Implications of Dirac's Formula

Once for every Planck's constant of time the entire universe undergoes the big bang. This is the implication of Dirac's theory. That is 9.1 x 10 ^ 35 times per second (roughly) the entire universe undergoes an oscillation phase between matter and non-matter. (More correctly, it is a two-pole oscillation that goes: matter, non-matter, anti-matter, non-anti-matter. And even more correctly it is a 3-pole oscillation.......) Every quanta disappears virtually and then randomly appears simultaneously, or at least to the perspective of a (human, 3D) conscious observer.
Does this mean I can walk through a wall? Why yes, but it is extremely unlikely. A body of matter, or specifically in this case a human body is composed of a vast and unthinkable quantity of quanta. The probability is relative to this quantity that a spontaneous Kt-event may occur (like passing through a wall) for a large body object as described.
Yet it is still a matter of fact that a particular quantum within the body will randomly reappear perhaps a half-universe away, the probability that another random quantum will replace it increases with the mass of the object being observed. That is to say as we "zoom-out" to a larger frame of reference the probability that entirety of the object will "wink" out of existence decreases proportionally. Every particle in the universe experiences this oscillation of coming into being and destruction, it is simply the average of this activity which we experience as a conscious reality. What is important about this phenomenon as conscious observers is not the how or the why but rather the fact that it happens at all. What exists from a certain perspective is not "to be or not to be" but a collection of probabilities of existence in a large enough sample as to outweigh the odds of non-existence.

Methods for enumerating dimensional intersections

The 51st crease of the 6th to 9th dimensional vector.
Assuming there are 10 dimensions and we are assuming direct vector connection
between the 6th and 9th dimensions, as an edge of a 10-dimensional
hypercubic manifold, there are 10 minus 2 (the number of components in the
selected vector) factorial (for all permutations of intersection with
other 8 remaining dimensions by iterating through binary selection yeilds
the 51st datum of 8! items) 51 of 40320 with a lower bound of 0. 0 means
no intersections. The 51st binary permutation of the set {1, 2, 3, 4,
5, 7, 8, 10}. By choosing binary modality we get the enumeration 110011
or 32+16+3+1=51. The right-most binary numeral corresponds to the left-most
element of the set in one-to-one correspondance an intersection results in
the Turing subset {1, 2, 5, 7}.
The second method is iterative order, that is, to take all possible
single member sets (8 of them). Then 51-8=43. Then take all possible two
member sets which is 8*7 or 56. Since 43<56 therefore the 43rd member of
the set of two dimensional intersections, a set of sets: {{1, 2}, {1, 3},
..., {8,10}}. So the iterative answer is another Turing method:
1 {1,2}
2 {1,3}
3 {1,4}
4 {1,5}
5 {1,7}
6 {1,8}
7 {1,10}
This gets us to number eight so we can subtract n-1 from 43 until we get
a remainder and a quotient: 43-8=35, 35-7=28, 28-6=22, 22-5=16, 16-4=12,
12-3=9, 9-2=7, 7-1=6. So we result with the two-dimensional where the
members are the number of operations, the 8th, and the remainder, the 6
sixth. The result equals {A(6), A(8)} or {7, 10}. The 51st crease is an
intersection of the 6th, 7th, 8th and 10th dimensions.