3
$\begingroup$

I am a starter at nuclear and particles physics. I am reading particles and nuclei an introduction to the physical concepts. There was a paragraph about Colour-neutral particles and how only colourless particles can be found as free particles. Later the book adresses to the $\pi^+$ meson and states that it has only three possible combinations: $$ \left\vert{\pi^+}\right\rangle = \left\{\begin{matrix} \left\vert u_r\bar d_{\bar r}\right\rangle\\ \left\vert u_g\bar d_{\bar g}\right\rangle\\ \left\vert u_b\bar d_{\bar b}\right\rangle \end{matrix}\right. $$

My question is why they didn't count $\left\vert u_{\bar r}\bar d_r\right\rangle$ or others like that as possible. Does antiquarks can only gain anti-colours? and if so why?

$\endgroup$
1

1 Answer 1

10
$\begingroup$

In quantum mechanics the operation of swapping the charge is called charge conjugation and this operation swaps a particle for its antiparticle. So for example if we take an electron and swap the charge from $-e$ to $+e$ the resulting particle is a positron. That means we cannot have an electron with a positive charge because that particle is a positron, and likewise we cannot have a positron with a negative charge because that particle is an electron.

And this applies to quarks as well. So to take your example, if we start with $u_r$ and swap the charge to $\bar r$ the result is an anti-up quark. We cannot have a "non-anti" quark with an anti-charge because that particle is by definition an anti-quark.

I guess what's confused you is the notation of putting bars on both the quark and the charge i.e. writing $\bar u_{\bar r}$ seems to suggest we could have other permutations like $u_{\bar r}$ or $\bar u_r$. But don't be mislead by this. Swapping the charge also swaps the particle to its antiparticle.

$\endgroup$
1
  • $\begingroup$ Thanks. I understood it $\endgroup$
    – Kid A
    Jan 9, 2023 at 8:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.