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From: Russell Standish <R.Standish.domain.name.hidden>

Date: Mon, 19 Mar 2001 11:17:40 +1100 (EST)

George Levy wrote:

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

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

----------------------------------------------------------------------------

Received on Sun Mar 18 2001 - 16:42:22 PST

Date: Mon, 19 Mar 2001 11:17:40 +1100 (EST)

George Levy wrote:

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

----------------------------------------------------------------------------

Received on Sun Mar 18 2001 - 16:42:22 PST

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