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Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0so far i did the multiplication of \[\left(\begin{matrix}a \\ \sqrt{1a^2}\end{matrix}\right)\left(\begin{matrix}b \space \space \sqrt{1b^2} \ \end{matrix}\right)\]= \[(ab)^2+2ab\sqrt{(1a^2)(1b^2)}+1a^2b^2+(ab)^2\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2The outcome of one measurement relative a different basis is the dot product of those two measurements you have not done the dot product correctly

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yep!! but the dead line is 27 feb ... so i haven't looked into it.

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2oh I see what you wrote, but It's backwards\[(a0\rangle+b1\rangle)\cdot(c0\rangle+d1\rangle)=ac+bd\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0so its just \[ab+\sqrt{1a^2}\sqrt{1b^2}\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2\[\left[\begin{matrix}a&b\end{matrix}\right]\left[\begin{matrix}\sqrt{1a^2}\\\sqrt{1b^2}\end{matrix}\right]\]yes, then find the probability from the sum of each product of the amplitudes

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0does that mean the result i just square it??

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2not 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

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0oh thats where i was confusiing it so we have \[(ab)^2,1b^2a^22(ab)^2\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2no, square each term individually to find the probability. remember that in\[a0\rangle+b1\rangle\]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\[\text{Pr}(a,bc,d)=a^20\rangle+b^21\rangle\cdot c^20\rangle+d^21\rangle=a^2b^2+c^2d^2\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2typo, I meant\[a^2c^2+b^2d^2\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0\[Pr(a,\sqrt{1a^2}b,\sqrt{1b^2})=a^20>+(1a^2)1>(b^2)0>+(1b^2)1>\] \[a^2b^2+(1a^2)(1b^2)\]

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1is this related to Week II Quantum Computing??

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2well, more formally\[\large P_\psi(u)=\\langle u\psi\rangle\^2\]\[=\(b\langle0+\sqrt{1b^2}\langle1)(a0\rangle+\sqrt{1a^2}1\rangle)\^2\]\[\ab+\sqrt{(1b^2)(1a^2)}\^2\]from which you get the same result

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0oh great first principle i get from the course thanks ca we continue with two more problems???

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0https://www.edx.org/c4x/BerkeleyX/CS191x/asset/hw2.pdf P 13 b and 14

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0oh great first principle i get from the course thanks ca we continue with two more problems???

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2I 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

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0okay but did you try and guess the answer for 13 b

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2yeah, I tried a few things, still don't quite understand how to get the second state since a00>+b00> was entangled and therefore not necessarily able to be decomposed... anyway shall we continue with P14 ?

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1it's pretty late around here ... 1:40 am ... i guess i'll have to start tomorrow ... is @TuringTest in the course?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2Heck yeah I am! I hope I survive it :) I really hope you join too @experimentX

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1lol ... 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
 2 years ago
Best ResponseYou've already chosen the best response.2Yeah, 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

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0lol yes thats good ,sry for the delay but is there time for us to do 14

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0there s also a openstudy specially for the course http://www.getstudyroom.com/course/43483#question43485

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0honestly 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
 2 years ago
Best ResponseYou've already chosen the best response.1I am with EM ... too

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0so for the edX the final exams it open book ???

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1yeah ... good think about it, And I am good at cheating.

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0sry i was actually asking if they allow

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0did you try the study app

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2Yes the final exams are openbook,but you are not supposed to discuss on OS for example. it's okay for homework though shall we do 14?

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2it'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

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0i tried to write \[\frac{3}{4}00+\frac{\sqrt {5}}{4}01\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0so we do thatdw:1361574751462:dw

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0since it is not normalised

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2rather, we divide the vector by \(\sqrt{a^2+b^2}\)

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0new state \[\frac{ 3 }{ 4 }\sqrt{\frac{7}{8}}00\frac{ 2\sqrt{5} }{ 7 }\sqrt{\frac{7}{8} }01\] P(00)=9/14

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0do we still call it 00,is this correct?

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0i will take this as notes lol

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0thanks @TuringTest @experimentX

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2yeah, i's still 00 and so far it is pretty easy; we'll see how long it stays that way

experimentX
 2 years ago
Best ResponseYou've already chosen the best response.1lol ... i didn't to a damn thing except troll around ... thanks to @TuringTest

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2:p well I stil want to know how 13b works... we'll have to find someone else I guess

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0well i guess we shud start with 13 a sincethe question lays ref to a

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0i see from the notes that this is the bell state \[\phi^+>=\frac{1}{\sqrt{2}}(00+11)\]

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2still not getting it :(

TuringTest
 2 years ago
Best ResponseYou've already chosen the best response.2\[\phi^+\rangle=\frac{1}{\sqrt2}(00\rangle+11\rangle\]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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