## k^0-\bar{k^0} ## mixing and gauge boson propagator

[Monaliza Smile]

Hi all ,

I'm studying $k^0-\bar{k^0}$ mixing from Cheng & Li's book "gauge_theory_of_elementary_particle_physics", page: 379 , I found that they consider the gauge boson propagator in the Feynman gauge, where
$i\Delta_{\mu\nu} = -i \frac{g_{\mu\nu}}{k^2-M^2_W}$ , but till my knowledge in this gauge the goldstone bosons remain in the theory, so why the contribution of the charged goldstone bosons for instance has not been taken into account in these calculation of the box diagram of $k^0-\bar{k^0}$ like $W^\pm$ ?

 

[AndyW]

Hello,

Thank you for bringing up this interesting question. The gauge boson propagator in the Feynman gauge is indeed given by $i\Delta_{\mu\nu} = -i \frac{g_{\mu\nu}}{k^2-M^2_W}$, but this is just a mathematical representation of the gauge boson's behavior in the theory. In reality, the gauge boson propagator is a combination of both the physical gauge boson and the unphysical Goldstone boson.

In the Feynman gauge, the Goldstone bosons are absorbed by the gauge bosons, giving them mass. So, while the Goldstone bosons are still present in the theory, they are no longer considered as individual particles. This is why their contribution is not explicitly taken into account in the calculation of the box diagram for $k^0-\bar{k^0}$ mixing.

I hope this helps clarify the issue. Let me know if you have any other questions. Happy studying!