Decays of the Z boson

Why does the Z boson only decay to fermion-antifermion pairs? I'd just like to understand the basic reason why something like Z --> anti-down, strange wouldn't work. This would conserve charge. It obviously wouldn't conserve strangeness, but the weak interaction doesn't, so I'm just wondering why this decay is forbidden.

Thanks!
 

Bill_K

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In a nutshell, because quarks and leptons occur in SU(2) doublets and because the CKM matrix is unitary. Key words to google are FCNC (Flavor Changing Neutral Currents) and the GIM Mechanism which suppresses them.

For a simple explicit demonstration that they exactly cancel in tree order, see pp 144-146 here.
 

Vanadium 50

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Because the Z is flavor-diagonal. Only the charged weak interaction violates flavor, because that links a quark to a different antiquark. The Z (and the photon) link it to the same antiquark.
 

jkp

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Indeed at tree-level the Z-boson couplings to quark-anti-quark are flavour-diagonal even after you rotate the quarks to their mass eigenstate basis. This is because the neutral Z must couple to two down-type quarks or two up-type quarks and the combined unitary rotations cancel. The charged W boson couples to one up-type quark and one down-type quark, the combination of the two unitary rotations of up and down quarks results in the CKM matrix.

Now I know that the process Bs --> muon + anti-muon, recently discovered at the LHC, comes mostly from a Z penguin with a W in the loop. This is a rare flavour changing process. Does this mean that at the one-loop level that the Z boson should also have flavour-changing decays to a strange + anti-bottom?
 
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Does this mean that at the one-loop level that the Z boson should also have flavour-changing decays to a strange + anti-bottom?
I would expect those, but according to the http://pdglive.lbl.gov/Rsummary.brl?nodein=S044&inscript=Y&fsizein=1&wholedec0=Y [Broken], no such decay was found yet.
 
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