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May 22, 2015

May 22, 2015

Posted by **qwerty** on Monday, March 25, 2013 at 10:32pm.

(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 =

- Quantum Physics -
**Anonymous**, Wednesday, March 27, 2013 at 12:19amplz post other answers

- Quantum Physics -
**Ku**, Wednesday, March 27, 2013 at 4:02amHelp!

- Quantum Physics -
**FLu**, Wednesday, March 27, 2013 at 4:05pmAnyone please?

- Quantum Physics -
**Anonymous**, Thursday, March 28, 2013 at 12:55ama) 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))

- Quantum Physics -
**FLu**, Thursday, March 28, 2013 at 4:05amThanks Anonymous!

- Quantum Physics -
**Anonymous**, Thursday, March 28, 2013 at 11:20amq5 pls

- Quantum Physics -
**Ar**, Thursday, March 28, 2013 at 6:14pmPLease problem 5?

- Quantum Physics -
**Gyanno**, Friday, March 29, 2013 at 4:40amproblem 13 pls?

- Quantum Physics -
**Bel**, Friday, March 29, 2013 at 5:30am12B pls???

- Quantum Physics -
**FLu**, Friday, March 29, 2013 at 5:36amClick on FLu and you see the answers guys.

Finally, anyone for Problem 5 please?

- Quantum Physics -
**FLu**, Friday, March 29, 2013 at 5:38am12)b)

Fourth Tick

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**Bel**, Friday, March 29, 2013 at 5:43amthanks

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**Mat**, Friday, March 29, 2013 at 9:00amProblem 5 please?

- Quantum Physics -
**Tun**, Friday, March 29, 2013 at 11:49amProblem 5 Please?

- Quantum Physics -
**Gyanno**, Saturday, March 30, 2013 at 3:06pmP5 plz??

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**My**, Sunday, March 31, 2013 at 11:28amConsider 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

- Quantum Physics -
**wat**, Sunday, September 29, 2013 at 2:56pmq6) 1