I have a very basic understanting of decoherence (i.e. I,ve read the Wikipedia page), but I was recently reading Heisenberg's The Physical Principles of the Quantum Theory and I came across a thought experiment which, I think, is decoherence, as it shows loss of interference and seems analogous to optical decoherence. According to Wikipedia decoherence was first developed in 1952 and this book is from 1930, so I don't know.
This is, not exactly but paraphrased, the experiment.
To begin with, imagine a beam of atoms of width $d$ is sent through a field $F$ which is inhomogeneous in the $x$-direction, and which can separate atoms into states $n$ and $m$ with energies $E_n$ and $E_m$ respectively (Like the Stern-Gerlach experiment). The energy of state $m$ depends on the field so the force in the $x$-direction experienced by an atom in that state will be $$-\frac{\partial E_m}{\partial x} \, .$$ Thus the beam will be deflected by an angle $$\frac{\partial E_m}{\partial x}\frac{T}{p}$$ where $T$ is the amount of time the atom spends in the field and $p$ is the atom's momentum. The angular separation of the beams in state $m$ and state $n$ is therefore $$\alpha=\left(\frac{\partial E_m}{\partial x}-\frac{\partial E_n}{\partial x}\right)\frac{T}{p} \, .$$ For the beam to be split into separate beams the angular separation must be larger than the natural scattering of the beam by diffraction, so $$\alpha\ge\frac{\lambda}{d}=\frac{h}{pd}$$ where $d$ is the beam width and $\lambda$ is the wavelength of the particles. Since $E_m$ depends on field strength, there will be a change in the phase of the the atom's state $|m\rangle$ associated with the atoms passing through the field, with $\varphi_m$ being the phase change and $\varphi=\varphi_m-\varphi_n$. However, since the beam has a width and the Field is not uniform, the phase change will vary depending on the atom 'passing' through different parts of the field. This will introduce an uncertainty $\Delta\varphi$ in the phase change difference. Since the phase change is $$\varphi_m=\frac{2\pi}{h}E_m T$$ And the uncertainty in position is the width of the beam $d$ the uncertainty in the energy will be $$\frac{\partial E_m}{\partial x}d$$ So we have $$\Delta\varphi=2\pi\left(\frac{\partial E_m}{\partial x}-\frac{\partial E_n}{\partial x}\right)\frac{Td}{h}=2\pi \frac{pd}{h}\alpha$$ So $\Delta\varphi\ge 2\pi$, so the phase is entirely undetermined and random.
Now imagine a beam of atoms in state $n$ is sent to a detector which detects if an atom is in a state $l$. The probability of measuring an atom in the state $l$ when it is in the state $n$ is $$|\langle l | n \rangle|^2=|\sum_m\langle l | m \rangle\langle m | n \rangle|^2=\sum_{m'm''}\langle l | m' \rangle\langle m' | n \rangle\langle m'' | l \rangle\langle n | m'' \rangle$$ Now imagine before the atoms get to the detector, we introduce the field seen previously which separates the atoms in their states $m$ now to the probability we must necessarily introduce the phase factor, averaged out over all the possible phase changes introduced by the field $$\sum_{m'm''}\langle l | m' \rangle\langle m' | n \rangle\langle m'' | l \rangle\langle n | m'' \rangle\langle e^{\varphi_{m'}-\varphi_{m''}}\rangle$$ And (assuming the phase difference has the same probability for all possible values, reasonable given its large uncertainty) $$\langle e^{\varphi_{m'}-\varphi_{m''}}\rangle=\delta_{m'm''}$$ And so the probability becomes $$\sum_m|\langle l | m \rangle|^2|\langle m | n \rangle|^2$$ The interference is lost, and the classical probability is obtained.
