Quantum measurement and qubits.

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Quantum measurement and qubits.

Physics
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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\[p_v=\text{Pr[we output }|v\rangle\text{|the qubit was in the state}|v\rangle]\]
(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?
@Jemurray3 @JamesJ Some basic QM help please!

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http://onlinephysicsbooks.blogspot.com/2009/09/modern-quantum-mechanics.html
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.
password:http://onlinephysicsbooks.blogspot.com
that is a link, the link itself is not the password, nor is the password on the link what is the password?
sure.that is the password.i have got it from the native blog
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
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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?
\[\langle u|u\rangle=\cos^2\left(\frac\pi4-\frac x2\right)\langle0|0\rangle+\sin^2\left(\frac\pi4-\frac x2\right)\langle1|1\rangle=1\]\[p_u=|\langle u|u\rangle|^2=1\]\[\langle u^\perp|v\rangle=-\sin\left(\frac\pi4-\frac x2\right)\cos\left(\frac\pi4+\frac x2\right)+\cos\left(\frac\pi4-\frac x2\right)\sin\left(\frac\pi4+\frac x2\right)\]\[=-\frac12[\sin\frac\pi2-\sin x]+\frac12[\sin\frac\pi2+\sin x]=\sin x\]\[p_v=|\langle u^\perp|v\rangle|^2=\sin^2x\]\[\frac12p_u+\frac12p_v=\frac{1+\sin^2x}2\]
b) \[\cos ^2(\frac{\pi}{4}-\frac{x}{2})\] isn't right?
but why?
Measurement II: Measure in the standard basis and interpret 0 as |u> and 1 as |v>. So \[=\cos(\frac{\pi}{4}-\frac{x}{2})<0|0>\] \[=\cos(\frac{\pi}{4}-\frac{x}{2})<1|1>\] \[p_u=||^2=\cos^2(\frac{\pi}{4}-\frac{x}{2})\] \[p_v=||^2=\cos^2(\frac{\pi}{4}-\frac{x}{2})\] \[\frac{1}{2}p_u+\frac{1}{2}p_v=\cos^2(\frac{\pi}{4}-\frac{x}{2})\] if i made a mistake, pls correct me! (sorry for my english, i'm italian) :-)
I just got it a different wy, but your answer is right what is \(u^+\) ?
u orthonormal... what different way?
\[\sin\left(\frac\pi4-\frac x2\right)=\cos\left(\frac\pi4+\frac x2\right)\]was all I think I needed
point c) i found \[\frac{1}{2}\] and \[\frac{1}{2}\] is it correct?
I'm not there yet, let me write my way to do b)
Ok.
\[|\langle0|u\rangle|^2=\cos^2\left(\frac\pi4-\frac x2\right)\]\[|\langle1|v\rangle|^2=\sin^2\left(\frac\pi4+\frac x2\right)=\cos^2\left(\frac\pi4-\frac x2\right)\]from which you get the same result not sure if my notation is kosher though :P
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?
Yes, and so far the answer 1/2 is all that makes sense to me :( let's try to think it thorugh
Ooops no... measurement 2: \[\frac{1}{2}+cx\]
oh I understand the second one...
It's not difficult... you have to substitute the approximation at the measurement...
yes, they both make sense now, we just made the mistake of actually subbing in x=0 at first :p
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...
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
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
ooohhh, i think you're right...
So point d), measurement II.
I hope that's the right logic. If you have another Idea let me know. Thanks for the help, I was really stuck!
Thank you!!! Anyway, all right but 1/2 +cx :-(
\[\cos^2\left(\frac\pi4-\frac x2\right)=\sin^2\left(\frac\pi4+\frac x2\right)\approx\frac12-c(-\frac x2)=\frac12+\frac{cx}2\]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.
perhaps*
Problem 12? How much degrees?
75 i found...
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...
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?
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 \(\cos^4\frac\pi8\)
will have a \(\cos^2\frac\pi8\) chance of getting though again*
Ok, now i have understand... thank you! :-)
Next week, next assignment :-) See you soon!
Welcome, and thank you :D good teamwork, look forward to working with you again!
Me too, bye! :-)
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
edX has a study group called the Discussion Forum!

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