3
$\begingroup$

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 again demand $\mathcal{M}_4^{EFT}= \mathcal{M}_4^{UV}+O(M^{-2})$.

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 $\mathcal{M}_4^{EFT}= \mathcal{M}_4^{UV}+O(M^{-2})$ 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...

$\endgroup$

1 Answer 1

1
$\begingroup$

I don't see how we can determine both coefficients with only the above information.

We can't.

We need one observable for each EFT quantity that we want to match to the UV. The physical mass $m_\text{phys}^2$ allows to 1-loop match the EFT mass $m$ to the UV quantities, and the 2-2 scattering amplitude allows to 1-loop match the EFT coupling $C_4$ to the UV quantities. If you wanted to 1-loop match the EFT coupling $C_6$, you'd need another equation, as you correctly observe, coming from an observable such as the 3-3 scattering amplitude.
The 3-3 scattering amplitude is very messy and the author doesn't even give the calculations for the tree-level matching with the $\lambda_1$ coupling turned on, but only the results (eq. $2.16$).

If you want to match the coupling $C_6$ without the messy calculations, the author provides in part 3 a functional method to obtain it far more easily: the result is in eq. $3.47$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.