Post by: Eternal Student on 18/12/2023 01:40:59
What is the current definition (or defining property) of a Fermion?
I thought Fermions could be defined as particles with half integer spins. I mean it turns out they also follow Fermi-Dirac statistics rather than Bose-Einstein but you did not have to define them as particles which follow the Fermi-Dirac statistics OR by some definition such as particles where no two Fermions could not have identical quantum numbers.
Just to be clear I though the following were all equivalent:
Particle has half integer spin <=> particle is a Fermion <=> no two of these particles can have the same Quantum numbers.
However, there seems to be some shift toward defining a Fermion purely as a particle which follows Fermi - Dirac statistics and/or the Pauli exclusion principle and a removal of the half-integer spin characteristic. Take this definition from Wikipedia:
In particle physics, a fermion is a particle that follows Fermi?Dirac statistics. Generally, it has a half-odd-integer spin: spin 1/2, spin 3/2, etc. [ From https://en.wikipedia.org/wiki/Fermion ]
What is the word "Generally" doing in there? Is there a Fermion with integer spin and similarly do we now know of a Boson with half-integer spin?
Best Wishes.
Post by: alancalverd on 18/12/2023 08:04:57
Post by: paul cotter on 18/12/2023 09:12:11
Post by: Zer0 on 18/12/2023 17:43:18
Post by: evan_au on 18/12/2023 21:18:05
Quote from: Zer0
Can Fermions have 0 spin?No; that would be a Boson.
The Higgs boson is the only elementary particle with spin zero (in the Standard Model).
Quote from: OP
Generally, it has a half-odd-integer spinElectrons have half-integer spin, and are fermions.
But at very low temperatures, in the right materials, electrons can couple to phonons, forming Cooper pairs, where 2 electrons effectively become a loosely-coupled spin 0 compound particle, producing the phenomenon of superconductivity.
In this state, electrons no longer follow the Pauli exclusion principle, and can all take on the same low energy level.
https://en.wikipedia.org/wiki/Cooper_pair
It seems, rather than defining the electron as a Fermion, they are saying that any particle that behaves like a Fermion is a Fermion (and if they stop behaving like Fermions, they are no longer Fermions).
- Of course, quantum field theory doesn't talk about particles at all, but of fields which evolve over time, following rules which produce behaviour equivalent to the Pauli Exclusion Principle.
You can construct spin 0 particles out of half-integer spins, but you can't construct half-integer spins out of integer spins.
However, the Supersymmetry hypothesis posits that for every Boson, there is a heavier Supersymmetric counterpart which has half-integer spin (and vice-versa). Physicists were hoping that the LHC would discover them (solving the Dark Matter mystery in the process) - but if they exist, they may have mass/energy too high to be produced in the LHC in measurable numbers.
https://en.wikipedia.org/wiki/Supersymmetry
Post by: Eternal Student on 20/12/2023 16:09:43
....electrons can ... become a loosely-coupled spin 0 compound particle.... It seems, rather than defining the electron as a Fermion, they are saying that any particle that behaves like a Fermion is a Fermion (and if they stop behaving like Fermions, they are no longer Fermions).I quite like that. Thank you for digging up the information.
Best Wishes.
Post by: varsigma on 23/12/2023 17:40:27
An electron in a vacuum isn't exactly the same thing as an electron in a copper wire or a transistor. Electrons are quasiparticles when they propagate in a physical medium.
Post by: evan_au on 02/01/2024 20:58:41
Quote from: Walter Reuther
If it looks like a duck, walks like a duck and quacks like a duck, then it just may be a duck
Post by: varsigma on 08/01/2024 22:57:23
If it looks like a duck, walks like a duck and quacks like a duck, then it just may be a duckIf it looks like a metallic conductor , it's probably a Fermi surface.
Post by: evan_au on 09/01/2024 10:01:24
Quote from: varsigma
If it looks like a metallic conductor , it's probably a Fermi surfaceIf we take a metallic superconductor (like Al, Ga, Hg, Zn, etc), the metallic lustre at room temperature is provided by electrons in the Fermi surface.
Once it becomes superconducting (at around 0.8-4K), I presume it retains its metallic lustre(?)
- Is this because most electrons are still at the Fermi surface, while a small fraction form Cooper pairs, and are no longer at the Fermi surface?
See: https://en.wikipedia.org/wiki/List_of_superconductors
Post by: varsigma on 09/01/2024 19:45:21
If we take a metallic superconductor (like Al, Ga, Hg, Zn, etc), the metallic lustre at room temperature is provided by electrons in the Fermi surface.I can go with that. Based on knowing that the bulk of electrons in a metal does not contribute to a current in general. The current (superconducting or otherwise), is a small fraction of the 'available' population in the Fermi gas.
Once it becomes superconducting (at around 0.8-4K), I presume it retains its metallic lustre(?)
- Is this because most electrons are still at the Fermi surface, while a small fraction form Cooper pairs, and are no longer at the Fermi surface?
So yeah, I would expect the lustre of a metal to be unaffected.
Post by: Eternal Student on 11/01/2024 02:03:52
If it looks like a metallic conductor , it's probably a Fermi surface.?
Everything, well everything with electrons in it, should have a Fermi Surface.
[You seem new to the forum, @varsigma. Hi and welcome. I've decided to put a "thank you" on one of your earlier posts. But I just can't begin to address @evan_au 's later post without identifying a potential error that may have started here].
...the metallic lustre at room temperature is provided by electrons in the Fermi surface....The shiny-ness of metals is usually explained under the assumption that the electrons at the Fermi surface are free to move (in particular it's about having the ability to change velocity which is equivalent to changing the wave vector number k that appears in their wave function).
Silicon, for example, has a well defined Fermi surface at low temperatures but it is NOT shiny. A Fermi surface does not grant metallic lustre, freedom of movement is important. It just so happens that the Fermi surface (for Silicon) is right at the boundary between the 1st and 2nd Brillouin zones of the reciprocal crystal lattice. What this means is that there is an energy gap or "band gap" for the electrons. There is a clear gap in energy that must be crossed if the electron is to change from a wave vector of k to (k + a bit more). This gap can be crossed if, for example, a high voltage is applied across the crystal or if the Silicon is at a high temperature and there is enough thermal energy available for the electrons to cross that gap. So Silicon is a semi-conductor, it's an insulator except under high voltage and becomes more conductive if you increase the temperature. Anyway, we weren't discussing conduction but just response to an applied E field. The electrons at this Fermi surface are not able to change their motion easily in response to a small applied electric field. They can do very litlle as an e-m wave (like visible light) passes over them.
Once.. [metal]... becomes superconducting (at around 0.8-4K), I presume it retains its metallic lustre(?)The general report from the internet is that a superconducting metal is a better reflector ("a better mirror" if you prefer) of visible light than they would be at room temperature.
The situation is complicated by considering incoming photons of different energies. Exotic types of quasi-particles can be formed inside a superconductor so that an incoming photon is just swallowed up into a component of that quasiparticle. Hence, at some frequencies (especially higher frequencies) the superconductor is not perfectly reflecting. At some frequencies it can actually abosrb more of the radiation than it would have done at room temperature.
Apparently you can get results like this:
Reflectivity.jpg (10.17 kB . 447x252 - viewed 492 times)The y-axis is reflectivity ( where 1 = 100% reflection).
The x-axis is a measure of the frequency of the incident light (in some arbitrary units)
The orange line is the result for the metal in the superconducting state. The blue line is for the metal at room temperature.
When superconductive, the reflectivity is (near as you can tell) 100% for frequencies below a certain freq. However, it's lower than the room temperature reflectivity at some higher frequencies.
[Diagram taken from a Physics Stack Exchange: https://physics.stackexchange.com/questions/214218/should-a-superconductor-act-as-a-perfect-mirror which also has some discussion].
Best Wishes.
Post by: varsigma on 11/01/2024 21:12:32
Everything, well everything with electrons in it, should have a Fermi Surface.Yeah. I was being elliptical. I understand that objects reflect light because of the electrons; in metals this reflection is complicated by the existence of conduction electrons. We didn't look at reflectivity of solids in a course I did a while ago, but I nonetheless based my opinionated reply on knowing that currents in conductors (or in any crystalline lattice) are a small fraction of the overall population.
In fact most metallic conductors have random inclusions or impurity atoms, these slow electrons down because suddenly there's a plateau or Landau level and the electrons can conserve angular momentum. Something like that (I think).
According to Bennett, electrons don't travel in straight lines but prefer to cycle around, although that might be just the concept he was getting across.
Post by: evan_au on 11/01/2024 21:40:23
Quote from: varsigma
electrons don't travel in straight lines but prefer to cycle aroundIn a magnetic field, moving electrons will cycle around the magnetic field lines
- That assumes that the magnetic field penetrates the material (eg Al, Cu)
- Magnetic fields do not penetrate a Type 1 superconductor
I assume for Fe or Co, the magnetic domains inside the material would affect the motion of electrons more than the Earth's magnetic field?
Post by: Eternal Student on 11/01/2024 23:07:10
In a magnetic field, moving electrons will cycle around the magnetic field linesYes, but may not be all that important in a finite lump of conductor (see below).
Magnetic fields do not penetrate a Type 1 superconductorAbsolutely. One of the main reasons why you wouldn't expect e-m radiation to penetrate into a superconductor. There can only be 0 magnetic field inside (and also 0 electric field) and hence no way to sustain an electro-magnetic wave. It makes sense for a superconductor to be 100% reflective (and at sensible frequencies of radiation it is). At higher frequencies something weird is going on and the e-m radiation that gets in can't be just a straightforward e-m wave or photon anymore. The internet articles I looked at suggested various quasi-particles must be considered and there is no simple photon that could be identified inside the superconductor.
I assume for Fe or Co, the magnetic domains inside the material would affect the motion of electrons more than the Earth's magnetic field?I can't answer that.
(i) I would have thought it depends on how magnetised the material is. A permanenet bar magnet has quite a strong Magentic field. Meanwhile, a recently heated and bashed lump of Iron has approx. 0 strength magnetic field over any finite length, at least until you get down to atomic lengths.
But I'm going to guess "no" and that the earths magentic field doesn't significantly cause spiralling either.
(ii) It's going to depend on whether you're sending a current along the lines of the magnetic field or perpendicular to it. So, on a macrosocopic scale we might consider sending a current along the long bit (North pole to South pole) of a permanent bar magnet.
Case 1: You would have thought that if you send the current parallel to the magnetic field then the electrons do not spiral around it. B x V = no force with B // V
Case 2: B orthogonal to V. The Hall effect has been studied and you're probably familiar with it. Unlike a cyclotron where charges may be in free space and can endless spiral around, a bar magnet is just a finite piece of material and it has definite edges where the electrons must stop. The Hall effect suggests that if you send a current across the short stubby width of a bar magnet then charges would accumulate on the sides until an E field is established that exactly counteracts the magnetic bit of the Lorentz force an electron travelling in the main current flow would experience. So, once again there is no force perpendicular to the main motion and no spiralling (at least not in the steady state once this pool of charges has been established).
Best Wishes.
Post by: alancalverd on 12/01/2024 07:48:15
But I'm going to guess "no" and that the earths magentic field doesn't significantly cause spiralling either.
It happens: https://en.wikipedia.org/wiki/Whistler_(radio) (https://en.wikipedia.org/wiki/Whistler_(radio))
Post by: evan_au on 12/01/2024 10:25:22
Quote from: ?
a recently heated and bashed lump of Iron has approx. 0 strength magnetic field over any finite length, at least until you get down to atomic lengths.
Quote from: Wikipedia
In most materials the domains are microscopic in size, around 10−4 - 10−6 mAlthough a cubic micrometer is small by human standards, it is still large on an atomic scale..
The ultimate human-made cycler had electrons or protons steered by a magnetic field in a vacuum:
https://en.wikipedia.org/wiki/Cyclotron
Post by: Eternal Student on 12/01/2024 15:15:54
I'm not sure I follow. The "whistler" looks like something that can happen in a lot of space, the atmosphere around planet earth.
The cyclotron was already mentioned and charged particles can take spiral paths because they are also in a lot of space - and even that ends if the charged particle takes a path with a radius too large and hits the edge of the device, or has z-axis velocity that will make it hit the top or bottom of the cyclotron.
In a small lump of material the Hall effect occurrs.
(https://cdnintech.com/media/chapter/39764/1512345123/media/image5.png)
[Image taken from a website called "IntechOpen" which in turn seems to have come from a book "Finite element Analysis" - I'm not endorsing the site or the book but the diagram was the best I could find]
.....when a magnetic field with a perpendicular component is applied, their (electrons or holes) paths between collisions are curved; thus, moving charges accumulate on one face of the material. This leaves equal and opposite charges exposed on the other face, where there is a scarcity of mobile charges. The result is an asymmetric distribution of charge density across the Hall element.... The separation of charge establishes an electric field that opposes the migration of further charge
[Words taken from https://en.wikipedia.org/wiki/Hall_effect because I know Wikipedia has a more relaxed approach to copyright].
In the steady state, we have no net force acting on the electrons orthogonal to their drift velocity and so no curved path.
Don't get me wrong, the electrons may move in some spiral manner through the conductor - but a sustained or constant magnetic field in the region wouldn't seem to be the cause of that.
In most materials the domains are... (bigger than atomic scales)I didn't know that and will concede that point. Thank you for doing the research to get the info.
Best Wishes.