On Wed, Jul 19, 2017 at 12:03:24PM +1000, Bruce Kellett wrote:
> On 6/07/2017 5:55 pm, Russell Standish wrote:
> >On Thu, Jul 06, 2017 at 04:18:49PM +1000, Bruce Kellett wrote:
> >>On 6/07/2017 2:33 pm, Russell Standish wrote:
> >>>Establishing linearity is key.
> >>Yes, and you haven't made progress with that.
> >All I ask is to give me some more time on this. I have some further
> >ideas in this regard, but need some dedicated time to think about it.
>
> I have been thinking about it as well. I think your problem is even
> more difficult that just establishing that the sum of two observer
> moments is also an observer moment. If OMs are to form a linear
> vector space, they have to satisfy some further axioms:
>
> The axioms of associativity and commutativity are fairly easy if you
> have additivity, but the existence of a zero vector, 0, such that V
> + 0 = V for any vector V in the space, and the existence of an
> inverse, -V, s.t. V + (-V) = 0, might be more difficult in terms of
> OMs. What is a null OM? What is an inverse (negative) OM?
>
> I think you need these properties as well as additivity in order to
> have a vector space. At least, these are among the axioms for a
> vector space as listed by Wikipedia.
>
> Bruce
>
0 will correspond to the universal set, either the everything/nothing
object, or maybe the reference OM in the reduced space described in my
appendix.
The inverse vector corresponds to the complement.
You would be right if the zero vector corresponds to the
everything/nothing object, as that is clearly not an OM. If it is the
reference OM, then I'm not sure the Born rule derivation goes
through as given.
At first blush, the complement is unlikely to be an OM, but given the
duality between a set and its complement, we can invoke Leibniz's
indiscernibles principle, and identify a set with its complement.
In any case, I think you have a valid point that a linear combination
of observer moments is not in general an observer moment - for one
thing, the linear combination of all observer moments (ie the "3p") isn't.
What I'm trying now is considering more primitive objects, namely all
the subsets of the everything/nothing (ie the set [0,1]∞) constructed
from a set of "atomic" subsets. I have found a natural encoding of the
sets as complex vectors (well rays, really, as multiplying by an
arbitrary complex prefix doesn't change the underlying subset).
The nice thing is that this formulation naturally requires a complex
field - the real field doesn't cut the mustard. This addresses my
biggest concern with my QM derivation. It also exhibits the
relationship:
\P_A + \P_B = \P_{A∪B} + \P_{A∩B}
I haven't formally proved it, but it works on a number of nontrivial
test example I tried by hand. I'd like to see if I can shoot it down
before spending a lot of time proving theorems.
It remains to be seen how the Born rule fares - the sets in question
have a natural measure, so one ought to be able to derive a formula
describing the Born rule. However, the vector space {\P_A} does not
have an obvious inner product defined...
Anyway, I'll post more on this topic to give you guys a chance to take
your own potshots when I get time - I'm still travelling for the next
couple of weeks, so even finding time to write this stuff up in a post
is daunting ATM.
--
----------------------------------------------------------------------------
Dr Russell Standish Phone 0425 253119 (mobile)
Principal, High Performance Coders
Visiting Senior Research Fellow [email protected]
Economics, Kingston University http://www.hpcoders.com.au
----------------------------------------------------------------------------
--
You received this message because you are subscribed to the Google Groups
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To post to this group, send email to [email protected].
Visit this group at https://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.
