# Re: on formally indescribable ....

From: Russell Standish <R.Standish.domain.name.hidden>
Date: Mon, 19 Mar 2001 11:17:40 +1100 (EST)

George Levy wrote:
>
>
>
> Marchal wrote:
>
> > [BM] And it shows that COMP entails SE is SE is correct.
>
> It shows that COMP entails the general shape of SE but not SE proper.
>
> > [BM] I also use the COMP hypothesis which help by giving both TEs, and
> > precise mathematical theories. It is frequent that focusing on point
> > and focusing on links are related by dualities. With the Z logics the
> > links are really topological. I think I have shown a way to derive
> > the thickness from the turing-tropic views.
>
> I don't understand, but if you actually found a way to derive the "thickness" (
> which I guess is Planck's constant) from purely philosophical arguments, you
> deserve a Nobel Prize. However, until I see the proof, I strongly doubt that you
> have achieved such a feat.
>

The numerical value of Planck's constant is merely an arbitrary result
of how we define our units of length, time, mass etc. It is quite
common in theoretical physics to rescale these fundamental quantities
by setting h (or \hbar), c and G to unity, in which case the units are
Planck length, Planck time and Planck mass.

When I was writing my Occam's paper, I mentioned it to a friend, who
aksed whether I could derive Heisenberg's uncertainty principle from
first principles, and I replied I think I can.

The fact that one has a linear time evolution equation (Schroedinger
Equation) in a complex hilbert space is established in "Occam's
razor", with observables being linear operators (in my version, I use
projection operators, but that not necessary - a projection operator
has had it's eigenvalues set to unity by means of rescaling). Now
obviously some of those operator form noncommuting pairs. In
particular, the position operator must be {\bf x}, and momentum is
that operator that is translation invariant, i.e. d/dx (the i\hbar
multiplying factor is fixed by convention, it is not significant).

Now
(\Delta A)^2(\Delta B)^2 = <(A-<A>)^2><(B-<B>)^2> \geq <AB> - <A><B>

by the triangle inequality and by symmetry,
(\Delta A)^2(\Delta B)^2 \geq <BA> - <A><B>

Now, we also have |a-b|\leq max{|a|,|b|}, so we can take the
difference of the above inequalities
(\Delta A)^2(\Delta B)^2 \geq |<[A,B]>|

so if [A,B]=1, then \Delta A \Delta B \geq 1, which is Heisenberg's
uncertainty principle.

I have just reproduced a proof, I first saw nearly 20 years ago. Try
as I might, I haven't been able to track down a source of this proof though.

Cheers

----------------------------------------------------------------------------
Dr. Russell Standish Director
High Performance Computing Support Unit, Phone 9385 6967
UNSW SYDNEY 2052 Fax 9385 6965
Australia R.Standish.domain.name.hidden
Room 2075, Red Centre http://parallel.hpc.unsw.edu.au/rks
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Received on Sun Mar 18 2001 - 16:42:22 PST

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