2
$\begingroup$

enter image description here

For the system of $K^-,\pi^+,\pi^+$, with the invariant mass spectrum peaking about 1.87 GeV, call this resonant peak $D^+$; we find its spin to be zero by experiment.

Using the quark model, how do we explain that there is no possibility that this $D^+$ meson be a strange particle?

I know that, for $D^+$ decaying into $K^-$$(\bar{u}s)$$\pi^+(u\bar{d})$$\pi^+$$(u\bar{d})$, if $D^+$ is a strange particle,it should be $c\bar{s}$, that quark flavor can change, and that strangeness is not conserved. So it's a weak interaction.

For the weak interaction, C-parity is not conserved. By drawing Feynman diagrams, I found it's possible to be strange. I believe I have considered all quantum numbers. But I still don't know how to use the quark model to show it cannot be a strange particle!

$\endgroup$
3
  • 3
    $\begingroup$ Please see this guidance about screenshots and this guidance about exercises. To edit into an on-topic conceptual question: have you written down all of the quarks and antiquarks in the initial and final states? Have you found all of the flavor quantum numbers in the initial and final states? $\endgroup$
    – rob
    Commented Dec 3, 2022 at 15:02
  • $\begingroup$ Could you please make it reopened? $\endgroup$
    – jun xiang
    Commented Dec 4, 2022 at 5:17
  • $\begingroup$ Ok,thank you very much! $\endgroup$
    – jun xiang
    Commented Dec 5, 2022 at 6:34

1 Answer 1

3
$\begingroup$

Actually, there is an extremely improbable possibility, $\Delta S=2$, so doubly weak, as you found, after all.

How unlikely? Every time you include an extra virtual W in your diagram, as you appreciated you must do for that logical possibility, you have an extra suppression factor of ~ $m^2 G_F\approx 10^{-5}$ in the amplitude, so a 10 order of magnitude suppression in the width, w.r.t. the singly weak decay modes.

So, as you can check (!) from the $c\bar d$ singly weak decays, $D^+(1870) \to K^- 2\pi^+$, and the also singly weak $c\bar s$ singly weak decays, $D_s^+(1968) \to K^- K^+ \pi^+$ , either of these is over a billion times more likely than your logical possibility; which is thus pragmatically dubbed "no possibility"...

The conclusion is that the $D^+$ has to be $c\bar d$.

$\endgroup$
2
  • $\begingroup$ "Every time" is somewhat an over-generalisation: the "box" or "penguin" diagrams have two virtual W but their rate is not as "extremely improbable" as here. Of course, my comment is more pedantic and not relevant for this particular problem. $\endgroup$
    – Martino
    Commented Dec 9, 2022 at 23:36
  • $\begingroup$ Of course. The box diagrams are doubly weak and are far less probable that singly weak processes, but there are none such of this type there. Every time, $G_F^2$ in the amp, $G_F^4$ in the rate . $\endgroup$ Commented Dec 10, 2022 at 1:35

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.