anonymous
  • anonymous
qubit measurement-quantum mechanics
Physics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • 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
  • 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
  • anonymous
|dw:1361565948497:dw|
experimentX
  • experimentX
yep!! but the dead line is 27 feb ... so i haven't looked into it.
TuringTest
  • 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
  • anonymous
so its just \[ab+\sqrt{1-a^2}\sqrt{1-b^2}\]
TuringTest
  • 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
  • anonymous
does that mean the result i just square it??
TuringTest
  • 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
  • anonymous
oh thats where i was confusiing it so we have \[(ab)^2,1-b^2-a^2-2(ab)^2\]
anonymous
  • anonymous
, or +
TuringTest
  • 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
  • TuringTest
typo, I meant\[a^2c^2+b^2d^2\]
anonymous
  • 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
  • anonymous
is this our case?
experimentX
  • experimentX
is this related to Week II Quantum Computing??
anonymous
  • anonymous
https://www.edx.org/c4x/BerkeleyX/CS191x/asset/3-1.pdf
anonymous
  • anonymous
yes lecture 3,4
TuringTest
  • 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
  • anonymous
oh great first principle i get from the course thanks ca we continue with two more problems???
anonymous
  • anonymous
https://www.edx.org/c4x/BerkeleyX/CS191x/asset/hw2.pdf P 13 b and 14
anonymous
  • anonymous
oh great first principle i get from the course thanks ca we continue with two more problems???
anonymous
  • anonymous
for 13a) i got 0.5
TuringTest
  • 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
  • anonymous
okay but did you try and guess the answer for 13 b
TuringTest
  • 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
  • experimentX
it's pretty late around here ... 1:40 am ... i guess i'll have to start tomorrow ... is @TuringTest in the course?
TuringTest
  • TuringTest
Heck yeah I am! I hope I survive it :) I really hope you join too @experimentX
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
  • 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
  • anonymous
lol yes thats good ,sry for the delay but is there time for us to do 14
anonymous
  • anonymous
there s also a openstudy specially for the course http://www.getstudyroom.com/course/43483#question-43485
anonymous
  • 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
  • experimentX
I am with EM ... too
anonymous
  • anonymous
so for the edX the final exams it open book ???
experimentX
  • experimentX
yeah ... good think about it, And I am good at cheating.
anonymous
  • anonymous
sry i was actually asking if they allow
anonymous
  • anonymous
did you try the study app
TuringTest
  • 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
  • anonymous
yes we can do it
TuringTest
  • 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
  • anonymous
i tried to write \[\frac{3}{4}|00+\frac{-\sqrt {5}}{4}|01\]
anonymous
  • anonymous
so we do that|dw:1361574751462:dw|
anonymous
  • anonymous
since it is not normalised
TuringTest
  • TuringTest
exactly
TuringTest
  • TuringTest
rather, we divide the vector by \(\sqrt{a^2+b^2}\)
anonymous
  • 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
  • anonymous
do we still call it 00,is this correct?
anonymous
  • anonymous
wow its really easy
anonymous
  • anonymous
i will take this as notes lol
anonymous
  • anonymous
thanks @TuringTest @experimentX
TuringTest
  • TuringTest
yeah, i's still 00 and so far it is pretty easy; we'll see how long it stays that way
experimentX
  • experimentX
lol ... i didn't to a damn thing except troll around ... thanks to @TuringTest
TuringTest
  • TuringTest
:p well I stil want to know how 13b works... we'll have to find someone else I guess
anonymous
  • anonymous
well i guess we shud start with 13 a sincethe question lays ref to a
anonymous
  • anonymous
i see from the notes that this is the bell state \[|\phi^+>=\frac{1}{\sqrt{2}}(|00+|11)\]
anonymous
  • anonymous
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TuringTest
  • TuringTest
still not getting it :(
TuringTest
  • 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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