I would like to renormalize $\phi^4$ theory with Lagrangian \begin{equation} \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m_0^2 \phi^2 - \frac{\lambda_0}{4!} \phi^4 \end{equation} up to first order in perturbation theory. Especially, I am interested in the explicit form of the $\Lambda$-dependent bare coupling $m_0(\Lambda)$. I am following the discussion from section 5.5.2 and 5.6 of "A modern introduction to quantum field theory" by M. Maggiore.

At this order, the $2 \rightarrow 2$ scattering amplitude is given by \begin{equation} \mathcal{M}_{2 \rightarrow 2} = - i \lambda \end{equation} while the $2$-point function is governed by the tadpole diagram. The latter is governed (after amputation of external legs and after Wick rotation of the integral) by the divergent expression ($D_F(0)$ in my notes corresponds to $B$ in Maggiores Book) \begin{equation} - i \lambda D_F(0) = \frac{(-i \lambda)}{16 \pi^2} \int_0^\infty dk \frac{k^3}{k^2 + m^2}. \end{equation} Introducing a momentum cutoff, we can write this as: \begin{equation} D_F(0;\Lambda) = \frac{\lambda}{32 \pi^2} \left(\Lambda^2 + m^2 \log\left(\frac{m^2}{\Lambda^2 + m^2} \right)\right). \end{equation} One now resums the tadpole-graphs and obtains for the resummed $2$-point function in momentum space \begin{align} \hat{D}_F^R(p; \Lambda) &= \hat{D}_F(p) \sum_{n=0}^\infty \left(- i \lambda D_F(0;\Lambda) \hat{D}_F(p)\right)^n\\ &= \frac{i}{p^2 -m^2_0 -D_F(0;\Lambda)} \tag{1} \end{align} where $\hat{D}_F(p)$ is the Feynman propagator in momentum space.

The renormalization condition for mass renormalization is then given by \begin{equation} \lim_{\Lambda \rightarrow \infty} \left[ m_0(\Lambda)^2 + D_F(0;\Lambda) \right] \overset{!}{=} m_R^2 \tag{2} \end{equation} or more explicitely by: \begin{equation} \lim_{\Lambda \rightarrow \infty}\left[ m_0(\Lambda)^2 + \frac{\lambda}{32 \pi^2} \left(\Lambda^2 + m_0(\Lambda)^2 \log\left(\frac{m_0(\Lambda)^2}{\Lambda^2 + m_0(\Lambda)^2} \right)\right)\right] \overset{!}{=} m_R^2. \tag{3} \end{equation}

Unfortunately I am not able to show, that a bare, real, $\Lambda$-dependent coupling $m_0(\Lambda)$ exists, which satisfies this renormalization condition. For example I tried the Ansatz $m_0(\Lambda)^2 = \alpha + \beta \Lambda^2$, but with this Ansatz the renormalization condition becomes \begin{equation} \Lambda^2 + \left(\alpha + \beta \Lambda^2\right) \mathrm{log}\left( \frac{\beta}{1+\beta}\right) = 32 \pi^2 m_R^2, \end{equation} which can only hold if \begin{equation} \beta \log\left(\frac{\beta}{1+\beta} \right) \overset{!}{=} 1 \end{equation} holds, which has unfortunately no real solution for $\beta$.

Do you have a clue, how to show (within this scheme), that an $m_0(\Lambda)$ exists, which satisfies above renormalization condition or do you have maybe an idea regarding its explicit form?

P.S. In Maggiores book, this problem is solved by just saying, that $m_0(\Lambda)$ is chosen, such that it cancels the divergency, i.e. by setting $m_R^2 = m_0^2(\Lambda) + D_F(0;\Lambda)$. Hence he just assumes, that above renormalization condition is satisfied by some $m_0(\Lambda)$ without showing it.

P.P.S. In https://www.physics.mcgill.ca/~jcline/qft1b.pdf (eq.40/41) the problem is "solved" by replacing somehow the $\Lambda$-dependent bare coupling in the logarithm by a mass scale. In my opinion this is not a sensible solution and just wrong.

P.P.P.S. In https://www.physics.umd.edu/courses/Phys851/Luty/notes/renorm.pdf it is solved by a nice way (see eq. 3.10), namely by utilization of a RG-differential equation. Unfortunately existence/uniqueness and stability behaviour of a solution wasn't shown, but I guess, that this is no major obstacle. But I don't want to show it with a RG-equation!


1 Answer 1


Your equation for $m_0^2(\Lambda)$ came from first order perturbation theory in $\lambda$ so you would only expect the solution to be valid to first order $\lambda$. It is trivial to solve to first order, since at zeroth order $m_0^2=m_R^2$.

$$m_0^2(\Lambda)= m_R^2- \frac{\lambda}{32 \pi^2} \left(\Lambda^2 + m_R^2 \log\left(\frac{m_R^2}{\Lambda^2 + m_R^2} \right)\right) +\mathcal{O}(\lambda^2).$$

(As a side note, your use of the limit $\Lambda\rightarrow 0$ does not make sense. The whole point is you are choosing a different $m_0^2(\Lambda)$ for each $\Lambda$ such that the physical pole $m_R^2$ stays fixed.)

  • $\begingroup$ Thanks for the answer. I think you are right with your main point: In the term of eq. (3) proportional to $\mathcal{O}(\lambda)$ we can replace $m_0(\Lambda)$ by a finite constant, since anything else would correspond to a higher order in perturbation theory. Thank you! Nevertheless, I think that my limit $\Lambda \rightarrow \infty$ makes sense: Requiring equality of the pole and $m_R^2$ just in the limit is just a weaker renormalization condition. But this makes equally good sense, when the regulator is removed. I mean: "Your" limit is just a special case of "my" limit. $\endgroup$
    – warpfel
    Jun 23, 2023 at 17:19

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