Is Conservation of Momentum Overlooked in Bells Inequality Violations

morrobay
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In the EPR scenario the correlation results are explained with the conservation laws of classical mechanics as applied to spin. The Bell type inequalities are derived on expected spin values.
But the violations of these inequalities are then explained with QM: That simultaneous knowledge of mutually non commuting observables is forbidden, collapse of wave function , non locality ...
If there were a more complete understanding of conservation of momentum as applied to the physics of spin with non- parallel detector settings could violations of the inequalities be explained classically ?
 
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morrobay said:
If there were a more complete understanding of conservation of momentum as applied to the physics of spin with non- parallel detector settings could violations of the inequalities be explained classically ?

What's incomplete about our understanding of conservation of momentum in these situations?
 
morrobay said:
In the EPR scenario the correlation results are explained with the conservation laws of classical mechanics as applied to spin. The Bell type inequalities are derived on expected spin values.
But the violations of these inequalities are then explained with QM: That simultaneous knowledge of mutually non commuting observables is forbidden, collapse of wave function , non locality ...
If there were a more complete understanding of conservation of momentum as applied to the physics of spin with non- parallel detector settings could violations of the inequalities be explained classically ?

the spins come opposite (in case of anti-correlated)...momentum is conserved

why do you feel it's overlooked?
 
Momentum is conserved with this inequality with perfect anti correlation at parallel detector settings
n[x-z+] + n[y+z-] ≥ n[x+y+] The spin values derived from entangled particles created at the source with total spin, when measured at A and B, equal to zero.
In accord with the conservation laws.
One of eight:
..A::::::::::::::B..
x y z :::::::::x y z
+ + -::::::::::- - +
But when the above inequality is violated with non parallel settings the total spin at A and B does not equal zero, in conflict with conservation laws. So the question is about the interaction at detector with particle at time of measurement. Given that particles were created with total spin of zero at source and the inequality was violated (spin change ) Then can spin change be related to an interaction with the detector and conservation of momentum for total system ? Again, with non parallel detector settings can the spin be affected by the conservation laws in such a way that the inequality is violated ? And if so then is a particle or classical physics mechanism the explanation?
 
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As an example for the above, with a 4 out of 8 table of spins with parallel detector settings
A:::::::::::::::B
x y z:::::::x y z
+ ++:::::::- - -
- + +:::::::+ - -
- - +:::::::+ + -
- - -:::::::+ + +
n[x+y-] + n[y-z+] ≥ n[x-y-] The Bell type inequality with expected spin values is applied in experiments when detector settings are not aligned. If only the spin on axis for x in first term of inequality:n[x+y-] changes spin to n[x-y-] during measurement due to particles interaction with magnetic field in detector and angle of detector, then the inequality is dis proven. Maybe a particle physicist can explain a mechanism for this spin change in terms of state of particle, detector angle/field interaction and the conservation laws.
 
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morrobay said:
... But when the above inequality is violated with non parallel settings the total spin at A and B does not equal zero, in conflict with conservation laws. ...

Ah, this is not quite correct. There is conservation for the system of A+B. This system no longer exists after one is measured, say you found A=-1 at 0 degrees. You now know B=+1 at that same angle. But instead you measure B at 15 degrees. The answer may or may not be +1, exactly as you imagine.

But now the question is: is there conservation of spin on a single particle? So you are really asking what happens to the spin of the observation apparatus when a particle's known spin is measured on a different basis. Ie is that somehow affected when it comes into contact with B? I would answer: sure, precisely *because* there is conservation.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
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