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It's always seemed to me that these types of question only exist because we're considering a choice between two imperfect models. If we had a better model of wh

It's always seemed to me that these types of question only exist because we're considering a choice between two imperfect models. If we had a better model of what a "particle" really is, then there would be no dualing models nor paradox. Do we really have to choose between wave and particle? What does the "particle" model bring to the table that a localized (wavelength-sized) wave/vibration could not?

gus_massa 2y ago
We have a better model, but it's an ecuation so horrible that nobody want to solve it. Luckly, sometimes the exact solution can be very accuately aproximated with a wave ecuation. Luckly, sometimes the exact solution can be very accuately aproximated with a particle ecuation. (Sometimes, the exact solution can be aproximated saying that the lowest energy state is an eigenvector of the Schoedinger equation. Is that a wave? It's not localized, but not very wavy.) But neither are the exact solution, just aproximations that solve tpgether 99% of the experiment. It's difficult to explain, because to explaing the detials you need like two years of algebra and calculus and then like another 2 years of physics, and now you get a degree in physics. It's possible to solve the difficult ecuation only in very simple cases like electron-electron colissions, if you allow some cheating and a tiny error. For more complicated systems like electron-muon there are some problems. And for more complicated systems, you get more technical problems and more aproximations.
- tim-kt 2y ago
  What is the name of the better model which you are talking about?
  - kadoban 2y ago
    I'm also not sure, but I'm thinking some variation of Quantum Field Theory.
    - gus_massa 2y ago
      Yes, in general https://en.wikipedia.org/wiki/Quantum_field_theory https://en.wikipedia.org/wiki/Quantum_field_theory but also the list of details in https://en.wikipedia.org/wiki/Standard_Model https://en.wikipedia.org/wiki/Standard_Model
tim-kt 2y ago
The photoelectric effect [0] can be explained if light behaves as discrete particles, but not when it's a wave since a higher amplitude does not imply a higher energy transfer. [0] https://en.m.wikipedia.org/wiki/Photoelectric_effect https://en.m.wikipedia.org/wiki/Photoelectric_effect
- rusk 2y ago
  If I emit a bass signal at a low amplitude, but then emit it at a higher amplitude, I can see the effect on a glass of water on the table. What’s happening here if amplitude does not carry power? My understanding is that theoretically energy transfer is a function of wavelength.
  - tim-kt 2y ago
    Sorry, my last sentence wasn't formulated well. Yes, a wave with higher amplitude (or one could say "intensity") has a higher energy. The photoelectric effect happens when you shine light with "enough" energy on some material such that the atoms of the material are ionized, i. e. electrons are freed. You need a minimal energy for this and if you use dim light with a low frequency, you will not see the effect. Now, if you increase the frequency of the light, you can measure electrons. If, instead, you make the light brighter, that is, increase the amplitude of the wave (if it were a wave), you don't see electrons. So at least in this experiment, light does not function as a wave.
    - deleted 2y ago
      [deleted]
    - HarHarVeryFunny 2y ago
      But once you've increased the light frequency (i.e. photon energy) above the required threshold, THEN making the light brighter (more photons) will increase the number of electrons emitted.
  - tsimionescu 2y ago
    The point here is that the total displacement of water caused by a sound wave depends on both the amplitude of the wave, and its frequency, with no limit: if the wave has high enough amplitude, it will displace water even if the wave is very low frequency. However, this is not true for EM interactions. If you shine infrared light on a solar panel, you'll see 0 current from it, even with an extremely powerful source of light (at some point the material might heat up enough it starts showing some thermo-electric effect, but that's a different thing). However, if you take even a very low intensity ultraviolet source, you'll see a measurable current right away. This is the unexpected behavior that quantized interactions have, which can't be reproduced with non-qunatized waves like sound waves.
fallingknife 2y ago
I also don't understand this. AFAIK "particle" in this context means quantized unit rather than contiguous solid object. And I see no reason why a quantized unit of a wave can't propagate through two slits simultaneously. But my level of understanding here is YouTube level so if you know more please correct me.
- ithkuil 2y ago
  I trust the physics works out. The problem in these discussions is how to build an intuition about the underlying physical model. I fail to have an intuition of how can a quantized unit of wave propagate through both slits. I know that the equations say that the probability of finding the particle at a given location is given by the amplitude squared of the wave function (Born rule). The image that a "quantized unit of wave propagates through two slits simultaneously" doesn't help me build any further intuition. Do the two parts going through the two different paths carry half the unit? Clearly that's not the case otherwise they wouldn't be quanta anymore. So does it mean that the entire wavefront is "one unit" no matter how spread out? But in that case, "one unit" of what?
  - HarHarVeryFunny 2y ago
    If one just sends photons through a narrow single slit, then the pattern that builds up on the screen (if you send multiple photons, and record their positions) will be a banded diffraction pattern. If you have two slits, with a detector to determine which slit the photon went thru, then it'll behave as if it only went thru one of the two slits, at random, and what'll build up on the screen will be the two (slit A + slit B) overlayed diffraction patterns. Finally, if you have two slits with NO detector, then what will build up on the screen is the interference pattern as if the photon had gone thru both slits simultaneously and the two resulting banded diffraction patterns interfered with each other. So, what SEEMS to be happening in this case is that the quantum state of the system post-slit is that of the photon simultaneously having gone thru both slits, each slit having diverted it per diffraction, and then these diffraction patterns (probabilities) interferering. Wave collapse can only be happening after this interference (if it was before then there would only be one diffraction pattern and no interference), presumably when quantum state interacts with the screen. So, yeah, it seems that the "photon" does "go" through both slits, but this is a quantum representation, not a classical one.
aqme28 2y ago
We do have a better model, but the mathematical model doesn’t analogize well to the real-life concepts we’re used to.
IshKebab 2y ago
Precisely. The question "is it a particle or a wave" is wrong. It's neither. It's a particle-wave. Something that behaves like a classical wave or particle depending on the situation, but it doesn't switch between them or anything like that. It's not a "particle that has interference" or a "wave with a location". Classic labelling issue.
jfengel 2y ago
Being pedantic about the language, there is only one model, and effectively every physicist agrees on it. What they differ about is the interpretation of that model. The equations are the same, but differ in what the variables refer to in the real world. It's really a matter of solving the equation for X vs Y, saying which one is independent and which is dependent. The purpose is to take the fact that none of the variables correspond directly to anything we have any experience with. The best we can hope for is to isolate part of it and say "this much is like this thing we understand, but there's an additional thing that we'll treat as a correction". We can try to take the whole thing seriously, and just call it "a quantum thingy" which is not like anything else. This is sometimes called "shut up and calculate", but even that makes assumptions about what things are feasible to calculate and which are hard. That skews your understanding even if you're trying to let it speak for itself.
- simpaticoder 2y ago
  >there is only one model There is one set of observations, and many many models to describe them: Schrödinger equation formulation, matrix mechanics (Heisenberg, Born, and Jordan), path integral formulation (Feynman), phase space formulation, density matrix formulation, QFT or second quantization, variational formulation, pilot wave theory aka de Broglie-Bohm theory, Hamilton-Jacobi formulation, PT-symmetric quantum mechanics, Dirac equation formulation (well, not really independent, just for spin 1/2 particles). They all give the same results, and are therefore mathematically equivalent, but different models tend to be associated with different interpretations: Schrödinger Equation : Copenhagen, Bohmian Mechanics, Many-Worlds Matrix Mechanics : Copenhagen Path Integral : Many-Worlds, Stochastic Density Matrix: Ensemble, Decoherence-based Second Quantization : Many-Worlds Pilot Wave Theory : Bohmian Mechanics Consistent Histories : Decoherence-based Relational QM : Relational Interpretation Stochastic Models : Stochastic Interpretations, GRW (Ghirardi–Rimini–Weber) Collapse
kromem 2y ago
In video games that have procedural generation, there's often a seed function that predicts a continuous geometry. But in order to track state changes from free agents, when you get close to that geometry the engine converts it to discrete units. This duality of continuous foundation becoming discrete units around the point of observation/interaction is not the result of dueling models, but a unified system. I sometimes wonder if we'd struggle with interpreting QM the same way if there wasn't a paradigm blindness with the interpretations all predating the advances in models in information systems.
canjobear 2y ago
> What does the "particle" model bring to the table that a localized (wavelength-sized) wave/vibration could not? A lot of the article is about this. Start with the section "The Wave Function of Two Particles and a Single Door". The wave packet view can't explain why you don't for example see a "particle" (that is, a dot on a detector) show up simultaneously having gone through two different doors. You have to think about it in terms of a wave in the space of possible joint particle positions.

gus_massa 2y ago

We have a better model, but it's an ecuation so horrible that nobody want to solve it. Luckly, sometimes the exact solution can be very accuately aproximated with a wave ecuation. Luckly, sometimes the exact solution can be very accuately aproximated with a particle ecuation. (Sometimes, the exact solution can be aproximated saying that the lowest energy state is an eigenvector of the Schoedinger equation. Is that a wave? It's not localized, but not very wavy.) But neither are the exact solution, just aproximations that solve tpgether 99% of the experiment. It's difficult to explain, because to explaing the detials you need like two years of algebra and calculus and then like another 2 years of physics, and now you get a degree in physics. It's possible to solve the difficult ecuation only in very simple cases like electron-electron colissions, if you allow some cheating and a tiny error. For more complicated systems like electron-muon there are some problems. And for more complicated systems, you get more technical problems and more aproximations.

tim-kt 2y ago
What is the name of the better model which you are talking about?
- kadoban 2y ago
  I'm also not sure, but I'm thinking some variation of Quantum Field Theory.
  - gus_massa 2y ago
    Yes, in general https://en.wikipedia.org/wiki/Quantum_field_theory https://en.wikipedia.org/wiki/Quantum_field_theory but also the list of details in https://en.wikipedia.org/wiki/Standard_Model https://en.wikipedia.org/wiki/Standard_Model

tim-kt 2y ago

The photoelectric effect [0] can be explained if light behaves as discrete particles, but not when it's a wave since a higher amplitude does not imply a higher energy transfer. [0] https://en.m.wikipedia.org/wiki/Photoelectric_effect https://en.m.wikipedia.org/wiki/Photoelectric_effect

rusk 2y ago
If I emit a bass signal at a low amplitude, but then emit it at a higher amplitude, I can see the effect on a glass of water on the table. What’s happening here if amplitude does not carry power? My understanding is that theoretically energy transfer is a function of wavelength.
- tim-kt 2y ago
  Sorry, my last sentence wasn't formulated well. Yes, a wave with higher amplitude (or one could say "intensity") has a higher energy. The photoelectric effect happens when you shine light with "enough" energy on some material such that the atoms of the material are ionized, i. e. electrons are freed. You need a minimal energy for this and if you use dim light with a low frequency, you will not see the effect. Now, if you increase the frequency of the light, you can measure electrons. If, instead, you make the light brighter, that is, increase the amplitude of the wave (if it were a wave), you don't see electrons. So at least in this experiment, light does not function as a wave.
  - deleted 2y ago
    [deleted]
  - HarHarVeryFunny 2y ago
    But once you've increased the light frequency (i.e. photon energy) above the required threshold, THEN making the light brighter (more photons) will increase the number of electrons emitted.
- tsimionescu 2y ago
  The point here is that the total displacement of water caused by a sound wave depends on both the amplitude of the wave, and its frequency, with no limit: if the wave has high enough amplitude, it will displace water even if the wave is very low frequency. However, this is not true for EM interactions. If you shine infrared light on a solar panel, you'll see 0 current from it, even with an extremely powerful source of light (at some point the material might heat up enough it starts showing some thermo-electric effect, but that's a different thing). However, if you take even a very low intensity ultraviolet source, you'll see a measurable current right away. This is the unexpected behavior that quantized interactions have, which can't be reproduced with non-qunatized waves like sound waves.

fallingknife 2y ago

I also don't understand this. AFAIK "particle" in this context means quantized unit rather than contiguous solid object. And I see no reason why a quantized unit of a wave can't propagate through two slits simultaneously. But my level of understanding here is YouTube level so if you know more please correct me.

ithkuil 2y ago
I trust the physics works out. The problem in these discussions is how to build an intuition about the underlying physical model. I fail to have an intuition of how can a quantized unit of wave propagate through both slits. I know that the equations say that the probability of finding the particle at a given location is given by the amplitude squared of the wave function (Born rule). The image that a "quantized unit of wave propagates through two slits simultaneously" doesn't help me build any further intuition. Do the two parts going through the two different paths carry half the unit? Clearly that's not the case otherwise they wouldn't be quanta anymore. So does it mean that the entire wavefront is "one unit" no matter how spread out? But in that case, "one unit" of what?
- HarHarVeryFunny 2y ago
  If one just sends photons through a narrow single slit, then the pattern that builds up on the screen (if you send multiple photons, and record their positions) will be a banded diffraction pattern. If you have two slits, with a detector to determine which slit the photon went thru, then it'll behave as if it only went thru one of the two slits, at random, and what'll build up on the screen will be the two (slit A + slit B) overlayed diffraction patterns. Finally, if you have two slits with NO detector, then what will build up on the screen is the interference pattern as if the photon had gone thru both slits simultaneously and the two resulting banded diffraction patterns interfered with each other. So, what SEEMS to be happening in this case is that the quantum state of the system post-slit is that of the photon simultaneously having gone thru both slits, each slit having diverted it per diffraction, and then these diffraction patterns (probabilities) interferering. Wave collapse can only be happening after this interference (if it was before then there would only be one diffraction pattern and no interference), presumably when quantum state interacts with the screen. So, yeah, it seems that the "photon" does "go" through both slits, but this is a quantum representation, not a classical one.

aqme28 2y ago

We do have a better model, but the mathematical model doesn’t analogize well to the real-life concepts we’re used to.

IshKebab 2y ago

Precisely. The question "is it a particle or a wave" is wrong. It's neither. It's a particle-wave. Something that behaves like a classical wave or particle depending on the situation, but it doesn't switch between them or anything like that. It's not a "particle that has interference" or a "wave with a location". Classic labelling issue.

jfengel 2y ago

Being pedantic about the language, there is only one model, and effectively every physicist agrees on it. What they differ about is the interpretation of that model. The equations are the same, but differ in what the variables refer to in the real world. It's really a matter of solving the equation for X vs Y, saying which one is independent and which is dependent. The purpose is to take the fact that none of the variables correspond directly to anything we have any experience with. The best we can hope for is to isolate part of it and say "this much is like this thing we understand, but there's an additional thing that we'll treat as a correction". We can try to take the whole thing seriously, and just call it "a quantum thingy" which is not like anything else. This is sometimes called "shut up and calculate", but even that makes assumptions about what things are feasible to calculate and which are hard. That skews your understanding even if you're trying to let it speak for itself.

simpaticoder 2y ago
>there is only one model There is one set of observations, and many many models to describe them: Schrödinger equation formulation, matrix mechanics (Heisenberg, Born, and Jordan), path integral formulation (Feynman), phase space formulation, density matrix formulation, QFT or second quantization, variational formulation, pilot wave theory aka de Broglie-Bohm theory, Hamilton-Jacobi formulation, PT-symmetric quantum mechanics, Dirac equation formulation (well, not really independent, just for spin 1/2 particles). They all give the same results, and are therefore mathematically equivalent, but different models tend to be associated with different interpretations: Schrödinger Equation : Copenhagen, Bohmian Mechanics, Many-Worlds Matrix Mechanics : Copenhagen Path Integral : Many-Worlds, Stochastic Density Matrix: Ensemble, Decoherence-based Second Quantization : Many-Worlds Pilot Wave Theory : Bohmian Mechanics Consistent Histories : Decoherence-based Relational QM : Relational Interpretation Stochastic Models : Stochastic Interpretations, GRW (Ghirardi–Rimini–Weber) Collapse

kromem 2y ago

In video games that have procedural generation, there's often a seed function that predicts a continuous geometry. But in order to track state changes from free agents, when you get close to that geometry the engine converts it to discrete units. This duality of continuous foundation becoming discrete units around the point of observation/interaction is not the result of dueling models, but a unified system. I sometimes wonder if we'd struggle with interpreting QM the same way if there wasn't a paradigm blindness with the interpretations all predating the advances in models in information systems.

canjobear 2y ago

> What does the "particle" model bring to the table that a localized (wavelength-sized) wave/vibration could not? A lot of the article is about this. Start with the section "The Wave Function of Two Particles and a Single Door". The wave packet view can't explain why you don't for example see a "particle" (that is, a dot on a detector) show up simultaneously having gone through two different doors. You have to think about it in terms of a wave in the space of possible joint particle positions.