in Quantum Field Theory

Renormalization Group Flow

The standard model contains three gauge couplings, which are very different in strength. This is not really a problem of the standard model, because we can simply put these measured values in by hand. However, Grand Unified Theories (GUTs) provide a beautiful explanation for this difference in strength. A simple group $G_{GUT}$ implies that we have only one gauge coupling as long as $G_{GUT}$ is unbroken. The gauge symmetry $G_{GUT}$ is broken at some high energy scale in the early universe. Afterwards, we have three distinct gauge couplings with approximately equal strength. The gauge couplings are not constant, but depend on the energy scale. This is described by the renormalization group equations (RGEs).

The RGEs for a gauge coupling depend on the number of particles that carry the corresponding charge. (This is discussed in detail below.) Therefore, we can use the known particle content of the standard model to compute how the three couplings change with energy. This is shown schematically in the figure below. unification-coupling-strengths

The couplings change differently with energy, because the number of particles that carry, for example, color ($SU(3)$ charge) or isospin ($SU(2)$ charge) are different. The known particle content of the standard model and the hypothesis that there is one unified coupling at high energies therefore provide a beautiful explanation why strong interactions are strong and weak interactions are weak.

This is not just theory. For example, the strong coupling “constant” $\alpha_S$ has been measured at very different energy scales. Some of these measurements are summarized in the following plot.

 

To understand how all this comes about, recall that in quantum field theory we have a cloud of virtual particle-antiparticle pairs around each particle. This situation is similar to the classical situation of an electron inside a dielectric medium. Through the presence of the electron, the electrical neutral molecules around it get polarized, which is illustrated in the figure below. As a result, the electrical charge of the electron gets partially hidden or screened. This is known as dielectric screening.

dielectric

Analogous to what happens in a dielectric medium, the virtual particle-antiparticle pairs get polarized and the charge of the particle screened. Concretely this means if we are close to a particle, we measure a different charge than from far away because there are fewer virtual particle-antiparticle pairs that screen the charge. This screening effect happens not only for electrical charge but for color-charge and weak-isospin, too. In particle physics, the notion of distance is closely related to the notion of energy. If we shoot a particle with lots of energy onto an electron it comes closer to the electron before it gets deflected than a particle with less energy. Therefore, the particle with more energy feels a larger charge. It may now seem that our three coupling strengths all get bigger if we measure them at higher energies. However, it turns out that gauge bosons have the opposite effect than fermions. They anti-screen and thus make a given charge bigger at larger distances. Recall that we have:

  • One gauge boson for $U(1)$.
  • Three gauge bosons for $SU(2)$, because the adjoint representation is $3$-dimensional.
  • Eight gauge bosons for $SU(3)$, because the adjoint representation is $8$-dimensional.

These numbers tell us that $SU(3)$ couplings are more affected by the gauge boson anti-screening effect, simply because there are more $SU(3)$ gauge bosons. In fact, it can be computed that their effect outweighs the screening effect of the quarks and thus the $SU(3)$ coupling constant gets weaker at smaller distances. For $SU(2)$ the gauge boson and fermion effect is almost equal and therefore the coupling strength is approximately constant. For $U(1)$ the ordinary screening effect dominates and the corresponding coupling strength becomes stronger at smaller distances.  Given the coupling strengths at some energy scale, we can compute at which energy scale they become approximately equal.

This energy scale is closely related to the mass of the GUT gauge bosons $m_X$. From an effective field theory point of view, at energies much higher than $m_X$ the breaking of the GUT symmetry has a negligible effect and therefore the gauge coupling constants unify. The mathematical description of this coupling strength change with energy, known as renormalization group equations, is the topic of the next section. The coupling constants change so slowly with the energy that the scale where they are approximately equal is incredibly high. This means the GUT gauge bosons are so heavy that it is no wonder they have not been seen in experiment yet.

The Renormlization Group Equations

To illustrate the arguments that lead to the famous renormalization group equations, we discuss shortly the arguably simplest example in quantum field theory: the Coulomb potential $ V(r) =\frac{e^2}{4 \pi r}$. In QFT it corresponds to the exchange of a single photon

prop1

and $\frac{1}{4 \pi r}$ is the Fourier transform of the propagator. A $1$-loop correction to this diagram is, for example,

prop2

which yields a correction of order $e^4$ to the Coulomb potential. In momentum space, the Coulomb potential then reads

\begin{equation}
\tilde{V}(p)=e^2 \frac{1-e^2 \Pi_2(p^2)}{p^2},
\end{equation}
where $\Pi_2(p^2) = \frac{1}{2\pi^2} \int_0^1 dx \, x(1-x)\left[ \frac{2}{\epsilon} + \ln \left(\frac{\tilde{\mu}^2}{m^2-p^2x(1-x)}\right) \right]$ (see for example the QFT book by Schwartz or the similar free chapter here). The usual problem in QFT is now that $\Pi_2(p^2)$ is infinite and we need to renormalize. For this reason, we demand that the potential between two particles separated by some distance $r_0$ should be $V(r_0) = \frac{e_R^2}{2\pi r_0}$, where $e_R$ denotes the renormalized charge. In momentum space this means $\tilde{V}(p_0)=\frac{e_R^2}{p_0^2}$. This defines the renormalized charge

\begin{equation}
e_R^2 := p_0^2 \tilde{V}(p_0) = e^2- e^4 \Pi_2(p^2) + \ldots \, ,
\end{equation}

where the dots denote higher order corrections. Equally, we can solve for the bare charge
\begin{equation} \label{eq:defbarecharge}
e^2 := e_R^2+e_R^4 \Pi_2(p_0^2) + \ldots
\end{equation}

At another momentum scale $p$, the potential reads

\begin{align}
\tilde{V}(p)& = \frac{e^2}{p^2} – \frac{e^4 \Pi_2(p^2)}{p^2} + \ldots \stackrel{\text{Eq. \ref{eq:defbarecharge}}}{=} \frac{e_R^2}{p^2} – \frac{e_R^4\left[\Pi_2(p^2)-\Pi_2(p_0^2) \right]}{p^2} + \ldots \notag \\
&=\frac{e^2}{p^2} \left( 1+ \frac{e_R^2}{2\pi^2} \int_0^1 dx \, x(1-x) \ln \left( \frac{p^2x(1-x)-m^2}{p_0^2x(1-x)-m^2} \right) \right) + \ldots
\end{align}

For large momenta $|p^2| \gg m^2$ the mass drops out and we have
\begin{equation}
\tilde{V}(p) \approx \frac{e_R^2}{p^2} \left( 1 + \frac{e_R^2}{12 \pi^2} \ln\left( \frac{p^2}{p_0^2}\right) \right) + \mathcal{O}(e_R^6) = \frac{e_{\text{eff}}^2(p)}{p^2} + \mathcal{O}(e_R^6) \, ,
\end{equation}

with
\begin{equation}
e_{\text{eff}}^2(p) := e_R^2 \left( 1 + \frac{e_R^2}{12 \pi^2} \ln\left( \frac{p^2}{p_0^2}\right) \right) \, .
\end{equation}
This means we introduce an effective charge $e_{\text{eff}}(p)$, such that the potential looks for momentum transfer $p$ like the usual Coulomb potential, but with charge $e_{\text{eff}}(p)$ instead of $e_R$. This describes exactly the screening effect discussed at the beginning of this chapter. For large momenta, which means at short distances, we have an effective charge $e_{\text{eff}}(p)$, which is larger than the renormalized $e_R$. In analogy to the dielectric medium discussed above, here the virtual $e^+ e^-$ pair acts like a dipole.

Including additional loops in the series, such as

prop3
yields analogously

\begin{align}
\tilde{V}(p) &= \frac{e_R^2}{p^2} \left( 1 + \frac{e_R^2}{12 \pi^2} \ln\left( \frac{p^2}{p_0^2}\right) + \left(\frac{e_R^2}{12 \pi^2} \ln\left( \frac{p^2}{p_0^2}\right) \right)^2 + \ldots \right) \notag \\
&= \frac{1}{p^2} \left( \frac{e_R^2}{1- \frac{e_R^2}{12\pi^2\ln\left( \frac{p^2}{p_0^2 } \right)}} \right) = \frac{e_{\text{eff}}^2(p)}{p^2} \, ,
\end{align}

with

\begin{equation}
\label{eq:effetoallorders}
e_{\text{eff}}^2(p) := \frac{e_R^2}{1- \frac{e_R^2}{12\pi^2\ln\left( \frac{p^2}{p_0^2 } \right)}} \,.
\end{equation}

It is convenient to rewrite Eq. \ref{eq:effetoallorders} as

\begin{equation}
\frac{1}{e_{\text{eff}}^2(p)} = \frac{1}{e_R^2} – \frac{1}{12\pi^2} \ln\left( \frac{p^2}{p_0^2} \right) \, .
\end{equation}

The main idea of the renormalization group is that the choice of the reference scale $p_0$ does not matter. What is actually measured in experiments is $e_{\text{eff}}$ and not $e_R$. For example, if we want that our renormalized charge $e_R$ corresponds to the macroscopic electric charge, we need to use $p_0=0$, which corresponds to $r_0 = \infty$. Thus $e_R=e_{\text{eff}}^2(0)$. In contrast, for $p_0=m_e$, we have

\begin{equation}
\frac{1}{e_{\text{eff}}^2(p)} = \frac{1}{e_R^2} – \frac{1}{12\pi^2} \ln\left( \frac{p^2}{m_e^2} \right)
\end{equation}

and therefore $e_R = e_{\text{eff}}(m_e)$. In general

\begin{equation}
\frac{1}{e_{\text{eff}}^2(p)} = \frac{1}{e_{\text{eff}}^2(\mu)} – \frac{1}{12\pi^2} \ln\left( \frac{p^2}{\mu^2} \right) \, .
\end{equation}

Taking the derivative with respect to the scale $\mu$ yields

\begin{equation}
0 = – \frac{2}{e_{\text{eff}}^3(\mu)} \frac{d e_{\text{eff}}(\mu)}{d\mu} + \frac{1}{12\pi^2} \ln\left( \frac{2}{\mu} \right) \, ,
\end{equation}

which we can rewrite as

\begin{equation}
\label{eq:incompleteRGE}
\mu \frac{d e_{\text{eff}}(\mu)}{d \mu} = \frac{e_{\text{eff}(\mu)}^3}{12\pi^2} \, .
\end{equation}

This is called a renormalization group equation (RGE) and it enables us to compute how $e_{\text{eff}}$ depends on the scale $\mu$, i.e. the screening of the charge through the vacuum polarizations. Eq. \ref{eq:incompleteRGE} is not the complete RGE for the electric charge, because we only considered a virtual $e^+ e^-$ pair in the loops, although other particles contribute, too. The derivation of the complete RGEs for various gauge couplings and models is the topic of the next section.

In general right-hand side is called $\beta-$function and for a gauge coupling $g$, we have

\begin{equation}
\mu \frac{d g(\mu)}{d \mu} = \beta (g(\mu) )
\end{equation}

In this post I describe how we compute the $\beta-$functions in practice and here how we can solve them.

One last thing: Maybe you wonder about the name “renormalization group”. Here’s how the book “Quantum Field Theory for the Gifted Amateur” explains it:

The renormalization group is a bit of a misnomer as it is not really a group. The name arises from the study of how a system behaves under rescaling transformations and such transformations do of course form a group. However, the “blurring” that occurs when we rescale and then integrate up to a cut-off, thereby removing fine structure (and this is the very essence of the renormalization group procedure) is not invertible (the fine details are lost and you can’t put them back). Thus the transformations consisting of rescaling and integrating up to a cut-off do not form a mathematical group because the inverse transformation does not exist.”

 

P.S. I wrote a textbook, which is, in some sense, the book I wished had existed when I started my journey in physics. It's called "Physics from Symmetry" and you can buy it, for example, at Amazon. And I'm now on Twitter too if you'd like to get updates about what I'm recently up to.

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