User talk:Vectorboson

--Vectorboson (talk) 13:58, 14 May 2008 (UTC){{helpme}} How do I edit a generated table like...

Hi! You'd have to edit the template for the table. {{particles}} is a template containing a table, so to edit that you'd have to go to Template:Particles. Bjelleklang - talk Bug Me 11:04, 12 May 2008 (UTC)

Or simply click on the e in the upper left corner. Headbomb (ταλκ · κοντριβς) 11:41, 12 May 2008 (UTC)

PROBLEM WITH THE GAMMA SYMBOL FOR A PHOTON

This is a "Y" not a gamma ...
γ

Should be this... $\gamma$

But fixing it turns out to be a journey... after looking up the SubatomicParticle template...

http://en.wikipedia.org/wiki/Template:SubatomicParticle/doc
It depends on
SubatomicParticle/symbol template
http://en.wikipedia.org/wiki/Template:SubatomicParticle/symbol
which in turn depends on
PhysicsParticleTemplate...
http://en.wikipedia.org/wiki/Template:PhysicsParticle

What exactly is the problem? The photon symbol is indeed a gamma (
γ
). Compare
γ
and Y. First is gamma, second is Y. Headbomb (ταλκ · κοντριβς) 14:25, 14 May 2008 (UTC)

I am in discussion with Hqb on this page...
http://en.wikipedia.org/wiki/Template_talk:PhysicsParticle
He thinks the problem is with my browser, and it might be. But the html source is different for the two and my browser shows them distinctly different. It never was a big deal, I will move on to other stuff.--Vectorboson (talk) 16:43, 14 May 2008 (UTC)

"LIST OF BARYONS" SECTION UNDER CONSTRUCTION:

Relation between isospin and up and down quark content

The third component of isospin for the up quark is 1⁄2; for the down quark it is - 1⁄2; and it is zero for the other quarks. The third component of isospin for a baryon is simply the sum of that of its quarks. Different baryons that have the same constituent quarks are distinguishable by their other characteristics (like spin).

Particles of isospin 3⁄2 can only be made by a mixture of three u and d quarks.
The four
Δ
s are:

Δ⁺⁺
(uuu) with Iz = 3⁄2

Δ⁺
(uud) with Iz = 1⁄2

Δ⁰
(udd) with Iz = - 1⁄2

Δ⁻
(ddd) with Iz = - 3⁄2

Particles of isospin 1 are made of one c, s, or b quark together with two u quarks or two d quarks or a u and a d quark.
The nine
Σ
s are:

Σ⁺
(uus),
Σ⁰
(uds),
Σ⁺
(dds), with Iz being +1, 0, -1 respectively

Σ⁺⁺
_c (uuc),
Σ⁺
_c (udc),
Σ⁰
_c (ddc), with Iz being +1, 0, -1 respectively

Σ⁺
_b (uub),
Σ⁰
_b (udb),
Σ⁻
_b (ddb) with Iz being +1, 0, -1 respectively

Particles of isospin 1⁄2 can be made of two u quarks and one d quark, or two d quarks and one u quark.
The two
N
s are:

p⁺
(uud),
n⁰
(udd)

They can also be made of a combination of two c, s, or b quarks together with one u or d quark.
The twelve
Ξ
s are:

Ξ⁰
(uss),
Ξ⁻
(dss),

Ξ⁺
_c (usc),
Ξ⁰
_c (dsc),

Ξ⁺⁺
_cc (ucc),
Ξ⁺
_cc (dcc),

Ξ⁰
_b (usb),
Ξ⁻
_b (dsb),

Ξ⁺
_cb (ucb),
Ξ⁰
_cb (dcb),

Ξ⁰
_bb (ubb),
Ξ⁻
_bb (dbb)

Particles of isospin 0 can be made of one u and one d quark plus one c, s, or b.
The three
Λ
s are:

Λ⁰
(uds),

Λ⁺
_c (udc),

Λ⁰
_b (udb)

Isospin 0 baryons can also be made of no u or d quarks at all.
The ten
Ω
s are:

Ω⁻
(sss),
Ω⁰
_c (ssc),
Ω⁺
_cc (scc),
Ω⁻
_b (ssb),
Ω⁰
_cb (scb),
Ω⁻
_bb (sbb)

Ω⁺⁺
_ccc (ccc},
Ω⁺
_ccb (ccb),
Ω⁰
_cbb (cbb),

Ω⁻
_bbb (bbb)

Rules for making baryons

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 I_z are additive numbers and that I = 1⁄2 for u and d quarks and that I_z = 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? Headbomb (talk · contribs) 13:32, 3 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 had 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)

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.Headbomb (talk · contribs) 01:23, 4 May 2008 (UTC)

Let's stick to the talk page in the list of baryons, else till will be a pain in the ass to keep things up to date here and there simultaneouslyHeadbomb (talk · contribs) 01:27, 4 May 2008 (UTC)

Handout

I've checked that handout and it's very confusing in some parts.

With one fermion, it's pretty simple. You have two spin possibilities: + and -. Wave functions are simply |+> and |-> and correspond to |S,S_z> |1⁄2,+1⁄2> and |1⁄2,-1⁄2> respectively.

So for 1 fermion:

$|{\frac {1}{2}},+{\frac {1}{2}}\rangle =|+\rangle$
$|{\frac {1}{2}},-{\frac {1}{2}}\rangle =|-\rangle$

And you can get from |-> to |+> with the ladder up operator very easily
$T_{+}|d\rangle =|u\rangle$

For two fermions, things get a bit messier, but are still are easy enough to handle. There are four possibilities ++,+-,-+,-- With S_z = 1, there is |++>, for S_z = 0 there are |+-> and |-+>, and for S_z = -1 there is |-->

Giving
$|1,+1\rangle =|+\rangle |+\rangle$
$|1,0\rangle ={\frac {1}{\sqrt {2}}}(|+\rangle |-\rangle +|-\rangle |+\rangle )$
$|1,-1\rangle =|-\rangle |-\rangle$
and
$|0,0\rangle ={\frac {1}{\sqrt {2}}}(|+\rangle |-\rangle -|-\rangle |+\rangle )$

Again from the |1,-1> it's pretty easy to get to use the ladder up operator to get the |1,0>

$|1,0\rangle =T_{+}|1,-1\rangle$
$=T_{+}(|-\rangle |-\rangle )$
$=T_{+}|-\rangle |-\rangle +|-\rangle T_{+}|-\rangle$
$=|+\rangle |-\rangle +|-\rangle |+\rangle$
which after normalization becomes
$|1,0\rangle ={\frac {1}{\sqrt {2}}}(|+\rangle |-\rangle +|-\rangle |+\rangle )$
And to get the |1,+1> $|1,+1\rangle =T_{+}|1,0\rangle$
$=T_{+}({\frac {1}{\sqrt {2}}}(|+\rangle |-\rangle +|-\rangle |+\rangle ))$
$={\frac {1}{\sqrt {2}}}(T_{+}|+\rangle |-\rangle +|+\rangle T_{+}|-\rangle +T_{+}|-\rangle |+\rangle +|-\rangle T_{+}|+\rangle )$
$={\frac {1}{\sqrt {2}}}(0+|+\rangle |+\rangle +|+\rangle |+\rangle +0)$
which after normalisation becomes
$|1,+1\rangle =|+\rangle |+\rangle$

Now for three fermions I'm almost completely confused. There are 8 possibilites +++,++-,+-+,-++,+--,-+-,--+,---. With S_z = 3/2 you have |+++>, with S_z = 1/2, you have |++->,|+-+>,|-++>, with S_z = -1/2 you have |--+>,|-+->,|+--> and with S_z = -3/2 you have |--->.

Now the way I would combine them would give this:

$|{\frac {3}{2}},+{\frac {3}{2}}\rangle =|+\rangle |+\rangle |+\rangle$
$|{\frac {3}{2}},+{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|+\rangle |+\rangle |-\rangle +|+\rangle |-\rangle |+\rangle +|-\rangle |+\rangle |+\rangle )$
$|{\frac {3}{2}},-{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {3}}}(|-\rangle |-\rangle |+\rangle +|-\rangle |+\rangle |-\rangle +|+\rangle |-\rangle |-\rangle )$
$|{\frac {3}{2}},-{\frac {3}{2}}\rangle =|-\rangle |-\rangle |-\rangle$

And you can go from |---> to |+++> with the T₊ operator as well, so I assume that I did things correctly here.

From here I don't get it. I don't know how to build a S_z = -1/2 state — any combination of two - and one + seem to work. There does not seem to be a systematic approach for finding wave functions. And the choice seems completely arbitrary. The handout lists

$|{\frac {1}{2}},-{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {6}}}(2|-\rangle |-\rangle |+\rangle -|-\rangle |+\rangle |-\rangle -|+\rangle |-\rangle |-\rangle )$
but if this one works, why don't
$|{\frac {1}{2}},-{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {6}}}(2|+\rangle |-\rangle |-\rangle -|-\rangle |+\rangle |-\rangle -|-\rangle |-\rangle |+\rangle )$ (A)
and
$|{\frac {1}{2}},-{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {6}}}(2|-\rangle |+\rangle |-\rangle -|-\rangle |-\rangle |+\rangle -|+\rangle |-\rangle |-\rangle )$ (B)
work also?

Inspection with the T₊ operator reveals that any of them works. And the 3 choices are not orthogonal states, so why isn't
$|{\frac {1}{2}},-{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {18}}}(2|-\rangle |-\rangle |+\rangle -|-\rangle |+\rangle |-\rangle -|+\rangle |-\rangle |-\rangle +2|+\rangle |-\rangle |-\rangle -|-\rangle |+\rangle |-\rangle -|-\rangle |-\rangle |+\rangle +2|-\rangle |+\rangle |-\rangle -|-\rangle |-\rangle |+\rangle -|+\rangle |-\rangle |-\rangle )$ (C)
listed as the |1/2,-1/2> symmetrical under 1<->2 exchange wave function?

Similar remarks apply for the antisymmetrical under 1<->2 exchange wave functions. Headbomb (ταλκ · κοντριβς) 20:07, 12 May 2008 (UTC)

(chuckle) I envy you -- you are working on understanding real physics or at least real math while I am working on understanding how to change a "Y" into a "gamma" inside a table in Wikiscript! I didn't see this post until just now -- let me absorb your question a bit and I'll try to get you a good answer by tomorrow. You ask good questions and I usually have to do a lot of remembering to come up with the answer you deserve.--Vectorboson (talk) 22:07, 12 May 2008 (UTC)

I'd gladly switch place. :P

I took me a while to understand how to edit those damn boxes, but certainly not as long as it takes to understand this topic. I think I'll stop after I figure out what the wave functions of baryons are, assuming a 6-flavour symmetry. My quantum physics classes were over two years ago and I've barely touched the topic since (at least the parts dealing with spin and the exclusion principle). I think I need to dust of that Shankar (Principles of Quantum Mechanics, excellent book if I recall correctly) again. Headbomb (ταλκ · κοντριβς) 01:36, 13 May 2008 (UTC)

I have labeled your two incorrect equations as (A) and (B). They are both incorrect because they are NOT symmetric under interchange of the first two quarks. At this point we are trying to build the basis for the baryon octet states. These baryon states are made up of the mixed symmetry states from flavor (isospin) and the mixed symmetry states of real spin. These mixed symmetry states ARE symmetric (or antisymmetric) under the interchange of the first two quarks. Interchanging the third quark with either of the first two yields a state with no definite symmetry.

So the only two states that we can make that are symmetric under the interchange of the first two quarks are...
$|{\frac {1}{2}},-{\frac {1}{2}}\rangle =-{\frac {1}{\sqrt {6}}}(2|-\rangle |-\rangle |+\rangle -|-\rangle |+\rangle |-\rangle -|+\rangle |-\rangle |-\rangle )$
and
$|{\frac {1}{2}},{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {6}}}(2|+\rangle |+\rangle |-\rangle -|+\rangle |-\rangle |+\rangle -|-\rangle |+\rangle |+\rangle )$

And the only states we can make that are antisymmetric under the exchange of the first two quarks are...
$|{\frac {1}{2}},-{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {2}}}(|+\rangle |-\rangle |-\rangle -|-\rangle |+\rangle |-\rangle )$
and...
$|{\frac {1}{2}},{\frac {1}{2}}\rangle ={\frac {1}{\sqrt {2}}}(|+\rangle |-\rangle |+\rangle -|-\rangle |+\rangle |+\rangle )$

From here you ought to be able to pick up the thread in the handout again--Vectorboson (talk) 15:31, 13 May 2008 (UTC)

Ok, I now see I did things a bit too quickly when speaking about (A) and (B), they aren't 1<->2 symmetric on their own. But (C) is (and so is (A)+(B)). Headbomb (ταλκ · κοντριβς) 15:43, 13 May 2008 (UTC)

Nevermind, (C) is just a fancy way of writing 0. Headbomb (ταλκ · κοντριβς) 16:02, 13 May 2008 (UTC)

And, of course, (A) + (B) IS the same as the correct equation from the handout.--Vectorboson (talk) 16:13, 13 May 2008 (UTC)

Yeah I see that now. I hate fancy 0s. I don't know how much time I've wasted in my life by applying operators on fancy 0s.Headbomb (ταλκ · κοντριβς) 16:27, 13 May 2008 (UTC)

Just to let you know, I found the systematic way to get the wavefunctions. You simply need to build a vector with all the permutation of particular total spin (a|+>|+>|->, b|+>|->+>, c|->|+>|+>), exchange 1<->2, impose either symmetry or antisymmetry, and orthogonalize. This puts some conditions on a,b,c. Do the same for 1<->3 exchange, and then you can find other conditions for a,b,c. You end up with the handout wavefunctions in 2 minutes. Very simple when you know how to do it.Headbomb (ταλκ · κοντριβς) 14:33, 14 May 2008 (UTC)

Cool!--Vectorboson (talk) 16:43, 14 May 2008 (UTC)

List of baryon has been significantly expanded.

Since you're more of an expert on this than me, I wondered if you could go over the overview and tell me if you spot any factual mistakes. I'm also currently looking for references for all of this stuff, so if you have any, feel free to mention them. Don't worry about format, just place the link/book/article between <ref></ref> and I'll fix it.

A million bunch of thanks Headbomb (ταλκ · κοντριβς) 06:23, 30 May 2008 (UTC)

WikiProject Physics participation

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On 2008-06-25, the WikiProject Physics participant list was rewritten from scratch as a way to remove all inactive participants, and to facilitate the coordination of WikiProject Physics efforts. The list now contains more information, is easier to browse, is visually more appealing, and will be maintained up to date.

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Unreferenced BLPs

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Women in Red World Contest

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