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TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1A 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\[p_v=\text{Pr[we output }v\rangle\text{the qubit was in the state}v\rangle]\]

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1(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?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1@Jemurray3 @JamesJ Some basic QM help please!

undikyumantoro
 one year ago
Best ResponseYou've already chosen the best response.0http://onlinephysicsbooks.blogspot.com/2009/09/modernquantummechanics.html

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1Thanks, 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.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1that is a link, the link itself is not the password, nor is the password on the link what is the password?

undikyumantoro
 one year ago
Best ResponseYou've already chosen the best response.0sure.that is the password.i have got it from the native blog

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1at 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

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1The example at the end seems to be my clue. Can anyone help me apply it to the problem at hand?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1@UnkleRhaukus QM help

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1\[\langle uu\rangle=\cos^2\left(\frac\pi4\frac x2\right)\langle00\rangle+\sin^2\left(\frac\pi4\frac x2\right)\langle11\rangle=1\]\[p_u=\langle uu\rangle^2=1\]\[\langle u^\perpv\rangle=\sin\left(\frac\pi4\frac x2\right)\cos\left(\frac\pi4+\frac x2\right)+\cos\left(\frac\pi4\frac x2\right)\sin\left(\frac\pi4+\frac x2\right)\]\[=\frac12[\sin\frac\pi2\sin x]+\frac12[\sin\frac\pi2+\sin x]=\sin x\]\[p_v=\langle u^\perpv\rangle^2=\sin^2x\]\[\frac12p_u+\frac12p_v=\frac{1+\sin^2x}2\]

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1b) \[\cos ^2(\frac{\pi}{4}\frac{x}{2})\] isn't right?

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1Measurement II: Measure in the standard basis and interpret 0 as u> and 1 as v>. So \[<uu>=\cos(\frac{\pi}{4}\frac{x}{2})<00>\] \[<u^+v>=\cos(\frac{\pi}{4}\frac{x}{2})<11>\] \[p_u=<uu>^2=\cos^2(\frac{\pi}{4}\frac{x}{2})\] \[p_v=<u^+v>^2=\cos^2(\frac{\pi}{4}\frac{x}{2})\] \[\frac{1}{2}p_u+\frac{1}{2}p_v=\cos^2(\frac{\pi}{4}\frac{x}{2})\] if i made a mistake, pls correct me! (sorry for my english, i'm italian) :)

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I just got it a different wy, but your answer is right what is \(u^+\) ?

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1u orthonormal... what different way?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1\[\sin\left(\frac\pi4\frac x2\right)=\cos\left(\frac\pi4+\frac x2\right)\]was all I think I needed

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1point c) i found \[\frac{1}{2}\] and \[\frac{1}{2}\] is it correct?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I'm not there yet, let me write my way to do b)

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1\[\langle0u\rangle^2=\cos^2\left(\frac\pi4\frac x2\right)\]\[\langle1v\rangle^2=\sin^2\left(\frac\pi4+\frac x2\right)=\cos^2\left(\frac\pi4\frac x2\right)\]from which you get the same result not sure if my notation is kosher though :P

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1point 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?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1Yes, and so far the answer 1/2 is all that makes sense to me :( let's try to think it thorugh

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1Ooops no... measurement 2: \[\frac{1}{2}+cx\]

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1oh I understand the second one...

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1It's not difficult... you have to substitute the approximation at the measurement...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1yes, they both make sense now, we just made the mistake of actually subbing in x=0 at first :p

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1Me too... ok, last question... d) Based on your answer to part (c), which measurement better distinguishes between u> and v> ? I don't understand...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I 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

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1the second measure preserves the sign of x when x is small, which is critical in knowing which side of 45 degrees we are on

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1ooohhh, i think you're right...

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1So point d), measurement II.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1I hope that's the right logic. If you have another Idea let me know. Thanks for the help, I was really stuck!

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1Thank you!!! Anyway, all right but 1/2 +cx :(

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1\[\cos^2\left(\frac\pi4\frac x2\right)=\sin^2\left(\frac\pi4+\frac x2\right)\approx\frac12c(\frac x2)=\frac12+\frac{cx}2\]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.

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1Problem 12? How much degrees?

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1The 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...

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1the 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?

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1that 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 \(\cos^4\frac\pi8\)

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1will have a \(\cos^2\frac\pi8\) chance of getting though again*

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1Ok, now i have understand... thank you! :)

unknonsense
 one year ago
Best ResponseYou've already chosen the best response.1Next week, next assignment :) See you soon!

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.1Welcome, and thank you :D good teamwork, look forward to working with you again!

Jonask
 one year ago
Best ResponseYou've already chosen the best response.1edX 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#group43487

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0edX has a study group called the Discussion Forum!
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