fanieh said:
Or can one teach decoherence by totally bypassing the density matrix
Hi fanieh, let me first off apologize in advance because I'm struggling to understand where the source of your difficulty is and so my comments may not hit the mark properly.
It looks to me like your mixing up (sorry for the unavoidable pun) 'density matrix' and 'decoherence' a bit. The first thing you've got to get clear in your head is the answer to the question "why do we need density matrices in QM at all?" Forget environments and 'classicality' for the moment - just focus on QM.
When we write down a pure state description for some entity, say ##| \psi \rangle##, then whatever we might interpret this state to mean (interpretation dependent) it is the most complete description available to us that describes the properties of that entity consistent with the label ##\psi##. There's no other mathematical object that does any 'better'.
But how does QM cope when we are less certain about the state? Suppose we're trying to prepare 2-level atoms in the excited state ##|e \rangle##, but our preparation procedure is imperfect and we get atoms in the ground state ##|g \rangle## 10% of the time - and for any given atom we just don't know without doing a measurement whether we have ##|e \rangle## or ##|g \rangle ##.
We can't now write down a pure state for any given atom - even though we know our preparation procedure has produced pure states every time - we simply don't know which of those pure states has been prepared. How are we to treat such an object, or set of objects, within the formalism of QM? In this situation the best we can do is to say something like "for each atom it is in the state ##| e \rangle## with probability 0.9 and in the state ##|g \rangle## with probability 0.1"
The best way to do that is to write down a density operator for each atom that looks like $$\rho = 0.1 |g \rangle \langle g | + 0.9 |e \rangle \langle e |$$ we can think of this as a way to deal with our 'ignorance' of the initial conditions. If you have this ignorance of the preparation conditions then this is the neatest way to describe the properties of your atoms.
You could also calculate expectation values for measurements assuming pure states ##| e \rangle## and then pure states ##| g \rangle## and then add them with the appropriate weightings 0.9 and 0.1 to get your final expectation value. When you're making a measurement you still need to apply the rules of QM (for example, the Born rule) to predict the results whether you use a density matrix approach or this latter approach of just weighting pure state results.
Now let's imagine a different situation involving 2-level atoms. This time I'm going to imagine Alice having a perfect preparation procedure that prepares every atom in its excited state ##|e \rangle##. She's now going to fire that atom through a high-Q (lossless) cavity which has nothing in it (the field inside is in its vacuum state, which is a pure state). If we've got the atomic transition frequency matching the cavity mode frequency then there's going to be an interaction between the atom and the field. We can tailor the cavity transit time such that there's going to be ##\frac 1 {2}## probability of finding the atom in its ground state after interaction with the cavity and ##\frac 1 {2}## probability of finding the atom in its excited state after it leaves the cavity.
Suppose Alice now sends these atoms that have been through the cavity on to Bob, but she doesn't make a measurement. What is the 'state' of the atoms that get to Bob? The combined state of the atom-field system is actually given by the pure state $$| \psi \rangle = \frac 1 { \sqrt 2 } ( |e,0 \rangle + |g, 1 \rangle)$$Conceptually now we cannot say that Bob's atoms are in pure states - there's no way that we could legitimately make the statement that they're actually in the pure state ##|e \rangle## with probability ##\frac 1 {2}## and the pure state ##|g \rangle## with probability ##\frac 1 {2}##
What we can say is that,
if we're interested in the properties of the atoms alone, then we can treat them
as if we had them in the pure state ##|e \rangle## with probability ##\frac 1 {2}## and the pure state ##|g \rangle## with probability ##\frac 1 {2}##
The beauty of the density matrix approach is that we can use the same formalism for either the first situation in which we had an imperfect preparation procedure AND the second situation. Now in the second situation Bob's 'ignorance' is coming about because he doesn't have access to the full atom-field system, he only has a subset, or a component, to work with. So if you think about it enough it's clear that we're going to have to have a formalism that looks the same (as far as Bob is concerned) for both types of situation above. So physically we can see why 'proper' and 'improper' mixtures have to be equivalent when we're only interested in the properties of subsets of larger systems.
Now - what's the substantive difference here between the fields and an environment? In the second situation above we ignore or 'trace out' over the field to get a description that works for the atoms alone. But that's exactly what we do when we trace out over environmental variables in the decoherence approach. The only difference is really that the field is a very, very 'small' environment. The entanglement between the atom and field is shared only between 2 objects. An 'environment' does not consist of just one other single entity but squillions of them and so the entanglement is shared out. We could think of the entanglement as a lump of butter - in the atom-field case it's just like spreading a big chunk of butter on a tiny drop scone. In the atom-environment case it would be like spreading the same chunk over many, many slices of toast - we'd get to the point with enough toast that we didn't even know it had been buttered!
