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I Can entropy drop in the Copenhagen interpretation?

  1. May 3, 2017 #1
    I've been reading some speculative articles about the possible quantum arrow of time which emerges through collapse and irreversibility.

    My question is: does collapse in the Copenhagen interpretation (or perhaps in a objective-collapse model) allow the spontaneous decrease of entropy where an improbable event like reforming of a broken glass can happen?

    As far as I know, the only thing that is irreversible in Copenhagen (which isn't in MWI or BM) is the loss of branches when measurement occurs. But does that imply that for some reason entropy cannot go in reverse in the branch we observe, if it's in fact the only one?

    My bet would be that - yes, the entropy can decrease and that is compatible with collapse and the Sch. equation, but wavefunctions cannot reinterfere. So in one way, entropy can decrease, but in a global sense considering lost branches - it cannot.

    Therefore, I don't understand the attempt to define the arrow of time through collapse since collapse doesn't disallow the spontaneous decrease of entropy.

    All opinions are welcome, thanks in advance.
     
  2. jcsd
  3. May 3, 2017 #2

    atyy

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    Collapse assumes an arrow of time, which could be an entropic arrow.

    It is still not clear, but it may be that one can use decoherence to help define when collapse occurs, and that decoherence results in a practical irreversibility that corresponds to an increase in entropy.

    https://arxiv.org/abs/1003.2133
    Proof of the Ergodic Theorem and the H-Theorem in Quantum Mechanics
    John von Neumann

    https://arxiv.org/abs/1507.06479
    Typicality of thermal equilibrium and thermalization in isolated macroscopic quantum systems
    Hal Tasaki

    https://arxiv.org/abs/1511.01999
    Quantum statistical mechanical derivation of the second law of thermodynamics: a hybrid setting approach
    Hal Tasaki
     
  4. May 3, 2017 #3

    PeterDonis

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    I don't think we have a model of collapse that is sufficiently well developed to answer this question. As the papers linked to by atyy show, this is still an open area of research.
     
  5. May 4, 2017 #4
    Thanks for the articles. So let me try to reach a conclusion. The collapse arrow of time and the entropic arrow of time may not overlap, therefore it is still possible that things collapse into lower entropy configurations - but it is not possible for the wavefunction to reinterfere with old phases if the collapse happens literally and not FAPP?
     
  6. May 4, 2017 #5

    atyy

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    Actually, what I was thinking is that the change in entropy from collapse is not so relevant, because collapse already assumes an arrow of time. In Copenhagen, the universe is divided into classical and quantum parts. This division is subjective and observer-dependent. Collapse only occurs when one registers an irreversible mark on the classical measuring apparatus. Thus collapse assumes an arrow of time. When I use quantum mechanics, I usually assume an arrow that is the thermodynamic arrow.
     
  7. May 4, 2017 #6
    So would you say that all flavors of Copenhagen allow spontaneous entropy decreasing, like a broken glass reforming etc.? I mean, the possibility is already there in the wavefunction - the only mechanism that is added is collapse. So I don't really understand why people assume that some irreversibility follows in a sense that the sucession of 2 measured outcomes can't be reversed, or that just by chance all particles in gas 'collapse' in the state of same velocity? Is this violation if the 2nd law allowed according to any interpretation, including Copenhagen?
     
  8. May 4, 2017 #7
    It's all reversible, just very unlikely once macroscopic objects become involved. In moving from the quantum to the classical collapse effectively occurs.
     
  9. May 4, 2017 #8

    atyy

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    But the point is not that irreversibility follows. Irreversibility is a precondition for collapse.
     
  10. May 4, 2017 #9
    Irreversibility of what?

    As far as I understand, the only irreversible thing is the loss of branches after the measurement.

    It doesn't imply that the dynamics of the measured outcomes is irreversible. The second law follows if we only focus on measured stuff (or in one branch).

    Would you agree?
     
  11. May 4, 2017 #10

    atyy

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    It assumes that the measurement outcome really happened in the classical world. This is a vague, subjective notion, but it is often assumed for something to really happen, it is irreversible. In the purest classical sense, things need not be irreversible for things to really happen. For example, it could really happen that someone dies and the returns to life. However, the dying and the returning to life must all be real events that really happen.

    In contrast, the wave function in Copenhagen has no definite status as reality. Thus the reversibility of the wave function (unitarity) is not related to reversibility of reality.
     
  12. May 4, 2017 #11

    tom.stoer

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    The Copenhagen interpretation assumes a Heisenberg cut separating the observed quantum system from the classical world. It is not defined how the interaction of the quantum system with the classical world does work or how information or entropy is exchanged. The interaction does not comply with the unitary quantum time evolution b/c this would not cause any collapse.

    Assigning (von Neumann) entropy to a quantum state ##|\psi\rangle## with density operator ##\rho = |\psi\rangle\langle\psi|## means to calculate

    $$S = -\text{tr}\,\rho\,ln\rho$$

    But in the course of a measurement the quantum system is always described by ##|\psi\rangle##, therefore ##\rho## is always a pure state: unitary time evolution evolves pure states into pure states, and the collaps reduces the pure state to some (pure) eigenstate. So the entropy of the quantum system remains

    $$S = 0 $$

    I do not see how to assign non-zero entropy i.e. mixed states to the quantum system b/c the collapse postulates the system to be in a pure state after the measurement, neither do I see a way to assign von Neumann entropy to the combination of the quantum system and the classical world. So for me the answer to your question is somewhere between "entropy remains zero" and "this question does not make sense in the Copenhagen interpretation".

    Considering the whole universe as the quantum system and denying any Heisenberg cut we end up with a pure unitary time evolution including the measurement itself. This automatically results in the "many-worlds-interpretation" which we may not like but which is at least logically sound.
     
  13. May 4, 2017 #12

    stevendaryl

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    Well, if you want entropy to be an objective property of a system, then there is no principled way to assign any value other than zero. However, if you allow entropy to be subjective, then the way that entropy changes is as follows:
    1. You start with a system in a pure state, so its von Neumann entropy is zero.
    2. With time, that system becomes entangled with the environment.
    3. After "tracing out" environmental degrees of freedom, then you get an effective density matrix for the system that is nonzero.
    The "tracing out" operation is subjective, because it relies on a modeling decision to split the universe into "system of interest" plus "the rest of the universe". The increase of entropy comes from this split.
     
  14. May 5, 2017 #13

    tom.stoer

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    How would the reduced density matrix of a mixed state (depending on the split) comply with the Copenhagen or other collaps interpretations postulating an eigenstate i.e. pure state of the system?
     
  15. May 5, 2017 #14

    atyy

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    Include the measuring apparatus (quantum) as part of the quantum world (quantum measuring apparatus + quantum system). Then have another measuring apparatus (classical) to measure the measuring apparatus. So the trace is done over the quantum measuring apparatus. But the entire quantum world is in a pure state.
     
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