HeavyWater said:
Thank you PeterDonis And Nugatory. Clearly, what I need is a DEFINITION of an entangled state and it looks like Nugatory has given this forum a definition. Let me think on this for a bit. I learn best by examples...Are you saying that Positronium is an entangled state? What is Bothering me is that at infinity I end up with two separate wave functions that don't overlap...so I no longer have an entangled state.
The good thing with science is that you have to make the definitions really clear. To say "I have an entangled state" is not accurate enough. It doesn't tell you what's entangled. So we have to say what is entangled!
The most simple way to define it is to consider two compatible independent observables ##A## and ##B##, represented by self-adjoint operators ##\hat{A}## and ##\hat{B}##. Then there is a complete orthonormal set of simultaneous eigenvectors ##|a,b \rangle## of ##\hat{A}## and ##\hat{B}##, where for simplicity I assume that both ##\hat{A}## and ##\hat{B}## have only true eigenvectors and eigenstates (i.e., both have a discrete spectrum like, the spin-##z## operators of two particles). The meaning, according to Born's rule is: If the quantum system is prepared in a state represented by the Statistical operator
##\hat{\rho}=|a,b \rangle \langle a,b|rangle,##
then both observables take the determined values ##a## and ##b##, respectively.
Now we define any state that is a proper superposition of these basis states as a state, where the observables ##A## and ##B## are entangled. For simplicity we assume that the common eigenstates of ##\hat{A}## and ##\hat{B}## are only one-dimensional, i.e., the determination of both observables with certain values determine the state completely. Then, if you define
$$|\psi \rangle=\sum_{a,b} c_{ab} |a,b \rangle,$$
where
$$\sum_{a,b} |c_{ab}|^2=1,$$
and at least two of the ##c_{ab}## are not 0, the state represented by
$$\hat{\rho}_{\psi}=|\psi \rangle \langle \psi|$$
is a state for which ##A## and ##B## are entangled.
The fascinating thing is that ##A## and ##B## can be any observables of this kind you can define within QT. It can be two compatible observables of one particle. E.g., you can use position and spin-##z## component of a non-relativistic particle. With a Stern-Gerlach apparatus you can get states, where the position and spin-##z## component are entangled, i.e., if you have a beam of particles entering the inhomogeneous field of the Stern-Gerlach apparatus, after some traveling there the beam will split in two parts (position), and each partial beam consists of particles with a definite spin-##z## component. Take the original setup with spin-1/2 silver atoms. Then the wave function after running through the magnetic field is a superposition of exactly this form
$$\psi(\vec{x})=\psi_1(\vec{x}) |\sigma_z=1/2 \rangle + \psi_2(\vec{x}) |\sigma_z=-1/2 \rangle,$$
where ##\psi_1## and ##\psi_2## are wave packets (e.g., Gaussians) which are peaked narrowly in clearly distinct regions of space. The wave function doesn't describe particles with a well-defined position and spin-##z## component, but you have 100% correlation between position and spin-##z## component, i.e., if you find a particle in region 1 (region 2), then with 100% probality the spin-##z## component will have the value ##+1/2## (##-1/2##).
The two entangled observables can also be related to two different particles. Often you have experiments with two photons, whose polarization states are entangled. The fascinating thing with such a case is that the two photons may be detected on very far distant places but still can be in a polarization-entangled state. As in the case of the Stern-Gerlach example above the single-photon polarization can be completely indetermined (i.e., each of both observers just sees a stream of unpolarized photons when you send them such prepared photon pairs), but on the other hand they are 100% correlated, i.e., if you put the polarization filters for each of the photons in the same direction, always one will go through and then necessarily the other will be blocked and vice versa, when the state is given by the polarization-entangled state
$$|\psi \rangle=\frac{1}{\sqrt{2}} (|HV \rangle-|VH \rangle).$$
