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What are the principal results of the decoherence in QM?

  1. May 17, 2005 #1
    Hi all,

    1)What are the principal results of the decoherence in QM?
    2) Is there a general theorem we can use to determinate the states of macroscopic bodies (huge number of particles)?
    3) if yes, what are the known limitations?

    Terra Incognita
     
  2. jcsd
  3. May 17, 2005 #2

    vanesch

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    A very good review in my opinion is given in quant-ph/0312059

    The main result of decoherence is the following: when you take into account the (tiny) interactions with the environment, and consider unitary evolution describing this, then the overall wavefunction evolves very quickly into a sum of states:

    |psi> = |someenvironmentstate> |pointerstate1> |sys1> + |someenvironmentstatebis> |pointerstate2> |sys2> +...

    where "pointerstate" corresponds to a classical state of the measurement apparatus, and "sys1", "sys2"... correspond to the eigenstates of the system under study, of the operator implemented by the measurement apparatus.

    As such, decoherence solves a part of the "measurement problem", namely the "preferred basis" problem.

    cheers,
    Patrick.
     
    Last edited: May 17, 2005
  4. May 17, 2005 #3
    Ok, seems a good paper. I've quiclky read it. I think they do not explain the preferred basis selection.

    I don't understand very well. Any |psi> state may be decomposed like that (we just select the eigenbasis of each hilbert space and write down the |psi> state). Therefore I do not understand why a particular eigenbasis for the |psi> vector is selected or if you prefer the particular eigenbasis of |pointerstate> for example (what makes this particular selection?: the rule).

    If one can select the eigenbasis as one likes, what additional information does decoherence add relatively to the collapse postulate?

    Thanks in advance,

    Terra Incognita :frown:
     
  5. May 17, 2005 #4

    vanesch

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    You cannot have read a paper of 40 pages in 5 minutes :-)

    cheers,
    Patrick.
     
  6. May 17, 2005 #5
    Hello Terra Incognita,

    Maybe you will find this article useful:
    http://physicsweb.org/articles/world/18/3/5/1

    It's about interference experiments with large molecules.
     
  7. May 17, 2005 #6
    You are right. But I already knew it (I've read it one year ago I think) :tongue2: . This paper explains very well the current status on the decoherence program (end 2003), better than zureck papers (external view, even if it is also a biased view). However it does not explain the rule to choose a particular eigen basis.
    In other words, if we have an hamiltonian H= HSO+HEO+Hint (where HSO is the free system hamiltonian, HEO the free environment and Hint the interaction). I do not know what eigen basis to select (the eigen basis of H, the one of HSO etc ...).

    Terra Incognita
     
  8. May 17, 2005 #7
    Thanks edgardo. This one I haven't already read it. I will check it and give a feed back.

    Terra Incognita
     
  9. May 17, 2005 #8
    Ok, after a brief review. I have the same problem (what selects the preferred basis).
    In this paper, the implicitly selected eigenbasis (the preferred basis) to study the decoherence is mainly the position (e.g. the position of the photons or molecules on the screen on the double slit experiment). However, No explanation is given on why the observed results of the experiment are in this basis (the preferred basis problem).
    If you prefer, if I construct a quantum system experiment, I would like to know, a priori, what basis I will see during my observation: the collapse postulate does not tell what the basis is.

    I may understand the entanglement and the unitary time evolution. However, to infer experimental results, if I have no rule to select the eigen basis, I stay with the collapse postulate of QM and I do not see what additional information brings decoherence.

    Terra Incognita :frown: .
     
  10. May 17, 2005 #9

    vanesch

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    That is not true for an arbitrary basis. If you have a tensor product of two hilbert spaces, H = H1 x H2, and if {|a_n>} is a basis of H1, and {|b_m>} is a basis of H2, then a basis of H is of course {|a_n> x |b_m> }. This means that any ket |psi> of H can be written as nxm terms. But this means, for instance, that a particular |a_k> appears several times in the decomposition (in principle it can appear m times). However, there is only ONE basis (apart from degeneracy) in H1 and H2 so that each |a_n> only appears ONCE. That is the theorem of Schmidt decomposition. This basis is dependent on |psi> of course.

    cheers,
    Patrick.
     
  11. May 17, 2005 #10

    vanesch

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    That's exactly the problem that decoherence takes up!

    Well, that should follow from the interaction hamiltonian of the measurement apparatus with the environment: only (rough) position states are robust against "mixing" under this hamiltonian (meaning: under this evolution, the Schmidt decomposition of |psi(t)> remains roughly the same).

    In more detail:

    if at t0, we have a certain psi and a Schmidt decomposition:

    |psi(t0)> = |environment state1> |pointer POSITION state1> + |environment state2> |pointer POSITION state 2> +...

    then at a later time, evolving under the interaction hamiltonian of measurement system and environment, we have of course another state |psi(t1)> with ANOTHER Schmidt decomposition |environment state B1> |pointer POSITION state B1> +...

    Well, it turns out that the pointer position state B1, although different from state1 (and maybe almost orthogonal in Hilbert space if there are many pointer positions), is still classically very close to it. And you will not find states that look like, say, pure momentum states of the pointer. The Schmidt decomposition of |psi(t)>, once the interaction hamiltonian with the environment is taken into account, is dominated by almost classical and almost stable position states for the pointer (in fact highly localized wave packets).

    This is in fact a property of the interaction hamiltonian with the environment, which is roughly diagonal in the position wavepacket basis.

    I have to say I'm myself still trying to understand all the details of this, but I can feel in my bones that this is a plausible explanation...

    cheers,
    Patrick.
     
  12. May 17, 2005 #11
    Thanks Patrick, but I have problems to understand.

    For every |psi>, you can write |psi>= sum_nm c_nm|a_n>|b_m>= sum_n |a_n>(sum_m cnm|b_m>)= sum_n |a_n>|c(a_n)>.
    Therefore, for each |a_n> (measurement result a_n), I know I have the associated state |c(a_n)>.
    If I choose another eigenbasis, I will have other possibilities (freedom of the collapse postulate).
    Therefore, what eigenbasis is the prefered one and why? Why do we need to choose the particular schmidt decomposition you proposed and not my decomposition:
    If I choose the double slit experiment, I think the interference pattern is not in the schmitt decomposition you proposed but in one of the decompositions, I have shown.

    Terra Incognita :frown:
     
  13. May 17, 2005 #12
    Therefore, you are saying that we are not measuring a position eigenbasis but another different eigenbasis that is "close" to the position eigenbasis. Ok. But why this eigenbasis and not another one? (collapse postulate always work, whatever eigenbasis we select).


    This is not the case for the double slit experiment, where the impacts of the photons on the screen are the energy of the photons time the sreen position.
    Why do we select the energy photon eigenbasis and the position screen eigenbasis? :frown:

    Once you have selected your eigenbasis, with the collapse postulate you have no more problems (what I understand). However, I do not know where one shows what eigenbasis is selected in a measurement experiment (if there exists a rule of selection).
    I may accept that QM does not describe this selection procedure (external from the theory): we have to learn (a posteriori) from the different experiments in order to know what eignen basis is selected during a particular experiment. If this is the case, frankly, I think QM has to be amended with an external rule to select this basis (if it is possible).

    Terra Incognita
     
  14. May 17, 2005 #13

    vanesch

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    You puzzled me for about 10 minutes, I felt woossshh !
    I reread the statement of the Schmidt decomposition theorem (which states that the decomposition is UNIQUE) and didn't find directly where your argument fails.

    I think I found it. There is no guarantee that |c(a_k)> and |c(a_l)> are ORTHOGONAL states.

    The Schmidt decomposition tells us that, requiring orthogonal states (a BASIS) in both spaces, you have a unique decomposition (except if the coefficients of certain terms are identical, in which case you can play your game ; that's the case of degeneracy).

    cheers,
    Patrick.
     
  15. May 17, 2005 #14

    dlgoff

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    Maybe I should start another thread for this question.

    From the article: "To avoid the seemingly decisive role played by the observer, physicists put forward many alternative theories and interpretations. Often this was done at the price of introducing as-yet-unobserved quantities into quantum mechanics called hidden variables.

    Decoherence theory, in contrast, is based firmly on the conventional framework of quantum mechanics. ..."

    Why can't hidden variables be introduced/included in Decoherence theory?

    Regards
    Don
     
  16. May 17, 2005 #15
    Sorry, It wasn't my intention :tongue2:

    Why do we need individual orthogonal states? (is there a good explanation, or just a common guess as the use of the position eigen basis as a preferred basis in most of the experiments).
    Collapse postulate just requires the global state to be orthogonal <a_k|<c(a_k)|a_l>|c(a_l)>= delta(k,l). And we have this result, with my proposed decompostion.

    Let's suppose a moment this case is the general case. What eigenbasis is selected in such experiments?

    Terra Incognita :frown:
     
  17. May 17, 2005 #16
    Reformulated with other words, we can say this is the problem of the eigenbasis selection (what I would like to know).
    If we know, a priori, using QM theory, what eigenbasis will be selected by a given experiment, no hidden variable is required (what eigenbasis we select to apply the collapse postulate).
    If QM theory is really independant of the eigenbasis selection (the collapse postulate), we therefore are free to imagine an external eigenbasis ad hoc selection to reflect what is given by all the possible experiments. This can be done through hidden variables (e.g. trhough the labelling of all the past, present and future experiments: experiment(a) => eigenbasis(a) or through more compact procedures: may be what decoherence tries to do - my current questions).

    TI.
     
  18. May 17, 2005 #17

    vanesch

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    Well, I'm also of the opinion that SOME extra postulate is necessary, and I think that the Schmidt decomposition enters in that postulate.
    Warning: this is my personal view (but I think I'm in the company of people like Wigner and Stapp).
    I think there needs to be an extra postulate that 1) requires us to write the state of the universe as a Schmidt decomposition in H_mybody x H_restofuniverse, 2) makes us experience consciously only ONE of those terms and 3) with a probability given by the Born rule.

    But what is interesting is that we do not have to postulate a particular basis by hand: requiring Schmidt decomposition naturally leads to terms which correspond to "classical" states, which most of the time come down to "position" states (wave packets).

    I'm also of the opinion that this is an interesting result but doesn't solve the entire issue, as some (like Zurek) seem to claim.

    cheers,
    Patrick.

    EDIT: however, if you somehow accept the Born rule, even in your basis you will then find a result, which is probably the most tangible result of decoherence: if you calculate the reduced density matrix of the measurement apparatus (by tracing out the states of the environment), you are NATURALLY lead to a Schmidt decomposition, because it is in the corresponding basis for the measurement apparatus that this reduced density matrix is diagonal. It is then very natural to consider that reduced density matrix as describing an (improper) mixture. If you work in another basis, your reduced density matrix will not be diagonal.
     
    Last edited: May 17, 2005
  19. May 17, 2005 #18
    Thanks a lot Patrick. But it is not a very satisfactory reply. (it is a sort of "do the experiment first then use the selected basis displayed by the experiment to make other predictions"). :frown:

    For example, can we build by hand an interaction between a system and a human being where in the schmitt basis decomposition, the human position is no more the the preferred basis?

    Or, don't you think that this schmitt decomposition may lead to false predictions on some particular huge systems (where macroscopic quantum effects are viewed)?

    TI.
     
  20. May 17, 2005 #19

    dlgoff

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    My thought was, maybe that's it. The hidden varable somehow being the act of experimenting?

    Does this sound as wierd to you as it does to me?
     
  21. May 17, 2005 #20

    vanesch

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    I think you have to consider the environment versus the human body, not the system under study and the human body, if the human body is to be part of the measurement apparatus.
    I think that what you get out of decoherence is that if that human body is in a room where there is air, EM radiation corresponding to a temperature of 20 degrees, some light etc... then the Schmidt decomposition in H_room x H_body+system very quickly leads to states where the body is in localised states. These are not pure position states. These are wavepackets that define position and momentum as accurately as possible (often called coherent states).

    Have you seen my edit concerning the reduced density matrix in my previous post ?

    cheers,
    Patrick.
     
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