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Jonask

  • one year ago

qubit measurement-quantum mechanics

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

  3. TuringTest
    • one year ago
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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

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

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

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

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

  8. TuringTest
    • one year ago
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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

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

  10. TuringTest
    • one year ago
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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

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

  12. Jonask
    • one year ago
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    , or +

  13. TuringTest
    • one year ago
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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\[\text{Pr}(a,b|c,d)=a^2|0\rangle+b^2|1\rangle\cdot c^2|0\rangle+d^2|1\rangle=a^2b^2+c^2d^2\]

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

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

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

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

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

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

  20. TuringTest
    • one year ago
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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

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

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

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

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

  25. TuringTest
    • one year ago
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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

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

  27. TuringTest
    • one year ago
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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 ?

  28. experimentX
    • one year ago
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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?

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

  30. experimentX
    • one year ago
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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

  31. TuringTest
    • one year ago
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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

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

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

  34. Jonask
    • one year ago
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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

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

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

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

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

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

  40. TuringTest
    • one year ago
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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?

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

  42. TuringTest
    • one year ago
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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

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

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

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

  46. TuringTest
    • one year ago
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    exactly

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

  48. Jonask
    • one year ago
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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

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

  50. Jonask
    • one year ago
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    wow its really easy

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

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

  53. TuringTest
    • one year ago
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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

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

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

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

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

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

  60. TuringTest
    • one year ago
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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?

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