text:
Talk:Fermi energy

Fermi temperature
Fermi temperature is a stub right now, and doesn't contain any information that couldn't be easily shoehorned in here, especially since the concept is mentioned in this article. I'm pretty new here, does anyone have thoughts? -- 7segment 01:49, 21 March 2006 (UTC)

 * Good idea, I have just redirected it. Rex the first 12:35, 5 April 2006 (UTC)

 * Hmm, I think this was the wrong move. The discussion of the Fermi temperature in this article is pretty well useless; you could probably derive the only result presented (that Tf=Ef/kB) simply by noticing that Boltzmann's constant was in units of joules per kelvin.  Personally, I think the Fermi temperature should have its own article.  I can contribute something to it...after I finish my physics assignment!  Ckerr 14:34, 26 August 2006 (UTC)

 * Agreed. A separate article would be useful. Zebas (talk) 17:50, 14 December 2009 (UTC)

Dimensions
The article presupposes that people are interested solely in 3-D fermi gases, which is a grave error. LeBofSportif 17:15, 5 June 2006 (UTC)

I am a little confused about the vector term in the 3d model. Based on the since the n value has to be a positive integer, why is it necessary to have an absolute value of the vector? If all components of the vector are positive, how could the vector be negative? Methinks I am missing something. Good article btw. -Hellkyte

You don't seem to understand the concept of a vector very well. A vector is not a number, so there cannot be a number of states less than or equal to a vector; the notion of less than or equal to a vector does not even make sense because there is no canonical ordering on vectors in 3-dimensional Euclidean space. "Absolute value" is also sort of a misnomer; the |s represent the operation of taking the magnitude of the vector, which is sort of like the absolute value of a number (the two coincide on the real numbers), but not quite the same. "[why?]" removed. — Preceding unsigned comment added by 75.161.171.91 (talk) 21:35, 4 April 2015 (UTC)

What?
What on earth is this article going on about? Can someone explain any of it so it makes sense to me? I have no idea about physics or anything like that.

 * Sorry; there's not a whole lot that can be done. The Fermi energy is not an easily graspable concept the way, say, momentum is, and I cannot off the top of my head see anything needlessly complicated in the way the article presents it.  Your best bet is to probably get an idea about physics, then try to read the article. Ckerr 14:37, 26 August 2006 (UTC)

 * though there is nothing needlessly complicated about the introduction it assumes familiarity with many terms and concepts that are not 'required' to understand the concept of fermi energy. will try to remedy. --V. 18:23, 15 February 2007 (UTC)

Very nice and easy to understand article
In my opinion the starting part of this article (the introduction) is clear and very easy to understand. This is in contrast with many (most?) other physics related articles I have read so far. Most of them generally suppose that the reader actually already knows everything about the matters discussed in the particular article. This article doesnt suppose any more than very basic prior knowledge about it's subject. Congratulations to the author(s). It would be very nice if more of the physics related articles were like this one.

--- Hey, I totally agree. Compared to the German graduate book I've read before this really helped me a lot! Especially the derivation of the three-dimensional potential is very well introduced and structured!

About the Nucleus
"Now since the fermi energy only applies to fermions of the same type, one must divide this density in two. This is because the presence of neutrons does not affect the fermi energy of the protons in the nucleus."

But in the following calculation, no difference can be seen. So what does "divide this density into two" actually mean? --Sandycx 08:09, 12 November 2006 (UTC)

Since it looks like the above was dealt with (area in the derived formula = .5 the norm), could someone please explain the justification for dividing by two, seeing as #protons != #neutrons in atoms? I feel like the derivation assumes that, which isn't true for many atoms. Doing a correction of the area for a specific nucleon based on the atomic makeup for an atom isn't terribly difficult, although it would add a couple extra lines. Just seems wrong to make this poor of an assumption to avoid a little extra derivation. Maybe I am missing something though, I am but a lowly chemist. -Hellkyte

Degeneracy isn't always 2, and more explanation of the derivation
In the derivation for 3 dimensions the degeneracy of the fermions has implicitly been assumed to be 2. Whilst this is the case for electrons, it is not true in general, so I think it should be mentioned. Also, I think the line $$N = \frac{1}{8} \times 2 \times \frac{4}{3} \pi n_f^3$$ is poorly explained. My thoughts are: As I'm new to wikipedia I thought I'd try to canvass opinion on whether adding this information would make the derivation too long before adding it. Uberdude85 01:30, 21 November 2006 (UTC)
 * the eighth is because we are in 3D and only want the eighth of the sphere in nx-ny-nz space where they are all positive,
 * the 2 is the degeneracy which we have assumed is 2, which is not always the case,
 * the four thirds pi nf is the volume of a spere of radius nf,
 * and most importantly nf is never defined! It is $$ \sqrt{ \left( n_x^2 + n_y^2 + n_z^2\right) }$$.

Highest filled state
I'd like to take issue with the statement that Fermi energy is the energy of the highest filled state at T=0. If the highest filled state just happens to fill a band, then the fermi energy (the electrochemical potential) is mid-gap. Should this be changed? —Preceding unsigned comment added by 163.1.18.226 (talk) 17:56, 7 February 2010 (UTC)
 * The fermi energy (as the term is used in this article) is not the same as the electrochemical potential. I don't know how to make it any more clear, it's the second sentence of the article. Is it phrased in a confusing way? Did you just not read the second sentence of the article? Any other suggestions? --Steve (talk) 18:23, 7 February 2010 (UTC)
 * Yes. It is commonly misused in, for example, semiconductor physics. Maybe that means that there should be a section to explain the misuse, but otherwise the article should do it right. Gah4 (talk) 20:07, 27 March 2024 (UTC)
 * Yes. It is commonly misused in, for example, semiconductor physics. Maybe that means that there should be a section to explain the misuse, but otherwise the article should do it right. Gah4 (talk) 20:07, 27 March 2024 (UTC)
 * Yes. It is commonly misused in, for example, semiconductor physics. Maybe that means that there should be a section to explain the misuse, but otherwise the article should do it right. Gah4 (talk) 20:07, 27 March 2024 (UTC)

Chemical potential
This article claims that the Fermi energy is identically equivalent to the chemical potential. I believe this is incorrect. The Fermi energy refers to the energy at the Fermi surface, which is equivalent to the chemical potential, in a non-interacting theory. In the presence of interactions, the Fermi surface can become fuzzed out, so that the Fermi energy is not well-defined.

The chemical potential, on the other hand, is a thermodynamic concept which is not concerned with any microscopic model for a system. It is defined even in classical statistical mechanics, in which there is no such thing as a "Fermi sea." -- CYD

 * You may be right, at least about some of that. I went looking through my thermodynamics textbook (Sears & Salinger) for some ammunition, and was surprised to find a formula giving the chemical potential in terms of the Fermi energy:

 * $$\mu = \epsilon _F \left[ 1- \frac{\pi ^2}{12} \left(\frac{kT}{\epsilon _F}\right) ^2 + \frac{\pi^4}{80} \left(\frac{kT}{\epsilon _F}\right)^4 + ... \right] $$

 * εF was introduced as an empirical constant and later defined in words as "the maximum energy of an electron at absolute zero". An F-D distribution function involving εF was given as a low-temperature approximation. However, the energy in the exact F-D distribution function is definitely the chemical potential from classical thermodynamics. Sears & Salinger gives a very detailed derivation of this. Although chemical potential was originally introduced in macroscopic thermodynamics, the relationship to microscopic theories (including Fermi seas!) is rigorously defined. You don't complain about temperature being invalid in microscopic theories, do you?

 * That said, in solid state physics, I have often seen the F-D distribution given involving EF instead of μ -- hence my confusion here and on Fermi-Dirac statistics. I'll think about fixing the articles, but I haven't really got it straight in my head yet. -- Tim Starling

 * after several days calculation, i still can't obtain the coefficient in the second term of the series mentioned above, and i can't find the derivation in Sear and Salinger also. Would any body suggest me for further reaing?tobywhc 13:11, 10 December 2006 (UTC)

 * Ashcroft and Mermin has a fine derivation of the first two terms of the expansion on pp. 42-47. I also agree that the relevant quantity in the F-D distribution is the chemical potential rather than the Fermi energy. Reasoning naively, it makes sense that the finite T value of $$\mu$$ is lower than the Fermi energy because the band structure has spherical symmetry. This means that there are more states slightly above the Fermi energy than slightly below, and consequently that the point of half-filling drops as you smear out the Fermi edge. Csmallw (talk) 22:11, 19 May 2009 (UTC)

 * actually the chemical potential at zero temperature is (almost by definition) equal to the Fermi energy. this is true even for a system with a more complicated potential, there the fermi surfave can become complicated but the fermi energy, the energy of the highest occupied state (@ T=0) is still unique. --V. 18:23, 15 February 2007 (UTC)

Fermi energy in semiconductor

 * The very first line of this article states "The Fermi energy is (...) the energy of the highest occupied quantum state in a system of fermions at zero temperature". This definition does not work at all of you consider a semiconductor, in which case the Fermi energy is in the band gap, so it is in fact an unoccupied level. As for the difference between the chemical potential and the Fermi energy, the confusion comes from the definition of the Fermi energy itself, which some varies according to author.  For example, in Eisberg & Resnick define it as the state with a probability of occupation of ½.  According to this definition, the Fermi energy does indeed vary with temperature.  However, the more accepted definition is that the Fermi energy is simple the limit of the chemical potential at T->0.   See for example Ashcroft & Mermin; look in the back of the book for Fermi energy, and you get a direct reference to the difference with the chemical potential.  Just don’t look in Kittel if you want to understand.

 * If you are advocating changing the first line to "lowest unoccupied state" Im willing to go with that. That seems to be consistent with defining the fermi energy as the limit of mu for T->0. I think it does not make sense to to start this article with a definition based on the chemical potential as this would probably not help anyone who does not already know what the fermi energy is. Also, the Fermi energy is independent of temperature in any definition (the fermi-dirac distribution function is 1/2 for E=Ef at any temperature). --V. 02:05, 21 July 2007 (UTC)

 * Actually, there is the same problem with "lowest unoccupied state". In an intrinsic semi-conductor, the lowest unoccupied state at 0K is the first level at the bottom of the conduction band, whereas the Fermi level is in the middle of the band gap.  That’s why in solid state physics the Fermi energy is defined as the chemical potential at 0K.  I don’t mind keeping the part about the highest occupied state, but it should be seen as a visualisation of what the Fermi energy is, not as a definition.  It should also be noted that this is only valid for metals or a fermions gas.  Maybe keeping the introduction the same, and add a note for a clarification later in the article?  Also, it is actually not true that that the Fermi distribution is always equal to ½ at E=Ef.  That is because it is the chemical potential which should be used in the Fermi-Dirac distribution, not the Fermi energy.  Since they are almost the same (see equation in article for the equation relating the two), a lot of text books just use the Fermi-energy, but this is wrong.  The reason why it is wrong is that by definition, as you mentioned, the Fermi energy is independent of temperature.  However, in many cases in a semi-conductor, the chemical energy, thus the energy where the fermi-dirac is equal to ½, changes with temperature.  This is why it can’t be said that the Fermi distribution is always equal to ½ at E=Ef.  --22 July 2007

 * I would probably start with the statement that it's where the f-d is half, and then clarify the problem for more complicated density of states later, as I don't know that the full explanation could fit clearly and concisely in the lead. — Laura Scudder ☎ 03:00, 26 July 2007 (UTC)

 * I see what you mean. So lets keep the intro the way it is, it seems to get the concept across. Feel free to add a section on "Fermi-Energy in Semiconductors" to explain some of the subtleties.

 * The problem is not unique to semiconductors. It occurs whenever you do not have a continuum of states, but rather a gap between energy levels.  I recommend specifying that Fermi energy is chemical potential at T=0, and explaining that this is equivalent to highest filled state in cases where there are a continuum of energy levels.   —Preceding unsigned comment added by 74.102.151.97 (talk) 11:08, 28 March 2011 (UTC)

 * What about this definition of the FE at finite Temperatures
 * $$N=\int_{-\infty}^{E_F(T)} g(E) dE=\int_{-\infty}^{+\infty} g(E) f(E,\mu,T) dE$$
 * i.e. the upper cutoff of the integral over the density of states. I saw this in several books on SSP... In the case of free electrons, this also exlains the expasion of the chemical potential in terms of EF at the end of the article (look for example at the Sommerfeld expansion in Ashcroft) ...

Incluse general spin states
I think article should be expanded to include any half integer fermion case, instead of concentrating only on spin 1/2 particles —Preceding unsigned comment added by 203.197.196.1 (talk) 23:27, 23 April 2008 (UTC)

Quasi Fermi levels
I have started an article on quasi Fermi level, it needs work and is not ready for a "prime time", but when it is I think a link would be appropriate. --Thorseth (talk) 21:40, 20 May 2008 (UTC)

Chemical Potential = Fermi Energy at 0K?
I think it's a serious error to state that chemical energy (but not electrochemical energy) is equal to Fermi energy at absolute temperature. Does the author mean ELECTROCHEMICAL energy?

Because that makes sense.

!! I see the error now. Throughout the article CHEMICAL potential is used for ELECTROCHEMICAL potential. But they are very different concepts.

Chemical potential has to do with doping, whereas electrochemical potential gives rise to the electromotive force.

I suggest we fix this error.

I see that my concern has been brought up before by CYD. He's right and he has put the problem better than me. If nobody is willing to correct this serious error, I am going to do it

128.46.213.219 (talk) 20:45, 22 August 2008 (UTC)

 * Well, the quantity that actually enters calculations is:
 * $$e^{(E-\mu)/k_BT}$$
 * where E is the (say) electron energy level is and μ is the (electro??)chemical potential. It's the energy difference that counts. So if E is measured relative to the energy of an electron in vacuum at that point, then μ also must be measured relative to the vacuum energy at that point, i.e. μ is the chemical potential. If E is measured relative to the energy of an electron in vacuum at a fixed point in space (or at infinity), then μ needs to be likewise, so μ would have to be the electrochemical potential.

 * Do you agree so far?

 * In my field (semiconductor materials science), E is virtually always measured relative to the vacuum energy at that point (or in a given material, relative to the conduction band edge at that point, which amounts to the same thing). So chemical potential, not electrochemical potential, would be the right thing to use. Perhaps other fields are different though. At the very least, this could be explained better. What do other people think? --Steve (talk) 02:58, 23 August 2008 (UTC)

 * Using the definition of Fermi energy in the first line of this article, the limit as T$$\rightarrow$$0K of the chemical potential $$\mu$$ is not always the Fermi energy. For example, in an intrinsic semiconductor, $$\mu$$ is in the gap between the valence band and the conduction band. In this case, the limit as T$$\rightarrow$$0K of $$\mu$$ is in the gap, which is not at the Fermi energy which is at the top of the valence band, according to the definition in this article. For the article to be self-consistent, this should be fixed.
 * However, I have come across at least one place where the definition of Fermi energy is the chemical potential at T=0K. It was explained that all states at or below the Fermi energy are occupied, which is true even if the T=0K definition of Fermi energy is in the gap and the highest energy state that is occupied is at the top of the valence band. But that is not the definition stated in the first line of this wiki. --Bob K31416 (talk) 00:16, 6 May 2009 (UTC)

 * Bob, this is incorrect. See my comment from April, 8 + refs here.--Evgeny (talk) 16:59, 7 May 2009 (UTC)

 * I think your point regarding an intrinsic semiconductor's chemical potential µ, is described by this excerpt from the Shigelski 2004 abstract that you mentioned,
 * "...the Fermi–Dirac distribution function ceases to be accurate for the thermal occupation probabilities when the temperature is so low that the number of electrons in the conduction band is of order unity. The use of the correct occupation numbers results in µ going to the bottom of the conduction band as T-->0."
 * Please note that this supports the basic point that I mentioned for the Fermi energy of an intrinsic semiconductor (although according to Shigelski 2004, I got the zero-temperature limit of µ wrong). According to the definition of Fermi energy in the first line of this wiki, the Fermi energy for an intrinsic semiconductor is at the top of the valence band. Thus, if we use the zero-temperature limit of µ at the bottom of the conduction band, from Shigelski 2004, the Fermi energy is still not the limit as T$$\rightarrow$$0K of the chemical potential $$\mu$$, for this case of an intrinsic semiconductor. --Bob K31416 (talk) 19:39, 7 May 2009 (UTC)

 * The approach I've seen most often is that the books initially define EF as the "highest occupied state / lowest unoccupied state", and then they get to semiconductors/insulators and say "oh wait, that definition is ambiguous when there's a band gap", and then they say "OK the real definition is mu at 0K". :-) --Steve (talk) 06:36, 6 May 2009 (UTC)

 * Thanks for contributing to the discussion. Regarding the Fermi energy EF, I presume you meant the energy of the "highest occupied state..." as T$$\rightarrow$$0, or equivalently, for lowest energy configuration of the system. I don't see how that "is ambiguous when there's a band gap". Are you sure that was in the books that you were recalling? --Bob K31416 (talk) 14:49, 6 May 2009 (UTC)

Relationship to Maxwell distribution
If I understand correctly, doing an analogous derivation at finite temperature for sufficiently low density would yield a Maxwell distribution. If this is correct, it seems worth mentioning. It connects it to something more familiar to many. It can be explained where that approximation fails, and why it matters that the particles are fermions. 72.75.67.226 (talk) 03:55, 8 October 2009 (UTC)

Even number of particles in the square well example
"...,all the energy levels up to n=N/2 are occupied and all the higher levels are empty." What if the number of particles N is not an even number? Andres.felipe.ordonez (talk) 00:58, 17 January 2012 (UTC)

Merge with Fermi gas
I'm proposing a merge between Fermi energy page and Fermi gas into one page (Fermi gas) as the former is a subproperty of the latter concept. This proposal also takes in fact the Fermi energy article has better equations and is better explained than the Fermi gas article so mostly the new merged article won't lose any generality. MaoGo (talk) 17:44, 4 January 2018 (UTC)
 * I think that the content of Fermi energy should be merged into Fermi gas. However I also think that a short stub should be kept at Fermi energy with just a formula and a brief explanation. Ruslik_ Zero 19:15, 4 January 2018 (UTC)
 * That is a good idea. While it is true that the Fermi energy is a concept based on a non-interacting Fermi gas, Fermi level and Fermi energy are often used interchangably and it's good to keep the discussion of the differences between the two as part of a separate small article. --Mark viking (talk) 19:44, 4 January 2018 (UTC)
 * Maybe we can leave a stub, but I'm not sure if it is a good idea. In think Fermi level does pretty well alone, while Fermi energy (and other related terms like Fermi temperature) can be redirected to a major section in Fermi gas. MaoGo (talk) 19:56, 4 January 2018 (UTC)

 * Weak oppose (but support Ruslik0 plan). I suggest (1) make Fermi gas into the article you're imagining, including by copying better sections from this article (or elsewhere). Then (2) strip Fermi energy down in places that it covers things that are better explained and more topical in the new Fermi gas article, either deleting sections entirely or putting a sentence or two plus "Main article: Fermi gas". Then (3) we can reassess whether there's enough Fermi energy article content left to make it worth keeping. I suspect that it will be worthwhile to keep a short (or maybe very short!) Fermi energy article with the definition, how it differs from Fermi level (see also this page of my website), and maybe the "typical Fermi energies" section, and not much else besides links to Fermi gas. That all seems to me to be a bit too much for a section of Fermi gas. But I don't feel strongly, it would probably be OK either way. --Steve (talk) 01:26, 5 January 2018 (UTC)
 * Oppose because many things apart from Fermi gases have a Fermi energy, which is the more general concept. Fermi energy is the Chemical potential of an electrically charged moiety. Its properties are described in that article. Xxanthippe (talk) 21:29, 4 January 2018 (UTC).
 * At risk of getting sidetracked, "Fermi energy is the chemical potential of an electrically charged moiety" is indeed one definition of Fermi energy, but it's not the only definition, and in particular it's not the definition currently used for the Fermi energy article. See this page of my website for example. Your definition, "the chemical potential of an electrically charged moiety", is instead covered in the Fermi level article. --Steve (talk) 14:45, 5 January 2018 (UTC)
 * All definitions reduce to the thermodynamic definition. Xxanthippe (talk) 23:27, 5 January 2018 (UTC).
 * As others suggested we can leave Fermi energy article alive to discuss this kind of definition problems, but this merge is not even about that, it is about getting Fermi gas calculations into the right article. — Preceding unsigned comment added by MaoGo (talk • contribs) 20:44, 6 January 2018 (UTC)

 * Oppose because the current-article is already a nice size, whereas the Fermi gas article is quite long. One problem with the long articles is that, sometimes, one just wants a quick-ref, just to quickly look up the handful of most prominent relations/values/statements/facts.  When faced with a long article, this becomes daunting and exhausting; a companion short article is actually nice.  And this is that short article. 162.204.250.21 (talk) 02:00, 12 August 2019 (UTC)

Typical values of fermi momentum?
For us lazy-bones, it would be nice to have typical values for the fermi momentum too. So, for example, for metals, I think, maybe, that the fermi momentum, is less than a few inverse angstroms, i.e. less than the interatomic spacing, right? But it would be nice to know without having to whip out my calculator, and googling around. (and what about the fermi temperature? What might that be?) and out of shear curiosty, same for white dwarves, nuclei. And neutron stars? 162.204.250.21 (talk) 02:05, 12 August 2019 (UTC)
 * Once you get there, next is Fermi velocity, which would also be nice to know. That is, typical Fermi energy, momentum, and velocity for electrons in a metal, and nucleons in a nucleus. Gah4 (talk) 20:10, 27 March 2024 (UTC)
 * Once you get there, next is Fermi velocity, which would also be nice to know. That is, typical Fermi energy, momentum, and velocity for electrons in a metal, and nucleons in a nucleus. Gah4 (talk) 20:10, 27 March 2024 (UTC)

Grammar rules and equations
Re: recent changes by with "we don't typeset grammatical commas in the equations" − who are "we" and why they don't? AFAIK, as a rule, scientific papers do use normal English grammar. E.g., https://www.e-education.psu.edu/styleforstudents/c4_p3.html Evgeny (talk) 18:13, 3 December 2021 (UTC)
 * Duly noted. I see this is an oft-debated topic (the link you posted, along with some other sources), and the usage depends on the exact structure of the sentence and typesetting. It looks confusing at times when the punctuation is in the formula (when the formula is not in line with prose, in which case there doesn't seem to be any confusion), so some authors circumvent it by changing the sentence structure. If I (unknowingly) went against a convention, which seems like common practice but not universal as some other rules, please revert my edits without further ado. ComplexRational (talk) 19:57, 3 December 2021 (UTC)
 * OK, thanks; frankly, I feel rather strongly about this, so will revert your edits. Evgeny (talk) 22:42, 3 December 2021 (UTC)

Uncertainty principle forbids zero kinetic energy vs 'In a Fermi gas, the lowest occupied state is taken to have zero kinetic energy'
The second sentence of the first paragraph currently states: 'In a Fermi gas, the lowest occupied state is taken to have zero kinetic energy'.

Is that a simplification? Wouldn't it in fact have a non-zero kinetic energy due to the uncertainty principle? MathewMunro (talk) 13:09, 6 January 2024 (UTC)
 * No, the uncertainty principle doesn't "forbid" zero kinetic energy. Do you mind citing a source that made you think so? Evgeny (talk) 08:42, 7 January 2024 (UTC)
 * Zero KE means zero momentum, and completely delocalized. If you want a particle to be somewhat nearby, then yes. Gah4 (talk) 22:59, 7 January 2024 (UTC)
 * Fermi gas is a model of ideal gas (of fermions). There is no notion of any finite volume in this model. So yes, all particles can be considered as completely delocalized as one wishes. Evgeny (talk) 09:24, 8 January 2024 (UTC)
 * OK, in the usual case one considers a sphere in momentum or k-space. The number is large enough that one can use a continuum approximation, instead of counting the discrete energy levels. And so with the continuum approximation, the need to distinguish 0 is small. I suppose for light nuclei it is wrong, but even so, close enough. Gah4 (talk) 20:16, 27 March 2024 (UTC)
 * Fermi gas is a model of ideal gas (of fermions). There is no notion of any finite volume in this model. So yes, all particles can be considered as completely delocalized as one wishes. Evgeny (talk) 09:24, 8 January 2024 (UTC)
 * OK, in the usual case one considers a sphere in momentum or k-space. The number is large enough that one can use a continuum approximation, instead of counting the discrete energy levels. And so with the continuum approximation, the need to distinguish 0 is small. I suppose for light nuclei it is wrong, but even so, close enough. Gah4 (talk) 20:16, 27 March 2024 (UTC)
 * OK, in the usual case one considers a sphere in momentum or k-space. The number is large enough that one can use a continuum approximation, instead of counting the discrete energy levels. And so with the continuum approximation, the need to distinguish 0 is small. I suppose for light nuclei it is wrong, but even so, close enough. Gah4 (talk) 20:16, 27 March 2024 (UTC)

quarks and gluons
Recent edits on proton remind me that quarks and gluons should also have a Fermi energy inside the nucleon. Maybe that can be given here, including an approximate Fermi energy. Gah4 (talk) 20:19, 27 March 2024 (UTC)
 * Do you have a source?--ReyHahn (talk) 22:25, 27 March 2024 (UTC)
 * No, that is why I am asking. OK, QCD matter says: (corresponding to the QCD energy scale ΛQCD ≈ 200 MeV), which suggests it might be about 200 MeV. Gah4 (talk) 02:36, 29 March 2024 (UTC)
 * No, that is why I am asking. OK, QCD matter says: (corresponding to the QCD energy scale ΛQCD ≈ 200 MeV), which suggests it might be about 200 MeV. Gah4 (talk) 02:36, 29 March 2024 (UTC)
 * No, that is why I am asking. OK, QCD matter says: (corresponding to the QCD energy scale ΛQCD ≈ 200 MeV), which suggests it might be about 200 MeV. Gah4 (talk) 02:36, 29 March 2024 (UTC)

title: Talk:Fermi energy


text:
Talk:Algebra over a field

Over a commutative ring
It's nice to see this separated out into its own thing as it should be. The problem is that there's really no good name for it. "algebra over a field" isn't quite right, since algebras can be defined over any commutative ring with no change in the concept, and one does study these things with the same terminology and notation in (for example) algebraic geometry. But "algebra over a commutative ring" isn't too hot either, being less familiar and (with "commutative") rather unwieldy.

One idea is "algebra over a base", since the commutative ring in question is called the base ring (or base field). One does hear phrases such as "Let A be an algebra over the base ring K.", for example. But I can easily imagine that this means something else too. Another idea is "algebra over a base ring", which has some of the problems of the version with "commutative" above, but is not so enormous.

Ideas?

-- Toby 06:39 Mar 7, 2003 (UTC)

Well, while editing this page, I decided that it's OK as it is. The fact the "vector space" and "module" are completely different terms is a much bigger problem (and much bigger than Wikipedia) that already prevents discussing the general case in the same breath as the more familiar. So let it be. -- Toby 07:46 Mar 7, 2003 (UTC)

Alternative algebra bug?
This page states that the only finite dimensional alternative algebras over the reals are R, C, the quaternions and the octonions. This doesn't seem correct, since any matrix ring is a finite dimensional alternative algebra over the reals, or any real vector space with xy := 0 as multiplication. I don't know how to correct this -- maybe any finite dimensional alternative division algebra? --Sven 217.9.27.15 00:36, 20 Apr 2005 (UTC)

 * Beats me. Maybe the statement is "any irreducible finite dimensional alternative algebra that is not an associative algebra"? (since a matrix ring is associative, and a real vector space with xy=0 but xx!=0 seems to be completely reducible, to me).linas 01:38, 20 Apr 2005 (UTC)

 * I see no one fixed this in 15 months. What was probably meant was the Bruck-Kleinfeld theorem specialized to the field of real numbers: the only finite dimensional real alternative division algebras are R, C, the quaternions and octonions. The original author might also have meant Hurwitz' theorem: the only finite dimensional real normed alternative algebras are R, C, the quaternions and octonions. But I'm guessing the Bruck-Kleinfeld result is what was meant. --Michael Kinyon 12:19, 31 July 2006 (UTC)

index free notation ???
I have big doubts about the validity of the section "index free notation".
 * first, say why is it called like this
 * second, should "k-algebra" be corrected to "K-algebra" everywhere ? (simultaneous use of K and k is very unfortunate ; in addition the prefix "k-" often means something else not related to any element "k" or set "K")
 * third, according to the introduction, we do not suppose all algebras to be associative, thus they are not rings in general
 * fourth, they should even less be supposed to be unital, and I think that eta(1) must be a unit if not eta=o

So I strongly suggest to move that paragraph (and the following, based upon it) away from the main page (e.g. into the Talk page) until these issues are clarified. — MFH:Talk 15:02, 2 June 2006 (UTC)

 * fifth, the range of η must lie in the centre of A to really make it a k-algebra. -- EJ (talk) 16:03, 25 February 2008 (UTC)

Simplification
I was wondering if the definition at the beginning be simplified for non-math majors. I found a very nice and simple definition that explained the definition of an algebra perfectly: ''A set together with a pair of binary operations defined on the set. Usually, the set and the operations include an identity element, and the operations are commutative or associative.'' Nice, plain English. Start talking about the Ks and As in the next section, not the first definition. It's quite intimidating. ForestAngel (talk) 01:43, 24 July 2008 (UTC)

 * That's not right. You have to have not only the ring structure (a set with two binary operations obeying certain algebraic laws) but also the vector-space structure (where the addition in the vector space is the same as the addition in the ring, but you've also got multiplication by scalars).  So it's really three operations:
 * (1) addition,
 * (2) multiplication of two members of the algebra, and
 * (3) multiplication of a member of the algebra by a scalar.
 * Michael Hardy (talk) 20:05, 8 December 2009 (UTC)

 * I think I've addressed this now. Mlm42 (talk) 02:43, 13 November 2008 (UTC)

Missing examples
I'm not an algebraist, but when I try to think of examples of this concept, the first things that come to my mind are (1) the algebra of polynomials over a field, and (2) the algebra of n×n matrices over a field. Yet I don't see those mentioned in the section titled "examples". Should they be there, or am I missing something? Michael Hardy (talk) 20:02, 8 December 2009 (UTC)
 * ...oh, I see there are more parts of the article with examples than the part I was looking at.....
 * to be continued..... Michael Hardy (talk) 20:13, 8 December 2009 (UTC)

Classification is incomplete
This article claims that there exist only two two-dimensional unital associative algebras over the complex numbers up to isomorphism, but the classification given does not include quaternions (the case aa=-1). It also misses the case 1≈1,0][0,1 (the 2-2 identity matrix) and a≈1,0][0,0 which would have aa=a. Both these have two complex dimensions and are associative and unital. Clearly the classification is inaccurate, but I don't have access to the source given. The source looks legit, so someone should double check it. —Preceding unsigned comment added by 98.149.109.69 (talk) 00:52, 17 February 2010 (UTC)

 * The result is correct (and would still be correct without the word "associative"). Note that quaternion multiplication isn't bilinear over C (though it is bilinear over R), so the quaternions do not form an algebra over the complex numbers. --Zundark (talk) 09:09, 17 February 2010 (UTC)

 * Reference to Eduard Study for the classification of algebras over C looks way back. The subject of classification of algebras stimulated the work of many mathematicians back then; see hypercomplex number for more context and results.Rgdboer (talk) 21:01, 6 December 2011 (UTC)

 * Working from first principles makes this clearer:
 * The case of aa=−1 does not generate the quaternions. Putting b=ia we get bb=1, so it is isomorphic to the case with aa=1.
 * The case of aa=a: Putting b=2a−1 we get bb=1, so it is also isomorphic to the case with aa=1. Quondum talkcontr 09:41, 7 December 2011 (UTC)

Improved, but.... (distributive ≠ bilinear)
The latest edit is a very neat, understandable nutshell, and thus I'm loathe to change it. However, it has a significant shortcoming: it is no longer perfectly correct, because it replaces bilinearity with distributivity, and the two are not equivalent. One can thus find examples that fit the new definition (distributivity) that do not satisfy the other axiom of (bi-)linearity: homogeneity (referred to in the article as compatibility with scalars). Since the standard definition does require homogeneity, can we find a wording that includes this requirement without destroying the elegance of the new wording? — Quondum 04:33, 16 October 2012 (UTC)

 * Why not: "In mathematics, an algebra over a field is a vector space equipped with a bilinear product that distributes over vector addition"? D.Lazard (talk) 07:57, 16 October 2012 (UTC)

 * Thanks, I have incorporated this suggestion; it is nicely compact. — Quondum 09:07, 16 October 2012 (UTC)

 * In an edit which was in conflict with the preceding one, I have crossed out "that distributes over vector addition" in my preceding post and added in comment:
 * In fact, bilinearity implies distributivity and mention of the distributivity is not needed here. By the way, the lead of bilinear map is rather unclear and deserve to be rewritten in the spirit of that bilinear form. Also, the section "Definition and motivation" gives only examples of unital associative algebras. For not being confusing, it should contain a less restricted example, like a 3-dimensional vector space equipped with the cross product.

 * I'll be bold and will remove the mention of distributivity in the lead. Please revert if I have missed something --D.Lazard (talk) 09:33, 16 October 2012 (UTC)
 * Yup, nice. I feel a bit silly for missing something that obvious. — Quondum 14:45, 16 October 2012 (UTC)

 * I like the new version, but I felt something really had to be said about the common usage of "algebra" for "associative algebra". I hope my attempt doesn't interfere too much with the lead. Rschwieb (talk) 13:06, 16 October 2012 (UTC)

 * "Because", "ubiquity", "importance" seem WP:OR. Moreover "algebra" does not appear only in textbooks, as suggested in the last paragraph of the lead. I suggest: 'In many texts, "algebra" is used for "unital associative algebra". It is almost always the case when "algebra" is used without further specification.' --D.Lazard (talk) 14:21, 16 October 2012 (UTC)
 * Or perhaps (forgive my nitpicking; feel free to ignore it): 'In many texts, "algebra" is used to mean "unital associative algebra". It is typically the case when "algebra" is used without further qual ification.'? Underlining is purely to highlight changes. — Quondum 15:03, 16 October 2012 (UTC)

I agree D.Lazard... let's take out "importance". "Texts" for textbooks is fine too. However, I don't know what to do about "ubiquity". How else do you describe the predominance of associative algebras over nonassociative ones in print? There is probably a better word. Q: Perhaps my wording is not the best for all English speakers... if there is a more straightforward version, have at it :) Rschwieb (talk) 15:13, 16 October 2012 (UTC)

 * To Rschwieb: the "predominance" is a matter of opinion that has not its place in this lead. Is there really a predominance of associative algebras over Lie algebras? The use of "almost always" or "typically" says essentially the same in a less controversial way.
 * To Quondum: I have used "specification" to include the case of the texts, like textbooks, that give a definition of "algebra". By "without specification", I meant "without definition nor qualification". The latter wording may be better. About "almost always" vs. "typically": we have to think to the reader who have read the word of "algebra" and want to know its meaning. "Almost always" means clearly that he has to read it as "unital associative algebra". My level of English is not enough to be sure that "typically" is as clear. --D.Lazard (talk) 15:47, 16 October 2012 (UTC)

I would like to propose that we create associative algebra over a field, either renaming this article or whatever else. As for the convention, the simplest solution is to skip the discussion on it altogether. I agree with R that "algebra" "often" refers to associative ones (and almost never say Lie or Jordan ones), but the point is that Wikipedia doesn't have to concern with such issue. -- Taku (talk) 17:28, 16 October 2012 (UTC)

 * I'm not entirely comfortable with the statement of "almost always" or "typically", as it actually seems to invite challenge (on the rest I tend to go along with D.Lazard). But is it the case that WP can adopt, without comment, a convention of using the term "algebra" in a specific sense (e.g. the most general mentioned here, namely not necessarily either unital or associative, alternatively the most restricted sense, being unital and associative)? — Quondum 17:44, 16 October 2012 (UTC)

 * IMO, the point is not WP conventions, there are fine as they are. The point is that, when someone consult WP because he has read "algebra" without explanation, he has to be informed that this probably means "unital associative algebra". If providing such an information breaks some WP rule, this is a case of WP:Ignore all rules. --D.Lazard (talk) 18:16, 16 October 2012 (UTC)
 * Thus I propose:

In many texts, "algebra" is used to mean "unital associative algebra". In particular, if "algebra" is used without any definition nor qualification, it means very probably "unital associative algebra".
 * --D.Lazard (talk) 18:30, 16 October 2012 (UTC)

 * This seems to repeat itself, somewhat unnecessarily. A reader will get the idea from the first of the two sentences. Indicating that the term can also be used in the general sense may be of value:
 * In many texts, "algebra" is used to mean "unital associative algebra". A more general algebra may be indicated by a definition or qualification.
 * — Quondum 19:35, 16 October 2012 (UTC)

Bilinear
There is no reason to put the bilinear condition on the algebra product. This restriction is not given in references: Hazewinkle has (p 3): An algebra over a field k (or k-algebra) is a set A which is both a ring and a vector space over k in such a manner that the additive group structures are the same and the axiom (λa)b = a(λ b) = λ(ab) is satisfied for all λ in k and a,b in A.

Bahturin (1993) has (p 29): A ring R which also is a vector space over a field F is called an algebra over F, or F-algebra, if for any λ in F and any x,y in R
 * λ(xy)=(λ x) y = x(λ y)

In Basic Structures of Modern Algebra, Kluwer Academic, ISBN 0-7923-2459-5.

The definition of Steven Kantz (2000) is wide enough to include Lie algebras: A vector space (over a field) on which is also defined an operation of multiplication. In Dictionary of Algebra, Arithmetic and Trigonometry, CRC Press, ISBN 978-1-4200-3802-6.

The term "bilinear" is commonly used for bilinear form, but not as an axiom for algebras. Usage here provides an unnecessary complication for someone trying to determine the exact requirements for this mathematical structure.Rgdboer (talk) 22:32, 24 October 2012 (UTC)

 * Your quotes from Hazewinkle and Bahturin (defining what I would call a unital associative algebra) give three separate requirements:
 * It is a vector space over a field K.
 * It is a ring.
 * The multiplication of the ring is homogeneous of degree 1 over K.
 * This implies bilinearity, even if one avoids the term.
 * What is defined first in the lead is the more general case (that is not required to be unital or associative), which may be given with the following axioms:
 * It is a vector space over a field K.
 * It has a multiplication operation that is closed on the vector space
 * The multiplication operation distributes over the vector space addition.
 * The multiplication operation is homogeneous of degree 1 over K.
 * The last two axioms are encapsulated in "bilinear", so the term cannot be removed without replacing it with something equivalent, such as the explicit axioms.
 * The definition by Steven Kantz would normally be regarded as deficient: it allows the class being defined to include cases that do not satisfy the axiom of homogeneity, and in particular includes objects that are neither associative algebras nor non-associative algebras. — Quondum 08:43, 25 October 2012 (UTC)
 * Since the mentioned axiom is precisely bilinearity, I wouldn't say it's misleading but maybe it does put another layer between the reader and understanding. In this case, Krantz's  definition appears to be vague to the point of incorrectness. If you exclude it, then you are omitting the fact that the copy of the scalar field is central in the algebra. In some texts, that is how algebra structure is defined: for a ring (or Lie ring) A, and a commutative ring R, A has an R algebra structure if there is a ring homomorphism (or Lie ring homomorphism) from R to the center of A. The axiom relating the module structure and the multiplication appears in all texts I have ever read on the topic in some form. Rschwieb (talk) 14:44, 25 October 2012 (UTC)
 * Actually while thinking about it now, I'm not sure the "central homomorphism" description of Lie algebras works that way. But I am sure it works that way for associative algebras. Rschwieb (talk) 14:46, 25 October 2012 (UTC)

Very valuable responses. Thank you. Continuing to look at reliable sources, there is JDH Smith and AB Romanowska in Post-modern Algebra (1999) who confirm use of bilinear, but in a slightly broader context. They write on page 158:
 * If K is a unital commutative ring, then a K-algebra (AK, ·, 1) is a unital K-module AK and a monoid (A, · , 1) such that the monoid multiplication A × A → A; (a,b) → ab is bilinear.

This terse text with little history or commentary includes the opinion:
 * The rather grandiose name [algebra] probably arose because early twentieth-century algebra was almost exclusively devoted to the study of "algebras" in this sense.

Algebras have considerable structure. Perhaps inclusion of bilinearity as a defining property of the product helps prepare students for tensor products of algebras, where the bilinear property comes to the fore.Rgdboer (talk) 20:48, 25 October 2012 (UTC)

 * I do not agree with Smith-Romanovska opinion for the origin of the name of "algebra". IMO, this name arose before than algebras began to be studied for themselves. At the beginning, an algebra was simply a framework where one can operate like in usual algebra, now called elementary algebra. Similar shifts from the name of an area of mathematics to the name of a structure among others arose also for geometry (non Euclidean geometries), arithmetic (floating point arithmetic, Presburger arithmetic) and even for (mathematical) logic. D.Lazard (talk) 07:41, 26 October 2012 (UTC)

@Rgdboer I agree with your sentiment in the third to last paragraph about bilinearity, but I want to emphasize that I think a lot of results for both associative and Lie algebras use the fact that the scalars commute with the multiplication operation (although I can't think of a specific description atm :.

Now I'm curious to know if sesquilinearity is ever substituted for the axiom we are talking about. I haven't seen it before, but it seems like it could be possible for some sort of weird complex algebra-like object. Rschwieb (talk) 13:08, 26 October 2012 (UTC)

 * Sesquilinearity seems to occur frequently in representations in physics as a sesquilinear form (i.e. onto complex scalars) rather than a sesquilinear mapping onto the whole algebra space. Nevertheless, it should not be weird: surely any sesquilinear mapping on a complex vector space can always be expressed as a real bilinear mapping on a real vector space, yielding a normal ℝ-algebra? — Quondum 14:46, 26 October 2012 (UTC)

Hans Zassenhaus addressed the definition of an algebra in his Theory of Groups (1936). First he identifies a semi-module as an additive abelian semi-group.(p 92). Then he identifies a semi-ring as a semi-module with a multiplication that is both left and right distributive (p 93). Then he writes "A quasi-ring is a semi-ring which under addition is a module. In other words, the axioms of a quasi-ring are obtained from the ring axioms by omitting the associative law of multiplication"(p 95). On page 103 the definition of an algebra is given:
 * A quasi-ring over a field as coefficient ring is called an algebra. An algebra always has a basis over its coefficient field. An associative algebra is an algebra satisfying the associative law of multiplication.

An advantage of Zassenhaus terminology is the integration of Lie rings and Lie algebras in the treatment (instead of making them exceptional in nomenclature as when the definition of algebra is framed too narrowly).Rgdboer (talk) 01:36, 10 November 2012 (UTC)

Hitsuji Santos (talk) 04:03, 6 November 2013 (UTC) I wish every mathematician could be so clear as you. Congratulations.

Unital zero algebra
In the section « unital zero algebra », i read « (λ+u) (μ+v) = λμ + (λv+μu) ». Am i the only one chocked by the fact that scalars are freely added to vectors ? The law + is suposed to be an operation between vectors. I think correction is needed. — Preceding unsigned comment added by 109.23.11.115 (talk) 16:44, 13 March 2015 (UTC)

 * It does make sense as it stands; we are taking the direct sum of scalars and vectors by construction. The notation is possibly confusing, since it is not distinguishing between a scalar and an element of the direct sum for which the vector part is zero, for example. This could be made more rigorous if it creates confusion, at the cost of verbosity. —Quondum 17:07, 13 March 2015 (UTC)

link in section « unital algebra »
The link « unital algebra » redirects to the article « Ring ». I disagree : an unital algebra is not necessarily associative, whereas it is the case for rings... — Preceding unsigned comment added by 109.23.11.115 (talk • contribs) 2015-03-13T16:50:48

 * This does seem to be a problem, and I agree., you changed the redirect; care to comment? —Quondum 17:01, 13 March 2015 (UTC)
 * In any case, previous redirect to Pseudo-ring was worse. I have edited this article (rm main template) and redirected  here for fixing the issue. D.Lazard (talk) 17:33, 13 March 2015 (UTC)
 * This seems to be the best route to go. There were still links to that redirect in this article, so I've removed them. —Quondum 20:46, 13 March 2015 (UTC)

Should we put that they are just monoid objects?
These algebras are just monoid objects in the category of vector spaces (with tensor product as monoidal product.) Also, algebra homomorphisms are just monoid object homeomorphisms. I am not sure where to put it though. TheKing44 (talk) 21:45, 12 May 2015 (UTC)

 * I'm not arguing definitively against inserting it, but my take is that it's a fairly isolated statement that might be awkward to fit in. We could add such a line to numerous other articles. The line rightly belongs (and already appears) in the monoid (category theory) article. Rschwieb (talk) 12:48, 13 May 2015 (UTC)

Assessment comment
Substituted at 01:44, 5 May 2016 (UTC)

Is the article's first sentence intended to be humorous?
At this moment the first sentence of this article reads: "In mathematics, an algebra is one of the fundamental algebraic structures used in abstract algebra." This sounds quite circular and rather funny. I suggest the sentence ought to be modified to remove the ha ha factor. — Preceding unsigned comment added by Dratman (talk • contribs) 01:00, 11 August 2016 (UTC)

 * I agree. I have edited the lead for fixing this and removing a WP:PEACOCK term ("fundamental"). D.Lazard (talk) 08:45, 11 August 2016 (UTC)

Thank you for the motivating examples
They are incredibly helpful to understand "why the heck would anyone care about this construction", and sorely missing or buried later in the text in too many Wikipedia math articles. 50.197.48.113 (talk) 14:43, 2 February 2024 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Talk:Algebra over a field