Comparison between quantum entanglement and a classical version

[Adel Makram]

You will have something that is in a superposition of ud + du, true enough. But only at special angles. Those being related to whatever angle you filtered at. Any other angle will be some combination of uu + ud + dd +du.

So does entangled state!
If u, d represent basis vector of spin-up and spin-down, respectively, then entangled state has 4 components of various combinations of spin up and down of two particles at Alice and Bob places.

 

[Adel Makram]

You will have something that is in a superposition of ud + du, true enough. But only at special angles.

Also it s true for the frame of reference at this angle. For other angles, rotational transformation matrix is applied and yield similar coefficients to entangled state with sin²(θ₂-θ₁/2) and cos²(θ₂-θ₁/2) , where θ₁ and θ₂ are angles of Alice and Bob detectors, respectively. If θ₁=θ₂, the equation will reduced to the usual entangled state.​

 

[Adel Makram]

It will not simulate an entangled state at all. The stats may match at selectively chosen angles, but that's it. One is product state stats, the other is entangled state stats.

I got it at last.

Two particles with opposite spins after being filtered out will not simulate the entangled state.

Consider particle-1 state, P₁= 1/√2(U₊+U_-)
and particle-2 state, P₂= 1/√2(V₊+V_-)
where U₊, U_-, V₊, V_- are states of spin up and down for particle-1 and up and down for particle-2, respectively.
The state of two particles in opposite momenta after exisiting a filtering device is a product state,
lP₁>⊗lP₂>=1/√2[U₊V_-+U_-V₊]
All four possible states of two particles can be also represented onto the space of spin-up and spin-down at Alice and Bob devices.
U₊=cos(θ₁/2) u₁ + sin(θ₁/2) v₁
U_-=sin(θ₁/2) u₁ - cos(θ₁/2) v₁
V₊=cos(θ₂/2) u₂ + sin(θ₂/2) v₂
V_-=sin(θ₂/2) u₂ - cos(θ₂/2) v₂

lP₁>⊗lP₂>=1/√2[(cos(θ₁/2) u₁ + sin(θ₁/2) v₁)(sin(θ₂/2) u₂ - cos(θ₂/2) v₂) + (sin(θ₁/2) u₁ - cos(θ₁/2) v₁)(cos(θ₂/2) u₂ + sin(θ₂/2) v₂)]

after multiplying and collecting like terms,
lP₁>⊗lP₂>=1/√2{ [(cos(θ₁/2)(sin(θ₂/2)+(sin(θ₁/2)(cos(θ₂/2)] u₁u₂ + [-(cos(θ₁/2)(cos(θ₂/2)+(sin(θ₁/2)(sin(θ₂/2)] u₁v₂ +[(sin(θ₁/2)(sin(θ₂/2)-(cos(θ₁/2)(cos(θ₂/2)] v₁u₂ - [(sin(θ₁/2)(cos(θ₂/2)+(cos(θ₁/2)(sin(θ₂/2)] v₁v₂}

Bob should set his angle θ₂=π in order to get his spin up.
In doing so, only coefficients of u₁u₂ and v₁v₂ are non-zero. This explains why at those angles only the state may simulate the entangled state. For all other angles, it is not. And even though, the probability of both Alice and Bob spins-up in this product state is not equal to its counterpart in the entangled state.

So the probability of getting spin-up for Alice (θ₁=0) and for Bob (θ₂=π),
P_up,up= (1/√2[(cos(θ₁/2)(sin(θ₂/2)+(sin(θ₁/2)(cos(θ₂/2))² = 1/2 at (θ₁=0) and for Bob (θ₂=π)
And the probability of getting spin-down for Alice and Bob;
P_down,down= (-1/√2[(sin(θ₁/2)(cos(θ₂/2)+(cos(θ₁/2)(sin(θ₂/2))² =1/2 at (θ₁=0) and for Bob (θ₂=π)