Quantum measurement and qubits.

- TuringTest

Quantum measurement and qubits.

- chestercat

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- 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

(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

@Jemurray3 @JamesJ Some basic QM help please!

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## More answers

- anonymous

http://onlinephysicsbooks.blogspot.com/2009/09/modern-quantum-mechanics.html

- 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

password:http://onlinephysicsbooks.blogspot.com

- TuringTest

that is a link, the link itself is not the password, nor is the password on the link
what is the password?

- anonymous

sure.that is the password.i have got it from the native blog

- 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

##### 1 Attachment

- TuringTest

The example at the end seems to be my clue. Can anyone help me apply it to the problem at hand?

- TuringTest

@UnkleRhaukus QM help

- 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

b) \[\cos ^2(\frac{\pi}{4}-\frac{x}{2})\]
isn't right?

- TuringTest

but why?

- 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

I just got it a different wy, but your answer is right
what is \(u^+\) ?

- anonymous

u orthonormal... what different way?

- TuringTest

\[\sin\left(\frac\pi4-\frac x2\right)=\cos\left(\frac\pi4+\frac x2\right)\]was all I think I needed

- anonymous

point c)
i found \[\frac{1}{2}\] and \[\frac{1}{2}\] is it correct?

- TuringTest

I'm not there yet, let me write my way to do b)

- anonymous

Ok.

- 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

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

Yes, and so far the answer 1/2 is all that makes sense to me :(
let's try to think it thorugh

- anonymous

Ooops no... measurement 2: \[\frac{1}{2}+cx\]

- TuringTest

oh I understand the second one...

- anonymous

It's not difficult... you have to substitute the approximation at the measurement...

- TuringTest

yes, they both make sense now, we just made the mistake of actually subbing in x=0 at first :p

- 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

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

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

ooohhh, i think you're right...

- anonymous

So point d), measurement II.

- 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

Thank you!!! Anyway, all right but 1/2 +cx :-(

- 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

perhaps*

- anonymous

Problem 12? How much degrees?

- anonymous

75 i found...

- 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

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

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

will have a \(\cos^2\frac\pi8\) chance of getting though again*

- anonymous

Ok, now i have understand... thank you! :-)

- anonymous

Next week, next assignment :-) See you soon!

- TuringTest

Welcome, and thank you :D
good teamwork, look forward to working with you again!

- anonymous

Me too, bye! :-)

- 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

edX has a study group called the Discussion Forum!

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