One might still be worried about Deutsch’s **Additivity**. What if it is actually necessary prove the Born rule? In this case one wouldn’t be able to use the Born rule in the Many-Worlds interpretation without committing oneself to stupid decisions, such as giving away all your money to take part in St. Petersburg’s lottery. Should one give up on the Many-Worlds interpretation then? Or start betting against the Born rule? If these thoughts are keeping you awake at night, then you need Wallace’s version of the Deutsch-Wallace theorem, that replaces Deutsch’s simplistic decision theory with a proper one that allows for bounded utilities.

The way he did it was to separate the construction of probabilities and utilities from a set of preferences, which is a standard but rather laborious part of decision theory, from the part of the argument that is quantum per se, establishing that these preferences are those that come from the Born rule. This is good for two reasons: the first is that it lets you ignore the nasty bits of decision theory until they’re needed1, and secondly it prevents you from using decision theory for doing the work that’s supposed to come from Many-Worlds; Deutsch’s **Additivity** does make life easier, too easy one might suspect.

He started with the quantum part, proving a theorem called **Equivalence**, which states that Amir must be indifferent between games that assign equal Born-rule weights to the same rewards.

**Equivalence**: Let

\[ \ket{G} = \sum_{ij} \alpha_{ij} \ket{M_{ij}}\ket{r_j}\quad\text{and}\quad \ket{G’} = \sum_{ij} \beta_{ij} \ket{M_{ij}}\ket{r_j}.\] If

\[ \forall j\quad\sum_i |\alpha_{ij}|^2 = \sum_i|\beta_{ij}|^2\] then $G \sim G’$.

Here $\sim$ means indifference. We have to use that, instead of saying that the games have equal values, as we are only talking about preferences, leaving values for later.

I want to emphasize that **Equivalence** is really everything we need from quantum mechanics. It says, for example, that the game (taken unnormalised for clarity) \[ \ket{G} = 2\ket{M_0}\ket{r_0}+\ket{M_1}\ket{r_1}\] is equivalent to the game

\[ \ket{G’} = \ket{M_0}\ket{r_0}+\ket{M_1}\ket{r_0}+\ket{M_2}\ket{r_0}+\ket{M_3}\ket{r_0}+\ket{M_4}\ket{r_1},\] as they both assign weight $4$ to reward $r_0$ and weight $1$ to reward $r_1$. Some alternative version of **Equivalence** that summed the modulus of the amplitudes instead, as it would be appropriate in classical probability theory, would claim that $G$ was actually equivalent to

\[ \ket{G’^\prime} = \ket{M_0}\ket{r_0}+\ket{M_1}\ket{r_0}+\ket{M_2}\ket{r_1},\] as they both would assign weight $2$ to reward $r_0$ and weight $1$ to reward $r_1$, a decidedly non-quantum result.

I’m not going to prove **Equivalence** here the way Wallace did, though, because I think his proof is a pain in the ass and I don’t like his axioms. They are too formal, making the proof needlessly complex, and thus fail at its main task of persuasion. More importantly, I think his axiom of **Diachronic consistency** is untenable, as I argue in my paper: Wallace defends it only on rationality grounds, but I argue it is not valid in some toy many-world theories, so it should be derived as theorem only in those where the physics warrant it.

Instead, I’ll show that it does follow from **Label indifference** and **Measurement indifference** (with $V(G) = V(G’)$ replaced with $G \sim G’$), in order to dissipate suspicions that **Additivity** is what is doing the real work. The argument turns out to be pretty much the same as in the previous post.

First we consider the case where $\beta_{ij} = \alpha_{ij}e^{i\theta_{ij}}$. Again, we can erase the phases via **Measurement indifference**, because we can apply the unitary $\sum_{ij} e^{-i\theta_{ij}}\ket{ij}\bra{ij}$ before the measurement without affecting which state $\ket{ij}$ leads to which result $\ket{M_{ij}}$. This allows us to consider only positive real amplitudes. We restrict further to amplitudes that are square roots of integers by continuity, which again we will not deal with explicitly.

Consider then the games

\[ \ket{G} = \sqrt{n}\ket{M_0}\ket{r_0} + \sqrt{m}\ket{M_1}\ket{r_1}\]and

\[ \ket{G’} = \sum_{i=0}^{n-1}\ket{M_i}\ket{r_0} + \sum_{i=n}^{n+m-1}\ket{M_i}\ket{r_1}.\]We do again the devious measurement scheme of the previous post on $\ket{G}$, by using an ancilla on the state $\ket{0}$, and a unitary $U$ such that

\[U\ket{0}\ket{0} = \ket{0}\frac1{\sqrt{n}}\sum_{i=0}^{n-1}\ket{i}\quad\text{and}\quad U\ket{1}\ket{0} = \ket{0}\frac1{\sqrt{m}}\sum_{i=n}^{n+m-1}\ket{i}.\]We measure the ancilla, recording the result in a measurement device $\ket{A_i}$, and from that we set the original measurement device to $\ket{M_0}$ or $\ket{M_1}$, accordingly. This results in the game

\[ \ket{G’^\prime} = \sum_{i=0}^{n-1}\ket{A_i}\ket{M_0}\ket{r_0} + \sum_{i=n}^{n+m-1}\ket{A_i}\ket{M_1}\ket{r_1},\] which is just a fancy relabelling of $\ket{G’}$. By **Measurement indifference** Amir is indifferent between $G$ and $G’^\prime$, and by **Label indifference** he is indifferent between $G’^\prime$ and $G’$, so **Equivalence** is proven for this pair of games. Now the general case follows obviously from this, but the notation becomes horrible, so I’ll not do it explicitly.

Off to the second part then, where we’ll join **Equivalence** with some decision theory and prove the Born rule.