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

• I
• Monaliza Smile
In summary, the gauge boson propagator in the Feynman gauge is represented by ## i\Delta_{\mu\nu} = -i \frac{g_{\mu\nu}}{k^2-M^2_W} ##, but in reality it is a combination of the physical gauge boson and the unphysical Goldstone boson. The Goldstone bosons are absorbed by the gauge bosons in this gauge, giving them mass. Therefore, their contribution is not explicitly taken into account in the calculation of the box diagram for ## k^0-\bar{k^0} ## mixing.
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## ?

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!

## 1. What is k^0-\bar{k^0} mixing?

K^0-\bar{k^0} mixing is a phenomenon in the field of particle physics where a neutral kaon, k^0, can transform into its antiparticle, \bar{k^0}, and vice versa. This mixing occurs due to the weak interaction and is a manifestation of the violation of CP symmetry.

## 2. How is k^0-\bar{k^0} mixing related to CP violation?

K^0-\bar{k^0} mixing is one of the key phenomena that provides evidence for the violation of CP symmetry in particle physics. CP violation occurs when the laws of physics do not behave the same way under the combined operation of charge conjugation (C) and parity (P) transformations. The mixing between k^0 and \bar{k^0} states is a direct consequence of this violation.

## 3. What are gauge boson propagators?

Gauge boson propagators are the mathematical expressions that describe the behavior and interactions of gauge bosons, which are the force-carrying particles in the Standard Model of particle physics. These propagators are used to calculate the probability of a particle interacting with another particle through the exchange of a gauge boson.

## 4. What is the role of gauge boson propagators in k^0-\bar{k^0} mixing?

In the context of k^0-\bar{k^0} mixing, gauge boson propagators are used to calculate the amplitude of the weak interaction that causes the transformation between k^0 and \bar{k^0} states. These propagators are essential for accurately predicting the rate of k^0-\bar{k^0} mixing and understanding the underlying physics behind this phenomenon.

## 5. Are there any experimental observations of k^0-\bar{k^0} mixing?

Yes, k^0-\bar{k^0} mixing has been observed in several experiments, including the CPLEAR experiment at CERN and the KTeV experiment at Fermilab. These experiments have confirmed the existence of k^0-\bar{k^0} mixing and also measured its rate, providing important insights into the violation of CP symmetry in particle physics.

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