Talk:List of baryons

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Let's bring this to Featured List/Article status

Let's get to work folks.Headbomb (talk) 17:47, 20 April 2008 (UTC)[reply]

List of suggested improvements that remains to be done (or to be thrown away)

~~Add references for section on Isospin and quark content.~~ (Moved the section to [Isospin]) Headbomb
Decide what stays here and what goes on the baryon page. Headbomb
~~Possible diagram update/removal.~~ (Diagrams were integrated in overview text). Headbomb
~~List the remainder of the baryons you can make from six quarks, possibly excluding the t-quark because they do not hadronize~~ (Did not list baryons containing t quarks). Headbomb
~~Add some well-known resonances.~~ Wing gundam

(Decided to remove the resonances from here and expand the Sigma, Xi... pages to contain them instead. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 02:58, 23 April 2008 (UTC))

~~List exotic baryons.~~ Headbomb
~~Find decay mode references for the Delta(1232)s.~~ (PDG reference contains them, albeit in a cryptic way.) Headbomb

List Progress Overview

Bold means currently not in the list. Blanks could be particles, or could be forbidden states. x are forbidden states. See Rules for making baryons - Take 3 down this page.

Particles and Isospins
Makeup	Isospin 0	Isospin 1/2	Isospin 1	Isospin 3/2
uuu	x	x	x	Delta++
uud	x	Proton	x	Delta+
uus	x	x	Sigma+	x
uuc	x	x	Sigma C++	x
uub	x	x	Sigma B+	x
uut	x	x	Sigma T++	x
udd	x	Neutron	x	Delta0
uds	Lambda0	x	Sigma0	x
udc	Lamba C+	x	Sigma C+	x
udb	Lambda B0	x	Sigma B0	x
udt	Lambda T+	x	Sigma T+	x
uss	x	Xi0	x	x
usc	x	Xi C+	x	x
usb	x	Xi B0	x	x
ust	x	Xi T+	x	x
ucc	x	Xi CC++	x	x
ucb	x	Xi CB+	x	x
uct	x	Xi CT++	x	x
ubb	x	Xi BB0	x	x
ubt	x	Xi BT+	x	x
utt	x	Xi TT++	x	x
ddd	x	x	x	Delta0
dds	x	x	Sigma-	x
ddc	x	x	Sigma C0	x
ddb	x	x	Sigma B-	x
ddt	x	x	Sigma T0	x
dss	x	Xi-	x	x
dsc	x	Xi C0	x	x
dsb	x	Xi B-	x	x
dst	x	Xi T0	x	x
dcc	x	Xi CC+	x	x
dcb	x	Xi CB0	x	x
dct	x	Xi CT+	x	x
dbb	x	Xi BB-	x	x
dbt	x	X BT0	x	x
dtt	x	Xi TT+	x	x
sss	Omega-	x	x	x
ssc	Omega C0	x	x	x
ssb	Omega B-	x	x	x
sst	Omega T0	x	x	x
scc	Omega CC+	x	x	x
scb	Omega CB0	x	x	x
sct	Omega CT+	x	x	x
sbb	Omega BB-	x	x	x
sbt	Omega BT0	x	x	x
stt	Omega TT+	x	x	x
ccc	Omega CCC++	x	x	x
ccb	Omega CCB+	x	x	x
cct	Omega CCT++	x	x	x
cbb	Omega CBB0	x	x	x
cbt	Omega CBT+	x	x	x
ctt	Omega CTT++	x	x	x
bbb	Omega BBB-	x	x	x
bbt	Omega BBT0	x	x	x
btt	Omega BTT+	x	x	x
ttt	Omega TTT++	x	x	x

Created by Headbomb (talk) 17:39, 21 April 2008 (UTC)[reply]
Last updated by [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 00:58, 27 April 2008 (UTC)

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)[reply]

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

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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)[reply]

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).

And thus there are 62 triquark baryons. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 00:53, 27 April 2008 (UTC)

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)[reply]

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
Ξ⁻
_bb (observed) could exist and that
Ξ⁰
_cb(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)[reply]

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)[reply]

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).

So what am I missing? [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 00:20, 4 May 2008 (UTC)

First, I-spin is NOT additive, I_z IS additive-- so forget about I-spin for a minute and concentrate on I_z. When constructing composite particles, I_z is the additive quantum number. A proton has two up quarks and one down quark -- the I_z values add to 1/2. The neutron has two down quarks and one up quark-- the I_z 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 I_z = 0) plus another quark with I_z = 0 -- total lambda I_z then = 0.

Did that help?--Vectorboson (talk) 01:01, 4 May 2008 (UTC)[reply]

I'm fine with I_z, 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 (I_z 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)[reply]

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

I_z = 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

I_z of Delta++ = +3/2

I_z of Delta+ = +1/2

I_z of Delta0 = -1/2

I_z 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)[reply]

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

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

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

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

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), I_z (as given above), S (
s
− s), C (c −
c
), B*prime; (
b
− b), T (t −
t
) 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, I_z, S, C, B, and T to u and d quark content with simple algebra. I'll remark that the I_z 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 −
u
) and D (
d
− 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

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)

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.

Expect an update from me soon. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 21:05, 5 May 2008 (UTC)

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)[reply]

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)[reply]

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.

BTW, that image was brought here by me because it had more particle than the baryon decuplet, and contained it. I felt it had a greater value than the decuplet image so I don't know why you refer to it as a "detritus that I rejected for inclusion". I would prefer figure 14.4 in the PDG review of the quark model containing udsc over the .jpg we just talked about, and make figures containing different mix of quarks to show that the pyramid and the decahedron work with any selection of 4 quarks. But you do not need isospin to make it, or understand it. It is simply every combination of quarks that is allowed in a certain spin state. The base of the udsc pyramid are made by choosing three quarks (let's say uds) and trying every combination to get a baryon decuplet. The first floor is made by imposing a c quark, and trying every combination of two uds and one c quark. Second floor is made by imposing two c quarks , and the tip is made by the remaining ccc baryon. It is a very beautiful figure and I would very much like to have a poster with all the six quarks arranged in such a fashion, but I don't know how to project 6D into a 3D. The pyramid is a clever way of showing 4D into 3D (on a 2D screen no less using a volume with perspective), but I don't know if you could show 6D into 3D. For the meantime I think I'll have to settle for a poster of what's on my talk page right now (6 octets (missing the Lambdas) and 6 decuplets corresponding to groups of uds, udc, udb, udt, scb, sct) [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 00:00, 6 May 2008 (UTC)

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 |I_z| 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 (I_z=1), udb (I_z = 0), ddb (I_z = -1). The greatest |I_z| 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 (I_z=1/2), udd (I_z = -1/2). The greatest |I_z| 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 (I_z=3/2), uud (I_z=1/2), udd (I_z = -1/2), ddd (I_z=-3/2). The greatest |I_z| 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.

Still can't figure out the Lambdas. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 23:27, 5 May 2008 (UTC)

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)[reply]

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)

Since the sigma decays into a lambda electromagnetically, you can think of the sigma as an excited state of the lambda if that helps.--Vectorboson (talk) 22:49, 6 May 2008 (UTC)[reply]

Not really. I agree that there is something different about the lambda, since it has a different mass, and different decay products. What I'm asking is why does it exist in the first place? Why are there two uds in spin j=1/2+ states, but not two uud in spin j=1/2+ state? [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 23:28, 6 May 2008 (UTC)

I think moving the isospin section to the isospin page is a great idea. --Vectorboson (talk) 22:49, 6 May 2008 (UTC)[reply]

Yeah, it didn't really belong here and it's a good addition to the isospin page. [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 23:28, 6 May 2008 (UTC)

Not sure I wanna answer your question about why there are two uds particles in spin j=1/2 octet... but here goes. The wave functions in the baryon octet are pretty complicated considering that all of the quantum numbers must combine in a normalized wave function symmetric under interchange of any two quarks. It turns out there are two independent ways to do this with a quark content of uds. One with isospin 0 and one with isospin 1. --Vectorboson (talk) 15:32, 7 May 2008 (UTC)[reply]

By now you're going to expect this question :P. How would you write the two ways (Dirac notation doesn't scare me BTW, so if you need to use that, go ahead)? [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 16:49, 7 May 2008 (UTC)

Baryon wavefunctions

Alright, I decided I'd give a shot at writing wave functions. Since the color charge is always anti-symmetric under interchange of any two quarks, then the non-color part of the wave function needs to be symmetric. I got this for the uds decuplet. + means spin 1/2, - means spin -1/2.

Decuplet

$|\Delta ^{++}\rangle =|uuu^{\frac {3}{2}}\rangle =|u+\rangle |u+\rangle |u+\rangle$
$|\Delta ^{+}\rangle =|uud^{\frac {3}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|u+\rangle |u+\rangle |d+\rangle +|u+\rangle |d+\rangle |u+\rangle +|d+\rangle |u+\rangle |u+\rangle )$
$|\Delta ^{0}\rangle =|udd^{\frac {3}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|u+\rangle |d+\rangle |d+\rangle +|d+\rangle |u+\rangle |d+\rangle +|d+\rangle |d+\rangle |u+\rangle )$
$|\Delta ^{-}\rangle =|ddd^{\frac {3}{2}}\rangle =|d+\rangle |d+\rangle |d+\rangle$

$|\Sigma ^{*+}\rangle =|uus^{\frac {3}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|u+\rangle |u+\rangle |s+\rangle +|u+\rangle |s+\rangle |u+\rangle +|s+\rangle |u+\rangle |u+\rangle )$
$|\Sigma ^{*0}\rangle =|uds^{\frac {3}{2}}\rangle ={\frac {1}{\sqrt {6}}}(|u+\rangle |d+\rangle |s+\rangle +|u+\rangle |s+\rangle |d+\rangle +|d+\rangle |u+\rangle |s+\rangle +|d+\rangle |s+\rangle |u+\rangle +|s+\rangle |u+\rangle |d+\rangle +|s+\rangle |d+\rangle |u+\rangle )$
$|\Sigma ^{*-}\rangle =|dss^{\frac {3}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|d+\rangle |s+\rangle |s+\rangle +|s+\rangle |d+\rangle |s+\rangle +|d+\rangle |d+\rangle |s+\rangle )$

$|\Xi ^{*0}\rangle =|uss^{\frac {3}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|u+\rangle |s+\rangle |s+\rangle +|s+\rangle |u+\rangle |s+\rangle +|s+\rangle |s+\rangle |u+\rangle )$
$|\Xi ^{*-}\rangle =|dds^{\frac {3}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|d+\rangle |s+\rangle |s+\rangle +|s+\rangle |d+\rangle |s+\rangle +|d+\rangle |s+\rangle |s+\rangle )$

$|\Omega ^{-}\rangle =|sss^{\frac {3}{2}}\rangle =|s+\rangle |s+\rangle |s+\rangle$

Octet

For the octet it is trickier, I got to this, but I'm not confident I got the right wavefunctions, especially for the Lambda0/Sigma0 or if I assigned the correct wavefunctions to the correct particle (I got inspired from this image [2]).

Also I don't quite understand how you'd tell the excited state from the fundamental state by looking at the ± sign. I also noticed that there were other particles with a ± in the wavefunction, would I be correct in saying that the
Λ⁰
is to the
Σ⁰
in the uds octet what the
Ξ′⁺
_c is to the
Ξ⁺
_c in the usc octet?

$|p^{+}\rangle =|uud^{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {9}}}(|u+\rangle |u+\rangle |d-\rangle +|u+\rangle |u-\rangle |d+\rangle +|u-\rangle |u+\rangle |d+\rangle +|u+\rangle |d+\rangle |u-\rangle +|u+\rangle |d-\rangle |u+\rangle +|u-\rangle |d+\rangle |u+\rangle +|d+\rangle |u+\rangle |u-\rangle +|d+\rangle |u-\rangle |u+\rangle +|d-\rangle |u+\rangle |u+\rangle )$

or more concisely

$|p^{+}\rangle ={\frac {1}{\sqrt {9}}}{\begin{pmatrix}|u\rangle |u\rangle |d\rangle ,&|u\rangle |d\rangle |u\rangle ,&|d\rangle |u\rangle |u\rangle \\\end{pmatrix}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

$|n^{+}\rangle ={\frac {1}{\sqrt {9}}}{\begin{pmatrix}|u\rangle |d\rangle |d\rangle ,&|u\rangle |d\rangle |d\rangle ,&|d\rangle |d\rangle |u\rangle \\\end{pmatrix}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

$|\Sigma ^{+}\rangle ={\frac {1}{\sqrt {9}}}{\begin{pmatrix}|u\rangle |u\rangle |s\rangle ,&|u\rangle |s\rangle |u\rangle ,&|s\rangle |u\rangle |u\rangle \\\end{pmatrix}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

$|\Sigma ^{0}\rangle ={\frac {1}{\sqrt {18}}}{\begin{pmatrix}|u\rangle |d\rangle |s\rangle ,&|u\rangle |s\rangle |d\rangle ,&|d\rangle |u\rangle |s\rangle ,&|d\rangle |s\rangle |u\rangle ,&|s\rangle |u\rangle |d\rangle ,&|s\rangle |d\rangle |u\rangle \\\end{pmatrix}}{\begin{pmatrix}&?&\\1&1&1\\1&1&1\\1&1&1\\1&1&1\\1&1&1\\1&1&1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

$|\Lambda ^{0}\rangle ={\frac {1}{\sqrt {18}}}{\begin{pmatrix}|u\rangle |d\rangle |s\rangle ,&|u\rangle |s\rangle |d\rangle ,&|d\rangle |u\rangle |s\rangle ,&|d\rangle |s\rangle |u\rangle ,&|s\rangle |u\rangle |d\rangle ,&|s\rangle |d\rangle |u\rangle \\\end{pmatrix}}{\begin{pmatrix}&?&\\1&1&1\\1&1&1\\-1&-1&-1\\-1&-1&-1\\1&1&1\\-1&-1&-1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

$|\Sigma ^{-}\rangle ={\frac {1}{\sqrt {9}}}{\begin{pmatrix}|d\rangle |d\rangle |s\rangle ,&|d\rangle |s\rangle |d\rangle ,&|d\rangle |d\rangle |s\rangle \\\end{pmatrix}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

$|\Xi ^{0}\rangle ={\frac {1}{\sqrt {9}}}{\begin{pmatrix}|u\rangle |s\rangle |s\rangle ,&|s\rangle |u\rangle |s\rangle ,&|s\rangle |s\rangle |u\rangle \\\end{pmatrix}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

$|\Xi ^{-}\rangle ={\frac {1}{\sqrt {9}}}{\begin{pmatrix}|d\rangle |s\rangle |s\rangle ,&|s\rangle |d\rangle |s\rangle ,&|s\rangle |s\rangle |d\rangle \\\end{pmatrix}}{\begin{pmatrix}1&1&1\\1&1&1\\1&1&1\\\end{pmatrix}}{\begin{pmatrix}|+\rangle |+\rangle |-\rangle \\|+\rangle |-\rangle |+\rangle \\|-\rangle |+\rangle |+\rangle \\\end{pmatrix}}$

You're gaining on it! I like your decuplet wave functions (there is a typo on the sigma-star-minus (it is dds) line that needs fixing and also the cascade*-). So far I don't like your octet wave functions. Starting with the proton, these are all mixtures of mixed symmetry spin and mixed symmetry isospin (or flavour if you prefer) states. These mixed symmetry states come about naturally when combining the quarks.

For example the mixed symmetry (symmetric for 1-2 interchange) isospin state 1/SQRT(6)(2ddu-udd-dud) joins the mixed symmetry spin -- and then you combine the mixed assymetric parts and finally join them all together into a proton to get...

proton = 1/SQRT(18) { 2(u+u+d-) + 2(d-u+u+) + 2(u+d-u+) - (u+u-d+) - (d+u-u+) - (u+d+u-) - (u-u+d+) - (d+u+u-) - (u-d+u+) }

Then you can use ladder operators to get the other states in the octet (TEDIOUS!) (sorry about the formatting) --Vectorboson (talk) 21:34, 8 May 2008 (UTC)[reply]

Footnote [a]

Can you tell me what footnote "a" means (about precision in mass units vs. MeV) please?--Vectorboson (talk) 22:49, 6 May 2008 (UTC)[reply]

From what I gather, it simply means that it is easier to calibrate the instruments in atomic mass unit than it is to calibrate in MeV/c^2.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 23:28, 6 May 2008 (UTC)

Resonance

"This is not always the case, as with Ξ⁰_c (J=1/2) and Ξ⁰_c (J=1/2), where the J=1/2 state is marked by Ξ′⁰_c, the prime in this case also indicating a resonance, but of the same spin."

Is that sentence really what is meant? Or is this better?

"This is not always the case, as with
Ξ⁰
_c (J=1/2) and
Ξ^∗0
_c (J=1/2), where the prime indicates a resonance, but of the same spin."Headbomb (talk) 04:31, 20 April 2008 (UTC)[reply]

I changed the sentence to what I felt was more accurate.Headbomb (talk) 17:25, 20 April 2008 (UTC)[reply]

I have removed the resonances from the list, and expanded the Sigma baryon,Xi baryon... pages to contain them instead.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 03:00, 23 April 2008 (UTC)

The resonances you removed represented unique particles of the 10+8+8'+4 20-plet, with the same difference as Delta+'s and protons, which is why they were explicitly left, and identified with a prime.Wing gundam (talk) 02:23, 29 April 2008 (UTC)[reply]

If you remember what they were, you can find them in the Sigma baryon, Xi baryons, etc... pages to save you the trouble of retyping all the code.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 04:14, 29 April 2008 (UTC)

Isospin, decay mode, resonance listings

I'm preparing to add a separate column for isospin, if there are no objections. I'm also going to list the mass after every particle's symbol which requires it, per PDG guidelines. I also strongly recommend against listing high energy resonances of baryons, say, for the Sigma0 J=1/2 and Sigma0 J=3/2, as in fact there exist numerous other values of J, even if they do not exist in any pyramid-like table. 68.10.32.199 (talk) 17:19, 17 April 2008 (UTC)[reply]

I suggest we include all baryons with 3 or 4 star status in the PDG, since this is the definitive source for information like this. That would mean lots of Ns, Δs, Λs and Σs, but none of the double and triple charm and bottom states that have not been observed.Mjamja (talk) 13:40, 18 April 2008 (UTC)[reply]

You understand that the majority of these are isomers, right? i.e. theyre not actual "distinct" baryons. though difficult to describe, they're basically energized variants of their ground states. Don't think of these resonances as separate baryons; rather, imagine them as 'energized variations' of a representative particle.
If we are attempting to make a list of baryons, which has the potential to become grossly over-complicated by including minor variations of almost every entry, and terribly confusing to anyone who isn't an expert on particle physics, I think it would be better to include only the aforementioned representative particle, although a few exceptions are necessary:
From the article, "Recursively, for each baryon group (Nucleon, Delta, Sigma, etc) and for each possible quark content structure possible within the group, data for the ground resonance state of the arrangement has been included. ... Currently, certain other well known resonances, namely those populating the decuplet of the primary SU(3) and the 20-plets of the SU(4) groups, are included below."Wing gundam (talk) 07:02, 19 April 2008 (UTC)[reply]

Oops forgot to log in. I'm also going to add a † marker and a note below the table to denote particles not yet observed. Wing gundam (talk) 17:22, 17 April 2008 (UTC)[reply]

According to Isospin#SU(2), "... isospin is described by two numbers, I, the total isospin, and I₃, the component of the spin vector in a given direction. The proton and neutron both have I=1/2, as they belong to the doublet. The proton has I₃=+1/2 or 'isospin-up' and the neutron has I₃=−1/2 or 'isospin-down'.". Please make clear in each instance in the article where you mention isospin whether you are talking about the total isospin or the component. JRSpriggs (talk) 14:54, 29 April 2008 (UTC)[reply]

It is always the modulus of the isospin (I) we're talking about, except when isospin projection are mentionned (I_z), but I guess that could be clarified.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|talk]] · [[::Special:Contributions/Headbomb|contribs]]) 15:06, 29 April 2008 (UTC)

Spin vs. Total angular momentum

I wondered if the J given was spin or rather total angular momentum. Upon browsing the PDG archives, I found particles with J=5/2 and J=7/2, which to my understanding would be impossible to achieve by spin alone (three quarks, so spin can be at most 3/2). Someone edited it to total angular moment a while ago, but I changed it because the particle were given in their fundamental state (aka J=S+L where L = 0). Should we keep J ( equal to S+L) or change it to S? [[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 02:31, 6 May 2008 (UTC)

J=(S+L) --Vectorboson (talk) 22:49, 7 May 2008 (UTC)[reply]

I'll edit in consequence.[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 22:55, 7 May 2008 (UTC)

Name of bottom Sigmas

Yes, they decay strongly, so their masses should be in the names. However, resonances are often grouped together, and for the name an approximate of the average mass value is chosen. They will probably end up as being
Σ
_b(5810), but as of now the values that goes in the brackets hasn't been decided and I would really be surprised if these two particles ended as
Σ
_b(5808) and
Σ
_b(5815).[[::User:Headbomb|Headbomb]] ([[::User talk:Headbomb|ταλκ]] · [[::Special:Contributions/Headbomb|κοντριβς]]) 14:44, 6 May 2008 (UTC)

Diagram Updates

I think it would be a good idea if we could fit something like theses diagram somewhere in the article

http://www.fnal.gov/pub/inquiring/physics/discoveries/images/Baryon%20Chart_MR.jpg
http://www.regenerating-universe.org/images/Baryon.gif
Headbomb (talk) 18:03, 20 April 2008 (UTC)[reply]

On second thought, the more I look at them, the less I see a need for them. They already are in the baryon article.Headbomb (talk) 07:02, 21 April 2008 (UTC)[reply]