6 ms·

This is incorrect. There are particles. They are excitations in the field. There still is the 'split particle paradox' because QFT does not solve the measureme

by cjfd 2y ago

This is incorrect. There are particles. They are excitations in the field. There still is the 'split particle paradox' because QFT does not solve the measurement problem. The 'some kind of interaction of graph nodes' by which I am guessing you are referring to Feynman diagrams are not of a fundamental nature. They are an approximation known as 'perturbation theory'.

movpasd 2y ago
I think what they must be referring to is the fact that particles are only rigorously defined in the free theory. When coupling is introduced, how the free theory relates to the coupled theory depends on heuristic/formal assumptions. We're leaving my area of understanding, but I believe Haag's theorem shows that the naïve approach, where the interacting and free theories share a Hilbert space, completely fails -- even stronger than that, _no_ Hilbert space could even support an interacting QFT (in the ways required by scattering theory). This is a pretty strong argument against the existence of particles except as asymptotic approximations. Since we don't have consensus on a well-defined, non-perturbative gauge theory, mathematically speaking it's difficult to make any firm statements about what states "exist" in absolute. (I'm certain that people working on the various flavours of non-perturbative (but still heuristic) QFT -- like lattice QFT -- would have more insights about the internal structure of non-asymptotic interactions.)
- codethief 2y ago
  > This is a pretty strong argument against the existence of particles except as asymptotic approximations. I think it's also a pretty strong argument against the mathematical well-definedness of typical (interacting) QFTs in the first place.
- westurner 2y ago
  Though it doesn't resolve whether a "quanta" is a particle or a measurable convergence of waves, Electrons and Photons are observed with high speed imaging. "Quantum microscopy study makes electrons visible in slow motion" https://news.ycombinator.com/item?id=40981054 https://news.ycombinator.com/item?id=40981054 There exist single photon emitters and single photon detectors. Qualify that there are single photons if there are single photon emitters: Single-photon source: https://en.wikipedia.org/wiki/Single-photon_source https://en.wikipedia.org/wiki/Single-photon_source QFT is not yet reconciled with (n-body) [quantum] gravity, which it has 100% error in oredicting. random chance. TOD IIRC, QFT cannot explain why superfluid helium walks up the sides of a container against gravity, given the mass of each particle/wave of the superfluid and of the beaker and the earth, sun, and moon; though we say that gravity at any given point is the net sum of directional vectors acting upon said given point, or actually gravitational waves with phase and amplitude. You said "gauge theory", "Topological gauge theory of vortices in type-III superconductors" https://news.ycombinator.com/item?id=41803662 https://news.ycombinator.com/item?id=41803662 From https://news.ycombinator.com/context?id=43081303 https://news.ycombinator.com/context?id=43081303 .. https://news.ycombinator.com/item?id=43310933 https://news.ycombinator.com/item?id=43310933 : > Probably not gauge symmetry there, then.
DiogenesKynikos 2y ago
Particles are an approximation to the actual behavior of the field, and are used in perturbation theory to calculate the more complicated field behavior. This works well when interactions are weak. Electrons do not couple strongly to the electromagnetic field, so it makes sense to view electrons as particles. However, quarks couple very strongly to the strong force (hence the name), so the perturbative approach breaks down, and it makes less sense to view quarks as particles.
- drpossum 2y ago
  So in a non-perturbative QFT calculation which has a well defined particle-number operator, that's just "an approximation" within the theory? What is it approximating?
  - GoblinSlayer 2y ago
    Energy capacity?
    - deleted 2y ago
      [deleted]
    - drpossum 2y ago
      I'll bite: Explain yourself. Also, for context, my question was posed because the idea of "particle number" as well as "quantum states of particles (which are countable) represented in a Fock space" and in general the idea of particles are, like, page 2 of any QFT textbook. It doesn't approximate anything in the theory. Creation and annihilation of particles (and hence the well-defined concept of a particle) is fundamental to the construction of the theory itself, perturbative or not.
      - DiogenesKynikos 2y ago
        Particles are page 2 of any QFT textbook because the free particle is the only system we can exactly solve. In practice, that solution is usually used as the basis for a perturbative expansion.
gus_massa 2y ago
Perhaps a better way to say it is that particles are not longer small balls of dirt [1], but a mathematical construction that is useful to generate an infinite serie [2] to calculate the results. Since in some conditions these mathematical tricks behave very similar to small balls of dirt, we reused the word "particle" and even the names we used when we thought they were small balls of dirt. [11] We probably never thought they were made of dirt, and in any case the magnetic moment is the double of the value of the small ball of dirt model. [2] That has so many infinites that would make a mathematician cry.
- cjfd 2y ago
  Note that particles are not just for perturbation theory. There is a particle whenever there exists a particle annihilation/creation field configuration. A proton is a particle so writing down its creation/annihilation field configuration is in theory possible, though maybe not in practice. Another point is that infinities do not necessarily make mathematicians cry. Abraham Robinson is quite pleased with them. It seems a possible hypothesis that at least some QFT are mathematically well-defined using non-standard analysis. Where 'some QFT' at least renormalizable and perhaps also asymptotically free. I don't know enough about it to know how the Haag theorem, mentioned in another comment impacts this.
  - nine_k 2y ago
    Another analogy (flawed as any of them). Sports teams "exist" in a sense. They meet one another in well-defined interactions, called matches, and such an interaction can be described as if teams were well-defined atomic entities, producing a score. But a sports team is not atomic, not a "final reality" entity. A sports team can pass through one gate, or through several gates, when entering a stadium. From a doctor's perspective, the team "does not exist", a doctor only operates in terms of individual players' organisms.

by cjfd 2y ago

movpasd 2y ago

I think what they must be referring to is the fact that particles are only rigorously defined in the free theory. When coupling is introduced, how the free theory relates to the coupled theory depends on heuristic/formal assumptions. We're leaving my area of understanding, but I believe Haag's theorem shows that the naïve approach, where the interacting and free theories share a Hilbert space, completely fails -- even stronger than that, _no_ Hilbert space could even support an interacting QFT (in the ways required by scattering theory). This is a pretty strong argument against the existence of particles except as asymptotic approximations. Since we don't have consensus on a well-defined, non-perturbative gauge theory, mathematically speaking it's difficult to make any firm statements about what states "exist" in absolute. (I'm certain that people working on the various flavours of non-perturbative (but still heuristic) QFT -- like lattice QFT -- would have more insights about the internal structure of non-asymptotic interactions.)

codethief 2y ago
> This is a pretty strong argument against the existence of particles except as asymptotic approximations. I think it's also a pretty strong argument against the mathematical well-definedness of typical (interacting) QFTs in the first place.
westurner 2y ago
Though it doesn't resolve whether a "quanta" is a particle or a measurable convergence of waves, Electrons and Photons are observed with high speed imaging. "Quantum microscopy study makes electrons visible in slow motion" https://news.ycombinator.com/item?id=40981054 https://news.ycombinator.com/item?id=40981054 There exist single photon emitters and single photon detectors. Qualify that there are single photons if there are single photon emitters: Single-photon source: https://en.wikipedia.org/wiki/Single-photon_source https://en.wikipedia.org/wiki/Single-photon_source QFT is not yet reconciled with (n-body) [quantum] gravity, which it has 100% error in oredicting. random chance. TOD IIRC, QFT cannot explain why superfluid helium walks up the sides of a container against gravity, given the mass of each particle/wave of the superfluid and of the beaker and the earth, sun, and moon; though we say that gravity at any given point is the net sum of directional vectors acting upon said given point, or actually gravitational waves with phase and amplitude. You said "gauge theory", "Topological gauge theory of vortices in type-III superconductors" https://news.ycombinator.com/item?id=41803662 https://news.ycombinator.com/item?id=41803662 From https://news.ycombinator.com/context?id=43081303 https://news.ycombinator.com/context?id=43081303 .. https://news.ycombinator.com/item?id=43310933 https://news.ycombinator.com/item?id=43310933 : > Probably not gauge symmetry there, then.

DiogenesKynikos 2y ago

Particles are an approximation to the actual behavior of the field, and are used in perturbation theory to calculate the more complicated field behavior. This works well when interactions are weak. Electrons do not couple strongly to the electromagnetic field, so it makes sense to view electrons as particles. However, quarks couple very strongly to the strong force (hence the name), so the perturbative approach breaks down, and it makes less sense to view quarks as particles.

drpossum 2y ago
So in a non-perturbative QFT calculation which has a well defined particle-number operator, that's just "an approximation" within the theory? What is it approximating?
- GoblinSlayer 2y ago
  Energy capacity?
  - deleted 2y ago
    [deleted]
  - drpossum 2y ago
    I'll bite: Explain yourself. Also, for context, my question was posed because the idea of "particle number" as well as "quantum states of particles (which are countable) represented in a Fock space" and in general the idea of particles are, like, page 2 of any QFT textbook. It doesn't approximate anything in the theory. Creation and annihilation of particles (and hence the well-defined concept of a particle) is fundamental to the construction of the theory itself, perturbative or not.
    - DiogenesKynikos 2y ago
      Particles are page 2 of any QFT textbook because the free particle is the only system we can exactly solve. In practice, that solution is usually used as the basis for a perturbative expansion.

gus_massa 2y ago

Perhaps a better way to say it is that particles are not longer small balls of dirt [1], but a mathematical construction that is useful to generate an infinite serie [2] to calculate the results. Since in some conditions these mathematical tricks behave very similar to small balls of dirt, we reused the word "particle" and even the names we used when we thought they were small balls of dirt. [11] We probably never thought they were made of dirt, and in any case the magnetic moment is the double of the value of the small ball of dirt model. [2] That has so many infinites that would make a mathematician cry.

cjfd 2y ago
Note that particles are not just for perturbation theory. There is a particle whenever there exists a particle annihilation/creation field configuration. A proton is a particle so writing down its creation/annihilation field configuration is in theory possible, though maybe not in practice. Another point is that infinities do not necessarily make mathematicians cry. Abraham Robinson is quite pleased with them. It seems a possible hypothesis that at least some QFT are mathematically well-defined using non-standard analysis. Where 'some QFT' at least renormalizable and perhaps also asymptotically free. I don't know enough about it to know how the Haag theorem, mentioned in another comment impacts this.
- nine_k 2y ago
  Another analogy (flawed as any of them). Sports teams "exist" in a sense. They meet one another in well-defined interactions, called matches, and such an interaction can be described as if teams were well-defined atomic entities, producing a score. But a sports team is not atomic, not a "final reality" entity. A sports team can pass through one gate, or through several gates, when entering a stadium. From a doctor's perspective, the team "does not exist", a doctor only operates in terms of individual players' organisms.