I don't fully appreciate what the discovery of the decoherence phenomenon adds to the Copenaghen interpretation of QM.

I will be more precise: the Copenaghen interpretation, if I am not wrong, is summarized by the following concepts:

  1. QM can, and has to, predict only the probabilities of measurement outcomes ("has to" because descriptions of nature such as the classical one, where every quantity always assumes a certain value is in disagreement with our current experimental observations, such as the violation of Bell's Inequality.).

  2. In order to do QM, we use state vectors. However, these must not be considered as a real physical entity, but rather tools in a similar way as probability distributions in classical physics (This means that in an EPR experiment, the change of the vector state describing Bob's qubit after Alice has measured hers in a galaxy far away is not in contrast with relativity.).

  3. The probabilities are computed via the Born rule: the state after measurement is an eigenstate of the observable measured, and the outcome is the corresponding eigenvalue.

Now the decoherence: if one considers the universe divided into a qubit (i.e. a two level system) system (S), apparatus (A) and the rest (E) can give a QM dynamical description of a measuring process on the qubit.

Initially, the universe is in the state $|\psi\rangle = (a |0 \rangle + b |1 \rangle)\ |A_0 \rangle | \text{rest} \rangle$.

If $|A_i \rangle $ are such that they can imprint themselves in the environment, i.e. they do not entangle themselves with it, after the unitary evolution the SA system is described by the matrix: $\rho = |a|^2 |0\rangle\langle0||A_1\rangle\langle A_1| + |b|^2 |1\rangle\langle 1||A_2\rangle\langle A_2|$.

This analysis explains why at macroscopic scales classical physics works, i.e. we don't see superposition of states and probabilities don't interfere: the environment interaction rapidly evolves the vector states into those states that can imprint in the environment, or in mixture of them.

This is surely interesting on its own, but is there anything else to understand from decoherence?

The fact that superpositions states turn into mixtures of the eigenstates of the measured observable seems to explain the "measurement problem" (that in the Copenaghen paradigm is not a problem actually, but the most important assumption, as stated in item 1 above) but it does not: after the above evolution the system is a classical mixture, but the probabilities are still there. Explaining the measurement problem means to find an evolution that deterministically bring the system into the outcome state (in the Copenaghen interpretation, this is rejected). So decoherence does not explain the Born reduction rule.

Additionally, in order to interpret the mixture above as something analogue to a classical mixture, we must have already adopted the Born probability rule, so decoherence does not explain it either.

To sum up: having adopted the Copenaghen interpretation of QM, decoherence explains in its framework the transition from quantum world (superposition always possible, quantum interferences) to classical world (no Schroedinger's cat states, bayes rule), but does not replaces or explain any of its axioms...or does it?

  • $\begingroup$ You are focusing on the Born rule as evidence that decoherence doesn't "solve the measurement problem", but I think this is off the mark. Consider a physicist in a box, doing an experiment on system $S$, and let's use Copenhagen. When the physicist makes an observation on $S$, $S$ collapses. But what about an alien observing the physicist and $S$? To the alien, the physicist+$S$ combined system's evolution should be unitary. So, did $S$ collapse or not? The real problem with Copenhagen is that it's not self consistent. The Born rule is just an empirical postulate, with or without decoherence. $\endgroup$
    – DanielSank
    Sep 10, 2014 at 20:19
  • $\begingroup$ I think that in the moment the alien start to observe the physicist+$S$, he is in the box too. Hence alien+phyisicist+$S$+"anything else left to make the system closed" follow an unitary dinamic, the one above, that will leave the alien+phys+$S$ subsystem in the above density matrix. This system will "collapse" with a certain probability. $\endgroup$ Sep 10, 2014 at 20:33
  • $\begingroup$ I don't understand the confusion. We all agree that decoherence leaves us with a classical probability distribution, correct? The problem is then equivalent to having coin flip as a system. The physicist inside the box flips the coin and looks, the probability distribution from his perspective becomes a delta on heads while from the perspective of the alien the full probability distribution is still there. Where is the inconsistency? It's just different knowledge about the system. $\endgroup$
    – Bubble
    Sep 10, 2014 at 21:16
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    $\begingroup$ @Bubble: Correct. However, some people call the question of how the universe decides which classical result to give you "the measurement problem". I would say that it's just a postulate (namely the Born rule), that you just just get one nobody knows why you get the one you get. In the picture you paint with decoherence and diagonalized density matrices, this is self-consistent and totally fine, as far as I can tell. The advantage of this over the so-called "Copenhagen" approach is that it's self-consistent, whereas Copenhagen is not, as I illustrated in my previous post. $\endgroup$
    – DanielSank
    Sep 10, 2014 at 23:12

2 Answers 2


No, decoherence is not a new fundamental feature of quantum physics. It is a phenomenon which occurs when you couple a system with a few degrees of freedom to one with a lot of degrees of freedom and which you can derive from the postulates of quantum physics.

There really is no measurement problem. Once you get a classical probability distribution (up to exponential precision) via decoherence what wave function "collapse" is becomes clear - it is merely gaining information about the realization of a classical probability distribution. If you throw a die it has a certain probability distribution associated with the outcome. You throw a die, now you know the probability distribution, and then you look at it - after which the die has "collapsed" to a 6? No! You just gained information about the realization of a probability distribution.

The real difference between classical stochastic physics and quantum physics comes from coherences, which in turn come from the fact that observables do not commute in general. Decoherence tells you why you don't observe coherent states in the macroscopic world and after that what is left is just looking on which side the die fell.

Quantum physics tells us that the world is inherently probabilistic and that there is no way around that. It also tells us that there is no "realism", but that's another issue and comes from the fact that observables do not commute.

  • $\begingroup$ "There really is no measurement problem." While I agree with the sentiment, this is maybe too strong a statement. $\endgroup$
    – DanielSank
    Sep 10, 2014 at 20:05
  • $\begingroup$ "The real difference between classical stochastic physics and quantum physics comes from coherences..." Well, the fact that the "noise" in quantum mechanics is truly fundamental is also a difference, and a really important one. $\endgroup$
    – DanielSank
    Sep 10, 2014 at 20:06
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    $\begingroup$ This is really a darn good answer. I'm glad this way of thinking is catching on. $\endgroup$
    – DanielSank
    Sep 10, 2014 at 20:07
  • $\begingroup$ Well, there is no fundamental noise in the initial condition in classical physics. That and that observables don't commute are the only differences I think. But unless you think that in classical physics you can reduce the entropy of everything to 0 you will always have some "noise" (and I do mean zero because of Lyapunov (in)stability). However, I tried to avoid this issue by saying "classical stochastic" implying the difference between a classical inherently stochastic theory and QM. $\endgroup$
    – Bubble
    Sep 10, 2014 at 20:35
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    $\begingroup$ Right, I get what you're saying and I think we agree. I was just trying to point out that in any real classical physical system, the "noise" is at least presumably due to lack of knowledge of some sub-part of the total system. In quantum, it's there by construction. $\endgroup$
    – DanielSank
    Sep 10, 2014 at 23:13

I will give you an easier assignment to start with: explain the origin of Newton's laws, using nothing but statistical mechanics of Newtonian systems. Can you do it? No. Statistical mechanics follows from Newton's laws PLUS a few assumptions about phase space averaging.

In the same way decoherence does not lead you beyond the framework of quantum mechanics, either. It can't explain anything about quantum mechanics that is not already expressed in the original framework, even though it makes the meaning of the Copenhagen Interpretation a little less scary.

Quantum mechanics PLUS phase space averaging rules explain why the classical world exists, and that's it. You can use Feynman's argument (or was it Dirac's?) about the recovery of the classical action from a path integral or relative-state arguments, or weak measurements that localize particle paths, whatever makes you more comfortable with physical reality, the results seem to be all the same and very little, if anything can be learned from it.

  • $\begingroup$ I read some of your answers and comments on this site from the last few days. Overall they seem quite reasonable, but the claim that there is no collapse of the wave function in the Copenhagen interpretation is unconventional. Even so I admit that the collapse postulate itself was introduced by von Neumann, some probabilistic element similar to Born's rule seems unavoidable to me. I would like to learn more about your views, but I'm not sure about the best way forward. Should I try to conduct a conversation with you in the comment, or talk in chat, or ask a specific question? $\endgroup$ Sep 13, 2014 at 21:25

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