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Decoupling

Last updated Aug 5, 2023

Heavy particles should decouple, which means that they do not contribute at lower energies. E.g. the top quark should only contribute to the beta function for $\mu \ll m_t$.

Consider the vacuum polarization function

$$\Pi(q^2) = \frac{\alpha}{3\pi} \mu^{-2\epsilon}\left(\frac{1}{\epsilon} + \log 4\pi - \gamma_E - 6\int^1_0\text{d}x x(1-x) \log\frac{m^2-q^2x(1-x)}{\mu^2} + \mathcal{O}(\epsilon)\right)$$

  1. Now in a mass-dependent renormalization scheme like the $\mu$ scheme we get the renormalized function $\Pi^R(\frac{q^2}{\mu^2})$

$$\Pi^R(\frac{q^2}{\mu^2}) = \frac{\alpha}{3\pi}(-6\int^1_0 \text{d}x x(1-x)\log(\frac{m^2 - q^2 x(1-x)}{m^2 + \mu^2 x (1-x)}))$$

One can obtain the contribution to the beta function (I think by derivative) $$\beta(\alpha) = 4\frac{\alpha}{\pi}\int^1_0\text{d}x \frac{\mu^2x^2(1-x)^2}{m^2+\mu^2x(1-x)}$$

For large masses $m^2 \gg \mu^2$ the beta function goes to zero $\beta \to 0$ for large masses $m \to \infty$. Therefore in the heavy mass limit the fermion decouples and does not contribute to the running of the coupling anymore. This allows to neglect heavy particles and describe our world as low energy EFT.

  1. In a mass-independent renormalization scheme like $\overline{\text{MS}}$ does not happen at the level of the beta function. It has to be enforced through the construction of a suitable EFT.

$$\Pi^R(\frac{q^2}{\mu^2}) = \frac{\alpha}{3\pi}(-6 \int^1_0\text{d}x x(1-x)\log\frac{m^2-q^2x(1-x)}{\mu^2})$$

contribution to the beta function $$\beta_\overline{\text{MS}} = \alpha(\frac{2\alpha}{3\pi}) + \mathcal{O}(\alpha^3)$$

Therefore beta function does not depend on the mass.

Constructing decoupling EFT with the EFT recipe:

  1. energy scales: We have a Lagrangian with light d.o.f. $m$ and one heavy d.o.f. $M$
  1. d.o.f.: light d.o.f.

  2. symmetries: Symmetries of fundamental theory

  3. Lagrangian: $$\mathcal{L} = \mathcal{L}_\text{only light d.o.f.} + \sum_i c_i \frac{\mathcal{O}_i}{M^{d_i - 4}}$$

  1. power counting: Operators scale with the energy of the light d.o.f. $\ll M$

  2. accuracy:

  3. matching: Require same amplitudes for incoming and outgoing light d.o.f. at $\mu = M$: Heaby mass threshold

Example: QCD with $n-1$ light quarks and one heavy quark of mass $M$.

$\mathcal{L}\text{EFT} = \mathcal{L}\text{QCD - light d.o.f.} + \sum_i c_i\frac{\mathcal{O}_i}{M^{d_i - 4}}$

When matching, we get a relation between the coupling constant of the full theory ($\alpha^{n_f}_S$) and the EFT (($\alpha^{n_f-1}_S$)) where we only have one flavor less. The couplings are renormalized in $\overline{\text{MS}}$ and have corresponding beta functions for their respective amount of active flavors $\beta^{(n_f)}$ and $\beta^{(n_f - 1)}$.

$$\alpha^{(n_f)}_S(\mu^2) = \alpha^{(n_f-1)}S(\mu^2)\left(1 + \sum^\infty{k=1} c_k \left(\frac{\alpha^{(n_f-1)}_S(\mu^2)}{\pi}\right)^k\right)$$

with $c_i = \sum_k c_{ik} \log^{k}\frac{\mu}{M}$

Differentiating by $\log\mu$ and using the running $\frac{\text{d}\alpha^{(i)}_S}{\text{d}\log\mu}=\alpha^{(i)}_S\beta^{(i)}(\alpha^{(i)}_S)$ we get $$-\frac{\alpha^{(n_f)^2}_S}{2\pi}\beta_0^{(n_f)} = -\frac{\alpha^{(n_f-1)^2}_S}{2\pi}\beta_0^{(n_f-1)} + \frac{\alpha^{(n_f - 1)^2}S}{\pi}c{11}$$

We can then get $c_{11}$ by inserting the expression of the beta function in terms of the number of colors and active flavors and using $\alpha^{(n_f)}_S = \alpha^{(n_f-1)}_S + \mathcal{O}(\alpha^2_S)$.

For $c_{10}$ compute loop with heavy fermion running inside. In $\overline{\text{MS}}$ $\Pi^R(q^2=0, \mu=M)=0$. This leads to $c_{10}=0$ and in turn $\alpha^{(n_f)}_S(M) = \alpha^{(n_f - 1)}_S(M) + \mathcal{O}(\alpha^3_S)$ ($\alpha_S$ continous at $M$ threshold).

Similar for two loop matching. Different results depending on the choice of the threshold mass: Pole-mass or $\overline{\text{MS}}$-mass

One finds $c_{20} \neq 0$ which leads to $$\alpha^{(n_f)}_S(M) = \alpha^{(n_f-1)}S(M)\left(1 + c{20}\left(\frac{\alpha^{(n_f - 1)(M)}_S}{\pi}\right)^2+\mathcal{O}(\alpha^2_S)\right).$$ So $\alpha_S$ is discontinous at $M$, which is okay since it is not a physical observable