I Carroll's Derivation of the Born Rule in MWI with Unequal Amplitudes

[jbergman]

TL;DR Summary

The equal probability case of the Born rule is relatively easy to grok in the MWI discussion. I wanted to have a focused discussion on the case where the magnitude of the amplitudes are unequal and derivations from some of the literature in the field.

In Caroll and Seben's paper, Many Worlds, the Born Rule, and Self-Locating Uncertainty, they present a derivation of the Born rule.

For the equal probability case their derivation is based on the following principles.

• Self Locating Uncertainty - "the condition of an observer who knows that the environment they experience occurs multiple times in the universe, but doesn’t know which example they are actually experiencing"
• The Principle of Indifference - "if all you know is that you are one of N occurrences of a particular set of observer data, you should assign equal credence, 1/N, to each possibility"

But I want to focus on the unequal probability case. We have state of the universe with spin prepared such that by the Born rule $P(\downarrow) = 2/3$.
$$| \Psi \rangle = | O \rangle (|\uparrow \rangle + \sqrt{2} | \downarrow \rangle)| E \rangle$$
According to Carroll, entanglement of the measurement happens before the observer registers so the state evolves to
$$| \Psi \rangle = | O \rangle (|\uparrow \rangle | E_{\uparrow} \rangle + \sqrt{2} | \downarrow \rangle | E_{\downarrow} \rangle)$$
Finally the observer sees the measurement ant the state of the observer changes on each branch.

$$| \Psi \rangle = | O_{\uparrow} \rangle |\uparrow \rangle | E_{\uparrow} \rangle + \sqrt{2} | O_{\downarrow} \rangle| \downarrow \rangle | E_{\downarrow} \rangle$$

Carroll himself points out that the principle of indifference might suggest that branch counting would give an equal probability for each outcome. However, Carroll adds another principle, "Epistemic Separability", which he argues that this isn't so.

Epistemic Separability is basically interpreted as saying that a unitary transformation that acts only on the environment doesn't affect the probabilities of the Observer/System subsystem. So in the unequal probabilities case they consider the state we described above which is after measurement but before the observer has registered the measurement (where we have relabeled $E_{\downarrow}$ as $E_1$, etc., because we will need to consider more orthonormal states of the environment besides up and down.

$$| \Psi \rangle = | O \rangle (|\uparrow \rangle | E_1 \rangle + \sqrt{2} | \downarrow \rangle | E_2 \rangle)$$

They then consider a unitary transformation,

$$U = | \hat{E}_1 \rangle \langle E_1 | + \frac{(| \hat{E}_2 \rangle + | \hat{E}_3 \rangle )}{\sqrt{2}} \langle E_2 | + \frac{(| \hat{E}_2 \rangle - | \hat{E}_3 \rangle )}{\sqrt{2}} \langle E_3 | + \sum_{\mu > 3} | \hat{E}_{\mu} \rangle \langle E_{\mu} |$$

Then

$$ U | \Psi \rangle = | O \rangle (|\uparrow \rangle | \hat{E}_1 \rangle + | \downarrow \rangle | \hat{E}_2 \rangle + | \downarrow \rangle | \hat{E}_3 \rangle)$$

So, in essence they have applied a unitary transformation that only affects the environment that results in three states with equal amplitude and hence equal probabilities for each of the worlds. This leads to their conclusion that since $\uparrow$ occurs in one of the worlds and $\downarrow$ occurs in the other 2 that,
$$P(\uparrow | \Psi) = 1/2 P(\downarrow | \Psi) = 1/3$$

This argument is similar to Zurek's and essentially can be seen as saying that their is a freedom to introduce extra state from the environment such that we can arrange to have a set of orthonormal states/worlds such that they all have equal amplitude.

Overall, the argument seems logical, but a few things bother me. For one, it implies that branch counting is ok when we have equal probabilities but not when we have unequal probabilities. It also suggests that every state that has unequal probabilities is equivalent to a state with equal magnitude amplitudes when looking at a subsystem. Lastly, it doesn't seem to fit nicely with the description of worlds as decoherent orthonormal states. We have somehow had to introduce this additional unitary operator which we applied after to get more decoherence to achieve the appropriate branch counts.

I feel like the Principle of Indifference is not solid enough and instead we need a more physical explanation that allows to rely only on branch counts.

 

[kered rettop]

The principle of indifference (PoI) is easy to abuse. This page explains the issues.

"Simply stated, it suggests that if there are n possible outcomes and there is no reason to view one as more likely than another, then each should be assigned a probability of 1/n."
"However, this principle has to be used with great care. [Example] ...this type of counterexample can be ruled out by referring to the structure of the problem and arguing that there is sufficient reason to find three outcomes more likely than one."
"More serious problems arise when we apply the principle to everyday problems, which are often not endowed with sufficient symmetries."

The fact that we do have sufficient reason to think that nature cares about amplitude means we cannot apply the PoI to naive branch-counting. Carroll and Sebens are quite clear about how their argument works: "This route to the Born Rule has a simple physical interpretation. Take the wave function and write it as a sum over orthonormal basis vectors with equal amplitudes for each term in the sum (so that many terms may contribute to a single branch). Then the Born Rule is simply a matter of counting – every term in that sum contributes an equal probability."

Note that they are not saying "just count the branches". They are saying "count the terms you have just written." These terms are not Everettian branches, they are terms in an appropriate decomposition of an Everettian branch.

Of course, "every term in that sum contributes an equal probability" is only true if the outcomes in each term are independent. This means we can't just divide an awkward term into two half-sized ones. The outcome - if such a thing is even meaningful - would be the same in both cases and they won't "contribute an equal probability."

So, to derive the Born Rule using counting methods we need a formula for how the probabilities add, and another formula for how amplitudes add. (As spelled out in the quote from Carroll and Sebens above. Except they don't mention the formulae explicitly.) The formulae will generally both depend on N, so N can be eliminated, leaving probability as a function of amplitude.

Consider four cases:
1 The terms are orthogonal. The total or branch amplitude is then √N times the "term" amplitude. The outcomes are mutually exclusive so their probabilities can simply be added. i.e. their total is N times the "term" probability. This gives the Born Rule, P=√A
2 The terms are parallel. The total or branch amplitude is then N times the "term" amplitude. The probabilities (if that's even a meaningful concept) are for 100% correlated outcomes, so the total probability is the same as a single "term" probability, i.e. 1 x the "term" probability. The Born Rule is replaced by P=A. That's also a good result. It shows that the proper Born Rule requires orthogonality - making decoherence an integral part of QM.
3 The terms have all sorts of phases. You might be able to use a statistical argument, but it wouldn't be counting.
4 The terms have some other statistics due to hidden variables or a dash of superdeterminism. You're probably sunk.


Edited for typos and spellos

 

[jbergman]

The fact that we do have sufficient reason to think that nature cares about amplitude means we cannot apply the PoI to naive branch-counting. Carroll and Sebens are quite clear about how their argument works: "This route to the Born Rule has a simple physical interpretation. Take the wave function and write it as a sum over orthonormal basis vectors with equal amplitudes for each term in the sum (so that many terms may contribute to a single branch). Then the Born Rule is simply a matter of counting – every term in that sum contributes an equal probability."

Note that they are not saying "just count the branches". They are saying "count the terms you have just written." These terms are not Everettian branches, they are terms in an appropriate decomposition of an Everettian branch.

I disagree with your last sentence, "These terms are not Everettian branches, they are terms in an appropriate decomposition of an Everettian branch".

In Saunders paper, he actually implies the opposite is true, i.e, that there are a large number of decoherent branches associated with each term.

The way I view the decomposition into orthonormal states is that we are decomposing into logically separate worlds which actually may result in many physical branches but that all have the same properties with respect to the subsystems of interest.

I have to look more closely at that quote you provided since it seems to suggest Carroll agrees with your reading.

 

[jbergman]

I disagree with your last sentence, "These terms are not Everettian branches, they are terms in an appropriate decomposition of an Everettian branch".

In Saunders paper, he actually implies the opposite is true, i.e, that there are a large number of decoherent branches associated with each term.

The way I view the decomposition into orthonormal states is that we are decomposing into logically separate worlds which actually may result in many physical branches but that all have the same properties with respect to the subsystems of interest.

I have to look more closely at that quote you provided since it seems to suggest Carroll agrees with your reading.

Looking more closely, I see what you are saying. For example looking at the state I gave of,

$$ U | \Psi \rangle = | O \rangle (|\uparrow \rangle | \hat{E}_1 \rangle + | \downarrow \rangle | \hat{E}_2 \rangle + | \downarrow \rangle | \hat{E}_3 \rangle)$$

We see that two of the terms could be considered as belonging to one branch and another term as belonging to another.

Looking more closely at Carroll and Seben's argument it actually seems to rely primarily on the principle of indifference and the principle of epistemic separability and hence doesn't really make much of a physical argument at all which still bothers me. Saunders has an interesting paper that looks like an interesting alternative that I am still digesting. https://arxiv.org/abs/2201.06087

 

[PeterDonis]

We see that two of the terms could be considered as belonging to one branch and another term as belonging to another.

That is my understanding of how the MWI defines branches: by macroscopically distinguishable outcomes. So any "branch" will in fact not be a single microscopic quantum state but a subspace of the Hilbert space containing a huge number of microscopic quantum states, all of which have the same macroscopic properties. In other words, the kets $\ket{\uparrow}$ and $\ket{\downarrow}$ are macroscopically distinguishable, but the kets $\ket{E_2}$ and $\ket{E_3}$ are not (and in fact the ket $\ket{E_1}$ would not need to be distinguishable from those two either, since we could take all of the distinguishable parts and fold them into $\ket{\uparrow}$, so that ket now describes all of the macroscopic properties that correspond to "spin up observed").

 

[jbergman]

That is my understanding of how the MWI defines branches: by macroscopically distinguishable outcomes. So any "branch" will in fact not be a single microscopic quantum state but a subspace of the Hilbert space containing a huge number of microscopic quantum states, all of which have the same macroscopic properties. In other words, the kets $\ket{\uparrow}$ and $\ket{\downarrow}$ are macroscopically distinguishable, but the kets $\ket{E_2}$ and $\ket{E_3}$ are not (and in fact the ket $\ket{E_1}$ would not need to be distinguishable from those two either, since we could take all of the distinguishable parts and fold them into $\ket{\uparrow}$, so that ket now describes all of the macroscopic properties that correspond to "spin up observed").

We have to be a bit careful. This is related to the preferred basis problem. There are many different ways to choose a basis for a Hilbert space. In some of those, a "world" might be the sum of many basis vectors and in others it might not.

 

[PeterDonis]

This is related to the preferred basis problem.

Sort of. Defining a subspace of the Hilbert space that consists of all states with the same macroscopic properties does require defining which macroscopic properties you are going to use. Doing that can be viewed as choosing a basis--for example, choosing the direction for the spin measurement in your example. But the subspace you obtain is basis independent. For example, the subspaces you get for the "up" and "down" results of a spin-z measurement are the same even if you choose to write the states in, say, the spin-x basis; the states in each subspace look different mathematically, but physically they're the same states.

 

[jbergman]

Sort of. Defining a subspace of the Hilbert space that consists of all states with the same macroscopic properties does require defining which macroscopic properties you are going to use. Doing that can be viewed as choosing a basis--for example, choosing the direction for the spin measurement in your example. But the subspace you obtain is basis independent. For example, the subspaces you get for the "up" and "down" results of a spin-z measurement are the same even if you choose to write the states in, say, the spin-x basis; the states in each subspace look different mathematically, but physically they're the same states.

A pure state is just a vector so I'm not sure why you bring subspaces into the discussion. What I would say is that the vector components will look different in a different basis, but as you say it is the same vector. So, I didn't quite get your comment about a large number of microstates corresponding to a macrostate.

 

[PeterDonis]

A pure state is just a vector so I'm not sure why you bring subspaces into the discussion

A ket can represent a single vector (more precisely, ray), or it can represent a subspace. In the cases under discussion, where we are explicitly including an environment with a huge number of degrees of freedom and we are talking about macroscopically distinguishable measurement results, the latter is always going to be the case, because a measurement result like "observed spin-z up" corresponds to a huge number of possible quantum states of the measuring device and its environment, not just one.

I didn't quite get your comment about a large number of microstates corresponding to a macrostate.

As above, for any macroscopic measuring device, a macrostate like "observed spin-z up" will correspond to huge number of quantum microstates, not just one.

 

[kered rettop]

Looking more closely, I see what you are saying. For example looking at the state I gave of,

$$ U | \Psi \rangle = | O \rangle (|\uparrow \rangle | \hat{E}_1 \rangle + | \downarrow \rangle | \hat{E}_2 \rangle + | \downarrow \rangle | \hat{E}_3 \rangle)$$

We see that two of the terms could be considered as belonging to one branch and another term as belonging to another.

There's no "could be considered" about it"! They do by the definition of "branch" as a macroscopic state corresponding to a macroscopic world with a particular outcome.

Looking more closely at Carroll and Seben's argument it actually seems to rely primarily on the principle of indifference and the principle of epistemic separability and hence doesn't really make much of a physical argument at all which still bothers me. Saunders has an interesting paper that looks like an interesting alternative that I am still digesting. https://arxiv.org/abs/2201.06087

ES reduces the slippery argument that "all my substates are equivalent so I assign equal probabilities to them" to the more plausible-sounding "probabilities should not depend on how two air molecules interact the other side of the room". Etc. That's a physical argument. And the principle of indifference no longer has to be applied to an ensemble of sub-worlds. If it's still required at all, it is only to an ensemble of environmental configurations.

 

[jbergman]

There's no "could be" about it. They do by the definition of "branch" as a macroscopic state corresponding to a macroscopic world with a particular outcome.

A branch is the state corresponding to macroscopic world. The particular outcome bit is false. In fact, I can have two spin down branches that are separately decoherent because of other entanglement like the air molecules you described. Those are not one macroscopic world, but instead two separate macroscopic worlds.

 

[PeterDonis]

A branch is the state corresponding to macroscopic world.

A "macroscopic world" does not correspond to a single quantum state. A given macroscopic state--for example, a macroscopic measuring device that registers the result "observed z-spin up"--can be realized by any of a huge number of microscopic quantum states of the atoms that compose the device, and which particular one of those states the macroscopic device is in will fluctuate from moment to moment. You have to take that into account in any proper understanding of what a "macroscopic world" is.

The particular outcome bit is false.

No, your counterclaim is false.

In fact, I can have two spin down branches that are separately decoherent because of other entanglement like the air molecules you described. Those are not one macroscopic world, but instead two separate macroscopic worlds.

If the air molecule interactions do not affect any macroscopic variable (such as temperature, pressure, etc.), then they do not result in any branching of "macroscopic worlds".

I strongly suggest that you take a step back and consider carefully what you are saying. As far as I can tell, your claims have no basis in any references that discuss the MWI, certainly not in the Carroll reference that is under discussion in this thread.

 

[kered rettop]

A branch is the state corresponding to macroscopic world. The particular outcome bit is false. In fact, I can have two spin down branches that are separately decoherent because of other entanglement like the air molecules you described. Those are not one macroscopic world, but instead two separate macroscopic worlds.

Not so. Worlds are characterised by the macroscopic phenomena, the observable outcomes. They are a single world. The objects you refer to are sub-branches. This is merely semantics and has no bearing on the validity of Carroll and Sebens's arguments.

 

[jbergman]

If the air molecule interactions do not affect any macroscopic variable (such as temperature, pressure, etc.), then they do not result in any branching of "macroscopic worlds".

I agree that macroscopic state must be affected. You edited out the part about entanglement before I mentioned the molecules. If a large enough # of other interactions occur to change some state like temperature then you have a different decoherent branch.

That may have nothing to do with whether the spin was measured up or down. I will cite Saunders paper later where he specifically talks about this scenario.

 

[jbergman]

Not so. Worlds are characterised by the macroscopic phenomena, the observable outcomes. They are a single world. The objects you refer to are sub-branches. This, by the way, is pure semantics and has no bearing on the validity of Carroll and Seben's arguments.

I agree about macroscopic phenomena, but you seem to suggest that all of the spin down parts of the wavefunction were in a single world. I was disputing that claim because you could have other macroscopic phenomena that changed besides that.

I also agree that it has no bearing on their argument.

 

[PeterDonis]

If a large enough # of other interactions occur to change some state like temperature then you have a different decoherent branch.

Yes, but "entanglement" by itself doesn't do that. For a macroscopic state like temperature to change, an interaction that exchanges energy between the environment and some macroscopic object needs to happen. Just internal interactions between air molecules doesn't do that.

you seem to suggest that all of the spin down parts of the wavefunction were in a single world.

In the context of that particular measurement being made, yes.

I was disputing that claim because you could have other macroscopic phenomena that changed besides that.

Not just "other macroscopic phenomena", but macroscopic phenomena that depend on quantum uncertainty, like a measurement result. Of course not all such phenomena are measurement results, but on the other hand the category of such phenomena is a lot narrower than "well, pretty much everything that happens at all".

 

[jbergman]

A ket can represent a single vector (more precisely, ray), or it can represent a subspace. In the cases under discussion, where we are explicitly including an environment with a huge number of degrees of freedom and we are talking about macroscopically distinguishable measurement results.

This is false. Please show an example where a ket is a subspace (of dimension > 1). Maybe you are thinking about density matrices.

As above, for any macroscopic measuring device, a macrostate like "observed spin-z up" will correspond to huge number of quantum microstates, not just one.

Yes. But at a single moment in time it will only occupy a single microstate.

 

[kered rettop]

I agree that macroscopic state must be affected. You edited out the part about entanglement before I mentioned the molecules. If a large enough # of other interactions occur to change some state like temperature then you have a different decoherent branch.

Yes, and if they don't then you don't. If you want to talk about probabilites then you need to specify which measured property(ies) you're talking about. If it's spin then all the states with spin-up belong in the spin-up world, regardless of temperature.

 

[PeterDonis]

Please show an example where a ket is a subspace (of dimension > 1).

Um, every ket that has been written down in this thread? All of those kets refer to systems with very large numbers of degrees of freedom, in which there are a very large number of microstates that all correspond to the same macroscopic parameter (for example, "observed z-spin up" for $\ket{\uparrow}$). None of them correspond to a simple single state of a simple system like a qubit.

 

[jbergman]

On the topic of branches, from Saunders' Branch Counting in the Everett Interpretation of Quantum Mechanics,

It thus follows from decoherence theory that huge numbers of branches are produced in a single run of any realistic experiment. For each outcome there are large numbers of branches, differing in ways irrelevant to the outcome.

 

[PeterDonis]

On the topic of branches, from Saunders' Branch Counting in the Everett Interpretation of Quantum Mechanics,

The claims there are, AFAIK, debatable (to say the least). I don't think they represent any kind of broad consensus even among MWI proponents.

(Note, btw, that this reference is not a peer-reviewed paper, it's just one particular person's opinion.)

 

[jbergman]

The claims there are, AFAIK, debatable (to say the least). I don't think they represent any kind of broad consensus even among MWI proponents.

(Note, btw, that this reference is not a peer-reviewed paper, it's just one particular person's opinion.)

It was published here.

 

[kered rettop]

I agree about macroscopic phenomena, but you seem to suggest that all of the spin down parts of the wavefunction were in a single world. I was disputing that claim because you could have other macroscopic phenomena that changed besides that.

That is exactly what I'm "claiming". Alice measures a spin and creates a spin-up and a spin-down world. Bob measure temperature and creates an infinite ensemble of temperature-worlds. The splits intersect to create an infinite ensemble of [spin, temperature] worlds. Seems simple enough to me!

 

[PeterDonis]

It was published here.

That was not at all evident from the link you originally posted. If you are posting a reference to a published paper, please post the reference to the actual publication. (If it's an arxiv link, post a link to the arxiv preprint page, not the PDF itself, since the PDF in this case had no publication information whatever. But a link to the actual journal article page, as you have given now, is even better.)

 

[jbergman]

That is exactly what I'm "claiming". Alice measures a spin and creates a spin-up and a spin-down world. Bob measure temperature and creates an infinite ensemble of temperature-worlds. The splits intersect to create an infinite ensemble of [spin, temperature] worlds. Seems simple enough to me!

Again from Saunder's paper, this suggests to me that the situation is more complicated, but I agree that Carroll's argument doesn't rely on anything. Instead his simple symmetry arguments justify his conclusion.

Subsequent landmark papers [17–20] barely mentioned the concept of branching, but showed that in scattering environments (gasses and radiation baths), coherent states for much smaller masses, going down to large molecules, obey quasi-classical equations (‘quasi-classical' as going beyond classical Hamiltonian mechanics—as including friction, for example). But then they are branches in our sense.7 It follows that a delocalized state of masses of the order of large molecules (and anything greater), in a scattering environment involving much lighter degrees of freedom, branches into localized states, each evolving quasi-classically, unable to interfere with each other. The suppression of interference is moreover extremely rapid; in gases at ordinary temperatures, it is much faster than thermal relaxation times.8

These examples are important because thermal environments are ubiquitous; for a tiny speck of dust, and anything larger, even the cosmic microwave background will do. They are called examples of ‘quantum Brownian motion', because the dissipation term is linear in the velocity. There are many other kinds of decoherence (see, for example, [23]), but they fit the same mould: the Everett interpretation does not just concern the macroscopic, it concerns the emergence of dynamics at near-microscopic scales as well. And in quantum Brownian motion, molecular degrees of freedom do not only couple to the environment, they couple with each other—whether by collisions, near-neighbour interactions, or exchanges of phonons. Superpositions of localized states will be produced all the time, for example, of reflected and transmitted states, produced in collisions, which since differing in velocities, rapidly decohere, hence produce branching; no special Hamiltonian (as in chaos) is needed. The implication is that for macroscopic numbers of degrees of freedom at ordinary temperatures, branching in Everett's sense is enormous, rapid and ubiquitous.

 

[kered rettop]

Again from Saunder's paper, this suggests to me that the situation is more complicated, but I agree that Carroll's argument doesn't rely on anything. Instead his simple symmetry arguments justify his conclusion.

How do you square that? If simple symmetry arguments justify Carroll and Sebens's conclusion, how can the situation be more complicated? The slab of text you quoted from Saunders doesn't seem to have any bearing on their argument: it just tells the reader that decoherence is everywhere.

 

[jbergman]

How do you square that? If simple symmetry arguments justify Carroll and Sebens's conclusion, how can the situation be more complicated? The slab of text you quoted from Saunders doesn't seem to have any bearing on their argument: it just tells the reader that decoherence is everywhere.

By situation, I mean physical branches are hard to understand and reason about. Carroll and Sebens have come up with an approach that makes that largely unnecessary by allowing one to basically ignore transformations to the environment outside of the subsystem of interest.

I was hoping that there might be another way to understand things more directly in terms of counts of physical branches, but that may not be possible.

 

[Quantumental]

Surely whether the change is macroscopic or microscopic does not have any bearing on whether or not there's several branches in MWI?

If I throw a tennis ball against the wall in virtually every world it will bounce back, it doesn't have tunnel through the wall for it to be counted as having created several branches.

 

[jbergman]

Surely whether the change is macroscopic or microscopic does not have any bearing on whether or not there's several branches in MWI?

If I throw a tennis ball against the wall in virtually every world it will bounce back, it doesn't have tunnel through the wall for it to be counted as having created several branches.

Not sure what the tennis ball is intended to illustrate. But I believe you are correct. That was basically what the quote from the Saunders article I mentioned above was supposed to illuminate.

 

[PeterDonis]

Surely whether the change is macroscopic or microscopic does not have any bearing on whether or not there's several branches in MWI?

"Macroscopic" isn't quite the right word; the right word is "decoherence", since that is actually what defines "branches" in the MWI. I should have been more precise about that earlier. (The original MWI was developed before decoherence theory, so it had to rely on vague terms like "macroscopic" and "measurement" to define when branching happened, but it doesn't need that now; "decoherence" is both more precise and better grounded in the actual dynamics.)

 

[PeterDonis]

this suggests to me that the situation is more complicated

Saunders is trying to describe it as more complicated, but I'm not convinced his description is actually correct.

For example, consider a gas in a box in thermal equilibrium. The gas molecules are constantly interacting, and according to Saunders, that means that countless decoherent branches are constantly being created. But is that really correct? Why do the interactions decohere? They don't change the temperature, pressure, density, volume, etc. of the gas. I get that the interactions mean we can't, in principle, treat the gas molecules as having definite microscopic positions and velocities which we just don't know, as classical statistical mechanics does--but as a matter of actual fact, classical statistical mechanics works just fine in this domain. Adding QM to the mix doesn't really change anything. So where is the decoherence?

Carroll and Sebens have come up with an approach that makes that largely unnecessary by allowing one to basically ignore transformations to the environment outside of the subsystem of interest.

But in many cases, the system is its own "environment". For example, consider the gas in the box above. Suppose we run a Schrodinger's cat type experiment, where we have a radioactive atom, and if it decays, we move a piston that changes the volume of the gas (and therefore also changes other thermodynamic parameters), otherwise we leave the gas alone. The two alternative outcomes (gas volume changed, gas volume not changed) will decohere, so we have two branches from that decoherence process. But that decoherence happens just within the gas; we don't need to invoke any interaction with the outside environment for it to happen. (We do have to interact with the gas from the outside to measure whether its volume and other properties have changed, but the decoherence happens whether we measure the gas or not.)

I think Carroll and Sebens would say that, in this case, we have two branches. But it seems like Saunders would say we have a huge number of branches, that fall into two categories that are all we can distinguish. And since no interaction with the environment is required, on Saunders' view, to create all those huge number of branches, I'm not sure how Carroll and Sebens would rebut Saunders' view in this case, unless they would take the position I take above, that there isn't any decoherence at all within each of the two branches--the gas molecule interactions, if we hold the thermodynamic variables as fixed, do not decohere anything beyond what was already decohered in each branch (volume changed, volume not changed).

 

[jbergman]

Saunders is trying to describe it as more complicated, but I'm not convinced his description is actually correct.

For example, consider a gas in a box in thermal equilibrium. The gas molecules are constantly interacting, and according to Saunders, that means that countless decoherent branches are constantly being created. But is that really correct? Why do the interactions decohere? They don't change the temperature, pressure, density, volume, etc. of the gas. I get that the interactions mean we can't, in principle, treat the gas molecules as having definite microscopic positions and velocities which we just don't know, as classical statistical mechanics does--but as a matter of actual fact, classical statistical mechanics works just fine in this domain. Adding QM to the mix doesn't really change anything. So where is the decoherence?

I am still working on fully understanding decoherence. This is probably a topic for another thread, because I think it is a key part in understanding MWI branches.

 

[PeterDonis]

I am still working on fully understanding decoherence. This is probably a topic for another thread

Or several. I believe there have been some previous PF threads on the topic, including some good references to the literature; it might be worth searching PF.

 

[kered rettop]

Saunders is trying to describe it as more complicated, but I'm not convinced his description is actually correct.

For example, consider a gas in a box in thermal equilibrium. The gas molecules are constantly interacting, and according to Saunders, that means that countless decoherent branches are constantly being created. But is that really correct? Why do the interactions decohere? They don't change the temperature, pressure, density, volume, etc. of the gas. I get that the interactions mean we can't, in principle, treat the gas molecules as having definite microscopic positions and velocities which we just don't know, as classical statistical mechanics does--but as a matter of actual fact, classical statistical mechanics works just fine in this domain. Adding QM to the mix doesn't really change anything. So where is the decoherence?


But in many cases, the system is its own "environment". For example, consider the gas in the box above. Suppose we run a Schrodinger's cat type experiment, where we have a radioactive atom, and if it decays, we move a piston that changes the volume of the gas (and therefore also changes other thermodynamic parameters), otherwise we leave the gas alone. The two alternative outcomes (gas volume changed, gas volume not changed) will decohere, so we have two branches from that decoherence process. But that decoherence happens just within the gas; we don't need to invoke any interaction with the outside environment for it to happen. (We do have to interact with the gas from the outside to measure whether its volume and other properties have changed, but the decoherence happens whether we measure the gas or not.)

I think Carroll and Sebens would say that, in this case, we have two branches. But it seems like Saunders would say we have a huge number of branches, that fall into two categories that are all we can distinguish. And since no interaction with the environment is required, on Saunders' view, to create all those huge number of branches, I'm not sure how Carroll and Sebens would rebut Saunders' view in this case, unless they would take the position I take above, that there isn't any decoherence at all within each of the two branches--the gas molecule interactions, if we hold the thermodynamic variables as fixed, do not decohere anything beyond what was already decohered in each branch (volume changed, volume not changed).

Carroll and Sebens use observable outcomes to structure the branching, as they are talking about many worlds. So they say there are two branches. However they do also create sub-branches like the ones Saunders apparently refers to as branches. So a different name is needed: it's just semantics.

I don't think distinguishability is relevant here. As I understand it, decoherence means loss of phase relationships, which is guaranteed by the orthogonality of the components after enough interactions have occurred. Distinguishability is a result in some situations, it is not a necessary condition for something to be called decoherence, neither is it sufficient.

But in that case everyone's branches are already decohered. Which means that there can be no further decoherence. So, yes, "there isn't any decoherence ]at all[ within each of the two branches". However, "at all" may be an overstatement. What happens is a microscopic interaction creates a coherent superposition. Which promptly decoheres. Does that spoil anything?

I hope that was coherent enough .

 

[PeterDonis]

What happens is a microscopic interaction creates a coherent superposition. Which promptly decoheres.

I'm not sure this is true. Microscopic interactions are what decohere things in the first place. For example, in my gas in the box Schrodinger's cat-type experiment, microscopic interactions within the gas on the two different branches (volume changed, volume not changed) are what decohere the branches. Those same microscopic interactions can't also be creating coherent superpositions.

 

[kered rettop]

I'm not sure this is true. Microscopic interactions are what decohere things in the first place. For example, in my gas in the box Schrodinger's cat-type experiment, microscopic interactions within the gas on the two different branches (volume changed, volume not changed) are what decohere the branches. Those same microscopic interactions can't also be creating coherent superpositions.

I don't see why not. What's the problem with an interaction creating new entanglements and at the same time spreading previous entanglements?

 

[PeterDonis]

What's the problem with an interaction creating new entanglements and at the same time spreading previous entanglements?

The same interaction can't be both coherent and decoherent. It has to be one or the other. (Technically it's a continuum, with "fully coherent" and "fully decoherent" as opposite endpoints, but the point is the same--the same interaction can't be at two different points on the continuum, or moving in two different directions. If the interaction is decohering--moving towards the "fully decoherent" endpoint--it can't also be moving towards the "fully coherent" endpoint. It has to be one or the other.)

 

[kered rettop]

The same interaction can't be both coherent and decoherent. It has to be one or the other. (Technically it's a continuum, with "fully coherent" and "fully decoherent" as opposite endpoints, but the point is the same--the same interaction can't be at two different points on the continuum, or moving in two different directions. If the interaction is decohering--moving towards the "fully decoherent" endpoint--it can't also be moving towards the "fully coherent" endpoint. It has to be one or the other.)

I'm not saying it's going in two directions at once, I'm saying there are two different terms for the change in coherence corresponding to different idealised models. Both apply at once in a more realistic model.

 

[PeterDonis]

I'm saying there are two different terms for the change in coherence corresponding to different idealised models. Both apply at once in a more realistic model.

I don't understand this. Can you be more specific? Math would help.

 

[kered rettop]

I don't understand this. Can you be more specific?

There are two idealised cases for the interaction between two molecules.

The first idealised case treats the molecules' joint state as separable and pure. After the interaction, the joint state is no longer separable, the molecules are entangled. The entanglement is a coherent state - which you interpreted as an increase in coherence of the whole system. I don't know how you quantify coherence but your claim sounds highly plausible to me. The interaction has increased the coherence.

The second idealised case treats the first molecule's state as entangled with a third. The first molecule interacts with the second, thus spreading the entanglement to include the second. This is the first step in decoherence - subsequent interactions propagate it through the entire system. The interaction has decreased the coherence.

The actual initial state will be a superposition of both these cases. So a single interaction will be a superposition of both, the decoherence of |> _1,3 occuring at the same time as the entangling of |>₁|>₂

Math would help.

I seriously doubt it!

 

[PeterDonis]

The first idealised case treats the molecules' joint state as separable and pure.

In this case the interaction is not decohering anything. Decoherence in the environment means at least one of the interacting particles in the environment is already entangled with something else (because the decoherence is spreading entanglement that originally came from an external interaction, such as the environment interacting with a measuring device that has recorded a result). If either of the molecules is already entangled with something else, their joint state prior to the interaction cannot be pure.

It is possible in principle for two of the gas molecules to not be entangled with anything else--in principle we could at least prepare the gas in such a state at the start of the experiment. But any such state will not last very long. Even in the absence of external interaction, the gas will decohere itself as its molecules interact. So the chances of any two interacting molecules not being entangled with anything else will decrease with time. Given the number of molecules involved, I would expect the decoherence time to be very short.

The second idealised case treats the first molecule's state as entangled with a third.

And as you note, this decreases coherence. See above.

The actual initial state will be a superposition of both these cases.

No, it won't. You can't have a quantum state of any system that is a superposition of pure and not pure. That is mathematically impossible. (This illustrates why I said math would help: trying to actually write down the math to back up your claim in the quote above would, or should, have showed you why the claim is wrong.)

I seriously doubt it!

Your doubt is misplaced. See above.

 

[Morbert]

Saunders justifies his notion of very many branches in section 6 using a consistent histories formalism. Specifically, he constructs branching histories using a fine-grained projective decomposition of the identity (PDI) and n-tuples as sequences of properties constructed from this PDI.

Generally speaking, this is not possible, so I do not believe this section. Branching histories are instead obtained by e.g. using a coarse-grained PDI at an early time $t_1$, and increasingly refined PDIs at later times. His argument might still be recoverable with some modification. I will see if this is the case when I have the some free time.

 

[jbergman]

Saunders justifies his notion of very many branches in section 6 using a consistent histories formalism. Specifically, he constructs branching histories using a fine-grained projective decomposition of the identity and n-tuples as sequences of properties constructed from this PDI.

Generally speaking, this is not possible, so I do not believe this section. Branching histories are instead obtained by e.g. using a coarse-grained PDI at an early time $t_1$, and increasingly refined PDIs at later times. His argument might still be recoverable with some modification. I will see if this is the case when I have the some free time.

What does PDI stand for?

 

[Morbert]

What does PDI stand for?

Projective decomposition of the identity. Updated the post to clarify

 

[Morbert]

Saunders constructs the states $C_{\underline{\alpha}_n}(t_n)|\phi;0\rangle$ for branch counting purposes, where $C_{\underline{\alpha}_n}$ is the ususal history operator: a chain of projectors
$$C_{\underline{\alpha}_n} = P_{\alpha_n}(t_n)\dots P_{\alpha_1}(t_1)$$He remarks that the states are orthogonal to one another if the $\underline{\alpha}_n$s differ at any of the times $t_1,\dots,t_n$. This is not generally true (see this paper for details). It is true if the history space has a branching structure, but this cannot be assumed for arbitrary n-tuples of properties. To describe a branching structure, it's useful to consider a truncated history operator $C_{\underline{\alpha}_k}$ where $t_k < t_n$ $$C_{\underline{\alpha}_k} = P_{\alpha_k}(t_k)\dots P_{\alpha_1}(t_1)$$$$C_{\underline{\alpha}_{k+1}} = P_{\alpha_{k+1}}(t_{k+1})C_{\underline{\alpha}_{k}}$$In a history space with a branching structure, if $||P_{\alpha_{k+1}}(t_{k+1})C_{\underline{\alpha}_{k}}|\phi;0\rangle||^2$ and $||P_{\alpha_{k+1}}(t_{k+1})C_{\underline{\alpha'}_{k}}|\phi;0\rangle||^2$ are nonzero, then $C_{\underline{\alpha'}_{k}} = C_{\underline{\alpha}_{k}}$ I.e. If the system has some property $\alpha_{k+1}$ at time $t_{k+1}$, then it must have arrived at this property via the unique history $C_{\alpha_{k}}$. Similarly, if you are walking out onto a branch of a tree, that branch only has one connection to the trunk (branches do not "recombine" when growing away from the trunk). One way to construct a branching history space is to use properties that are appropriate refinements of previous properties. I.e. $$\sum_{\alpha_{k+1};\alpha_{k+1}\subseteq\alpha_k}P_{\alpha_{k+1}} = P_{\alpha_k}$$One process that satisfies this is a process where at each time $t_k$ the property $\alpha_k$ includes an irreversible record by some environmental degrees of freedom. These environmental states would serve as the "memory states" Saunders mentions.

 

[kered rettop]

Posted in error, deleted, decluttered and reposted. Apologies to @weirdoguy for breaking the threading model.

In this case the interaction is not decohering anything. Decoherence in the environment means at least one of the interacting particles in the environment is already entangled with something else (because the decoherence is spreading entanglement that originally came from an external interaction, such as the environment interacting with a measuring device that has recorded a result). If either of the molecules is already entangled with something else, their joint state prior to the interaction cannot be pure.
...
No, it won't. You can't have a quantum state of any system that is a superposition of pure and not pure. That is mathematically impossible.

I can't believe I'm reading this! You can't just replace a state that I defined to be pure with one that is not and then imply that I'm trying to do the mathematically impossible with your alternative.

(This illustrates why I said math would help: trying to actually write down the math to back up your claim in the quote above would, or should, have showed you why the claim is wrong.)

I did not say that I hadn't done the maths to the limit of my abilities. Of course I had. I am not a complete idiot. What I said was that I didn't think it would be helpful to post - per your post #39 - any maths in order to explain what I was saying.

 

[PeterDonis]

You can't just replace a state that I defined to be pure with one that is not and then imply that I'm trying to do the mathematically impossible with your alternative.

I can't believe I'm reading this. You appear to not grasp my actual point at all. You can't just "define" a state to be pure whenever you like. If you want to claim that a particular state has certain properties and is relevant to what we're discussing, you have to show that those things are the case. And to do that, you need to first explicitly write down the state. Just waving your hands does not work.

What I said was that I didn't think it would be helpful to post - per your post #39 - any maths in order to explain what I was saying.

And, as I said, your doubt about this was misplaced. If you had actually tried to write down explicitly the state you were waving your hands about, instead of just talking about it in words, it would have become clear to you that you were waving your hands about a state that is, in fact, mathematically impossible.

 

[PeterDonis]

a state that I defined to be pure

There is also a confusion of terminology in our discussion. The relevant distinction is not between "pure" and "mixed" but between non-entangled and entangled. I should have clarified that in post #41. An entangled state is still a pure state of the joint system. (If we trace over part of the system, we get a mixed state for the rest of the system, but we don't need to do that here.)