Quantum measurement limit of decoherence and scattering: why are position pointer said classical?

Question

In decoherence theory, we explain the decoherence by hamiltonian evolutions between a system and its environment. Calling $H$ the total hamiltonian, I have:

$$H=H_S + H_E + H_{SE} $$

A pointer state $|s\rangle$ is a state of the system $S$ for which the associated observable $|s\rangle \langle s |$ will commute with the total Hamiltonian. Then, coherences between pointer states will be killed under the Hamiltonian evolution.

One particular limit is the quantum-measurement limit for which $H \approx H_{SE}$. As typical interactions between a system and its environment are of the form:

$$H_{SE} = X \otimes E$$

Where $X$ is the position operator of the system, we figure out that the pointer states are the states $|x\rangle$, and thus the coherence between position eigenstates will be killed.

My question:

Now, a particular example of this situation if I understood is the quantum scattering. Basically our system $S$ (a molecule for example) will interact with environmental molecules under $H_{SE}$. And this interaction will kill superposition in the position basis as I explained. And from this we say that it matches the classical behavior in which the molecule doesn't have position superposition.

However I am puzzled by this. Indeed for me the classical limit should be a packet that is narrow both in position and momentum, like a coherent state.

Thus I am not sure to really understand why quantum scattering, as it is a particular case of quantum measurement limit, would fit as a good quantum to classical correspondance. If the particle is localized in space it is spread in momentum and thus highly non classical.

Does that mean that experimentally, we really see that molecules in a gaz for example are well defined in position, but very poorly in momentum (and then this explanation would match the decoherence theory) ?

A paper in which is explicitly said that collisional decoherence is a particular case of the quantum measurement limit (i.e hamiltonian dominated by interacting part) is The quantum-to-classical transition and decoherence on page 5, beginning of second column.

Answer 1

the classical limit should be a packet that is narrow both in position and momentum, like a coherent state.

That's correct. The approximation $H\approx H_{SE}$ can be used to show the tendency for pointer states to be localized in position, but that's just an approximation. The kinetic term $H_S$ tends to make a particle's wavefunction spread out, which competes with the tendency of $H_{SE}$ to make it localized. The balance of these two trends results in the particle being somewhat localized in position, but still spread out enough that it can have a somewhat well-defined momentum, too, so its location can move like a classical particle's location does.

The "measurement limit" is the limit in which the decoherence in one observable's eigenbasis is so rapid and so complete that competing terms don't have a chance. This is the defining feature of a perfect measurement. Natural measurements, like the mutual "measurements" incurred by the molecules in a macroscopic sample of gas, are far enough away from being perfect that both position and momentum can be approximately well-defined (without violating the uncertainty principle $\Delta x\Delta p\gtrsim\hbar$, of course).