What, in essence, distinguishes light from matter?

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Main Question or Discussion Point

Before quantum mechanics, light was generally seen as a wave and matter as particles (biliards). From e.g. the discovery of the photoelectric effect, one saw that light can also be seen as a particle. From e.g. the double slit experiment, one makes the interpretation that matter can also be seen as a wave.

One of the early conclusions of quantum mechanics is then that waves and particles are the same thing. Nevertheless, we would refer to the 'classical limit' of light to be a wave (i.e. a coherent state) and the 'classical limit' of matter to be particles (i.e. a fock state). Where does this distinction actually show up?
I would think that having a finite mass or not has at least something to do with it, but I'm not sure where the distinction enters precisely.

A standard explanation from introductionary quantum mechanics is that the de broglie wavelength scales inversely to mass. But to me this seems quite ad hoc given our current knowledge of quantum mechanics.

The transitions from quantum mechanics to the classical world occur trough decoherence. So why does decoherence act differently on light than on say electrons? It means that both have different pointer states?
 

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  • #2
atyy
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The classical limit of light is a wave because quantum light is a boson. There are not (simple) classical counterparts of electrons because they are fermions. One can see this in the Feynman path integral, in which bosons can be described by classical trajectories or classical field configurations, whereas fermions require Grassmann numbers.

However, there are interesting suggestions about how fermions could arise from classical statistics: https://arxiv.org/abs/1006.4254v2.

While the mass does not directly affect the ability of a boson to have a classical counterpart, the masslessness of the photon and graviton means that they are longer ranged forces, which is why light and gravity are the two quantum fields that are also known classically.
 
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Thanks for your answer.
In some sense, it seems indeed logical to me that this is about bosons vs fermions. But in another sense, not really.

Bosonic and fermionic atoms -except at very low temperatures- behave quite similar. Likewise, I wouldn't be able to tell apart a bosonic from a fermionic car. So I was always thinking that the distinction between bosons and fermions was entirely quantum mechanical, and they become the same as ##\hbar\rightarrow 0##.
 
  • #4
vanhees71
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Before quantum mechanics, light was generally seen as a wave and matter as particles (biliards). From e.g. the discovery of the photoelectric effect, one saw that light can also be seen as a particle. From e.g. the double slit experiment, one makes the interpretation that matter can also be seen as a wave.
There's no other way to understand quantum theory than to learn the mathematics needed to formulate it. It is very hard if not impossible to explain quantum theory without the only adequate language to express it, which is the math.

The idea that light is something like a flow of billiard-ball like particles goes back to Newton, the discovery of wave-like features to Young. The final breakthrough as far as classical physics is concerned was Maxwell's theory of the electromagnetic field and Hertz's discovery of the predicted electromagnetic waves, i.e., the finding that light is nothing else than an electromagnetic wave field at certain wave lengths our eyes are sensitive to.

Then, in 1905, came Einstein's paper on the particle nature of light, but one should be very well aware that this is not consistent theory but a "heuristic point of view" as Einstein carefully titled his paper. The modern quantum theory (in its nonrelativistic formulation) has been discovered 20 years later, independently by three groups/people: Heisenberg, Born, and Jordan in terms of "matrix mechanics", Schrödinger in terms of "wave mechanics", and by Dirac in terms of "transformation theory". All these were different formulations of the same theory, and it was soon very clear that the photoelectric effect on the level modeled by Einstein does not prove any particle-like properties of light. It is sufficient to treat the electrons with quantum theory but letting the electromagnetic field still be classical (semi-classical approximation). This is also not the case for the leading-order approximation of the Compton effect (which is also errorneously claimed in some textbooks, although Klein and Nishina derived it in the semiclassical approximation, where only the electrons were quantized but not the electromagnetic field).

Almost at the same time as non-relativistic QT was discovered also relativistic QFT of the electromagnetic field was formulated by Born (already in one of the first papers by Born, Heisenberg, and Jordan). As it turned out somewhat later, relativistic QFT is the adequate formulation for relativistic QT of particles either, because in collisions at relativistic energies particles can be destroyed and created (as long as all conservation laws are fulfilled), and QFT deals by construction with such situations, where the particle number is not conserved.

Applied to the electromagnetic field, relativistic QFT tells us that photons cannot in any way be interpreted as billiard-ball like particles. It's not even possible to define a position for such a would-be particle. The reason is that the electromagnetic field is a "massless field", and the photons, i.e., certain kinds of excitations of the em. field (socalled one-quantum Fock states) are thus massless. The only thing they have thus in common with what would be classical massless particles is the energy-momentum relation, ##E=|\vec{p}| c##.

Also the "wave description" of particles, a la Schrödinger, which is possible in the non-relativistic limit is not in any way a classical field theory since the meaning of the wave function is not to describe some continuous directly observable quantity but, properly normalized, the square of the wave function is the probability density to find the particle at a given place.
 
  • #5
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There's no other way to understand quantum theory than to learn the mathematics needed to formulate it. It is very hard if not impossible to explain quantum theory without the only adequate language to express it, which is the math.

The idea that light is something like a flow of billiard-ball like particles goes back to Newton, the discovery of wave-like features to Young. The final breakthrough as far as classical physics is concerned was Maxwell's theory of the electromagnetic field and Hertz's discovery of the predicted electromagnetic waves, i.e., the finding that light is nothing else than an electromagnetic wave field at certain wave lengths our eyes are sensitive to.

Then, in 1905, came Einstein's paper on the particle nature of light, but one should be very well aware that this is not consistent theory but a "heuristic point of view" as Einstein carefully titled his paper. The modern quantum theory (in its nonrelativistic formulation) has been discovered 20 years later, independently by three groups/people: Heisenberg, Born, and Jordan in terms of "matrix mechanics", Schrödinger in terms of "wave mechanics", and by Dirac in terms of "transformation theory". All these were different formulations of the same theory, and it was soon very clear that the photoelectric effect on the level modeled by Einstein does not prove any particle-like properties of light. It is sufficient to treat the electrons with quantum theory but letting the electromagnetic field still be classical (semi-classical approximation). This is also not the case for the leading-order approximation of the Compton effect (which is also errorneously claimed in some textbooks, although Klein and Nishina derived it in the semiclassical approximation, where only the electrons were quantized but not the electromagnetic field).

Almost at the same time as non-relativistic QT was discovered also relativistic QFT of the electromagnetic field was formulated by Born (already in one of the first papers by Born, Heisenberg, and Jordan). As it turned out somewhat later, relativistic QFT is the adequate formulation for relativistic QT of particles either, because in collisions at relativistic energies particles can be destroyed and created (as long as all conservation laws are fulfilled), and QFT deals by construction with such situations, where the particle number is not conserved.
Thanks for your answer, but I knew most of these things. I do have a working knowledge of QM (not all of it of course, as I am a mortal) now and simplified my introduction in the question for the sake of brevity, just started rethinking the very basics because of further knowledge. Someone changed the question label to 'B', though I personally don't think this is trivial matter.

Applied to the electromagnetic field, relativistic QFT tells us that photons cannot in any way be interpreted as billiard-ball like particles. It's not even possible to define a position for such a would-be particle. The reason is that the electromagnetic field is a "massless field", and the photons, i.e., certain kinds of excitations of the em. field (socalled one-quantum Fock states) are thus massless. The only thing they have thus in common with what would be classical massless particles is the energy-momentum relation, ##E=|\vec{p}| c##.
Yes, eigenstates of the photon particle number operator are massless just like all optical states (except for an effective mass for the transverse motion in cavities). Does this fact restore a symmetry between density and phase (more or less the conjugate quantity) or so, so that the pointer states become coherent states (eigenstates of the annihilation operator)?

Also the "wave description" of particles, a la Schrödinger, which is possible in the non-relativistic limit is not in any way a classical field theory since the meaning of the wave function is not to describe some continuous directly observable quantity but, properly normalized, the square of the wave function is the probability density to find the particle at a given place.
If the wavefunction is a plane wave, the 'continuous directly observable quantity' seems the momentum of the particle to me. But aside from that, what intrigues me a bit is that this 1st quantisation picture you are refering to here implicitly assumes the amount of particles is fixed. Why is this assumption often valid anyway (it seems to me that this validity in the 'qm but not too qm' regime should be justified iff fock states with fixed particle number and complete phase uncertainty are the pointer states)? Is this the case because the particles are massive and hence their amount is "continuously measured" by all other gravitational objects in the universe so that particle number superpositions decohere quickly ?
 
  • #6
PeterDonis
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Bosonic and fermionic atoms -except at very low temperatures- behave quite similar.
That's because they are both bound states of fermions. The comparison you should be making is, for example, between atoms and light--just as in your OP of this thread. Those are very different things (@atyy gave some good reasons why in post #2 of this thread), and the comparison between them is a much better illustration of the difference between fermions and bosons.

I wouldn't be able to tell apart a bosonic from a fermionic car.
No, you wouldn't be able to make a bosonic car; for example, you can't make a car out of light.

I was always thinking that the distinction between bosons and fermions was entirely quantum mechanical, and they become the same as ##\hbar\rightarrow 0##.
Then you thought wrong. The stability of ordinary macroscopic objects, like tables and chairs and human bodies, depends crucially on the fact that the atoms out of which they are made are bound states of fermions.
 
  • #7
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That's because they are both bound states of fermions. The comparison you should be making is, for example, between atoms and light--just as in your OP of this thread. Those are very different things (@atyy gave some good reasons why in post #2 of this thread), and the comparison between them is a much better illustration of the difference between fermions and bosons.

No, you wouldn't be able to make a bosonic car; for example, you can't make a car out of light.

Then you thought wrong. The stability of ordinary macroscopic objects, like tables and chairs and human bodies, depends crucially on the fact that the atoms out of which they are made are bound states of fermions.
Thanks for your answer, but I do not agree with your premise that atoms or cars cannot be bosonic. If the total spin of an object is integer, it is a boson by definition. And then it is a matter of level of description, without any further knowledge about the constitution of bosonic atoms one perfectly describes phenomena as bose-einstein condensation. If fermionic atoms or electrons pair up, they also become composite bosons in e.g. superconductivity.

What happens at the level of atoms or cars, as opposed to photons, is that the bosonic objects feel a hard-core interaction (which can microscopically be attributed to the fermionic constituents indeed).

So probably, I should have been a bit more precise in my wording. Non-interacting fermions and non-interacting bosons don't become equivalent in the classical limit, but non-interacting fermions must become equivalent with hard-core bosons (since my car will not change behaviour if it loses a single electron).
 
  • #8
PeterDonis
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I do not agree with your premise that atoms or cars cannot be bosonic. If the total spin of an object is integer, it is a boson by definition.
This is not correct. Atoms with integral total spin do share some properties with fundamental bosons, but not all of them. And one of the most important properties they do not share is that an atom with integral spin still takes up much more space than it would if it were a fundamental boson.

For example, Helium-4, which is an integral spin atom, can form a superfluid at a low enough temperature; this is a Bose-Einstein condensate and indicates that Helium-4 shares this particular property with fundamental bosons. However, the density of Helium-4 does not drastically decrease when it becomes a superfluid. That is because Helium-4 atoms, even when superfluid, are still bound states of fermions, and those fermions obey the Pauli exclusion principle, which forces individual Helium-4 atoms to take up much more space than they would if they were really fundamental bosons.

So probably, I should have been a bit more precise in my wording.
Yes, by not using the term "boson" to refer to things that don't have all of the properties of fundamental bosons. Note that I also did not say that Helium-4 atoms were fermions. I said they were bound states of fundamental fermions. Composite objects like these really aren't "fermions" or "bosons" since they don't share all of the properties of either. They're composite objects.
 
  • #9
A. Neumaier
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Before quantum mechanics, light was generally seen as a wave and matter as particles
No. The particle view and the wave view both had their advocates, and for geometric optics, the particle view is actually the more useful one - used to make lenses and cameras.
waves and particles are the same thing.
No. They are extremal limiting views of the same underlying stuff described in more abstract terms.
we would refer to the 'classical limit' of light to be a wave (i.e. a coherent state) and the 'classical limit' of matter to be particles (i.e. a fock state).
No. Coherent states and Fock states are quantum concepts. The classical limit of multiparticle quantum mechanics can be done in different ways and either gives a classical multiparticle theory or a classical field theory, or a mixture of both.
The classical limit of light is a wave because quantum light is a boson.
Not really. A hydrogen atom is also a boson, but in an effective quantum description, seen from far (so that it behaves like a point), its classical limit is not a wave. It is the long range nature (masslessness) mentioned also by you that makes photons conceived as classical fields. But one can also get geometric optics as a classical limit of photons.
Composite objects like these really aren't "fermions" or "bosons" since they don't share all of the properties of either. They're composite objects.
They are composite bosons or composite fermions. thephystudent is right, the term fermion refers to half-integral spin and nothing else.
not using the term "boson" to refer to things that don't have all of the properties of fundamental bosons.
But ''fundamental boson'' is a concept quite different from that of ''boson''! It may even be a vacuous concept. We don't know which particles are truly elementary and which ones are composites of still unknown particles. At some point in time, atoms were elementary, then electrons, protons, and neutrons. Possibly everything is composite.
 
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PeterDonis
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the term fermion refers to half-integral spin and nothing else.
And by this reasoning, the term "boson" refers to integral spin and nothing else. Which is fine as long as you then resist the temptation to apply all other properties traditionally associated with "bosons" to objects like Helium-4 atoms that have integral spin but don't behave like "bosons" in other ways (such as the amount of space they take up).
 
  • #11
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No. The particle view and the wave view both had their advocates, and for geometric optics, the particle view is actually the more useful one - used to make lenses and cameras.

No. They are extremal limiting views of the same underlying stuff described in more abstract terms.

No. Coherent states and Fock states are quantum concepts. The classical limit of multiparticle quantum mechanics can be done in different ways and either gives a classical multiparticle theory or a classical field theory, or a mixture of both.

Not really. A hydrogen atom is also a boson, but in an effective quantum description, seen from far (so that it behaves like a point), its classical limit is not a wave. It is the long range nature (masslessness) mentioned also by you that makes photons conceived as classical fields. But one can also get geometric optics as a classical limit of photons.

They are composite bosons or composite fermions. thephystudent is right, the term fermion refers to half-integral spin and nothing else.

But ''fundamental boson'' is a concept quite different from that of ''boson''! It may even be a vacuous concept. We don't know which particles are truly elementary and which ones are composites of still unknown particles. At some point in time, atoms were elementary, then electrons, protons, and neutrons. Possibly everything is composite.
Thank you very much for your clarification, A. Neumaier. Can you recommend a good reference for the part in bold?
 
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A. Neumaier
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Thank you very much for your clarification, A. Neumaier. Can you recommend a good reference for the part in bold?
In the first quantized form, the classical limit gives a classical multiparticle theory.
In the second quantized form, the classical limit gives a classical field theory.
If you second-quantize only some particles, you get a mix of both.
I don't have a convenient reference, but this is not difficult to see once you think about it.
 
  • #13
A. Neumaier
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all other properties traditionally associated with "bosons"
What are these properties? There are Bose-Einstein condensates and Cooper pairs, both of which show that composite bosons behave very similar to elementary bosons, just because of their integral spin. Whereas densities are not directly related to the statistics but to compositeness. A hydrogen atom made of spin zero protons and electrons is sizeable although composed of bosons only.
 
  • #14
PeterDonis
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densities are not directly related to the statistics
I'm thinking of things like the classic paper by Dyson & Lenard on the stability of matter, which shows, roughly, that for a macroscopic piece of matter to take up an amount of volume proportional to the number of particles, the particles must be fermions.

A hydrogen atom made of spin zero protons and electrons is sizeable although composed of bosons only.
But such an "atom" will have a very different response than an actual hydrogen atom, composed of a spin-1/2 proton and a spin-1/ electron, when a very large number of them are put together into a large object. At least, that's how I'm reading results like those of the Dyson & Lenard paper. A macroscopic piece of matter composed of bosons will have a volume that is not proportional to the number of particles, but is much smaller.

Another way of looking at it is that a gas of non-relativistic fermions has an adiabatic index of 5/3, but a gas of non-relativistic bosons has a much smaller adiabatic index, so it resists compression much less strongly.

But, for example, Helium-4 has an adiabatic index of 5/3 (in the non-relativistic regime), yet it has integral spin. So obviously it does not have all of the properties of "bosons", if those properties include the adiabatic index; its adiabatic index is what we would expect for a gas of fermions, not bosons.
 
  • #15
A. Neumaier
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So obviously it does not have all of the properties of "bosons", if those properties include the adiabatic index; its adiabatic index is what we would expect for a gas of fermions, not bosons.
This is a property not of the single boson but a property of a macroscopic collection of bosons, interacting in a specific way. If you modify the Coulomb interaction to some arbitrary law, stability of matter is no longer guaranteed for nuclei plus fermionic electrons. And for noninteracting particles, stability is not dependent on statistics, I believe.
 
  • #16
PeterDonis
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If you modify the Coulomb interaction to some arbitrary law, stability of matter is no longer guaranteed for nuclei plus fermionic electrons.
The stability of individual atoms, yes, because the relevant interaction for individual atoms, considered by themselves, is Coulomb. But the atoms in a gas of Helium-4 with an adiabatic index of 5/3 don't have a Coulomb interaction between them; to a very good approximation they don't interact at all (the goodness of that approximation is shown by the very low boiling point).

for noninteracting particles, stability is not dependent on statistics, I believe.
AFAIK the reasoning in the Dyson & Lenard paper about the stability of bulk matter (i.e., matter consisting of a large number of particles) does not require interacting particles, it just depends on statistics.
 
  • #17
A. Neumaier
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AFAIK the reasoning in the Dyson & Lenard paper about the stability of bulk matter (i.e., matter consisting of a large number of particles) does not require interacting particles, it just depends on statistics.
You don't remember correctly. The paper is
Dyson, F. J., & Lenard, A. (1967). Stability of matter. I. Journal of Mathematical Physics, 8(3), 423-434.
and assumes from the outset the Coulomb interaction between a large collection of charged particles.
But the atoms in a gas of Helium-4 with an adiabatic index of 5/3 don't have a Coulomb interaction between them; to a very good approximation they don't interact at all (the goodness of that approximation is shown by the very low boiling point).
All neutral atoms attract each other with a residual van der Waals force of order ##O(r^{-6})## that results from the exchange energy of fermions. This is the dominant interaction, which is relevant even macroscopically for the equation of state of helium. As a consequence, helium forms weakly bound states (helium clusters).
 
  • #18
PeterDonis
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The paper is
Dyson, F. J., & Lenard, A. (1967). Stability of matter. I. Journal of Mathematical Physics, 8(3), 423-434.
and assumes from the outset the Coulomb interaction between a large collection of charged particles.
I see. This assumption seems a bit strange when talking about something like a tank of Helium-4 gas, since, as you note:

All neutral atoms attract each other with a residual van der Waals force of order ##O(r^{-6}##) that results from the exchange energy of fermions.
Yes, and this is certainly not a Coulomb interaction, nor can it be an exchange energy between Helium-4 atoms, since those are bosons (using your terminology). So what fermions does this exchange energy refer to?
 
  • #19
vanhees71
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In the first quantized form, the classical limit gives a classical multiparticle theory.
In the second quantized form, the classical limit gives a classical field theory.
If you second-quantize only some particles, you get a mix of both.
I don't have a convenient reference, but this is not difficult to see once you think about it.
Hm, I find it a bit difficult to see, but maybe that's because I've another idea about what's obvious than you have ;-)).

For me "first quantization" is quantum mechanics as comes out by "canonical quantization" of a classical single or multi-particle system with a fixed number of particles. To see what "classical limit" may mean, let's restrict ourselves to a single particle (maybe in an external potential). Then one way to get the non-relativistic limit is the Sommerfeld-Wenzel-Kramers-Brillouin method (usually it's called WKB, but it goes all back to Sommerfeld) applied to Schrödinger wave mechanics, which leads in leading order to the Hamilton-Jacobi partial differential equation which is equivalent to the usual way to describe the particle in phase space since the characteristics of the HJPDE are simply these trajectories. It's, by the way how Schrödinger found his wave equation, i.e., by asking about which wave equation what lead in the eikonal approximation (which is nothing else than the leading-order SWKB approximation) to the HJPDE. This is in analogy how you get from the full wave optics (i.e., Maxwell's equations) to geometrical optics, where you can reinterpret in some sense the characteristics of the eikonal (the light rays) as particle trajectories. In this sense you have a classical particle limit out of Schrödinger wave mechanics or even classical electrodynamics.

What's not so clear is, what you mean by "In the second quantized form, the classical limit gives a classical field theory." Of course, if you take QED and investigate coherent states with a large average photon number you get classical electrodynamics for the expectation values of the field operators.

To really get the mathematical details is not so easy to see, at least not for me. E.g., take interacting fermionic QED. I haven't seen any way how to derive classical elecron theory a la Lorentz from it, i.e., describing electrons (and positrons) as particles and the photon field as classical waves. I think it would be great to try to solve the old puzzle of a consistent classical theory of point particles including radiative reactions. From experience I think the state today is that the best approximation is the Landau-Lifshitz approximation. Do you know a paper about this? I'd be very interested!
 
  • #20
A. Neumaier
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Yes, and this is certainly not a Coulomb interaction, nor can it be an exchange energy between Helium-4 atoms, since those are bosons (using your terminology). So what fermions does this exchange energy refer to?
My terminology is everyone's except yours. https://en.wikipedia.org/wiki/Boson: ''Some composite particles are also bosons, such as mesons and stable nuclei of even mass number such as deuterium (with one proton and one neutron, mass number = 2), helium-4, or lead-208; as well as some quasiparticles (e.g. Cooper pairs, plasmons, and phonons).''

The exchange energy is between the electrons of the helium gas.

The Helium 4 atoms wouldn't be covered by their analysis if they were treated as weakly interacting effective bosonic particles, because (i) as effective particles they are bosons, hence would be unstable, but (ii) the argument doesn't apply since the force between them is not Coulomb and since they are uncharged.

But when a helium gas is treated as a collection of N alpha particles and 4N electrons interaction via Coulomb forces, their analysis applies and proves stability - and indeed, this is the case.
 
  • #21
PeterDonis
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The exchange energy is between the electrons of the helium gas.
Ok, that's what I figured.
 
  • #22
PeterDonis
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The Helium 4 atoms wouldn't be covered by their analysis if they were treated as weakly interacting effective bosonic particles, because (i) as effective particles they are bosons, hence would be unstable, but (ii) the argument doesn't apply since the force between them is not Coulomb and since they are uncharged.

But when a helium gas is treated as a collection of N alpha particles and 4N electrons interaction via Coulomb forces, their analysis applies and proves stability - and indeed, this is the case.
Ok, this makes sense.

To get back to the OP question of this thread, though: this would seem to indicate that the key difference between light and matter is that the Dyson/Lenard analysis applies to matter, whereas it doesn't to light. The reason for that is not just that matter contains fermions (even though many of its composite entities are bosons), but that it contains charged fermions; a macroscopic body made of uncharged fermions would not satisfy the conditions of the Dyson/Lenard analysis, correct?
 
  • #23
A. Neumaier
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What's not so clear is, what you mean by "In the second quantized form, the classical limit gives a classical field theory."
Since QED has not been rigorously constructed, there are only nonrigorous arguments. the simplest is to take the path integral and there the limit ##\hbar\to 0##. This produces a classical variational principle. of course the Fermions are there represented by Grassmann variables, which are not yet classical. But the observables are the even expressions in the fermionic fields, e.g. Fermionic currents, but also nonlocal fields. The latter can be Wigner transformed and then satisfy classical Vlasov-type equations. The end result is the relativistic Vlasov-Maxwell system. For ##c\to\infty## one obtains the nonrelativistic Vlasov-Maxwell system, which can be derived rigorously from nonrelativistic approximations of QED.
E.g., take interacting fermionic QED. I haven't seen any way how to derive classical electron theory a la Lorentz from it, i.e., describing electrons (and positrons) as particles and the photon field as classical waves. [...] Do you know a paper about this? I'd be very interested!
This is the mixed case, and I haven't seen it discussed in the literature. In principle it should work in an analogous way, at least in the nonrelativistic setting. Second-quantize only the electromagnetic field, then take the classical limit in the resulting path integral formulation, and Wigner-transform the Grassmann part. But of course, the limit is formal, and there is no guarantee that the resulting dynamical system is free from the paradoxes of classical electrodynamics.
 
  • #24
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a macroscopic body made of uncharged fermions would not satisfy the conditions of the Dyson/Lenard analysis, correct?
It wouldn't. Uncharged + Coulonb interaction implies no interaction, and huge noninteracting collections of quantum particles are ideal quantum gases and hence stable, irrespective of their statistics, which only affects the details of the equation of state.
 
  • #25
PeterDonis
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huge noninteracting collections of quantum particles are ideal quantum gases and hence stable, irrespective of their statistics
An ideal Bose gas is stable, even at zero temperature? Won't it have zero pressure and therefore collapse?
 

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