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 worlds1A world must include the position degree of freedom, which has an infinite-dimensional Hilbert space, so this is fine., 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 measure2Note that there exists measures for which it is false. I wanted to show that the Born measure is the unique one that gives the correct law of large numbers, but it turns out this is false, there are plenty of measures that do work..

Now, to the axioms. Besides the ones defining what a measure is, Short assumes3His formulation is slightly different, but mathematically equivalent. 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.