Does Gleason's Theorem Imply Born's Rule?

Question

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 1

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 $F$.

Answer 2

I think Gleason's theorem needs the extra hypothesis of non-contextuality to imply the Born rule. One could, in principle, introduce other measures of probability, but then one violates non-contextuality. See this paper for instance.

Answer 3

Gleason's theorem (GT) says that any measure on the space of states that obeys the rules of the probability calculus is given by the Born rule for some state. This does not imply the Born rule for several reasons.

GT doesn't pick any particular state as you noted. There is another related problem. There are circumstances under which the "probabilities" predicted by the Born rule break the rules of the probability calculus, e.g. - during interference experiments:

https://arxiv.org/abs/math/9911150.

So an explanation is needed of when number given by the Born rule respects the rules of probability. That explanation involves decoherence, which also picks out the set of possible states:

https://arxiv.org/abs/1404.2635

There are other explanatory problems with the use of probability in physics in general:

https://www.youtube.com/watch?v=wfzSE4Hoxbc

and GT does nothing to address such problems.

One particular issue with the use of probability is that the Born rule is just postulated in all collapse interpretations of quantum mechanics, which means that none of them can give any explanation of it.

Answer 4

No. I think few extra assumptions are required to move from Gleason's theorem to Born rule. I will first state the theorem as given in wikipedia.

Gleason's Theorem: Assume a finite dimensional Hilbert space $\mathcal H$ with dimension $d>2$. Let $f$ be a function from projection operators to the unit interval with the property that for any projector $\Pi$, $$0 \leq f(\Pi) \leq 1$$

and if a set $\{\Pi_i\}$ of projection operators sum to the identity matrix (that is, if they correspond to an orthonormal basis), then $$\sum_i f(\Pi_i) = 1$$

Then Gleason's theorem says that there exists some density matrix $\sigma$ such that

$$f(\Pi_i)=tr[\Pi_i \sigma]$$

......

Now, I will give a contradicting example for probability assignment for state that agrees with Gleason's theorem but not with Born rule.

Let $P(\rho, \Pi_i)$ denote the probability of getting the outcome $i$ when the system state is $\rho$. Then we know from Gleason's theorem, we know that for some density matrix $\sigma$,

$$P(\rho, \Pi_i)=tr[\Pi_i \sigma]$$

The question is, this would be Born rule only if for every choice of $\rho$, $\sigma=\rho$ is chosen. Imagine what would happen if we instead choose the following rule for probability assignment-

$$P(\rho, \Pi_i)=tr[\Pi_i U(\rho)\rho U^{\dagger}(\rho)]\tag{1}\label{genprob}$$

where $U(\rho)$ is an arbitrary Unitary function of $\rho$. Gleason's theorem does not rule out this kind of probability assignment.

What extra assumption can give us the Born rule? I can think of two assumptions that do the job.

1. If $Tr[\Pi \rho]=0$ then $P(\rho, \Pi)=0$.
2. $P(\alpha_1\rho_1 + \alpha_2\rho_2, \Pi)=\alpha_1 P(\rho_1, \Pi) + \alpha_2 P(\rho_2, \Pi)$, where $\alpha_i$ are non-negative real numbers with $\alpha_1 + \alpha_2=1$

Now, for pure state $\rho$ in \eqref{genprob} (i.e. $\rho= |\psi\rangle \langle \psi|$), assumption (1) causes $U(\rho)$ to become trivial. i.e. $U(\rho)\rho U^{\dagger}(\rho)=\rho$. This is because if $U(\rho)\rho U^{\dagger}(\rho)=\rho'\neq\rho$, then consider a basis that contains $\Pi_1=|\psi\rangle \langle \psi|$. Then for $j\neq 1$ there exists a projector $\Pi_j$ orthogonal to $\Pi_1$ in this basis such that $P(\rho, \Pi_j)=tr[\Pi_j \rho']\neq0$ although $Tr[\Pi_j \rho]=0$. This contradicts assumption (1).

Hence Born rule follows from assumption (1) and Gleason's theorem for pure states $\rho$. Assumption (2) takes care of the impure state case.