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I know that the question "does Born's rule follow from Gleason's theorem" has already answers on the website: see here, and here. I am not satisfied with the answers given (one cannot rule out that I have not understood them) and think my question is not a duplicate (if it is, please show me).

I think that a (pretty obvious and rather trivial) derivation of Born's rule can be made from Gleason's theorem and a few assumptions; so, I think it would be easy for you to point out which assumption you would reject, based on physical or philosophical reasons.

Let us consider a fixed system.

  1. Assume that a (finite dimensional of dimension greater or equal than $3$) Hilbert space $\mathcal{H}$ is given.

  2. Assume that for every measurement device that has output set a finite set $S$, a family $(H_s)_{s \in S}$ of pairwise-orthogonal subspaces of $\mathcal{H}$ is given, called the eigenspaces of the measurement device.

  3. Assume that every finitely-indexed family of pairwise-orthogonal subspaces of $\mathcal{H}$ is associated to a measurement device.

  4. Assume that to each measurement device with output set $S$, is assigned a family $(p_s)_{s \in S}$ (that are thought as probabilities) of real, nonnegative real numbers that sum to $1$.

  5. Assume that whenever a subspace $H$ of $\mathcal{H}$ appears as an eigensubspace of two measurement devices, the probabilities in 4. are equal.

Consider, for each subspace $H$ of $\mathcal{H}$, define $\phi(H)$ to be the probability given by 4. to the corresponding output of any measurement device that has $H$ as eigenspace (there is at least one, from assumption 3.). I claim that this map $\phi$ is well-defined, because of 5., and from 3., it satisfies the hypotheses of Gleason's theorem.

Indeed:

  • Let $H$ a subspace. If $H$ is an eigenspace of two measurements devices, then the assigned probabilities are the same. So $\phi$ is well-defined.
  • For every subspace $H$, $\phi(S) \in [0,1]$ by definition.
  • Let us prove that $\phi$ is additive. Let $(H_1,\cdots,H_n)$ be a family of pairwise-orthogonal subspaces. We have to show that $\sum^n_{i=1} \phi(H_i) = \phi(\oplus^n_{i=1} H_i)$. Let $H := \oplus^n_{i=1} H_i$. For that, consider two measurement devices: the first has $\left(H_1,\cdots,H_n,H^\perp\right)$ as eigensubspaces, and the second has $\left(H, H^\perp\right)$. According to 4. and 5., $1 = \phi(H^\perp) + \sum^n_{i=1} \phi(H_i)$ and $1 = \phi(H) + \phi(H^\perp)$. Therefore, $\sum^n_{i=1} \phi(H_i) = \phi(H)$.

Therefore, there is a unique density matrix $A$ such that for each subspace $H$, $\phi(H) = Tr(AP_H)$; so, we have Born's rule.

I think that 1. and 2. are not the core of the problem; 3. is a technical assumption, and some strengthening of Gleason's theorem could allow a weaker assumption (but this is a math problem); 4. can be thought as an experimental assumption (these probabilities can be given as limits of experimental frequencies) or an ontological assumption; to me, the only assumption that stands out is 5 and I feel it is the only one that one should reject, but I am not sure everybody thinks this way.

My question is: can you indicate me which of these assumptions is controversial enough to make people think that Born's rule does not follow from Gleason's theorem? Why? Or do you think this argument is missing the point?

EDIT: I forgot an important point. When stating Born's rule as "the probability of such outcome of such measurement is given by such formula", do we assume that "probability" has a definite meaning? Of course, if "probability" has no meaning at all, then Born's rule is just a definition of probability, and there's nothing to derive. However, I do not find this very interesting. For me, all what is worth proving is the following statement:

Any assignment of real numbers to outcomes of measurements that can rightfully be called "probabilities", in the sense that they respect some formal conditions, must be given by Born's rule.

This is such a statement that "my" derivation proves.

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  • $\begingroup$ I believe there are two things here. First, it is not clear that in general (in fact, very likely a counterexample can be conceived with some effort) the assignment of the probabilities according to (4) and (5) extends to a well-defined assignment of probabilities to the entire family of the projection operators on your Hilbert space (what you'd need to have in order to start with the conventional Gleason theorem). But, suppose you get lucky and it is well-founded, somehow, say using the data collected from a real physical experiment(s). $\endgroup$ Commented Feb 15, 2022 at 17:52
  • $\begingroup$ Then, second, I believe you implicitly used the Born rule when you identified the probabilities (defined somehow, or collected from the physical experiment) with projection operators in (4) and (5). So, even if in the end you have a well-defined probability measure on the family of the projection operators that you know admits the representation according to the Gleason theorem, then saying $\phi(H) = Tr \, A \, P_H$ proves the Born rule is just saying what you had already assumed. $\endgroup$ Commented Feb 15, 2022 at 17:53
  • $\begingroup$ I don't understand your first objection: in my post, I claim to give a proof that 4. and 5. give an assignment of probabilities on the set of projection operators. If you want, I can develop a bit. $\endgroup$
    – Plop
    Commented Feb 15, 2022 at 17:59
  • $\begingroup$ I don't understand your second objection either. The numbers $(p_s)_{s \in S}$ associated to outcomes of a measurement are just numbers that are part of the data. Whether they really are "probabilities" or how where do they come from is not relevant, to my opinion. I'll edit my question to explain further this point. $\endgroup$
    – Plop
    Commented Feb 15, 2022 at 18:04
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    $\begingroup$ I haven't carefully looked at whether your question is a dupe or not (will do so later) but the point is that once you postulate that the measurement outcomes are probabilistic, it's not at all an issue to arrive at the conclusion that the Born rule is the only consistent probability measure to be assigned. The issue is that you can't get this postulate of probabilistic measurement outcomes without inserting it by hand -- because the rest of the quantum mechanics is unitary/deterministic. $\endgroup$
    – user87745
    Commented Feb 15, 2022 at 19:19

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