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

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
dw:1361565948497:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yep!! but the dead line is 27 feb ... so i haven't looked into it.
 one year ago

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

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

TuringTest Group TitleBest 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
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
does that mean the result i just square it??
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
no, 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\[\text{Pr}(a,bc,d)=a^20\rangle+b^21\rangle\cdot c^20\rangle+d^21\rangle=a^2b^2+c^2d^2\]
 one year ago

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

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
is this our case?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
is this related to Week II Quantum Computing??
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
https://www.edx.org/c4x/BerkeleyX/CS191x/asset/31.pdf
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
yes lecture 3,4
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
well, 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
 one year ago

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

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

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
for 13a) i got 0.5
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
okay but did you try and guess the answer for 13 b
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
yeah, 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 ?
 one year ago

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

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
I am with EM ... too
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
so for the edX the final exams it open book ???
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
yeah ... good think about it, And I am good at cheating.
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
sry i was actually asking if they allow
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
did you try the study app
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
Yes 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?
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
yes we can do it
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
so we do thatdw:1361574751462:dw
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
since it is not normalised
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
exactly
 one year ago

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

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

Jonask Group TitleBest ResponseYou've already chosen the best response.0
do we still call it 00,is this correct?
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
wow its really easy
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
i will take this as notes lol
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
thanks @TuringTest @experimentX
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
lol ... i didn't to a damn thing except troll around ... thanks to @TuringTest
 one year ago

TuringTest Group TitleBest 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
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
well i guess we shud start with 13 a sincethe question lays ref to a
 one year ago

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

TuringTest Group TitleBest ResponseYou've already chosen the best response.2
still not getting it :(
 one year ago

TuringTest Group TitleBest 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?
 one year ago
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