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Bell's Theorem in CHSH Form: An Interactive Derivation

Building

A proof against the Einstein Podolsky Rosen Paradox, including Bell's Theorem, plus subsequent work by CHSH and Tsirelson.

The ultimate goal of this project is to create an interactive webpage in the style of a Distill.pub article that rules out Local Realism in Quantum Mechanics. The article will use similar logical approaches to Bell’s Theorem and the works of John Clauser, Michael Horne, Abner Shimony, and Richard Holt (CHSH). Subsequently, we will go on to prove Tsirelson’s Bound.

Note: The website you're currently on is not the actual project. Once finished, the project will be linked at the top and bottom of the project proposal.

Learning Goals§

My aim with this project is to improve on various skills related to Math, Physics, and Computer Science:

  1. Formal mathematical writing skills
  2. Deepen understanding of different Quantum Mechanics concepts
  3. Simulation coding
  4. Derivation Skills
  5. Overall public writing skills

Derivation Goes Here:§

Classical:

Ai(an,λ)A_i(a_n, \lambda), Bi(bn,λ)B_i(b_n, \lambda)

E(an,bn)=1Ni=1NAiBiE(a_n,b_n) = \frac{1}{N} \sum_{i=1}^{N} A_iB_i

S=E(a1,b1)E(a1,b2)+E(a2,b1)+E(a2,b2)S = E(a_1,b_1) - E(a_1,b_2) + E(a_2, b_1) + E(a_2,b_2)

s=ABAB+AB+ABs = AB - AB' + A'B + A'B'

s=A(BB)+A(B+B)s = A(B - B') + A'(B + B')

Looking at BB and BB':

If B=BB = B', then BB=0B - B' = 0 and B+B=±2B + B' = \pm2

If BBB \neq B' then BB=±2B - B' = \pm2 and B+B=0B + B' = 0

In both scenarios, s=2|s| = 2

S=1Ni=1NAiBi1Ni=1NAiBi+1Ni=1NAiBi+1Ni=1NAiBiS = \frac{1}{N} \sum_{i=1}^{N} A_iB_i - \frac{1}{N} \sum_{i=1}^{N} A_iB'_i + \frac{1}{N} \sum_{i=1}^{N} A'_iB_i +\frac{1}{N} \sum_{i=1}^{N} A'_iB'_i

S=1Ni=1N(ABAB+AB+AB)S = \frac{1}{N} \sum_{i=1}^{N} (AB - AB' + A'B + A'B)

S=1Ni=1NsS = \frac{1}{N} \sum_{i=1}^{N} s

Since s=2|s| = 2, S=2|S| = 2

Quantum:

E(an,bn)=1Ni=1NAiBiE(a_n,b_n) = \frac{1}{N} \sum_{i=1}^{N} A_iB_i

E(an,bn)=limN1Ni=1NAiBiE(a_n,b_n) = \lim_{N \to \infty} \frac{1}{N} \sum_{i=1}^{N} A_iB_i

E(an,bn)=A,B±1AiBiP(A,Ba)E(a_n,b_n) = \sum_{A,B \in \pm 1} A_iB_i * P(A,B| a)