# Can the $0$-norm represent determinism?

In Scott Aaronson's Quantum Computing since Democritus, he presents classical probability theory as based on the $1$-norm, and QM as 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.

Now, he seems to claim that the $1$-norm and the $2$-norm are the only options here, although earlier in the book he allowed for three choices:

(1) determinism, (2) classical probabilities, or (3) quantum mechanics.

My question is: Would it be natural to let determinism be simply based on the $0$-norm? Is this problematic? Wouldn't the definition $0^0=0$ (for this purpose) and $p_i\in\{0,1\}$ suffice?

If so, what would be the equivalent to "probability vector" and "amplitude vector" be called? ("Indicator vector"??)

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Scott Aaronson's answer to a related question: physics.stackexchange.com/a/63756/17609. –  Glen The Udderboat May 8 '13 at 6:24

Yes, all of this works out. You just get finite sets and functions between them.

Incidentally, if you want an even more abstract approach to probability that unifies classical and quantum probability into one framework, check out noncommutative probability. I've written about this here and the specialization to the cases Aaronson describes is given here.

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Jeez, you're amazingly quick! And thanks for the extra links. Just to be sure, is the conventional name for a vector consisting of one $1$ and $0$'s otherwise indeed "indicator vector"? (Although I do realise that it isn't even necessary anymore, given your answer.) –  Glen The Udderboat May 1 '13 at 23:15
"[Y]ou define $x^0$ to be equal to $1$ if $x=0$ and $0$ otherwise." You meant this exactly the other way around, right? –  Glen The Udderboat May 1 '13 at 23:25
@Gugg: oops. Yes. I would use "indicator function." –  Qiaochu Yuan May 1 '13 at 23:32

Another way to view the 0-norm or determinism condition is as the intersection of the 1- and 2-norm conditions. In that respect it is not a different type of theory, only a special case, and it can be considered a matter of convention whether you count it as distinct given the other two.

Positivity assumptions are crucial in limiting the set of possibilities to two or three.

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Thanks! In order to understand your remark on positivity assumptions, should I be looking at positive-definite, positive, or totally positive? –  Glen The Udderboat May 2 '13 at 6:23
Or do you mean $|\cdot |^{\ p}$ should be constrained to being a positive real number? –  Glen The Udderboat May 2 '13 at 6:44
The second is automatic, I was thinking of $p_i$ being non-negative so as to be interpretable as frequencies, and the related condition that the sum is $1$. –  zyx May 2 '13 at 6:45
I'm (at least) slightly confused. $\alpha_i$ isn't constrained to being non-negative, right? And isn't $|\cdot |$ by definition non-negative and real? –  Glen The Udderboat May 2 '13 at 6:54
For example, you could form composite systems under the same rule as in QM (tensor product) but starting from solutions of $\sum p_i = 0$. –  zyx May 2 '13 at 17:44