Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A question on MWI (for Dmitry67)

  1. May 14, 2009 #1

    Demystifier

    User Avatar
    Science Advisor

    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?
     
  2. jcsd
  3. May 14, 2009 #2
    Of course, ALL initial conditions!
    I can just agree with Max Tegmark

    For more details, check page 10
    http://arxiv.org/PS_cache/arxiv/pdf/0704/0704.0646v2.pdf
     
  4. Oct 13, 2009 #3

    Demystifier

    User Avatar
    Science Advisor

    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.
     
  5. Oct 13, 2009 #4
    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.
     
  6. Oct 13, 2009 #5
    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 :)
     
  7. Oct 14, 2009 #6

    Demystifier

    User Avatar
    Science Advisor

    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.
     
  8. Oct 14, 2009 #7

    Demystifier

    User Avatar
    Science Advisor

    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.
     
  9. Oct 14, 2009 #8
    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.
     
  10. Oct 14, 2009 #9

    Demystifier

    User Avatar
    Science Advisor

    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.
     
  11. Oct 14, 2009 #10
    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).
     
  12. Oct 14, 2009 #11
    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?
     
  13. Oct 14, 2009 #12
    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>
     
  14. Oct 15, 2009 #13

    Demystifier

    User Avatar
    Science Advisor

    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.
     
  15. Oct 15, 2009 #14

    Demystifier

    User Avatar
    Science Advisor

    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".
     
  16. Oct 15, 2009 #15

    Demystifier

    User Avatar
    Science Advisor

    See some literature on the theory of decoherence.
     
  17. Oct 15, 2009 #16
    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.
     
  18. Oct 15, 2009 #17

    Demystifier

    User Avatar
    Science Advisor

    I agree that the definition of a branch is not so precise as, e.g., the definition of the Hilbert space.
     
  19. Oct 15, 2009 #18
    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:
     
  20. Oct 15, 2009 #19

    Demystifier

    User Avatar
    Science Advisor

    There are no observers in MWI. Just the wave function.

    There is no collapse in MWI. Just a determinisic equation (Schrodinger, Wheeler-DeWitt, or whatever) for the wave function.
     
  21. Oct 15, 2009 #20

    Demystifier

    User Avatar
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A question on MWI (for Dmitry67)
  1. Qm => Mwi? (Replies: 13)

  2. Proof of MWI? (Replies: 29)

  3. Splitting in MWI (Replies: 6)

  4. Another MWI question (Replies: 2)

Loading...