# Talk:Isospin

## Paragraph on SU(6) and symmetry breaking

• I disagree with removing my paragraph on SU(6) and symmetry breaking. It's an extra paragraph (and one I thought was decently written) and isospin always confused me until I knew that information (I'm a physicist). It's important to also consider the readers of the page, high-school kids are not going to be learning about isospin. This is a page most likely visited by undergraduate physics majors, who are looking for the connections to the big picture. Jason Quinn 00:51, 9 Jul 2004 (UTC)

Hello, Jason. Here's is some kind of justification for removing that paragraph
1. Connections to the big picture are already available as wikilinks to articles on flavor and strong interaction etc. And while it definitely helps to connect it to the big picture more explicitly, the content of that paragraph more rightly belonged to the article on flavor, or quark.
2. Secondly, it isn't too clear what you mean by adjectives like "good symmetry", "the Eightfold Way is okay" and "badly broken". While I am sure they are part of the jargon which you use as a practising physicist, they are IMHO not the sort of sentences that should live in an encyclopaedic article.
3. Thirdly, and IMextremelyHO again, sentences like "a results of just the up and down quarks", and "flavor includes all known quarks" are both obfuscatory in meaning, and also physically incorrect (of course, based on what little physics I know).
4. Incidentally, but no less importantly, it isn't "your" paragraph anymore - you sort of gave it away to be beaten up and kicked around around by everyone on Wikipedia when you pressed that submit button
5. And, very incidentally, there is no reason why high school kids should not be interested in reading about isospin - I did, when I was in that category.
Regards, [[User:AmarChandra|Amar | Talk]] 05:56, Jul 9, 2004 (UTC)
Exorcised paragraph reproduced below for convenience
In high-energy physics, isospin is considered a subgroup of a larger symmetry group, the Eightfold Way, which is itself a subgroup of SU(6) flavor. Every new quark introduces a larger symmetry group. Isospin was a results of just the up and down quarks (u and d), which are nearly the same mass. The Eightfold Way includes the strange quark (s), which is somewhat heavier. Flavor includes all known quarks, adding charm, bottom, and top (c, b, and t). However the masses of the other quarks are significantly different than the up and down quarks so the strong force is not symmetric with respect to interchange of particles. The symmetry is said to be more and more broken the more the masses differ. Isospin is a good symmetry, the Eightfold Way is okay, and flavor is badly broken.

## So... is it still "used"?

If I'm reading the article correctly -- and I should point out this is the first math/physics article I've come across that I consider even remotely readable -- it seems that isospin is not "real" under a modern treatment, but considered to be a consequence of other physics -- specifically the parton/quark model. Is this correct? If so, I think I/we should add another paragraph at the end, noting this.

Maury 22:37, 28 November 2005 (UTC)

Not true, since quarks carry isospin. Up quark is isospin +1/2 and down quark is isospin -1/2. Pairs of half-integer isospin quarks in a meson sum up to integer isospin values, and so on, according to the representations of SU(2). linas 04:17, 29 November 2005 (UTC)
Ok, but is it the _spin_ of the quarks, or _iso_spin? The article makes it seem that the conservation law in question is not "real", and simply a consequence of the quarks other properties. IE, do quarks carry "isospin charge", or display it in groups. Is there an actual isospin conservation/symmetry at the quark level, or is this a side effect of the dynamics of the strong force (containment)?
There are lots of conservation laws that we no longer consider fundamental, but a side effect of other physics. My question is whether this is one of those examples.
Maury 13:03, 29 November 2005 (UTC)
Yes, quarks carry isospin charge. Its called "iso"spin because its mathematically similar to spin, in that its an SU(2) symmetry. Isospin is the simplest flavour symmetry, and its the closest to being "conserved" in that the mass of the up quark is nearly identical to the mass of the down quark (both of which are believed to have small masses, on the order of half-a-dozen MeV). Isospin is "still" a fundamental symmetry, as there is no other viable explanation at this time. For example, the preon theory has bitten the dust, and there is no experimentally workable string theory.
If you look for mesons made out of u or d and the next-heavier quark, the strange quark, you find that you can arrange them on the corners of a hexagon, called the Eightfold way, which corresponds to the adjoint representation of SU(3). Although the quarks can be arranged in this table (think of it as being a kind of "periodic table"), the mass symmetry is less exact, since the strange quark is much heavier. However, the fact that there is a table, and the fact that the table resembles the adjoint rep of su(3), is one of the ways in which physicists "discovered" quarks: The table makes a strong prediction: namely "all" particles must fit into the table (and to a point, they do). Since the octet is the adjoint rep, which is reducible, the search was on for particles that belong to the fundamental representation, and these turned out to be the quarks.
The table in fact makes much much stronger predictions than just mass symmetries: it also states which particle decays are allowed, and which are forbidden. This is actually the true strength of the table, and the whole notion of flavour symmetry.
The first particle that failed to fit into the su(3) flavour symmetry table was j/psi, which lead to the charmed quark and a Nobel prize. The table was extended to su(4). We now have top and botton, which make up an su(6) flavour symmetry. There are strong theoretical reasons to understand that things stop there; there won't be a 7th or 8th quark.
As mass symmetries, the su(3) flavour symmetry is worse than sub-space that is su(2) isospin symmetry, because the mass of the strange quark is not so small. The mass symmetry gets worse when the charmed quark is included to make su(4). In short, mas flavour symmetry is "badly broken". however, the actual prediction that "all mesons must fit into the adjoint (or higher) representation of su(6)" still stands. This prediction makes strong statements about which particles can decay into what, so in that sense, there really is a real, honest-to-goodness su(6) something in there (with isospin su(2) symmetry being and important, low energy subpiece of it all). Isospin, and more generally flavour symmetry will remain "fundamental" until something is found to replace the Standard Model. linas 15:38, 29 November 2005 (UTC)
Ok, so I understand this. But in the article on hypercharge it says the flavors are strangeness, charm, bottomness and topness. Are the later two really isospin? If so, should that article be changed? Or is hypercharge simply a mathematical construct of some utility that doesn't really have a physical reality?
One interesting comment in the weak isospin article is that it seems to be implying it is the presense of isospin that stops a quark decaying into itself. It is not clear in that article why this is, and it is not mentioned in "this" article at all. Perhaps this is something to add, if it is the case?
I also gather, from your notes above, that the reason they thought their was an isospin was due to the fact that the proton and neutron are otherwise so similar. Normally this would make them degenerate in some sense? Only by introducing the isospin, and having different values, did they "become" separate particles. If I understand this correctly, it would seem a sentance to that effect would be useful as well.
Maury 22:10, 29 November 2005 (UTC)

Don't confuse up and down with top and bottom. In order of increasing mass, the six quarks are: d,u,s,c,b,t. Isospin refers to u,d only, not t,b.

Just a minor correction, the order of increasing mass is u,d,s,c,b,t, as the d is more massive than the u.

When dealing with physics, there is damned litle difference between "mathematical construct" and "physical reality", one cannot perceive the second without employing the first. Isopsin is a construct that mirrors reality.

This article is about "strong" isospin, the isospin that applies to the mass eigenstates for the strong interactions. Unfortunately, these are not eigenstates of the weak interaction, and no one knows why, and there is no convincing theory that talks about this. The mixing between these is given by the CKM matrix. I'll add a sentence to this effect shortly.

Yes, the proton and neutron are (almost) degenerate; they have nearly the same mass. The "only" difference between them is that one is charged. I'll add a sentence emphasizing this. linas 00:12, 30 November 2005 (UTC)

OK, so this article basically sucked. I completely re-wrote it and expanded it, and I hope that now your questions are clearly addressed. linas 02:28, 30 November 2005 (UTC)
If you have any other comments, questions, criticisms, etc. please do state them, as now would be a good time to polish this article and make it as clear as I can make it.linas 04:08, 30 November 2005 (UTC)

## strong theoretical reasons

What are the "strong theoretical reasons" to expect that things stop at flavour SU(6)? As far as I know, there is no theoretical input on the number of generations in the standard model, only experiemental reasons like Z-decay, which only rules out light extra generations. -lethe talk 13:32, 19 December 2005 (UTC)

Not sure. The argument I remember was indeed the one-loop amplitude for Z-decay, which was explictly proportional to N_f, the number of flavour generations. The argument, as I remember it, was that, even for large multi-loop corrections, there was no way to fit the experimental data with N_f greater than 3. But I don't remember it well; presumably it said something like "with extra generations whose mass was below the GUT scale". It would be nice to have a wiki aricle that stated the one-loop Z decay amplitude, and clarified this. Hmm. The wiki article on flavour doesn't even mention this (that N_f is almost certainly 3). linas 18:48, 10 January 2006 (UTC)

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## Corrections to wave functions

I recived the following email, which I replied to:

Hi Quilbert,

On Sat, Jan 13, 2007 at 04:20:48PM +0000, Quilbert wrote:
> Hi!
> To me, the proton and neutron wave functions in the article "Isospin"
> don't make sense. What you state (for the proton) would just be
> (|duu>  + |udu> + |uud>)/sqrt(3), wouldn't it? But that, in my eyes,
> is the Delta+ wave function. I wanted to change the article but then
> saw you're a PhD and seem to be expert on the field. So maybe I am wrong.
> Why do you think that is the correct state? Just apply the I+ operator,
> it doesn't yield zero, and thus N wouldn't be an isospin doublet.
> With kind regards,
> Quilbert

The right place to ask this question would be on the talk page
associated with the article.

I suppose what is written there is confusing, or at least
misleading. -- where it says "+ perms", I did not mean to imply
that a + sign should be taken for all permutations. Yes, the all-plusses
case would be a Delta. A minus sign is needed, one would need
something like (|uud> - |udu> + |duu>) or something
like that; I'm too lazy to check and verify that this gives
all the right quantum numbers for all the different cases.
(And matches the sign conventions used by popular textbooks).

If you are a grad student, and you are studying this, then
please do compare this to what the textbooks say, and make
the needed corrections.

--linas


linas 21:13, 15 January 2007 (UTC)

Hi, yes that was my email, and now I have changed the wave functions in the way I think they are correct. Does everyone agree? At least now, if you apply ${\displaystyle I_{+}}$ or ${\displaystyle S_{+}}$, you get zero, since the sum of all entries is zero for all rows and columns. The other possibility for this would be the matrix ${\displaystyle \left({\begin{array}{ccc}0&-1&1\\1&0&-1\\-1&1&0\end{array}}\right)}$, but I think that is another doublet state, maybe the N(1440)? From what I've read in different places I'm pretty sure the matrix I put into the article is the correct one.

Hi, As a 4th year physics undergrad trying to get some reasonable understanding of isospin, the presentation in this section is not really sufficient. Could you maybe elaborate on how exactly you arrive at this formulation, or at least provide a link to a more detailed explanation?

Thanks very much.

Hi, I'll try to clarify: The states presented are elements of the product space of an eigenspace of ${\displaystyle I_{3}}$ (proton: eigenvalue ${\displaystyle +{\frac {1}{2}}}$) and an eigenspace of ${\displaystyle S_{3}}$ (here: eigenvalue ${\displaystyle +{\frac {1}{2}}}$ chosen). There are nine basis vectors in this space, which correspond to the nine matrix cells. Now we can elaborate some restrictions for the matrix of the actual state:
Given all these restrictions (${\displaystyle 1+3+3+2=9}$ for both cases; actually, the "1" is only a real equation, allowing for an arbitrary complex phase on the final results), we arrive at two possible matrices: one given in the article with symmetrical spatial part, which for some wave function is the ground state of the proton, and one given above (normalization factor ${\displaystyle {\frac {1}{\sqrt {6}}}}$ to be added) with antisymmetrical spatial part, which corresponds to some excited states.
Note that you can also derive the matrix by symmetrizing ${\displaystyle |u\rangle |\uparrow \rangle ={\frac {1}{6}}(2|duu\rangle -|udu\rangle -|uud\rangle )(2|\downarrow \uparrow \uparrow \rangle -|\uparrow \downarrow \uparrow \rangle -|\uparrow \uparrow \downarrow \rangle )}$ with matrix ${\displaystyle {\frac {1}{6}}\left({\begin{array}{ccc}4&-2&-2\\-2&1&1\\-2&1&1\end{array}}\right)}$. Please confirm whether this explanation was sufficient. --Quilbert 07:36, 10 July 2007 (UTC)

## Isospin symmetry states of the pion

The isospin properties of antiquarks have not been explained. This is probably quite important, since it leads to a different form for the pion triplet states than you might expect in the analogy to dynamic spin. The assignment of isospin to the antiquarks is such that it allows quark/antiquark combination states to have the same isospin transformations.

I haven't verified, but the assignment given in Halzen and Martin which allows this is ${\displaystyle I_{3}(+{\frac {1}{2}})=-\vert {\bar {d}}\rangle }$ and ${\displaystyle I_{3}(-{\frac {1}{2}})=\vert {\bar {u}}\rangle }$.

Perhaps this needs to be clarified?

OlekG 17:37, 17 May 2007 (UTC)

## Isospin symmetry of quarks

Hi I wonder if anyone can help? In the section "Isospin Symmetry of Quarks"(part shown below), The Proton (duu, udu, uud) the Neutron (udd, dud, ddu) this makes sense but the values given in the statement to the right, are identical for both the proton & neutron, also the sets of up and down arrows appear to be identical for p & n is this an error? or am I missing something fundamental.

"In the framework of the Standard Model, the isospin symmetry of the proton and neutron are reinterpreted as the isospin symmetry of the up and down quarks. Technically, the nucleon doublet states are seen to be linear combinations of products of 3-particle isospin doublet states and spin doublet states. That is, the (spin-up) proton wave function, in terms of quark-flavour eigenstates, is described by

and the (spin-up) neutron by

Here, is the up quark flavour eigenstate, and is the down quark flavour eigenstate, while and are the eigenstates of Sz. Although the above is the technically correct way of denoting a proton and neutron in terms of quark flavour and spin eigenstates, this is almost always glossed over, and these are more simply referred to as uud and udd."

If anyone could help, I would be grateful Rowly (talk) 19:26, 7 February 2008 (UTC)

This is not an error. The states described are a mixing of z-component isospin eigenstates and z-component spin eigenstates. The way the states are mixed (the matrix) is identical for protons and neutrons. Both are considered spin-up, so the z-component spin eigenstates are also identical. Only the z component of the isospin is different (eigenspace to the left of the matrix), thus the quark composition. Please come back if you still have questions. Greetings --Quilbert (talk) 04:24, 10 February 2008 (UTC)

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This article has been delisted from GA due to a lack of inline citation. Note that the good article criteria have changed since this article was promoted and inline citations did not used to be required. I've passed the article on to the unreferenced GA task force; once it is adequately sourced, it should be renominated.--jwandersTalk 00:32, 13 February 2008 (UTC)

Two neutrons have not Coulombic repulsion. But they have attraction through the strong nuclear force when they approach each other in a distance of the order of 2x10-15m. So, as there is not repulsion, but there is attraction between two neutrons when they interact in distances of the order of 2fm, it means that two neutrons would have to form a dineutron 0n2, and would never have to separate. And the Universe would have to be fulfilled by a great quantity of dineutros.

However a dineutron does not exist in Nature, because the dineutron is never formed. Why?

## Heisenberg’s solution

Heisenberg tried to explain it by proposing the concept of isospin. Unfortunatelly the isospin is a pure mathematical concept, and Heisenberg’s proposal actually is unacceptable. Indeed, as two neutrons in the structure of the dineutron would agglutinated by a force of attraction, obviously such a force of attraction must be surpassed by a 'force of repulsion, in order to separate the two neutrons. Because only a force of repulsion can neutralize the effect of a force of attraction.

A pure mathematical concept, as the isospin, cannot produce a force of repulstion. Heisenberg’s mathematical solution only describes what happens between two neutrons. His solution does explain why it happens. In short: Heisenberg’s solution does not show what is the cause responsible for the fact that a dineutron is never formed.

## An acceptable solution

An acceptable solution capable to explain why the dineutron is not formed must consider a model of neutron in which, in the distances of the range of 2fm, there appears a force of repulsion between the two neutrons. Such model of neutron must be able to point out the physical cause capable to produce a force of repulsion.

A neutron model, to be acceptable, must be able to explain the origin of the force of repulsion, responsible for the fact why the dineutron does not exist in Nature.

## Heisenberg’s scientific criterion

Heinseberg tried to expell any metaphysical idealism from the process of scientific discovery:

“But later Heisenberg is very clear about avoiding any metaphysical idealism. In "The Copenhagen Interpretation of Quantum Theory" in Physics and Philosophy he states explicitly that quantum theory does not contain genuinely subjective features, since it does not introduce the mind of the physicist as part of the atomic event, and that the transition from possible to actual in the act of observation is in the physical and not the psychical act of observation”. http://64.233.169.104/search?q=cache:SgRagcpjAYQJ:www.philsci.com/book4-3.htm+heisenberg+metaphysical+viepoint&hl=pt-BR&ct=clnk&cd=8&gl=br

According to Heisenberg, it’s not the aim of science to discover the hidden mechanisms underlying the phenomena, as one could wish, motivated by a personal idealism supported by a trust in the causality principle. He stated that to look for physical mechanisms is metaphysics, and must be eliminated from the process of scientific research. From his viewpoint, only observable quantities are of the interest of science.

Paradoxically, Heisenberg’s criterion sometimes brings results disagree to experimental results, as for instance the case of the electron’s orbits: while the experience of the chamber fog shows us that electron’s trajectory indeed exists, Heisenberg proposed to eliminate the concept of trajectory in Quantum Mechanics, because he did not believe in the existence of electronic orbits within the atom.

Such Heisenberg’s hypothesis (that there no exist electron’s orbits within the atom) is disagree not only to the chamber fog experiment. It’s disagree to the experiment made by the Nobel Laureate Hans Dehmelt. His experiment shows that the electron peruses the space between to orbits into the electrosphere of atoms, a result disagree to Quantum Mechanics (disagree to Heisenberg’s criterion).

## Discussion with Einstein in 1926

A book( 1 ) reproduces a Heisenberg’s lecture in 1968, where he spoke about his scientific criterion. And he tells his discussion with Einstein in 1926, when the father of relativity asked him to explain the philosophy of Quantum Mechanics:

- What was the underlying philosophy to such strange theory? The theory seems to be very beautiful, but what do you mean with observable quantities?

- I don’t believe so long in the existence of electronic orbits, in spite of the traces within the chamber fog.

And Heisenberg said that he felt the need to go back to the quantities that, indeed, could be observed. Because that was the sort of philosophy used in the relativity too: Einstein abandoned the absolute time and introduced only the time in the own co-ordenate system.

Einstein laughed, and said:

- But you have to understand that you are completely wrong

And Heisenberg replied:

- But why? Isn’t true that you did use such same philosophy?

- Oh, yes – answered Einstein – I may have used it, but even so it’s a nonsense.

## Repercussions of Heisenberg’s scientific criterion

Since the begginning, Heisenberg’s scientific criterion was disagree to some experimental findings, as for instance the existence of electron’s trajectory within the chamber fog. Also, it’s also obvious that something is missing in his solution concerning the dineutron, because from an abstract mathematical concept as the isospin one cannot explain the origin of the repulsion force that does not allow a dineutron to be formed from the interaction by the strong force.

Heisenberg’s criterion was successful for the development of technology. However probably it can fail in a most deep level or research. Such conclusion is obvious, because since the begginning Heisenberg’s criterion did not consider some fundamental mechanisms existing in Nature, as for instance the electron’s trajectory, and the repulsive force within the neutron.

But along the 20th Century all the theories of Modern Physics have been developed by following the Heisenberg’s criterion. Among them we can mention the prevailing theories, as the Standard Model, the Supersymmetry and the String Theory. Today the physicists hope to get experimental confirmations for those prevailing theories, when the LHC will start to work.

The question is: will the Heisenberg’s scientific criterion succeed in such depper level of research?

Or there will need to start everything again, from the zero, by considering a new scientific criterion?

## Successor of Quantum Mechanics

As consequence of the application of Heisenberg’s criterion to the development of Quantum Mechanics, of course would be impossible to avoid several inconsistences in the theory. It’s easy to understand why. After all, it makes no sense to hope that a theory, born in contrast with some experimental findings (as for instance the existence of electron’s trajectory within the chamber fog) could be entirelly successful.

So, along the 20th Century the theorists tried to eliminate some inconsistences of Quantum Mechanics. However they have applied the same method responsible for the inconsistences: the Heisenberg’s method. So, instead of to look for new fundamental concepts missing in Quantum Mechanics, the theorists developed its successor, the Quantum Fields Theory, by keeping the fundamental concepts established for Quantum Mechanics from the Heisenberg’s criterion.

In another words: instead of looking for new fundamental concepts that Heisenberg’s criterion was unable to discover, the theorists try to eliminate the inconsistence of Quantum Mechanics from the manipulation of the mathematics, believing that the mathematics is able to make better a theory from which are missing some fundamental principles existing in Nature, in the same way as Heisenberg tried to solve the problem of the dineutron, simply by neglecting the causality principle and proposing a pure mathematical concept, as the isospin.

The question is:

1- Keeping the Heisenberg’s criterion, as the theorists did up to now, is there chance to eliminate the inconsistence of Quantum Mechanics?

2- Or is there need to start everything again, by a different scientific criterion?

The book Quantum Ring Theory( 2 ) tries to respond these two questions.

## Rules for making baryons

I'm kinda confused about how baryons are made up. All I can find is that baryons are made of up three quarks. Since there are 6 kinds of quarks, shouldn't there be 6^3=729 distinct baryons (going by quark composition)? Not only that, but the delta+ and the proton have the same quark make-up, the only difference are the different spin states. Wouldn't this mean that there are even more distinct particles since we have to keep tract of the spin alignments of the baryons with u and d quarks? Headbomb 20:43, 21 March 2008 (UTC)

Don't you mean 6^3=216 baryons? I don't know the answer. SkyLined (talk) 20:50, 21 March 2008 (UTC)

Alright I've looked into this. It seems (from what I can gather) that I was on the right track. It looks like baryons are made from 3 quarks, any quarks. Since there are 6 different quarks, then we have 6^3 combinations of 3 quarks. However, from the Delta+ and the proton, each with quark composition u/u/d, it looks like the spin orientation have to be taken into account. Since each quark can be in +1/2 or -1/2 isospin state, then we have 12 different quarks/quarkstates possible for each of the three quarks, which gives us 12^3 different combinations of three quarks. If we remove the degeneracies (such as ssd (3/2),sds(3/2),dss(3/2)), then we have 364 (12+11+10...+11+10+9...+10+9+8+...3+2+1+2+1+1) distinct combination of quarks/quarkstates.

Now I'm not sure of this, but I think that it is the modulus of the spin that is important, so particles with spin -3/2 and -1/2 really are the same than the particles with spin 3/2 and spin 1/2. Removing these degeneracies leaves us with half the particles, and thus there are 182 distinct baryons that can be made from three quarks.

Did I understand it correctly? Headbomb 21:41, 22 March 2008 (UTC)

It's far more complicated than that. Saying that the quark content of a baryon is xyz is shorthand that hides a lot of detail. For the uds system, there are three orthonormal states: ½(usd - sud + dsu - sdu), 1/sqrt(6)(uds - usd - dus + sud - sdu + dsu), and 1/sqrt(12)(usd - sud + sdu - dsu + 2dus - 2uds). Counting hadrons is most easily done with SU(n) multiplets, but I don't fully understand them, so I can't explain it. In addition there are all the excited states as well. Finally, it's a bit academic to count the top quark in these calculations—it will decay to W+b long before it has a chance to hadronize.Mjamja (talk) 13:40, 18 April 2008 (UTC)
Yes, I've read a bit on the topic since, and the update by Wing Gundam help me understand things better. Isospin is what matters when differentiating particles of the same quark makeup. Now to understand what the hell Isospin is...
In any cases, for baryons made of 3 quarks of 6 flavors, there are (6+5+...)+(5+4+3+...)+(2+1)+(1)=56 distinct quark makeup. And while counting the t-quark baryons may be academic, I'm trying to understand the rules for making baryons are, rather than what baryons may be found in nature.

Also would be interesting if someone updated the list of baryons to take into account all the possibilities, with placeholders for the undiscovered particles Headbomb 22:12, 22 March 2008 (UTC)

I don't think it's scientific to declare a baryon before one is actually observed in an acclerator... Wing gundam (talk) 17:25, 17 April 2008 (UTC)

Are you saying is that we should not write about anything that is predicted/expected/suspected to exist but which has not been proven through laboratory testing? That doesn't sound very scientific to me ;) -- SkyLined (talk) 08:08, 24 April 2008 (UTC)
I agree in a certain way to wing gundam: Probably we should have different tables for observed and predicted (and not yet observed) SM baryons.. Tatonzolo (talk) 12:41, 24 April 2008 (UTC)
You are right in that it should be clear which particles have been proven to exist and which are predicted to exist, but have not been seen. Having two tables is a valid option, but putting everything into one table and marking them in some way is a valid alternative. I think the choice between these two options is a trade off between making it clear which baryons are similar (=one table, putting simililar baryons close together) and making it clear which baryons have been proven to exist (=two tables, seperating proven and non-proven baryons). I prefer one table, with a clear marker to indicate which particles are proven and which are not, but I have no argument other than personal preference. If we go with that option, we should put a remark ABOVE the table, so people know what to expect, rather than as a footer (which may not get read). -- SkyLined (talk) 14:27, 24 April 2008 (UTC)
concerning the clarity of the Baryon table I would suggest to make some "multilines" for the names, having Delta repeated lots of times is not beatiful, and possibly the p/n/n+ and n/n/N0 notations are quite unused... this was on the aesthetical side.. on practical side it would be useful to group the Baryons "a-la PDG" grouping them by flavour quantum numbers.. The mass scale for the Baryons would be mmore understandable to the less experts and the resonances would be better fitted... Tatonzolo (talk) 15:13, 24 April 2008 (UTC)
I think dividing them into unobserved and observed tables would clutter up the page and we'd lose part of the benefits of grouping particles together in the table. Also unobserved particles have daggers next to names to indicate exactly that, and have their masses, decay, lifetimes, and references missing. We could always add (unobserved) next to their names, I guess (removing daggers). Or we could add the move dagger note at the top of the table, (this option gets my vote for now).
Also I've list the p/p+/N+ because even though they are rarely used, they are still used. Perhaps we could add a note that p and n are the most commonly used symbols, with p+ and n0 trailing behind. As a side note, I've used p+ and n0 everywhere to indicate the charge in decays and whatnot, so it's easier to see charge conservation.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 16:40, 24 April 2008 (UTC)

### Take 3

Alright. I think I have it.

u and d quarks each carry isospin 1/2. This is why isospin can be either 3/2 (3 aligned u and/or d quarks), 1 (2 aligned u and/or d quarks), 1/2 (2 aligned u with one unaligned d, 2 aligned d with one unaligned u, or 1 u or d quarks) or 0 (unaligned u and d quarks, or 0 u or d quarks). Three unaligned u or three unaligned d is forbidden by Pauli, and so is two unaligned u or two unaligned d.

Isospin 3/2 baryons are the 4 Deltas (uuu, uud, udd, ddd) Isopin 1 baryons are the 12 Sigmas (uus, uuc, uub, uut, uds, udc, udb, udt, dds, ddc, ddb, ddt) Isospin 1/2 baryons are the two nucleons (uud, udd), and the 20 Xis (uss, usc, usb, ust, ucc, ucb, uct, ubb, ubt, utt, dss, dsc, dsb, dst, dcc, dcb, dct, dbb, dbt, dtt). Isospin 0 baryons are the Lambda (uds, udc, udb, udt) and the 20 Omega (sss, ssc, ssb, sst, scc, scb, sct, sbb, sbt, stt, ccc, ccb, cct, cbb, cbt, ctt, bbb, bbt, btt, ttt).

wrong. again. (don't mean to be harsh, but...). isospin is a number unique to baryons. however, the biggest mistake i notice is that you're listing baryons containing top quarks, which don't even exist! (see the note i left on your talk). There are no true rules for 'baryon making', whatever the hell that even is. The list, at this moment, is actually complete, provided no-one attempts to stuff it with theorical baryons, some of which may not exist. Really, the only thing this article needs is to lose some dead fat.
For example, the section "Relation between isospin and up and down quark content" is completely irrelevent to the present subject matter, by which i mean to say an explication of the fine details of isospin has no place in a list of baryons. On a side note, yet equally valid point, this section is full of gross errors in its basic understanding of particle physics, making elementary mistakes of comprehension in its explanation of isospin. This is meaningless, however, as it already does not belong on a list. The section currently labeled overview suffices to perform its function of acting as a legend, and, in truth, i think it looked much better before it was even split from the introductory paragraph. This is, after all, a LIST.Wing gundam (talk) 02:16, 29 April 2008 (UTC)

Well, I know that PDG lists that u quarks have isospin 1/2 and that d quarks have isospin -1/2, so isospin cannot be unique to baryons. Especially considering that the pi mesons form an isospin triplet. As for t quarks, I've sort of giving them a "magnetic monopole" treatment (see my talk page). I'm not opposed to not listing them, but it would be nice to know where they would fit, were they to have a bigger lifetime.
And there must be a rule for what baryons can exist, else they wouldn't show patterns such as [1]. Stuff like SU(3)xSU(2)xU(1) or SU(6) represents something and obey some rules. I'm trying to figure what the hell those rules are, and there are no books out there that seem to care to explain things to people that don't know Lie algebra, and the books on Lie algebra aren't written to be understood. At least tell me where I'm wrong, or how I'm wrong. All I know is with the rules I gave, I can reproduce every baryon diagram I encountered, and I can "predict" every baryon listed in the PDG. I can think of no reason why Template:SubatomicParticle (observed) could exist and that Template:SubatomicParticle(unobserved) could not. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 15:11, 29 April 2008 (UTC)

Perhaps I can offer a historical explanation of Isospin to help in general understanding. Mesons and Baryons occur in multiplets. Members of the same multiplet have similar masses and differ in charge number by unit steps. Also the interactions of different members of a multiplet do not depend strongly on their charge. For these reasons each member of a given multiplet can be regarded as a different charge state of a single particle which has an extra degree of freedom in an internal space -- isospin space. The number of possible orientations of a particle in isospin space is 2I + 1. Thus the isospin quntum number of a particle may be determined simply by finding out how in how many charge states it can exist.--Vectorboson (talk) 20:30, 1 May 2008 (UTC)

Also, on the issue of rules for a baryon to exist. The first rule is that it will exist unless there is a reason it cannot. In the case of a baryon containing the top quark, calculations have been done to estimate the time it takes for a baryon to form and the result is that it takes longer for a baryon to form than it takes for the top quark to decay (about 10 to -24 seconds). --Vectorboson (talk) 20:30, 1 May 2008 (UTC)

### Old rules vs. new rules

[copied from Vectorboson's talk page]

With my old rules, I could make every baryons out there with no extra baryons. The rules were quarks of the same flavor must have their isospin aligned, and quarks of different flavor can, but need not, have their isospin aligned. See Talk:List of baryons#List Progress Overview for the list of particles and their corresponding isospin values it gave me.

Now if I go with the PDG rules; that I and Iz are additive numbers and that I = 1⁄2 for u and d quarks and that Iz = 1⁄2 for u and −1⁄2 for d, then I can't account for nucleons (can't get isospin 1⁄2 with three u or d quarks, and Lambda's (can't get isospin 0 with a u and d quark).

First, I-spin is NOT additive, Iz IS additive-- so forget about I-spin for a minute and concentrate on Iz. When constructing composite particles, Iz is the additive quantum number. A proton has two up quarks and one down quark -- the Iz values add to 1/2. The neutron has two down quarks and one up quark-- the Iz values add to -1/2. I-spin doesn't have a direction in real space, so I-spin CANNOT "align" as can spin.
Lambda's have one up and one down quark (total Iz = 0) plus another quark with Iz = 0 -- total lambda Iz then = 0.
Did that help?--Vectorboson (talk) 01:01, 4 May 2008 (UTC)

I'm fine with Iz, it's the isospin itself I have a problem with (BTW PDG lists the Isospin as an additive number, table 14.1 PDG quark model, if that's a mistake we'll need a reference for the article).

I thought isospin was a vector, just like spin is, and that as such it had a length (I) and a direction that could only be probed on one of the component (Iz by convention, sometimes noted I3) because the components did not commute.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 01:23, 4 May 2008 (UTC)

I noticed that JRSpriggs tried to bring tensor algebra and possibly some group theory in this, so just as a note, I don't get tensor algebra, tensor products, lie algebra, group theory etc... at ALL. So any attempt to explains things going these way will be lost on me.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 13:45, 4 May 2008 (UTC)

Certainly, the mathematical formalism for dealing with isospin parallels that of real spin and that is what led to its unfortunate name. But emphasize that any rotations and projections you imagine are NOT in x-y-z space but in some internal space. I will try and find you a reference for only the z-component of isospin being additive.--Vectorboson (talk) 15:10, 4 May 2008 (UTC)
Now some thoughts about the latest changes to this section. It is factually correct, and I like the table idea, but it should express the POINT of isospin. Each isospin singlet, doublet, triplet or other multiplet is, as far as the strong interactions are concerned, the same particle in a different electric-charge state.
In presentation, these multiplets should be shown together (perhaps, each should have its own table?). Adjacent states always differ by one unit of electric charge. And the number of states defines the isospin with this relationship
I = 1/2(NumStates -1 )
And according to the generalized Gell-Mann / Nishijima relationship, the z-component of isospin is
Iz = Q – 1/2( BaryonNum + S+ C + B + T)
So for example, the delta particle comes in 4 charge states, so it has isospin equal to 3/2
And
Iz of Delta++ = +3/2
Iz of Delta+ = +1/2
Iz of Delta0 = -1/2
Iz of Delta- = -3/2
At this point let’s emphasize what spin (and parity) has to do with this.
If we consider three-quark combinations of the up and the down quark. Since the quarks have spin = ½ (real spin, NOT isospin) the three-quark combinations can have spin =1/2 or spin = 3/2.
The Delta baryon has spin = 3/2 while the nucleons have spin = ½. So REAL spin is used to help classify to which family a charge state belongs.
Also, let’s point out that since the quarks have a color charge, and inside a baryon, the three quarks all have different color, so the Pauli Exclusion Principle is not violated. Historically, it was the delta++ with its three up quarks – all in the same spin state – that made us realize there were three colors in the strong force.--Vectorboson (talk) 15:10, 4 May 2008 (UTC)

Yes that's fine an all, but that still hasn't answer the question. What are the rules of baryon making? Why can't there be a I=0 uus baryon? Why can't there be a uuu baryon with I=1/2?

Perhaps if I ask my question different.

In terms of quark content (u is numbers of u quarks, d is number of d quarks etc...) The rule for charge of a particle is

${\displaystyle Q=-{\frac {1}{3}}(d+s+b)+{\frac {2}{3}}(u+c+t)+{\frac {1}{3}}({\bar {d}}+{\bar {s}}+{\bar {b}})-{\frac {2}{3}}({\bar {u}}+{\bar {c}}+{\bar {t}})}$

The rule for the baryon number of a particle is

${\displaystyle B={\frac {1}{3}}(u+d+s+c+b+t)-{\frac {1}{3}}({\bar {u}}+{\bar {d}}+{\bar {s}}+{\bar {c}}+{\bar {b}}+{\bar {t}})}$

The rule for the z component of isospin of a particle contains at least this term

${\displaystyle I_{z}={\frac {1}{2}}(u-d)}$

And because I strongly suspect that there should be some symmetry with the above equations, I think that the full equation is

${\displaystyle I_{z}={\frac {1}{2}}(u-d)-{\frac {1}{2}}({\bar {u}}-{\bar {d}})}$

Writing this in terms of quantum numbers B (as given above), Iz (as given above), S (Template:SubatomicParticle − s), C (c − Template:SubatomicParticle), B*prime; (Template:SubatomicParticle − b), T (t − Template:SubatomicParticle) is a bit tricky since u and d quark content dependance is not explicit, but really isn't all that hard to do since since you can get from B, Iz, S, C, B, and T to u and d quark content with simple algebra. I'll remark that the Iz looks rather artificial detracts from the fundamental understanding of things; it really looks to me as nothing more than a historical leftover from the particle zoo. It would seem infinitely more natural to have defined U (u − Template:SubatomicParticle) and D (Template:SubatomicParticle − d) quantum number, and while we're at that, we might as well have defined positive U,D,S,C,B,T quantum numbers to reflect the quark content, rather than quark for u-type and antiquark for d-type quark, but I disgress. Anyway, when that's done, you end up with

${\displaystyle Q=I_{z}+{\frac {1}{2}}(B+S+C+B^{\prime }+T)}$ (Gell-Mann–Nishijima formula‎)

or in terms of the hypercharge (Y=S+C+B′+T)

${\displaystyle Q=I_{z}+{\frac {1}{2}}(B+Y)}$

Now in all of this, we never tackled isospin (either its length or the vector). So what is it? What's the rule in terms of quark content? What does the axis represent? What does isospin 3/2 mean? What does isospin 1/2 mean?[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 16:31, 5 May 2008 (UTC)

After friggin' around with combinations of three quarks (uds, udc, udb, udt, scb, sct) I finally got what isospin was. And all I have to say is ... wow. That people still use it is beyond me. It's completely useless, and it's completely counter intuitive once you understand the quark model. Altough I shouldn't rejoice too quickly because I can't make the Lambdas fit.

You are being very harsh with history. Working from the bottom of your last entries toward the top... ANY conserved quantum number was (and is) a valuable insight into clues about how things work. Before we knew about quarks, we knew that Isospin and z-component were both conserved by the strong interactions. That explained branching ratios of decays and explained the symmetries in experiments where we collided pi+ with neutrons and pi- with protons. Your criticism is much like asking why do we need Newton's ideas now that we have Einstein's. That helped us understand that nucleons and other hadrons had a symmetry in an inner space that could be mathematically dealt with with a rotational formalism -- that was BIG! Internal symmetries are actually much easier to deal with than external symmetries which must be MEASURED to know. Now that we know about quarks, the only thing that has changed is that instead of referring to nucleons, isospin applies directly to quarks.

Your criticism about what "inner-space" or "axis" we are talking about and what does isospin 3/2 or 1/2 MEAN is actually to answer. When we predicted which particles should exist, we needed a scheme -- a grouping -- that showed which discovered particles fit in where and which missing particles we needed to search for. The diagrams that demonstrate these families are among the detritus that you have already rejected for inclusion in this article...
http://www.fnal.gov/pub/inquiring/physics/discoveries/images/Baryon%20Chart_MR.jpg
The formalism is beautiful and perfect and it does still need isospin much as the periodic table still needs the spdf nomenclature that defines energy shells

Finally your first questions "Why can't there be a I=0 uus baryon? Why can't there be a uuu baryon with I=1/2?" are also easily answered. As I indicated before, START by adding up the Iz components and you will get your answers. A uus baryon would have z-component = 1/2+1/2+0 = 1 ...so it CANNOT be part of an I=0 family. Similarly a uuu baryon has Iz=3/2 so it CANNOT be part of an I=1/2 grouping.--Vectorboson (talk) 23:05, 5 May 2008 (UTC)

I applaud this (and any) effort to explain difficult science in layman's terms. But while it is usually possible to explain the "facts" this way, it is often impossible to explain how we know what we know without resorting to difficult mathematical formalism. The Pauli Exclusion Principle for example is easily stated in words, but trying to explain why an anti-symmetric wave function vanishes for fermions and what that has to do with the matter is much more difficult. This article describes the general quark content of the baryons, but it does not address the mixing ratios of the various quark states involved or the symmetrization of the wave function which, by the way MUST take into account the internal symmetries of the quarks.--Vectorboson (talk) 23:19, 5 May 2008 (UTC)

I am not harsh with history. The way I finally figured out was by going through history and putting myself through the minds of those who didn't have the c,b,t quarks to work around with. It made sense back then to think that the neutron and proton were "variation of the same particle". It made sense to think that all deltas were the same particle, and that the different charges were the result of being in different states. Isospin had merits back then, and whoever came up with it gave nature a hell of a good shot and deserve a pat on the back for coming with a simple way of organizing particles and interpreting the meaning of the organization with the current knowledge at the time. However now, in light of the quark model, it doesn't make sense to think of the proton and neutron as the same particle, and isospin doesn't help to understand how particles are related. So I am harsh towards this generation of particle physicists because they did not have re-written the formalism in a more natural and comprehensible way.

And I also object to your characterization of my criticism of isospin as being analogous to a criticism of Newton. Newton is an approximation of Einstein that applies at low gravity and at low speeds, so it is worth more than simply being an old way of understanding things, it is also useful and connects with the everyday world. The meaning of "speed, energy, momentum" etc.. isn't hidden in Newton.

Isospin however, hides the true nature of things in favor of a artificial constructs. It is very unnatural to express charge in terms of baryon number and projection of isospin rather than in terms of quark content. It is equally unnatural to classify particles in groups of isospin rather than in groups of quark content. Doing so is neither an "approximation" of reality nor does it help anyone to understand anything even approximetaly. An equivalent analogy of me criticizing the concept of isospin would be someone else criticizing the classification of the the chemical characteristics of elements in terms of atomic mass, neutron number and a new quantity called "chemi-spin" that would be defined in a Chemispin (C)= Atomic number(A) - number of electrons + 12" fashion rather than dealing in terms of electronic configurations directly.

Imagine the kind of mess we'd be dealing in chemistry if we spoke of Element A=71/N=40/C=14 that shows similarity to Element A=69/N=38/C=14 rather than speaking of Gallium-69++ and Gallium-71++. We could rewrite all the classification of chemistry using linear combinations of any number of quantities and it would work. But what a damn mess it would be. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 01:20, 6 May 2008 (UTC)

### Take 4

Isospin: Definition the greatest value of |Iz| within a group of specific non u and d quark content and of specific spin state.

Example 1: Baryons of spin 1/2 and with a single bottom quark as non-u/d quark content. There are three particle in this group, uub (Iz=1), udb (Iz = 0), ddb (Iz = -1). The greatest |Iz| is 1, and thus these three particles have isospin 1. Isospin 1 particles containing a bottom and a mix of two u or d quark content are bottom Sigmas.

Example 2: Baryons of spin 1/2 with no non-u/d quarks. There are two particle in this group, uud (Iz=1/2), udd (Iz = -1/2). The greatest |Iz| is 1/2, and thus these two particles have isospin 1/2. Isospin 1/2 particles made of a mixture of three u or d quarks are nucleons.

Example 3: Baryons of spin 3/2 with no non-u/d quarks. There are four particle in this group, uuu (Iz=3/2), uud (Iz=1/2), udd (Iz = -1/2), ddd (Iz=-3/2). The greatest |Iz| is 3/2, and thus these four particles have isospin 3/2. Isospin 3/2 particles made of a mixture of three u or d quarks are Deltas.

But you are almost there! -- What's hard to figure -- all the lamdas have Iz=0 and are singlet states. so by your own rules (please don't put your definitions on the article page), the lambda's are I=0.--Vectorboson (talk) 00:48, 6 May 2008 (UTC)

I can tell the difference between a proton and a delta (different spin state), and I understand why they have different isospin (Pauli removes uuu and ddd from spin 1/2, so isospin is reduced by 1). However, I can't tell the difference between a sigma0 and a lambda0, both are uds, and both have spin 1/2.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 01:24, 6 May 2008 (UTC)

## References

1- A. Salam, P. Dirac, W. Heisenberg, Unification of Fundamental Forces, Syndicate of the Press- University of Cambridge, 1990

2- W. Guglinski, Quantum Ring Theory-foundations for cold fusion, Bäuu Press, 2006

## False Claims in "Symmetry" Section

There are several overtly false statements in the section titled "Symmetry." Here is a quote of the problematic section:

Isospin was introduced by Werner Heisenberg to explain several related symmetries:

• The mass of the neutron and the proton are almost identical: they are nearly degenerate, and are thus often called nucleons. Although the proton has a positive charge, and the neutron is neutral, they are almost identical in all other respects.
• The strength of the strong interaction between any pair of nucleons is the same, independent of whether they are interacting as protons or as neutrons.
• The mass of the pions which mediate the strong interaction between the nucleons are the same. In particular, the mass of the positive pion (and its antiparticle) is nearly identical to that of the neutral pion.

In his 1932 article "Über den Bau der Atomkerne" (Zietschrift für Physik 77: 1-11), Heisenberg indeed introduced a formalism in which the proton/neutron distinction is treated by adding a two-dimensional "isospin" component to nucleon wavefunctions (although he doesn't use the term isospin, which was introduced by Wigner in 1937--Physical Review 51: 106-119). However, in this article Heisenberg does not relate isospin to the mass degeneracy of protons and neutrons, at all. Nor does Heisenberg even emphasize proton/neutron symmetry: he introduces a speculative Hamiltonian for a nucleus, constructed using Pauli matrices over isospin space, in which three of the five terms are not symmetric under proton/neutron exchange; in fact, he argues that this asymmetry explains the preponderance of neutrons over protons in large nuclei. Heisenberg's Hamiltonian is not isospin invariant, and he makes no suggestion that the strong force Hamiltonian should be isospin invariant (he doesn't even refer to the strong force). Furthermore, Heisenberg clearly has nothing to say about pions in this article, as the meson hadn't been predicted nor observed at this point (the prediction came in 1935, observation in 1947); there was simply no notion of a pion in 1932.

This section of the article contains inaccurate content and should certainly be updated. Historically, Heisenberg's introduction of isospin is separable from the development of the modern usage of isospin, i.e. to explain mass degeneracies by postulating an invariance of the strong force Hamiltonian. The latter was developed soon after Heisenberg's original work...the relevant contributions here would be articles by Cassen and Condor in 1936 (Physical Review 50: 846-849), the Wigner article mentioned above, and several articles by Nicholas Kemmer in 1937-1938 which extended the notion of isospin to mesons, and properly predicted the existence of a neutral pion.

Bdeen (talk) 08:08, 22 May 2008 (UTC)

Please feel free to make these corrections. I had added the incorrect text long ago, based on my faulty memory of schoolwork. Prior to my additions, the article was only a few sentences long, and waseven more incorrect. Sorry. linas (talk) 01:44, 31 May 2008 (UTC)

## Modern understanding of isospin

The entire section titled "Modern understanding of isospin" makes me very nervous, in part because it does not agree with Bdeen's comments immediately above, and partly because it includes pictures of octets etc. which, strictly speaking are about su(3) flavor symmetry i.,e. strangeness, and not about isospin. Yes, of course isospin is a subgroup/subalgebra of su(3) but this section gives me the willies by sort of confusing these multiple distinct notions. linas (talk) 01:44, 31 May 2008 (UTC)

Under the understanding that the isospin of Heisenberg is distinct from the modern usage of isospin in nuclear physics, I don't see the problem. And these pictures very clearly relate the relation between Iz and particles, which in turns is essential to understand to define what I is. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 05:52, 31 May 2008 (UTC)

I wrote it without reference, locking myself for over a month putting myself in the situation that at first there were only uds baryons around, but no knowledge of quarks, then quarks were discovered, then the c quark was found, etc. This is what I reconstructed. So history-wise this could be less than accurate.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 05:52, 31 May 2008 (UTC)

# Particle Physics Bias

It would be useful to include in this article a discussion of the distinction between isospin as a notational convenience and isospin as a concept which has dynamical implications. This is discussed, for example, in Griffiths .[1]. The example he uses is that the Hamiltonian for the deuteron can have terms that depend on the product of the isospins of the two nucleaons e.g. ${\displaystyle {I^{(1)}}\cdot {I^{(2)}}}$ so long as it is unchanged under isospin rotations (e.g. SU(2)). Without some kind of example like this introduction of the symmetry group SU(2) to describe isospin is very perplexing, especially to the novice. — Preceding unsigned comment added by Luriol (talkcontribs) 19:47, 6 April 2013 (UTC)