Calculating effective action at two loops

[saadhusayn]

Homework Statement

I'm trying to derive the effective action at two loops for the quartic vertices from the attached paper. For quartic vertices, e.g. $$-\frac{g}{2}\epsilon^{abx}\epsilon^{cdx} A_{a} Y^{i}_{b}A_{c}Y^{i}_{d} \tag{5.6}.$$

According to the paper,the explicit expressions for quartic vertices are given by

$$ \int \lambda_{4} \Delta_{1}(\tau, \tau | m_{1}) \Delta_{2}(\tau, \tau | m_{2}) \text{ }(5.1)$$

Here, $$\lambda_{4}$$ is a quartic vertex and the $$\Delta_{i}$$s are propagators with masses $$m_{i}$$.






[1]: https://arxiv.org/abs/hep-th/9705091

Relevant Equations

The Feynman diagrams are given in figure 1 in the paper.

Does this mean that the expression for the above vertex is

$$ -\frac{g}{2}\epsilon^{abx}\epsilon^{cdx}\int d\tau \langle A_{a} (\tau) A_{c} (\tau)\rangle \langle Y^{i}_{b} (\tau)Y^{i}_{d}(\tau) \rangle $$

 

[mark726]

Hello,

I can confirm that the expression for the above vertex is correct. This expression is derived from the Feynman rules for calculating scattering amplitudes in a quantum field theory. The $\epsilon^{abx}$ terms represent the polarization vectors of the external particles, while the $A_a$ and $Y^i_b$ are the corresponding field operators. The integral over $\tau$ represents the time evolution of the particles.

In this expression, the $g$ term represents the coupling constant, which determines the strength of the interaction between the particles. The $\langle A_{a} (\tau) A_{c} (\tau)\rangle$ and $\langle Y^{i}_{b} (\tau)Y^{i}_{d}(\tau) \rangle$ terms are the propagators for the particles, which describe the probability of the particles propagating from one point to another.

Overall, this expression is a mathematical representation of the interaction between two particles, taking into account their polarization and time evolution. It is an important tool for calculating scattering amplitudes and understanding the behavior of particles in quantum field theory. I hope this helps clarify any confusion.