Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Is it possible to create (m)any theoretically workable framework(s) - that do(es) produce probabilities - by taking QM and replacing the Born(-like) rule(s) with something that is not equivalent to it (them)? And, only if so, would that still be called a quantum theory (not necessarily physical such as QM) or does one call that something else?

share|improve this question
1  
What would you mean by "not equivalent to it"? If that means giving different answers, then you're in trouble, given the experimental evidential support that the current quantum theory enjoys. And if doesn't mean giving different answers, then why bother? –  twistor59 Jan 10 '13 at 20:40
    
If you take a rule and you replace it with something not equivalent by definition you won't get the same result. One can simply multiply all probability distribution by some normalised function and obtain something that makes different predictions... wrong ones. :-) –  Ebenezer Sklivvze Jan 10 '13 at 20:43

2 Answers 2

up vote 2 down vote accepted

I would imagine that this is not possible, at least not without making such radical changes to the other postulates that it can no longer be reasonably said to be "a quantum theory". The reason is that if you express quantum mechanics in terms of the density matrix formalism (rather than in terms of pure-state wavefunctions) then the Born rule becomes "the probabilities are the diagonal entries of the density matrix." (Or, more generally, "probabilities are the results of linear operators that project the density matrix onto $[0,1]$.")

The trace of a Hermitian matrix is preserved under unitary transformations, and it is this that leads to the conservation of probability in this formalism. I suppose you could try to change this rule to something else, like "the probabilities are the (normalised) cube roots of the diagonal entries of the density matrix", but the problem with this is that these new "probabilities" will no longer combine like probabilities when you combine density matrices.

So it seems that if you want to keep the ability to formulate quantum theory in terms of density matrices, it probably isn't possible to change the Born rule without messing with either (a) the rules of classical probability theory, or (b) unitary evolution. Changing either of these would be extremely radical, and would probably result in needing to change almost everything else as well.

share|improve this answer
    
OK, thanks! From your answer, I take it then that any variation with respect to the Born rule would (at least) require a change in the dimensionality of the amplitudes? Are real and complex numbers the only kinds of numbers which would then allow for matrix formulations, while preserving (the notion of) probability? Is there one more way, or even many more ways? What would such theories be called? –  Glen The Udderboat Jan 11 '13 at 8:50
    
I don't know much about it, but I know there are such things as "real quantum mechanics" and "quaternion (or quaternionic) quantum mechanics" (the latter is somewhat easier to Google for!). I guess in my answer I was thinking that you wanted to keep everything complex-valued, but only change the Born rule. I'm not 100% sure but I think in real and quaterionic QM, the Born rule is still the square of the magnitude, it's just a different thing that's being squared. –  Nathaniel Jan 11 '13 at 9:59
    
You're right, my comment was a follow-on question. I'll take your original "no" for an answer (at least until somebody disagrees convincingly). –  Glen The Udderboat Jan 11 '13 at 10:55

I object to the people who say if a theory doesn't give different results, "why bother"? It's become fashionable in physics to snicker at the idea of looking for physical meaning behind the equations; this same point is made as the ultimate identifying mark of a crackpot in John Baez's semi-satirical "crackpot index".

If this were really the mark of worthlessness in physics, why would we not snicker at Lagrangian Mechanics? Does it make any different "predictions" than Newtonian Mechanics? And if not, does that make it useless?

You cannot tell if one person is smarter by another by looking at the things those people claim to find interesting. In particular, the statement that you do not find a new theory "interesting" because it fails to make different predictions does not make you smarter than me.

Although I don't really understand it, I think that Bohmian mechanics strives to explain the evolution of systems without invoking the Born postulate. I'm not a fan of Bohmianism because I don't find it physically convincing; I lost interest in it when a Bohmian told me that one possible configuration of the hydrogen atom had the electron sitting stationary a fixed distance from the proton. But that's just me.

I believe that ultimately we have to be able to explain things in physics by the causal time-evolution of some kind of wave function in good old four-dimensional space time. I've succeded in creating causal wave-theoretic explanations of a handful of phenomenon which are traditionally held to be examples of wave-function collapse, including the appearance of flecks of silver on a photographic plate illuminated by the light of a distant star. You can read my explanation on my blog under the title "Quantum Siphoning".

share|improve this answer
1  
(Most of) this answer does not address the question as I meant to pose it. The question is about the mathematical possibility of certain quantum theories that do give different results, but which are still probabilistic. The question is also not about whether such theories might be scientific, only whether they would be consistent ("theoretically workable"). [The "quantum-interpretations" tag is not mine and maybe implies an unfortunate suggestion.] If BM doesn't invoke the Born postulate, how could it be varied with respect to it? [That is not necessarily a rhetorical question.] –  Glen The Udderboat Jan 11 '13 at 7:53

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.