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Jonask Group Title

qubit measurement-quantum mechanics

  • one year ago
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  1. Jonask Group Title
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    • one year ago
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  2. Jonask Group Title
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    so far i did the multiplication of \[\left(\begin{matrix}a \\ \sqrt{1-a^2}\end{matrix}\right)\left(\begin{matrix}b \space \space \sqrt{1-b^2} \ \end{matrix}\right)\]= \[(ab)^2+2ab\sqrt{(1-a^2)(1-b^2)}+1-a^2-b^2+(ab)^2\]

    • one year ago
  3. TuringTest Group Title
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    The outcome of one measurement relative a different basis is the dot product of those two measurements you have not done the dot product correctly

    • one year ago
  4. Jonask Group Title
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    |dw:1361565948497:dw|

    • one year ago
  5. experimentX Group Title
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    yep!! but the dead line is 27 feb ... so i haven't looked into it.

    • one year ago
  6. TuringTest Group Title
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    oh I see what you wrote, but It's backwards\[(a|0\rangle+b|1\rangle)\cdot(c|0\rangle+d|1\rangle)=ac+bd\]

    • one year ago
  7. Jonask Group Title
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    so its just \[ab+\sqrt{1-a^2}\sqrt{1-b^2}\]

    • one year ago
  8. TuringTest Group Title
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    \[\left[\begin{matrix}a&b\end{matrix}\right]\left[\begin{matrix}\sqrt{1-a^2}\\\sqrt{1-b^2}\end{matrix}\right]\]yes, then find the probability from the sum of each product of the amplitudes

    • one year ago
  9. Jonask Group Title
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    does that mean the result i just square it??

    • one year ago
  10. TuringTest Group Title
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    not the result, but each amplitude, so in this case each term individually gets squared we could have squared each probability amplitude (the coefficients of |0> and |1>) before taking the dot product

    • one year ago
  11. Jonask Group Title
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    oh thats where i was confusiing it so we have \[(ab)^2,1-b^2-a^2-2(ab)^2\]

    • one year ago
  12. Jonask Group Title
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    , or +

    • one year ago
  13. TuringTest Group Title
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    no, square each term individually to find the probability. remember that in\[a|0\rangle+b|1\rangle\]the probabilities of measuring 1 or 0 are \(a^2\) and \(b^2\), respectively. So the *probability* of measuring the outcome, as we go from one coordinate system to the other is\[\text{Pr}(a,b|c,d)=a^2|0\rangle+b^2|1\rangle\cdot c^2|0\rangle+d^2|1\rangle=a^2b^2+c^2d^2\]

    • one year ago
  14. TuringTest Group Title
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    typo, I meant\[a^2c^2+b^2d^2\]

    • one year ago
  15. Jonask Group Title
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    \[Pr(a,\sqrt{1-a^2}|b,\sqrt{1-b^2})=a^2|0>+(1-a^2)|1>(b^2)|0>+(1-b^2)|1>\] \[a^2b^2+(1-a^2)(1-b^2)\]

    • one year ago
  16. Jonask Group Title
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    is this our case?

    • one year ago
  17. experimentX Group Title
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    is this related to Week II Quantum Computing??

    • one year ago
  18. Jonask Group Title
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    https://www.edx.org/c4x/BerkeleyX/CS191x/asset/3-1.pdf

    • one year ago
  19. Jonask Group Title
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    yes lecture 3,4

    • one year ago
  20. TuringTest Group Title
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    well, more formally\[\large P_\psi(u)=\|\langle u|\psi\rangle\|^2\]\[=\|(b\langle0|+\sqrt{1-b^2}\langle1|)(a|0\rangle+\sqrt{1-a^2}|1\rangle)\|^2\]\[\|ab+\sqrt{(1-b^2)(1-a^2)}\|^2\]from which you get the same result

    • one year ago
  21. Jonask Group Title
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    oh great first principle i get from the course thanks ca we continue with two more problems???

    • one year ago
  22. Jonask Group Title
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    https://www.edx.org/c4x/BerkeleyX/CS191x/asset/hw2.pdf P 13 b and 14

    • one year ago
  23. Jonask Group Title
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    oh great first principle i get from the course thanks ca we continue with two more problems???

    • one year ago
  24. Jonask Group Title
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    for 13a) i got 0.5

    • one year ago
  25. TuringTest Group Title
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    I actually didn't get 13b either :p but I can show you 14 I think, let me redo it so I am sure how I did the problem

    • one year ago
  26. Jonask Group Title
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    okay but did you try and guess the answer for 13 b

    • one year ago
  27. TuringTest Group Title
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    yeah, I tried a few things, still don't quite understand how to get the second state since a|00>+b|00> was entangled and therefore not necessarily able to be decomposed... anyway shall we continue with P14 ?

    • one year ago
  28. experimentX Group Title
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    it's pretty late around here ... 1:40 am ... i guess i'll have to start tomorrow ... is @TuringTest in the course?

    • one year ago
  29. TuringTest Group Title
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    Heck yeah I am! I hope I survive it :) I really hope you join too @experimentX

    • one year ago
  30. experimentX Group Title
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    lol ... I managed to finish the first assignment on the last day ... I am always later runner .. i hope i will be able to complete it :) ... well nice to know there are other people who's doing it. looks like we can cooperate :D

    • one year ago
  31. TuringTest Group Title
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    Yeah, between us that makes 5 I hve found. I'm sure it will take more than one mind here to do all the assignments :P

    • one year ago
  32. Jonask Group Title
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    lol yes thats good ,sry for the delay but is there time for us to do 14

    • one year ago
  33. Jonask Group Title
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    there s also a openstudy specially for the course http://www.getstudyroom.com/course/43483#question-43485

    • one year ago
  34. Jonask Group Title
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    honestly i also have hard tym doing this courses chemistry,qphysics and electricity i have to stay in touch with my college lessons and still do the online courses its like triple major students so its a bit challenging 4 me

    • one year ago
  35. experimentX Group Title
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    I am with EM ... too

    • one year ago
  36. Jonask Group Title
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    so for the edX the final exams it open book ???

    • one year ago
  37. experimentX Group Title
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    yeah ... good think about it, And I am good at cheating.

    • one year ago
  38. Jonask Group Title
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    sry i was actually asking if they allow

    • one year ago
  39. Jonask Group Title
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    did you try the study app

    • one year ago
  40. TuringTest Group Title
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    Yes the final exams are open-book,but you are not supposed to discuss on OS for example. it's okay for homework though shall we do 14?

    • one year ago
  41. Jonask Group Title
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    yes we can do it

    • one year ago
  42. TuringTest Group Title
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    it's pretty simple really, after the first measurement we can eliminate the possibilities |10> and |11>, leaving the other two amplitudes now to make sure that their probabilities still add to one we have to normalize the remaining vector. after normalization just square the amplitude of |00> to find the probability of it being the outcome

    • one year ago
  43. Jonask Group Title
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    i tried to write \[\frac{3}{4}|00+\frac{-\sqrt {5}}{4}|01\]

    • one year ago
  44. Jonask Group Title
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    so we do that|dw:1361574751462:dw|

    • one year ago
  45. Jonask Group Title
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    since it is not normalised

    • one year ago
  46. TuringTest Group Title
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    exactly

    • one year ago
  47. TuringTest Group Title
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    rather, we divide the vector by \(\sqrt{a^2+b^2}\)

    • one year ago
  48. Jonask Group Title
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    new state \[\frac{ 3 }{ 4 }\sqrt{\frac{7}{8}}|00-\frac{ 2\sqrt{5} }{ 7 }\sqrt{\frac{7}{8} }|01\] P(00)=9/14

    • one year ago
  49. Jonask Group Title
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    do we still call it 00,is this correct?

    • one year ago
  50. Jonask Group Title
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    wow its really easy

    • one year ago
  51. Jonask Group Title
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    i will take this as notes lol

    • one year ago
  52. Jonask Group Title
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    thanks @TuringTest @experimentX

    • one year ago
  53. TuringTest Group Title
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    yeah, i's still 00 and so far it is pretty easy; we'll see how long it stays that way

    • one year ago
  54. experimentX Group Title
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    lol ... i didn't to a damn thing except troll around ... thanks to @TuringTest

    • one year ago
  55. TuringTest Group Title
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    :p well I stil want to know how 13b works... we'll have to find someone else I guess

    • one year ago
  56. Jonask Group Title
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    well i guess we shud start with 13 a sincethe question lays ref to a

    • one year ago
  57. Jonask Group Title
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    i see from the notes that this is the bell state \[|\phi^+>=\frac{1}{\sqrt{2}}(|00+|11)\]

    • one year ago
  58. Jonask Group Title
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  59. TuringTest Group Title
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    still not getting it :(

    • one year ago
  60. TuringTest Group Title
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    \[|\phi^+\rangle=\frac{1}{\sqrt2}(|00\rangle+|11\rangle\]is just one possible Bell state. To quote the notes: "In fact, there are states *such as* \(|\phi^+\rangle=\frac{1}{\sqrt2}(|00\rangle+|11\rangle\) which cannot be decomposed in this way as a state of the first qubit and that of the second qubit." So how can we assume ours is able to be decomposed since we don't know what \(\alpha\) and \(\beta\) are?

    • one year ago
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