Why is there no consensus about the meaning of probability in MWI?

[PeroK]

No, this is not correct. You have relative frequencies in each world, but each world only has one set of relative frequencies. If I am in the world with six heads, I have no way of knowing that there is only one world like mine, while there are multiple worlds with three heads and three tails.

We have a calculable probability of getting three heads and three tails from six tosses. You toss the coin six times to confirm this. It all works out.

In fact, many elementary probability problems in the homework section are solved this way. They involve only counting, as opposed to explicit probabilities. Most problems involving fair coins, dice or cards succumb to this technique where probability reduces to counting.

 

[pines-demon]

Even in classical mechanics, we have to make some assumptions (like ergodic hypothesis) and then show that those assumptions allow us to define probabilities.

Now we are getting at it. So what are those assumptions? Is the ergodic hypothesis that goes wrong in MWI?

It is not the claim that "we cannot define probabilities". But you have to define the meaning of probabilities in your scenario, especially if you want to prove something about those probabilities, like that they follow BR.

This is circular, I am trying to understand the claim that the probabilities are not well definable in MWI and for that I have to define my own? I am trying to compare it with classical mechanics to see what is different here. Is it the many outcomes? Is is the deterministic aspect? Is it that a omniscient observers that can see all worlds do not see the Born rule?

 

[PeterDonis]

We have a calculable probability of getting three heads and three tails from six tosses.

You have a calculable probability assuming that the coin is fair, i.e., that for a single toss the probability is 1/2 for heads and 1/2 for tails.

But if I live in a world where that coin has come up heads six times in a row, I am going to question whether that coin is fair. I am not going to say, well, there must be other worlds where the coin came up three heads and three tails and my world is just very improbable. It is no response to that to just assert that the coin is fair. In the real world we assess whether coins are fair by looking at the data we see on the relative frequency of heads and tails. If we get data that is evidence against the coin being fair, we take that data into account.

The application of this same logic to the MWI should be clear.

 

[gentzen]

Sure I agree that you can replace collapse with Bohmian arguments but that does not tell me anything about the Born rule. Where in that link discusses the Born rule derivation?

In 9. Quantum Randomness

In both cases it seems natural to try to justify these equilibrium distributions by means of mixing-type, convergence-to-equilibrium arguments (Bohm 1953; Valentini & Westman 2005). It has been argued, however, that in both cases the ultimate justification for these probability distributions must be in terms of statistical patterns exhibited by ensembles of actual subsystems within a typical individual universe (Bell 1981b, reprinted 1987c: 129; Dürr, Goldstein, & Zanghì 1992a). And in both cases the status of, and justification for, equilibrium distributions remains controversial (Dürr & Struyve 2020; see also Goldstein 2012).

It is also perhaps worth noting that the typicality-grounded account of quantum randomness in Bohmian mechanics is extremely similar to Everett’s account (Everett 1957) of quantum randomness for “many worlds”, despite the huge metaphysical differences that exist between these two versions of quantum theory.

• Dürr, Detlef and Ward Struyve, 2020, “Typicality in the Foundations of Statistical Physics and Born’s rule”, in V.Allori, A. Bassi, D. Dürr and N. Zanghì (eds), Do Wave Functions Jump? Perspectives of the Work of GianCarlo Ghirardi, Cham: Springer, pp. 35–43. [Dürr & Struyve 2020 preprint available online]
• Goldstein, Sheldon, 2012, “Typicality and Notions of Probability in Physics”, in Y. Ben-Menahem and M. Hemmo (eds), Probability in Physics, Berlin: Springer-Verlag, pp. 59–71. [Goldstein 2012 preprint available online]

 

[pines-demon]

All of my posts in this thread before your post #43.

Please point to one. Since post #20 I have tried to ask you to clarify more.

I'm not sure I accept the framework you are using in post #43 anyway. But in any case that post doesn't address any of the issues I raised, as I said.

So if I understand correctly you are against the two assumptions. In #43 I ask if (1) the multiple outcomes of MWI and (2) the fact that MWI is deterministic if you considers all the worlds, are the source of the problem to define probabilities in MWI?

 

[kered rettop]

Your answer doesn't make sense to me. Let me phrase it another way. Why do we experience outcomes in agreement with the Born rule given that there is only unitary evolution?I

As others have already explained much research has been made to try and answer this with Zurek, Carroll, Vaidman and Wallace all trying to answer this question.

To just assert this is so doesn't really offer any explanation.

If you carefully go through what Zurek, Carroll, Vaidman and Wallace
are saying you'll see that they all take the same approach which can be simplied as follows:
edit: maybe not Vaidman, I haven't knowingly read him.

Write the global state as a decomposition into a sum-of-products.
Manipulate the expression to create a set of equal-amplitude, orthonormal product terms
Using vector addition, add the amplitudes. Set the total to the original amplitude.
Justify an equal probability assumption of some kind for the equal-amplitude terms.
Using the rule for adding probabilities, add the probabilities.
Normalise the total probability if you haven't already.
Optional. Conjecture that the discrete model here can be generalised to the continuous case.
This gives you the Born Rule.

There doesn't seem to be much room for doubt that it's a valid derivation as long as the equal probability assumption is justifiable. Different authors differ in the way they justify it. And this leads to different meanings for probability. But there isn't so much as a sniff of the Born Rule before it appears as the conclusion.

 

[pines-demon]

In 9. Quantum Randomness

Thanks. I do not think these papers derive it, but instead give arguments on how it could emerge from statistical grounds, so at least they solve the problem of the meaning of probability in Bohmian mechanics, not really a proof of the Born rule.

 

[gentzen]

Now we are getting at it. So what are those assumptions? Is the ergodic hypothesis that goes wrong in MWI?

I think even in the classical case it was slightly controversial, at least the hypotheses. But then QM came along, and it was no longer important to resolve those controversies. But people still joke about the related suicides. (I am no expert on that stuff, so my description of the situation might be completely wrong.)

What goes wrong for MWI is just that there is no agreement, not even among the MWI proponents. And many actual derivations are controversial, just like Sheldon Goldstein described the situation for the typicality argument with respect to Bohmian mechanics.

 

[PeterDonis]

Please point to one.

I only made a few posts before your post #43. Read all of them.

So if I understand correctly you are against the two assumptions.

You don't understand correctly. I said I wasn't sure I accept them. I made my posts before your post #43, so you should not expect my posts to take your assumptions into account. And because I'm not sure I accept your assumptions in post #43, I don't see the point of trying to recast my arguments specifically to address them. Either read my posts before your post #43 and respond to them, or don't.

 

[PeterDonis]

Since post #20 I have tried to ask you to clarify more.

I already responded to your post #20 in my posts #21 and #22.

 

[PeroK]

You have a calculable probability assuming that the coin is fair, i.e., that for a single toss the probability is 1/2 for heads and 1/2 for tails.

But if I live in a world where that coin has come up heads six times in a row, I am going to question whether that coin is fair. I am not going to say, well, there must be other worlds where the coin came up three heads and three tails and my world is just very improbable. It is no response to that to just assert that the coin is fair. In the real world we assess whether coins are fair by looking at the data we see on the relative frequency of heads and tails. If we get data that is evidence against the coin being fair, we take that data into account.

The application of this same logic to the MWI should be clear.

If you could perceive all worlds, you would see there are no probabilities in the branching. That you are confined to one experience means you cannot distinguish between MWI and a fundamentally probabilistic single world.

If there are $N$ worlds, and you have the experience of only one of them, then there is a probability of $1/N$ that you are experiencing a given world out of the $N$. That applies whether you know about the other worlds or not.

PS that, in a nutshell, is where the probabilities arise: the probability you are experiencing a given world.

 

[kered rettop]

I think even in the classical case it was slightly controversial, at least the hypotheses. But then QM came along, and it was no longer important to resolve those controversies. But people still joke about the related suicides. (I am no expert on that stuff, so my description of the situation might be completely wrong.)

I had an idea that suicide driven by despair at ever agreeing with your peers was slightly different from quantum suicide, but I may be wrong. I usually am.

 

[PeterDonis]

If you could perceive all worlds

Which we can't, and the MWI says we can't, so this is irrelevant.

That you are confined to one experience means you cannot distinguish between MWI and a fundamentally probabilistic single world.

That is what the MWI claims, yes. It has to in order to claim that it is a valid interpretation of QM and therefore must make the same predictions for experimental results as standard QM does.

If there are $N$ worlds, and you have the experience of only one of them, then there is a probability of $1/N$ that you are experiencing a given world out of the $N$.

No. You cannot just help yourself to this assertion since it is precisely the point at issue: what concept of "probability" is the MWI using to make this claim? As far as I know, there isn't one that is both generally accepted and justifies the claim.

 

[mattt]

No, this is not correct. You have relative frequencies in each world, but each world only has one set of relative frequencies. If I am in the world with six heads, I have no way of knowing that there is only one world like mine, while there are multiple worlds with three heads and three tails. (Note, though, that these worlds are not identical: the sequence of individual results is different in each one. They just all have three heads and three tails.) So I have no way of making sense of the statement that "three heads and three tails are more probable than six heads". I only know that in my world there were six heads.

In fact, if I am in the world with six heads, my natural conclusion from that data will not be that there must be other worlds with different relative frequencies and my world is very improbable. My natural conclusion will be that the coin is biased towards heads. Which points to another issue with formulating a meaningful concept of probability in the MWI: observers in different worlds with different relative frequencies will not agree on what the probabilities are. And since they can never detect the other worlds, let alone compare data with them, there is no way to even make sense of the notion that they should take other worlds into account in making their probability calculations.

Exactly. That's why I said in another threat about the many worlds interpretations, that in many of those worlds, if they have human-like creatures at all, they will develop different scientific theories/models (to account for what they really observe).

 

[PeterDonis]

If there are $N$ worlds, and you have the experience of only one of them, then there is a probability of $1/N$ that you are experiencing a given world out of the $N$.

You cannot just help yourself to this assertion since it is precisely the point at issue: what concept of "probability" is the MWI using to make this claim? As far as I know, there isn't one that is both generally accepted and justifies the claim.

Let me expand on this. Consider a "quantum coin flip" scenario, where we do spin measurements that standard QM says have a 50-50 chance of each result on many qubits, one at a time, and record the results.

The MWI claim is basically this: the statement that each measurement has a 50-50 chance of each result is justified because that is what the weightings in the wave function say. And when we unpack that, we find that it is saying that, at each measurement, there is a split into two worlds, one for each result, and each world has equal weighting, so there is a 50-50 chance of "we" ending up in each world. (As I have said in previous posts, this is already a misstatement, since "we" end up in every world. But I'll put that aside for this post.)

But how did we get that wave function in the first place? Well, we prepared lots of qubits in the same way and made the same measurement on them, and the recorded statistics tell us the weighting is 50-50. But that raises two issues. The first is one I've already raised in previous posts: the relative frequency of results in one world is not the same thing as the relative weighting of multiple worlds. The MWI takes a wave function that we get from measuring the former, and says that it tells us the latter, but at the very least it should justify that switch.

The second issue is the one I have been trying to articulate in this particular sub-discussion. If we are going to use the observed statistics to tell us what the wave function is, we can't then turn around and claim that the wave function tells us that there are worlds in which the observed statistics are very different, it's just that those worlds are very improbable because they have a much lower weighting. We should expect the interpretation to give back the same observed statistics that were used to construct the wave function on which the interpretation is based. Otherwise there is an obvious issue of consistency.

 

[pines-demon]

Read all of them.

I will try to recreate your argument.

In post #18 you say that:
(1) We never measure branch weightings only weights relative to our own branch.
(2) The ignorance interpretation relies of not having sufficient knowledge. But in MWI all outcomes occur. MWI is not about initial conditions.

Then you explain that (1) can be justified, but (2) cannot.
New question: why is so important 2? Is 2 what makes MWI probabilities ill-defined? Why do we need to invoke initial conditions?

I already responded to your post #20 in my posts #21 and #22.

In post #21, you seem to be responding to my use of the terms but not clarifying.

In #22, you expand a bit more. You argue that:
(3) In classical mechanics, possibility is epistemic: the dynamics have a single outcome but we do cannot have full knowledge
(4) In MWI, possibility is ontic: "the deterministic dynamics is that all of the "possible" outcomes, i.e., all the outcomes that have nonzero amplitudes in the wave function, actually occur. "
Then you argue that in MWI we end in all the branches but there is still a "we" in each branch that cannot detect the rest. You conclue by repeating that standard ignorance interpretation of probability doesn't even make sense.

You seem to affirm that we need a meaningful concept of probability to start with. My question is what in all these elements is what you find key in making probabilities ill-defined?

There are two questions here how to define the Born rule and how to define probabilities. I would argue that if you forget about (1) (sometimes justified as you say), we can come up with probabilities for MWI that would predict different branching that would correspond to our relative weight according to own branch. If this is enough to derive the Born rule is another question.

 

[jbergman]

If you carefully go through what Zurek, Carroll, Vaidman and Wallace
are saying you'll see that they all take the same approach which can be simplied as follows:
edit: maybe not Vaidman, I haven't knowingly read him.

Write the global state as a decomposition into a sum-of-products.
Manipulate the expression to create a set of equal-amplitude, orthonormal product terms
Using vector addition, add the amplitudes. Set the total to the original amplitude.
Justify an equal probability assumption of some kind for the equal-amplitude terms.
Using the rule for adding probabilities, add the probabilities.
Normalise the total probability if you haven't already.
Optional. Conjecture that the discrete model here can be generalised to the continuous case.
This gives you the Born Rule.

There doesn't seem to be much room for doubt that it's a valid derivation as long as the equal probability assumption is justifiable. Different authors differ in the way they justify it. And this leads to different meanings for probability. But there isn't so much as a sniff of the Born Rule before it appears as the conclusion.

Well the doubt creeps in for me in the following case. Imagine you have a quantum observable with two states and probability of 1/4 and 3/4 via the born rule respectively.

All the arguments I have seen that the probability isn't 1/2 for being in either branch require adding additional state and thus creating more worlds and then arguing that 1/4 of the worlds will contain one of our measurements and 3/4 will have the other measurement.

I haven't been quite able to see the justification for this step.

 

[pines-demon]

Well the doubt creeps in for me in the following case. Imagine you have a quantum observable with two states and probability of 1/4 and 3/4 via the born rule respectively.

All the arguments I have seen that the probability isn't 1/2 for being in either branch require adding additional state and thus creating more worlds and then arguing that 1/4 of the worlds will contain one of our measurements and 3/4 will have the other measurement.

I haven't been quite able to see the justification for this step.

I do not think there is one. It is circular. But if all we care is to define probabilities, a method such as this one does the work.

 

[PeterDonis]

I will try to recreate your argument.

I'm not sure I can make sense of your attempted recreation. However, I'll try to respond as best I can.

Then you explain that (1) can be justified, but (2) cannot.

No, that's not what I said in post #18. I said that there are various attempts in the literature to justify (1), but I also said I didn't find any of them convincing. And I said I was not aware of any attempts in the literature to justify (2).

why is so important 2?

They're both important.

what in all these elements is what you find key in making probabilities ill-defined?

Everything.

There are two questions here how to define the Born rule and how to define probabilities.

Yes, I agree, and the second question is logically prior to the first, since the Born Rule assumes that there is a meaningful concept of probability for it to rely on. I have already made this comment in previous posts.

if you forget about (1) (sometimes justified as you say)

No, that's not what I said. See above.

 

[kered rettop]

Well the doubt creeps in for me in the following case. Imagine you have a quantum observable with two states and probability of 1/4 and 3/4 via the born rule respectively.

All the arguments I have seen that the probability isn't 1/2 for being in either branch require adding additional state and thus creating more worlds and then arguing that 1/4 of the worlds will contain one of our measurements and 3/4 will have the other measurement.

I haven't been quite able to see the justification for this step.

The assumption that the probabilities ought to be equal only holds in a scenario where the states are equivalent in all relevant ways and you can invoke the principle of indifference. Otherwise you don't have a clue what they are. MWI goes to great lengths to contruct a set of equiprobable, orthonormal, "microstates" for that purpose. And yes, you do need "more microstates": three times as many in order to get the probability ratio. And three times as many orthogonal vectors add up to sqrt(3) time the amplitude. Hence the Born Rule.

 

[PeterDonis]

MWI goes to great lengths to contruct a set of equiprobable, orthonormal, "microstates" for that purpose.

Does it? I understand that MWI proponents claim this. But where in the math are these "microstates"? There are amplitudes for each term in the wave function, but amplitudes are not states.

 

[kered rettop]

Edit: funny enough the difference between interpretations is not sharp, it even leads to bold claims like MWI= Bohmian mechanics or MWI=superdeterminism

I've heard of Bohmiam mechanics being called "MWI in denial" but why on earth would anyone say "MWI=superdeterminism"?

 

[jbergman]

Does it? I understand that MWI proponents claim this. But where in the math are these "microstates"? There are amplitudes for each term in the wave function, but amplitudes are not states.

I believe they assume there other states that you can tensor product with what you are observing to give you a much larger number of total worlds.

 

[kered rettop]

Does it? I understand that MWI proponents claim this. But where in the math are these "microstates"?

I dont know what you mean by "where in the maths".

There are amplitudes for each term in the wave function, but amplitudes are not states.

True, but why say so? I do not understand your point at all.

 

[gentzen]

I've heard of Bohmiam mechanics being called "MWI in denial" but why on earth would anyone say "MWI=superdeterminism"?

Probably because he misunderstands superdeterminism. Historically, the name "superdeterminism" arose when somebody (I can lookup who it was) pointed out a tricky loophole in Bell's theorem. Bell admitted the existence of that loophole, but tried to dismiss it nevertheless by badmouthing it. I had always admiration for people like Jarek Duda that came up with proposals like maximal entropy random walk, which "accidentally" violated the superderminism loophole.

 

[kered rettop]

I believe they assume there other states that you can tensor product with what you are observing to give you a much larger number of total worlds.

I believe that too. I also believe that, even if they mean something completely different - like "mushrooms talk in the dark", it would still have been a correct assumption. The tensor product one, not the mushroom one.

 

[PeterDonis]

I dont know what you mean by "where in the maths".

I mean just what I say. If I make a spin measurement on a qubit with, say, probability 2/3 for one outcome and 1/3 for the other, the math says there are two outcomes with unequal weights. It does not say there are 3 outcomes, two of which happen to be identical. At least, that's the math I'm aware of. So for anyone who wants to claim there are in fact 3 outcomes, I am asking what math they are using and where in that math the 3 outcomes are, because I don't see them in any math I'm aware of; I only see two outcomes.

why say so?

See above.

 

[pines-demon]

Jarek Duda that came up with proposals like maximal entropy random walk, which "accidentally" violated the superderminism loophole.

Please tell, what is this about?

 

[PeterDonis]

I believe they assume there other states that you can tensor product with what you are observing to give you a much larger number of total worlds.

What states? Where do they come from? And how can it possibly be legitimate to just assume they are there, instead of actually doing the math and showing that they are there?

 

[kered rettop]

Please tell, what is this about?

Well it certainly isn't about the topic of the thread. Which I started because I wanted to know. (I still don't.) Maybe everyone thinks it's been done to death and are now throwing snowballs at each other to relax?

 

[gentzen]

Please tell, what is this about?

Just a concrete example why "superdeterminism" should just means to violate that tricky loophole in Bell's theorem. Of course, it does points to a certain kind of defect, like an effective violation of the arrow of time. But maybe you still can get something good or interesting in exchange for that defect.

 

[pines-demon]

Just a concrete example why "superdeterminism" should just means to violate that tricky loophole in Bell's theorem. Of course, it does points to a certain kind of defect, like an effective violation of the arrow of time. But maybe you still can get something good or interesting in exchange for that defect.

I do not want to get too far from the topic. But so Duda came up with a classical stochastic system that violates statistical independence?

Sorry @kered rettop I will go back to topic soon.

 

[PeterDonis]

the topic of the thread. Which I started because I wanted to know. (I still don't.)

Well, the thread's title question, if not quite answered to your satisfaction, is certainly illustrated by the discussion we've had, in which no consensus has been achieved.

 

[kered rettop]

I mean just what I say. If I make a spin measurement on a qubit with, say, probability 2/3 for one outcome and 1/3 for the other, the math says there are two outcomes with unequal weights. It does not say there are 3 outcomes, two of which happen to be identical. At least, that's the math I'm aware of. So for anyone who wants to claim there are in fact 3 outcomes, I am asking what math they are using and where in that math the 3 outcomes are, because I don't see them in any math I'm aware of; I only see two outcomes.

Well, you've switched to talking about outcomes of the initial measurement. I was talking about the product terms when you do a fine-grained decomposition of the decohered state. (Down to every degree of freedom if necessary.) Is not such a term a state?

 

[kered rettop]

Well, the thread's title question, if not quite answered to your satisfaction, is certainly illustrated by the discussion we've had, in which no consensus has been achieved.

True. But I never doubted the lack of consensus.
Anyway, I still want to digest your post #18 so I'm not giving up on the thread yet.