- #1

- 2

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter steven.noyce
- Start date

- #1

- 2

- 0

- #2

malawi_glenn

Science Advisor

Homework Helper

- 4,795

- 22

name?

- #3

- 160

- 0

Sorry that actually I don't quite understand your question.

But I would explain what I know and may help you.

First of all, the gauge fields is in the adjoint representation of gauge groups. Take the group to be SU(N), the number of generators is N^2 - 1, so that we have 8 gluons in QCD.

However, there is no need to give those gluons specific names.

First of all, in fact we have given those gluon fields an adjoint index, which allows us to manipulate them. Moreover, there is no calculation and no experiments that can directly observe the specific gluons due to the

That's why if you take a look at the table of masses of gauge bosons, you would see there is a bound for photon, say mass of photon < 10^{-10} kg (I just write it as an example, not a true number in the literature.), but you would see the mass of gluons to be

Hope this helps you.

- #4

CarlB

Science Advisor

Homework Helper

- 1,238

- 33

I would love to see such a list, but I do not see how one can eliminate a single name from the list of 9 gluons that is so simply conceived by any novice learning of color charge before learning that there are indeed only 8 gluons.

I think I know what you're getting at.

If X and Y are colors, then a gluon should exist that takes X and changes it to Y. Such a gluon would be an X/Y, for example, R/G. Since there are three choices of X and three for Y, there should be 9 gluons.

The counting argument works when X is not equal to Y. There are 6 such gluons:

R/G, R/B, G/R, G/B, B/R, B/G.

Gluons follow the adjoint representation of SU(3). In this representation, these 6 gluons are represented by 3x3 matrices with a single 1, either above or below the diagonal, like this:

[tex]\left(\begin{array}{ccc}

0&1&0\\0&0&0\\0&0&0\end{array}\right)[/tex] = R/G

where the matrix entries are "R", "G", "B" instead of "1", "2", "3". Note that the above matrix converts the "G" ket: (0,1,0) into the "R" ket: (1,0,0) so it is called the R/G. It is at least possible that my notation is reversed from everybody else and this should be the G/R.

You'd think that there would also be gluons represented by a single 1, on the diagonal, but sadly this is not the case. Instead, the diagonal gluons look like:

[tex]\left(\begin{array}{ccc}

+1&0&0\\0&-1&0\\0&0&0\end{array}\right)[/tex] = R/R - G/G

[tex]\left(\begin{array}{ccc}

0&0&0\\0&+1&0\\0&0&-1\end{array}\right)[/tex] = G/G - B/B

[tex]\left(\begin{array}{ccc}

-1&0&0\\0&0&0\\0&0&+1\end{array}\right)[/tex] = B/B - R/R

There are three of these, but first, no one of them is R/R, and second, they are not linearly independent; they sum to zero. And since they are not linearly independent, we leave one of them off. And for technical reasons, instead of taking the first two of the above, the custom is to take the first one, plus:

[tex]\left(\begin{array}{ccc}

+1&0&0\\0&+1&0\\0&0&-2\end{array}\right)[/tex] = R/R + G/G - 2 B/B

Last edited:

- #5

- 2

- 0

CarlB, I was wondering how you chose three of the six diagonal gluons. Would there not also be

[tex]

\left(\begin{array}{ccc}

-1&0&0\\0&+1&0\\0&0&0\end{array}\right)

[/tex]

[tex]

\left(\begin{array}{ccc}

0&0&0\\0&-1&0\\0&0&+1\end{array}\right)

[/tex]

and

[tex]

\left(\begin{array}{ccc}

+1&0&0\\0&0&0\\0&0&-1\end{array}\right)

[/tex]

?

How did you chose three from the six? Is there not a difference between things such as R/R-G/G and G/G-R/R?

Also, I am super confused as to what the ¨technical reasons" would be for using that last matrix instead of two of the others.

Thanks for the help!

- #6

CarlB

Science Advisor

Homework Helper

- 1,238

- 33

Thanks, guys, it is beginning to make more sense now.

CarlB, I was wondering how you chose three of the six diagonal gluons. Would there not also be

[tex]

\left(\begin{array}{ccc}

-1&0&0\\0&+1&0\\0&0&0\end{array}\right)

[/tex]

Yes! However, quantum mechanics always allows arbitrary complex phases to states. The action of the operator R/R - G/G on a state is the same as the action of the operator

[tex]\exp(i\theta)(R/R - G/G)[/tex]

for any theta, except for that arbitrary complex phase.

Also, I am super confused as to what the ¨technical reasons" would be for using that last matrix instead of two of the others.

It's really not that technical. I'll try explaining. However, I'm likely to get some terms wrong as I am just writing this from my memory and it's not a subject I care much about. (I work with Clifford algebra stuff rather than symmetry stuff.) Hopefully someone will step in and fix this.

When you analyze a Lie group, the first thing you do is to try to find a maximal Abelian subgroup. Which means it is as big as possible and all its elements commute with each other. This a Cartan subgroup, and in terms of a matrix representation, you can arrange for the Cartan subgroup elements to all be diagonal. For SU(3), these are the gluons that are confusing, R/R - G/G, etc.

Now there are an infinite number of ways of describing that Abelian subgroup. For the case of SU(3) it is 2-dimensional because, as mentioned before, the three objects (R/R-G/G), etc., all sum to zero so there really are only two. So you're going to have to pick two objects out to represent the subgroup with a "basis".

For a basis, one of the things you want is for them to be orthogonal. But (R/R - G/G) and (G/G - B/B) are not orthogonal. To make them orthogonal, you have to replace that second one by (G/G-B/B) - (B/B - R/R) = (R/R + G/G - 2B/B). To see this, you have to know what "orthogonal" means for the basis elements of the Cartan subalgebra.

I'm not sure if this is right, but the obvious way to define orthogonal is to take the trace of the product. This works, but I'm not sure if it's the right way of defining it.

Don't worry about thanking the people who help here. It's our way of either learning the material or trying to not forget it.

Share: