Charge-current interaction: linear terms in a given quark

[spaghetti3451]

I am trying to figure out the interaction terms in the Lagrangian which are linear in the top quark.I find that these terms can only be found in the charged-current interaction since the neutral-current interaction and the gluon-fermion couplings are quadratic in the top quark.

Now, I find two interaction terms in the charged-current interaction which are linear in the top quark:

$ie_{W} \left( V_{mn}\bar{u}_{m}\gamma^{\mu}(1+\gamma^{5}) d_{n} + (V^{\dagger})_{mn}\bar{d}_{m}\gamma^{\mu}(1+\gamma^{5})u_{n} \right),$

where $\displaystyle{e_{W} = g_{2}/2\sqrt{2}}$.

Do both of these terms cause the top quark to decay. My hunch is that, since the top quark must be an initial state, only the second term is allowed, because the second term does not have a bar on the top quark.

 

[AndyW]

I would like to clarify that the terms in the Lagrangian that are linear in the top quark do not necessarily correspond to decay processes. These terms represent the interactions of the top quark with other particles, such as the W and Z bosons.

In the case of the charged-current interaction, the first term in the equation you provided represents the interaction of the top quark with a down-type quark (d) via the exchange of a W boson. This process does not involve the decay of the top quark, but rather its interaction with another particle.

The second term in the equation represents the interaction of a down-type quark with the top quark, again via the exchange of a W boson. This process can be seen as the decay of a down-type quark into a top quark and a W boson.

To summarize, the terms in the Lagrangian that are linear in the top quark represent its interactions with other particles, and do not necessarily correspond to decay processes. Only when the top quark is an initial state (not when it is exchanged between particles), can it decay into other particles.