A Note on Deutsch s Quantum Theory of Probability and Decisions

Transcription

1 A Note on Deutsch s Quantum Theory of Probability and Decisions Malcolm R. Forster University of Wisconsin-Madison Draft, June 12, 1999 ABSTRACT: David Deutsch has a forthcoming article called Quantum Theory of Probability and Decisions" in which he claims to derive the standard probabilistic interpretation of the wavefunction from non-probabilistic considerations. In this note, I try to develop his argument in a classical context, to see whether the same ideas can generate probabilistic conclusions from nonprobabilistic assumptions in a more general way. While my answer is negative, the attempt pinpoints the parts of Deutsch s quantum mechanical argument that must rely on special features of the formalism (assuming that his proof is correct). 1. No Probabilities In, No Probabilities Out? You cannot derive an ought from an is. And I believe that it was Popper (1959) who once said, in response to those who think that randomness can be defined nonprobabilistically, that one cannot derive a probabilistic conclusion from non-probabilistic premises. Yet this is exactly what Deutsch (to appear) claims for quantum mechanical probabilities. Historically in the development of quantum theory, the Schrödinger equation came first, and Born s probabilistic interpretation of the wave function came later, as an additional postulate. If Deutsch is right, then from a few reasonable axioms, that any rational agent must act as if Born s postulate is true. More specifically, Deutsch considers each wavefunction, together with a particular experimental setup (formally, an observable about to be measured), to characterize a game, whose payoff is be proportional to the value of the measured observable. Prior to the measurement, when the outcome is unknown, Deutsch postulates that every quantum mechanical game has a value, which determine a rational agent s preferences. His aim is to show from three non-probabilistic axioms, which characterize the rationality of a player in non-probabilistic terms, it follows that the value of a game is equal to the

2 expected value of outcomes as determined by Born s postulate. 1 According to Deutsch, previous attempts to do this (e.g. Everett (1957), Finkelstein (1963), Hartle (1968), DeWitt and Graham (1973), Ohkuwa (1993)) applied only to infinite sets of measurements (which do not occur in nature), and not to the outcomes of individual measurements (which do). If the argument works in quantum mechanics, why shouldn t it work more generally? That is, why not apply Deutsch s three axioms to any situation in which there is uncertainty about the outcome? What is so special about quantum mechanical uncertainty? Deutsch does not do this, but if it could be done, it seems to me that it would have important foundational implications for decision theory. For example, consider the game in which we toss a fair coin and received $1 if the coin lands heads up, and $0 if the coin lands heads down. Denote this game by G 0 (1, 0). This game is different from the game in which the coin is biased ¾ towards heads. Denote this game by G 1 (1, 0). Finally, consider the case when the coin is biased ¾ towards tails, again with the same payoffs for each outcome. Denote this by G 2 (1, 0). Then, under standard utility maximization theory, if we have choice of which game to play, then it is rational to choose G 1 (assuming that the dollar amounts reflect the true value of the payoffs) because it has an expected payoff of 75 cents, which is higher than the 50 cent and the 25 cent expected payoffs of the other games. A successful application of Deutsch s argument promises to reach the same conclusion from non-probabilistic assumptions. The idea would be to construe any action as a playing a game, and the choice amongst possible actions as a choice about which game to play. Every game has a value, and the fundamental decision rule is to choose to play the game with the highest value. If Deutsch s three axioms were to imply that the value of a game is its expected payoff, then we have an argument not only for expected utility maximization, but for determining the probabilistic weights of possible game outcomes from non-probabilistic assumptions. 1 Actually, his argument is restricted to the case in which there is no degeneracy that is, he assumes that all eigenvalues of the observable are different. 2

3 I plan to show that Deutsch s argument does not go through in this more general case. Nevertheless, my conclusion is not entirely negative. First of all, my argument does not impugn anything that Deutsch does in his original paper, since it is exclusively concerned with the quantum mechanical context. Second, if Deutsch s argument is correct, then my analysis will pinpoint the crucial points at which the quantum formalism is playing a role. More specifically, I will argue that with two additional axioms, which I call the Sure-Thing axiom and the Symmetry axiom, Deutsch s argument does generalize to a broad class of instances in the everyday contexts. Therefore, if Deutsch s argument is valid, it is because the quantum mechanical formalism somehow ensures the truth of these axioms. That promises to be helpful in understanding his analysis. The Sure-Thing axiom is relatively uncontroversial as a constraint on rationality. It just says that if all outcomes of a game lead to the same payoff, then the value of game is equal to that payoff. On the other hand, the Symmetry axiom is a probabilistic assumption, rather like an inverse to Laplace s principle of indifference. Laplace s principle of indifference says that if we are indifferent about two possible events, then they are equally probable, whereas Symmetry states the converse if two events are equally probable, then we are indifferent with respect to them (in a sense to be made precise). So in the general situation, Popper is still right no probabilities in, no probabilities out. Nevertheless, it might show how symmetry considerations help to determine probability values (see also Jaynes, 1971). 2. The Axioms and the Argument The project begins with the assumption that a rational agent, who has a choice of games to play, should choose the game with the highest value. The existence of these values ensure that preferences follows standard properties, such as transitivity: If game A has a higher value than B, and B has a higher value than C, the A has a higher value than C. However, it is not assumed that these values are equal to the expected payoffs of the games. That is to be proved from a set of axioms, which will be non-probabilistic as long as possible. 3

4 Let G(x 1, x 2,, x n ) be a game with n mutually exclusive and exhaustive payoffs x 1, x 2,, x n. The different outcomes need not be equally probable. The value of the game is denoted by V(G(x 1, x 2,, x n )). In the special case in which the payoffs are equally probable, then I will write: V(G(x 1, x 2,, x n )) = V(x 1, x 2,, x n ). So whenever you see the notation V(x 1, x 2,, x n ), you know that the outcomes are equally probable. A composite game is one in which one or more of the payoffs is a game. Thus, for example, G(G 1 (a, b), c) is a game in which the first outcome is to play game G 1 (with possible payoffs a and b) and the second outcome is to receive the payoff c unconditionally. Axiom 1: Additivity. If a constant payoff k is added to every outcome of a game, then the value of the game is the same as the original game with an unconditional payoff k added on. That is, V(G(a + k, b + k)) = k + V(G(a, b)). Axiom 2: Zero-Sum. Consider two games with the same number of outcomes, where for every outcome, the payoff in one game is the negative of the payoff in the other. Then the sum of the values of these games is zero. That is, V(G(a, b)) +(G( a, b)) = 0. Deutsch describes this axiom as concerning two-player zero-sum games. But his application has nothing to do with two players. The axiom is used only to justify a single players substitution of one game for another. Axiom 3: Substitutability. If two games have the same value, then substituting one game for the other in a composite game will not change the value of the composite game. So, for example, if V(G 1 (a, b)) = d, then G(G 1 (a, b), c) and G(d, c) have the same value. Note that we must, in general, preserve the ordering of the payoffs. We cannot substitute G(c, d) for G(d, c), except in special cases. Substitutability is the contentious axiom because, at the first sight, it appears to assume what we need to prove. In the proofs to follow, it sometimes appears to assume that two games are interchangeable because they have the same expected payoff. However, let it be clear at this point that Substitutability only sanctions substitution if the values of the games are equal. 4

5 For this reason, I believe that we need to add the following axiom, which does not appear in Deutsch s argument, either because of an omission, or because it is automatically true in the quantum mechanical argument. Axiom 4: Sure-Thing. If the payoffs for all possible outcomes of a game are the same, then value of the game is equal to that payoff. That is, V(G(a, a)) = a. The outcomes need not be equiprobable (the special case in which the outcomes are equiprobable does follow as a corollary to Theorem 1, below). Although Deutsch proceeds on the assumption that Axiom 4 follows from the previous assumptions, I don t see how to prove it. However, this is not an significant issue because the assumption is unproblematic whether it is redundant or not. The fifth axiom is one that I have added, not because Deutsch overlooked it, but because it is automatically taken care of in the quantum mechanical context. Axiom 5: Symmetry. If outcomes are equally probable, then the value of a game is preserved if we swap the payoffs around. That is, V(a, b) = V(b, a). For example, if we toss a fair coin, then the game in which you get $1 for heads up and $0 for heads down has the same value as a game in which you get $0 for heads up and $1 for tails down. This would not be true if the coin were biased. The point of the argument is to show how these axioms entail that the value of a game is its expected payoff. It is not obvious even in the simplest case. Consider a simple game in which there are two equally probable outcomes with payoffs a and b, respectively. In that case we know that the expected payoff is ½(a + b). The problem is to use the 5 axioms to show that the value of the game is exactly equal to this amount. Theorem 1: If axioms 1 to 5 are true, then V(a, b) = ½(a + b). Proof: From the Zero-Sum axiom, V(a, b) + V( a, b) = 0, But, V( a, b) = V(b (a + b), a (a + b)) By Additivity, V(b (a + b), a (a + b)) = (a + b) + V(b, a), and hence, V( a, b) = V(b, a) (a + b). By Symmetry, V(b, a) = V(a, b). 5

6 Therefore, 2V(a, b) = (a + b), from which the theorem easily follows. Notice that the use of Symmetry is crucial here, for otherwise the result would follow even when the outcomes were not equally probable, and that would give the wrong answer. We made no use of the Sure-Thing axiom in proving Theorem 1. The next task is to generalize this result to an arbitrary number of equally probable outcomes. Theorem 2: If Axioms 1 through 5 are true, then V(x 1, x 2,, x n ) = 1/n(x 1 + x x n ). Proof: The proof is in two parts. Part (a). First prove that the theorem is true for n = 2, 4, 8, 16, This part of the proof is by mathematical induction on the integers m, where n = 2 m. Case m = 1 is true by Theorem 1. Now assume that the proposition is true for m = k. We need to prove that it is true for m = k + 1. So, assume V(x 1, x 2,, x r ) = 1/r(x 1 + x x r ), where r = 2 k. We need to show that V(x 1, x 2,, x 2r ) = 1/2r(x 1 + x x 2r ). The key idea is to divide the game into two subgames: G 1 (x 1, x 2,, x r ) and G 2 (x r+1, x r+2,, x 2r ), and consider the composite game G(G 1 (x 1, x 2,, x r ), G 2 (x r+1, x r+2,, x 2r )). Because each subgame has the same number of equally probable outcomes, each subgame is equally probable. Let the subgames have values a and b respectively. Then Substitutability tells us that we may replace each subgame by a simple game in which one receives its value unconditionally. Therefore, V(x 1, x 2,, x 2r ) = V(a, b) = ½(a + b), where the last step follows from Theorem 1. We may now substitute for a and b using the inductive hypothesis to prove the result. Part (b). We need to prove the result for intermediate cases for which n 2 m. To do this it is sufficient to prove that if the result is true for n = k, then it is true for n = k 1. This is a kind of backwards induction, but given for any n, Part (a) establishes the theorem for some higher value of n, so this will work. 6

7 Consider any game with k 1 equally likely outcomes. We want to prove that V(x 1, x 2,, x k 1 ) = (1/k 1)(x 1 + x x k 1 ). Form the composite game G(G 1 (x 1, x 2,, x k 1 ), x k ), in which all k outcomes are equally probable. (This implies that G is a composite game in which G 1 has probability (k 1/k) and x k has probability 1/k.) The trick is that we are free to set the value of x k anything we want. Set it at V(x 1, x 2,, x k 1 ). This means that G(G 1 (x 1, x 2,, x k 1 ), x k ) can be replaced by the game G(x k, x k ) in which the two payoffs are the same (even though they are not equally likely). The Sure-Thing axiom implies that V(G(x k, x k )) = x k. From the way we defined G, it is also true that V(x 1, x 2,, x k ) = V(G(G 1 (x 1, x 2,, x k 1 ), x k )). Therefore, V(x 1, x 2,, x k ) = x k. In other words, for this particular choice of x k, V(x 1, x 2,, x k 1 ) = V(x 1, x 2,, x k ). The inductive hypothesis now tells us that, V(x 1, x 2,, x k ) = (1/k)(x 1 + x x k ), form which it follows that, V(x 1, x 2,, x k 1 ) = (1/k)(x 1 + x x k 1 + V(x 1, x 2,, x k 1 )). To obtain the final result, collect common terms and simplify. The proof of Theorem 2 illustrates the crucial role played by Substitutability. There is a natural suspicion here that probabilistic considerations play a role independently of the Symmetry axiom. The suspicion is that in order to know whether games can be substituted for one another you have to know the probabilities. However, its role in the proof of Theorem 2 does not go beyond a knowledge of equiprobability. For, it follows from the fact that x 1, x 2,, x k 1, x k are equally probable, that x 1, x 2,, x k 1 are equally probable. The part that goes beyond equiprobability is the step that appeals to the Sure-Thing axiom. Symmetry does not apply here. However, the idea that G(a, a) can be substituted for an unconditional payment of a does not depend on the relative probabilities of the outcomes in any way. 7

8 Nevertheless, in the next step, in the generalization from Theorems 1 and 2, Substitutability plays an even more critical role. (I will only sketch how the generalization from these theorems proceeds, and refer the reader to Deutsch (to appear) for the details.) The goal is to prove that V(G(a, b)) = Pr(a).a + Pr(b).b, for any values of the probabilities Pr(a) and Pr(b). The key idea is that we can substitute G(a, b) with a game G 0 (a, a,, a, b, b,, b), where the subscript 0 indicates that all the outcomes are equally probable. We substitute as many a s and b s as are needed so that their relative frequencies approximate Pr(a) and Pr(b) as closely as we desire. Then we apply Theorem 2 to G 0 (a, a,, a, b, b,, b). I think it is clear enough from this brief sketch that the crux of the proof is justifying the substitution of the right number of a s and b s without somehow assuming that they are substitutable because they yield the same expected payoffs. For me, this is the part of Deutsch s argument that I find the most suspicious. Why should G(a, b) have the same value as G 0 (a, a,, a, b, b,, b). Perhaps this is another special assumption that falls out of the quantum mechanical formalism? 3. Conclusion My attempt at extending Deutsch s argument to the classical domain has forced the addition of at least two axioms which I called the Sure-Thing axiom and the Symmetry axiom. From these, I am fairly confident that Deutsch s proof goes through at least as far as Theorem 2. However, beyond that, I suspect that additional axioms will be needed. None of this impugns Deutsch s argument in the quantum mechanical context. In fact, the analysis should deepen his argument by bringing out the points at which the quantum mechanical formalism is doing real work. It is only in this way that we will really understand how it is that one can get (effectively) probabilistic conclusions from nonprobabilistic premises. 8

9 References Deutsch, David (to appear). Quantum Theory of Probability and Decisions. Proc. R. Soc. Lond. A. The preprint is available at DeWitt, B.S. and Graham, N in The Many-Worlds Interpretation of Quantum Mechanics (Princeton University Press). Everett, H Rev. Mod. Phys Finkelstein, D Trans. N.Y. Acad. Sci Hartle, J.B Am. Jour. Phys. 36, Jaynes, E. T. (1971), The Well-Posed Problem. In V. P. Godambe and D. A. Sprott (eds.), Foundations of Statistics, Toronto, Holt, Rinehart and Winston, pp Ohkuwa, Y. Phys. Rev D Popper, Karl (1959): The Logic of Scientific Discovery. London: Hutchinson. 9