# Wallace’s version of the Deutsch-Wallace theorem, part 2

To conclude the proof of Wallace’s version of the Deutsch-Wallace theorem, we shall add the Equivalence theorem from the previous post to a pretty weak decision theory, and show that if you are rational and live in a universe described by the Many-Worlds interpretation, you must bet according to the Born rule.

The first rationality assumption we need is pretty minimal: we only demand Amir to have preferences between games that are in a precise sense coherent. They must be transitive, in the sense that if he would rather vote for Strache than to vote for Kern, and would prefer to vote for Kern over Kurz, than he must choose to vote for Strache over Kurz. He must also have definite preferences about any pair of games: either he thinks that Strache is better than Kurz, or that Kurz is better than Strache, or he is indifferent between them. He is not allowed to say that they are not comparable. Note that we are not judging whether his preferences are politically coherent, or whether voting for Strache is at all a good idea. The axiom is then:

• Ordering: Amir’s preferences between games, written as $G \succeq G’$, define a total order in the set of games: if $G \succeq G’$, and $G’ \succeq G”$, then $G \succeq G”$. Moreover, for any two games $G$ and $G’$, either $G \succ G’$, or $G \sim G’$, or $G \prec G’$.

This means that the $\succeq$ behaves like the usual $\ge$ relation between real numbers1.

The second and last rationality assumption we shall use is rather stronger, but I think still pretty well-justified. We demand that Amir’s preferences between games must remain consistent while he plays: if he prefers game $G$ to game $G’$, he cannot change his mind if $G$ and $G’$ are offered as rewards inside another game:

• Consistency: Let $\alpha \neq 0$, and consider the games $\ket{F} = \alpha\ket{M_0}\ket{G} + \beta\ket{M_1}\ket{z}$ and $\ket{F’} = \alpha\ket{M_0}\ket{G’} + \beta\ket{M_1}\ket{z},$ that differ only on the game given as a reward when the measurement result is $M_0$. Then $F \succeq F’$ iff $G \succeq G’$.

It is easy to check that Consistency actually implies all three relations $F \succ F’$ iff $G \succ G’,$ $F \sim F’$ iff $G \sim G’,$ and $F \prec F’$ iff $G \prec G’$.

These assumptions, together with Indifference and Substitution, are enough to imply the

• Born rule: Suppose you are rational, and consider the games
$\ket{G} = \sum_i \alpha_i\ket{M_i}\ket{z_i}\quad\text{and}\quad\ket{G’} = \sum_i \beta_i\ket{D_i}\ket{w_i}.$ Then there exists a function $u$ such that
$u(G) = \sum_i |\alpha_i|^2 u(z_i)$ and $G \succ G’ \iff u(G) > u(G’)$ Moreover, $u$ is unique modulo the choice of a zero and a unity.

This theorem says that you are free to decide your preferences between the rewards: these will define their utility. Your freedom ends here, however: the probabilities that you assign to obtaining said rewards must be given by the Born rule, on pain of irrationality.

A comment is also in order about the uniqueness: the choice of a zero and a unity is analogous to the one that must be done in a temperature scale. In the Celsius scale, for example, the zero is chosen as the freezing point of the water, and the unity as $1/100$ the difference between the freezing point and the boiling point. In the Fahrenheit scale, the zero is chosen as the coldest temperature in Gdańsk’s winter, and the unity as $1/96$ the difference between the temperature of Gdańsk’s winter and the temperature of the blood of a healthy male. In any case, the choice of these two values define the temperature scale uniquely, and the same is true for utility, as implied by the following theorem:

• Uniqueness: If $u$ is a utility, then $\mathcal F(u)$ is a utility if and only if $\mathcal F(u) = au+b$ for some real numbers $a,b$ such that $a>0$.

The proof of the ‘if’ direction is easy: just note that $\mathcal F(u(G)) = a\sum_i |\alpha_i|^2 u(z_i) + b = au(G)+b,$ and that such positive affine transformations preserve the ordering of real numbers. The proof of the ‘only if’ direction is not particularly hard, but it is a bit longer and I shall skip it2. Since the choice of a value for the utility at two rewards $x$ and $y$ is enough to fix $a$ and $b$, the claim follows.

But enough foreplay, now we need to start proving the Born rule theorem in earnest. We’ll build it out of two lemmas: Slider, that says that the weights of a game with rewards $x$ and $y$ behave like a tuner for the preferences, and Closure, that says that as we move this slider we are bound to hit any reward between $x$ and $y$.

• Slider: Let $x$ and $y$ be rewards such that $x \succ y$, and consider the games
$\ket{G} = \sqrt{p}\ket{M_0}\ket{x} + \sqrt{1-p}\ket{M_1}\ket{y}$ and
$\ket{G’} = \sqrt{q}\ket{M_0}\ket{x} + \sqrt{1-q}\ket{M_1}\ket{y}.$ Then $G \succ G’$ iff $p > q$.

Proof: suppose $p > q$. Then we can define the games
$\ket{F} = \sqrt{q}\ket{M_0}\ket{x} + \sqrt{p-q}\ket{M_1}\ket{x} + \sqrt{1-p}\ket{M_2}\ket{y}$ and
$\ket{F’} = \sqrt{q}\ket{M_0}\ket{x} + \sqrt{p-q}\ket{M_1}\ket{y} + \sqrt{1-p}\ket{M_2}\ket{y}.$
Note that the weights of rewards $x$ and $y$ in the game $F$ are $p$ and $1-p$, and in the game $F’$ they are $q$ and $1-q$, so by Equivalence we have that $F \sim G$ and $F’ \sim G’$. Since Consistency implies that $F \succ F’$, transitivity gives us $G \succ G’$. To prove the other direction, note that $p = q$ implies directly that $G \sim G’$, and $p < q$ implies $G \prec G'$ by the flipped argument.

• Closure: Let $x,y$, and $z$ be rewards such that $x \succ y$ and $x \succeq z \succeq y$, and let
$\ket{G_p} = \sqrt{p}\ket{M_0}\ket{x} + \sqrt{1-p}\ket{M_1}\ket{y}.$
Then there exists a unique $p_z$ such that$z \sim \ket{G_{p_z}}.$

Proof: since $\succeq$ is a total order, for all $\rho$ it must be the case that either $z\succ \ket{G_p}, \quad z \sim \ket{G_p},\quad\text{or}\quad z \prec \ket{G_p}.$Moreover, Slider tells us that there exists a critical $p_z$ such that
\begin{align*}
p < p_z \quad &\Rightarrow \quad \ket{G_p} \prec z \end{align*} Then some continuity argument will conclude that $z \sim \ket{G_{p_z}}$.

Now for the main proof: Let $x$ and $y$ be fixed rewards such that $x \succ y$. Set $u(x)$ and $u(y)$ to be any real numbers such that $u(x) > u(y)$, defining the unity and the zero of the utility function3. Now because of Closure for every reward $z$ such that $x \succeq z \succeq y$ there will be a unique number $p_z$ such that
$z \sim \sqrt{p_z}\ket{M_0}\ket{x} + \sqrt{1-p_z}\ket{M_1}\ket{y}.$ Define then
$u(z) = p_z u(x) + (1-p_z) u(y).$ We want to show that the utilities so defined do represent the preferences between any two rewards $z$ and $w$ in the sense that $z \succ w$ iff $u(z) > u(w)$. Suppose that $u(z) > u(w)$. This is the case iff $p_z > p_w$, which by Slider is equivalent to $\sqrt{p_z}\ket{M_0}\ket{x} +\sqrt{1-p_z}\ket{M_1}\ket{y} \succ \sqrt{p_w}\ket{M_0}\ket{x} + \sqrt{1-p_w}\ket{M_1}\ket{y},$ which is equivalent to $z \succ w$.
Now we want to show that for any game $\ket{G} = \sqrt{q}\ket{M_0}\ket{z} + \sqrt{1-q}\ket{M_1}\ket{w}$ its utility is given by $u(G) = q u(z) + (1-q) u(w),$ as advertised. By Consistency, we can replace $z$ and $w$ in $G$ by their equivalent games, and we have that
\begin{multline}
\ket{G} \sim \sqrt{q p_z}\ket{M_0}\ket{M_0}\ket{x} + \sqrt{q(1-p_z)}\ket{M_0}\ket{M_1}\ket{y} + \\ \sqrt{(1-q)p_w}\ket{M_1}\ket{M_0}\ket{x} +\sqrt{(1-q)(1-p_w)}\ket{M_1}\ket{M_1}\ket{y}. \end{multline} By Equivalence,
$\ket{G} \sim \sqrt{\lambda p_z + (1-q)p_w}\ket{M_0}\ket{x} + \sqrt{q(1-p_z)+(1-q)(1-p_w)}\ket{M_1}\ket{y},$
and since $x \succeq G \succeq y$, its utility is given by the above formula, so
\begin{align*}
u(G) &= (q p_z + (1-q)p_w)u(x) + (q(1-p_z)+(1-q)(1-p_w))u(y)\\
&= q u(z) + (1-q) u(w),
\end{align*} as we wanted to show.

With this argument we have proved the Born rule theorem for any game inside the interval $[y,x] = \{z: x\succeq z \succeq y\}$. This would be enough if we were to assume that the set of rewards was something so lame, but since we want to be deal with more interesting reward sets – like $\mathbb R$ – we cannot stop now. It is fortunately not hard to complete the proof: consider a sequence of intervals $[y_i,x_i]$ such that all of them contain $[y,x]$ and that their union equals the set of rewards. By the above proof, in each such interval there exists a utility function $f_i$ that satisfies the requirements. We want to show that these functions agree with each other, and as such define a unique utility over the whole set of rewards. For that, consider a reward $z$ in $[x_i,y_i]\cap [x_j,y_j]$ for some $i,j$. Then it must be the case that either $x\succeq z \succeq y$, or $x\succ y \succ z$, or $z \succ x \succ y$. By Closure, there exists unique $p_z,p_y$, and $p_x$ such that
\begin{align*}
z &\sim \sqrt{p_z}\ket{M_0}\ket{x} + \sqrt{1-p_z}\ket{M_1}\ket{y}, \\
y &\sim \sqrt{p_y}\ket{M_0}\ket{x} + \sqrt{1-p_y}\ket{M_1}\ket{z}, \\
x &\sim \sqrt{p_x}\ket{M_0}\ket{z} + \sqrt{1-p_x}\ket{M_1}\ket{y}. \\
\end{align*}Since $f_i$ and $f_j$ are utilities over this interval, we must have that for $k=i,j$
\begin{align*}
f_k(z) &= p_zf_k(x) + (1-p_z)f_k(y), \\
f_k(y) &= p_yf_k(x) + (1-p_y)f_k(z), \\
f_k(x) &= p_xf_k(z) + (1-p_x)f_k(y). \\
\end{align*}Now, we use our freedom to set the zero and the unity of the utilities to choose $f_k(y) = u(y)$ and $f_k(x) = u(x)$, taking these equations to
\begin{align*}
f_k(z) &= p_zu(x) + (1-p_z)u(y), \\
u(y) &= p_yu(x) + (1-p_y)f_k(z), \\
u(x) &= p_xf_k(z) + (1-p_x)u(y), \\
\end{align*}which uniquely define $f_k(z)$ in all three situations, implying that $f_i(z)=f_j(z)$. Setting $u(z)$ to be this common value, we have defined a unique utility function over the whole set of rewards, and we’re done.

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### 8 Responses to Wallace’s version of the Deutsch-Wallace theorem, part 2

1. Davi Barros says:

I guess if this formalism could be useful (if this hasn’t be done before) to make sense of the problem of the spatial distribution of the electromagnetic field.

Since there is the famous non-existence of the position operator — on the other hand there are sensible definitions of ‘photon wavefunctions’ — it’s not obvious for me that a probability distribution can be obtainable from the Born’s rule.

2. Mateus Araújo says:

I don’t know which problem are you talking about. Could you please explain?

3. Davi Barros says:

I think this discussion starts with a paper from Newton and Wigner – https://doi.org/10.1103/RevModPhys.21.400 – where they give a reasonable definition for a position operator in a relativistic setting, but only for massless particles of spin 0 and 1/2, and for massive particles.

Some people then argued that there is no photon wavefunction.

The discussion about this subject that I like most is in this paper here – https://doi.org/10.1088/2040-8986/aa98b6 – where Birula and Birula present a review on their definition of a photon wavefunction. A particularly interesting piece of text is this:

‘The photon wave functions in the position representation do not have all the
properties of the wave functions of massive particles and some physicists even denied
their existence. Edwin Power wrote [21]: “Strictly speaking there are no such
functions!”. These controversies arose due to the absence of the standard probabilistic
interpretation of the wave functions since there is no local photon density.’

And then it says that although this wavefunction have a local modulus squared, it doesn’t have the dimension of the probability density but of the energy density and conclude the paragraph saying that the photon is where its energy is.

4. Mateus Araújo says:

My knowledge of particle physics is quite limited, but I’m pretty sure that there are no massless particles of spin 0 and 1/2.

But in any case, I doubt that this formalism can be of any help; the starting point of the Deutsch-Wallace setup is that you already know the wavefunction of the particle at hand, and which components of the wavefunction give rise to which measurement results. It is just too abstract.

5. Davi Barros says:

The content of the article of Wigner and Newton is also very abstract, is talks about elementary particles with given properties. It doesn’t matter if they are on the standard modeln or not.

But the starting point of the Deutsch-Wallace setup is the reason why I suspect that this formalism may be of use. By solving Maxwell’s equations you can find the ‘wavefunction’ and from it you can know how the photon interacts with matter. In particular you can find how the electrons on a CCD reacts to the incident light by a measurement following the prescription of the MWI.

Given that, the argument of the theorem could, perhaps, give precise meaning of the mod squared of this function as a probability distribution under normalisation. Without the need to start from a position operator.

6. Mateus Araújo says:

No, what gives it meaning is the interaction between the CCD and the photons. After that you can run the Deutsch-Wallace argument or simply postulate the Born rule, it doesn’t make a difference.

7. Davi Barros says:

So it’s settled. Nice

8. Mateus Araújo says:

¬¬