# Typical size of QCD bound states

Consider the Hydrogen atom and its quantum mechanical description.

The excited states of the electron are at energies $E = -E_0/n^2$ with $E_0 = 13.6 \,$eV. This is a weakly coupled bound state ($\alpha_{em}$ is small) and its typical size with the electron in the $n$-th level is parametrized in terms of the Bohr radius $$r_n \sim \frac{n^2}{\alpha_{em} \mu_e}$$ where $\mu_e$ is the electron rest mass. The same result can be obtained by looking at the average of the radius $\langle|\textbf{r}|\rangle$ computed with the wave functions that we got once we solve the Schrodinger equation for the Hydrogen atom.

On the other hand, hadrons are strongly coupled bound state of a different dynamics (QCD) and a analytic description of these bound states is a challenge. However, begin $\Lambda_{QCD}$ the only scale involved in the process, their typical size is often quoted to be

$$r_{had} \sim a\, \Lambda_{QCD}^{-1}$$ where $a$ is coefficient.

Question: What is the value of $a$ for most of the hadrons? Is it expected to be $O(1)$? Does the size of a bound state depend somehow on the binding energy in a clear way? Do we know the value of $a$ by lattice computations or experiments?

QCD has heavy quark bound states (charmonium, bottomonium) which are approximately Coulombic, with sizes $r\sim (\alpha_s m_Q)^{-1}$, where $\alpha_s$ is the running coupling at the scale $m_Q$.

Light quark and purely gluonic bound states have sizes of order $\Lambda^{-1}$, because this is the only scale in the problem. Note that

1) The parameter $a$ in $r=a\Lambda^{-1}$ is scheme dependent (because $\Lambda$ is scheme dependent).

2) $a$ can be determined in lattice QCD (or experimentally). Note that we have to define precisely what we mean by $r$. What is usually quoted for the proton is the root mean square electric charge radius. Other radii, like the weak radius, or the baryon charge radius are slightly different.

3) For typical hadrons and standard schemes (like MS-bar) $a\sim 1$. For example, the proton radius is roughly 0.7 fm, and the MS-bar scale parameter is of order 200 MeV=1/fm.

4) Some QCD bound states are larger. For example, the lead nucleus is a QCD bound state which is quite a bit bigger than $\Lambda^{-1}$.

• Thanks for the answer. Why $r\sim (\alpha_s m_Q)^{-1}$ for heavy quark bound states? Should I look at the $q\bar{q}$-potential? – apt45 Jul 22 '18 at 9:00
• @apt45 Heavy quark bound states are parametrically small ($r\sim m_Q^{-1}\ll \Lambda_{QCD}^{-1}$), and the dominant interaction is the perturbative one gluon exchange potential $V\sim \alpha_s/r$, – Thomas Jul 22 '18 at 15:00
• thanks. Can you give me a reference to understand this fact? Thanks. – apt45 Jul 22 '18 at 15:07
• @apt45 Heavy quarks are non-relativistic, and their motion is described by a Schrodinger equation with a color-Coulomb potential. This is described in standard text books on particle physics (it's chapter 5.7 in my copy of Griffith's book) – Thomas Jul 22 '18 at 15:58

The value of $a$ is undefined. There is no distinct reason for the author for choosing $a$ that would considerably differ from 1.

The expected value is, 1 as you say.

Varying $a$ means a varying RGPEP ( renormalization group procedure for effective particle) parameter. In exact RGPEP calculations, the bound-state mass eigenvalues are independent of the RGPEP parameter.

Changing $a$ would lead to miniscule changes in the interaction and eigenvalues.

To avoid changes in the eigenvalues, the coupling constant needs to be changed, but the required changes of $a$ are very small and can be ignored.