A satisfactory derivation of the Born rule

Earlier this year I wrote a post complaining about all existing derivations of the Born rule. A couple of months later yet another derivation, this time by Short, appeared on the arXiv. Purely out of professional duty I went ahead and read the paper. To my surprise, it’s actually pretty nice. The axioms are clean and well-motivated, and it does make a connection with relative frequencies. I would even say it’s the best derivation so far.

So, how does it go? Following Everett, Short wants to define a measure on the set of worlds, informally speaking a way of counting worlds. From that you can do everything: talk about probability, derive the law of large numbers, and so on. Let’s say your world branches into a superposition of infinitely many worlds1, indexed by the natural numbers $\mathbb{N}$:
\[ \ket{\psi} = \sum_{i\in \mathbb{N}} \alpha_i \ket{\psi_i}. \] Then the probability of being in the world $i$ is understood as the fraction of your future selves in the world $i$, the relative measure
\[ p_{\ket{\psi}}(i) = \frac{\mu_{\ket{\psi}}(i)}{\mu_{\ket{\psi}}(\mathbb{N})}.\]The law of large numbers states that most of your future selves see frequencies close to the true probability. Mathematically, it is a statement like
\[ p_{\ket{\psi}^{\otimes n}}\left(|f_i\,-\,p_{\ket{\psi}}(i)| > \varepsilon \right) \le 2e^{-2n\varepsilon^2}, \]which you can prove or disprove once you have the measure2.

Now, to the axioms. Besides the ones defining what a measure is, Short assumes3 that if $\alpha_i = 0$ then $\mu_{\ket{\psi}}(i) = 0$, and that if a unitary $U$ acts non-trivially only on a subset $S$ of the worlds, then $\mu_{U\ket{\psi}}(S) = \mu_{\ket{\psi}}(S)$. The first axiom is hopefully uncontroversial, but the second one demands explanation: it means that if you mix around some subset of worlds, you just mix around their measures, but do not change their total measure. It corresponds to the experimental practice of assuming that you can always coarse-grain or fine-grain your measurements without changing their probabilities. I think it’s fine, I even used it in my own derivation of the Born rule. It is very powerful, though.

It immediately implies that the total measure of any quantum state only depends on its 2-norm. To see that, consider the subset $S$ to be the entire set $\mathbb{N}$; the second axiom implies that you can apply any unitary to your quantum state $\ket{\psi}$ without changing its measure. Applying then $U = \ket{0}\bra{\psi}/\sqrt{\langle \psi|\psi\rangle} + \ldots$ we take $\ket{\psi}$ to $\sqrt{\langle \psi|\psi\rangle}\ket{0}$, so for any quantum state $\mu_{\ket{\psi}}(\mathbb{N}) = \mu_{\sqrt{\langle \psi|\psi\rangle}\ket{0}}(\mathbb{N})$.

It also implies that if a unitary $U$ acts trivially on a subset $S$ then we also have that $\mu_{U\ket{\psi}}(S) = \mu_{\ket{\psi}}(S)$, because $U$ will act non-trivially only on the complement of $S$, and
\[ \mu_{U\ket{\psi}}(S) + \mu_{U\ket{\psi}}(\mathbb{N} \setminus S) = \mu_{U\ket{\psi}}(\mathbb{N}) = \mu_{\ket{\psi}}(\mathbb{N}) = \mu_{\ket{\psi}}(S) + \mu_{\ket{\psi}}(\mathbb{N} \setminus S).\]

We can then see we don’t need to consider complex amplitudes. Consider the unitary $U_{i_0}$ such that $U_{i_0}\ket{i_0} = \frac{\alpha_{i_0}^*}{|\alpha_{i_0}|}\ket{i_0}$ for some $i_0$, and acts as identity on the other $\ket{i}$. The second axiom implies that it doesn’t change the measure of $i_0$. Repeating the argument for all $i_0$, we map $\ket{\psi}$ to $\sum_i |\alpha_i|\ket{i}$ without changing any measures.

Now we shall see that to compute the measure of any world $i$ in any state $\ket{\psi}$, that is, $\mu_{\ket{\psi}}(i)$, it is enough to compute $\mu_{\alpha_i\ket{0}+\beta\ket{1}}(0)$ for some $\beta$. Consider the unitary
\[U = \Pi_i + \frac{\ket{i+1}\bra{\psi}(\id-\Pi_i)}{\sqrt{\bra{\psi}(\id-\Pi_i)\ket{\psi}}} + \ldots,\]where $\Pi_i$ is the projector onto world $i$. It maps any state $\ket{\psi}$ into
\[U\ket{\psi} = \alpha_i \ket{i} + \beta\ket{i+1},\]where $\beta=\sqrt{\bra{\psi}(\id-\Pi_i)\ket{\psi}}$, and we have that $\mu_{\ket{\psi}}(i) = \mu_{U\ket{\psi}}(i)$. Now consider the unitary
\[ V = \ket{0}\bra{i} + \ket{i}\bra{0} + \ldots \] It takes $U\ket{\psi}$ to
\[VU\ket{\psi} = \alpha_i\ket{0} + \beta\ket{i+1}\] It acts trivially on $i+1$, so $\mu_{VU\ket{\psi}}(i+1) = \mu_{U\ket{\psi}}(i+1)$. Since the total measure of $U\ket{\psi}$ and $VU\ket{\psi}$ are equal, we have that
\[ \mu_{VU\ket{\psi}}(0) + \mu_{VU\ket{\psi}}(i+1) = \mu_{U\ket{\psi}}(i) + \mu_{U\ket{\psi}}(i+1),\] so $\mu_{VU\ket{\psi}}(0) = \mu_{U\ket{\psi}}(i)$. Doing the same trick again to map $i+1$ to $1$ we reduce the state to $\alpha_i\ket{0}+\beta\ket{1}$ as we wanted.

This reduction does all the heavy lifting. It implies in particular that if two worlds have the same amplitude, they must have the same measure, so if we have for example the state $\ket{\psi} = \alpha\ket{0} + \alpha\ket{1}$, then $\mu_{\ket{\psi}}(0) = \mu_{\ket{\psi}}(1)$. Since $\mu_{\ket{\psi}}(\mathbb{N}) = \mu_{\ket{\psi}}(0) + \mu_{\ket{\psi}}(1)$, we have that
\[ p_{\ket{\psi}}(0) = p_{\ket{\psi}}(1) = \frac12.\]A more interesting case is the state $\ket{\psi} = \alpha\sqrt{p}\ket{0} + \alpha\sqrt{q}\ket{1}$ for positive integers $p,q$. We apply to it the unitary $U$ such that
\[U\ket{0} = \frac1{\sqrt p}\sum_{i=0}^{p-1} \ket{i}\quad\text{and}\quad U\ket{1} = \frac1{\sqrt q}\sum_{i=p}^{p+q-1} \ket{i},\]taking $\ket{\psi}$ to $\alpha\sum_{i=0}^{p+q-1}\ket{i}$. Now all amplitudes are equal, and therefore all the measures are equal, call it $x$. Then the total measure is $(p+q)x$, the measure of the original world 0 is $px$, and the measure of the original world 1 is $qx$. Therefore the probability of the original world 0 is
\[p_{\ket{\psi}}(0) = \frac{p}{p+q}.\] Since the argument is valid for all $\alpha$, we have proven the Born rule for all worlds where the ratio between the amplitudes is the square root of a positive rational. Since such amplitudes are dense in the set of amplitudes, we only need a continuity argument to get the complete Born rule.

Normally I don’t care about the continuity argument, as one usually needs a separate postulate to get it, and the continuum is just a convenient fiction anyway. Here the situation is a bit more interesting, because the axioms we have are already strong enough to get it, there’s no need for an extra continuity axiom. Unfortunately I couldn’t find an elegant proof, so I’ll refer you to the original paper for that.

To conclude, I’m still skeptical about this proving the Born rule business, in the sense of replacing it with a better set of axioms to be included in the axioms of quantum mechanics. I don’t think we’ll ever get something better than simply postulating the measure of worlds to be $\mu(\ket{\psi}) = \langle\psi|\psi\rangle$. It’s a similar situation with the other axioms: there are tons of good arguments why one should use complex numbers, or tensor products, or unitary evolution. But when it comes down to writing down the axioms of quantum mechanics, nobody uses the arguments, they write the axioms directly. If what you want is an argument why we should use the Born rule, though, then this is a pretty good one.

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10 Responses to A satisfactory derivation of the Born rule

  1. Abitbol says:

    have you read this paper?


    Do you have any comments?

    I must say I do not belong to the set of authors of this paper and don’t really like this paper.

    I wish you luck for the new year.

  2. Mateus Araújo says:

    Not really, I skimmed it when it appeared online. I don’t think the modal approach to quantum mechanics makes much sense to start with, and their axioms are rather contrived. As far as reconstructions go, I find the oldest one by Hardy pretty nice, and the other by Chiribella also quite good.

    Happy new year!

  3. Philippe Grangier says:

    As an author of the quoted paper, I must say it is somehow outdated. A better version is https://arxiv.org/abs/1910.13738 , published as Found Phys 50, 1781 (2020) .
    We keep the same axioms (sorry if Mateus think they are contrived), but use them as a physical basis to justify the hypotheses of Gleason’s theorem, which gives Born’s rule.

    More recently we use the same axioms to justify the hypotheses of Uhlhorn’s theorem, which gives unitary transforms, see https://arxiv.org/abs/2111.10758 just published in Entropy https://www.mdpi.com/1099-4300/24/2/199

  4. Mateus Araújo says:

    Gleason’s theorem doesn’t give you the Born rule. It only says that the probability rule must be the trace of the projector with a density matrix, but it is silent about which density matrix. You can therefore completely specify your preparation and measurement procedure, and Gleason’s theorem won’t be able to tell you what the probabilities will be.

    Also, it doesn’t hold for Hilbert spaces of dimension 2.

  5. Philippe Grangier says:

    To make sense of Gleason’s theorem you must provide a physical justification of its hypotheses, and in particular of the projectors themselves. Note that if the probability 1 is realized, the density matrix is just another projector (pure state), coming thereby from another measurement result; this is the basic Born’s rule as a the square modulus of a scalar product. The generalization to a density matrix follows naturally.
    Also Gleason theorem works perfectly with Hilbert spaces of dimension 2, provided that they are subspaces of a larger Hilbert space of dimension at least 3 : one qubit in a pair is a nice example.
    I really don’t understand why so many bad things are written about Gleason’s theorem, given that it gets a huge result from very modest hypotheses. Certainly it is difficult to demonstrate and thus to grasp, but with some effort even the proof is not that bad.

  6. Mateus Araújo says:

    Gleason’s theorem is a great result. It just doesn’t give you the Born rule. It’s not about saying bad things or good things about it, it’s just about describing it precisely.

  7. Philippe Grangier says:

    If you consider that the only way to get Born’s rule is by counting how many universes take such and such branches, you are certainly right. I was just telling that there may be other views, for which Gleason’s theorem does give you the Born rule.

  8. Mateus Araújo says:

    Where is this aggression coming from? I did say why Gleason’s theorem doesn’t give the Born rule, and it’s not related to Many-Worlds, but rather that the theorem is silent about which density matrix you should use in the non-contextual measure over orthogonal projections. Nothing forces you to use the density matrix corresponding to the physical preparation. You can even use a fixed one for all Gleason cares. If you disagree you could address that instead of insulting me.

  9. Philippe Grangier says:

    Sorry I did not feel what I wrote was aggressive, but maybe I just (mis?)understood that MWI was your preferred approach to get Born’s rule.
    As for my position, I just repeat that if you want to use Gleason’s theorem for physics, not only for mathematics, the preliminaries are essential : why looking for probabilities ? why projectors ? why non-contextual assignment of probabilities ? how to go from pure states to density matrices ? The answers to these questions must come from physics, not from the theorem itself, this is just what I was writing before.

  10. Mateus Araújo says:

    Indeed, I think Many-Worlds is the best way to do it. But I find insulting the insinuation that I would mischaracterise Gleason’s theorem just because it’s not a Many-Worlds argument.

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