Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I How to treat quark color pairs mathematically

  1. Jun 13, 2018 #1
    I am trying to work through a problem in the textbook "Particle Physics in a Nutshell." However, I am realizing how little I actually understand about working through problems involving quark color pairs.

    Given in the problem is the meson color singlet [tex]1/\sqrt{3}(r\bar{r}+g\bar{g}+b\bar{b})[/tex] which I understand. The problem asks to rotate this state by θ with the first Gell-Mann matrix as the generator.

    I understand that I should be operating with a rotation operator [tex]e^{-i\theta\lambda_1}[/tex]
    but I am not sure what I should be operating on, exactly.

    I have not been able to find anything that might point me in the right direction in the textbook or online. Once I know how to work with these color/anticolor components I should be able to progress.

    Last edited: Jun 13, 2018
  2. jcsd
  3. Jun 13, 2018 #2


    User Avatar
    Science Advisor
    Gold Member

    Your particle state is in a tensor product space. You will need to express the representation of the group element (the exponential of the Gell-Mann matrix) in the appropriate basis, or you could try to see the component parts and expand in terms of eigen-vectors starting with the representation of the Gell-Mann Matrix itself.

    Here's the action of the Gell-Mann generator on the state you gave:
    \delta \psi = 1/\sqrt{3}\left( [\lambda_1r]\otimes\bar{r} + r\otimes[\lambda_1^\dagger \bar{r}] +[\lambda_1g]\otimes\bar{g} + g\otimes[\lambda_1^\dagger \bar{g}] +[\lambda_1b]\otimes\bar{b} + b\otimes[\lambda_1^\dagger \bar{b}] \right) [/tex]
    Where the [itex]\lambda_1[/itex] is the fundamental rep. Now since the first Gell-Mann generator basically swaps r and g colors and maps b to zero (if the color ordering is the same as I recall) then this infinitesmal action will become:
    [tex]\delta \psi = 1/\sqrt{3}\left( g\otimes\bar{r} + r\otimes \bar{g} +r\otimes\bar{g} + g\otimes \bar{r} \right) = 2/\sqrt{3}\left( g\otimes\bar{r} + r\otimes \bar{g}\right) [/tex]
    (Ok I'll now stop being explicit with the tensor products and just use the same dyadic notation you began with...)
    Now you'll be expanding the exponential as a power series so you need to see the next term by acting on this again with [itex]\lambda_1[/itex].
    [tex]\delta^2\psi = 4/\sqrt{3}\left(r\otimes\bar{r} + g\otimes\bar{g} + g\otimes\bar{g}+r\otimes\bar{r} \right) = 4/\sqrt{3} \left(r\otimes\bar{r} + g\otimes\bar{g}\right)[/tex]
    So we're repeating some of the terms. So separate out the b component and then work out the power expansion of the exponential action. You should see real and imaginary terms that resolve as cosine times some terms plus i sine times others plus the blue terms that go away after the 0th order term in the expansion.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?