Anyways. Suppose you want to characterize what observables you can extract from the correlations among the three observables each has access to. Based on what I just said, that amounts to characterizing the vector (C(a,b),C(b,c),C(a,c)) where a,b,c are the three allowed measurement directions
If I label that as (x,y,z) for short, it turns out that they must satisfy $x^2+y^2+z^2\leq 2xyz+1$
Now you can ask what correlations you could produce classically, if you assume that all three spin observables had well-defined values even when you don’t observe them
And it works out to be nice: it’s just a tetrahedron. Specifically, it’s got vertices at (1,1,1), (-1,-1,1), (-1,1,-1), and (1,-1,-1)
Which you can check is embedded nicely inside that quantum set
Now, suppose you want to generalize this setup to involve particles of higher spin eg s=1
On the quantum side, nothing much changes: the same proportionality result I gave earlier occurs, just with a different constant
Things are a bit dicier on the classical side: you need to impose some extra restrictions to make sure you get the same variances as in the quantum case
But once the dust settles (and this is what I’ve been working on showing) you get another polytope embedded in the quantum set
It’s a polytope with more facets, but it’s still a polytope
hold up, before the classical version would have been "my spin is a point on $S^2$, i send a probablitly distribution p(x_1,x_2) with domain $S^2\times S^2$"
So that’s 2^3 options. However, you require that each observable be unbiased ie +1/2 as often as -1/2. So you effectively just need to choose four probabilities
Ok, for spin 1/2 you have classically: A (symmetric around $0$) probability distribution on $\{\pm\}^3\times \{\pm\}^3$ which you can tune as you like. The most possible correlation is given by XXX
It probably should start at $1/2^n$ and end with $1/2^{m-1}$, @maths, but you skipped the important part. So you know how to get that as an upper bound on $|s_m-s_n|$? Oh, you're assuming $m<n$ instead of $m>n$. I see.
Oh, ok, so you do understand that. Now you're almost done. How do you find the geometric series $\sum\limits_{k=m}^{n-1} 1/2^k$? Don't you know a formula?
maybe, but thats subjective. I understand how to derive the formula for the geometric series, but i'm just confused about the limits of the sum thats all
@TedShifrin yeah. It's the convex hull of eight points: $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1),(-1/2,-1/2,-1/2),(-1/2,1/2,1/2),(1/2,-1/2,1/2),(1/2,1/2,-1/2)$
so basically you take a tetrahedron and add an extra point just above the center of each base, splitting each tetrahedral face into three triangular facets
The trouble is that the way I did the facet enumeration step was rather laborious. I need to get a version of it which I can just plug-and-play in mathematica
There's some cool stuff on projective geometry and computer graphics, some differential equations, some complex matrices, and some applications (like graph theory). It's meant for a less sophisticated audience.
If I tell you I have the identity map on $\Bbb R^3$, I can write down lots of matrices and you won't know the bases (unless you know they're the same in both domain and range).
Start with my (free) text, @maths. Freely linked in my profile.
