Sum of the probabilities equals 3 in bipartite covariance ?

[jk22]

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 ?

 

[Jilang]

Can you be a bit more specific? What is p(+A) for example?

 

[jk22]

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 A1=A\otimes\mathbb{1}.

If I take A=diag(1,-1) B=\left(\begin{array}{cc}cos(\theta)&sin(\theta)\\sin(\theta)&-cos(\theta)\end{array}\right) 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 \langle A\otimes B\rangle - \langle A\rangle\langle B\rangle

Of course the average of A is p(+A)-p(-A)=0= average of B.

 

[naima]

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?

 

[amitbashyal]

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 occurring 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.

 

[jk22]

if 3 events are sure this means they happen as a whole and nothing else so we normally write p(abc) is 1 ?

 

[amitbashyal]

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.

 

[jk22]

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.