Hidden Variable - How to calculate the straight line

Edgardo
Messages
706
Reaction score
17
There are plots that compare the expectation value E(a,b) of
(i) a hidden variable theory
(ii) quantum mechanics

For example here:
1. Talk given by Alain Aspect, video at 2m28s
2. PF Thread: The Unfair Sampling Assumption & Bell Tests

For the hidden variable theory the expectation value E(a,b) or E(theta) looks like a line. My question: How do you get this line?

--------

Let me further explain:

From what I understand the hidden variable is constructed as follows:
E(a,b) is given by:

E(\textbf{a},\textbf{b}) = \int \rho(\lambda) A(\lambda, \textbf{a}) B(\lambda,\textbf{b}) d\lambda

(Let's call this expression HV-Expectation)

where the vectors a and b are the directions of the polarizers.
(See Aspect talk at 0m54s)


Aspect asks us at 2m28s to construct a hidden variable and comes up with:

A(\lambda, \textbf{a}) = sign( \{cos(2(\theta_a-\lambda )\}
B(\lambda, \textbf{b}) = sign( \{cos(2(\theta_b-\lambda )\}
(Let's call them HV-equations)

where \theta_a and \theta_b are the angles of the vectors a and b with respect to some axis.

Now, I suppose that \lambda is the angle of a vector \vec{\lambda}. This vector \vec{\lambda} represents the hidden variable (sort of like a classical polarization). At least that's what I understood after reading Bell's original paper, see Eq (4).
(I also don't understand why Bell introduces the vector a').

Anyways, Aspect then explains that we get the line after plugging the HV-equations into HV-Expectation.

How is the calculation done?
 
Physics news on Phys.org
I have meanwhile found a nice explanation in the book Quantum theory: concepts and methods by Asher Peres, see page 161-162.

The hidden variable expectation value is:

E(\theta) = -1 + 2\theta/\pi

where \theta is the angle between the vectors a and b. This is indeed a line.
 
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!
Back
Top