2
$\begingroup$

The masses of $Z$ or $H$ bosons is less than the sum of the mass of two $t$ quarks. Why do $Z$ or $H$ bosons decay to a pair of quarks $t^+t^−$ with mass larger than boson in the inertial reference frame of boson?

$\endgroup$

2 Answers 2

3
$\begingroup$

What you are alluding to is the concept of on-shell and off-shell particles. The basic idea is that we require on-shell particles to have exactly their masses. This is formally stated by saying we require these particles to hold the classical equations of motion, in the case of free-particles this would be either the KG (scalar) or Dirac (fermion) equations, which imply exactly this mass condition. In QFT we introduce a new non-intuitive concept of off-shell particles which do not have to have their exact masses!

The way this works is that if we can (theoretically) directly measure the particles we require them to be on-shell (also called external particles or incoming/outgoing S-matrix states) and if we cannot measure them then they are allowed to be off-shell.

As your question correctly states on-shell masses of the Higgs and Z bosons are not enough to produce two on-shell top quarks. However, this is not to say we cannot observe processes where the top quarks are not final states like $H/Z \rightarrow t^+ t^- \rightarrow \text{stuff}$. Semantically speaking some would call this a "decay" into top quarks. However, I prefer the definition of decays as processes where an on-shell particle goes to other on-shell particles, in which case this is not a decay rather an interaction through top quarks.

$\endgroup$
2
$\begingroup$

The Z/H particles that decay to a pair of top quarks are virtual particles, namely they are intermediate states that do not correspond to free ("on-shell") particles. As such, they are not constrained to have the Masses of the free particles. enter image description here

You can think of it in terms of the Feynman path integral: in QM you have to consider all possible paths and not just the ones that satisfy the classical equation of motion. The intermediate states in this process corresponds to similar 'paths' that do not have to obey the classical constraints.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.