12
$\begingroup$

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.

$\endgroup$

4 Answers 4

6
$\begingroup$

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 $F$.

$\endgroup$
3
  • 2
    $\begingroup$ Your old answer has some interesting maths, but this gets much closer to the heart of the question as I understood it. The Born rule has nontrivial physical content, i.e. it is at heart a postulate and not a result, and no amount of mathematical jiggling will do away with the need to put in a nontrivial assertion about how the world works. $+1$. $\endgroup$ Dec 13, 2017 at 9:59
  • $\begingroup$ @Emilio: that's exactly what the paper I quoted in my answer below tried to prove ! The Born rule cannot be derived solely from "unitary quantum mechanics" but needs an extra postulate of SOME sort. $\endgroup$
    – entrop-x
    Dec 14, 2017 at 6:39
  • $\begingroup$ @entrop-x Indeed I upvoted your answer! $\endgroup$ Dec 22, 2017 at 15:23
2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Can you pleases explain what is this "hypothesis of non-contextuality"? $\endgroup$
    – Lior
    Dec 13, 2017 at 15:44
  • 1
    $\begingroup$ To a certain degree, quantum theory is of course contextual: the outcome of a measurement depends on the kind of measurement one does. But there is also a part that is non-contextual: if I do two "compatible" measurements, one which is coarse-grained and another one which is finer-grained, for instance, the outcome probability of a coarse-grained measurement is supposed to be the sum of the probabilities of the corresponding finer-grained measurements. This doesn't have to be so, but it is (implicitly) postulated. That's the hypothesis of non-contextuality of compatible measurements. $\endgroup$
    – entrop-x
    Dec 13, 2017 at 15:52
0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ As presented in standard axiomatical approach, the Born's rule assumes no experimental setup or interpretation (ensemble or individual systems, one measurement or more), just a definite probability measure obeying Kolmogorov's axioms. This rule is simply a meaning to a mathematical expression. As of such, "GT --> Born's rule" can be a valid mathematical theorem - see the answer by Valter Moretti above. $\endgroup$
    – DanielC
    Dec 12, 2017 at 13:09
  • $\begingroup$ The approach you describe has nothing to do with physics, which describes and explains how the world works. Physics can't concern itself just with mathematical expressions and has to explain how they are relevant to the real world. so any answer that takes the approach you mention is wrong. $\endgroup$
    – alanf
    Dec 12, 2017 at 13:59
  • $\begingroup$ I stand corrected to part of my comment, Valter explained that the Kolmogorov axioms are unrelated. Born's rule is the correspondence between the mathematical axioms/assumptions and the "real world", for it speaks of "values of observables which can be experimentally obtained". That is what I meant by "this rule is a meaning to a mathematical expression". $\endgroup$
    – DanielC
    Dec 12, 2017 at 16:36
  • 1
    $\begingroup$ As far as I can tell, your first sentence is missing a predicate: "Gleason's theorem (GT) says that [the only measure on the space of states that (obeys the rules of the probability calculus if (the measure is the same as that (given by the Born rule for some state)))]." The entire portion in brackets is a noun phrase with no associated verb; all of the portion in the outer parentheses forms a relative clause, not predicate, for that noun phrase. $\endgroup$ Dec 12, 2017 at 16:43
  • $\begingroup$ @Acccumulation I made a correction. $\endgroup$
    – alanf
    Dec 13, 2017 at 10:31
0
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ I don't really get your point. If 1. is not satisfied, why should we even think that $\rho$ is the state of the system? I think that one should define the state of a system to be the unique density matrix given by Gleason's Theorem. To me, postulating that the state of a system is a density matrix is a vacuous assumption if one does not add an operational definition such as the Born rule... I think the right way to think of this is just to define a state to be a function $f$ as in Gleason's theorem. Then (and this is Gleason's theorem), a state HAS TO BE a unique density matrix. $\endgroup$
    – Plop
    Mar 4 at 12:35
  • $\begingroup$ @Plop I would agree with your point if in eqn (1) of my answer, the unitary $U(\rho)$ was in fact not $\rho$ dependent. In that case, every $\rho$ and $\sigma$ would be related by a fixed unitary transformation $U$ and we may as well treat our density matrix to be the rotated unique $\sigma$ given by Gleason's theorem, as you have mentioned. But the $\rho$ dependence of $U(\rho)$ in eqn (1) gives rise to an admittedly exotic theory that is not ruled out by Gleason's theorem. But as I will explain, this exotic theory has other issues. It will lead to superluminal signaling. $\endgroup$ Mar 5 at 16:51
  • $\begingroup$ I think I had not noticed that it was $U(\rho)$ instead of just $U$; but it makes even less sense to me now. When you say "the system is in state $\rho$", does this sentence have an operational meaning or is it just an empty sentence? To me, the sentence "the system is in state $\rho$" should (by definition) mean that the experimental probabilities are correctly predicted by the Born rule with right $\rho$ inside. $\endgroup$
    – Plop
    Mar 5 at 16:58
  • $\begingroup$ The reason for superluminal signaling is that effectively, at the operational level, a state $\rho$ is actually $U(\rho) \rho U^\dagger (\rho)$. This $\rho$ dependent transformation will generally render the unitary evolution nonlinear. Then, due some results due to Gisin and Davies, superluminal signaling will be possible. (See- 'Dynamical Reduction Models' by Angelo Bassi, GianCarlo Ghirardi, section 5.3). Hence, Gleason's Theorem together with no superluminal signaling condition gives rise to Born rule. $\endgroup$ Mar 5 at 16:58
  • $\begingroup$ Calling my mother "John Lennon" on my phone doesn't give me the possibility to contact the dead Beatle at all; and of course we can give any name to the state of a system, including a name in the form of a unit vector, but this does not mean that this is the "right name". And to me, Gleason's theorem proves that there is only one "good" way to name a state as the data encoding the experimental probabilities: the name has to be a density matrix. $\endgroup$
    – Plop
    Mar 5 at 17:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.