imsmooth said:
I know there are numerous threads on this and I have read quite a bit such as EPR and Bell's inequality. I hope I can ask this the right way:
A particle has 0 spin and gives off two children particles with spins -1/2 and +1/2 (we don't know which is which yet, or they have to end up this way when we know one of them). If you measure one then you know the value of the other, so where is the mystery? They both had to have opposite values from the start. Whatever value I get for particle 1 then particle 2 must have the opposite value.
Whatever way I orient my detectors, and even if I get a higher than 50/50 value for the first, by orienting my detector the same way for particle 2 I should get the opposite value, so what am I missing?
You are completely right. There's nothing mysterious, and EPR were flawed. It's one of the most overrated papers in the history of physics, and Einstein himself didn't like the way Podolski has written it ("burying the main argument under erodition"). Einstein was more concerned about "inseparability", implied by entanglement than anything else. Today we know from very accurate observations that any local determinstic hidden-variable theory, implying what's known as "Bell's inequalities" and related other relations of subsequent work, are wrong, and QT is right.
Your case is particularly illuminating and goes back to Bohm as an example for the typical EPR setup. Consider, e.g., the decay of a neutral pion (pseudoscalar particle) at rest to an electron-positron pair. Neglecting the inevitable uncertainty in momentum the idealized state of this electron-positron pair is an entangled state of the form
$$|\Psi \rangle = \frac{1}{\sqrt{2}} (|e^-,\vec{p},\sigma_z=1/2 \rangle \otimes |e^+,-\vec{p},\sigma_z=-1/2 \rangle - |e^-,\vec{p},\sigma_z=-1/2 \rangle \otimes |e^+,-\vec{p},\sigma_z=+1/2 \rangle).$$
If you wait long enough (and supposed you are working in empty space so that nothing interacts with the electron or positron before hitting the detectors) you can measure the spin of the electron in some direction (usually the ##z## direction) and that of the positron in the same direction at as far distant places as you like. In the ideal case you have a big spherical detector around the decaying pion so that you register all electron-positron pairs.
The above state indeed implies that if you find the electron to have ##\sigma_z=+1/2## then necessarily you'll measure ##\sigma_z=-1/2## for the positron. There's nothing mysterious in this but very simply explained by the fact of angular-momentum conservation: The pseudoscalar pion at rest must decay into a state of total angular momentum 0, and that uniquely determines the above given spin-singlet state for the electron-positron system. The same is true for momentum.
Note that though the spin states of both the electron and the positron are maximally undetermined, there's still 100% correlation between the ##\sigma_z##-spin components of the particles, and this correlation is stronger than possible within any local deterministic hidden-variable model. Measuring spin components in different directions choosing the right relative angles between the corresponding spin directions shows the violation of Bell's inequality, valid for any such hidden-variable model, by QT, and this violation is very accurately observed (I'm not sure whether this has ever been done with entangled particles from particle decays, but it's regularly done with polarization-entangled photons).
Also there's no problem with the original EPR argument since what they describe is a measurement of the two compatible observables ##\vec{x}_{e^-}-\vec{x}_{e^+}## and ##\vec{P}=\vec{p}_{e^-} +\vec{p}_{e^+}##. Due to the entanglement in momentum (which transfers to an entanglement in position, if you reexpress the above state in position rather than momentum eigenstates) you need to measure the momentum (position) of one of the particles and this implies the momentum (and position) of the other particle.