The decoherence process has allowed us to explain various (classical and decoherence) sources of measurement noise in quantum systems. I intuitively understand this physical concept of decoherence-inducing-noise.

I am trying to understand this concept from an information theoretic perspective. We know that the environment-induced-decoherence takes away state information from a system. That would imply reduction in entropy of the system. Now analyzing the modeling process, the measurement process of a (environment induced) decohering system can be modeled by adding "noise" terms. That would imply addition of entropy to the system. Which is contradicting.

So my question is, from an information perspective, how can the (environment-induced-)decoherence process consistently (theoretically) describe the occurrence of (non-classical) noise? Or, what is the nature of such a noise?

Edit This is an explanation on why I think that decoherence might lead to reduction in entropy. I am using an information theoretic interpretation on entropy for this purpose, which can be interpreted as the amount of information that you lack about the exact physical state of the system (see https://physics.stackexchange.com/q/119515).

The measurement problem has 3 parts (Schlosshauer book on Decoherence). For this discussion only first 2 are relevant: 1) the problem of preferred basis and 2) the problem of non observability of interference.

Consider an unknown system that is initially isolated. Now I (or the environment) make measurements on this system and that makes the system vulnerable to decoherence. So it happens that through einselection I can find a set of robust states that are least prone to decoherence (pointer states) Let's call them $|\Psi_1\rangle$ and $|\Psi_2\rangle$. So this process allowed me to gain more information about the system, then I had before. Hence, the reduction in entropy! Now of course, initially the system was in a pure superposition of these states say $|\Psi\rangle = \alpha |\Psi_1\rangle + \beta |\Psi_2\rangle$, with a definite $\alpha$, $\beta$, therefore zero entropy. But as an observer with no initial knowledge of the system (an alien observer on system earth!) I model my belief on the system as an arbitrary superposition, say $|\Phi\rangle = a|\Phi_1\rangle + b|\Phi_2\rangle$ with unknown $a$, $b$, therefore a positive entropy. Now ofcourse, through the measurement process (and thanks to the physical process of einselection) I can find relationships between $(\alpha, \beta, \Psi_1, \Psi_2)$ and $(a,b,\Phi_1,\Phi_2)$. Hence, this entire process of measurement has 2 competing entropy elements, 1) First entropy element that is reduced by gaining knowledge on the pointer states, hence the basis and coefficients, 2) second entropy element that increases due to the maping of pure states to mixed states.

  • $\begingroup$ I'm a little confused about your first point regarding a reduction of entropy. Decoherence is the coupling of a system to a much larger environment and of course adds entropy. $\endgroup$
    – AGML
    Nov 10, 2015 at 19:58
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    $\begingroup$ "Taking away information" = reducing distinguishability = increasing entropy. $\endgroup$ Nov 10, 2015 at 21:06
  • $\begingroup$ I think "Taking away information" is-not-always-equal-to reducing indistinguishably. Following on the edit above, taking away information also allows the observer (in this case the environment) to know more about the system. Hence, I think there are 2 competing elements here! $\endgroup$ Nov 10, 2015 at 21:45
  • $\begingroup$ *comment edit: -in- distinguishability $\endgroup$ Nov 10, 2015 at 21:52
  • $\begingroup$ @PeeyushKumar Information is a measure that is always calculated with reference to a particular observer. So in a measurement process, the information of the observer is increased (ignorance/entropy decreased). In a decoherence process, the information of the observer is decreased (ignorance/entropy increased). I usually find that talking about information as if it is some intrinsic property of a system inevitably leads to confusion. $\endgroup$ Nov 10, 2015 at 22:02

2 Answers 2


Better calibrated response:

I'm a little confused about your first argument, that decoherence should somehow reduce entropy. Decoherence is the coupling of a system to a much larger system and of course adds entropy.

Another way of putting it: if one ignores the environmental degrees of freedom, decoherence essentially maps pure states (zero entropy) to mixed states (nonzero entropy). Mixed states can be viewed as pure states plus classical noise. So decoherence adds noise, and thus entropy.

  • $\begingroup$ You make a great point here. So if I view 'decoherence' only as a mapping from pure states to mixed states (ignoring the environmental dof) the explanation completely fits. But there is an inherent assumption on the knowledge of a particular basis. If we suppose an inf-theor-interpretation of entropy which is the amount of information that you lack about the exact physical state of the system ,seelink, then I think there is more to the above question (and my first argument). I will add an edit to the question, to elaborate on this point further. $\endgroup$ Nov 10, 2015 at 20:50
  • $\begingroup$ You might be confusing uncertainty about the state with uncertainty about the outcomes of measurements. Given a pure state $\psi$ we have complete information about the exact physical state: it is $\psi$. The entropy is zero, because we know the state. Then, we take information away, leaving a mixed state. This can be viewed as multiple pure states with some uncertainty. Now we have only incomplete information about the state. Entropy has been added. $\endgroup$
    – AGML
    Nov 10, 2015 at 21:18
  • $\begingroup$ Yes, I agree with that. I am not confusing these 2 uncertainties. In fact on the contrary, I am thinking about a scenario where both these uncertainties are important and how that relates to noise addition. As you assume in the comment given a pure state ... now suppose there is no initial information on that pure state! Please see the edit on my question, hopefully that will elaborate on this point! $\endgroup$ Nov 10, 2015 at 21:33

The quantum state is a vector in a Hilbert space. Each basis vector in the Hilbert space is a possible observation one can make. For example, when modelling the position of a free particle, we assign one basis vector to each point in space. The length of the shadow that the state casts upon each vector is the (square root of) the probability that a measurement will detect the state at a given position.

Crucially, if the state is (very nearly) aligned with one of the basis vectors, the measurement is (very nearly) deterministic.

If we had two particles we would need two basis vectors per possible position: one for each particle. Thus, very roughly speaking, the dimension of the Hilbert space associated with a given property is

$N_{properties} * N_{particles}$.

The above equation is obviously nonsense when $N_{properties}$ is infinite (which is usually the case), but the intuition here still holds.

Now suppose we take an initially isolated particle which is in some state very far from the relevant basis vectors. Then measurements on the particle are very undetermined: we have good odds of getting very many different outcomes.

Suppose further we suddenly thrust the poor isolated particle into a system with macroscopic particle number. For example, suppose we open the door on Schrodinger's cat, bringing it into contact with the air in the room, etc. We have suddenly vastly increased $N_{particles}$ in the above equation, and therefore vastly increased the dimensionality of the Hilbert space.

Now think about our original state vector. It will behave classically if it is orthogonal, or nearly orthogonal, to any of the new vectors in the Hilbert space. But we have suddenly added $N_{properties} * N_{particles}$ new vectors - at random orientations w.r.t the state - to be 'almost orthogonal' to!

It is a curious fact about high-dimensional spaces that, as the dimension $D$ of the space is increased, the number of 'nearly' orthogonal vectors to a given vectors scales as $exp(D)$. (See: https://mathoverflow.net/questions/25983/intuitive-crutches-for-higher-dimensional-thinking). Therefore the probability that our state will line up with a basis - that it will behave classically - scales exponentially with particle number. This is decoherence.

  • $\begingroup$ Ah, I didn't read carefully enough - you wanted an information-theoretic analysis. Sorry. $\endgroup$
    – AGML
    Nov 10, 2015 at 19:53
  • $\begingroup$ I like this explanation, gives a great insight. Although I think it's only a partial answer! $\endgroup$ Nov 10, 2015 at 22:47
  • $\begingroup$ I have been thinking about this... so is in-distinguishability of the environment (or macrosystem)states cause for noise? $\endgroup$ Nov 12, 2015 at 0:36

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