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Sum of the probabilities equals 3 in bipartite covariance ?

  1. Apr 6, 2014 #1
    If we consider a bipartite system as in EPRB experiment we get the probabilities :

    p(++)=p(--)=1/4*(1-cos(theta))
    p(+-)=p(-+)=1/4*(1+cos(theta))

    p(+A)=p(+B)=p(-A)=p(-B)=1/2

    Thus the sum of all the probabilities equals 3...

    How does that come ? Is it because in fact there are only double events out of which we consider the averages of A and B sides too, thus making 3 sample set out of one experiment ?
     
    Last edited: Apr 6, 2014
  2. jcsd
  3. Apr 6, 2014 #2
    Can you be a bit more specific? What is p(+A) for example?
     
  4. Apr 6, 2014 #3
    p(+A) is the probability of measuring + for the A operator. A and B are just spin 1/2 operators and we measure a singlet state Psi=1/Sqrt(2)(0,1,-1,0)

    measuring A in this bipartite system means of course measuring [tex]A1=A\otimes\mathbb{1}[/tex].

    If I take A=diag(1,-1) [tex]B=\left(\begin{array}{cc}cos(\theta)&sin(\theta)\\sin(\theta)&-cos(\theta)\end{array}\right)[/tex] then A1 just has two eigenvalues 1 and -1. p(+A) is the probability of measurement for the eigenvalue 1 of A1.

    All this was to compute the covariance [tex]\langle A\otimes B\rangle - \langle A\rangle\langle B\rangle[/tex]

    Of course the average of A is p(+A)-p(-A)=0= average of B.
     
    Last edited: Apr 6, 2014
  5. Apr 6, 2014 #4

    naima

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    If 3 events A B and C are certain to occur we have
    p(A) = p(B) = p(C) = 1 and we have p(A) + p(B) + p(C) = 3.
    No problem.
    What do you want to do with that?
     
  6. Apr 6, 2014 #5
    I am not sure about specific probability of these operators but if I remember the theory of probability correctly,
    P(A)+P(B)+P(C) = 1 means the probability of occuring event A or event B or event C is 1. If probability of these events are interconnected it is always 1. If A, B and C are independent, dividing by total events will give 1 anyway.
     
  7. Apr 6, 2014 #6
    if 3 events are sure this means they happen as a whole and nothing else so we normally write p(abc) is 1 ?
     
  8. Apr 6, 2014 #7
    I would say P(A)*P(B)*P(C) = 1. It means the probability of occurring event A and event B and event C is 1 which means all three events will surely happen and as you can see each of these events need to have probability 1 individually in this case.

    The P(A) + P(B) + P(C) = 1 means either of these events will surely happen.
     
  9. Apr 7, 2014 #8
    but the axioms of probability say p(omega) is 1 where omega is the set of all events.
    I got it :
    In our case event A is +-,++,--,-+
    B is +a,-a
    C is +b,-b

    but B and C are not real events we deduce them from A, for example +a is the reunion of ++ and +-
    aso.
    Thanks, i have often dumb analyses.
     
    Last edited: Apr 7, 2014
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