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TuringTest

  • 3 years ago

Quantum measurement and qubits.

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  1. TuringTest
    • 3 years ago
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    A qubit is either in the state\[|u\rangle=\cos\left(\frac\pi4-\frac x2\right)|0\rangle+\sin\left(\frac\pi4-\frac x2\right)|1\rangle\]or\[|v\rangle=\cos\left(\frac\pi4+\frac x2\right)|0\rangle+\sin\left(\frac\pi4+\frac x2\right)|1\rangle\]and we want to determine which is the case by measuring it. One of the following two measurements is optimal in terms of the probability of success. |dw:1361119926837:dw| Measurement I: Measure in the\[|u\rangle=\cos\left(\frac\pi4-\frac x2\right)|0\rangle+\sin\left(\frac\pi4-\frac x2\right)|1\rangle\]\[|u^\perp\rangle=-\sin\left(\frac\pi4-\frac x2\right)|0\rangle+\cos\left(\frac\pi4-\frac x2\right)|1\rangle\]basis and interpret the outcome \(u\) as \(|u\rangle\) and \(u^\perp\) as \(|v\rangle\). Measurement II: Measure in the standard basis and interpret \(0\) as \(|u\rangle\) and \(1\) as \(|v\rangle\). The probability of success is defined as \(\frac12p_u+\frac12p_v\), where\[p_u=\text{Pr[we output }|u\rangle\text{|the qubit was in the state}|u\rangle]\]and\[p_v=\text{Pr[we output }|v\rangle\text{|the qubit was in the state}|v\rangle]\]

  2. TuringTest
    • 3 years ago
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    (a) What is the probability of success of Measurement I as a function of x? (b) What is the probability of success of Measurement II as a function of x?

  3. TuringTest
    • 3 years ago
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    @Jemurray3 @JamesJ Some basic QM help please!

  4. undikyumantoro
    • 3 years ago
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    http://onlinephysicsbooks.blogspot.com/2009/09/modern-quantum-mechanics.html

  5. TuringTest
    • 3 years ago
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    Thanks, but I'm having trouble getting the password to open that file. also I was really hoping that someone already knew enough QM to help me here rather than just referring me to a book.

  6. undikyumantoro
    • 3 years ago
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    password: http://onlinephysicsbooks.blogspot.com

  7. TuringTest
    • 3 years ago
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    that is a link, the link itself is not the password, nor is the password on the link what is the password?

  8. undikyumantoro
    • 3 years ago
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    sure.that is the password.i have got it from the native blog

  9. TuringTest
    • 3 years ago
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    at the end of this set of notes is a similar example. I don't know if you can open it without being enrolled in edX https://www.edx.org/c4x/BerkeleyX/CS191x/asset/chap1.pdf still the example is a bit too different from the question here to really clear it up for me... or maybe I'm just being stupid

  10. TuringTest
    • 3 years ago
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  11. TuringTest
    • 3 years ago
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    The example at the end seems to be my clue. Can anyone help me apply it to the problem at hand?

  12. TuringTest
    • 3 years ago
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    @UnkleRhaukus QM help

  13. TuringTest
    • 3 years ago
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    \[\langle u|u\rangle=\cos^2\left(\frac\pi4-\frac x2\right)\langle0|0\rangle+\sin^2\left(\frac\pi4-\frac x2\right)\langle1|1\rangle=1\]\[p_u=|\langle u|u\rangle|^2=1\]\[\langle u^\perp|v\rangle=-\sin\left(\frac\pi4-\frac x2\right)\cos\left(\frac\pi4+\frac x2\right)+\cos\left(\frac\pi4-\frac x2\right)\sin\left(\frac\pi4+\frac x2\right)\]\[=-\frac12[\sin\frac\pi2-\sin x]+\frac12[\sin\frac\pi2+\sin x]=\sin x\]\[p_v=|\langle u^\perp|v\rangle|^2=\sin^2x\]\[\frac12p_u+\frac12p_v=\frac{1+\sin^2x}2\]

  14. unknonsense
    • 3 years ago
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    b) \[\cos ^2(\frac{\pi}{4}-\frac{x}{2})\] isn't right?

  15. TuringTest
    • 3 years ago
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    but why?

  16. unknonsense
    • 3 years ago
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    Measurement II: Measure in the standard basis and interpret 0 as |u> and 1 as |v>. So \[<u|u>=\cos(\frac{\pi}{4}-\frac{x}{2})<0|0>\] \[<u^+|v>=\cos(\frac{\pi}{4}-\frac{x}{2})<1|1>\] \[p_u=|<u|u>|^2=\cos^2(\frac{\pi}{4}-\frac{x}{2})\] \[p_v=|<u^+|v>|^2=\cos^2(\frac{\pi}{4}-\frac{x}{2})\] \[\frac{1}{2}p_u+\frac{1}{2}p_v=\cos^2(\frac{\pi}{4}-\frac{x}{2})\] if i made a mistake, pls correct me! (sorry for my english, i'm italian) :-)

  17. TuringTest
    • 3 years ago
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    I just got it a different wy, but your answer is right what is \(u^+\) ?

  18. unknonsense
    • 3 years ago
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    u orthonormal... what different way?

  19. TuringTest
    • 3 years ago
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    \[\sin\left(\frac\pi4-\frac x2\right)=\cos\left(\frac\pi4+\frac x2\right)\]was all I think I needed

  20. unknonsense
    • 3 years ago
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    point c) i found \[\frac{1}{2}\] and \[\frac{1}{2}\] is it correct?

  21. TuringTest
    • 3 years ago
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    I'm not there yet, let me write my way to do b)

  22. unknonsense
    • 3 years ago
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    Ok.

  23. TuringTest
    • 3 years ago
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    \[|\langle0|u\rangle|^2=\cos^2\left(\frac\pi4-\frac x2\right)\]\[|\langle1|v\rangle|^2=\sin^2\left(\frac\pi4+\frac x2\right)=\cos^2\left(\frac\pi4-\frac x2\right)\]from which you get the same result not sure if my notation is kosher though :P

  24. unknonsense
    • 3 years ago
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    point c) maybe i was wrong... correct is measurement 1: \[\frac{1+x^2}{2}\] measurement 2: \[\frac{1}{2}-cx\] is the same for you?

  25. TuringTest
    • 3 years ago
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    Yes, and so far the answer 1/2 is all that makes sense to me :( let's try to think it thorugh

  26. unknonsense
    • 3 years ago
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    Ooops no... measurement 2: \[\frac{1}{2}+cx\]

  27. TuringTest
    • 3 years ago
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    oh I understand the second one...

  28. unknonsense
    • 3 years ago
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    It's not difficult... you have to substitute the approximation at the measurement...

  29. TuringTest
    • 3 years ago
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    yes, they both make sense now, we just made the mistake of actually subbing in x=0 at first :p

  30. unknonsense
    • 3 years ago
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    Me too... ok, last question... d) Based on your answer to part (c), which measurement better distinguishes between |u> and |v> ? I don't understand...

  31. TuringTest
    • 3 years ago
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    I think that because the first estimation gives an x^2 that loses the sign of x (which way from 45 to u or v) so that is not a good way to distinguish between u and v

  32. TuringTest
    • 3 years ago
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    the second measure preserves the sign of x when x is small, which is critical in knowing which side of 45 degrees we are on

  33. unknonsense
    • 3 years ago
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    ooohhh, i think you're right...

  34. unknonsense
    • 3 years ago
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    So point d), measurement II.

  35. TuringTest
    • 3 years ago
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    I hope that's the right logic. If you have another Idea let me know. Thanks for the help, I was really stuck!

  36. unknonsense
    • 3 years ago
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    Thank you!!! Anyway, all right but 1/2 +cx :-(

  37. TuringTest
    • 3 years ago
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    \[\cos^2\left(\frac\pi4-\frac x2\right)=\sin^2\left(\frac\pi4+\frac x2\right)\approx\frac12-c(-\frac x2)=\frac12+\frac{cx}2\]I would have thought that the 2 would get absorbed into c since c is unknown anyway, but the stupid grader seems to be picky, or perha are making a point.

  38. TuringTest
    • 3 years ago
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    perhaps*

  39. unknonsense
    • 3 years ago
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    Problem 12? How much degrees?

  40. unknonsense
    • 3 years ago
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    75 i found...

  41. unknonsense
    • 3 years ago
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    The only one i don't know how find is problem 13 c)... i found problem 13 a)0.5, b)67.5 but c) i don't know to calculate...

  42. TuringTest
    • 3 years ago
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    the probability that it is translated through the first lens is 1 the probability that it goes through the second lens is \(\cos^2\theta\) where \(\theta\) is the angle between the two lenses, in this case \(\frac\pi8\) make sense so far?

  43. TuringTest
    • 3 years ago
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    that is what I'm getting to so by passing through the third lens, the polarized photons from the second lens will have a \(\cos\frac\pi8\) chance of getting though again the total probability for the photon to pass though all three lenses is then \(\cos^4\frac\pi8\)

  44. TuringTest
    • 3 years ago
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    will have a \(\cos^2\frac\pi8\) chance of getting though again*

  45. unknonsense
    • 3 years ago
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    Ok, now i have understand... thank you! :-)

  46. unknonsense
    • 3 years ago
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    Next week, next assignment :-) See you soon!

  47. TuringTest
    • 3 years ago
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    Welcome, and thank you :D good teamwork, look forward to working with you again!

  48. unknonsense
    • 3 years ago
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    Me too, bye! :-)

  49. Jonask
    • 3 years ago
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    edX group is it illegal to form the group on OS or atleast discuss already edX has a study group http://www.getstudyroom.com/course/43483#group-43487

  50. JamesJ
    • 3 years ago
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    edX has a study group called the Discussion Forum!

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