Adel Makram said:
I don`t understand this. For if Alice has to imply that, it would mean the particle has a definite spin at the time of creation which contradicts the collapse theory which says the particle has no definite state of spin until it is measured.
Okay, let me try to get more rigorous. The way that QM predicts probabilities is this: The probability of getting a particular result is the absolute square of the amplitude for getting that result. So how does QM predict amplitudes? It's like this:
The amplitude for a result R is given by: \sum_\alpha C_\alpha \psi(R|\alpha)
where C_\alpha is the amplitude for being in initial state |\chi_\alpha\rangle, and \psi(R|\alpha) is the amplitude for getting result R given that the system starts off in initial state |\chi_\alpha\rangle. (|\chi_\alpha\rangle is any complete set of states)
You have complete freedom for choosing your complete set of states |\chi_\alpha\rangle. I might as well choose the following basis:
- |\chi_1\rangle = |\vec{a'}\rangle |\vec{a'}\rangle (Alice's particle and Bob's particle are both initially spin-up in the a'-direction)
- |\chi_2\rangle = |\vec{a'}\rangle |-\vec{a'}\rangle (Alice's particle is initially spin-up in the a'-direction Bob's particle is initially spin-up in the negative a'-direction.)
- |\chi_3\rangle = |-\vec{a'}\rangle |\vec{a'}\rangle (Alice's particle is spin-up in the negative a'-direction and Bob's particle is spin-up in the a'-direction)
- |\chi_4\rangle = |-\vec{a'}\rangle |-\vec{a'}\rangle (Alice's particle and Bob's particle are both initially spin-up in the negative a'-direction)
Because for spin-1/2 EPR, the total spin is zero, we can immediately compute the amplitudes C_\alpha:
C_1 = 0
C_2 = \frac{1}{\sqrt{2}}
C_3 = -\frac{1}{\sqrt{2}}
C_4 = 0
(The overall phase is unobservable, so we're free to choose C_2 to be real and positive. This forces C_3 to be negative and real in order for the total spin to be zero.) So our probability amplitude simplifies to:
\frac{1}{\sqrt{2}} \psi(R|\alpha=2) - \frac{1}{\sqrt{2}} \psi(R|\alpha=3)
Now, if the result R is that Alice measures spin-up along axis \vec{a}, while Bob measures spin-down along axis \vec{b}, then we can write:
\psi(R|\alpha=2) = \psi_A(\vec{a}|\vec{a'})\psi_B(\vec{b}|-\vec{a'})
\psi(R|\alpha=3) = \psi_A(\vec{a}|-\vec{a'})\psi_B(\vec{b}|\vec{a'})
where \psi_A(\vec{a}|\vec{z}) is the probability that Alice's particle will have spin-up in the \vec{a} direction given that it initially had spin up in the \vec{a'} direction, and \psi_B(\vec{b}|-\vec{a'}) is the probability that Bob's particle will have spin-up in the \vec{b} direction given that it initially had spin up in the -\vec{a'} direction. And similarly for the other terms.
What I was assuming in my first post was that there was a unique direction \vec{a'} such that \psi_A(\vec{a}|\vec{a'}) = 1 and \psi_A(\vec{a}|-\vec{a'}) = 0
