# Solving Newcomb’s paradox for classical and quantum predictors

A recent HN post reminded me of Newcomb’s paradox, which goes as follows (from Wiki):

There is a reliable predictor, another player, and two boxes designated A and B. The player is given a choice between taking only box B, or taking both boxes A and B. The player knows the following:

• Box A is clear, and always contains a visible $1,000. • Box B is opaque, and its content has already been set by the predictor: • If the predictor has predicted the player will take both boxes A and B, then box B contains nothing. • If the predictor has predicted that the player will take only box B, then box B contains$1,000,000.

The player does not know what the predictor predicted or what box B contains while making the choice.

The question is whether the player should take both boxes, or only box B.

I first saw this problem many years ago but didn’t have a strong opinion. Now it seems clear that the controversy is about the definition of “reliable predictor”. This is usually left vague, leading to many unreliable philosophical and game-theory arguments. As usual, I will try to solve the problem using physics. Interestingly, the analysis is different for a classical versus quantum predictor, and also depends on the interpretation of quantum mechanics.

## Classical predictor

Assume it is a classical supercomputer that, at prediction time, takes the state of the player and all the objects that they interact with until the decision. Call this state $S_i$. By running the physics forward, it arrives at either a state $S_{AB}$ or $S_B$, corresponding to the decision to take both boxes or only box B, respectively. In this case, one should obviously take only box B.

## Quantum predictor

In the quantum case, the initial wavefunction of the player/etc is $\psi_i$. The computer cannot measure the wavefunction directly due to the no-cloning theorem. Instead, one way to make the prediction is as follows. The decision to take both boxes corresponds to a set of orthonormal states $\{\psi_{AB}\}$, and likewise for $\{\psi_B\}$. These two sets are mutually orthonormal and form a complete basis, since there are only two choices. Given these sets, the computer can run Schrödinger’s equation back to prediction time to obtain the sets $\{\psi_{ABi}\}=e^{i H t}\{\psi_{AB}\}$ and $\{\psi_{Bi}\}=e^{i H t}\{\psi_B\}$, respectively. These are also mutually orthonormal due to unitarity. At prediction time, it can measure the projection operator $\displaystyle P_{B}=\sum_a |\psi_{Bi}^a\rangle \langle\psi_{Bi}^a|.$

The measurement gives 1 (take box B) with some probability $p$, and 0 (take both boxes) with probability $1-p$. This collapses the player’s wavefunction to one of the states in $\{\psi_{ABi}\}$ or $\{\psi_{Bi}\}$, which then evolves into a state in $\{\psi_{AB}\}$ or $\{\psi_B\}$. Thus, from the predictor’s perspective, the predictor is always right.

The player models this measurement as the predictor becoming entangled with the player, so that the total wavefunction is something like $\displaystyle \sqrt{p}(\psi_{Bi}\otimes \psi_\text{predictB}) + \sqrt{1-p}(\psi_{ABi}\otimes\psi_\text{predictAB}).$

If the player only makes a measurement at decision time, they will collapse the wavefunction to a state in $\{\psi_{B}\}$ with probability $p$, or a state in $\{\psi_{AB}\}$ with probability $1-p$. We assume that this is the measurement basis since the player’s state should not become a superposition of (take B only) and (take both). The expected value is then simply: $\displaystyle E[p] = p B + (1-p)A = A+p(B-A)$

where $A=\text{\1,000}$, $B=\text{\1,000,000}$. This is maximized at $p=1$, so the best decision is to take only box B, just as in the classical case.

Where we go from here depends on the interpretation of quantum mechanics. For many-worlds, there is only unitary evolution. The player ends up in the branch $\psi_{B}\otimes \psi_\text{predictB}$ with probability $p$, giving the expected value above.

However, for Copenhagen-type interpretations where different observers can use different wavefunctions, the player can do better, since they are free to make any measurements between prediction and decision time, while the predictor assumes unitary evolution1. In fact, they can make the predictor predict (take B only) with certainty, while they actually take both with certainty. One way is as follows. Assume the player makes the decision based on measuring a qubit at decision time, where $|\uparrow\rangle$ means take B only and $|\downarrow\rangle$ means take both. The state of the qubit oscillates between $|\uparrow\rangle$ and $|\downarrow\rangle$ with period $T$, where $T$ is the time between prediction and decision. At prediction time, assume the state is $|\uparrow\rangle$, so the predictor predicts (take B only). At time $T/2$, the player can make repeated measurements very quickly until decision time. The qubit stays in the $|\downarrow\rangle$ state due to the quantum Zeno effect. Thus, at decision time, the player takes both boxes. The extra $1,000 can then contribute to funding the delicate and expensive equipment needed for the qubit. We can take this one step further in some cases. For human players, the knowledge of the measurement protocol is classically encoded in the player’s brain in some way. If the supercomputer can decode this information instead of merely running the time evolution, they can also predict which measurements the player makes, and the probabilities of the subsequent results. We arrive back to the original case, where the best solution is to pick B only. This is not required by the postulates of quantum mechanics. The observer’s decision to make measurements on its state does not necessarily have to be encoded in its state itself. ## Real predictor In the real world, there are no such supercomputers, and no entity would risk$1,000,000 on a meaningless game. The best answer is to take both boxes.

1 In practice, a human’s measurements of their own state occur long after decoherence, so they have no control of their wavefunction in this way. However, if we are assuming all-powerful supercomputers, we may as well go all the way.