TuringTest
  • TuringTest
Quantum measurement and qubits.
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TuringTest
  • 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\[p_v=\text{Pr[we output }|v\rangle\text{|the qubit was in the state}|v\rangle]\]
TuringTest
  • 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?
TuringTest
  • TuringTest
@Jemurray3 @JamesJ Some basic QM help please!

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

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