How does quantum decoherence occur? My question is how does quantum decoherence happen. What happens with a quantum system when "observed?" Can you give a mathematical explanation that is simple, precise, and easy to understand?
 A: When a particle passes through an ideal pair of slits, it's wave function can be written in the form $$\psi(x) = \alpha\, \psi_1(x) + \beta \,\psi_2(x)$$ where $\psi_{1,2}(x)$ are the wave-functions you would get by closing either slit 2 or slit 1, respectively. We say that the wavefunction is in a coherent superposition. 
What is coherent about this superposition? Well, let's place a detector screen at $x$ and measure the rate at which particles accumulate there. This rate will be proportional to $P(x) = \left|\psi(x)\right|^2$ by the usual rules of quantum mechanics. Coherence comes about because the two components of the wavefunction can interfere:
$$P(x) = |\alpha|^2 |\psi_1(x)|^2 +  \alpha^*\beta\,\psi^*_1(x) \psi_2(x) + \alpha\beta^*\,\psi_1(x) \psi^*_2(x) + |\beta|^2 |\psi_2(x)|^2$$
Here the middle two terms are interference terms, and give a behaviour that is counter-intuitive from a classical probability point of view. Classically, the rate should just be proportional to the sum of the probabilities of the particle going through either slit 1 or slit 2.
Now suppose our particle met a mote of dust on its way to the screen. The wavefunction of the system, particle + dust, should now be written as $$\psi(x_p,x_d) = \alpha\, \psi_1(x_p) \phi_1(x_d) + \beta \,\psi_2(x_p) \phi_2(x_d)$$
where now $\phi_1(x_d)$ is the wavefunction of the dust mote if the particle went through slit 1, and $\phi_2(x_d)$ is its wavefunction if the particle went through slit 2. Again, we place a detector at $x_p$ to measure the probability of detecting a particle there. The rules of quantum mechanics instruct us to sum over the position of the dust mote when squaring the wavefunction:
$$P(x) = \int\!dx_d\, |\alpha|^2 |\psi_1(x_p)|^2  |\phi_1(x_d)|^2+  \alpha^*\beta\,\psi^*_1(x_p) \psi_2(x_p) \phi_1^*(x_d) \phi_2(x_d)\\ + \alpha\beta^*\,\psi_1(x_p) \psi^*_2(x_p) \phi_1(x_d) \phi^*_2(x_d)+ |\beta|^2|\psi_2(x_p)|^2 |\phi_2(x_d)|^2$$
Firstly, the wavefunctions $\phi_{1,2}(x_d)$ are normalized. Secondly, depending on how the dust mote encountered the particle, it may have radically different final wavefunctions, so in general $\phi_1$ and $\phi_2$ do not have much overlap: $\int\!dx_d \phi^*_1(x_d)\phi_2(x_d) \ll 1$. Putting this together, we find
$$P(x) = |\alpha|^2 |\psi_1(x_p)|^2 + |\beta|^2|\psi_2(x_p)|^2$$
Compare this probability to the one we found when there was no interaction with the dust mote, and the immediate difference is that the interference terms have vanished. There is no longer a coherent superposition but rather an incoherent sum, and the probabilities add classically. One speaks of decoherence due to an interaction with the environment.
