There is some part called quantum decoherence, and I am not sure why it is so sophisticated area. (according to the definition I see, what it seems to say is some probability turning into what we observe, some probability de-phased and interact with environment. So what equations would we further need to explain?


A quantum system (for example a particle) is normally described by a wavefunction, which is a vector $|\psi\rangle$ from its Hilbert space $H$. But this description is not complete when the system is entangled with other system. In this case, it is more appropriate to use the density matrix (or density operator). For $|\psi\rangle$, the density matrix is $\rho=|\psi\rangle\langle\psi|$. In general, it has the form $\rho=p_i\sum_i|\psi_i\rangle\langle\psi_i|$, in a basis made of orthogonal states $|\psi_i\rangle$ from $H$. The basis can be chosen differently, and this will give other values for the coefficients $p_i$. But in general, a different choice would yield non-diagonal terms, that is, terms of the form $p_{ij}|\psi'_i\rangle\langle\psi'_j|$, $i\neq j$. The density matrix has some properties like being Hermitian (self-adjoint), positive semi-definite, and has the trace $tr(\rho)=1$. A density matrix which can be put in the form $\rho=|\psi\rangle\langle\psi|$ iff $\rho^2=\rho$.

A density matrix can equally represent a statistical ensemble, for example an ensemble of state vectors $|\psi_i\rangle$, so that the result can be $|\psi_i\rangle$ with probability $p_i$. If we could interpret the density matrix as a statistical ensemble for any system, then quantum mechanics would be very much like classical mechanics. But it is not like this. For example, if we measure an observable which has as eigenstates the vectors $|\psi_i\rangle$, we obtain $|\psi_i\rangle$ with probability $p_i$. But if we choose to measure another observable, which has another basis of eigenstates $|\psi'_i\rangle$, it is possible that the density matrix has mixed, or nondiagonal terms $p_{ij}|\psi'_i\rangle\langle\psi'_j|$, $p_{ij}\neq 0$, $i\neq j$.

Any density matrix can be diagonalized in a proper basis, but the problem is that the observer is free to choose the basis, by choosing the observable to measure. Hence, the density matrix may have non-diagonal terms in the basis corresponding to the measured observable. Yet, the measurement yields only one eigenstate, as if the density matrix has, in that basis, only one non-vanishing entry, which is on the diagonal. How can we come from the density matrix of a general form to only one element, on the diagonal, and all of the others being zero?

This is supposed to happen in two steps. In the first step, the density matrix becomes diagonal with respect to the basis of eigenstates of the observable. This process is called decoherence. The step two is that, being in a diagonal form, we can interpret it statistically, and we claim that it is a statistical ensemble, having the probability $p_i$ to obtain the $i$-th eigenvalue as outcome.

Quantum decoherence is then the process by which the density matrix evolves so that its nondiagonal terms vanish. The decoherence program tries to prove that for any observable we choose to measure, the observed system evolves so that the density matrix is diagonal with respect to the basis of the observed system. This is a long term program, and there are some progresses which indicate how it may take place for special cases. It is considered that if we add into the equation not only the measurement device, but also the environment, the interactions would be complex enough to carry out the non-diagonal terms. It is "sophisticated" because it involves the environment in a very quantum way. It is considered by many that the decoherence program answers all major questions concerning the measurement and the transition from quantum to classical. But there are also opinions which deny this (see for example Penrose, "The Road to Reality").

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    $\begingroup$ I think that introducing the density matrix is not necessary with a first explanation of decoherence, and can confuse things. The important point (as you know) is that there is a classically favored basis (so that diagonal vs. off-diagonal makes sense). The density matrix formalism occludes this on a first reading. Just my two cents about expository style though. This answer is certainly correct. $\endgroup$ – Ryan Thorngren Sep 17 '12 at 8:12
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    $\begingroup$ @user404153: But the meaning of favored becomes clear only when one sees that it means that in this basis the density matrix becomes diagonal. This cannot be expressed in terms of wave functions! $\endgroup$ – Arnold Neumaier Sep 17 '12 at 9:34
  • $\begingroup$ @ArnoldNeumaier That's not quite true. In terms of wavefunctions, just expand in the favored basis. The off-diagonal data is the relative phases. $\endgroup$ – Ryan Thorngren Sep 17 '12 at 9:44
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    $\begingroup$ Dear user, well, morally speaking, the relative phases carry the same information as the off-diagonal elements of the density matrix. However, there is a technical glitch if you interpret this thing too literally: relative phases are always numbers whose absolute value is one while the off-diagonal elements of the density matrix may have (and in decoherence, they ultimately have) much tinier absolute value. This tiny absolute value or zero comes from averaging over many possible relative phases. And the classical averaging over microstates is only possible in the mixed ensemble (dens. mat.). $\endgroup$ – Luboš Motl Sep 17 '12 at 9:52
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    $\begingroup$ @user404153: Off-diag terms contain relative phases and "relative amplitudes". When $\rho$ is diagonalized, the diagonal contains the clean probabilities. If we want to avoid using $\rho$, we have to discuss in terms of a set of orthogonal wavefunctions, and their relative phases, and the "relative amplitudes". What vanish are the "relative amplitudes", the relative phases turn out to be irrelevant. The density matrix of a statistical ensemble is independent on the phases. So I agree with ArnoldNeumaier and Luboš Motl. But, if you know a simpler description, it would really make a good answer. $\endgroup$ – Cristi Stoica Sep 17 '12 at 10:12

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