# Quantum Physics

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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:

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I have done 6A:
(1/2 5/2)
(5/2 1/2)

6C and 6D: 0

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6B is
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))

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6B: exp(-3it/sqrt(2)|+> and exp(2it/sqrt(2)|->

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Problem 5 plz

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Thank you guys!!

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4b= 4

someone for the first??plzz

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Any kind of help is well received ;)

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3C=0

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1st is part c):

-z-h-
-x-h-

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3B is Z

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I meant answer to question 1 is part c.

-z-h-
-x-h-

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problem 4 a):

the eigenvalues are 1 and -1

then the lowest eingevalue (ground energy) is : -1

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4A is -1

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3D yes or no ?

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3d no

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Problem 7a
theta = pi/4
phi = 5*pi/2

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Problem 7b
1/2+i/2
1/sqrt(2)

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wrong

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7A and 7B wrong

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SOMEBODY KNOW ANSWER FOR: 3A? 5? 7? 2? 4C?

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hikikomori, do you have answer for the 3A? 5? 7? 2? 4c?

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hikikomori,Sorry, do you have answer for the 3A? 2? 4c?

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7A) pi/2 , 5*pi/4

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7B) 0.707 , -0.5-(0.5*i)

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5?

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4C (001, 010, 100, 111)

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anyone question 5?

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3A: C option

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Thanks all guys.
We now just need Q5's answers.

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We need Q.2 also

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for 2, last option is correct

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Thanks all guys
well-done

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desperately need answer for question 5. can someone explain a little how we get last option as correct for problem 2. as in the original cct if we give input 1> and 0> then apply cnot then we get 11> as the target bit flips.after that once apply Z gate (which will now be 4x4 matrix)we obatain (0,-1,0,1). how last option satisfy the same with same input as 0> once pass Z gate we get same 0> later as control bit is 1> it flips it after application of cnot and we get 11> as output. how are both equivalent??plz help

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Anyone??? Pls Q5... guys do soething pls

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5A: (2/K - 1), (2/K)

5B: (1- 2/K), (-2/K)

5C: (2/K - 1), (2/K)

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5A: (2/K), (1 - 2/K)

5B: (-2/K), (2/K - 1)

5C: (2/K - 1), (-2/K)

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Wrong? It worked for me!

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qwerty urs answer is also showing wrong dude

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All wrong?

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YES ALL WRONG

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the above answers to 5th question is of assignment 6 problem 5 don't get misguided

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WORKING ON IT AR . DISTRACTED BY SOME PERSONAL PROBLEMS IN LIFE ... NOT ABLE TO CONCENTRATE

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I think it should look like

(k-2)^x + ...
-------------
k^x

where x = m+1
but i have failed for find oput more :-(

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I am trying my best. Cant seem to get an answer. I will keep trying. Till then, I request others to try and post solutions here as well. Thanks.

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In the assignment 6, when you look to the posted answer it was written -answer to the part (c)- "Note that this is exactly the negation of the answer to part (a)". I think the following:
the state after " m+p steps + phase inversion" is equal to minus (-) the state after m-p steps! I checked this statement and it turns out to be true: the state after "2m step + phase inversion" must be equal to minus (-) the initial state...however when I enter the answer which becomes obvious when you apply the above observation, the grader says always "wrong"!...so guys to be honest with you: I still have only one shot for the three questions of problem 5: so or I get them all right (and for sure I will forward the right answer to all of you) or I will get them wrong...bye bye

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problem 3b,4c and 5

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well guys, i need few drinks (i'm doing my best MF) and we will be all right..cheer MF

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Some ppl think there is a simple answer to the problem.
To be honest I do not beleive it. For instance let m=1 and k=171. Why not? The original state was not defined in the problem, so it just might be that way. Will the "answer" work? I am sure it won't.

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Hi! I have just done 5B: ay=-1, ax=0, so, cheers and PhysTech, it must something about a cycle of the states from m-2, m-1, m, m+1, m+2

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Hi J. If the grader thinks it is the proper answer it doesn't mean it is. I tried it for a several k and there is no cicle except fo k=2. Try for k=7, and goes and goes without end, never in cicle.

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3B: Z (3rd option)
4C: Last option
5B: ay = -1; ax = 0
5A, 5C: ---Not yet solved correctly---

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Hi PhysTech, I am reading an article from twistedoakstudios(at)com in /blog/Post2644_grovers-quantum-search-algorithm and doing some calculations... Seems cyclic, in a geometric view.

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Ok, 5C is 1/sqrt(K) for both of them. I cannot figure out what is 5A... :(

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5A is not -1/sqrt(N) or any combinations with minus sign.

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@J
I input 1/sqrt(K) as 5C's answer, but they are wrong.

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@J
No they're right, I misspell the answer

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dudes there is something wrong with the system: i did not answer to 5) c) and as I mentioned before I have a last shot (only one shot) and guess what? I gave the answer to 5)b) (which was -1 and 0) and automatically the grader gave me the answer for 5)c)which is (1/sqrt(k),1/sqrt(k))...!!!!! there is something wrong here...don't you think so???

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wow! Strange behaviour @cheers xD Nevertheless good news for 5C :D Now, only 5A remains behind the Fortress of Solitude...

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@J "sorry I did not sleep the whole night"...I apologise to everybody...I'm still working on a) please forgive my "swearing" I did not sleep for 24h...booze make you thinking...cheers

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C'mon @cheers!! get some rest :) 5A is not (0,-1), or (0,1/sqrt(K)) or (0,-1/sqrt(N)) xD It must be something very close to it... I think the answer is translating the grover algorithm to a Bloch Sphere and see how it moves around the surface :) At least, that is what 5B,c suggest me...

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what is the answer of 6B?
(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))

or

6B: exp(-3it/sqrt(2)|+> and exp(2it/sqrt(2)|->

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Anonymous its the first.

(1/sqrt(2))*(e^(-3*i*t)) and (1/sqrt(2))*(e^(2*i*t))

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Anyone for Problem 5 a)?

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Dear friends, I tried this:
5c)(1/sqrt(k),1/sqrt(k))
but doesnt work...
Anyone for 5c and 5a???

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uppercase K

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5) a)?

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Thank you!!

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ANyone for 5)c)?

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Meant 5) a)?
Thank

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5 a)?!

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Gyus, 5 is easier than you think!
It takes exactly m steps from starting superposition (ay = 1/sqrt(K), ax = 1/sqrt(k)) to get to the solution (ay = 1, ax = 0), then it takes exactly m+1 steps to get back to initial state, but after each cycle the sign changes!
So it looks like this:
0: ay = 1/sqrt(K), ax = 1/sqrt(k)
m: ay = 1, ax = 0
2m+1: ay = -1/sqrt(K), ax = -1/sqrt(k)
3m+1: ay = -1, ax = 0
4m+2: ay = 1/sqrt(K), ax = 1/sqrt(k)
5m+2: ay = 1, ax = 0
6m+3: ay = -1/sqrt(K), ax = -1/sqrt(k)
........

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guess the right answer for 5a
and don't forget to write CAPITAL K

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It is this, thanks Andy!
5)a)
ay = -1/sqrt(K)
ax = -1/sqrt(K)

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thanks Andy,Flu and everyone worked for this subject!

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thank you!!!

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thx all!!

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Well done Guys... Cheers 2 all...

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5A:
ay=-1/sqrt(K)
ax=-1/sqrt(K)

5B:
ay=-1
ax=0

5C:
ay=1/sqrt(K)
ax=1/sqrt(K)

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hey guys,i want the answer of Problem 2

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Eisteinos

------*------
|
--Z---0------

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Suppose we have a quantum circuit that takes the input |0> and outputs |+>, and also takes the input |1> and outputs −|−>. If we input √2*i/√3|+> + 1√3|−>, what does the circuit output?
In the form a│0> + b│1>

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what is the answer for q1?

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1a. Z
1b. X
1c. X
1d. D=Z
1d. E=Z
1e. Z
1f. G=X
1f. G'=X

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8a- pi/3,o

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Anonymous, did you get all solutions for Q 7 and 8 except 8(D).

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@abc did you got question 8D?

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thnx, all of you for your help

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Esta resposta para 2 esta correta? Como posso inserir esta questÃ£o?

------*------
|
--Z---0------

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The answer to 2 is correct? How do I insert this?

------*------
|
--Z---0------

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