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

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

TuringTest
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0sure.that is the password.i have got it from the native blog

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

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

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

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

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

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

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

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

TuringTest
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1ooohhh, i think you're right...

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

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

TuringTest
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Problem 12? How much degrees?

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

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

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

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

Jonask
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0edX has a study group called the Discussion Forum!
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