7

I'm puzzled since several years on this basic aspect of quantum mechanics. Quantum theory is supposed to describe particle-wave symmetry of our world. It also describes our universe in term of bosons and fermions.

Now the experimental facts:

  • It seems rather easy to observe wave particle of bosons (think about photon of course) whereas they particle behaviour is pretty complicated to observe. You need stringent experimental condition to observe the particle behaviour of photons for instance.

  • It seems rather easy to observe particle behaviour of fermions (think about electron of course) whereas they wave behaviour is pretty complicated to observe. You need stringent experimental condition to observe the wave behaviour of electrons for instance.

I was just wondering if some people figure out the wave-particle behaviour in terms of fermion vs. boson classification, or if the above experimental facts are just pure coincidence.

I admit it would require a less stringent dichotomy boson/fermion, perhaps like in supersymmetry theories, that I do not know enough. Any comment are welcome.

FraSchelle
  • 10,493
  • 1
    There is also massless vs. massive. Massive quanta could arguably be considered usually more particle-like and massless more usefully thought of as wave-like. – DarenW Sep 11 '13 at 17:32
  • @DarenW Thanks a lot for this enlighten argument about the mass of the particle. Is there a fundamental reason why mass destroys interference? I know a bit about an argument by Penrose using Newton gravitation law to prove decoherence. Is it a well accepted argument ? Thanks again for your comment. NB: This is the exact same comment I did to John Rennie's answer, too: http://physics.stackexchange.com/a/76972/16689 – FraSchelle Sep 12 '13 at 09:35
  • In the specific case of photon vs electron, another key difference is that photons do not interact with each other (to first approximation), while electrons do. If electrons were bosons, I imagine it would still not be easy to make a macroscopic electron wave (that is, a BEC) because of Coulomb repulsion. – Rococo Jan 26 '15 at 02:33
  • @Rococo Well, a similar-to-BEC of electron is the BCS-condensate, which can be understood (up to some extend) as a BEC of Cooper-pairs. And yes the attractive interaction has to overcome the Coulomb repulsion. Thanks for this comment, you're perfectly right about the intrinsic interaction between fermions (better to talk about exchange interaction for instance) whereas bosons do not necessarily interact with each other. – FraSchelle Jan 27 '15 at 09:09

6 Answers6

9

Our current best experimentally verified theory, quantum field theory, isn't based on matter being particles or waves - all matter consists of excitations in quantum fields. The interactions of the quantum fields may appear particle like or wave like, so the wave-particle duality is a duality in the way the fields interact not a duality in the matter itself. The wave-particle duality is just a consequence of using approximate descriptions like the Schrödinger equation, and if we had discovered QFT before the Schrödinger equation generations of physics students would have been spared the confusion.

So wave-particle duality is not down to the fermion-boson distinction. You're quite correct that it's usually experimentally hard to see wave behaviour with fermions, but this is because it's hard to make coherent waves from any massive particles and all known fermions are massive. It would be just as hard to see wave behaviour with bosons, though of course it is routinely done with composite bosons like atoms or even buckyballs.

As Vibert points out, it's no harder to see particle like behaviour with photons than it is with electrons.

John Rennie
  • 355,118
  • @JohnRennie Thanks a lot for this enlighten argument about the mass of the particle. Is there a fundamental reason why mass destroys interference? I know a bit about an argument by Penrose using Newton gravitation law to prove decoherence. Is it a well accepted argument ? Thanks again for your comment. – FraSchelle Sep 12 '13 at 09:31
  • @JohnRennie By the way, I'm just thinking that I can revert the question as well. Is there a fundamental reason why all the fermions we know are massive ? (Sorry, this is also a comment to the Vibert's answer, see http://physics.stackexchange.com/a/76967/16689 ). Thanks in advance for trying to answer these extra questions :-) – FraSchelle Sep 12 '13 at 09:46
  • 1
    One reason is that there are no symmetries to protect fermion masses in the Standard Model from the Higgs mechanism. Roughly elementary fermion masses are $\lambda v$ where $v$ is the Higgs vacuum expectation value and $\lambda$ is some constant. Unless there is a compelling reason (such as a symmetry) why $\lambda$ should vanish, we expect it to be $O(1)$. – Vibert Sep 12 '13 at 10:13
3

Some remarks about both statements:

"You need stringent experimental condition to observe the particle behaviour of photons for instance."

I feel that you don't need to look very far. Counting photons is quite common in experimental physics. And you likely know about the double slit experiment, where you get those interference patterns - when the photon hits the screen, that's particle behaviour. (Or when a photon hits your eye and you observe it.) Also, many "real" particles are bosons - any atom with an even number of fermions is one. Rubidium-87 might be the most famous compound boson, since it's used to make Bose-Einstein condensates, but these compound bosons behave as any other matter particle.

"You need stringent experimental condition to observe the wave behaviour of electrons for instance."

That statement is more true, and it doesn't help that all fermions that we know are massive. There are of course many historical experiments [double slit experiment for electrons, or http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/davger2.html or scanning electron microscopy for a more modern example of electron diffraction] but you seem to be looking for more practical evidence. What about neutrinos? For all intents and purposes, neutrinos propagate through space as plane waves, oscillating slowly between different flavours.

Abhimanyu Pallavi Sudhir
  • 6,140
Vibert
  • 2,617
  • I'm always confused by what people called common in experimental physics. To my knowledge, photons have been proposed as particles at the beginning of the 20-th century, and the convincing experiments for their existence as particles traced back to the 80's. Is that what you coined you don't need to look very far ? Or maybe you know earlier experiments than I ? Ok, I'm just kidding. Thanks a lot for the mass argument. See also the John Rennie's answer who gave more argument about that http://physics.stackexchange.com/a/76972/16689 – FraSchelle Sep 12 '13 at 09:43
  • 1
    I was actually thinking: these days anyone can go to SRS, Thorlabs or Hamamatsu and buy over-the-counter tabletop equipment to count single photons. That's much more refined than the photoelectric effect or any other early experiment probing the particle nature of light. – Vibert Sep 12 '13 at 10:10
2

The wave behavior of particles like electrons and neutrons (whether fermions or bosons in their spin characterization) can be explained classically too. The electromagnetic interaction force that manifests itself as f=k/r^2, have an asymptotic behavior of f=k r, i.e a space spring force, when the forces involved are large and the displacement r is very small.. ie when we have a particle in a bound state.. this will include freely charged as well as neutral dipole particles. The result of this is that the whole space around a slit, for example, becomes governed by a force field that has a wave characteristics.. This way, a particle is guided by such a field resulting in a wave behavior and interference at the screen. To see this, take three, equally spaced, particles on a line. Fix the outer two particles and move the middle a very small distance. Assuming it is equally attracted by the two outer particles find the net force on it and you find it is kx, where x is the displacement normal to the line joining the three. To do a more elaborate proof, use k/r^2 force, then fire individual charged particles at a wall of charged particles with a gap to represent the slit and you will see the interference when collecting on a screen. Not only this, when I did it, I found that you can get the condition of 'total reflection' of the electrons despite the presence of the gap.. just like waves and also the disappearance of the interference pattern, when you play with the counting a little, to resemble the observer effect phenomenon.

Riad
  • 466
0

Electrons are fermions and photons are bosons. Both photons and electrons are cause and effect because when photon particle falls on semiconductor, free electrons are generated. Sometimes electrons are in higher states are unstable, so they try to come to down state where electron and hole pair recombination occurs resulting generation of photon.

When photon particle travels it constitutes a light wave, when electron particle travels, it constitutes a current wave.

Kyle Kanos
  • 28,229
  • 41
  • 68
  • 131
soumyakanta pradhan
  • 9
  • I'm not sure I understand the link with my question, but your point is correct, all excitations in the quantum world follow the particle/wave duality. Anyway I appreciate you tried to answer this question your first day on SE, so 10 points for you :-) – FraSchelle Jan 27 '15 at 09:06
-3

Bosons are not atoms, They are "force particles" in gauge theory and are massless and do not obey fermion statistics other than in weak field theory where W+ W- and Z take on their masses.

Daniel Park
  • 19
  • 1
    -1 because: a) this is a comment to either Vibert or John Rennie's answer, there is no mention of bosonic atoms in the OP's post b) this is completely wrong, atoms can be and frequently are bosons, an enormous literature has been written on the applications of bosonic atoms (e.g. this paper happens to spring to mind), see also this SE question – Mark Mitchison Sep 11 '13 at 13:20
  • Perhaps 'completely wrong' is a bit strong, given some of the apparently unresolved discussions on the SE question I linked above. Irrelevant to the OP's post it is, however. – Mark Mitchison Sep 11 '13 at 13:29
  • Thanks for the input. It gets a little philosophical because in materials science we are looking for "elementary" particles in their most stable hierarchy and that's third generation in standard model. I'm assuming this literature redirect is about Bose Einstein Condensates which is a fascinating topic for others to post. – Daniel Park Sep 11 '13 at 14:23
-4

Yes there is link.When the massive charged particles fully express themselves in quantum fields they would behave like-bosons but are not exclusively bosons. It is the property of both the field and the intrinsic property of the charged particles. So matter can behave like wave in quantum fields and away from the quantum fields they become matter. Therefore it can be better said matter wave duality but not particle wave duality.

Fr. Justin C
  • 1
  • 2
    This doesn't make much sense. Leptons, for instance, are charged fermions, not bosons. – ACuriousMind Aug 30 '15 at 19:22
  • @Fr.JustinC Thanks a lot for spending time answering this question, but your answer definitely makes no sense to me. In addition, it seems that the other answers above are much more interesting and right-to-the-point than yours, whatever yours tries to say. – FraSchelle Sep 01 '15 at 09:22