## TuringTest 3 years ago Quantum measurement and qubits.

1. TuringTest

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

(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

@Jemurray3 @JamesJ Some basic QM help please!

4. anonymous
5. TuringTest

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. anonymous
7. TuringTest

8. anonymous

sure.that is the password.i have got it from the native blog

9. TuringTest

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

11. TuringTest

The example at the end seems to be my clue. Can anyone help me apply it to the problem at hand?

12. TuringTest

@UnkleRhaukus QM help

13. TuringTest

$\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. anonymous

b) $\cos ^2(\frac{\pi}{4}-\frac{x}{2})$ isn't right?

15. TuringTest

but why?

16. anonymous

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

I just got it a different wy, but your answer is right what is $$u^+$$ ?

18. anonymous

u orthonormal... what different way?

19. TuringTest

$\sin\left(\frac\pi4-\frac x2\right)=\cos\left(\frac\pi4+\frac x2\right)$was all I think I needed

20. anonymous

point c) i found $\frac{1}{2}$ and $\frac{1}{2}$ is it correct?

21. TuringTest

I'm not there yet, let me write my way to do b)

22. anonymous

Ok.

23. TuringTest

$|\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. anonymous

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

Yes, and so far the answer 1/2 is all that makes sense to me :( let's try to think it thorugh

26. anonymous

Ooops no... measurement 2: $\frac{1}{2}+cx$

27. TuringTest

oh I understand the second one...

28. anonymous

It's not difficult... you have to substitute the approximation at the measurement...

29. TuringTest

yes, they both make sense now, we just made the mistake of actually subbing in x=0 at first :p

30. anonymous

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

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

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. anonymous

ooohhh, i think you're right...

34. anonymous

So point d), measurement II.

35. TuringTest

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

36. anonymous

Thank you!!! Anyway, all right but 1/2 +cx :-(

37. TuringTest

$\cos^2\left(\frac\pi4-\frac x2\right)=\sin^2\left(\frac\pi4+\frac x2\right)\approx\frac12-c(-\frac x2)=\frac12+\frac{cx}2$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

perhaps*

39. anonymous

Problem 12? How much degrees?

40. anonymous

75 i found...

41. anonymous

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

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

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

will have a $$\cos^2\frac\pi8$$ chance of getting though again*

45. anonymous

Ok, now i have understand... thank you! :-)

46. anonymous

Next week, next assignment :-) See you soon!

47. TuringTest

Welcome, and thank you :D good teamwork, look forward to working with you again!

48. anonymous

Me too, bye! :-)

49. anonymous

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

edX has a study group called the Discussion Forum!