 please?
Finally, anyone for Problem 5 please?
Answered by
FLu
12)b)
Fourth Tick
Fourth Tick
Answered by
Bel
thanks
Answered by
Mat
Problem 5 please?
Answered by
Tun
Problem 5 Please?
Answered by
Gyanno
P5 plz??
Answered by
My
Consider a deuteron in a cyclotron with field strength 0.5T. The deuteron is accelerated twice per rotation by a potential of V=25 kV. (a) If the radius of the cyclotron is 2 meter, what is the maximum energy of the deuteron? Express your answer in Joules (the deuteron mass is 3.34×10−27kg) b)Starting from a negligibly small velocity, how many full rotations does the deuteron need before it reaches this maximum energy? c) What is the time it takes for the deuteron to make one complete rotation when its energy is about 500 keV and when it is about 5 MeV? Ignore possible relativistic effect
Answered by
wat
q6) 1
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