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

TuringTest Group TitleBest 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?
 one year ago

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

undikyumantoro Group TitleBest ResponseYou've already chosen the best response.0
password:http://onlinephysicsbooks.blogspot.com
 one year ago

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

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

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

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
but why?
 one year ago

unknonsense Group TitleBest ResponseYou've already chosen the best response.1
Measurement 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) :)
 one year ago

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

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

TuringTest Group TitleBest 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
 one year ago

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

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

TuringTest Group TitleBest 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
 one year ago

unknonsense Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

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

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

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

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

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

unknonsense Group TitleBest ResponseYou've already chosen the best response.1
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...
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

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

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

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

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

TuringTest Group TitleBest 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.
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.1
perhaps*
 one year ago

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

unknonsense Group TitleBest ResponseYou've already chosen the best response.1
75 i found...
 one year ago

unknonsense Group TitleBest ResponseYou've already chosen the best response.1
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...
 one year ago

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

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

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

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

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

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

unknonsense Group TitleBest ResponseYou've already chosen the best response.1
Me too, bye! :)
 one year ago

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

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