Feynman diagram for a strong interaction decay

2 replies · 4K views · last reply
Gal Winer
Messages
4
Reaction score
0
hey,

i want to find the lowest order strong interaction feynman diagram for the following process:
[tex]\Sigma^{++}_c \rightarrow \Lambda^+_c \pi^+[/tex]

the quark composition is:

[tex]\Sigma^{++}_c = (c u u) \ \Lambda^+_c = (c u d). \pi^+ =( \bar{d} u )[/tex]
an image of a correct process is attached .

my question is: why can't the C quark be the one emitting the gluon?
my initial thought was Isospin conservation, as C doesn't carry isospin and the [tex]d \bar{d}[/tex] pair may have a [tex]I = 1[/tex]. but the pair may also have a [tex]I=0[/tex].

thanks for any help
 
Attachments
  • Feynman diagram with three incoming quark lines labelled c, u, u on the left and three outgoing lines c, u, d plus a d-bar and u on the right; a gluon exchanged between an up quark line and a down quark line creates a d-bar and u pair.
    fey.jpg
    6.4 KB · Views: 654
I don't think it's Isospin, cause in the case of u, it's isospin is 1/2, we look for conservation of isospin only in total of all the process, i.e isospin in LHS should be the same as the RHS, not part of RHS with part of LHS.

But who knows...

I think that the case of c transmitting gluon would have different products, as in

[tex](u u d)+ (\bar{d} c)[/tex]
which is a proton and some Kaon I believe.
 
Any of the quarks can emit a gluon and result in the same final state.