Understanding the Debate: Many Worlds vs Modalism in Everettian QM"

[Infrared]

The general form of the claim is of the sort that @Demystifier gave in post #27.

Disclaimer: I know next to nothing about quantum mechanics, so please let me know if I've gotten something horribly wrong.

What @Demystifier said looks right to me. Almost all (in the sense of @Elias1960) $\psi\in H$ satisfy $\langle\psi,\text{sun}\rangle\neq 0$. The set of such $\psi$ is open, as it is the preimage of the open set $\mathbb{C}^\times$ under the continuous map $\psi\mapsto\langle\psi,\text{sun}\rangle$, and it is dense, because if $\langle\psi,\text{sun}\rangle= 0$, then $\langle\psi+\varepsilon\cdot\text{sun},\text{sun}\rangle\neq 0$ for arbitrarily small $|\varepsilon|.$

I think it is a little imprecise to rephrase this as saying the complement of the set of such $\psi$ has measure zero though, since infinite-dimensional Hilbert spaces do not have measures with the properties we would like to require (https://en.wikipedia.org/wiki/Infinite-dimensional_Lebesgue_measure)

Edit: I realize now that I should have specialized to the subset of vectors with length $1$, but I think the above arguments hold. The intersection of an open set in $H$ with the unit ball is open in the unit ball, and the density argument is the same, as long as renormalize $\psi+\varepsilon\cdot\text{sun}$ to have length $1$.

 

[Elias1960]

MWI claims that there is a universal wave function, yes. But that's not the same as claiming that we know what it is and can show that it contains nonzero amplitudes for processes like the Sun turning into a pumpkin. Not all MWI proponents make the latter sort of claim.

I did not claim that we know what it is. Instead, I said

So, if we assume a general position for the wave function, the wave function will be non-zero in some arbitrary small environment of that pumpkin.

So, I made an assumption. This assumption is a very weak one, given that it is fulfilled by almost all wave functions.

Once making such an extremely weak assumption leads to absurdity, the other side, the MWI proponents, should explain that in MWI only those few exceptions are allowed, or at least (for whatever notion of probability) most probable (which would require some additional structure, thus, destroying the main claimed advantage of MWI). And, moreover, that such a subset of exceptional wave functions remains stable in time. (Which is highly unlikely, try it out with a wave function with initial values localized inside a box and then following the Schroedinger equation without an infinite potential which forces it to remain inside the box.)

Ah, I see: I should have said that your claim was misworded to begin with. The claim that needs to be shown is a claim about transition amplitudes, not wave functions. The general form of the claim is of the sort that @Demystifier gave in post #27.

It is not misworded, because I mean it this way, and my argument relies on this.

There is no reason for me to care about transition amplitudes, because they are only tools for computing the evolution of the wave function. I care about what (according to MWI) really exists, and this is the wave function of the whole universe.

Just to reformulate my point in MWI language (which is hand-waving, sorry, not well-defined, at least I have never seen a mathematical definition of branches of a given general wave function).

So, there is the branch of the Sun, and there is the branch of the pumpkin. Each branch has a wave function localized around the Sun resp. pumpkin, and the Schroedinger evolution preserves this localization sufficiently well.

Now, if the wave function of the universe is that around the Sun, and if the Born rule holds (claimed to be proven, but IMHO the proof makes no sense), then we can predict that we remain around the Sun. But can we make such a claim if the wave function of the universe is $\sqrt{2}^{-1}(\psi_{Sun} + \psi_{pumpkin})$? No. In this case, all we can say according to the Born rule is that the probability of being in the pumpkin branch is $\frac12$.

But once we are now in the Sun branch, isn't it obvious that we will remain here? It may be obvious. But nothing in MWI (beyond hand-waving of MWI proponents) can tell us that it is really so.

In dBB, we have the trajectories, which are continuous. Based on these trajectories, we can make a reasonable claim that we remain around the Sun. MWI does not have such trajectories, and they are even proud of not having them. So what could prevent us from switching to the pumpkin universe? Which non-existing additional structure of MWI does this job?

 

[PeterDonis]

There is no reason for me to care about transition amplitudes, because they are only tools for computing the evolution of the wave function.

The evolution of the wave function is not the issue. The issue is whether the "pumpkin" and "Sun" states are orthogonal or not.

If your claim is a general claim that, given any quantum state (ray in a Hilbert space), the set of quantum states which are orthogonal to it are a set of measure zero, then yes, it would be relevant. I'm not sure whether that statement is equivalent to your claim or not.

there is the branch of the Sun, and there is the branch of the pumpkin

No, these are not branches. They are states in the Hilbert space of the universal wave function. More precisely, we are picking out a particular subset of all of the degrees of freedom in the universal wave function, and hypothesizing that these degrees of freedom have, among their possible states, states that we call "Sun" and "pumpkin" (this is the hypothesis, which had not been explicitly stated until I just stated it, that I find highly doubtful). Then we are asking whether those two states are orthogonal.

 

[Elias1960]

The evolution of the wave function is not the issue. The issue is whether the "pumpkin" and "Sun" states are orthogonal or not.

No, not in my argument. For my argument, this question is completely irrelevant.

No, these are not branches. They are states in the Hilbert space of the universal wave function. More precisely, we are picking out a particular subset of all of the degrees of freedom in the universal wave function, and hypothesizing that these degrees of freedom have, among their possible states, states that we call "Sun" and "pumpkin" (this is the hypothesis, which had not been explicitly stated until I just stated it, that I find highly doubtful). Then we are asking whether those two states are orthogonal.

This makes no sense to me. I do not ask that question, instead I assume they are orthogonal. I have never seen a precise definition of branches which would be applicable to a general wave function, beyond cases like Schroedinger's cat where the two branches are identified (with vague words) quite easily.

I think it is time for me to give up and to accept that you will not understand my argument. As well, it seems impossible for me to understand your point, and I have to give up here too. This should not be an unexpected result, given that MWI is nothing close to be precisely defined, but nothing but hand-waving.

That hand-waving is supported by applying common sense as if it would be an axiom of MWI whenever the conflict between any precise version of MWI and common sense would become obvious (so, to "prove" the Born rule one assumes common sense rules giving Bayesian probability, as if that would make sense in MWI). No wonder that many end up interpreting that hand-waving with a different version of MWI which seems not in contradiction with their own common sense. My version of MWI is, in particular, based on taking some of these hand-waving claims as if they were facts: That there exists the wave function of the universe, but nothing else, and that one does not have to postulate any additional structures. One could start with others, which would give a quite different but also meaningless result.

If you would have a text which gives precise definitions of the terms and proving precise theorems about MWI, proving me wrong, this would be nice, but I have never seen such a thing.

 

[Demystifier]

Which measure?

Lebesgue.

 

[Infrared]

There isn't an analogue of the Lebesgue measure on infinite-dimensional Hilbert spaces. See the link in my post 36.

 

[Demystifier]

I disagree. What is your basis for this statement?

Let $\{|\psi_1\rangle, |\psi_2\rangle, \ldots\}$ be a complete orthogonal basis chosen such that
$$|\psi_1\rangle=|{\rm Sun}\rangle$$
An arbitrary state is a state of the form
$$|\psi\rangle=c_1|\psi_1\rangle+c_2|\psi_2\rangle+\ldots$$
subject to the constraint $|c_1|^2+|c_2|^2+\ldots=1$. Therefore
$$\langle\psi| {\rm Sun}\rangle=c_1^*$$
depends only on $c_1$, and not on $c_2,c_3,\ldots$. Hence the condition
$$\langle\psi| {\rm Sun}\rangle=0$$
is satisfied if and only if $c_1=0$. For all other values of $c_1\neq 0$ we have
$$\langle\psi| {\rm Sun}\rangle\neq 0$$

Concerning the associated Lebesgue measure, it is a technical issue which for practical purposes can be resolved by truncating the Hilbert space to a space with a very large but finite dimension.

 

[Demystifier]

There isn't an analogue of the Lebesgue measure on infinite-dimensional Hilbert spaces. See the link in my post 36.

Thanks for pointing this out, see the last sentence in my #42.

 

[PeterDonis]

I do not ask that question, instead I assume they are orthogonal.

Yes, you're right, "orthogonal" was the wrong word for what I was thinking of. See my response to @Demystifier coming in a moment.

 

[PeterDonis]

the condition
$$
\langle\psi| {\rm Sun}\rangle=0
$$
is satisfied if and only if $c_1=0$.

Yes, I see what you are saying now, and I also see that I was misstating the concern I have.

First, to give a concrete example of your argument, take the Hilbert space of a qubit. If I pick a particular qubit state, say spin-z up, the set of qubit states that are exactly orthogonal to that state is a set of measure zero, yes: it's the set containing the single state spin-z down. All other states have a nonzero spin-z up component.

Now, to restate my argument using the same concrete example: suppose I have a qubit in the spin-x up state. That state is not orthogonal to spin-z up, but that doesn't mean that a spin-x up qubit can magically turn into a spin-z up qubit any time it wants. The qubit has to undergo some process that changes its state, and that process has some Hamiltonian associated with it. (For example, we could put the qubit through a magnetic field.) This is true regardless of what QM interpretation we adopt, so it's true for the MWI. The MWI does not say that if we have a spin-x up qubit, there is some "branch" of the wave function in which it is a spin-z up qubit. All the MWI says is that if we measure the spin of this spin-x up qubit in the spin-z direction, we will get a spin-z up branch and a spin-z down branch. But the measurement is an interaction--it's a process that has a Hamiltonian associated with it.

Now please show me the process that takes a subspace of the universal Hilbert space containing a huge number of degrees of freedom in the "Sun" state and puts that subspace into the "pumpkin" state. Just showing me that "Sun" and "pumpkin" have a nonzero wave function overlap (as spin-z up and spin-x up qubit states do) is not enough; that by itself does not mean the MWI says the "Sun" state can magically turn into the "pumpkin" state.

 

[AlexCaledin]

A toy universe consisting of a single qubit or pumpking or whatever is something quite different from the real enormous universe having the same thing entangled with all the rest.

 

[martinbn]

Concerning the associated Lebesgue measure, it is a technical issue which for practical purposes can be resolved by truncating the Hilbert space to a space with a very large but finite dimension.

According to your terminology, any finite dimensional subspace will be of measure zero.

 

[Demystifier]

Now please show me the process that takes a subspace of the universal Hilbert space containing a huge number of degrees of freedom in the "Sun" state and puts that subspace into the "pumpkin" state.

If you want, I can write down a Hamiltonian that would induce such a process. (Hint: Consider e.g. the operator $|{\rm pumpkin}\rangle\langle{\rm Sun}|+h.c.$). But you will be happy to know that such a Hamiltonian cannot be realized in practice in the laboratory.

 

[PeterDonis]

you will be happy to know that such a Hamiltonian cannot be realized in practice in the laboratory.

The question is not whether it can be realized in the laboratory, but whether it is part of the Hamiltonian that actually governs the dynamics of the Sun. I don't see any reason for thinking that it is.

 

[Demystifier]

The question is not whether it can be realized in the laboratory, but whether it is part of the Hamiltonian that actually governs the dynamics of the Sun. I don't see any reason for thinking that it is.

Neither do I.