vanesch said:
I don't understand this. I would think that |left_photon_state> and |right_photon_state> are (almost) orthogonal ! (as wavepackets centered around different positions).
Or do you mean by "left photon state" the photon state from the left hole ?
Even in that case, they are orthogonal (or almost so, no ?)
Sorry, I was talking about "first order" interference (i.e. based on single photon events): left and right is the wave packet extension on the different areas of the same photon (approximated by spherical waves relatively to the slits sources).
In this case they are not orthogonal because we can see the interference results when they reach the screen. In fact it hilights the problem:
a) at the vicinity of the holes we have <left_photon_state|right_photon_state>= 0
b) at longer distances :
<left_photon_state|right_photon_state>=/=0 (inteferences)
Just by choosing the distance between the screen and the slits plate, I may have or not orthogonal states. Hence, I have problems with the Schmidt decomposition rule.
vanesch said:
But this has nothing to do with the Schmidt decomposition: you have to apply that one after the plate has interacted with the photons and with the environment. And then you're supposed to find something like:
|someenv> |brightleftscreen>|leftphotonwavepacket> + |otherenv> |brightrightscreen>|rightphotonwavepacket>
With |leftphotonwavepacket> about (|lefthole> + |righthole) and |rightphotonwavepacket> about (|lefthole> - |righthole>),
and |someenv> orthogonal to |otherenv> ; |brightleftscreen> orthogonal to |brightrightscreen> ...
I may simplify the problem:
H= H_slit_plate + H_photon + Hint
Where H_slit_plate describes the state evolution of the slit plate and Hint the interaction between the plate and the photons (we can replace photons by electrons if it is simpler to understand).
Now just suppose that we have [h_slit_plate, Hint]= 0.
In the case of electrons, Hint is a simple quantum wall potential with holes (the slits) at given positions.
If we now have @ to |psi(to)>= |slit_plate(to)>|photons(to)>
Due to this Hamiltonian, we will have at any t> to:
|psi(t)>= |slit_plate(t)>|photons(t)>
Where |photons(t)> is the state described in previous posts after the slits.
And |slit_plate(t)> is, for example, one eigenvalue of the Hamiltonian H_slit_plate.
=> In this case, we have no entanglement of the slit plate with the photons. You may notice that this example just highlights the implicit hypothesis used in the double slit experiment to show the QM results of collapse postulates and so on: no entanglement with the environment is required to get the known results.
vanesch said:
However, you are right that apparently the Schmidt decomposition in a strict sense does not always give us the correct basis if there is (near) degeneracy. I wasn't aware of this, just read up on it last night after this recent exchange of posts. In that case one also has to resort to some kind of stability criterion. That doesn't invalidate the "coarse" Schmidt decomposition (in which the reduced density matrix is almost diagonal). But there are some difficulties here I wasn't aware of.
Very interesting. May you detail?
My problem: I don’t see (or I don’t understand) how the Schmitt basis has been brought (the logical reasons) into the decoherence problem. And may be this is the source of my problems. Saying that the Schmitt decomposition is good just because it gives a single solution to the eigenbasis problem seems a very week argument (the same as saying the position is the preferred basis in most of problems). If this the case, I prefer to say that we only know the preferred eigenbasis after the experiment.
I would like to know how an observer infers a single eigenbasis for the observed results. Or if the eigenbasis is not important (independence): it’s only a matter of view (like the optical illusions analogy in my previous post). In this case, how to explain the different point of views an observer may have of an experiment (especially the ones concerning incompatible observables).
vanesch said:
That's where I think AN EXTRA POSTULATE is necessary (but which comes down to the traditional Born rule when we, intuitively use the correct basis: in that, when we buy a "momentum measurement apparatus" (because the vendor told us so, and it is written on the documentation), that we use the eigenstates of the P operator, and if we use an "angular momentum measurement apparatus" (again, because the salesman told us so), that we use the eigenstates of the L operator.
Do you know (or someone else) an apparatus that really measures the momentum p ?
I only know apparatuses that do position measurements (where we apply the born rules) and therefore deduce the momentum through the positions results (e.g. trough relations like <p>=d<r>/dt).
I also do not understand very well this point. Can we directly measure the momentum in QM?
vanesch said:
In fact, if we are only given the interaction hamiltonian of the apparatus with the system and with the environment, we are not naturally led to consider P or L, unless we introduce an extra postulate, for instance, which postulates that we should take, as preferred basis, what comes out of the Schmidt decomposition. But we can also work in the opposite way: we can say that we constructed quantum mechanics the way it is, because it works, when the salesman tells us that we have a momentum measurement apparatus, that things work out correctly when we do use the P eigenbasis etc... So maybe salesmen have a divine inspiration which is not reducible to physical laws: they _define_ our physical laws
cheers,
Patrick.
So what you are saying is « we just know the eigen basis after the experiment results ». If this is the case, you are saying that QM theory does not explain, a priori, why we see experimental results in a given eigenbasis. Rather it is an a posteriori selection (the salesperson has already tested its product: he knows what eigenbasis its product selects and we have to learn how its product works in order to “see” the results in the given eigenbasis ).
TI.
