Wednesday

October 7, 2015
Posted by **helpless** on Sunday, March 24, 2013 at 2:53am.

(a) What was the state after the (m−1)th step? Note that you can describe the superposition by specifying two numbers, αy and αx for x≠y. Use K to denote the total number of elements. Please fully simply your answer.

Answer in this format: αy: αx for x≠y:

(b) Now, if we run one more step (total of m+1 steps), what is the resulting superposition?

Answer in this format: αy: αx for x≠y:

(c) What if you now apply another phase inversion?

Answer in this format: αy: αx for x≠y:

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**Yu**, Sunday, March 24, 2013 at 6:48amHelp Please!

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**qwerty**, Monday, March 25, 2013 at 10:39pmHelp please!

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**s**, Tuesday, March 26, 2013 at 2:20amPLS HELP

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**Anonymous**, Wednesday, March 27, 2013 at 12:20amhelp guys

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**My**, Wednesday, March 27, 2013 at 4:00amAnyone please?

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**Anónimo**, Wednesday, March 27, 2013 at 6:02amSo the state is at |y> after m steps. So the probability of getting that is 1 and the rest is 0. So What is the mean and what if we move 1 step forward? Is the the same as moving 1 step back?

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**horseman**, Wednesday, March 27, 2013 at 9:28amI have tried to solve it through the trigonometrics way. If in the iteration m you have probability 0 for the rest of the situations, then cos[(2k+1)g] = 0 becasuse of that k=(pi/(4*g))-(1/2) and you know that cos(g)=sqrt(k-1)/sqrt(k) so you have the value of k and the value of cos(g) so you can apply it to k-1 and try to solve the values of coefficients with the aid of the trigonometrics functions. However the marker gives me a red cross, I don´t know what I´m doing wrong! Please help.

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**Anonymous**, Wednesday, March 27, 2013 at 2:19pmProblem 4,5,10,11 answers guys

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**Anonymous**, Wednesday, March 27, 2013 at 2:20pm12 a and 13 a also pls guys

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**FLu**, Wednesday, March 27, 2013 at 4:03pmProblem 10

1; -1

-1; 1

Problem 11

0000

0000

0010

0001

Anyone for Problem 4 and 5 please?

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**Anonymous**, Wednesday, March 27, 2013 at 4:13pmThanks flu

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**FLu**, Wednesday, March 27, 2013 at 4:37pmNo prob anynomous.

Problem 13 a

1; 0

0; 5

Problem 12 a

-1, 3

Would be nice if someone could figure out 4 and 5.

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**Maverick**, Wednesday, March 27, 2013 at 4:41pm12(a) The answer is −1,3

The states of definite energy and their energy are given by the eigenvectors and eigenvalues of the Hamiltonian. In this case, the eigenvectors are |+> and |−> with eigenvalues 3 and −1 respectively.

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**Anonymous**, Thursday, March 28, 2013 at 7:05amThank you all.

And does anyone can help me with p1,p5 and p13

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**FLu**, Thursday, March 28, 2013 at 7:11amProblem 1

a)No

b)1

c)a; -b

Problem 13

a)

1; 0

0; 5

b)Last tick for multiple question

c)0

d)0

Anyone for Problem 5 though please?

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**Mur**, Thursday, March 28, 2013 at 3:06pmYes, anyone for Problem 5 please!

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**rih**, Thursday, March 28, 2013 at 5:05pmProblem 6 please!

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**Anne**, Thursday, March 28, 2013 at 7:55pmFor the problem 6 the solution is "1"

Anyone tried the problem 5?

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**plz help**, Thursday, March 28, 2013 at 10:24pmplease guys problem 5

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**Thank you so much!!**, Thursday, March 28, 2013 at 11:32pmhi guys solution for Problem 4 and 5? PLease

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**rih**, Friday, March 29, 2013 at 2:17amproblem 4

- 1/sqrt(8) ,1/sqrt(8)

- - 1/sqrt(8) , 1/sqrt(8)

- 5/(2*sqrt(8), 1/(2*sqrt(8)

- -5/(2*sqrt(8)) , 1/(2*sqrt(8))

- 11/(4*sqrt(8)), -1/(4*sqrt(8))

problem 5 ???????

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**FLu**, Friday, March 29, 2013 at 4:11amYes guys, Problem 5 please?

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**Gyanno**, Friday, March 29, 2013 at 5:04am12b please and 5?

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**FLu**, Friday, March 29, 2013 at 5:37am12)b

fourth tick

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**FLu**, Friday, March 29, 2013 at 5:38amAnyone for Problem 5?

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**Anonymous**, Friday, March 29, 2013 at 8:09amproblem 5 plz

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**rih thks!!!**, Friday, March 29, 2013 at 11:08am4 ?

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**rih thks!!!**, Friday, March 29, 2013 at 11:09amanswer to P5 ?

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**Mat**, Friday, March 29, 2013 at 11:47amAnswer for Question 5 please?

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**Tun**, Friday, March 29, 2013 at 11:47amYes please question 5 anyone?

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**Anonymous**, Saturday, March 30, 2013 at 2:24amhelp

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**Anonymous**, Saturday, March 30, 2013 at 3:39amqwerty this is for you.

5a......

5b 1-2/K -2/K

5c 2/K-1 -2/K

I have no idea what i am doing wrong with the a part. If u do get it let me know. Thanks....WU

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**Anonymous**, Saturday, March 30, 2013 at 3:55amLOL qwerty for you

cant believe i just got it

5a -2/K+1

2/K

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**dal**, Saturday, March 30, 2013 at 4:03amI don't know who are you, but thank you!

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**Anonymous**, Saturday, March 30, 2013 at 4:15am@ dal You can just call me Wu

And you are welcome.

If you have made it this far you deserve the joy.

@ qwerty please make sure u got this too.

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**FLu**, Saturday, March 30, 2013 at 6:04amThanks, that's great!

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**FLu**, Saturday, March 30, 2013 at 6:07amAnonymous check circuit question, I have given the answer!

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**xaad**, Sunday, March 31, 2013 at 3:57amlove u anonymous ..

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**FLu**, Sunday, March 31, 2013 at 6:57amDoes anyone know this?

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

I have b) 500

Anyone for a) and c)?

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**Im Anononous**, Sunday, March 31, 2013 at 10:55am13 c) 0 13 d) 0

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**My**, Sunday, March 31, 2013 at 11:27amConsider 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?

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**Uer**, Sunday, March 31, 2013 at 11:35amA current I travels counterclockwise through a closed copper wire loop which has the shape of a rectangle with sides a and b.

What is the magnitude of the magnetic field at the center, C , of the rectangle? Express your answer in terms of a, b, I and mu_0. (Enter mu_0)

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**Rad**, Sunday, March 31, 2013 at 12:25pmThanks FLu!

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?

Can somebody help with a) and c) please?

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**FLu**, Sunday, March 31, 2013 at 4:21pmFound c)

Anyone for a) please?

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**chepe**, Sunday, March 31, 2013 at 8:04pmproblem 6 plz

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**Anonymous**, Sunday, March 31, 2013 at 8:44pm6) 1

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**elviria**, Monday, April 1, 2013 at 5:47amdoes anyone have the answer for problem 1b and c

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**TriCk**, Monday, April 1, 2013 at 7:02amAnswer of problem 1b is 1

and 1c is a,-b

and thanku very much anonymous and Flu....

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**Stud**, Thursday, April 18, 2013 at 2:51amHey, Wu (Anonymous). Need help with qwerty's question here: site name/display.cgi?id=1366086762

Any help will be appreciated :)

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**rare**, Tuesday, September 24, 2013 at 11:39amsomeone please ans 12a..-1,3 is showing wrong..please help

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**Anonymous**, Wednesday, September 25, 2013 at 12:43pm@rare: i have the same problem..

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**ss01**, Wednesday, September 25, 2013 at 8:50pmConsider a qubit subject to the Hamiltonian (1 4

4 1).

Calculate the states of definite energy. What are the energies of these states?

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**Anonymous**, Thursday, September 26, 2013 at 4:32am12

-3,5