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[I]f you want a universe with certain very generic properties, you seem forced to one of three choices: (1) determinism, (2) classical probabilities, or (3) quantum mechanics. [My emphasis.]

Scott Aaronson, Quantum Computing since Democritus

Aaronson then proceeds by arguing that there are only two theories that are "like" probability theory: probability theory itself and quantum mechanics. Probability theory is based on the $1$-norm, whereas quantum mechanics is based on the $2$-norm.

Call $\{v_1,\ldots,v_N\}$ a unit vector in the $p$-norm if $|v_1|^{\ p}+\cdots+|v_N|^{\ p}=1.$

The slide below is from a presentation of his.

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Now, it seems that there is yet another candidate out there: the $0$-norm. For this purpose one would have to define $0^0=0$, and, if one does, one has a theory "like" probability theory that is determinism.
($0^0$ is indeterminate, which, I understand, means that you can do as you please, as long as you do it consistently.)

It was remarked that the $0$-norm can be seen as a special case of both the $1$ and $2$-norms, in fact, that it can even be defined as the intersection between the two.

From there on I'm sort of questioning the original trilemma posed by Aaronson. It seems one can choose one of three possible realities; not the three that were mentioned, but rather options 1, 2 and 3:

$$\begin{array}{|l|c|c|c|} \hline &\text{Determinism}&\text{Classical probabilities}&\text{Quantum mechanics}\\ \hline \text{Option 1}&\color{green}{\text{True}}&\color{green}{\text{True}}&\color{green}{\text{True}}\\ \hline \text{Option 2}&\color{red}{\text{False}}&\color{green}{\text{True}}&\color{red}{\text{False}}\\ \hline \text{Option 3}&\color{red}{\text{False}}&\color{red}{\text{False}}&\color{green}{\text{True}}\\ \hline \end{array}$$

Scientific evidence favours quantum mechanics. That seems to rule out Option 2 (by, e.g., Bell's theorem), but, mathematically, doesn't seem to point to Option 3 only (meaning, of course, that I don't see it ruling out Option 1). (Bell's theorem is sort of empty under determinism.)

How does one rule out that reality can be described by Option 1, notwithstanding the fact that it's certainly easier (and more practical) to describe it by Option 3? In what sense is Option 3 the scientific one, if it is indeed the scientific one? (Some might say that quantum mechanics is deterministic, but then, according to the table, it is also classically probabilistic.)

Taking it possibly further then necessary: If, on scientific grounds, we can't make a distinction between Option 1 and Option 3, isn't the whole thing (i.e., the statuses of determinism and classical probability) just exclusively philosophical?

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Related question here, a little more elaborated on terminology and interpretations. –  Nikos M. Jun 9 at 19:13
    
Short answer, classical "kolmogorov-type" probabiility can handle and be used in quantum calculations (as in fact in other areas where dependent or in-dependent events and relations are studied in various combinations) –  Nikos M. Jun 9 at 19:15

3 Answers 3

Different types of theories are more suitable for describing different physical phenomena. The fact that you cannot use Newton's laws to describe an atom does not mean that you should not use them to describe your bicycle. Of course, some theories are more powerful than others. In fact, the three theories you mention form a hierarchy: $$ \text{deterministic} \subset \text{probabilistic} \subset \text{quantum} $$ Since quantum mechanics (in principle) can explain a wider range of phenomena, it is more powerful than the other two theories. However, it is also clear that quantum mechanics is not the final answer, since eventually it has to be somehow merged with general relativity. Thus, strictly speaking, none of the three theories tells you the whole truth, and you should include a fourth option that has False in all three columns.

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If the $0$-norm is the intersection of the $1$ and $2$-norms, how can the $1$-norm simultaneously be i) a proper subset of the $2$-norm, and ii) a proper superset of the $0$-norm? The hierarchy mentioned by you seems somewhat challenged by this. –  Glen The Udderboat May 2 '13 at 14:02
    
If you use only probability distributions where each probability is either 0 or 1, then that's just a deterministic state (indicator function) -- this explains the first inclusion. If you do quantum mechanics with diagonal density matrices, then that's just probability theory -- that's the second inclusion. Actually, I don't understand your table of options. Was it in the book or you came up with it yourself? –  Māris Ozols May 3 '13 at 11:52
    
(Actually, those two explanations only point to non-empty intersections, not inclusions.) It wasn't in the book, I cooked it up. But I was inspired by the first line of this answer. –  Glen The Udderboat May 3 '13 at 12:04
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I should have said: any deterministic state can be thought of a probability distribution and any probability distribution can be thought of a diagonal density matrix. That gives you the inclusions. –  Māris Ozols May 3 '13 at 13:12
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Not every probability distribution has only zeroes and ones; not every quantum state is diagonal. –  Māris Ozols May 3 '13 at 14:30

If one understands the 0-norm to be the intersection of the 1 and 2-norms, shouldn’t the trilemma really be the following (for any universe with very generic properties)?

1. Determinism, classical probabilities and quantum mechanics are all true.

2. Classical probabilities is true; the other two are false.

3. Quantum mechanics is true; the other two are false.

Thanks, that’s a very interesting and amusing way to put it! But I suppose it ultimately boils down to a matter of definition. Many people would interpret “Classical probabilities is true” to mean, not merely that all your states are probability vectors, but also that some states are nontrivial probability vectors. Likewise, they would interpret “Quantum mechanics is true” to mean that some states are nontrivial quantum superpositions.

One technical comment/correction: in case 2, the probabilistic case, you also shouldn’t rule out quantum mechanics being true—since maybe your states are “really” quantum mixed states whose density matrices just happen to be diagonal!

http://www.scottaaronson.com/blog/?p=1385#comment-73670

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It is certainly possible to scientifically distinguish between option 1 and 3 by either finding a deterministic hidden-variable theory underlying QM or proving that such a theory cannot be constructed. For example, if Bohmian interpretation can be extended to the relativistic regime (I know that there are proposals but I am not sure if they are accepted by the physics community as correct) option 1 is true and 3 is false.

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