Suppose that I accept that there is wave function collapse in quantum mechanics, and that the probabilities associated with each orthogonal subspace are a function of the wave function $\psi$ before the collapse.
I've seen some references that claim that in this case, Gleason's theorem implies that the probabilities are given by Born's rule, that is, by the squares of the absolute values of the amplitudes of $\psi$ (here is one such reference).
Loosely speaking, Gleason's theorem asserts that for any probability measure $\mu$ on a Hilbert space $\mathcal{H}$ (I mean, in the quantum sense, where $\mu$ is defined on subspaces of $\mathcal{H}$, and is additive under the sum of orthogonal subspaces) there is a state $\phi\in\mathcal{H}$ (more correctly: a density matrix) such that $\mu$ can be expressed by Born's rule using $\phi$.
I'm trying to understand how Gleason's theorem implies Born's rule. In other words, why is the $\phi$ in the theorem identical to $\psi$? Would there be any contradiction if for a state $\psi$ the probabilities were given by the forth powers of the amplitudes of $\psi$? I understand that in this case, $\psi\neq\phi$, but is there any problem with this?
Here is a related question, but it seems to me that it discusses a different issue - of how probabilities emerge in the many worlds interpretation.
Answer
I think I have understood your question now (and I deleted my previous answer since it actually referred to the wrong question). Let me try to summarize.
On the one hand we have a wavefunction $\psi$ in the Hilbert space $L^2(\mathbb R)$ for a given quantum system $S$ and we know that $\psi$ determines the state of $S$ in some (unspecified) sense: it can be used to extract transition probabilities and probabilities of outcomes when measuring observables.
($\psi$ could arise from some analogy optics - mechanics and can have some meaning different from that in Copenaghen intepretation, e.g. a Bohmian wave.)
On the other hand we know, from Gleason's theorem, that an (extremal to stick to the simplest case) probability measure associated to $S$ can be viewed as a a wavefunction $\phi \in L^2(\mathbb R)$ and Born's rule can be now safely used to compute the various probabilities of outcomes.
You would like to understand if necessarily $\psi=\phi$ up to phases as a consequence of Gleason's theorem.
Without further requirements on the procedure to extract transition probabilities (you only say that transitions probabilities can be extracted from $\psi$ with some unspecified procedure) it is not possible to conclude that $\psi=\phi$ up to phases in spite of Gleason's theorem.
We can only conclude that there must be an injective map $$F : L^2(\mathbb R) \ni \psi \to [\phi_\psi] \in L^2(\mathbb R)/\sim$$ where $[\cdot]$ denotes the equivalence class of unit vectors up to phases.
A trivial example of $F$ is $$\phi_\psi := \frac{1}{||\psi+ \chi||}(\psi + \chi)\quad\mbox{and} \quad \phi_{-\chi} := -\chi$$ where $\chi$ is a given (universal) unit vector.
This map is evidently non-physical since there is no reasonable way to fix $\chi$ and, assuming this form of $F$, some argument based on homogeneity of physical space would rule out $\chi$. However much more complicated functions $F$ (not affine nor linear) can be proposed and in the absence of further physical requirements on the correspondence $\psi$-$\phi_\psi$ (e.g. one may assume that some superposition principle is preserved by this correspondence) or on the way to extract probabilities from $\psi$, Gleason's theorem alone cannot establish the form of $F$.
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