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"also implies the existence of a Tbb state, with a bbud quark content, that should be stable except with regard to weak decays" Can someone explain this to me?
by addaon 2y ago
"also implies the existence of a Tbb state, with a bbud quark content, that should be stable except with regard to weak decays"
Can someone explain this to me?
Tcc(3875)+ can decay to a D0 and a D+, yes? And this is a strong decay?
I guess the reason Tbb doesn't have an equivalent strong decay to B mesons because of the sign difference -- that is, B0 and B+ would have anti-bs, not bs; and anti-B0 and anti-B+ would have negative charge?
And so the only major decay pathway is for the b itself to decay to a K+ (plus lepton noise), giving a temporary bu\s\u\d pentaquark, that then has uninhibited decays?
I guess what I'm asking is... is this the right way to think about this?
- adrian_b 2y agoIn strong decays, the products will contain the same quarks and antiquarks that have existed in the original particle, possibly with the addition of one or more quark-antiquark pairs that have been generated during the decay. In weak decays, one or more of the original quarks or antiquarks will be converted in a quark or antiquark with a different flavor, which is a process that has a low probability of happening, so the weak decays happen less frequently, therefore the hadrons that can decay only through weak decays have a lifetime that is many orders of magnitude greater than the hadrons that can decay through strong decays (or electromagnetic decays, i.e. annihilation of quarks with the corresponding antiquarks). D+ is c quark + d antiquark, D0 is c quark + u antiquark Tcc(3875)+ is 2 c quarks + d antiquark + u antiquark Therefore the 4 quarks/antiquarks in Tcc(3875)+ are the same as the 4 quarks/antiquarks in D0 + D+. So this is a strong decay, because no quark or antiquark is converted into another kind of quark or antiquark. For the Tbb- tetraquark, its composition would allow a similar strong decay into two b-quark + u or d antiquark hadrons, except that its binding energy is so great that the mass of the Tbb- tetraquark is smaller than the sum of the masses of the hadrons that would be produced during a strong decay (it is also smaller than the sum of masses of the hadrons that could be produced by an electromagnetic decay, see https://www.sciencedirect.com/science/article/pii/S0370269319307956 https://www.sciencedirect.com/science/article/pii/S037026931... ). This forbids the strong decay and the electromagnetic decay, so the only admissible decay must be weak, where one of the b quarks will be converted into another kind of quark.
- dukwon 2y agoThe strong decay would just be forbidden from conservation of energy. If the mass of the Tbb state is less than the sum of the B+ and the B0 masses, then that decay isn't allowed.