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I am thinking the $z$ boson would remain with the combined mass of both the $Z$ boson and the Higgs boson. This is because the Higgs boson is a spin-0 boson and the $Z$ boson is a spin-1 boson.

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The Standard Model allows two possibilities: The $Z$ can absorb the Higgs, or they can scatter off each other. The first corresponds to a three-particle $HZZ$ vertex in a Feynman diagram and the second to a four-particle $HHZZ$ vertex. You can see these two possible vertices in the answer to this question.

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You seem to believe particles absorb/trade masses in reactions. They only trade energy and momentum.

For a real Z absorbing a real h to end up with a real Z, you'd need both energy and momentum to balance across the reaction. In the center-of momentum frame of the input particles Z and H you'd then have a total energy $$ E=\sqrt{m_Z^2 + p^2} + \sqrt{m_h^2 + p^2}, $$ and zero total momentum, by frame choice. The output Z would then be at rest, with total energy $E=m_Z$.

However, for absolutely any p, including 0, $$ m_Z\neq \sqrt{m_Z^2 + p^2} + \sqrt{m_h^2 + p^2}, $$ so your reaction will not go.

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  • $\begingroup$ The masses count as energy. I mean energy in the masses get absorbed. $\endgroup$ – Roghan Arun May 8 at 17:14
  • $\begingroup$ What about 2 z bosons $\endgroup$ – Roghan Arun May 8 at 17:14
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    $\begingroup$ There is no HZZZ vertex in the SM, and I believe there is no tree diagram you can cook up for this, as per here. $\endgroup$ – Cosmas Zachos May 8 at 18:03
  • $\begingroup$ @Roghan Appreciative of the key point? $\endgroup$ – Cosmas Zachos May 13 at 14:01

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