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

[Quanundrum]

There seems to be as many branches of Everettian QM as there are branches in Everettian QM. One of the things that never seem to be clarified by anyone is whether MWI is really just Modalism or not. I.E. whether *all worlds* happen *all the time*, or if there are specific conditions necessary. I re-read a thread I was part of regarding improbability in MWI and noticed a stark contrast between the participants.

One school of thought seems to advocate for modalism I.E. that everything happens all the time from all branches, and others indicate there are parameters.

To put this into perspective: everyone seems to agree that if MWI is true then a quantum measurement branches the world into all non-zero probabilities, making them actual. However, some say that since the entire universe is fundamentally made up of subatomic particles this means that right now on this branch we inhabit reading this sentence will branch further onto at least 1 where the sun spontaneously goes supernova or turns into a giant pumpkin. The latter seems absurd, but isn't it just a natural consequence of taking MWI *really* seriously?

 

[PeterDonis]

One of the things that never seem to be clarified by anyone is whether MWI is really just Modalism or not.

What is "modalism"? Do you have a reference?

 

[Quanundrum]

What is "modalism"? Do you have a reference?

https://en.wikipedia.org/wiki/Modal_realism

 

[PeterDonis]

re-read a thread I was part of regarding improbability in MWI

Which thread? Please give a link.

 

[PeterDonis]

everyone seems to agree that if MWI is true then a quantum measurement branches the world into all non-zero probabilities, making them actual

No, that's not what MWI says. MWI says that the wave function is real. The wave function already contains all the "branches"; nothing has to be "split" when a measurement takes place. All that happens during a measurement, according to the MWI, is a unitary evolution that entangles subsystems. Nothing is "made" actual that wasn't already.

 

[PeterDonis]

some say that since the entire universe is fundamentally made up of subatomic particles this means that right now on this branch we inhabit reading this sentence will branch further onto at least 1 where the sun spontaneously goes supernova or turns into a giant pumpkin.

There might be a small nonzero amplitude for the first (sun spontaneously going supernova) in the wave function, and if so, yes, MWI says the unitary evolution of the wave function would include it.

There is no amplitude for the second (sun turning into a giant pumpkin) since it violates conservation laws. So the MWI does not say the unitary evolution of the wave function would include it.

 

[Quanundrum]

There might be a small nonzero amplitude for the first (sun spontaneously going supernova) in the wave function, and if so, yes, MWI says the unitary evolution of the wave function would include it.

There is no amplitude for the second (sun turning into a giant pumpkin) since it violates conservation laws. So the MWI does not say the unitary evolution of the wave function would include it.

Why though? I fail to see how they are physically different on a fundamental level. While the pumpkin sounds more ludicrous, it's just a reconfiguration of subatomic particles, just like a sun going supernova?

 

[PeterDonis]

While the pumpkin sounds more ludicrous, it's just a reconfiguration of subatomic particles

No, it isn't, it's a huge change in chemical composition which will violate conservation laws.

 

[Quanundrum]

No, it isn't, it's a huge change in chemical composition which will violate conservation laws.

And a stable sun going supernova in the split of a second isn't?

 

[PeterDonis]

a stable sun going supernova in the split of a second isn't?

If it can go supernova spontaneously, it's not stable.

In any case, a supernova, while it certainly induces reactions that change the chemical composition, does not do so in a way that violates any conservation laws.

 

[Quanundrum]

If it can go supernova spontaneously, it's not stable.

In any case, a supernova, while it certainly induces reactions that change the chemical composition, does not do so in a way that violates any conservation laws.

No, but I mean we know right now that the sun is stable, so for it to become unstable right this second branching off from this current branch, I don't see how that is inducing less radical changes than a pumpkin would?

 

[PeterDonis]

we know right now that the sun is stable

If we actually know that with certainty, then there is zero amplitude in the wave function for it to go supernova spontaneously. That means it can't "become unstable".

If there is in fact a nonzero amplitude in the wave function for the sun to spontaneously go supernova, then we are mistaken if we currently believe the sun is stable.

Since we don't know the exact wave function of the sun, we don't know for sure which of the two alternatives above is in fact the case.

In the case of the sun turning into a pumpkin, however, we do know with certainty that the first of the two alternatives above (zero amplitude in the wave function) is the case. We know that not because we know the exact wave function of the sun, but because we know that turning the sun into a pumpkin would violate conservation laws, and the wave function cannot contain nonzero amplitude for any process that violates conservation laws.

 

[Quanundrum]

If we actually know that with certainty, then there is zero amplitude in the wave function for it to go supernova spontaneously. That means it can't "become unstable".

If there is in fact a nonzero amplitude in the wave function for the sun to spontaneously go supernova, then we are mistaken if we currently believe the sun is stable.

Since we don't know the exact wave function of the sun, we don't know for sure which of the two alternatives above is in fact the case.

In the case of the sun turning into a pumpkin, however, we do know with certainty that the first of the two alternatives above (zero amplitude in the wave function) is the case. We know that not because we know the exact wave function of the sun, but because we know that turning the sun into a pumpkin would violate conservation laws, and the wave function cannot contain nonzero amplitude for any process that violates conservation laws.

So then going back to the example Michael Price gave: how would he become president in at least one branch? That whole argument was based around the particles reconfiguring spontaneously and putting him in the white house while also reconfiguring the particles making up the brains of everyone to produce false memories of him indeed as president?

 

[PeterDonis]

going back to the example Michael Price gave

In what thread? As I asked before, please give a link.

 

[Quanundrum]

In what thread? As I asked before, please give a link.

https://www.physicsforums.com/threads/improbability-of-the-many-worlds-interpretation.906532/page-2

 

[PeterDonis]

Thanks for the link.

That whole argument was based around the particles reconfiguring spontaneously and putting him in the white house while also reconfiguring the particles making up the brains of everyone to produce false memories of him indeed as president?

I'm not entirely sure what process he was thinking of or whether such a process would actually have a nonzero amplitude in the wave function. My basic objection to such scenarios is precisely that: instead of actually trying to figure out what the relevant wave function is and what processes have nonzero amplitudes, people just wave their hands and assume that any process they can imagine will have a nonzero amplitude in the wave function, with no justification whatever for that assumption.

 

[AlexCaledin]

The sun may be suddenly swallowed by a black hole passing by, and then the cosmic rays may well synthesize a pumpkin.

 

[Quanundrum]

The sun may be suddenly swallowed by a black hole passing by, and then the cosmic rays may well synthesize a pumpkin.

Wouldn't this require that said black hole already preexisted on the branch? Or are you saying that it has a non-zero probability of just suddenly "happening"?

 

[PeterDonis]

The sun may be suddenly swallowed by a black hole passing by, and then the cosmic rays may well synthesize a pumpkin.

Please show me the wave function that has a nonzero amplitude for this process.

 

[Demystifier]

And a stable sun going supernova in the split of a second isn't?

There are two types of physical laws:
A) Exact, always valid laws.
B) Statistical laws, the violation of which is very unlikely.

An example of an A-law is energy conservation. An example of a B-law is the 2nd law of thermodynamics. The Sun suddenly transforming into a supernova is forbidden by a B-law, not by an A-law.

 

[Morbert]

The MWI is distinct from modal realism. A modal realist concerns themself with a broader kripkean notion of 'world'. For example:

To put this into perspective: everyone seems to agree that if MWI is true then a quantum measurement branches the world into all non-zero probabilities

A modal realist might also consider a world where the MWI is false to be just as actual as a world where it is true.

Or a world where everything is made up of classical billiard balls.

Or a world where there are no regularities to be codified by laws of physics.

 

[Quanundrum]

There are two types of physical laws:
A) Exact, always valid laws.
B) Statistical laws, the violation of which is very unlikely.

An example of an A-law is energy conservation. An example of a B-law is the 2nd law of thermodynamics. The Sun suddenly transforming into a supernova is forbidden by a B-law, not by an A-law.

I am still struggling a bit with this. While I am of course well aware of the laws of physics in general, I still struggle to see how the usual understanding of them don't go out the window with MWI. Naturally anything that violates energy conservation is impossible, like a perpetuum mobile, but I don't see why rearrangement of the mass of the sun into another state of matter violates it.

Let's use another example:

Take this tweet by Sean Carroll



Given that the brain (based on all known physics at the moment) is a heat bath which kills all decoherence too fast, and the fact that one nucleus decaying would not be sufficient to alter behavior: why is this possible? I.E. it would take millions of atoms all decaying at once for his claim to be real. Sure, this is statistical, but why wouldn't the sun suddenly changing into a pumpkin (it surely has enough energy) also just be statistical?

 

[Demystifier]

Naturally anything that violates energy conservation is impossible, like a perpetuum mobile, but I don't see why rearrangement of the mass of the sun into another state of matter violates it.

Who said that such an rearrangement violates the conservation of energy? It doesn't.

 

[Elias1960]

There might be a small nonzero amplitude for the first (sun spontaneously going supernova) in the wave function, and if so, yes, MWI says the unitary evolution of the wave function would include it.

There is no amplitude for the second (sun turning into a giant pumpkin) since it violates conservation laws. So the MWI does not say the unitary evolution of the wave function would include it.

Wait. Is the wave function of the universe zero for a giant pumpkin at the place of the Sun?

If yes, why? Almost all of the wave functions are nonzero almost everywhere. Zeros are in general position a subset of codimension 2 in the configuration space. 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.

If it is non-zero now, it will probably remain non-zero.

Ok, assume you are right about the amplitudes. But so what? We nonetheless live in a universe with a wave function which tomorrow has a non-zero probability with that pumpkin.

And in MWI we have only that wave function. The amplitudes do not define any reality, they are only tools to compute the evolution of that wave function. If we are today in that branch with the Sun, fine. Tomorrow we may be in that branch with the pumpkin.

Not? That means, there should be some sort of structure which prevents us from jumping into that pumpkin universe. Bohmian trajectories in disguise?

 

[PeterDonis]

Is the wave function of the universe zero for a giant pumpkin at the place of the Sun?

If it isn't, I would like somebody to at least make an argument for why.

Almost all of the wave functions are nonzero almost everywhere.

This statement is way, way, way, way too strong.

We nonetheless live in a universe with a wave function which tomorrow has a non-zero probability with that pumpkin.

Why?

That means, there should be some sort of structure which prevents us from jumping into that pumpkin universe.

You have it backwards. If the Sun turning into a pumpkin is possible, there must be a structure--a set of interactions in the Hamiltonian--that allows it to happen.

 

[Demystifier]

This statement is way, way, way, way too strong.

What's strong about it? It is valid even for "typical" wave functions that physicists like to work with, such as momentum eigenstates and Gaussians.

 

[Demystifier]

If the Sun turning into a pumpkin is possible, there must be a structure--a set of interactions in the Hamiltonian--that allows it to happen.

No. All what is required is that $\langle {\rm pumpkin}|{\rm Sun}\rangle\neq 0$. Most states $|\psi\rangle$ satisfy $\langle\psi|{\rm Sun}\rangle\neq 0$. Those that don't are a subset of measure zero.

 

[Elias1960]

No. All what is required is that $\langle {\rm pumpkin}|{\rm Sun}\rangle\neq 0$.

Even if this would be zero, how do we know that our wave function is $|{\rm Sun}\rangle$ instead of
$\alpha_1|{\rm Sun}\rangle + \alpha_2|{\rm pumpkin}\rangle$?

This statement is way, way, way, way too strong.

No. If a mathematician says "almost all" this has a precise and strong meaning. Say, in differential geometry it means general position: A set defines a general position in B if it is an open subset (that means, there exists a small environment which is completely in this set) and its closure is the whole set B. Assume we have a space of smooth functions $X\to Y$, and Y has dimension n. Let's fix some point $y_0\in Y$, in our case we need $0\in \mathbb{C}$. Then for a map of general position the preimage of $y_0$ has codimension n. That means, it is a set of measure 0.

Ok, the theorem works for smooth maps, while wave functions are in $\mathcal{L}^2(Q,\mathbb{C})$. But let's not forget that we have to care only about wave functions with well-defined energy, and the energy operator contains second derivatives of the wave function. A pure mathematician will, therefore, not yet been satisfied by this short consideration, but I think for a physics forum it is sufficient.

You have it backwards. If the Sun turning into a pumpkin is possible, there must be a structure--a set of interactions in the Hamiltonian--that allows it to happen.

The structure which allows this to happen is the objectively existing wave function of the universe. There is no such animal in MWI as the Sun turning into a pumpkin. As the Sun, as the pumkin are simply points in the configuration space of all possible universes, $q_{Sun}, q_{pumpkin}\in Q$. All what exists is that wave function $\Psi(q)$. It exists objectively, we do not prepare it like in Copenhagen. So, $\Psi(q_{Sun})\neq 0$ seems a reasonable assumption, once there is a Sun around us (or let's hope so, actually I do not see a Sun). But about $\Psi(q_{pumpkin})$ we know nothing. And we cannot simply assume it is zero, because this is something which objectively exists according to MWI, it is nothing we can simply set as we like.

If you think all this makes no sense - ok, I agree, MWI makes no sense.

 

[martinbn]

No. All what is required is that $\langle {\rm pumpkin}|{\rm Sun}\rangle\neq 0$. Most states $|\psi\rangle$ satisfy $\langle\psi|{\rm Sun}\rangle\neq 0$. Those that don't are a subset of measure zero.

Which measure?

 

[martinbn]

No. If a mathematician says "almost all" this has a precise and strong meaning. Say, in differential geometry it means general position: A set defines a general position in B if it is an open subset (that means, there exists a small environment which is completely in this set) and its closure is the whole set B. Assume we have a space of smooth functions $X\to Y$, and Y has dimension n. Let's fix some point $y_0\in Y$, in our case we need $0\in \mathbb{C}$. Then for a map of general position the preimage of $y_0$ has codimension n. That means, it is a set of measure 0.

And if they are infinite dimensional?

 

[PeterDonis]

All what is required is that $\langle {\rm pumpkin}|{\rm Sun}\rangle\neq 0$.

Okay, then please show me how that is true.

Most states $|\psi\rangle$ satisfy $\langle\psi|{\rm Sun}\rangle\neq 0$.

I disagree. What is your basis for this statement?

 

[PeterDonis]

The structure which allows this to happen is the objectively existing wave function of the universe.

Which nobody has ever come within many, many orders of magnitude of actually writing down. So when I hear talk of "the objectively existing wave function of the universe", I translate that to "I can't be bothered to actually do any math, so I'm just going to wave my hands instead".

If you think all this makes no sense - ok, I agree, MWI makes no sense.

I do too. But I also think MWI, even treating it as a valid interpretation for the sake of argument, does not justify the kinds of claims I see being made in this thread. At any rate, I would like to see more justification for those claims than just a hand-waving invocation of "the wave function of the universe".

 

[PeterDonis]

If a mathematician says "almost all" this has a precise and strong meaning.

Then please show me the mathematical theorem that justifies, with that level of rigor, the statement you made.

 

[Elias1960]

Which nobody has ever come within many, many orders of magnitude of actually writing down. So when I hear talk of "the objectively existing wave function of the universe", I translate that to "I can't be bothered to actually do any math, so I'm just going to wave my hands instead".

This nicely characterizes MWI. It is not my invention that MWI claims that this strange entity exists. (The existence of this monster is the reason why I don't like dBB that much, and prefer Caticha's entropic dynamics where the wave function is epistemic.)

But if some interpretation postulates the existence of such an absurd entity, I'm free to use it to show the absurdity of this interpretation, not?

Then please show me the mathematical theorem that justifies, with that level of rigor, the statement you made.

Ok, no problem. The claim was:

Almost all of the wave functions are nonzero almost everywhere.

To say that almost all elements of some Banach space H belong to some subset $S\subset H$ means that
1.) for every element $h\in H$ and every $\varepsilon>0$ there exist some $s\in S$ with $\left\|h-s\right\|<\varepsilon$. (The closure of $S$ is $H$.)
2.) for every element $s\in S$ there exists some $\varepsilon>0$ so that for all $h\in H$ with $\left\|h-s\right\|<\varepsilon$ we have $h \in S$. ($S$ is an open subset of $H$.)
(This is a quite strong variant, there are weaker ones, which, for example, would allow to say that almost all real numbers are irrational. This claim would not fulfill condition (2).)

The space H we will consider is the subspace of elements of $\mathcal{L}^2(Q,\mathbb{C})$ which also belong to $\mathcal{C}^1(Q,\mathbb{C})$.

Let's specify now the "nonzero almost everywhere". $Q$ is assumed to have a metric $d(q_1,q_2)$. Let's fix an arbitrary $q_0\in Q$ and some $\varepsilon_0>0$. Then, let $N^{q_0,\varepsilon_0}$ be the subset of H so that for $\psi \in N^{q_0,\varepsilon_0}$ for all $q\in Q$ with $d(q,q_0)<\varepsilon_0$ we have $\psi(q)=0$.

Theorem: If $\psi$ is not nonzero almost everywhere, then $(\exists q_0\in Q)(\exists \varepsilon_0>0) \psi \in N^{q_0,\varepsilon_0}$.

Theorem: $(\forall q_0\in Q)(\forall \varepsilon_0>0)$ almost all $\psi\in H$ belong to $H\setminus N^{q_0,\varepsilon_0}$.

And if they are infinite dimensional?

Even if $Q$ is infinite-dimensional, the set of zeros of the wave function will, in the general case, have codimension 2.

 

[PeterDonis]

It is not my invention that MWI claims that this strange entity exists.

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.

The claim was

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.

 

[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.