gptejms said:
I am very skeptical about this consciousness thing,but I don't want to enter into a discussion on it right now.
I can understand that. It isn't really essential in the discussion, in fact. Replace it by the "observing experience" or something of the kind. Indeed, in a relative-state view, there are no objective observations, and what is experienced as such depends on what entity is supposed to experience the observation. In the rawest state, it is a kind of memory of experienced events.
I repeat my earlier question which went unanswered.An atom goes from an excited state |e> to the ground state |g>---its state while in transit is a(t)|e> + b(t) |g>,where a(0)=1,b(0)=0.So the atom goes into a superposition(from the initial state |e>) and then leaves the superposition---all via a unitary transformation.So your argument(earlier post) that a unitary transformation can't break a superposition(which though looks reasonable) doesen't seem to hold---kindly comment.
That is because you view the atom as a system, while it is interacting with something else. Indeed, you cannot have that an atom in isolation goes from an excited state to a ground state. By definition, a stationary state (such as an excited state) remains forever in that state.
You get transitions because of the coupling to the EM quantum field, and what YOU are describing looks like the local density matrix description, limited to the atom, of the bigger system "atom + EM".
So, "in the beginning" you need at least a photon in your EM field, and an atom in the ground state. What I will write here is "in principle". Nobody (knows how to) do that this way, and uses approximations to do the calculations.
|start> = |1 photon> |g>
Note that this is not a stationary state of the coupled system, otherwise there would be no interaction, and hence no excitation or decay.
So the above state must be rewritten in stationary states which are entanglements between the atom and the field (because you have to diagonalize your total hamiltonian: the |start> state is not an eigenstate of the total hamiltonian of atom + EM + interaction).
So we rewrite this as:
|start> = A |energy 1> + B |energy 2> + C |energy 3> + ...
Where |energy 1> is an entangled state of all the |n photon> states and all the ground and excited states of the atom, and |energy 2> also (but in different proportions)
A = <start| energy 1> ; B = <start | energy 2> ...
This can then evolve into:
|intermediate t> = A exp (i t E1) |energy 1> + B exp(i t E2) |energy 2> ...
and it is NOT (as you seem to suggest) necessarily factorizable in a product of an atom state (such as u |g> + v |e>) and an EM field state. I'm pretty sure you remain in a kind of entangled state, but if you wait long enough, and if the excited state is "rather" stable, you might hope that this evolves into something close to:
|intermediate 2> ~ |0 photon> |e> + small terms.
Now the atom is "excited".
The desexcitation will then be simply a further evolution of the state:
A exp (i t E1) |energy 1> + B exp(i t E2) |energy 2> ...
and for very long times, you will end up again in a state that is close to:
|1 photon> |g>
But note that the whole time, we've never left the superposition:
A exp (i t E1) |energy 1> + B exp(i t E2) |energy 2> ...
If I can give you a rough analogy: if you consider a pulse, which is Fourier transformed, it looks a bit as if you're saying: hey, a pulse cannot be composed of sine waves, because a long while before it, I have no signal, so where are your sine waves, and a long while after the pulse, the same. Nevertheless, during my pulse, I can understand that you build it up with sine waves. Well, here it is the same. The sine waves are the stationary states of the overall system, and their composition, all along time, remain the same. Your "initial, intermediate and final" views are particular stretches of your function in the time domain. It can be flat (state |g> or |e>) or a pulse ("superposition a|g> + b|e>"). But all this is just a result of the different phase factors in front of the stationary states of the overall system.
cheers,
Patrick.