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|dw:1361565948497:dw|

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

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

does that mean the result i just square it??

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

, or +

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?

is this related to Week II Quantum Computing??

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

yes lecture 3,4

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

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

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

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

I am with EM ... too

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

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

yes we can do it

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

exactly

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

do we still call it 00,is this correct?

wow its really easy

i will take this as notes lol

thanks @TuringTest @experimentX

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

lol ... i didn't to a damn thing except troll around ... thanks to @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
\[|\phi^+>=\frac{1}{\sqrt{2}}(|00+|11)\]

still not getting it :(