AI.Mayank@MCC.ARPA (07/18/85)
From: Mayank Prakash <AI.Mayank@MCC.ARPA>
Doug,
I now have a very general demonstration of why your setup cannot lead to the
kind of paradox that you describe. We have the following situation -
A <--(1) E (2)--> B
-----> Spaceship
E emits two particles 1 and 2, which proceed towards observers A and B
respectively. We have an observable M of particle 1, N of particle 2, and an
observable Q of the entire system of the two particles. In order for the values
of M and N to be correlated, the following conditions must be met -
(1) Q is aconstant of motion. This is required because if the value of Q can
change over time, then the correlation is lost in time.
(2) it should satisfy a relation of the form Q = f(M,N), where Q,M,N are
regarded as operators in the Hilbert space of the system, and f is an operator
function.
Now, suppose we make a measurement of Q on the system before they are emmited
from E, and suppose we obtain the value q. Since Q is a constant of motion, its
value will remain the same during the experiment. Since M and N are observables
of two non-interacting particles, they commute with each other. Since M and N
commute, they can be diagonalized simultaneously. Let u(m,n) denote the
simultaneous eigenstate of M and N in which M has the value m (on particle 1)
and N has the value n (on particle 2). Then, a general state of the system is
of the form
sum over all (m,n) of a(m, n) u(m, n)))
(unfortunately, I cannot type in a sigma). The state in which the systenm was
found to be in the state with value q for the operator Q is givenby
PSI = sum over all (m,n) satisfying (f(m, n) = q) of a(m, n)u(m, n)
where a(m, n) are some coeficients such that the state PSI is normalized. Now,
we need some more notation - let S(m, q) and T(n,q) be the sets
S(m, q) = {n: f(m, n) = q}
and
T(n, q) = {m: f(m, n) = q}
Note that the two sets have the following relationship - if m is in T(n, q),
then n is in S(m, q), and vice versa. What we want are your distributions D1
and D2. First, assume that B makes the measurement of N on particle 2 first,
i.e., in your parlance, from the spaceship's point of view. Suppose the value
obtained is n0. The probability of this is
D2(n0) = sum over all m in T(n0, q) of ABS(a(m, n0))^2
Now, suppose that A makes the observation first (in your parlance, in the rest
frame's point of view), and obtains the value m0. Then, B can only obtain
values in the set S(m0, q). Suppose it is n0. The probability for that is
ABS(a(m0, n0))^2. But the value that A will obtain cannot be predicted
beforehand. In order that B obtain the value n0, A must again obtain values in
the set T(n0, q), and the probability for that happenning is simply
D1(n0) = sum over all m in T(n0, q) of ABS(a(m, n0))^2
Voila! The same as D2!
Hope that clears the confusion.
- mayank.
==========================================================================
II Mayank Prakash AI.Mayank@MCC.ARPA (512) 834-3441 II
II 9430 Research Blvd., Echelon 1, Austin, TX 78759. II
==========================================================================
-------
-------