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#58: Quantum Mysteries - II

You remember the video I made about the gravitron of the earth direction? 

And how it can match even with the "systemic gravitron" ...?


And .. you do recall all the assertions made (by the "established kairos elites", the "only legit quantum 7" etc.)... 

about how "h is insignificant, i is unreal?"

Update on all that!

I was just solving a question today (for money....  :o ), that I stumbled across the exact meaning of "entanglement", about how i = Q, h = P (or i = P, h = Q) is possible (real, significant), i.e. a stable earthlings' gravitron that can help earthlings break on through to the other side, past the middlemen siege.

(It exploits the strong Q/17 (as in Quora symbol, Q ) red CT and the "P" (Cain CT), so it involves Q and P).

So here is the question:


And here is the answer I wrote, about how the "classical (quantum) assumption" is wrong, 

... there's no strict requirement of "instruction sets", and thus the i = 9 pairs cannot be "ruled out"..


Answer:

According to the classical assumption:
The four possible types of particles mentioned in C, are the ones that give the following test outcomes:


  P  ......  A: 1  B: 1
  Q  ......  A: 1  B: -1
  R  ......  A: -1  B: 1
  S  ......  A: -1  B: -1

The 16 possible pairings in the 2 particle system, are:

Particle
 12

 PP
 PQ
 PR
 PS
 QP
 QQ
 QR
 QS
 RP
 RQ
 RR
 RS
 SP
 SQ
 SR
 SS

However, experimental result show that Q (where B = -1) and S (where B = -1)  do not occur together.
Because (4) holds:
prob (B(1) = -1) /\ (AND) prob (B(2) = -1 .... is 0.
Thus we can rule out

QS
SQ

Because
B(1) = -1 if particle 1 is Q or S
B(2) = -1 if particle 2 is Q or S

Similarly, from (3), B(1) = 1 /\ A(2) = 1 is not observed in the experiment results, it has zero probability. Thus we can rule out

PP
RP
PQ
RQ

Because
B(1) = 1 if particle 1 is P or R
A(2) = 1 if particle 2 is P or Q


Similarly, from (2), A(1) = 1 /\ B(2) = 1 is not observed in the experiment results, it has zero probability. Thus we can rule out

PP
PR
QP
QR

Because
A(1) = 1 if particle 1 is P or Q
B(2) = 1 if particle 2 is P or R


So according to the classical assumption C, we can rule out 9 pairs, namely:
QS
SQ
PP
RP
PQ
RQ
PR
QP
QR


But the first experimental result says, prob (A(1) = 1) /\ (A(2) = 1 has a greater than 0 probability.

A(1) = 1 if particle 1 is P or Q
A(2) = 1 if particle 2 is P or Q

This is satisfied by only 2 pairs:  PQ and QP

However, both PQ and QP belong to the 9 pairs which are "ruled out" by the classical assumption C, thus should have 0 probability; but (1) holds PQ and QP are observed, thus C is logically inconsistent with observed results.


Thus!

... there's no strict requirement of "instruction sets", and thus the i = 9 pairs cannot be "ruled out" (at least 2, QP and PQ, so this makes the 2 forms of twin flames, out of which the sky girl, earth male type is rarer but more solid! ).

 

Oh by the way, the trend on the last post continues :) 

See the "16 types of pairs" above....  so are there "some 16 quantum pairs"? I mean a coincidence, 16 before her... 

 



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