# How to determine matching coefficient in Effective Field Theory?

• I
• Markus Kahn
In summary: Alternatively, you can solve for one coefficient and then substitute it into the other equation, giving a single equation with one unknown.

#### Markus Kahn

TL;DR Summary
Given the amplitudes of a scattering process at a fixed order of the EFT and full theory, I don't really understand how one is supposed to match them..
Assume that I have the Lagrangian
$$\mathcal{L}_{UV} =\frac{1}{2}\left[\left(\partial_{\mu} \phi\right)^{2}-m_{L}^{2} \phi^{2}+\left(\partial_{\mu} H\right)^{2}-M^{2} H^{2}\right] -\frac{\lambda_{0}}{4 !} \phi^{4}-\frac{\lambda_{2}}{4} \phi^{2} H^{2},$$
where ##\phi## is a light scalar field with mass ##m_L## and ##H## a heavy one with mass ##M##. Let the Lagrangian of the effective field theory (EFT) be
$$\mathcal{L}_{EFT} = \frac{1}{2}\left[\left(\partial_{\mu} \phi\right)^{2}-m^{2} \phi^{2}\right]-C_{4} \frac{\phi^{4}}{4 !}-\frac{C_{6}}{M^{2}} \frac{\phi^{6}}{6 !}.$$

Assume that I have calculated the ##4##-point function up to ##1##-loop order and regularized it correctly (renormalization scale ##\mu##). The results are:
\begin{align*} \mathcal{M}_{4}^{\mathrm{EFT}} &=-C_{4}+\frac{C_{4}^{2}}{32 \pi^{2}}[f(s, m)+f(t, m)+f(u, m)] \\ &+\frac{3 C_{4}^{2}}{32 \pi^{2}}\left(\log \left(\frac{\mu^{2}}{m^{2}}\right)+2\right)+\frac{C_{6} m^{2}}{32 \pi^{2} M^{2}}\left(\log \left(\frac{\mu^{2}}{m^{2}}\right)+1\right)\\\\ \mathcal{M}_{4}^{\mathrm{UV}} & \approx-\lambda_{0}+\frac{3 \lambda_{0}^{2}}{32 \pi^{2}}\left(\log \left(\frac{\mu^{2}}{m^{2}}\right)+2\right)+\frac{3 \lambda_{2}^{2}}{32 \pi^{2}}\left(\log \left(\frac{\mu^{2}}{M^{2}}\right)\right)+\frac{m^{2} \lambda_{2}^{2}}{48 \pi^{2} M^{2}} \\ &+\frac{\lambda_{0}^{2}}{32 \pi^{2}}[f(s, m)+f(t, m)+f(u, m)]. \end{align*}

The matching at tree-level resulted in:
$$m^2=m_L^2,\qquad C_4 = \lambda_0,\qquad C_6=0.$$
I would now like to perform the matching at one-loop, i.e. we demand ##\mathcal{M}_4^{EFT}= \mathcal{M}_4^{UV}+O(M^{-4})##.

Problem
We have two unknowns, ##C_4## and ##C_6##, that need to be expressed in terms of ##\lambda_0, \lambda_2, m, M,## etc. But ##\mathcal{M}_4^{EFT}= \mathcal{M}_4^{UV}+O(M^{-4})## gives us only one equation.. I don't see how we can determine both coefficients with only the above information.

Notes
I'm reading Adam Falkowski's lecture notes, see here. In section 2.3, p. 24, he performs the matching with only the above information and determines ##C_4##... I don't see how that is supposed to work.

Markus Kahn said:
Summary:: Given the amplitudes of a scattering process at a fixed order of the EFT and full theory, I don't really understand how one is supposed to match them..

Assume that I have the Lagrangian
$$\mathcal{L}_{UV} =\frac{1}{2}\left[\left(\partial_{\mu} \phi\right)^{2}-m_{L}^{2} \phi^{2}+\left(\partial_{\mu} H\right)^{2}-M^{2} H^{2}\right] -\frac{\lambda_{0}}{4 !} \phi^{4}-\frac{\lambda_{2}}{4} \phi^{2} H^{2},$$
where ##\phi## is a light scalar field with mass ##m_L## and ##H## a heavy one with mass ##M##. Let the Lagrangian of the effective field theory (EFT) be
$$\mathcal{L}_{EFT} = \frac{1}{2}\left[\left(\partial_{\mu} \phi\right)^{2}-m^{2} \phi^{2}\right]-C_{4} \frac{\phi^{4}}{4 !}-\frac{C_{6}}{M^{2}} \frac{\phi^{6}}{6 !}.$$

Assume that I have calculated the ##4##-point function up to ##1##-loop order and regularized it correctly (renormalization scale ##\mu##). The results are:
\begin{align*} \mathcal{M}_{4}^{\mathrm{EFT}} &=-C_{4}+\frac{C_{4}^{2}}{32 \pi^{2}}[f(s, m)+f(t, m)+f(u, m)] \\ &+\frac{3 C_{4}^{2}}{32 \pi^{2}}\left(\log \left(\frac{\mu^{2}}{m^{2}}\right)+2\right)+\frac{C_{6} m^{2}}{32 \pi^{2} M^{2}}\left(\log \left(\frac{\mu^{2}}{m^{2}}\right)+1\right)\\\\ \mathcal{M}_{4}^{\mathrm{UV}} & \approx-\lambda_{0}+\frac{3 \lambda_{0}^{2}}{32 \pi^{2}}\left(\log \left(\frac{\mu^{2}}{m^{2}}\right)+2\right)+\frac{3 \lambda_{2}^{2}}{32 \pi^{2}}\left(\log \left(\frac{\mu^{2}}{M^{2}}\right)\right)+\frac{m^{2} \lambda_{2}^{2}}{48 \pi^{2} M^{2}} \\ &+\frac{\lambda_{0}^{2}}{32 \pi^{2}}[f(s, m)+f(t, m)+f(u, m)]. \end{align*}

The matching at tree-level resulted in:
$$m^2=m_L^2,\qquad C_4 = \lambda_0,\qquad C_6=0.$$
I would now like to perform the matching at one-loop, i.e. we demand ##\mathcal{M}_4^{EFT}= \mathcal{M}_4^{UV}+O(M^{-4})##.

Problem
We have two unknowns, ##C_4## and ##C_6##, that need to be expressed in terms of ##\lambda_0, \lambda_2, m, M,## etc. But ##\mathcal{M}_4^{EFT}= \mathcal{M}_4^{UV}+O(M^{-4})## gives us only one equation.. I don't see how we can determine both coefficients with only the above information.

Notes
I'm reading Adam Falkowski's lecture notes, see here. In section 2.3, p. 24, he performs the matching with only the above information and determines ##C_4##... I don't see how that is supposed to work.
Did you take into consideration the fact that the log terms must match and that the constant pieces must also match? This gives two independent relations.