Given a Yukawa coupling as a function of scale $\mu$ and a vev, therefore $m_R(μ)=Y(μ)⟨ϕ⟩$, how can I compute the corresponding pole mass $m_p$? Relations I was able to find are (page 39) $$m_p=m_R−Σ(m_P)$$

or specifically for the electron (page 17)

$$ m_P =m_R - \frac{e_R^2 }{16 \pi^2}\left[ 2 (m_P-m_R) + \int dx (4m_R-2m_Px)\log( \frac{\mu^2 }{(1-x)(m_R^2-xm_P^2)}) \right] .$$

Now, what I don't understand is how these equations can be used, in practice, to compute $m_P$, if $m_R(\mu)$ is given. Any tip or reference to an example computations would be much appreciated! As the pole mass should be indendent of $\mu$, I'm confused which $\mu$ I should use in the formulas above in order to compute $m_P$.

  • $\begingroup$ Do you mean, is there a closed form solution? $\endgroup$ – innisfree Feb 25 '16 at 13:17
  • $\begingroup$ Also, for clarityI think $m_R(0)$ should be just $m_R$ (the argument is momentum slash $k=0$ not renormalization scale $\mu=0$) $\endgroup$ – innisfree Feb 25 '16 at 13:20
  • $\begingroup$ @innisfree My problem is that I do not understand how these formulas can be used to compute $m_P$, if $m_R(\mu)$ is known. In other words what $\mu$ do I use to calculate $m_P$? (Some value must be used, because I have only 1 equation for two unknown quantities $m_P$ and $\mu$... ?! $\endgroup$ – JakobH Feb 25 '16 at 13:29
  • $\begingroup$ physics.stackexchange.com/questions/64613/… $\endgroup$ – innisfree Feb 25 '16 at 13:33
  • $\begingroup$ physics.stackexchange.com/questions/80972/… $\endgroup$ – innisfree Feb 25 '16 at 13:34

You can't solve exactly, but you can solve order by order in perturbation theory. I'm essentially following chapter 27 of Srednicki. The relation $m_P = m_R - \Sigma(m_P)$ is the one you want, but $m_P$ can only be determined implicitly. Since $\Sigma$ begins at order $\alpha$, the difference between $m_R$ and $m_P$ is $\mathcal{O}(\alpha)$, so up to $\mathcal{O}(\alpha^2)$ you can replace $m_R$ by $m_P$ inside it. So we get:

$$m_P = m_R - \Sigma(m_R) + \mathcal{O}(\alpha^2)$$

If you want to go to a higher order, suppose you compute two-loop corrections to $\Sigma$ and get $\Sigma = \alpha \Sigma_1 + \alpha^2 \Sigma_2 + \mathcal{O}(\alpha^3)$. We have the relation

$$m_P = m_R - \alpha \Sigma_1(m_P) - \alpha^2 \Sigma_2(m_P) + \mathcal{O}(\alpha^3)$$

Now inside $\Sigma_1$ we can replace $m_P$ by its $\mathcal{O}(\alpha)$ expression in terms of $m_R$, while inside $\Sigma_2$ we just set $m_P = m_R$ like we did earlier, since it's already multiplied by $\alpha^2$:

$$\begin{align} m_P &= m_R - \alpha \Sigma_1(m_R - \alpha \Sigma_1(m_R)) - \alpha^2 \Sigma_2(m_R) + \mathcal{O}(\alpha^3) \\ &= m_R - \alpha [\Sigma_1(m_R) - \Sigma_1'(m_R)\alpha\Sigma_1(m_R)] - \alpha^2 \Sigma_2(m_R) + \mathcal{O}(\alpha^3)\\ &= m_R - \alpha \Sigma_1(m_R) + \alpha^2 [\Sigma_1(m_R)\Sigma_1'(m_R) - \Sigma_2(m_R)] + \mathcal{O}(\alpha^3) \end{align}$$

And so on.

  • $\begingroup$ Thanks a lot for your answer! I think my problem then is: What is $m_R$ here? If it is the running mass, i.e. $m_R=m_R(\mu)$ at what energy scale $\mu$ do we evaluate it? $\endgroup$ – JakobH Mar 15 '16 at 13:12
  • $\begingroup$ @JakobH: You can use whatever $\mu$ you want, because the pole mass must be independent of it. $m_R$ depends on $\mu$ in just the right way so that at each order in $\alpha$, $m_P$ is independent of $\mu$. $\endgroup$ – Javier Mar 15 '16 at 13:56

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