10:00 AM
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A: Why do valence electrons not push each other away?

Matt HansonFirst of all, the electrons do exert a repulsive force on one another. This is inherently accounted for in the electrostatic repulsion terms in the atomic Hamiltonian that is used to solve the Schrödinger equation numerically. The fact that electrons will pair up is just a result of the fact that...

 
Thank you for your answer. It makes sense. Another question: how can we consider electrons 'close to each other' (as in Lewis structures), in the Schrodinger model?
 
@MattHanson: Why is it a "problem" at all? What is wrong with just treating any measurement device as a special kind of environment that actively localizes the particle wavefunction nearer to certain distinguished spatial locations, just like lightning rods localize lightning strikes? There is no need or reason to believe so-called 'collapse' in order to explain measurements... Similarly, I don't see any 'missing piece' related to your post; it's simply that we cannot have a full accurate model because there are too many parts, and it happens that we can approximate it using thermodynamics.
 
Great discussion! If I may add, a measurement always does something to a system. For instance, photons bounce off an apple when you observe it (or are emitted by the apple). So how is the quantum measurement problem formulated? Is it: the interactions that we measure on a quantum system (e.g. the photons from the apple) are continuosly happening and when we 'register' the interactions, then the system changes? Or is it: nothing happens to a quantum system, and suddenly interactions occur that we then register? In the second case, we can explain the problem by the interactions.
 
@KoendeJong: Your 'cases' seem to suffer the same conceptual problem; why do you assume (wrongly) that the system only changes suddenly or when you 'register' interactions? Just think of the particle system you want to measure as an overhead thundercloud, and your measurement device as a regular grid of lightning rods that you raise towards the cloud. The higher it goes, the more it will have changed the lightning strike distribution. There is no discontinuity; no sudden change and nothing to do with any observer.
 
Well you have to wonder what measurement is. In the case of the lightning rod, it is just a stream of electrons choosing the optimal path. Why then, does any measurement localize the wave function? Plus, can you please provide some evidence for your claims? You seem quite confident, which can be justified with enough evidence.
 
10:00 AM
I already gave sufficient explanation; any measurement device is clearly part of the environment. And you are wrong about lightning; it never chooses a truly optimal path. But it's just an analogy, don't miss the forest for the trees.
 
The question remains: why is every measurement device (which is just in and of itself a quantum system) part of the environment? Still there are some two-body quantum systems for which the wave function does not collapse. What property(or plural) distinguishes these two? On the case of lighting, sure, I’m wrong, but I wanted to illustrate the following: words such ‘measurements, register, observer, etc.’ are abstractions and experiments are not done with abstractions, but with interactions between systems, so it is reasonable to ask for a physical explanation.
(the other question also remains: please provide evidence for your claims. It seems logical, but why does it not work with two-body quantum systems where no wave function collapses?)
 
By definition the particle wavefunction is determined by its environment, which is literally everything else! So it makes no sense to exclude the measurement device. And there is no such thing as a 2-body quantum system in reality, because there are many particles in reality. Furthermore, you are the only one using the term 'collapse' as if it is meaningful; my point from the beginning was that it is just meaningless. You should know that purely deterministic systems can exhibit sensitivity to initial conditions (e.g. Galton's board), and wavefunctions are sensitive to measurement devices too!
 
Yes, I agree. The measurement device should not be excluded. But the question is, when does the wave-function collapse (by which I mean the ‘cloud’ becomes a ‘point’)? Take the apple. Does the wave function collapse when the photon bounces off of it, or when the photon lands on the detector? You seem to assume the second case. I wonder why?

On 2-body quantum systems: I mean, electron-electron interaction for instance. It makes no sense to say: the moon and earth is not a two-body system because it is a complete system. We are not talking Bose-Einstein condensates; we consider general quant
 
@KoendeJong: I thought it was very clear that I explicitly denied 'collapse'!
I explicitly said "There is no discontinuity", which implies that 'collapse to a point' is simply false.
 
Replace ‘collapse’ with ‘localization’. Please remain civil.
 
10:08 AM
I'm not being uncivil; I'm emphasizing that you're misinterpreting me from the beginning. That's all.
The point (pun not intended) is that you (and many other people) keep talking about 'collapse to a point', but I always denied that.
 
Fine. Then I simply do not understand what is wrong with the (long) message I just typed. Can you please rephrase?
 
Once you remove that point-collapse assumption, then all the 'paradoxes' vanish. Let's look at the apple example.
You mistakenly assumed that there is a 'time of collapse', but that is because of the discrete nature of a point-collapse. If you don't make that assumption, then of course you wouldn't even think of any idea of 'time of collapse' because everything is changing continuously and there is absolutely no distinguished point in time.
(Let's stick to non-relativistic QM for the time being.)
 
So you do not accept the Schrödinger model of an atom?
 
It works, but not the way you think.
 
Enlighten me
 
10:12 AM
First, have you understood the Galton board and the sensitivity to initial conditions?
 
Yes.
 
So you know that despite the completely deterministic nature of the setup, tiny perturbations in the initial conditions lead to different outcomes.
 
Yes.
 
In a QM system, the initial conditions are literally the starting environment (say the potential function).
This potential is never exactly the same no matter how carefully you set up a QM system.
 
Okay.
 
10:14 AM
It varies slightly depending on the jiggling of every particle in your experiment, especially the detectors!
If you isolate the QM system greatly, and don't attempt to measure anything, then you are improving the reproducibility of the initial conditions.
But once you put in any measurement device, you are literally making reproducibility impossible!
This doesn't imply we can't get useful things from a QM system. It just implies we can't assume that the different measurement outcomes is due to some mythical 'collapse' rather than simply unreproducibility of our measurement device configuration at the quantum level.
 
This is where I do not follow. Take the apple-analogy. The photon changes the apple, not the detector. How do you know that the detectors in QM are of such influence?
 
Give me a few min I need to do something first.
 
Go ahead.
 
I'm back. Here's another analogy (which is not meant to be accurate).
Suppose that you manage to generate a particle with a spherically expanding wavefunction.
Suppose you have a spherical detector centred at the same point you generate the particle.
Suppose that (unlike in real QM) this detector simply causes the wavefunction to localize sharply (but not to a point) to the first atom of the detector that the particle reaches.
Then simply having unpredictable slight jiggling of the detector atoms would result in unpredictable measurements of the 'location' at which the particle hits the detector.
It's all deterministic, yet you might erroneously believe that it 'underwent a point-collapse to a uniformly random location'.
Which would be magical!
 
10:32 AM
How do I interpret this? As: at different times you measure one quantum system, or different parts of the quantum systems? So like temperature seems probabilistic but when you map the particles it is deterministic?
 
Yes to the latter. No to the former. You are never measuring the same thing twice. You set up many QM systems hoping that they are very close in some aspects, but you know very well that they can't be identical.
 
The most important question I have is: where is the evidence?
 
The evidence is in the fact that if you apply this viewpoint to analyze any QM experiments you will never get wrong answers and you will never need the fictitious magical 'collapse' idea.
The reason people came up with the 'collapse' idea was because they made the wrong assumption that the wavefunctions of the particles they generate are exactly identical.
 
Okay. Then I support your view.
 
By the way, there are some experiments that provide weak philosophical evidence. The quantum eraser experiment, for example, is no longer surprising with this view.
 
10:38 AM
But there still is no proof. It would be wrong to discard the other views as wrong at this stage of research.
 
There is no such thing as proof in science.
This isn't mathematics.
 
I quit. All I have to say is: your views make sense, but to be condisending
 
I'm not condescending.
I'm just stating a fact.
 
Sorry, it cut off.
 
Proofs are a mathematical thing.
Nothing else.
Every real-world statement cannot be 'proven', only justified with the help of some interpretation.
Even if it's based on some applied mathematics, you still need an interpretation.
 
10:39 AM
You instantly state that all the other views are wrong without enough evidence to support your claim.
 
And no interpretation can be formally precise.
 
Again, I quit. Have a nice day!
 
@KoendeJong I have the right to state my opinion as fact as long as I strongly believe it.
You wouldn't want people to criticize you for saying the sun rose today.
Or that the Earth orbits the Sun.
If it helps you, simply assume that everything I say is my opinion.
And naturally it's your choice whether to take it seriously or not!
But it's definitely correct to say that some things are wrong, such as the one I said above (i.e. "the wavefunctions of the particles they generate are exactly identical").
There is no need for evidence; this is beyond reasonable doubt.
You can't find a single physicist who would insist that they can generate two particles with identical wavefunction to infinite precision.
@KoendeJong You are also free to quit any discussion. I just advise you to not be so sensitive to words that were at no point meant the way you took it.
If I looked down on you, I would not be wasting my time writing detailed explanations here.
My actions speak for themselves.
 
Okay. I misinterpreted it, given your explanations. I’m sorry!
 
No problem at all.
I'm glad it is cleared up.
Feel free to ask further on any specific point if you wish. And if you haven't come across the quantum eraser, you can look that up.
I need to go for a while again.
See you later!