I see that it appears as a constant in the relation for the running of the strong coupling constant. What is its significance? Does it have to be established by experiment? Is it somehow a scale for quark confinement? If yes, how? I ask because I saw this in Perkins' Particle Astrophysics

After kT fell below the strong quantum chromodynamics (QCD) scale parameter ∼ 200 MeV, the remaining quarks, antiquarks, and gluons would no longer exist as separate components of a plasma but as quark bound states, forming the lighter hadrons such as pions and nucleons.


2 Answers 2


Dear dbrane, $\Lambda_{\rm QCD}$ is the only dimensionful parameter of pure QCD (pure means without extra matter).

It is dimensionful and replaces the dimensionless parameter $g_{\rm QCD}$, the QCD coupling constant. The process in which a dimensionless constant such as $g$ is replaced by a dimensionful one such as $\Lambda$ is called the dimensional transmutation:


The constant $g$ isn't quite constant but it depends on the characteristic energy scale of the processes - essentially logarithmically. Morally speaking, $$ \frac{1}{g^2(E)} = \frac{1}{g^2(E_0)} + K \cdot \ln (E/E_0), $$ at least in the leading approximation. Because $g$ depends on the scale, it is pretty much true that every value of $g$ is realized for some value of the energy scale $E$. Instead of talking about the values of $g$ for many specific values of $E$, one may talk about the value of $E$ where $g$ gets as big as one or so, and this value of $E$ is known as $\Lambda_{\rm QCD}$ although one must be a bit more careful to define it so that it is 150 MeV and not twice as much, for example.

Yes, it is the characteristic scale of confinement and all other typical processes of pure QCD - those that don't depend on the current quark masses etc. In most sentences about the QCD scale, including your quote, the detailed numerical constant is not too important and the sentences are valid as order-of-magnitude estimates. However, given a proper definition, the exact value of $\Lambda_{\rm QCD}$ may be experimentally determined. With this knowledge and given the known Lagrangian of QCD - and the methods to calculate its quantum effects - one may reconstruct the full function $g(E)$.

  • 4
    $\begingroup$ Since $g(\Lambda_{QCD})\sim 1$, one can say that for $E\ll\Lambda_{QCD}$ the theory is strongly coupled ($\Rightarrow$ confinement), and for $E\gg\Lambda_{QCD}$ it is weakly coupled ($\Rightarrow$ asymptotic freedom) $\endgroup$
    – Stan
    Commented May 4, 2014 at 15:57
  • $\begingroup$ Notice though, the reasoning that lets you write $g(E)$ (or $\alpha$ in the other answer) that way, and thus brings you to $\Lambda_{\text{QCD}}$, somehow assumes perturbative regime, so stricly speaking it doesn't hold for $E \approx \Lambda_{\text{QCD}}$ (you just know QCD is non-perturbative there). All this just to say, you can't prove there indeed is a very strong coupling (confinement) just noting that $g(\Lambda_{\text{QCD}})$ calculated that way is not $\ll 1$ (you can deduce asymptotic freedom for $E \gg \Lambda_{\text{QCD}}$). $\endgroup$ Commented Nov 5, 2016 at 11:55
  • 1
    $\begingroup$ Well, energies $E\approx \Lambda_{QCD}$ are exactly the fuzzy boundary between the (higher) energies where the perturbative expansions are good, accurate, and useful, and the (lower) energies where the perturbative approaches don't work well. $\endgroup$ Commented Nov 7, 2016 at 18:38
  • $\begingroup$ LubošMotl I agree (altough I would call it the boundary between the higher energies where perturbative approaches don't work well, and the lower ones where they're hopelessly ruled out), but I wanted to specify that 'non perturbative' doesn't imply 'infinitely strong, that is confinement'. It then happens that confinement indeed takes place, and at that scale, but as far as I know the $\Lambda_{\text{QCD}}$ thing is a nice start for guesses, there, not an argument. Sorry, my comment was more a reply to @Stan's one than to your answer by itself, I should have said that before. $\endgroup$ Commented Nov 8, 2016 at 21:14
  • 1
    $\begingroup$ It's dangerous to identify "confinement" with "infinitely strong". The coupling is only infinitely strong formally at "infinite length scales". Whether confinement occurs depends on physics of arbitrarily long length scales longer than the QCD length scale. $\endgroup$ Commented Nov 11, 2016 at 16:21

Working in dimensional regularization, in the MS bar scheme consider the renormalization group equation for the strong coupling


Reordering terms we get an expression we can integrate


for concreteness' sake let's consider the one loop case (my reasonnig will apply in general but the formulas get cumbersome) where $\beta_1=\beta_2=\ldots=0$


integration yealds


$C_1$ and $C_2$ are integration constants. We can simplify matters merging them simply doing $C\equiv{}C_1-C_2$ which leaves us with


You might be wondering about the $r$. We have integrated a first order differential equation, which should leve us only with one integration constant. The $r$ comes from the fact that $\mu$ is a dimensionful parameter, so in order to write the logarithm that appears in the integral we must introduce a dimensionful parameter. It is important to notice that $r$ can be ANY dimensionful parameter (with the same sign as $\mu$ for the log to make sense) since


for ANY value of $r$. $r$ just dissapears after the derivative. Thus, while $C$ will be specified by some intial conditions on $\alpha$ on some point $\mu_0$ with $r$ we have complete freedom to choose whatever we want (as long as it has the same sign as $\mu$).

Of course, different $r$-s will give us different $C$-s. To simplify matters we could wonder if we can always choose an appropriate $r$ such that $C=0$. In order to see if this is indeed possible, assume that $C'$ and $r'$ are such that the initial conditions are satisfied. Then if we wish $C$ to be zero the following equation must be satisfied


whose solution is


The moral of the story is that for any $r'$ and $C'$ we can always pick $r$ such that $C=0$. This way the solution of the differential equation in the first equation of this answer can be written (at one loop in the beta function)

$$\ln{\frac{\mu}{r}} = \frac{1}{\alpha\beta_0} \quad \Rightarrow \quad \frac{\mu}{r} = e^{1 / \alpha\beta_0} $$

This $r$ is $\Lambda_{QCD}$. Changing names and rearranging terms we get the famous formula (you might see some $2\pi$-s floating around in some places, it's just a matter of how you define the $\beta_i$-s in the first equation of the answer)

$$\Lambda_{QCD}= \mu e^{\frac{-1}{\beta_0\alpha(\mu)}}$$

One point I would like to make. You may have heard sometimes that $\Lambda_{QCD}$ is renormalization group invariant. The analysis above should make this statement crystal clear, after all we have seen that $\Lambda_{QCD}$ is, by construction, a mere constant!

Solving for $\alpha$


Notice that $\mu=\Lambda_{QCD}$ makes the coupling become singular. Therefore we see that $\Lambda_{QCD}$ is the scale at which nonperturbative effects take over.

In order to establish its numerical value, from the equation above we readily see that $\alpha$ defines $\Lambda_{QCD}$. Therefore we can get it using the experimental value for $\alpha$.

  • $\begingroup$ What's the experimental value of $\alpha$? What energy scale does it correspond to, i.e. what $\mu$ gives $\alpha(\mu) = \alpha_{exp}$? $\endgroup$
    – JamieBondi
    Commented Jul 14, 2017 at 18:44
  • $\begingroup$ @JamieBondi Check the pdg $\endgroup$
    – Yossarian
    Commented Jul 18, 2017 at 13:03

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.