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- 1 bose einstein condensate
- 2 Pauli exclusion principle
- 3 Definition of Fermions as particles requiring antisymmetric states
- 4 Fermions with odd integer spin
- 5 Bosonic field and Fermionic field articles
- 6 Elementary particles??
- 7 Please Explain
- 8 Confusing paragraph
- 9 New image
- 10 Experimental fact
- 11 Right-handed fermions
- 12 Elementary fermions
- 13 Nothing in Wikipedia on Weyl fermions?
- 14 All fermions obey Fermi Dirac statistics - true or false? (definition question)
bose einstein condensate
So a grain of sand prevents another grain of sand from occupying the same location.
Unless it's cold enough to form a bose einstein condensate!
BECs are formed out of bosons! Hence "bose." However, fermions can form condensates -- but they are condensates in momentum space only, not physical space. (i.e., the condensate has lots of fermions with the same momentum, but they are not all in the same place), so the Pauli ex. principle is not violated.
- Can you explain this or point to a reference? As far as I'm aware, there's no difference in this respect between position space and momentum space -- antisymmetry in both cases requires the wavefunction to vanish at equal position/equal momentum. Fpahl 06:15, 8 Oct 2004 (UTC)
Pauli exclusion principle
- The result of that discussion was that the textbook meant not that it doesn't always work, but that in some circumstances it's a safe approximation to ignore it in calculations. Tim agreed, so I guess the comment above is just a leftover. Fpahl 06:15, 8 Oct 2004 (UTC)
PEP works all the time for all the massive fermions in a bound state (I haven't heard about massless half-integer spin particles, as neutrinos are considered in Standard Model of Elementary Particle, to form a bound state). serbanut —Preceding signed but undated comment was added at 07:39, 15 September 2007 (UTC)
Definition of Fermions as particles requiring antisymmetric states
In the first line it is stated, and in the definition it is repeated, that fermions obey Fermi-Dirac statistics by definition. What is meant here is perhaps Fermi-Dirac statistics which to my knowledge codes for anticommutation relations for second quantized operators, thus antisymmetric wavefunctions. This is something different from the momentum distribution referred to by Fermi-Dirac statistics / distribution of many free fermions. I propose to change fermi-dirac into fermi in the two indicated places because it spreads confusion to those not familiar with the subject.
The subtlety is purely semantic, because clearly interacting electrons aren't in general described by fermi-dirac statistics. For instance, the fermi-liquid isn't merely described by a stepfunction in momentum space at zero temperature. —Preceding unsigned comment added by Danielabel (talk • contribs) 14:39, 16 September 2009 (UTC)
I'm wondering whether the article's definition of fermions as particles requiring antisymmetric states is the best approach. According to the spin-statistics theorem, this is equivalent with having half-integer spin. But if a particle were found to violate that theorem, say a particle with integer spin and antisymmetric wavefunction (I'm aware that it's not clear how this could be and it would probably require a major change in quantum field theory), would we say "we found a fermion whose spin isn't half-integer", or would we say "we found a boson whose state isn't symmetric"? I think the latter, which would imply that the definition of a fermion is a particle with half-integer spin. This is also how Eric Weisstein's World of Physics defines it. (The same applies, obviously, to bosons; I added a link to this discussion on their talk page.) Fpahl 06:15, 8 Oct 2004 (UTC)
- I disagree with you. For me, a fermion is an anticommuting operator (in "second quantization" formalism) and an expression like "spinless fermion" is not self-contradictory. _R_ 14:54, 8 Oct 2004 (UTC)
- You're right. A survey of my QFT textbooks yielded overwhelming evidence for this view, with two books explicitly stating the spin-statistics theorem as saying that fermions have half-integer spin, not that they anticommute. I could be forgiven, though, since Kaku in "Quantum Field Theory: A Modern Introduction", one of the standard texts, writes: "To demonstrate the spin-statistics theorem, let us quantize bosons with anticommutators and arrive at a contradiction. ... Likewise, one can prove that fermions quantized with commutators violates microcausality." This only makes sense if "fermions" is taken to mean "particles with half-integer spin".
- Do I understand correctly that in the above admittedly unlikely scenario you would be inclined to say "we found a fermion whose spin isn't half-integer"? Fpahl 15:24, 8 Oct 2004 (UTC)
- I would disagree with anticommutation rules - they do not always define fermions. What you say is true if you consider non-interacting systems. If you take bosons on a loop interacting via Dirac-delta potential, then 1) they have ,,Fermion comutation rules (they cannot occupy the same quantum state because of interaction) 2) they have still symmetric wave function.
- A new twist to the story: I emailed mathworld about this since your view suggests that their definitions are wrong, and they sent me this:
The Particle Data Group controls these definitions. Here are the definitions from Particle Data Group: Fermion: Any particle that has odd-half-integer (1/2, 3/2, ...) intrinsic angular momentum (spin), measured in units of h-bar. All particles are either fermions or bosons. Fermions obey a rule called the Pauli Exclusion Principle, which states that no two fermions can exist in the same state at the same time. Many of the properties of ordinary matter arise because of this rule. Electrons, protons, and neutrons are all fermions, as are all the fundamental matter particles, both quarks and leptons. Boson: A particle that has integer intrinsic angular momentum (spin) measured in units of h-bar (spin =0, 1, 2, ...). All particles are either fermions or bosons. The particles associated with all the fundamental interactions (forces) and composite particles with even numbers of fermion constituents (quarks) are bosons.
- Has the Particle Data Group got it wrong?! Fpahl 17:07, 8 Oct 2004 (UTC)
- I would argue that the PDG is primarily concerned with particle phenomenology, and thus intends their definition to discriminate solely between existing particles (for which the distinction between the definitions in question is moot). In QFT, it's certainly the case that when we say "fermion" we mean "fields represented by anticommuting Grassman numbers". If we want to talk about a particle that happens to have spin-1/2 but commutes, we'd call it a "pseudofermion". So one might argue that both characteristics are necessary for a particle to be a fermion, and thus that there are really four classes of particle: fermion, boson, pseudofermion and pseudoboson; only two of which actually exist. -- Xerxes 18:53, 2004 Oct 8 (UTC)
- (Note: I wrote the following before seeing Xerxes' answer) Of course, it's got it wrong! What else do you expect from particle physicists? Seriously though, the flaw in this definition is the following: how do you count the number of fermion constituents in a phonon?
- The more I think of it, the more I believe it depends on whether you focus on individual particles (in which case fermion=half-integer spin) or on their collective behaviour (in which case fermion=anticommuting operator=antisymmetric wavefunction). _R_ 19:05, 8 Oct 2004 (UTC)
Fermions with odd integer spin
We have all noticed that spin is described as being a multiple of hbar/2. I thought that it would be better to set this value to a constant giving,
hdot = hbar/2 = 5.2728584118222738157569629987554e-35 J.s
But now the equations for spin did not work with hdot, so I had to correct them.
Here are the corrected equations,
|sv| = sqrt(s(s + 2)) * hdot
Sz = ms.hdot
sv is the quantized spin vector,
|sv| is the norm of the spin vector,
s is the spin quantum number, which can be any non negative integer,
Sz is the spin z projection,
ms is the secondary spin quantum number, ranging from -s to +s in steps of two integers
For spin 1 particles this gives:
|sv| = sqrt(3).hdot and Sz = -hdot, +hdot
For spin 2 particles this gives:
|sv| = sqrt(8).hdot and Sz = -hdot, 0, +hdot
Now that the spin equations have been corrected, the definitions for fermions and bosons are incorrect, and must be redefined as follows.
Fermions are particles that that have an odd integer spin.
Bosons are particles that have an even integer spin.
Would these redefinitions have any other effects on the Standard Model? Can these redefinitions explain any currently unexplained phenomena? Are there any experiments that could confirm or refute these claims?
I would like eveyone to have a good think about this, and give me your objections to it, or even data to support it.
- The change you're proposing is a trivial renormalization; physicists are perfectly happy with the convention as it stands, tho mathematicians studying lie groups tend to use a normalization with steps of two. -- Xerxes 17:47, 2005 Jun 8 (UTC)
- Actually, the above explanation is no different than the Standard Model except in notation. In spin-statistics theorem, the only difference is the word "net" where fermions have a "net" energy rotation spin and bosons are particles where the "net" energy rotation cancels in the composite matter structures. The notation of the Standard Model was developed through the evolution of theory. Bosons were discovered first so assigned integer values beginning with 1 as is the common mathematical convention. Had fermions been discovered first, then fermions would have been denoted 1 and no doubt, bosons, would have taken even numbers to show that the net spin effect cancels. Scientific and mathematical notation often simply follows the legacy of the evolving discoveries i.e. s,p,d for azimuthal quantum numbers makes no real sense except it is legacy notation.--Voyajer 19:56, 21 December 2005 (UTC)
Bosonic field and Fermionic field articles
I am an ignorant of physics, so I do not understand these two sentences in the article, which look contradictory to me: "[fermions] are sometimes said to be the constituents of matter."
"All observed elementary particles are either fermions or bosons."
The first sentence seems to exclude bosons, and it is stated the same in other articles in Wikipedia ("quark", for example). Will anybody be so kind to explain? Thanks 220.127.116.11 23:42, 31 August 2006 (UTC) Nahuel
There are two groups of particles in elementary particle physics:
1. elementary particles described by non-interacting fields;
2. interaction carriers describing the interacting fields.
The first category contains fermions, and the second contains bosons. For example, electrons, muons, taons, neutrinos, quarks are fermions (spin one half) and they are considered elementary particles, and photons, Z0, W+/-, gluons are interaction carriers and they are bosons. serbanut —Preceding signed but undated comment was added at 07:48, 15 September 2007 (UTC)
This is the dilemma of Wikipedia: that reasonable questions about clarity and accuracy can persist for years without being resolved. This is an encyclopedia and the "non-expert" can be expected to be the "primary customer", not a particle physicist. I find the following ambiguity in 2012: in the introduction there is the statement: " Fermions are usually associated with matter, whereas bosons are generally force carrier particles; although in the current state of particle physics the distinction between the two concepts is unclear." I think that the distinction is much more clear than than the reader is led to believe. Moreover, the deuteron (spin =1), which is a boson, is as much a particle of matter as the proton (spin=1/2), a fermion. So the statement that "bosons are generally force carrier particles can hardly be correct. I suspect that the author meant to say that force carrier particles are generally (always?) bosons. In other words, this appears to me to be lax English (I hope!). In any case, a clarification by a knowledgable expert is over due: you can help save Wikipedia from scientific perdition. Tachyon 13:09, 8 October 2012 (UTC) — Preceding unsigned comment added by Janopus (talk • contribs)
- The statement that "fermions are usually associated with matter, whereas bosons are generally force carrier particles", which, following the discussion presented in this section, appears to have been present in the article since at least 2007 (!), cannot reasonably be said to be theoretically correct in current form. As the recent poster mentions, the deuteron -- as well as any meson, electronic cooper-pair, hydrogen molecule, helium-4 atom, or water molecule, among many, many, 'many' others falling clearly under any intuitive concept of the theoretically ill-defined term "matter" -- is also a bosonic system. There are two possible "fixes" for the issue; the first, which surely better represents the author's original intentions but is in my opinion the worse fix, would be to add the qualifier "elementary" in front of "fermions" and "bosons", making clear that the author is not referring to composite particular systems. However, since the proton and the neutron, which are 'not' elementary particles (as the article correctly indicates), are given in the article as "key building blocks of matter", further confusion may ensue. The better fix, in my opinion, would be to remove the statement entirely (as well as the similar associated statement from the Boson article), as "matter" in any case is not a term that has a robust theoretical definition in the Standard Model. I suppose some combination of the two, perhaps adding both the "elementary" qualifier and some disclaimer emphasising that "matter" is not a theoretically robust term in particle physics, might even be best in an article aimed generally at the layperson. At any rate, given the unfortunate behaviour of many Wikipedia science-section editors, I won't attempt to make any change myself, but since lay visitors have presumably been confused by the unfortunate wording in the article for some five years now, I suspect it would serve the article well to finally address the issue.18.104.22.168 (talk) 06:23, 15 October 2012 (UTC)
The paragraph begining "Of course..." is not helpfull to the uninitiated (I speak as someone visiting the page without a prior background in the subject matter). If there is a more appropriate topic on why "determining the fermionic or bosonic behavior of a composite particle (or system) is only seen at large (compared to size of the system) distance", perhaps there should be a link to it.
The article has the following paragraph:
All observed elementary particles are either fermions or bosons. A composite particle (made up of more fundamental particles) may either be a fermion or a boson, depending only on the number of fermions it contains:
- Composite particles containing an even number of fermions are bosons, such as a meson, or the nucleus of a carbon-12 atom.
- Composite particles containing odd number of fermions are fermions, such as a baryon, or the nucleus of a carbon-13 atom.
I could not understand how an even number of fermions are bosons, while all elementary particles are either fermions or bosons. Also I could not understand how the fermions have only odd number of fermions. It doesn't make sense.
I guess it could be poor choice of words. Otherwise please explain further about how these particles behave. —The preceding unsigned comment was added by Tenri (talk • contribs) 04:10, 10 March 2007 (UTC).
I think the wording is confusing--fermion simply describes any particle with half integer spin so it can describe both elementary as well as composite particles. Primary fermions are quark and leptons. 3 quarks together is a baryon which is a composite fermion (because it has half spin). 2 quarks together makes a meson (having integer spin). I will edit the text to try to clarify. Can someone please double check and see if it makes sense? ThanksTensegrity 23:45, 2 April 2007 (UTC) Ah, thank you for correcting my brain fart re: the atomic weight/nbr of nucleons. :) Tensegrity 18:10, 4 April 2007 (UTC)
Please comment of the new table of elementary particles at Wikipedia:Graphic_Lab/Images_to_improve#String Theory. Thanks. Dhatfield (talk) 22:51, 28 June 2008 (UTC)
Don't you think it is necesary for non physicist to know some fundamental experimental facts to explain why it is postulated the existence of fermions? Paranoidhuman (talk) 00:44, 18 August 2008 (UTC)
Chirality (physics) says there are some left-handed fermions (the ones that feel the weak nuclear force) implying there are right-handed ones that don't. Should this article on fermions mention chirality or refer somewhere ? Weak interaction has a table of left-handed fermions (and implies their antiparticles are right-handed and feel the weak interaction). Standard_Model seems to say W and Z behave differently regarding L/R-handedness. Are there any right-handed fermions (as opposed to antifermions) ? - Rod57 (talk) 17:57 26 November 2012 (UTC)
- Yes, there are plenty of right handed fermions - except for the neutrino, which only feels the weak force.-- cheers, Michael C. Price talk 20:43, 26 November 2012 (UTC)
When the text says a Dirac (massive) fermion can be treated as a "combination" of two Weyl (massless) fermions, the word combination must mean something other than a composite particle, which if it were made of two fermions would be a boson. I suspect it means superposition. If so it would be good say superoposition and for some knowledgeable person to add, or point to, a simple explanation of how a superposition of two massless particles can have mass.CharlesHBennett (talk) 13:54, 15 May 2013 (UTC)
Nothing in Wikipedia on Weyl fermions?
Wikipedia has articles on Dirac fermions and Majorana fermions but nothing at all on Weyl fermions beyond the parenthetical "massless" in this article and a mysterious link to Spinors whose only mention of any kind of fermion is in the phrase "electrons and other fermions". The article on Fermionic fields refers to Weyl spinors without however saying what they are. Moreover Wikipedia lists 38 topics named after Hermann Weyl with no mention of Weyl fermions. Yet a web search for "Weyl fermion" turns up 9,740 results. Why this omission? --Vaughan Pratt (talk) 17:53, 9 October 2013 (UTC)
All fermions obey Fermi Dirac statistics - true or false? (definition question)
At the moment the first sentence of the article says "In particle physics, a fermion [...] is any particle characterized by Fermi–Dirac statistics and [...]". I'd like to ask a question about definitions here, as I'm fairly sure that this sentence is wrong and misleading.
As far as I understand, when people say "Fermi Dirac statistics" they have in mind a result from statistical thermodynamics of non-interacting fermions in thermal equilibrium. This very specific and idealized result is associated with the Fermi occupation number . If this is the intended meaning of Fermi-Dirac statistics, then I'd like to point out that typically, fermions do not obey Fermi Dirac statistics: fermions that we can get into equilibrium generally have significant interactions (e.g. electrons); fermions with weak interactions are usually not in equilibrium (e.g. neutrinos). Fermi Dirac statistics are often a good approximation for electrons in metals / semiconductors, but by no means are they exact and violations are often found.
On the other hand, if "Fermi Dirac statistics" is intended to refer to the general quantum mechanical principles of Pauli exclusion / exchange antisymmetry / etc., then indeed all fermions follow Fermi Dirac statistics. However, if that is so, then the article Fermi–Dirac statistics needs to be massively reworked for this more general meaning.