How do probabilities emerge in the many-worlds interpretation?

Question

My understanding is that at each quantized unit of time that a split occurs, every possible recombination of particles occurs in the 'objective' universe. If this is the case, what relevance to probabilities hold to the behavior of the objective universe, and why do we observe these probabilities in the subjective universe?

I'm way beyond my educational depth here to understand the technical explanations available, and since I couldn't find a non-technical explanation online to help with developing some intuitive grasp, I was hoping someone here might be able to provide an explanation of this question.

It seems like if every moment every recombination occurs at each quantized branch split in the objective universe, then probability would be meaningless to the objective unfoldment of the universe, as every possible combination should occur exactly once, right? Why then, would distinct probability models hold from one quantized moment to the next.

For example, why is it anymore likely that from one moment to the next my computer continues to exist and I can type this post, rather than my computer turning into a purple elephant than my body being transported to mars and then to the andromedas galaxy and then to Bangladesh, then split into a billion pieces and reform as another creature, etc. etc. if all of these possibilities have already unfolded in the objective universe?

If all possible universes occur and are equally likely, how could probability emerge in a single branch? If they are not objectively all equally likely and probability does apply, how many times does the most likely possibility outcome occur relative to the least likely but quantifiable possibility in a single split?

Reiterated another way, how does probability dictate subjective reality so apparently if it is non-existent in a many-worlds interpretation of quantum mechanics that fulfills every possible particle combination of the universe? Alternatively, what am I failing to understand?

'How do probabilities emerge within many-worlds?' @ http://www.hedweb.com/manworld.htm#probabilities is unfortunately beyond me at this time, but perhaps holds the answer.

Answer 1

If you have a quantum state in which more than one of the possible outcomes of a particular measurement has non-zero amplitude (an unsharp state, as opposed to a sharp state in which there is only one outcome), then the MWI says that there will be multiple versions of you, and each version will see one possible outcome.

In the standard (non-MWI) way of looking at quantum mechanics, only one of those outcomes actually happens and it occurs with probability equal to the square of the amplitude. If the state is $$ \tfrac{1}{\sqrt{3}}|0\rangle+\sqrt{\tfrac{2}{3}}|2\rangle, $$ this means that if you do an observation, you will either see a 0 or a 2 on your measurement device. You won't see $\tfrac{1}{\sqrt{3}}$ or $\sqrt{\tfrac{2}{3}}$. Rather, the probability of getting 0 is $\left(\tfrac{1}{\sqrt{3}}\right)^2 = 1/3$ and the probability of 2 is $\left(\sqrt{\tfrac{2}{3}}\right)^2=2/3$. It's important to note that it's rather unclear how to understand that statement in the standard account. After all, only one thing happens in any particular experiment, so why not just say its probability is 1? Or why not assign the same probability to all of the possibilities regardless of their amplitude?

Somebody might say the probability is the relative frequency for many identical experiments. But this creates problems. Even for a very large number of experiments the probability and the relative frequency won't match exactly. And even the probability that they will approximately match is not one, it's one minus some small number. So then why can you neglect the small probability of an aberrant result? You might try to fix this by going for the limit of the relative frequency in an infinite number of experiments, but this would make the conclusion irrelevant for any finite number of experiments even if it gets rid of the problem in some formal sense.

Some people have said that probability is a measure of ignorance, but this also makes no sense. If you're ignorant then you don't know something and can't attach precise numbers to it. And in any case, why would those numbers be the square amplitudes, which act like things you can control in a very precise way by fiddling with magnetic fields around atoms and that sort of thing?

So if we couldn't explain anything about probability in the MWI we would be no worse off than the standard view of quantum mechanics, which also can't explain probabilities. We could just say "we guess this is what the probabilities are, but we don't know why." But it is possible to do a bit better than this. the square amplitudes do actually mean something in the MWI. The amplitudes are objectively real. So you might say the following.

If the state was sharp and I know the state, then I would be willing to bet one penny on the outcome in return for getting two pennies if I was right about the outcome. But is there a way of deciding whether I should bet a penny on an unsharp state, like the one given above? To decide that you would have to assign a number to each state $V(state)$ that represents how much you should bet. Let's look at a simpler case, in which the state is $$ \tfrac{1}{\sqrt{2}}|0\rangle+\tfrac{1}{\sqrt{2}}|2\rangle, $$ and the 0 and 2 stand for how many pennies you will win. The full argument can be read here:

http://arxiv.org/abs/quant-ph/9906015

but what it boils down to is the following. You could swap the 0 and the 2 and that won't change the state, so you should treat the two outcomes as having equal weight and act as if the value is $1/2(0+2) = 1$. (The argument is not quite that simple, but it does depend on the symmetry.) With some other assumptions about how to bet, you can get all of the probabilistic predictions of quantum mechanics.

So advocates of the MWI are no worse off than advocates of other interpretations of quantum mechanics when explaining probability, and arguably they are better off.

Answer 2

Your linked article ('How do probabilities emerge…') only seems to explain why each universe is internally consistent and acts according to what we would statistically expect. The argument pretty much goes that it is of course entirely possible to get weird behavior, but it is precisely as likely to occur as in a single universe obeying the known physical laws. The 'handwaving' part then usually goes something like 'Of course we might not be around to observe a universe where humans spontaneously decompose because the observer might decompose as well' or some variation of the anthropic principle.

I have to add though that this is merely the argument you tend to hear in popular discussions, I'm sure there are more technical explanations as well, and I'm looking forward to them ;)

Answer 3

There exists a valid derivation the Born rule, but here you start from a weaker assumption, it's not the sort of an ab initio derivation from only the other postulates as the MWI advocates would like to have. This argument works as follows. We start by asking how we know that the Born rule is in fact valid. The answer must involve doing experiments, collecting the statistics and testing if the Born Rule prediction is consistent with the measured statistics. This means that you can also consider a thought experiment where you would do such a experiment in a quantum coherent way, so that you have an observable A that corresponds to a measure of how far the statistics deviate from what the Born rule predicts.

The Born Rule is then equivalent to saying that there exists a sequence of such observables corresponding to collecting larger and larger amounts of statistics, such that in the limit of an infinite amount of statistics, the system will always be found to be in an eigenstate of A corresponding to an eigenvalue of zero.

So, for the Born rule to hold we only need to assume a weaker variant of the Born rule that merely says that if a system is known to be in some eigenstate of an observable, then measuring that observable will yield the corresponding eigenvalue of that eigenstate with certainty.