28 minutes ago, Killtech said:

If i naively take the Hilbert space as my state space Ω of some classical Markov process without assuming anything about its structure, then how would my probability vectors look for that scenario? Basically i have to model it as a Markov chain on a continuous state space - which adds quite a bit of complexity because my 'probability vector' has uncountable infinite dimension, that is, it is a density function on a continuous space. as you can see a probability vector in that case requires far more information to be uniquely specified then a density matrix.

Ok, let's stop right here and do a sanity check. In what sense do you see evolution of the quantum state (never mind pure states evolving via Schrödinger or density matrices evolving via V. Neumann eq.) as a Markovian process? I fail to see how they are related.

Schrödinger eq. allows for evolution of superpositions. V. Neumann too, if only perhaps more obscurely, because they are integrated over in the p's (the statistical weights).

You must mean something like results of measurements somehow occur "internally" as Markovian processes that must be implemented in the evolution of the quantum state, completing it, quite unpredictably. For some reason to be ascertained later, they comply with Born's rule that the odds follow the square of the wave function, or bilinears of it.

Otherwise, I don't know what you mean.

Just now, Killtech said:

to prevent a misunderstanding, can you specify what you mean by a point function? did you mean a quantile function? note that this only exist for state spaces compatible with numeric operations because in order to write \(Pr(X \geq x)\), one requires that \(x\) is something that understand what \( \geq \) means. if \(x \epsilon \{red, green, blue\}\) this does not work. quantiles do not exist in general case of probability theory but are specific to numeric valued random variables usually.

A point function is one that has one and only one value at any point in the coordinate system.

quantiles ae not points.

A point is specified by the coordinate system, the value is specified by the function.

Pretty basic really.

But in QM you cannot identify a specific point and assign a specifc value to it. By specific I mean exact.

You can do one or the other.

8 hours ago, Killtech said:

hidden variables are also a thing of interpretation. the Hilbert state space stores a lot of non-observable information that you can just as much name a hidden variable - and quantum mechanics does not work without it.

Example of what you think is such a hidden variable?

If the variable isn’t observable, how do you determine its state?

20 hours ago, joigus said:

Ok, let's stop right here and do a sanity check. In what sense do you see evolution of the quantum state (never mind pure states evolving via Schrödinger or density matrices evolving via V. Neumann eq.) as a Markovian process? I fail to see how they are related.

Schrödinger eq. allows for evolution of superpositions. V. Neumann too, if only perhaps more obscurely, because they are integrated over in the p's (the statistical weights).

You must mean something like results of measurements somehow occur "internally" as Markovian processes that must be implemented in the evolution of the quantum state, completing it, quite unpredictably. For some reason to be ascertained later, they comply with Born's rule that the odds follow the square of the wave function, or bilinears of it.

Otherwise, I don't know what you mean.

perhaps something more general. if you have a state space and some deterministic equation fully describing the time evolution of a state, then this alone is sufficient to view it as a stochastic process which in this special case is deterministic. This is simply the generalization from a equation describing the time evolution of a single state to one which describes the time evolution of a distribution of states (in QM called a mixed state). One example for this would be Liouville equation in classical Hamiltonian mechanics.

The generalization from single state to a distribution of states takes a bit of additional formalism but you can always apply it. If the time evolution equation depends only on the state and not its history, then this naturally holds for its stochastic process, too, that is it is Markovian.

If we start by focusing on the part of QM which is the deterministic evolution of states only, e.g. Schrödinger equation, then we can naively apply this approach here as well (lets disregard for now that the quantum state is not itself actually measurable). But that of course is what Von Neumann equation does already. Digging deeper we can figure out the former is a transformed way to write the latter using some additional simplifying assumptions about the state space and its time evolution. The reason why this yields a Markov process is again, neither Schrödinger nor Von Neumann need to know anything abouts a states history to predict how it will evolve.

If you can follow this aspect, we can go into measurement.

22 hours ago, studiot said:

Pretty basic really.

But in QM you cannot identify a specific point and assign a specifc value to it. By specific I mean exact.

You can do one or the other.

it my be basic, but i am not native English speaker and googling 'point function' turned out it is also a term used for the quantile function.

22 hours ago, swansont said:

Example of what you think is such a hidden variable?

If the variable isn’t observable, how do you determine its state?

In case of Bohm de Broglie theory, there is a hidden position and momentum variables which is referred to by the name. their values are however partially revealed by a single measurement. In Kochen-Specker and Bell's it is more general a quantity referred to by lambda without further specification - anything your calculation of predictions may depend on and is not obviously available information. Their terminology is technically general enough to question whether that involves something like the wave function.

The wave function is non-observable, as QM prohibits to measure it directly. You cannot determine it in a single measurement. but what you can do is repeat an experiments with an well prepared ensemble many times and obtain a distribution of data from which you can reconstruct the wave function. for details about this you can refer to standard literature on QM.

Edited June 17Jun 17 by Killtech

9 minutes ago, Killtech said:

perhaps something more general. if you have a state space and some deterministic equation fully describing the time evolution of a state, then this alone is sufficient to view it as a stochastic process which in this special case is deterministic.

Aha! So deterministic is a special case of non-deterministic...

14 minutes ago, Killtech said:

The reason why this yields a Markov process is again, neither Schrödinger nor Von Neumann need to know anything abouts a states history to predict how it will evolve.

They do. QM is non-Markovian. Markovian systems don't keep memory of their previous history. Quantum mechanical systems do. Except when you measure. When you measure you do erase a big chunk of the past history of the system. So you have to be very specific and very precise about when this whole Markovian thing comes into play.

It's not impossible. I didn't say it's nonsense. It's just you haven't convinced me. Others don't seem convinced either.

18 minutes ago, Killtech said:

If you can follow this aspect, we can go into measurement.

So far, I'm afraid I can't follow your 'aspect'.

Edited June 18Jun 18 by joigus

2 hours ago, Killtech said:

In case of Bohm de Broglie theory, there is a hidden position and momentum variables which is referred to by the name. their values are however partially revealed by a single measurement. In Kochen-Specker and Bell's it is more general a quantity referred to by lambda without further specification - anything your calculation of predictions may depend on and is not obviously available information. Their terminology is technically general enough to question whether that involves something like the wave function.

Position and momentum are observable

The hidden variable in the proofs isn’t specified because it’s a general proof; it doesn’t just forbid just one specific variable

2 hours ago, Killtech said:

The wave function is non-observable, as QM prohibits to measure it directly. You cannot determine it in a single measurement. but what you can do is repeat an experiments with an well prepared ensemble many times and obtain a distribution of data from which you can reconstruct the wave function. for details about this you can refer to standard literature on QM

So it’s not measurable but you can measure it? (And I disagree that it can’t be known with one measurement. e.g. if I have a hydrogen atom and it emits a 1420 MHz photon, we know it’s in the lower hyperfine level of the ground state. And you can prepare systems in specific quantum states.)

I don’t see how it counts as a hidden variable, though, since having a definite wave function doesn’t necessarily get rid of probability. A system in a superposition, be it known or unknown, will give you multiple possible results.

9 hours ago, joigus said:

They do. QM is non-Markovian. Markovian systems don't keep memory of their previous history. Quantum mechanical systems do. Except when you measure. When you measure you do erase a big chunk of the past history of the system. So you have to be very specific and very precise about when this whole Markovian thing comes into play.

It's not impossible. I didn't say it's nonsense. It's just you haven't convinced me. Others don't seem convinced either.

You are partially right about the history but indeed the devil is in the detail of how that is exactly defined on a mathematical level.

It makes a huge difference about what random variable we talk about. For example the momentum operator is a function of the Hilbert space to the real numbers^3 and in our case is a valid random variable in this model. We can apply it at every time step of the states evolution and get the momentum process associated with it. is it Markovian? no, because indeed its probabilities depends on the history of its previous states. But 3 values are barely enough to characterize a quantum state, hence no surprise there. In fact no set of observables is able of produce a Markov process.

Now let's look at the identity operator of the Hilbert space. Let's call it \(S\) because it gives us the current quantum state of the system. This is the default random variable for any state space \(\Omega\). For this one we have
\[P(S_t = \psi(x,t) | S_{t-1} = \psi(x,t-1), S_{t-2} = \psi(x,t-2)) = P(S_t = \psi(x,t) | S_{t-1} = \psi(x,t-1))\]
This is the Markov property for this specific process (the time discrete variant for simplicity). It can be proven directly from the Schrödinger equation which guarantees it by having no dependence on prior states of \(\psi\) other then its time derivative at time \(t\). Even though we introduce a random variable \(S\), it does not necessary mean we can measure it. It just means it is an object we are interested in and therefore need a random variable to track it. \(P(S_t = \psi(x,t))\) only means we have a theory that can make theoretical predictions about what state a quantum system may be in, reflecting our knowledge of the system.

8 hours ago, swansont said:

So it’s not measurable but you can measure it? (And I disagree that it can’t be known with one measurement. e.g. if I have a hydrogen atom and it emits a 1420 MHz photon, we know it’s in the lower hyperfine level of the ground state. And you can prepare systems in specific quantum states.)

you can make predictions about it, sure, but not measure it.

you know that in order to measure \(\psi(x,0)\) you would need to experimentally obtain its value for every \(x\), and that for a single particle this wave function belongs to.

if it were measurable - i.e. an observable, QM would require that a linear operators exist that corresponds to its measurement. in case of a function, you need infinitely many of such to extract the value of the function at each \(x\).

Edited June 18Jun 18 by Killtech

Just now, Killtech said:

if it were measurable - i.e. an observable, QM would require that a linear operators exist that corresponds to its measurement. in case of a function, you need infinitely many of such to extract the value of the function at each x.

careful here.

'measurable has a different specific meaning from observable in mathematics, esp in respect of Hilbert spaces.

Just now, Killtech said:

It just means it is an object we are interested in and therefore need a random variable to track it.

I quite agree.

But again some care is needed as you also need a physical mechanism to pass from one state to another. That mechanism must have defining or describing (mathematical0 statements.

What they cannot be is a statement of the form that the system passes from state A to state B etc by some random jump process that take zero time and/or passes directly from one point in space to another without passing through the intervening points.

Unfortunately, this is exactly what you see if your model involves taking successive 'snapshots' of the position of a 'point particle' , say an electron in an orbital.

The way I see it (physically) is the electron cloud or density picture that is built up is not a patchwork of jumps but more like a mist in front of you,, which thins and thickens in patches, as you watch.

That is in accordance with the fact that the wave function occupies the entire region under consideration, rather as the mist occupies the entire region around you.

1 hour ago, Killtech said:

you can make predictions about it, sure, but not measure it.

Wait...you were the one who said you could measure it. You claimed you just couldn’t do it in a single measurement.

“The wave function is non-observable, as QM prohibits to measure it directly. You cannot determine it in a single measurement.”

1 hour ago, Killtech said:

you know that in order to measure ψ(x,0) you would need to experimentally obtain its value for every x, and that for a single particle this wave function belongs to.

That’s not the wave function I referred to, though.

1 hour ago, Killtech said:

if it were measurable - i.e. an observable, QM would require that a linear operators exist that corresponds to its measurement. in case of a function, you need infinitely many of such to extract the value of the function at each x.

I didn’t claim it was an observable, but since Bell wasn’t referring to the wave function, that makes this whole thing a distraction and moot. The wave function is not an “additional variable”

“THE paradox of Einstein, Podolsky and Rosen [1] was advanced as an argument that quantum mechanics could not be a complete theory but should be supplemented by additional variables.”

https://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf

1 hour ago, Killtech said:

For example the momentum operator is a function of the Hilbert space to the real numbers^3 and in our case is a valid random variable in this model.

No. The momentum operator is an endomorphism (a function from square-integrable functions to square-integrable functions). It is no random function.

It's the measured values are random.

1 minute ago, joigus said:

No. The momentum operator is an endomorphism (a function from square-integrable functions to square-integrable functions). It is no random function.

right, sorry, i meant \(\psi \rightarrow \langle \psi | O | \psi \rangle\) is a random variable for an observable \(O\)

however, for \(S\) we use \(\psi \rightarrow \psi\) as a random variable. it's a function valued random variable.

Edited June 18Jun 18 by Killtech

1 hour ago, Killtech said:

however, for S we use ψ→ψ as a random variable. it's a function valued random variable.

Not really. The wave function is not considered to be a random variable, for good reasons. It's a representative of infinitely many valid wave functions that all embody the totality of statistical properties of the system. All these wave functions differ from each other in a constant phase factor. You said yourself the wave function cannot be observed. Random variables can be measured. Otherwise it doesn't make much sense to call them random variables, does it?

12 minutes ago, joigus said:

Not really. The wave function is not considered to be a random variable, for good reasons. It's a representative of infinitely many valid wave functions that all embody the totality of statistical properties of the system. All these wave functions differ from each other in a constant phase factor. You said yourself the wave function cannot be observed. Random variables can be measured. Otherwise it doesn't make much sense to call them random variables, does it?

Have you ever heard of a Hidden Markov Model (HMM)?

in a HMM the underlying Markov process is not observable but we we have many observable random variables that do depend on the hidden process. The goal of this concept is to learn about the underlying process from the available observations. It would seem like quantum mechanic behavior may be a prime example for it. So no, random variables cannot be considered observable in general. A model may freely specify which are and which aren't.

Edited June 18Jun 18 by Killtech

6 minutes ago, Killtech said:

Have you ever heard of a Hidden Markov Model (HMM)?

in a HMM the underlying Markov process is not observable but we we have many observable random variables that do depend on the hidden process. The goal of this concept is to learn about the underlying process from the available observations. It would seem like quantum mechanic behavior may be a prime example for it. So no, random variables cannot be considered observable in general. A model may freely specify which are and which aren't.

Random variables have probability distributions. As such, they must be observable. Otherwise what does it even mean for a hidden variable to have a certain probability to adopt a particular value, if that value cannot be observed?

AFAIK, in HMM, probabilities are assigned to the Y's (the measured variables), not to the X's (the hidden states). The X's are there to provide conditional probabilities P(Y|X). Someone might have called these X's "random". If that's the case, I think it's a misnomer.

Anyway, the wave function is not a random variable, except for mixture states, in which it is. But its status as such is a little bit flaky (there is a huge arbitrariness in the choice of wave function).

The probabilistic nature od QM is not truly random in that the outcomes are not truly independent, the condition for a random variable.

At all times the wave function 'probability' must obey the normalisation condition.

This means that any change (motion) must affect the entire function, not just its value at a coordinate point.

This leads you back to my mist analogy.

12 minutes ago, joigus said:

Random variables have probability distributions. As such, they must be observable. Otherwise what does it even mean for a hidden variable to have a certain probability to adopt a particular value, if that value cannot be observed?

AFAIK, in HMM, probabilities are assigned to the Y's (the measured variables), not to the X's (the hidden states). The X's are there to provide conditional probabilities P(Y|X). Someone might have called these X's "random". If that's the case, I think it's a misnomer.

In order to use X in context of any random event to calculate probabilities for (including conditional probabilities), it must be random variable. a random variable is merely a measurable function (in the sense of measure theory). You fundamentally need that property otherwise you produce unmeasurable event sets outside your sigma algebra which would prevent to calculate anything.

While that technically means that P(X) is defined and you could theoretically calculate such properties, their interpretation is left open if X is not itself observable. You can consider this to reflect our knowledge about which state X may be in from the indirect observations we have, that is something like a Bayesian interpretation rather then actual measurement. This interpretation of probabilities most often goes along with non-observable random variables.

But you are trying to apply additional restriction from your interpterion which aren't required. you are trying to force some realist interpretation onto random variables which is not part of their math. your approach may be understandable from a physical point of view but within probability theory it doesn't make sense. as a mathematical theory, anything that is fits in the axiomatic framework of the theory is valid and may be used. interpretation is an issue left for others to solve.

21 minutes ago, joigus said:

Anyway, the wave function is not a random variable, except for mixture states, in which it is. But its status as such is a little bit flaky (there is a huge arbitrariness in the choice of wave function).

In the model we are talking about, wave functions are merely the states of the hidden process, that is elements of \(\Omega\). \(S: \psi \rightarrow \psi\) is a random variable in this space.

2 hours ago, Killtech said:

Have you ever heard of a Hidden Markov Model (HMM)?

in a HMM the underlying Markov process is not observable but we we have many observable random variables that do depend on the hidden process. The goal of this concept is to learn about the underlying process from the available observations. It would seem like quantum mechanic behavior may be a prime example for it. So no, random variables cannot be considered observable in general. A model may freely specify which are and which aren't.

It’s applied to classical physics, but I’m not aware of it being applied to quantum.

Perhaps you could work through a simple hidden variable example of spin-1/2 particle pairs with a total spin of zero, measured along three different axes 120 degrees apart (standard example) and get the QM result.

39 minutes ago, swansont said:

Perhaps you could work through a simple hidden variable example of spin-1/2 particle pairs with a total spin of zero, measured along three different exes 120 degrees apart (standard example) and get the QM result.

can do. would do it for a simple qubit to not mess up with spin operators. takes a bit of time to write it all down formally with some latex, so not during the week. does not involve any additional hidden variables that QM does not have itself. its just a reframing into classical probability framework.

but i think it makes sense to first work out a common understanding of what a random variable is, as this is crucial for the construction.

1) Wave functions are, by definition, solutions of the wave equation. There is no other choice. Perhaps joigus means where is choice of wave eqaution (which there is).

2) No hidden variables are needed and the wave function again cannot be 'hidden' by definition.

3) Solutions can be complex, but we only want the real part.

3) The way to introduce probabilistic maths into this is to introduce a 'statistical weighting function', This has to be a function of a real variable, and again cannot be hidden.

On 6/18/2025 at 9:02 PM, studiot said:

1) Wave functions are, by definition, solutions of the wave equation. There is no other choice. Perhaps joigus means where is choice of wave eqaution (which there is).

What I mean --hopefully-- more precisely is that, as you imply, wave functions are determinations from a range of physical conditions (examples: hydrogen atom's ground state, free particles, etc) whose only undetermined co-factor are global or even local (in cases where the dynamical theory is a gauge theory) phases that can be chosen locally at will (not measured or indeed measurable at all). Summarising:

Wave funcions have,

1) A factor that is completely determined

2) Another co-factor that is a huge arbitrariness

Mind you, "huge" here is, if anything, an understatement.

It is futile to keep discussing here if there isn't a least common denominator of what is known to be the case for quanta.

Having said that, I sympathise with attempts to "better understand" what possible sub-reality[?] quantum mechanics is telling us about.

But variables half of which merit the qualifier of "completely determinable" and the other half "completely arbitrary", to me at least, cannot qualify as random, as @Killtech seems to purport.

Sorry for my tardiness in answering, but life-changing events are taking place for me lately.

[?] Less-stringent reality than what our intuitive criterion would have it?

Edited June 23Jun 23 by joigus
correction

3 hours ago, joigus said:

Sorry for my tardiness in answering, but life-changing events are taking place for me lately.

feel you there. quite a lot on my mind as well. reduces the time i can spend on the interesting matters.

3 hours ago, joigus said:

But variables half of which merit the qualifier of "completely determinable" and the other half "completely arbitrary", to me at least, cannot qualify as random, as @Killtech seems to purport.

this is not relevant for the question if we can model given experiments via classic probability theory. even if you include completely arbitrary information in your state space & model that do have no impact on the results, it just that: surplus info you could drop. it just makes it more tedious to deal with objects full of irrelevant information.

however, there is a reason i did chose this approach to start of with a rather large state space and that is because QM already implies that almost all of it is irreducible information to make correct predictions, so no model can afford to drop it and still get the correct predictions.

4 hours ago, joigus said:

Wave funcions have,

1) A factor that is completely determined

2) Another co-factor that is a huge arbitrariness

consider some pure state of an bound electron in a H-atom and its wave function. Let's take it as a singular system and not an enable of identically prepared ones and probe it using scattering experiments with some (hypothetical) idealized test particles that are so light and marginally charged (compared to the electron) that we can scatter many of them without collapsing or disturbing the target wave function.

Scattering theory says in the first Born approximation the electron will actually interact as if its charge was in fact physically distributed according to \(\rho = |\psi|^2\). Scattering amplitude in this first order approx will be the Fourier transform of that, meaning most of the information contained in the wave function will make a difference for the outcome, especially if we can freely choose the incoming angle and energy of the test particles. Higher order approx of this experiments will also give us info about the magnetic moments and so on.

Interestingly, even if the wave function is a superposition of two energy eigenstates, then in the Born approx its charge distribution is not stationary (unlike for energy states) but instead oscillates with a known frequency. so if our scattering experiments has some time resolution, we would also be able to distinguish such pure states as well. however, thinking classically, such a oscillation would naturally cause an EM-emission (with exactly the same wavelength as QM predicts) and loss of energy collapsing the state to the next lower stable solution (and only energy eigenstates are stationary and thus classically stable) - that is even classically one would expect quite the same behavior as QM predicts. just saying that such a state would be very short-lived and hence difficult to observe.

through such gedanken-experiments one can boil it down that is is only the global phase factor of the wave function that is truly irrelevant for any prediction, hence almost all of the information contained in the wave function does seem irreducible.

this line of thought is a bit of a brute force expansion of weak measurements if we had some test particles that could do that. i mean you could technically do it with very low energy photons but uff, measuring those will be a challenge.
A realistic experiments on this topic is of course https://www.nature.com/articles/nature10120 or https://www.nature.com/articles/nature12539

Edited June 23Jun 23 by Killtech

6 minutes ago, Killtech said:

Scattering theory says in the first Born approximation the electron will actually interact as if its charge was in fact physically distributed according to ρ=|ψ|2. Scattering amplitude in this first order approx will be the Fourier transform of that, meaning most of the information contained in the wave function will make a difference for the outcome, especially if we can freely choose the incoming angle and energy of the test particles. Higher order approx of this experiments will also give us info about the magnetic moments and so on.

I think you're conflating here Born's rule with the first Born approximation (scattering theory). Very different things, even if both bear the name "Born".

Born's rule is the general assumption that probabilities are bilinears of amplitudes.

Born's approximation, OTOH, is the simplification that scatterers can be considered as given, and unaffected by the scattered.

Further, the Fourier transform of any x-dependent quantum distribution is yet another thing. Namely, the momentum representation of said local distribution.

So you're conflating three different things here. Two of them bear the name 'Born". The other one was born at about the same time. The relation of none of them to the problem under discussion is born out by what I know about quantum mechanics.

Edited June 23Jun 23 by joigus
minor addition

19 minutes ago, joigus said:

I think you're conflating here Born's rule with the first Born approximation (scattering theory). Very different things, even if both bear the name "Born".

i haven't mixed them up, just applied the first order Born approx to a target that is not a classic potential, but a quantum system, a wave function itself.

In that case, this approximation uses an integral over \(\rho(x)V(x)\) where V is the classic point charge potential to calculate the effective scattering potential. And indeed here rho is calculated according to Born rule. But that's the game in scattering theory and needed to get predictions matching experimental results.

Normally, physics goes into the high energy regimes of deep inelastic scattering. but it gets quite interesting going the other way towards the shallow elastic scattering regime as well and that does not disturb the wave function so much making it open for repeated measurements. that's where devil comes in.

42 minutes ago, joigus said:

Further, the Fourier transform of any x-dependent quantum distribution is yet another thing. Namely, the momentum representation of said local distribution.

sure, this scenario represents an (1-electron, n-test paticles) system where we have n measurements. However, because all n test particles are by assumption perfectly prepared (i.e. we know their exact wave function), all n measurements effectively extract the only not-known information about the system and that is from the singular electron.

if you account that the test particle has spin (or is a photon with prepared polarization), then in the second approximation we can extract all about the target wave function excluding only the global phase factor. at least according to the theory.

sorry, i tend to skip over many details because i assume we all know here the details of quantum mechanics in an out, so that we do not need to get into the details how all that knowledge was derived exactly and can get quicker to the interesting stuff. but i tend to forget that what i say is not exactly trivial. i am getting used to talking to AI which has all the knowledge at hand to catch my drift and intension directly without the need to do a lot of explaining.

Edited June 23Jun 23 by Killtech

2 hours ago, Killtech said:

i haven't mixed them up, just applied the first order Born approx to a target that is not a classic potential, but a quantum system, a wave function itself.

In that case, this approximation uses an integral over ρ(x)V(x) where V is the classic point charge potential to calculate the effective scattering potential. And indeed here rho is calculated according to Born rule. But that's the game in scattering theory and needed to get predictions matching experimental results.

Yes, you have mixed them up. Scattering theory is about an incoming state that comes from \( t=-\infty \) and evolves towards an outgoing state that evolves towards \( t=+\infty \) .

Quite different from \[ \left|\psi\right|^{2} \]. Born's rule is about one and the same state. Scattering is about an incoming and outgoing states, both close to plane waves, and infinitely distant in time.

Do you really know about quantum mechanics? Doesn't sound like you do.

2 hours ago, Killtech said:

sorry, i tend to skip over many details because i assume we all know here the details of quantum mechanics in an out, so that we do not need to get into the details how all that knowledge was derived exactly and can get quicker to the interesting stuff. but i tend to forget that what i say is not exactly trivial. i am getting used to talking to AI which has all the knowledge at hand to catch my drift and intension directly without the need to do a lot of explaining.

Oh, come on, drop the attitude, will you? I know quite a bunch of details about quantum mechanics, and I don't need prosthetics for my intelligence. AI has failed to answer some of my deepest questions. Miserably so.

No wonder, really. AI works on the logical span of what humans have already thought. It's clueless about what's next. If you could paraphrase what it's trying to do (sometimes to astonishing perfection, I'll give you that), it is: How could I convince myself this is what human interlocutors would want to hear?

Edited June 24Jun 24 by joigus
Latex editing+nuance

6 minutes ago, joigus said:

Yes, you have mixed them up. Scattering theory is about an incoming state that comes from t=−∞ and evolves towards an outgoing state that evolves towards t=+∞ .

Quite different from |ψ|2 . Born's rule is about one and the same state. Scattering is about an incoming and outgoing states, both close to plane waves, and infinitely distant in time.

Do you really know about quantum mechanics? Doesn't sound like you do.

to keep this short look up the atomic form factor then https://en.wikipedia.org/wiki/Atomic_form_factor - you'll see a rho popping in that formula.

It's similar to what you do in the Hartee-Fock method for many electron atoms where you also calculate effective potentials.