The Sciences

# Quantum Mechanics and Decision Theory

Cosmic VarianceBy Sean CarrollApr 16, 2012 12:20 PM

xi

>, and we have a system described by a state \$latex |psirangle = a |x_1rangle + b |x_2rangle , \$ then a measurement of X is going to return either

x1

or

x2

. But we don't know which, and at this stage of the game we certainly don't know that the probability of

x1

is |a|^2 or the probability of

x2

is |b|^2; that's what we'd like to prove. In fact let's just focus on a simple special case, where \$latex a = b = frac{1}{sqrt{2}} . \$ If we can prove that in this case, the probability of either outcome is 50%, we've done the hard part of the work -- showing how probabilistic conclusions can arise at all from non-probabilistic assumptions. Then there's a bit of mathematical lifting one must do to generalize to other possible amplitudes, but that part is conceptually straightforward. Deutsch refers to this crucial step as deriving "tends to from does," in a mischievous parallel with attempts to derive ought from is. (Except I think in this case one has a chance of succeeding.) The technique used will be decision theory, which is a way of formalizing how we make rational choices. In decision theory we think of everything we do as a "game," and playing a game results in a "value" or "payoff" or "utility" -- what we expect to gain by playing the game. If we have the choice between two different (mutually exclusive) actions, we always choose the one with higher value; if the values are equal, we are indifferent. We are also indifferent if we are given the choice between playing two games with values V1 and V2 or a single game with value V3 = V1 + V2; that is, games can be broken into sub-games, and the values just add. Note that these properties make "value" something more subtle than "money." To a non-wealthy person, the value of two million dollars is not equal to twice the value of one million dollars. The first million is more valuable, because the second million has a smaller marginal value than the first -- the lifestyle change that it brings about is much less. But in the world of abstract "value points" this is taken into consideration, and our value is strictly linear; the value of an individual dollar will therefore depend on how many dollars we already have. There are various axioms assumed by decision theory, but for the purposes of this blog post I'll treat them as largely intuitive. Let's imagine that the game we're playing takes the form of a quantum measurement, and we have a quantum operator X whose eigenvalues are equal to the value we obtain by measuring them. That is, the value of an eigenstate |x> of X is given by \$latex V[|xrangle] = x .\$ The tricky thing we would like to prove amounts to the statement that the value of a superposition is given by the Born Rule probabilities. That is, for our one simple case of interest, we want to show that \$latex Vleft[frac{1}{sqrt{2}}(|x_1rangle + |x_2rangle)right] = frac{1}{2}(x_1 + x_2) . qquadqquad(1)\$ After that it would just be a matter of grinding. If we can prove this result, maximizing our value in the game of quantum mechanics is precisely the same as maximizing our expected value in a probabilistic world governed by the Born Rule. To get there we need two simple propositions that can be justified within the framework of decision theory. The first is:

Given a game with a certain set of possible payoffs, the value of playing a game with precisely minus that set of payoffs is minus the value of the original game.

Note that payoffs need not be positive! This principle explains what it's like to play a two-person zero-sum game. Whatever one person wins, the other loses. In that case, the value of the game to the two participants are equal in magnitude and opposite in sign. In our quantum-mechanics language, we have: \$latex Vleft[frac{1}{sqrt{2}}(|-x_1rangle + |-x_2rangle)right] = - Vleft[frac{1}{sqrt{2}}(|x_1rangle + |x_2rangle)right] . qquadqquad (2)\$ Keep that in mind. Here's the other principle we need:

If we take a game and increase every possible payoff by a fixed amount k, the value is equivalent to playing the original game, then receiving value k.

If I want to change the value of a playing a game by k, it doesn't matter whether I simply add k to each possible outcome, or just let you play the game and then give you k. I don't think we can argue with that. In our quantum notation we would have \$latex Vleft[frac{1}{sqrt{2}}(|x_1+krangle + |x_2+krangle)right] = Vleft[frac{1}{sqrt{2}}(|x_1rangle + |x_2rangle)right] +k . qquadqquad (3)\$ Okay, if we buy that, from now on it's simple algebra. Let's consider the specific choice \$latex k = -x_1 - x_2 \$ and plug this into (3). We get \$latex Vleft[frac{1}{sqrt{2}}(|-x_2rangle + |-x_1rangle)right] = Vleft[frac{1}{sqrt{2}}(|x_1rangle + |x_2rangle)right] -x_1 - x_2. \$ You can probably see where this is going (if you've managed to make it this far). Use our other rule (2) to make this \$latex -2 Vleft[frac{1}{sqrt{2}}(|x_1rangle + |x_2rangle)right] = -x_1 - x_2 , \$ which simplifies straightaway to \$latex Vleft[frac{1}{sqrt{2}}(|x_1rangle + |x_2rangle)right] = frac{1}{2}(x_1 + x_2) , \$ which is our sought-after result (1). Now, notice this result by itself doesn't contain the word "probability." It's simply a fairly formal manipulation, taking advantage of the additivity of values in decision theory and the linearity of quantum mechanics. But Deutsch argues -- and on this I think he's correct -- that this result implies we should act as if the Born Rule is true if we are rational decision-makers. We've shown that the value of a game described by an equal quantum superposition of states |

x1

> and |

x2

> is equal to the value of a game where we have a 50% chance of gaining value

x1

and a 50% chance of gaining

x2

. (In other words, if we acted as if the Born Rule were not true, someone else could make money off us by challenging us to such games, and that would be bad.) As someone who is sympathetic to pragmatism, I think that "we should always act as if A is true" is the same as "A is true." So the Born Rule emerges from the MWI plus some seemingly-innocent axioms of decision theory. While I certainly haven't followed the considerable literature that has grown up around this proposal over the years, I'll confess that it smells basically right to me. If anyone knows of any strong objections to the idea, I'd love to hear them. But reading about it has added a teensy bit to my confidence that the MWI is on the right track.

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