A question on MWI (for Dmitry67)

  • #1
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Dmitry, what do you think about the following generalization of MWI?

Wave function of the universe depends on the initial condition. Even if one chooses a very simple initial condition, one can still ask - why this initial condition and not some other? So why not ALL initial conditions? Why advocates of MWI (with the exception of Tegmark) do not say that ALL wave functions (that satisfy appropriate equations of motion) are realized in parallel universes?
 

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  • #2
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Of course, ALL initial conditions!
I can just agree with Max Tegmark

Whereas the traditional notion of initial conditions entails that our universe “started out” in some particular state, mathematical structures do not exist in an external space or time, are not created or destroyed, and in many cases also lack any internal structure resembling time. Instead, the MUH leaves no room for “initial conditions”, eliminating them altogether. This is because the mathematical structure is by definition a
complete description of the physical world. In contrast, a TOE saying that our universe just “started out” or “was created” in some unspecified state constitutes an incomplete
description, thus violating both the MUH and the ERH.

For more details, check page 10
http://arxiv.org/PS_cache/arxiv/pdf/0704/0704.0646v2.pdf
 
  • #3
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From another, now locked, thread:

Demystifier says:
It's easy to say so. But can you show by mathematics that such branches really appear? Can you point to a reference where it is shown?

Dmitry67 says:
just a standard decoherence stuff.
look at it this way: everythere in CI you get something random, in MWI you get a branch.
every 'random' event in CI 'injects' new information into system making it more complex.
The same in MWI

--------------------------

Now let us continue our discussion.

My objection is that there is no decoherence if the initial condition is perfectly symmetric.
For example, if the initial state is the eigenstate of the Hamiltonian (such as the ground state), then the state will not change during the evolution. No new branches will appear during the evolution.
 
  • #4
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It is an interesting point. You can also object that there were no systems with many degrees of freedom - such systems are used as 'observers' in the QD. In vaccuum with only 1 possible state there were no systems - and no decoherence.

So I agree with you that the theory of decoherence require some clarification regarding the behaivor at the BB. But in general I dont see any problems.

Say, there was super-heavy vacuum, which spontaneously converted into lower energy vacuum+lots of particles. It can start randomly at any place (CI) = it can start everywhere forming individual branches where vacuum started to decay in different places (MWI)

As there were less degrees of freedom, these branches could communicate much more efficiently then now.
 
  • #5
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This question wouldn't have arisen if you both hadn't forgotten about God the observer in CI.
Sorry if this was meant to be a serious/semi-serious discussion and i took it the wrong way, but come on, the initial state and MWI :)
 
  • #6
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Say, there was super-heavy vacuum, which spontaneously converted into lower energy vacuum+lots of particles.
But such a super-heavy "vacuum" is probably not unique. In other words, nature (God) still has to choose some particular initial condition among many others.

But that's not the problem. The problem is the following. If this "vacuum" still has a large degree of symmetry, then decoherence will not take place, despite the fact that there are many particles. In a sense, the particles will be arranged too symmetrically.
 
  • #7
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... where vacuum started to decay in different places (MWI)
But if the initial vacuum is very symmetric, and if the evolution is given by the Schrodinger equation only (as MWI asserts), then decay CANNOT start at different places. It will start at one place only, strictly determined by the initial symmetric state.
 
  • #8
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I think that it is unnatural to have a global time evolution at all. The wavefunction of the universe is static and satisfies the equation:

H|psi> = 0

Observers subjectively experience a time evolution, but what is really going on is that observers have "time evolved" copies that are all part of the same "eternal" wavefunction. Time does not really exist.
 
  • #9
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I think that it is unnatural to have a global time evolution at all. The wavefunction of the universe is static and satisfies the equation:

H|psi> = 0

Observers subjectively experience a time evolution, but what is really going on is that observers have "time evolved" copies that are all part of the same "eternal" wavefunction. Time does not really exist.
That seems irrelevant for the problem discussed by Dmitry67 and me.

The problem we discuss can be rephrased as follows:
If
(i) a very special (highly symmetric) state |psi> of the whole universe is chosen, and
(ii) |psi> represents the whole reality in the universe, and
(iii) the collapse of the state does not exist (MWI),
then
can nontrivial structures (like inhomogeneous universe, life, etc. ...) emerge?

Dmitry67 says yes. I say no.
 
  • #10
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That seems irrelevant for the problem discussed by Dmitry67 and me.

The problem we discuss can be rephrased as follows:
If
(i) a very special (highly symmetric) state |psi> of the whole universe is chosen, and
(ii) |psi> represents the whole reality in the universe, and
(iii) the collapse of the state does not exist (MWI),
then
can nontrivial structures (like inhomogeneous universe, life, etc. ...) emerge?

Dmitry67 says yes. I say no.

Suppose then that such nontrivial structures can evolve when you leave out condition (i). So, you start with a state that is not highly symmetrical. This would mean that you have some symmetry operator S(alpha) that does not leave |psi> invariant. But then you can symmetrize |psi> w.r.t. to S:

|phi_S> = Integral of S(alpha)|psi> d alpha

Then we see that under time evolution |psi_S> will be a superposition of the different sectors that do contain the nontrivial structures, the different sectors are related to each other by the symmetry operator
S(alpha).
 
  • #11
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From another, now locked, thread:

Demystifier says:
It's easy to say so. But can you show by mathematics that such branches really appear? Can you point to a reference where it is shown?

Dmitry67 says:
just a standard decoherence stuff.
look at it this way: everythere in CI you get something random, in MWI you get a branch.
every 'random' event in CI 'injects' new information into system making it more complex.
The same in MWI

The answer of Dmitry67 does not contain any mathematics, as the question of Demystifier required. We have precise mathematical definitions of Hilbert spaces and wave functions, of self-adjoint operators, of eigenstates and eigenvalues and so on. Can somebody provide an analogous precise mathematical definition of a "branch" of a wave function?
 
  • #12
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You need to define the observer first. If you consider the observer's conscious experiences to be formally describable using N bits of information that are represened physically in some way, you can introduce the ket vectors:

|b1,b2,b3,...bN>

that correspond to the observer having exactly that conscious experience that is described by the bit string b1,b2,b3,...,bN. The representation of the states |b1,b2,b3,...bN> in terms of wavefunctions of particles in the position representation is, of course, extremely compicated. The point is simply that these states exist.

The branch of a state |psi> in which the observer has the experience described by the bit string b1,b2,b3,...bN is obtained by applying the projection operator:


|b1,b2,b3,...bN><b1,b2,b3,...bN|

to |psi>
 
  • #13
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Suppose then that such nontrivial structures can evolve when you leave out condition (i). So, you start with a state that is not highly symmetrical. This would mean that you have some symmetry operator S(alpha) that does not leave |psi> invariant. But then you can symmetrize |psi> w.r.t. to S:

|phi_S> = Integral of S(alpha)|psi> d alpha

Then we see that under time evolution |psi_S> will be a superposition of the different sectors that do contain the nontrivial structures, the different sectors are related to each other by the symmetry operator
S(alpha).
Nice try! However, a superposition of nontrivial structures does not necessarily need to contain nontrivial structures. Indeed, in the case above it doesn't.

Take for example a constant function
f(x)=1 for all x
You can write this function as a sum of two (or more) very complicated functions. These functions may have nontrivial structures. Yet, the constant function above does not contain nontrivial structures. It is perfectly clear when f(x) represents some classical quantity. And the idea of MWI is that the wave function is also, in a certain sense, a "classical" quantity.
 
  • #14
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You need to define the observer first. If you consider the observer's conscious experiences to be formally describable using N bits of information that are represened physically in some way, you can introduce the ket vectors:

|b1,b2,b3,...bN>

that correspond to the observer having exactly that conscious experience that is described by the bit string b1,b2,b3,...,bN. The representation of the states |b1,b2,b3,...bN> in terms of wavefunctions of particles in the position representation is, of course, extremely compicated. The point is simply that these states exist.

The branch of a state |psi> in which the observer has the experience described by the bit string b1,b2,b3,...bN is obtained by applying the projection operator:


|b1,b2,b3,...bN><b1,b2,b3,...bN|

to |psi>
In my opinion, such a view of a "branch" within MWI is somewhat obsolete. A modern view of a branch within MWI is based on decoherence and does not need a notion of a "conscious observer".
 
  • #15
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The answer of Dmitry67 does not contain any mathematics, as the question of Demystifier required. We have precise mathematical definitions of Hilbert spaces and wave functions, of self-adjoint operators, of eigenstates and eigenvalues and so on. Can somebody provide an analogous precise mathematical definition of a "branch" of a wave function?
See some literature on the theory of decoherence.
 
  • #16
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See some literature on the theory of decoherence.

Demystifier,

in Reed Simon, Method of Modern Mathematical Physics, Vol. I, page 39, one can find the following definition:

Definition: a complete inner product space is a Hilbert space.

The notions of "inner product space" and of "completeness" are previously defined with analogous precision. Is there, in the literature about decoherence, a definition of the type:

Definition: a branch of a wave function [tex]\Psi[/tex] is ...

where other possible notions used in the definition are also precisely defined?

My position is that such a definition does not exist, and moreover there is no general consensus in the physics community about what a branch is, even if a vague definition were accepted. For example, I note that you suggest me to read the literature about decoherence, but in the previous definition of Count Iblis decoherence is not even mentioned.
 
  • #17
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My position is that such a definition does not exist, and moreover there is no general consensus in the physics community about what a branch is, even if a vague definition were accepted.
I agree that the definition of a branch is not so precise as, e.g., the definition of the Hilbert space.
 
  • #18
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In my opinion, such a view of a "branch" within MWI is somewhat obsolete. A modern view of a branch within MWI is based on decoherence and does not need a notion of a "conscious observer".

That's true. But then, if you let the dynamics of the system select the branches and not the observers that are contained in the wavefunction, you won't be able to fix the symmetry problem you pointed out here.

I think that (at least in principle) observers come first. The dynamics of a typical system is expected to be consistent with the existence of "decohered" observers. This means that you can just as well define branches in the way it is usually done (decohered consistent histories or whatever it is called). However, this definition will be flawed for the most general case.

E.g. in Deutsch' thought experiment where he reverses the collapse of a wavefunction, two branches recombine. No decoherence can have happened (or just a limited one which could be reversed). So, a definition of the sector in which an observer finds herself in cannot, in general, be based on any notion of decoherence.

In Deutsch's thought experiment it goes wrong locally, which is hard to realize in practice. But from a global perspective when you consider the wavefunction of the universe, it will also go wrong if the wavefunction is highly symmetrical, as you pointed out, or if there is no global time evolution.

So, just throw away the idea that you can always avoid doing the hard work of defining the observer. But that's just my opinion. :smile:
 
  • #19
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I think that (at least in principle) observers come first.
There are no observers in MWI. Just the wave function.

E.g. in Deutsch' thought experiment where he reverses the collapse of a wavefunction, two branches recombine.
There is no collapse in MWI. Just a determinisic equation (Schrodinger, Wheeler-DeWitt, or whatever) for the wave function.
 
  • #20
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But then, if you let the dynamics of the system select the branches and not the observers that are contained in the wavefunction, you won't be able to fix the symmetry problem you pointed out here.
If you are saying that branches cannot be defined without defining the observers, then I disagree. For example, you can take the position basis as the preferred basis and then define branches as parts of the wave function (in the configuration space) that do not overlap. This is analogous to the fact that branches of a wooden tree are also defined without observers.
 
  • #21
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By observers I don't mean "external observers" but simply "internal observers". Clearly they do exist, because otherwise I could not be typing this message right now.

What I'm saying is that the definition of the sector of an observer cannot always be obtained by looking at decoherence.
 
  • #22
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By observers I don't mean "external observers" but simply "internal observers". Clearly they do exist, because otherwise I could not be typing this message right now.
Internal observers certainly exist, but they do not play any fundamental role according to MWI. Therefore, during a discussion on MWI, it is better to avoid using the word "observer".

What I'm saying is that the definition of the sector of an observer cannot always be obtained by looking at decoherence.
I disagree. I think that all "classical" structures in MWI emerge from decoherence, including tables, chairs, bacteria, and humans (that, if you wish, may be thought of as "observers").
 
  • #23
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... For example, you can take the position basis as the preferred basis and then define branches as parts of the wave function (in the configuration space) that do not overlap. This is analogous to the fact that branches of a wooden tree are also defined without observers.

This kind of branches are usually considered in the context of Bohmian mechanics, and this definition can be made rather precise. Moreover, it does not require a (more or less arbitrary) decomposition of the wave function into system and environment.

In the decoherence-based approach the definition of branches is different, in my opinion not very clear, and however based on the decomposition into system and environment. See for example http://arxiv.org/abs/0904.0958" [Broken]. This is the reason why I argue that we have not yet a unique and universaly accepted notion of "branch".
 
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  • #24
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This kind of branches are usually considered in the context of Bohmian mechanics, and this definition can be made rather precise. Moreover, it does not require a (more or less arbitrary) decomposition of the wave function into system and environment.

In the decoherence-based approach the definition of branches is different, in my opinion not very clear, and however based on the decomposition into system and environment. See for example http://arxiv.org/abs/0904.0958" [Broken]. This is the reason why I argue that we have not yet a unique and universaly accepted notion of "branch".
You are probably right. However, the Bohmian interpretation is my favored interpretation, so you should not be surprised that I view decoherence from the point of view of a Bohmian. :wink:

Anyway, even without the Bohmian interpretation, the position basis is a preferred basis because, due to locality of interactions among wave functions, decoherence implies that density matrix is (approximately) diagonal in the position basis.
 
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  • #25
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I disagree. I think that all "classical" structures in MWI emerge from decoherence, including tables, chairs, bacteria, and humans (that, if you wish, may be thought of as "observers").

In principle, observers do not necessarily need to be decoherent classical structures. Counterexample: Consider a quantum computer that simulates a human brain and is able to perform the thought experiment by Deutsch in which a reversible measurement is performed.
 
  • #26
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In principle, observers do not necessarily need to be decoherent classical structures. Counterexample: Consider a quantum computer that simulates a human brain and is able to perform the thought experiment by Deutsch in which a reversible measurement is performed.
But can the result obtained (by such a quantum computer) eventually be seen by a true human observer?
If not, then such a quantum computer is at least useless.
If yes, then eventually decoherence takes place at some step.
 
  • #27
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Interesting.
Question is deeper than I expected.

So Demistifier is right, without any systems at all there is no decoherence. First moment after BB can not be analyzerd from the QD point of view.

But: no decoherence does not mean no complicated systems. Uranium nuclei is a supercomplex system oof 239*3 quarks and nobody knows how many gluons. And unless it decas and we detect it we observe unitary evolution of it.

To answer the question precisely we need to find an exact solution of the equation of the wavefunction for the whole universe close to t=0
 
  • #28
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But can the result obtained (by such a quantum computer) eventually be seen by a true human observer?
If not, then such a quantum computer is at least useless.
If yes, then eventually decoherence takes place at some step.

Assuming the answer is "yes", then it is still wrong to tie the definition of the sector for an observer to what you get by looking at decoherence. The two things just happen to coincide for most practical cases. Also, it is not clear why the brain implemented by the quantum computer would not qualify as a "true human observer".

A analogy. In practice almost all physical systems we encounter are close to local thermal equilibrium. Assuming thermal equilibrium identities can be derived that are not valid in general. It would then be wrong to use such a relation valid only in thermal equilibrium as a definition and as a result throw away one fundamental definition.
 
  • #29
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You are probably right. However, the Bohmian interpretation is my favored interpretation, so you should not be surprised that I view decoherence from the point of view of a Bohmian.

I am not surprised at all, because I too view decoherence from a Bohmian point of view, even if I am not sure that in this case it would have to be called "decoherence".


Anyway, even without the Bohmian interpretation, the position basis is a preferred basis because, due to locality of interactions among wave functions, decoherence implies that density matrix is (approximately) diagonal in the position basis.

It is possible that if the density matrix of the system is diagonal in the position basis then the wave function of system + environment decomposes into non-overlapping parts. However for me this is not evident, and I think it would be interesting to prove it in a rigorous manner.
 
  • #30
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Also, it is not clear why the brain implemented by the quantum computer would not qualify as a "true human observer".
It is even less clear why a simulation of a human observer (not sharing all physical properties of a true human observer) WOULD qualify as a true human observer.
 
  • #31
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To answer the question precisely we need to find an exact solution of the equation of the wavefunction for the whole universe close to t=0
Completely agree. :approve:
 
  • #32
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It is even less clear why a simulation of a human observer (not sharing all physical properties of a true human observer) WOULD qualify as a true human observer.

If you wish you can consider a thought experiment in which you do simulate all the physical degrees of freedom using a quantum computer. The computational states of the quantum computer are then related to the precise physical states of a human being living in some isolated box via a unitary transformation.
 
  • #33
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If you wish you can consider a thought experiment in which you do simulate all the physical degrees of freedom using a quantum computer. The computational states of the quantum computer are then related to the precise physical states of a human being living in some isolated box via a unitary transformation.
But then you also simulate decoherence existing in a true human observer. But I don't think that you can simulate decoherence without actual decoherence. So there is decoherence in this quantum computer. Of course, the computer as a whole is unitary and described by a pure state, while decoherence refers to subsystems only. In fact, I don't see any essential difference between this simulation of a human and an actual human. So what is the point of making this simulation anyway?
 
  • #34
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The definition of a branch is closely tied with slicing up of the Hilbert space of the universe into smaller Hilbert spaces of its interacting constituents (which is, obviously, non-unique). If you view the big Hilbert space as a tensor product of a number of smaller spaces, that gives you a natural set of basis states and an evolution Hamiltonian, which, when applied to any basis state, will likely turn it into a superposition, thus "branching" it.

There is a preferred slicing that identifies conscious beings as basis states. It is convenient because these states are relatively stable. Under the action of Hamiltonian, a conscious being can remain the same (be an eigenstate), evolve in a slow and continuous manner (move), or occasionally undergo discrete branchings when it acquires new knowledge.
 
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  • #35
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But then you also simulate decoherence existing in a true human observer. But I don't think that you can simulate decoherence without actual decoherence. So there is decoherence in this quantum computer. Of course, the computer as a whole is unitary and described by a pure state, while decoherence refers to subsystems only. In fact, I don't see any essential difference between this simulation of a human and an actual human. So what is the point of making this simulation anyway?


Suppose that the quantum computer simulates everything that happens inside a hypotetical closed box containing a human on which certain boundary conditions are imposed. There is then no decoherence of the box as a whole. This box can be simulated exactly by a quantum computer. If the box is seen to be unphysical, then you can replace it with an entire closed universe.

Then, the point is that the quantum computer that simulates it can have any preferred basis. For the quantum compoutation to work the decoherence time scale must be much longer than the simulation time. So, any "real" decoherence is irrelevant; you cannot define the sector of the observer by looking at decoherence.

However, the quantum computer does simulate correctly the effective decoherence. If the observer in the box were to make some quantum coherent system, then that system would be represented by some qubits and the environment by other qubits. They would then interact according to a unitary time evolution, leading to effective decoherence of the subsystem.
 

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