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 unitary 1 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:

  • 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 between receiving reward $r_0$ or playing a game that gives reward $r_0$ independently of the measurement result, even when this reward was part of a previous game:

  • Substitution: Amir must value equally the game \[\ket{G} = \alpha\ket{M_0}\ket{r_0} + \beta\ket{M_1}\ket{r_1}\] and the composite game
    \ket{G’} &= \alpha\ket{M_0}\ket{G’^\prime} + \beta\ket{M_1}\ket{r_1} \\
    &= \alpha\gamma\ket{M_0}\ket{D_0}\ket{r_0} + \alpha\delta\ket{M_0}\ket{D_1}\ket{r_0} + \beta\ket{M_1}\ket{r_1},
    \end{align*} where instead of receiving $r_0$ Amir plays the trivial game \[\ket{G’^\prime} = \gamma\ket{D_0}\ket{r_0} + \delta\ket{D_1}\ket{r_0}.\]

Now, to the proof. Consider the games
\ket{G} &= \frac1{\sqrt2}(\ket{M_0}\ket{r_0} + \ket{M_1}\ket{r_1}) \\
\ket{G’} &= \frac1{\sqrt2}(\ket{M_0}\ket{r_1} + \ket{M_1}\ket{r_0}) \\
\ket{G’^\prime} &= \frac1{\sqrt2}(\ket{M_0}\ket{r_0+r_1} + \ket{M_1}\ket{r_0+r_1}) \\
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 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 quantum states with amplitude $1/\sqrt{2}$ click with probability $1/2$.

We can easily extend this argument to show that games involving a uniform superposition of $n$ states \[ \ket{G} = \frac{1}{\sqrt{n}}\sum_{i=0}^{n-1}\ket{M_i}\ket{r_i}\] must have value3 \[ V(G) = \frac1n\sum_{i=0}^{n-1}r_i.\] Now we need to deal with non-uniform superpositions. Consider the games
\[ \ket{G} = \sqrt{\frac{2}{3}}\ket{M_0}\ket{r_0} + \frac{1}{\sqrt3}\ket{M_1}\ket{r_1}\] and
\[ \ket{G’} = \frac1{\sqrt2}(\ket{D_0}\ket{r_0} + \ket{D_1}\ket{r_0}). \] By Constancy the value of $G’$ is $r_0$, and by Substitution the value of $G$ must be equal to the value of
\ket{G’^\prime} &= \sqrt{\frac{2}{3}}\ket{M_0}\ket{G’} + \frac{1}{\sqrt3}\ket{M_1}\ket{r_1} \\
&= \frac{1}{\sqrt3}\ket{M_0}\ket{D_0}\ket{r_0} + \frac{1}{\sqrt3}\ket{M_0}\ket{D_1}\ket{r_0} + \frac{1}{\sqrt3}\ket{M_1}\ket{r_1}.
But $G’^\prime$ is just a uniform superposition, so from the previous argument we know that
\[ V(G’^\prime) = \frac13(r_0+r_0+r_1),\] and therefore that
\[ V(G) = \frac23r_0+\frac13r_1.\] Using analogous applications of Substitution we can show that for any positive integers $n$ and $m$ the value
\[ \ket{G} = \sqrt{\frac{n}{n+m}}\ket{M_0}\ket{r_0} + \sqrt{\frac{m}{n+m}}\ket{M_1}\ket{r_1}\] 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 assumption4, and to extend it to arbitrary complex amplitudes we can do a little trick: consider the game with a single outcome \[\ket{G} = e^{i\phi}\ket{M_0}\ket{r_0}.\] By Constancy the value of this game is $r_0$, independently of the phase $e^{i\phi}$. Now consider the game
\[ \ket{G} = \sqrt{\frac{n}{n+m}}e^{i\phi}\ket{M_0}\ket{r_0} + \sqrt{\frac{m}{n+m}}e^{i\varphi}\ket{M_1}\ket{r_1}.\] By Substitution we can replace the rewards $\ket{r_0}$ and $\ket{r_1}$ with the single outcome games $e^{-i\phi}\ket{D_0}\ket{r_0}$ and $e^{-i\varphi}\ket{D_1}\ket{r_1}$ without changing its value, so the phases play no role in determining the value of the game.

To summarize, we have show 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 \[ V(G) = |\alpha|^2r_0 + |\beta|^2r_1,\]which is just the Born rule.

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2 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 α=β=1 and γ=δ=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) = |α|r₀ + |β|r₁ 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.

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