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Colour Octet States

by Sekonda
Tags: colour, octet, states
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Sekonda
#1
Feb21-13, 01:03 PM
P: 209
Hey,

I'm having an issue seeing how these octet states are reproduced via SU(3) transformations, in my notes it is written:

"Now, the remaining 8 states in (25) mix into each other under SU(3) transformations. For example just interchange two labels such as R<->B and you'll see these mix"

I'm not exactly sure what he means by interchanging the labels, equation (25) reads:

[tex]R\bar{R},G\bar{G},B\bar{B},R\bar{G},R\bar{B},B\bar{G},B\bar{R},G\bar{R} ,G\bar{B}[/tex]

and one of these states is a singlet (i.e. colour neutral state) which is given by

[tex]\frac{1}{\sqrt3}(R\bar{R}+B\bar{B}+G\bar{G})[/tex]

anyway he interchanges the labels R and B and finds you get the octet states

[tex]\frac{1}{\sqrt2}(R\bar{R}-B\bar{B}),\frac{1}{\sqrt6}(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G},G\bar{R},G\bar{B}[/tex]

But I can't exactly see how we get these states by an interchange (but then again I'm not exactly sure what is meant by an interchange of labels) - it's using the SU(3) transformations obviously but I'm not sure how I'd set up the above 'mathematically' or 'matrix-ically'!

Anyway probably obvious but I'm slow,
Thanks in advance!
SK
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Sekonda
#2
Feb21-13, 01:17 PM
P: 209
I was thinking maybe an interchange of R<-->B means this:

[tex]\begin{pmatrix} R\\ G\\ B \end{pmatrix}\rightarrow \begin{pmatrix} B\\ G\\ R \end{pmatrix}[/tex]

but even if this is true, I'm getting confused on how to act our generator matrices on the wavefunction in order to reproduce the states listed in the last post!

Thanks!
SK
Bill_K
#3
Feb21-13, 05:52 PM
Sci Advisor
Thanks
Bill_K's Avatar
P: 4,160
Yes, the interchanges are particular elements of SU(3) and will look for example like R ↔ G, under which a state RB → GB, etc. It's obvious that all 9 states will transform into each other this way, but the point the author is making is the split between the singlet state (which transforms into itself) and the remaining octet.

The octet also transforms only into itself. Looking at the states you've listed, and calling the first one a, the second one b, under R ↔ G what happens is a → -a, and b → b - 3a/√3.

Sekonda
#4
Feb27-13, 10:55 AM
P: 209
Colour Octet States

I'm still having trouble trying to see how exactly these 8 states transform into the second list of states under this interchange. I think I would of understood it better if the author had explicitly shown this transformation though I'm aware this is something which is probably obvious to most but I can't figure it out... Though I have just had an idea so I'll give it a try!
Sekonda
#5
Feb27-13, 10:55 AM
P: 209
Thanks as well for clearing up the octet transformation stuff!
naima
#6
Mar1-13, 10:36 AM
PF Gold
P: 375
Quote Quote by Sekonda View Post
[tex]\frac{1}{\sqrt2}(R\bar{R}-B\bar{B}),\frac{1}{\sqrt6}(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G},G\bar{R},G\bar{B}[/tex]
But I can't exactly see how we get these states
you get the eight 3*3 matrices of the representation of SU (3) by exponing the eight generators of the su(3) algebra's adjoint representation.
Those generators are traceless hermitian matrices.
six are "Pauli like" matrices and two are diagonal:
[tex]\begin{pmatrix} 1&0&0 \\ 0&0&0 \\ 0&0& -1 \end{pmatrix} \\ and \begin{pmatrix} 1&0&0 \\ 0&1&0\\ 0&0& -2 \end{pmatrix}[/tex]
The first two terms come from them.
(And the 3*3 identity gives the singlet).
naima
#7
Mar6-13, 03:07 AM
PF Gold
P: 375
http://www.physicsforums.com/showthread.php?t=251070It is more usual to write the diagonal generators as
[tex]\begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0& 0 \end{pmatrix} \\ and \begin{pmatrix} 1&0&0 \\ 0&1&0\\ 0&0& -2 \end{pmatrix}[/tex]
So the first two terms would be proportional to
[tex] R\bar R - G\bar G[/tex] and [tex] R\bar R + G\bar G -2B \bar B[/tex]
read this for colour hypercharge and colour isospin
Sekonda
#8
Mar13-13, 04:33 PM
P: 209
Thanks naima for your input - didn't see this till just now.

The 3*3 matrices of SU(3) represent the transformations or rotations of colours in colour space that are allowed, the octet states represent the 8 gluons which are permanently confined and the singlet state represents a free particle (this state linked to the transformation matrix of the identity 3*3).

However I'm still confused on exactly what is meant when the author says 'Interchanging two labels such as R and B gives you'll see these states mix'

I'm not really sure what is meant physically by interchanging two labels and how this results in the states given - however I can see the link between the matrices you have stated above and the two states which are sums of colour-anticolour states.

I keep trying to show it mathematically by acting the Gell-Mann matrices on the RGB wavefunctions and see that these states arise for each SU(3) matrix but I'm not having any luck! (this is when I've interchanged the labels so that the state would be BGR with anti (RGB))

I'll keep on looking at it!
naima
#9
Mar14-13, 05:58 PM
PF Gold
P: 375
Quote Quote by Sekonda View Post
Hey,

I'm having an issue seeing how these octet states are reproduced via SU(3) transformations, in my notes it is written:

"Now, the remaining 8 states in (25) mix into each other under SU(3) transformations. For example just interchange two labels such as R<->B and you'll see these mix"

I'm not exactly sure what he means by interchanging the labels, equation (25) reads:

[tex]R\bar{R},G\bar{G},B\bar{B},R\bar{G},R\bar{B},B\bar{G},B\bar{R},G\bar{R} ,G\bar{B}[/tex]

and one of these states is a singlet (i.e. colour neutral state) which is given by

[tex]\frac{1}{\sqrt3}(R\bar{R}+B\bar{B}+G\bar{G})[/tex]

anyway he interchanges the labels R and B and finds you get the octet states

[tex]\frac{1}{\sqrt2}(R\bar{R}-B\bar{B}),\frac{1}{\sqrt6}(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G},G\bar{R},G\bar{B}[/tex]

But I can't exactly see how we get these states by an interchange (but then again I'm not exactly sure what is meant by an interchange of labels) - it's using the SU(3) transformations obviously but I'm not sure how I'd set up the above 'mathematically' or 'matrix-ically'!

Anyway probably obvious but I'm slow,
Thanks in advance!
SK
I think you misunderstand what the author wrote:
He starts from (25) a list of 9 states which generate a vector space.
one state (the singlet) is a sum of the first three. he says that this state is invariant
by colour permutation and then writes "Now the eight remaining states ..."
in fact the singlet is not belonging to the basis (25). he chooses another one (25b):
[tex](R\bar{R}+G\bar{G}+B\bar{B}),(R\bar{R}-G\bar{G}),(R\bar{R}+G\bar{G}-2B\bar{B}),R\bar{G},R\bar{B},B\bar{R},B\bar{G}, G \bar {R},G\bar{B}[/tex]
When he deletes the first one he gets the 8 remainig ones (he does not say he uses colour transformation to get them) he only looks at them and says:
"the remaining 8 states in (25b) mix into each other under SU(3) transformations" then he asks you to see what (25b) becomes when you exchange R <--> G
and do not gives the answer. (Bill_K gave you the answer)
naima
#10
Mar17-13, 04:34 AM
PF Gold
P: 375
I am reading Georgi's book "Lie algebras in particle physics" and i found a curious identity (Chapter 16): for all x,y,z
[tex](x*R\bar{R}+y*G\bar{G}+z*B\bar{B}) =
\\ (x + y + z)*(R\bar{R}+G\bar{G}+B\bar{B}) / 3 +
\\ (x - y )*(R\bar{R}- G\bar{G}) / 2 +
\\ (x + y -2 z)*(R\bar{R}+G\bar{G}-2B\bar{B}) / 6
[/tex]
x,y,z and the colors are symmetric and we see it uses
3 = 12 + 12 + 12
2 = 12 + (-1)2
6 = 12 + 12 + (-2)2
I know that this is just mathematics but is there a reason why it appears here like that?
samalkhaiat
#11
Mar17-13, 12:24 PM
Sci Advisor
P: 924
See
http://www.physicsforums.com/showthr...25#post3693225

Sam
Sekonda
#12
Mar26-13, 08:13 AM
P: 209
Thanks Naima, I thought I was misinterpreting the author. Thanks for being patient and explaining this to me!

Very much appreciated!
SK


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