The prisoner's dilemma and physics
In this electronic version of the prisoner's dilemma, two electrons are each trying to minimize their own energy. Each electron wishes to be close to the nucleus, but does not want to experience too much Coulomb repulsion from the other electron. The pay-offs are chosen to represent how well the energy is minimized in relative terms, but are not supposed to be quantitatively accurate. Suppose the red electron chooses the inner orbit (B) while the blue electron chooses the outer orbit (A). The red electron experiences a large attraction to the nucleus but a small repulsion from the blue electron. Hence the pay-off for the red electron is large (i.e. 3). However, the blue electron is far from the attractive nucleus and is also shielded by the red electron: hence the blue electron's pay-off is 0. A similar argument applies to the pay-off (0,3). If both electrons choose the inner shell, the attraction is large but the repulsion is also large, hence the pay-off is (1,1). If both electrons choose the outer shell, both the attraction and repulsion are smaller (2,2). Given that the attraction and repulsion will in general have different functional dependencies, the pay-offs shown are physically realizable. Although (B,B) is a Nash equilibrium (purple box), having the two electrons in the outer shell (A,A) would have given a larger, so-called Pareto optimal pay-off (green box). Physics tells us that (A,A) will be the one adopted by the electrons. At first sight, it seems that Pareto optima are therefore preferred over Nash equilibria. However, since the two electrons interact with each other, their wavefunctions are entangled, which renders the game co-operative. Consequently, co-operative game theory can be used to show that the Nash equilibrium and the Pareto optimum coincide in this scenario.