Fermion Self-Energy: Calculation and Analysis

[Safinaz]

Hi all,

In Peskin's book, Chapter 7, the self energy of electron has been calculated. In Equation (7. 28) $p\!\!\!/$ set to equal the mass of the electron $m_0$. What if I calculate the self energy of a massless fermion mediated by a loop of another massless fermion and a scalar, like the following diagram:

I got at the end this formula for the mass matrix:

$\Sigma_{ij}(k) = \frac{ y_{jm} y_{im} \Gamma(2)^{-1}}{(16\pi^2)} P_R^2 ~ p\!\!\!/ \int^1_0~ dx~ (1-x) ~ \log\Big( \frac{x\mu^2}{(1-x)m^2-xp^2} \Big) \\ = \frac{ y_{jm} y_{im} }{(32\pi^2)}~ P_R^2 ~ p\!\!\!/ ~ (-1+\log\frac{\mu^2}{m^2})$

Now what will be the value of $p\!\!\!/$ ? also what can be the renormalization scale $\mu$ ? or the cut off scale of the theory ..

In addition if I want to evaluate the amplitude of the process, this means I will square the previous formula and get $p^2$ which will equal zero on- shell, which means I have to calculate this process off-shell, but in this case how to find out the value of $p^2$ ?

 

[AndyW]

Hello,

It's great to see that you are exploring self energy calculations in Peskin's book. Your question about calculating the self energy of a massless fermion mediated by a loop of another massless fermion and a scalar is an interesting one. In this case, the value of $p\!\!\!/$ would depend on the specific momentum of the fermion in the loop and the scalar field. As for the renormalization scale $\mu$, it can be chosen arbitrarily as long as it is consistent throughout the calculation. The cut off scale of the theory would depend on the specific theory you are working with and the parameters involved.

When evaluating the amplitude of the process, it is important to note that the value of $p\!\!\!/$ will not equal zero on-shell. In fact, the on-shell condition for the fermion in the loop will affect the overall result of the calculation. To find the value of $p\!\!\!/$, you can use the momentum conservation law and solve for it in terms of the other momenta involved in the process.

I hope this helps answer your questions. Keep up the good work in your studies!