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Very often when people talk about decay they use branching ratio to describe a particular decay channel, for example:

$$Z\rightarrow e^{+}+e^{-}$$

If both Z and the electron and positon are one-shell I can understand it. Because this branching ratio one can measure in the lab.

But sometions people talk about decay into virtual particles, for example, the Higgs decay:

$$H \rightarrow Z+Z^{*}$$

My knowledge tells me this is not a physical process since one Z-boson is off-shell. But how does one calculate this braching ratio and how does one measure it?

My thought is that one can actually define $B(H \rightarrow Z+Z^{*})$ as: $$B(H \rightarrow Z+Z^{*})== B(H \rightarrow Z+e^{+}+e^{-})+B(H \rightarrow Z+\mu^{+}+\mu^{-})+...$$

where the "..." stands for all the possible "decay" the virtual $Z^{*}$ can happen. Is this right?

Another question also confuses me is that for the other Z boson without star, it will eventually decay into some other particles. Can one regard it as being virtual? I guess no, because it "was" onshell before it decays.

For example, in this picture: enter image description here

Should one interpret this feynman diagram as higgs decay? How does one distinguish this process where the higgs is being virtual with the real higgs decay?

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  • $\begingroup$ Isn't the other $Z$ virtual too? It decays into lighter particles before it reaches our detectors, after all, so it's an internal line in the Feynman diagram and is not required to be on-shell. $\endgroup$ – probably_someone May 17 at 18:03
  • $\begingroup$ @probably_someone thank you for your reply! This is what also confuses me. The onshell Z is just unstable but is not virtual? One use "*" to indicate the particle being virtual. So I guess $z$ is not virtual because it "was" onshell for sometime and then it decays? $\endgroup$ – Universe Maintainer May 17 at 18:09

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