atyy said:
But doesn't the definition of A and B being independent hold regardless of the tensor product structure and the Born rule?
Fredrik said:
Independence is the result P(a & b)=P(a)P(b). This result follows from the tensor product stuff. Perhaps it also follows from something else. This is why this argument doesn't prove that we need to use tensor products, and that's why the Aerts & Daubechies argument is so appealing. It does prove that (given some assumption that are rather difficult to understand), we have to use tensor products.
Judging from your quotes it seems you think there is a way to do probability without bilinear forms? But any probability P(a)
is a bilinear form, e.g.:
https://www.physicsforums.com/attachment.php?attachmentid=71661&stc=1&d=1406376963
(
Parthasarathry - Quantum Stochastic Calculus P. 1)
and because of this bilinear form-ness, P(a) = g(a,a) = <a,ψ(a)> immediately implies an isomorphism to a tensor product
https://www.physicsforums.com/attachment.php?attachmentid=71662&stc=1&d=1406377319
(Lang Linear Algebra 2nd Edition appendix)
so it looks to me like you're just unavoidably using tensor product structure no matter what you do, i.e. if you do something else or ignore the tensor product structure, it's there anyway (no matter what symbols you use), no?
I could be wrong, but it looks to me like you just can't use probabilities without using bilinear forms and tensor products. Once that is established, we add more structure when exhibiting something like
P(a)P(b) = g(a,a)h(b,b) = <a,ψ(a)><b,φ(b)> = ... = P(a&b)
it's going to
have to use more tensor products to give some meaning to whatever the & is supposed to mean (in the case of independence you exploit Pythagoras, but this special case lives in a structure built into bilinear forms which are an unavoidable consequence of introducing the very notion of probability).
If you admit the notion of probability exists, Parthasarathry's little calculation (or it in the limit) makes it completely obvious the Born rule will exist (if you've decided to base your mechanics on the notion of a state vector, which just falls out of the notion of probability...), and a similar point is made by Lubos:
Again, we need to know that mutually exclusive states are orthogonal and the probability has something to do with the length of a state vector (or its projection to a subspace).
That's everything we need to assume if we want to prove Born's rule.
...
That's the real reason why Born's rule works. The probabilities and mutual exclusiveness has to be expressed as a mathematical function or property of state vectors and the totally general rules for probabilities (like the additive behavior of probabilities under "or") heavily constrain what the map between the "human language" (probability, mutual exclusiveness) and the "mathematical properties" can be. The solution to these constraints is basically unique. The probabilities have to be given by the second powers of the moduli of the complex probability amplitudes. It's because only such "quadratic" formulae for the probabilities obey the general additive rules, thanks to the Pythagorean theorem.