# Deutsch’s version of the Deutsch-Wallace theorem

With the decision theory from the previous post already giving us probabilities, all that is left to do is add the Many-Worlds interpretation and show that these probabilities must actually be given by the Born rule. Sounds easy, no?

But we don’t actually need the whole Many-Worlds interpretation, just some stylized part of it that deals with simple measurement scenarios. We only need to say that when someone makes a measurement on (e.g) a qubit in the state $\alpha\ket{0} + \beta\ket{1},$ what happens is not a collapse into $\ket{0}$ or $\ket{1}$, but rather a unitary1 evolution into the state $\alpha\ket{0}\ket{M_0}+\beta\ket{1}\ket{M_1},$ which represents a macroscopic superposition of the measurement device showing result $M_0$ when the qubit is in the state $\ket{0}$ with the device showing result $M_1$ when the qubit is in the state $\ket{1}$.

We want to use these measurements to play the decision-theoretical games we were talking about in the previous post. To do that, we just say that Amir will get a reward depending on the measurement result: reward $r_0$ if the result is $M_0$, and reward $r_1$ if the result is $M_1$. We can represent this simply by appending the reward into the correct branch of the above macroscopic superposition, taking it to $\alpha\ket{0}\ket{M_0}\ket{r_0}+\beta\ket{1}\ket{M_1}\ket{r_1}.$ Since this state has all the information we need to define the game – the amplitudes, the measurement results, and the rewards – we can use it as the representation of the game. So when we need to write down a game $G$, we shall do this by using the state2
$\ket{G} = \alpha\ket{M_0}\ket{r_0} + \beta\ket{M_1}\ket{r_1}.$ And this is pretty much all we need from quantum mechanics.

Now we need to state two further rationality axioms, and we can proceed to the proof. The first one is that Amir must not care about what do we call the measurement results: if $0$ and $1$, or $\uparrow$ and $\downarrow$, or $H$ and $V$, it doesn’t matter. If two games are the same thing but for the labels of the measurement results, Amir must value these games equally:

• Label indifference: If two games $G$ and $G’$ differ only by the labels of the measurements, then $V(G) = V(G’)$.

The other axiom says that Amir must be indifferent to how exactly the measurement is made; as long as state $\ket{0}$ leads to measurement result $\ket{M_0}$ and state $\ket{1}$ leads to measurement result $\ket{M_1}$, anything goes.

• Measurement indifference: If two games $G$ and $G’$ differ only by the details of the measurement procedure, then $V(G) = V(G’)$.

Now, to the proof. To emphasize that the normalisation of the quantum states plays no role in the proof, I’ll use explicitly unnormalized states. Consider then the games
\begin{align*}
\ket{G} &= \ket{M_0}\ket{r_0} + \ket{M_1}\ket{r_1} \\
\ket{G’} &= \ket{M_0}\ket{r_1} + \ket{M_1}\ket{r_0} \\
\ket{G’^\prime} &= \ket{M_0}\ket{r_0+r_1} + \ket{M_1}\ket{r_0+r_1} \\
\end{align*}By Additivity, from the previous post, we have that
$V(G’^\prime) = V(G) + V(G’),$ and from Constancy that $V(G’^\prime) = r_0 + r_1$, so we already know that $V(G) + V(G’) = r_0 + r_1$. But the games $G$ and $G’$ are just relabellings of each other, so by Label indifference we must have $V(G) = V(G’)$, so we can conclude that $V(G) = \frac12(r_0+r_1)$ or, in other words, that uniform superpositions of two states click with probability $1/2$ each.

We can easily extend this argument to show that games involving a uniform superposition of $n$ states $\ket{G} = \sum_{i=0}^{n-1}\ket{M_i}\ket{r_i}$ must have value$V(G) = \frac1n\sum_{i=0}^{n-1}r_i.$To see that, consider the $n$ cyclic permutations $\ket{G_k} = \sum_{i=0}^{n-1}\ket{M_i}\ket{r_{i\oplus k}},$use Additivity and Constancy to show that $\sum_{k=0}^{n-1}V(G_k) = \sum_{i=0}^{n-1}r_i,$and invoke Label indifference to show that $V(G_k) = V(G)$.

Now we need to deal with non-uniform superpositions. Consider the game
$\ket{G} = \ket{M_0}\ket{r_0} + \sqrt2\ket{M_1}\ket{r_1},$ which is obtained by measuring the state $\ket{0}+\sqrt2\ket{1}$ the regular way. We can instead do it in a devious way: first we prepare an ancilla in the state $\ket{0}$, entangle it with the original state via a unitary $U$ that acts as
$U\ket{0}\ket{0} = \ket{0}\ket{0}\quad\text{and}\quad U\ket{1}\ket{0} = \ket{1}\frac{\ket{1}+\ket{2}}{\sqrt2},$ measure the ancilla with another device giving results $\ket{A_0},\ket{A_1}$, and $\ket{A_2}$, and finally set the original device to $\ket{M_0}$ if the other device is in the state $\ket{A_0}$, and to $\ket{M_1}$ if the other device is either in the state $\ket{A_1}$ or $\ket{A_2}$.

This results in the game
$\ket{G’} = \ket{A_0}\ket{M_0}\ket{r_0} + \ket{A_1}\ket{M_1}\ket{r_1}+\ket{A_2}\ket{M_1}\ket{r_1},$ and by Measurement indifference we have that $V(G)=V(G’)$. But $G’$ is just a fancy relabelling of
$\ket{G’^\prime} = \ket{M_0}\ket{r_0} + \ket{M_1}\ket{r_1}+\ket{M_2}\ket{r_1},$
so by Label indifference we have $V(G’)=V(G’^\prime)$. Now $G’^\prime$ is a uniform superposition of states, which we already know how to evaluate, so
$V(G) = \frac23r_0+\frac13r_1.$ Using analogous applications of Measurement indifference and Label indifference we can show that for any positive integers $n$ and $m$ the value of
$\ket{G} = \sqrt{n}\ket{M_0}\ket{r_0} + \sqrt{m}\ket{M_1}\ket{r_1}$ is
$V(G) = \frac{n}{n+m}r_0+\frac{m}{n+m}r_1,$ and we are pretty much done. To extend the argument to any positive real amplitudes one only needs a continuity assumption3, and to extend it to arbitrary complex amplitudes we use Measurement indifference again: before measuring a state $e^{i\phi}\sqrt{n}\ket{0} + e^{i\varphi}\sqrt{m}\ket{1}$ we apply the unitary $e^{-i\phi}\ket{0}\bra{0}+e^{-i\varphi}\ket{1}\bra{1}$ to it. Since $\ket{0}$ still goes to $\ket{M_0}$ and $\ket{1}$ still goes to $\ket{M_1}$ the value of the game doesn’t change.

To summarize, we have shown that the value of a game $\ket{G} = \alpha\ket{M_0}\ket{r_0} + \beta\ket{M_1}\ket{r_1}$ must be given by $V(G) = \frac1{|\alpha|^2+|\beta|^2}(|\alpha|^2r_0 + |\beta|^2r_1),$which is just the Born rule.

The argument given here is Deutsch’s version only in essence, not in detail. For starters, the notation is completely different, and I changed the decision theory. The important differences, though, is that I’ve postulated Measurement indifference explicitly, whereas Deutsch used it implicitly, and that I haven’t used Deutsch’s “principle of substitutibility”, as it is not necessary for the proof, and as I argue in my paper it is untenable from a purely decision-theoretical perspective.

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### 3 Responses to Deutsch’s version of the Deutsch-Wallace theorem

1. Ben R-G says:

I don’t see why a rational agent should accept Substitution. For example, take $\alpha=\beta=1$ and $\gamma=\delta=100$.

If you’re assuming some sort of normalization condition on games, shouldn’t it be mentioned somewhere?

Without it, I think you can prove pretty much anything. For example, I think your argument would prove $V(G) = |\alpha|r_0 + |\beta|r_1$ if you replaced all your coefficients by their squares.

2. Mateus Araújo says:

You’re right, Deutsch does assume normalization. I didn’t write it here explicitly because Deutsch didn’t write it explicitly in his paper. I know, lame excuse. I’m not proud of this blog post. I should rewrite it at some point.

But the deeper issue is that normalization is not actually required for the argument to go through. This is just a flaw in Deutsch’s specific proof. I wrote a paper that proves the Born rule via essentially the same argument, but uses explicitly non-normalized states, thus making it clear that normalization is unnecessary.

3. Mateus Araújo says:

Ben R-G,

I’ve read Deutsch’s paper again, and I think I was unfair, he was actually careful to never assume normalisation. It was only needed in my flawed formalisation of his proof. I’ve rewritten the proof here to make this explicit.