# Why should the observed probability distributions in quantum mechanics always align with the pointer basis of decoherence?

It has always been claimed decoherence solves the problem of the preferred-basis for observed probability distributions, but why should this be the case? If there is only one world, and there are probabilities for certain outcomes, why should the basis in which the probabilities are observed coincide with the pointer basis determined dynamically by decoherence?

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Or why don't we see superposed cats? – Raskolnikov Jun 6 '11 at 10:52
If we consider the heisenberg uncertainty principle a cat is a superposition of many cats, as we all are superpositions of many us. It is the size of hbar that makes this superposition irrelevant because we live in a world many orders of magnitude larger than this fuzziness. – anna v Jun 6 '11 at 14:15
@Raskolnikov the comment above should have been addressed to you. – anna v Jun 6 '11 at 15:44
@anna I do not see how it addresses the issue, but why don't you make it a reply to the OP? – Raskolnikov Jun 7 '11 at 7:47

It's because the probability that a given mixed state $\rho$ is found in a particular normalized state $\psi$ is only "sharp" and well-defined if $\psi$ is actually an eigenstate of $\rho$. So whatever basis in which $\rho$ is diagonal the decoherence produces, is also the basis of the states that have well-defined probabilities.

For all their general linear superpositions, one may calculate the expectation values of the probabilities but they're not probabilities that can be measured or "perceived". The prescription above was described e.g. at

http://motls.blogspot.com/2011/06/density-matrix-and-its-classical.html

It treats the density matrix in a similar way as the observables - even though the density matrix is not an observable. The rule that only the eigenstates of the density matrix may be "perceived" is pretty much equivalent to the "consistent histories" approach to quantum mechanics that tells you which questions may be asked in quantum mechanics and which can't.

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If I understand you correctly, the preferred basis is the eigenbasis of the density matrix and has thus a priori nothing to do with decoherence. Except that a posteriori, decoherence has produced a certain density matrix, and therefore its eigenbasis is the basis of pointer states. Correct? – Raskolnikov Jun 7 '11 at 8:07
If I understand you correctly, then yes! ;-) Decoherence is just a particular process, and/or a calculation imitating this process, that just allows one to trace what the eigenstates of the density matrix will be. But the fact that the basis in which $\rho$ is diagonal is preferred for defining "alternative histories or outcomes" whose probabilities may be nicely checked and "perceived" is an independent fact. – Luboš Motl Jun 8 '11 at 6:47

Let's hear it from the great quantum philosopher Niels Bohr. During his time, no one used the term preferred basis, but the term complementary observables covers pretty much the same issue. If you wish to object, please point out why. So let's say we have an electron floating around. Do we use the position basis, or the momentum basis? Bohr's answer is very ingenious and cuts right to the heart of the matter. Search and search into the dynamics of the electron in itself, and you will never find the answer. All experiments have to include the description of the experimental apparatus as part of the complete description. It is the choice of apparatus and its settings which determines the preferred basis. Is an electron passing through a double slit a particle or a wave? If you don't measure which slit it went through, the wave basis is preferred. If you do measure which slit it went through, the particle basis is preferred. Is the spin of an electron aligned along the z-axis, or the x-axis? The answer is not found in the electron itself. The answer is found in the orientation of the Stern-Gerlach magnetic field of the experiment. The so-called pointer basis is none other than the basis that the apparatus picks out. The pointer basis is very sensitive to the nature of the apparatus and its settings.

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Most people misunderstand what decoherence is all about. Decoherence is not really about how the environment affects the system. Decoherence is about how the system imprints itself irreversibly in the environment. Obviously, you can't observe a system unless information about the system has already been imprinted within the environment for you to pick up. The pointer information of decoherence are precisely those information about the system which becomes entangled with the environment for you to pick up. When stated in this manner, the OP's question becomes almost tautological.

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It is an utter misunderstanding that the preferred basis is given by the eigenstates of the reduced density matrix. If this false definition is correct, decoherence is tautological. Decoherence is defined to be a suppression of off-diagonal components in the basis of choice, but the basis is chosen to be the eigenstates? tautological. Besides, taking this definition seriously, as Motl does, the basis depends upon where the boundary of the system is drawn. Take Schrodinger's cat. Place the boundary at Pluto when the signals haven't had time to reach it yet and there is no entanglement assuming stosszahlansatz and the eigenstates form some funny bizarre unphysical states. When the signal reaches Pluto, it turns into a live or dead cat. Place the boundary around the cat so that the rest of the box is in the environment and the basis is more well behaved. When two eigenvalues nearly match, the eigenstates can mix into unphysical vectors. Also, the nondominant eigenvalue eigenstates are always funny, but stick to the dominant probable ones for the moment.

It is also sensitive to how thick or thin the boundary wall is. Too thin, not enough smearing, and the eigenstates also become funny and bad. Too much focus upon quantum fluctuations across the boundary. There is an art to choosing boundaries.

No, the true tenet of decoherence is maximal predictivity. What basis, such that if we know which basis vector allows us to predict the future with greatest accuracy, i.e. least increase in entropy? Informational patterns which survive over time, maybe in a transmuted encoding form. The basis comes from the interaction Hamiltonian term factors, and are the most stable over time. Many seemingly good bases suck because they mix up rapidly in time with no stability. The better the survival of information over time and the longer it survives, the higher the predictability and the better the basis. Then check if in this maximally predictable basis, the off diagonal terms are suppressed. This is decoherence.

Of course, some people would just claim the preferred basis is the basis of consciousness, but this is highly controversial.

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Without collapse all information survives. Unitarity guarantees this. – wnoise Nov 23 '11 at 19:42