## Friday, July 25, 2014

### Sleeping beauty and quantum mechanics

My newest favorite philosophical dilemma is the Sleeping Beauty problem.  The experiment goes as follows:

1. Sleeping Beauty is put to sleep.
2. We flip a coin.
3. If the coin is heads, then we wake Sleeping Beauty on Monday, and let her go.
4. If the coin is tails, then we wake Sleeping Beauty on Monday.  Then, we put her to sleep and cause her to lose all memory of waking up.  Then we wake her up on Tuesday, and let her go.
5. Now imagine Sleeping Beauty knows this whole setup, and has just been woken up.  What probability should she assign to the claim that the coin was tails?

There are two possible answers.  "Thirders" believe that Sleeping Beauty should assign a probability of 1/3 to tails.  "Halfers" believe that Sleeping Beauty has gained no new relevant information, and therefore should assign a probability of 1/2 to tails.  The thirder answer is most popular among philosophers.

This has deep implications for physics.

There is an argument that Everettian Quantum Mechanics (aka Many Worlds Interpretation, henceforth EQM) requires that you be a halfer.  Say that you tell Sleeping Beauty that you will wake her up on Monday and Tuesday, with a memory wipe in between.  The argument goes that this is exactly analogous to telling her that you will cause her wavefunction to branch, and in one branch she will wake up on Monday and in the other on Tuesday.  In both cases, the Sleeping Beauty on Monday and Sleeping Beauty on Tuesday are both real, and neither has access to the other (either because of the memory wipe or because they are in separate branches).

Therefore, the standard sleeping beauty problem ("Two-Branch-Beauty" on left) is equivalent to the quantum sleeping beauty problem ("Three-Branch-Beauty" on right).

From "Self-Locating Uncertainty and the Origin of Probability in Everettian Quantum Mechanics".  In this paper, they also interpret the very first coin flip as a quantum measurement which causes the wavefunction to branch.

In the Three-Branch-Beauty problem, EQM straightforwardly says that Sleeping Beauty should assign a probability of 1/2 to being in the "tails" branch.  If the Three-Branch-Beauty and Two-Branch-Beauty problems are indeed analogous, then she should also assign a probability of 1/2 in the Two-Branch problem.  And therefore, if we accept EQM, we must accept the unpopular halfer solution.

The problem with this argument, I think, is that assigning probabilities in EQM is... not at all straightforward.  This is actually the major disadvantage of EQM, is that it's not clear why we should interpret the branching worlds with probabilities.  How can we say one branch is more likely than another, if both branches are equally real?

The paper, "Self-Locating Uncertainty and the Origin of Probability in Everettian Quantum Mechanics" purports to answer this question (also see Sean Carroll's blog about it).  When we assign probabilities to branches, these are interpreted as "self-locating" probabilities.  That is, even if all branches are real, we still have to ask the question, which branch am I in now?  You might naively take the principle of "indifference", arguing that all branches are equally likely.  But that gets you the wrong probabilities.

The paper argues for a different principle, the "Epistemic Separability Principle", the idea that Sleeping Beauty assigns probabilities on the basis of only what she observes.  So if there is a second measurement device that Sleeping Beauty does not look at, then the probability she assigns to the first measurement should be independent of the result of the second measurement.  There's a simple argument on page 4 which shows that if there are two branches of the wavefunction with equal amplitude, then we should assign equal probabilities to each branch.

So let's say we have the Three-Branch-Beauty experiment, and Sleeping Beauty has just woken up.  By the Epistemic Separability Principle, she should assign equal probability to waking up in the first branch, and waking up in the second branch, because those two branches have equal amplitude.  The third branch does not have equal amplitude, and based on more mathematical arguments, you can show that the third branch is twice as likely as the other two.  Therefore, she would assign probabilities 1/4, 1/4, and 1/2 to the three locations.

Let's say we have the Two-Branch-Beauty experiment, and Sleeping Beauty has just woken up.  By the Epistemic Separability Principle, she should assign equal probability between waking up on Monday in the first branch, and waking up on Monday in the second branch, since those two branches have equal amplitudes.  Likewise, she should assign equal probability to waking up on Tuesday in the first branch, and waking up on Monday in the second branch.  Therefore, she should assign probabilities of 1/3, 1/3, and 1/3 to the three locations.  This recovers the popular thirder answer without giving up EQM.

In another post I will have further comments on Sleeping Beauty.  (Update: it's written)

miller said...

Hey, im unsure why the 'thirder' option does not apply to the three-branch-beauty as well as the two-branch one if they are equivalent (im guessing its got do to with the fact that there's not really a second measurement happening in the two-branch-beauty).

miller said...

Peter Lewis argued that the two problems are equivalent, and therefore the thirder argument also applies to the three-branch-beauty. Carroll and Sebens, however, argue that they are not equivalent. The key difference is that in the two-branch problem, waking up on Monday with tails has the same amplitude as waking up on Monday with heads. In the three-branch-problem, these cases do not have the same amplitude.

miller said...

Imagine a slight change to the SB problem: We wake her both days, using the memory wipe in between. We move her into a red room to ask for her probability assignment UNLESS it is Tuesday and the coin landed Heads; in that case, we use a blue room.

Before we move her, she knows that P(Heads&Mon.) = P(Tails&Mon.) = P(Heads&Tue.) = P(Tails&Tue.) = 1/4. So if we move her into a red room, she can deduce P(Heads|Red) = P(Heads&Red)/P(Red) = P(Heads&Mon.)/[1-P(Heads&Tue.)] = (1/4)/(3/4) = 1/3.

It does not matter how she would have known it was Tuesday after Heads rolled when she knows that it isn't. All that is important is that she knows it isn't Tuesday after Heads rolled, which is the exact same knowledge she has in the original problem. The answer is 1/3.

Halfers confuse themselves by treating the observation of an outcome with the occurrence of the outcome a part of that outcome. Tuesday happens after Heads is rolled, even in the original experiment. How, or Whether, Sleeping Beauty would observe it is irrelevant. The Quantum Mechanics issue is whether this is still true, not what the answer to the Sleeping Beauty Problem is.