qubit measurement-quantum mechanics

- anonymous

qubit measurement-quantum mechanics

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- schrodinger

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- anonymous

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- anonymous

so far i did the multiplication of
\[\left(\begin{matrix}a \\ \sqrt{1-a^2}\end{matrix}\right)\left(\begin{matrix}b \space \space \sqrt{1-b^2} \ \end{matrix}\right)\]=
\[(ab)^2+2ab\sqrt{(1-a^2)(1-b^2)}+1-a^2-b^2+(ab)^2\]

- TuringTest

The outcome of one measurement relative a different basis is the dot product of those two measurements
you have not done the dot product correctly

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- anonymous

|dw:1361565948497:dw|

- experimentX

yep!! but the dead line is 27 feb ... so i haven't looked into it.

- TuringTest

oh I see what you wrote, but It's backwards\[(a|0\rangle+b|1\rangle)\cdot(c|0\rangle+d|1\rangle)=ac+bd\]

- anonymous

so its just
\[ab+\sqrt{1-a^2}\sqrt{1-b^2}\]

- TuringTest

\[\left[\begin{matrix}a&b\end{matrix}\right]\left[\begin{matrix}\sqrt{1-a^2}\\\sqrt{1-b^2}\end{matrix}\right]\]yes, then find the probability from the sum of each product of the amplitudes

- anonymous

does that mean the result i just square it??

- TuringTest

not the result, but each amplitude, so in this case each term individually gets squared
we could have squared each probability amplitude (the coefficients of |0> and |1>) before taking the dot product

- anonymous

oh thats where i was confusiing it
so
we have \[(ab)^2,1-b^2-a^2-2(ab)^2\]

- anonymous

, or +

- TuringTest

no, square each term individually to find the probability.
remember that in\[a|0\rangle+b|1\rangle\]the probabilities of measuring 1 or 0 are \(a^2\) and \(b^2\), respectively. So the *probability* of measuring the outcome, as we go from one coordinate system to the other is\[\text{Pr}(a,b|c,d)=a^2|0\rangle+b^2|1\rangle\cdot c^2|0\rangle+d^2|1\rangle=a^2b^2+c^2d^2\]

- TuringTest

typo, I meant\[a^2c^2+b^2d^2\]

- anonymous

\[Pr(a,\sqrt{1-a^2}|b,\sqrt{1-b^2})=a^2|0>+(1-a^2)|1>(b^2)|0>+(1-b^2)|1>\]
\[a^2b^2+(1-a^2)(1-b^2)\]

- anonymous

is this our case?

- experimentX

is this related to Week II Quantum Computing??

- anonymous

https://www.edx.org/c4x/BerkeleyX/CS191x/asset/3-1.pdf

- anonymous

yes lecture 3,4

- TuringTest

well, more formally\[\large P_\psi(u)=\|\langle u|\psi\rangle\|^2\]\[=\|(b\langle0|+\sqrt{1-b^2}\langle1|)(a|0\rangle+\sqrt{1-a^2}|1\rangle)\|^2\]\[\|ab+\sqrt{(1-b^2)(1-a^2)}\|^2\]from which you get the same result

- anonymous

oh great first principle i get from the course thanks ca we continue with two more problems???

- anonymous

https://www.edx.org/c4x/BerkeleyX/CS191x/asset/hw2.pdf
P 13 b and 14

- anonymous

oh great first principle i get from the course thanks ca we continue with two more problems???

- anonymous

for 13a) i got 0.5

- TuringTest

I actually didn't get 13b either :p
but I can show you 14 I think, let me redo it so I am sure how I did the problem

- anonymous

okay but did you try and guess the answer for 13 b

- TuringTest

yeah, I tried a few things, still don't quite understand how to get the second state since a|00>+b|00> was entangled and therefore not necessarily able to be decomposed...
anyway shall we continue with P14 ?

- experimentX

it's pretty late around here ... 1:40 am ... i guess i'll have to start tomorrow ... is @TuringTest in the course?

- TuringTest

Heck yeah I am! I hope I survive it :)
I really hope you join too @experimentX

- experimentX

lol ... I managed to finish the first assignment on the last day ... I am always later runner .. i hope i will be able to complete it :) ... well nice to know there are other people who's doing it. looks like we can cooperate :D

- TuringTest

Yeah, between us that makes 5 I hve found. I'm sure it will take more than one mind here to do all the assignments :P

- anonymous

lol yes thats good ,sry for the delay but is there time for us to do 14

- anonymous

there s also a openstudy specially for the course http://www.getstudyroom.com/course/43483#question-43485

- anonymous

honestly i also have hard tym doing this courses chemistry,qphysics and electricity i have to stay in touch with my college lessons and still do the online courses its like triple major students so its a bit challenging 4 me

- experimentX

I am with EM ... too

- anonymous

so for the edX the final exams it open book ???

- experimentX

yeah ... good think about it, And I am good at cheating.

- anonymous

sry i was actually asking if they allow

- anonymous

did you try the study app

- TuringTest

Yes the final exams are open-book,but you are not supposed to discuss on OS for example. it's okay for homework though
shall we do 14?

- anonymous

yes we can do it

- TuringTest

it's pretty simple really, after the first measurement we can eliminate the possibilities |10> and |11>, leaving the other two amplitudes
now to make sure that their probabilities still add to one we have to normalize the remaining vector. after normalization just square the amplitude of |00> to find the probability of it being the outcome

- anonymous

i tried to write
\[\frac{3}{4}|00+\frac{-\sqrt {5}}{4}|01\]

- anonymous

so we do that|dw:1361574751462:dw|

- anonymous

since it is not normalised

- TuringTest

exactly

- TuringTest

rather, we divide the vector by \(\sqrt{a^2+b^2}\)

- anonymous

new state \[\frac{ 3 }{ 4 }\sqrt{\frac{7}{8}}|00-\frac{ 2\sqrt{5} }{ 7 }\sqrt{\frac{7}{8} }|01\]
P(00)=9/14

- anonymous

do we still call it 00,is this correct?

- anonymous

wow its really easy

- anonymous

i will take this as notes lol

- anonymous

thanks @TuringTest @experimentX

- TuringTest

yeah, i's still 00
and so far it is pretty easy; we'll see how long it stays that way

- experimentX

lol ... i didn't to a damn thing except troll around ... thanks to @TuringTest

- TuringTest

:p
well I stil want to know how 13b works... we'll have to find someone else I guess

- anonymous

well i guess we shud start with 13 a sincethe question lays ref to a

- anonymous

i see from the notes that this is the bell state
\[|\phi^+>=\frac{1}{\sqrt{2}}(|00+|11)\]

- anonymous

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- TuringTest

still not getting it :(

- TuringTest

\[|\phi^+\rangle=\frac{1}{\sqrt2}(|00\rangle+|11\rangle\]is just one possible Bell state. To quote the notes:
"In fact, there are states *such as* \(|\phi^+\rangle=\frac{1}{\sqrt2}(|00\rangle+|11\rangle\) which cannot be decomposed in this way as a state of the ﬁrst qubit and that of the second qubit."
So how can we assume ours is able to be decomposed since we don't know what \(\alpha\) and \(\beta\) are?

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