Is there any mainstream or conventional view of what 'orthogonality' denotes for the functional equation, 'orthogonal additivity'.
Correct me if I am wrong but the only bases or rather events, which are logically orthogonal to one another on spin 1/2 systems are events on the very same basis.So that would be nothing more then probabilism
No other bases commute (there are plenty of geometrically orthogonal bases, and complementary ones, all of them being in the former camp)? So i presume that addition is literally going on between non-commutting bases, so that it must incorporate some kind or norm or geometric orthgonality as well as the logical notion (which appears to me to be just probabilism)?
I am mainly concerned as to whether orthogonal additrivity ; if orthogonality just means disjoint, THen i presume that its just an presumption of probabilism. BUt i presumned that repsresntability was not the issue for n=2 , but unique-ness. So i cant see how assuming probabilism a second time is going to lead to some form cauchy additivity. Unless I am mistaken about which events are allowed to add under orthogonal addivity (presumably not just events on the same basis) ; on spin 1 systems one can infer this due to the fact that every event is intertwined, through equalities with every other event the 1/3 1/3 1/3 events for instancce appears along 1/3 1/6 1/2,
but in a two outcome system/simplex , totally ranked and infinite in the vertical sense , the set of ranked element in the unit 1 simplex, quantum or not for any fixed probability p there is only one other event it can be connected to it via additivty to, the 1-p event. It can be connected to other events through equiprobability relations but they will only lie on the same space when the two probabilites are a half. And at best normalization will generalize to symmetry. which is stronger. FOr at least three outcomes complement additivity and normalization generalization to symmetry and the any arbitary three events (disjoint ) or not, same world/basis or not, which sum to one must have a function sum which sums to one
\forallx,y,z,in [0,1]=dom(F)x+y+z=1 iff F(x)+F(z)+F(y)=1
\forallx,y,z,in [0,1]=dom(F) x+y =1 iff F(x)+F(z)=1
where F"[0,1] to [0,1] ;
this is not the same thing as probabilism (additivity in the sense of kolmogorovov) it entails cauchy additivity F(x+y)=F(x)+F(y) with F(0)=0 F(1)=1 over the unit triangle where F is non negative. As far I can see the identity ; see gleason explained but the proof is relatively straight forward, one get the first equation due to matching off equalities between atomic events and disjunctive events or due to degenerate space such as (A1/3, B2/3 C0) and
F(0)=0 F(1)=1 and with x=y iff F(x)=F(y) and normalization so that A , B , C are disjoiint
F(1/3)+F(2/3)+F(0)=0=1 implise F(1/3)+F(2/3)=1 which gives you the analogue of complement additivity to get to symmetry any arbitrary two events whether on the triple or not which sum to one, not that due to the
ranking x=y iff F(x)=F(y )
F(0)=0as the impossible event ranked equal to every space natural empty set in chance , 0, x(0)=0 x(emptyest)=0 so x(0)=x(emptyset) thus due to the ranking equal chance iff equal frame frunction/credence x(0)=x(emptyset) iff F(emptyset)=F(0) due to the ranking ning but probabilism assigns the empty set , a value of0
ie F(emptyset)=0 and asx(0)=x(emptyset) and x(0)=x(emptyset) iff F(0)=F(emptyset),
F(0)=0 as well,
0 to
and thus x=0 = x=empty, by rnak F(emptyset)=F(0)m
and normalization of frame function probabilitis when disjoint and domain probabilities when disjunct that for whenever
x1+y1 =1 F(x1)+F(y1)=1
aso y=1-x,one knows that the sum of the two chances on each on the x space