## Jonask 2 years ago qubit measurement-quantum mechanics

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

3. 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

|dw:1361565948497:dw|

5. experimentX

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

6. TuringTest

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

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

8. 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

does that mean the result i just square it??

10. 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

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

, or +

13. 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$\text{Pr}(a,b|c,d)=a^2|0\rangle+b^2|1\rangle\cdot c^2|0\rangle+d^2|1\rangle=a^2b^2+c^2d^2$

14. TuringTest

typo, I meant$a^2c^2+b^2d^2$

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

is this our case?

17. experimentX

is this related to Week II Quantum Computing??

yes lecture 3,4

20. 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

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

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

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

for 13a) i got 0.5

25. 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

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

27. 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 ?

28. experimentX

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

29. TuringTest

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

30. 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

31. 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

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

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

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

35. experimentX

I am with EM ... too

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

37. experimentX

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

sry i was actually asking if they allow

did you try the study app

40. 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?

yes we can do it

42. 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

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

so we do that|dw:1361574751462:dw|

since it is not normalised

46. TuringTest

exactly

47. TuringTest

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

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

do we still call it 00,is this correct?

wow its really easy

i will take this as notes lol

thanks @TuringTest @experimentX

53. TuringTest

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

54. experimentX

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

55. TuringTest

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

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

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

59. TuringTest

still not getting it :(

60. 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?