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

[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.)