Redshift Academy


Bell's Theorem
---------------



Let N be the number of objects.

N(A not B) + N(B not C) ≥ N(A not C)

This is generaly written as:

N(A ~B) + N(B ~C) ≥ N(A ~C)

Therefore,

N(1 + 2) + N(4 + 7) ≥ N(1 + 4)


Consider electrons 1 and 2 in the singlet state (1/√2)(|ud> - |du>.
Consider 3 possible orientations 0°, 45° and 90°

(z)
0°   45°
|   /
|  /
| /
 -------- 90° (x)


If the electrons are entangled we can view the spin of 1 always
being the opposite of 2 and vice versa.  Thus, if electron 1 is in
the up state then this is the same as saying that 2 is in the down
state and vice versa.

Returning to the above Venn diagram, we can define.

A is 1 up along 0° 

B is 1 up along 45°  ~B is 2 up along 45°

C is 1 up along 90°  ~C is 2 up along 90°

Substituting into Bell's theorem gives

N(1 up 0° 2 up 45° ) + N(1 up 45° 2 up 90°) ≥ N(1 up 0° 2 up 90°)

We can easily equate N with a probability (i.e. P = n/N_Total).  
In order tocompute these probabilities it is helpful to introduce 
the concept of PROJECTION OPERATORS.

Projection Operators
--------------------

A projection operator projects a vector onto a sub space.  Consider 
the following:



The vector, u, is defined in 3D space.  We want to find the linear 
operator that returns the projection of this vector onto the subspace 
with the orthonormal basis of {x,y}.

     - -         - -
    | 1 |       | 0 |
x = | 0 |   y = | 1 |
    | 0 |       | 0 |
     - -         - -

The projection operator that projects u onto the x-y plane is

P_xy = Σ|n><n| where n are the subspace basis vectors.
      ⁿ
So,

P_xy = |x><x| + |y><y|

      - -  -     -     - -  -     -
     | 1 || 1 0 0 | + | 0 || 0 1 0 |
   = | 0 | -     -    | 1 | -     -
     | 0 |            | 0 |
      - -              - -
      -     -     -     -
     | 1 0 0 | + | 0 0 0 |
   = | 0 0 0 |   | 0 1 0 |
     | 0 0 0 |   | 0 0 0 |
      -     -     -     -

      -     -
     | 1 0 0 |
   = | 0 1 0 |
     | 0 0 0 |
      -     -
                              - -
So clearly, if have a vector | x | then the projection onto 
                             | y |
                             | z |
                              - -
the x-y plane is

 -     -   - -     - -
| 1 0 0 | | x |   | x |
| 0 1 0 | | y | = | y | as expected
| 0 0 0 | | z |   | 0 |
 -     -   - -     - - 

If the space and subspace are the same, the projection operator 
is the unitary matrix, I, and the vector returned is just the 
original vector.

<b| => (b₁* b₂* b₃)  
       -  -  
|a> = | a₁ |
      | a₂ |
      | a₃ |
       -  -
          -  -  -          -      -               -
|a><b| = | a₁ || b₁* b₂* b₃* | =  |a₁b₁* a₁b₂* a₁b₃* |
         | a₂ | -          -   |a₂b₁* a₂b₂* a₂b₃* |
         | a₃ |                |a₃b₁* a₃b₂* a₃b₃* |
          -  -                    -               -   

Thus,
          -     -
|n><n| = | 1 0 0 |  where n is a basis vector of unit length
         | 0 1 0 |
         | 0 0 1 |
          -     -

|ψ> = Σa_n|n> where n are mutually orthogonal unit basis vectors
      ⁿ

<m|ψ> = Σa_n<m|n>     ...  <m|n> = δ_ij
        ⁿ

      = a_m

|ψ> = Σ<m|ψ>|n>
      ⁿ

First the bra vector dots into the state, giving the coefficient 
of |n> in the state, then it is multiplied by the unit vector |n> 
turning it back into a vector, with the right length to be a 
projection. 

We can write the above as,

|ψ> = Σ|n><m|ψ>
      ⁿ

An operator maps one vector into another vector, so this is an 
operator.

Finally we can rewrite as

Σ|n><n| = I when m = n and 0 otherwise.
ⁿ

The projection operators for spin can be written in terms of 
the σ matrices as follows:

                    -   -      -   -     -   -
P_+z = (I + σ_z)/2 = | 1  0 | + | 1  0 | = | 1 0 |
                | 0  1 |   | 0 -1 |   | 0 0 |
                    -   -      -   -     -   -
                    -   -      -   -     -   -
P_-z = (I - σ_z)/2 = | 1  0 | - | 1  0 | = | 0 0 |
                | 0  1 |   | 0 -1 |   | 0 1 |
                    -   -      -   -     -   -
                    -   -      -   -     -       -
P_+x = (I + σ_x)/2 = | 1  0 | + | 0  1 | = | 1/2 1/2 |
                | 0  1 |   | 1  0 |   | 1/2 1/2 |
                    -   -      -   -     -       -
                    -   -      -   -     -         -
P_-x = (I - σ_x)/2 = | 1  0 | - | 0  1 | = |  1/2 -1/2 |
                | 0  1 |   | 1  0 |   | -1/2  1/2 |
                    -   -      -   -     -         -
                    -   -      -   -     -        -
P_+y = (I + σ_y)/2 = | 1  0 | + | 0 -i | = | 1/2 -i/2 |
                | 0  1 |   | i  0 |   | i/2  1/2 |
                    -   -      -   -     -        -
                    -   -      -   -     -         -
P_-y = (I - σ_y)/2 = | 1  0 | - | 0 -i | = |  1/2  i/2 |
                | 0  1 |   | i  0 |   | -i/2  1/2 |
                    -   -      -   -     -         -

The projection operators provide an alternate way of calculating
the probability associated with a particular state.  Consider:
                    -  -
<ψ|P_+z|ψ> where ψ = | α |
                  | β |
                    -  -

Expanding we get:
      -   -  - -
[α^* β^*]| 1 0 || α | = α^*α 
     | 0 0 || β |
      -   -  - -

This is the probability associated with the +z direction.

Return to the singlet state and compute the probability P(A ~B)
(1 up 0° 2 up 45°).

<singlet state|(σ_z/2 + 1)(τ.n/2  + 1/2)|singlet state>

Where σ operates only on electron 1 and τ only operates on electron 2.
τ.n = τ_x/√2 +  τ_z/√2

<singlet state|(σ_z/2 + 1/2)((τ_x + τ_z)/2√2  + 1/2)|singlet state> 

Likewise, compute the probability P(B ~C) (1 up 45° 2 up 90°).

<singlet state|((σ_x + σ_z)/2√2  + 1/2)(τ_x/2 + 1/2)|singlet state> 

and for P(A ~C) (1 up 0° 2 up 90°).

<singlet state|(σ_z/2 + 1/2)(τ_x/2 + 1/2)|singlet state> 

The math to figure this out is somewhat tedious and will be
skipped in the interest of brevity.  What we find is:

P(A ~B) ~ 0.075

P(B ~C) ~ 0.075

P(B ~C) ~ 0.25

So this is a clear violation of Bell's Theorem!  A consequence of this
is that the hidden variables theory that says that each particle carries 
with it all of the required information at the time of separation,  and 
nothing needs to be transmitted from one particle to the other at the
time of measurement, cannot be true.