Color of gluon mediating quark-antiquark process?

  • Thread starter Stalafin
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  • #1
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Main Question or Discussion Point

I have the following process:
[itex]q(R) + \bar{q}(\bar{B}) \rightarrow q(R)+\bar{q}(\bar{B})[/itex]

In words: a quark with red color-charge and an antiquark with an anti-blue color-charge are incoming, and a red quark and anti-blue antiquark are emerging. Since I am not sure how else to draw that, I try to do some ASCII art here:
Code:
q(R)          q(R)
   1\           3/
     \          /
      \        /
       \      /
          ~
          ~ G (?,?)
          ~
          ~
        /   \
       /     \
      /       \
    2/        4\
q-(B-)      q-(B-)

----------------------> t
q are the quarks, q- the antiquarks, R is red, B- is antiblue, time runs from left to right.


The question is: which gluon(s) can participate in this exchange?

I have read Griffiths "Introduction to Elementary Particle Physics" on that topic and on p.290 Example 8.1 he more or less states exactly that problem. He talks about a "typical octet state" [itex]R\bar{B}[/itex]. What is that supposed to mean? I thought the octet states are always a superposition of two states, in this case:
[tex]|1\rangle = (R\bar{B} + B\bar{R})/\sqrt{2}[/tex]

I was thinking along the lines of color conservation. The only thing that made sense to me is that the mediating gluon has to be [itex]B\bar{B}[/itex]: the incoming red quark sends out a [itex]B\bar{B}[/itex] pair. The incoming anti-blue antiquark combines with the blue charge, and what remains is an anti-blue charge... But I somewhat feel that this logic is flawed. :(

What's the right answer here?
 

Answers and Replies

  • #2
Bill_K
Science Advisor
Insights Author
4,155
194
Isn't it just q(R)+qˉ(Bˉ) → g(RBˉ) → q(R)+qˉ(Bˉ)?
 

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