Is the QCD potential really monotonic ? How does it prevent two quarks from meson to annihilate? The QCD potential is made of two terms
-(4/3) * alpha_s / r
that describes the short distance
and 
the term 
+k*r
that describes the long distance
Of course, alpha is a function of energy, so it is a function of radius.
But in the measurements of QCD potential, I always see that the graphical potential VQCD=f(r) is monotonic, so I would think that it only attracts the two quarks.
So what does prevent the two quarks of the meson to annihilate from the QCD potential. Remark : in the word 'QCD potential', there is Q=quantum, so normally this potential should take into account the quantum effects. Does it take it into account ?
Does the potential stops decreasing at a low radius value ? At which value ? Why the measurements have not been made down to this low r value.
Thank you for your help.
 A: This monotonic quark-antiquark potential is phenomenological, not fundamental. It can be understood as arising from gluon exchange.
The potential doesn’t prevent the quark and antiquark in, for example, the neutral pion from annihilating. This annihilation typically produces two photons. A $\pi^0$ decays after a mere 84 attoseconds.
The QCD potential doesn’t stop decreasing as $r$ decreases, any more than the electrostatic potential does. (In both cases, the $1/r$ dependence arises from the exchange of a massless gauge boson. The confining $r$ term in QCD is due to direct gluon-gluon interactions, and has no counterpart in electromagnetism, which does not have direct photon-photon interactions.) However, the strong coupling constant $\alpha_s$ decreases slowly to zero at small distances / large energies. This is called asymptotic freedom.
A: I'd like to re-emphasize and elaborate on G. Smith's statement that the potential we're discussing is not a fundamental aspect of QCD.  (Perhaps these remarks should be comments to his answer, but I suspect they'll end up too long and unwieldy for that.)
I believe the main source of confusion here is an assumption that the potential ought to capture all the dynamics of QCD.  I think this is illustrated by the remark, "in the word 'QCD potential', there is Q=quantum", which currently appears in the question.
First of all, I would not use the term "QCD potential" for $V(r) = \sigma r - \frac{A}{r}$.  It's better to think of this as an "effective potential", and I'll mention some reasons for that below.  While you can get some quantum effects out of this potential (most notably a rough spectrum of charmonium [and bottomonium] states predicted by the Cornell crew almost 45 years ago), for the full picture you really need to work with the fundamental QCD field theory lagrangian itself.
Now, someone most familiar with non-relativistic quantum mechanics might object that the lagrangian is $L = T - V$, so we might expect the potential $V$ provide pretty much the same information as the lagrangian itself.  The key is that this form for the lagrangian does not hold in relativistic contexts, and relativistic effects are often crucial in the settings where we consider QCD.  I believe the potential description as a whole is only really appropriate for non-relativistic regimes of the theory (one big reason why it's merely "effective"), though that's not an issue I've thought about deeply.  (I don't do much with non-relativistic QCD in my lattice field theory research.)
The question indicates familiarity with the fact that short distances correspond to high energies, which we can use to qualitatively understand the obstruction to experimental measurements of the potential at arbitrarily short distances $r$.  In order to resolve smaller distances $r$, higher-energy probes have to be used (for example in deep inelastic scattering), eventually far exceeding the mass-energy of the quarks.  That is, we are now in a relativistic regime whose description in terms of the potential $V(r)$ is questionable at best.
Rather than talking about quark annihilation within mesons (which does occur, as G. Smith pointed out), perhaps it would be useful for me to rephrase this part of the question along the following lines: "Why doesn't the monotonic potential cause the quarks to collapse to $r \to 0$?"  Here is a sleazy qualitative way to look at that: If a quark is restricted to a small region $\Delta r$ around $r = 0$, the uncertainty relation $\Delta r \Delta p \gtrsim \hbar$ implies a large quantum uncertainty in its momentum $\Delta p \gtrsim \frac{\hbar}{\Delta r}$ which will very quickly kick it back out to a larger radius.  This is a purely quantum effect that has nothing to do with the QCD coupling or asymptotic freedom.
[As an aside, the QCD coupling does become relevant once that quark is kicked out to a large enough distance---in round numbers, an appropriate 'large' distance would be the charge radius of a typical hadron, approximately a femtometer (fm).  Since $\hbar c \approx 200$MeV$\cdot$fm, this $\Delta r \approx 1$fm implies a typical hadronic energy scale $c\Delta p \sim \frac{\hbar c}{\Delta r}$ of a few hundred MeV, and hence hadron masses far larger than the MeV-scale masses of the up and down quarks.]
While I'm ranting about the potential, let me wrap up by mentioning that there are subtleties at long distances, too.  The major complication here is known as "string breaking" and corresponds to a sufficiently massive/energetic meson decaying into two lighter mesons (for example, the process $\rho \to \pi\pi$).  [This phenomenon is the historical origin of string theory, though the "string" in question is now understood to be an effective description of a "gluonic flux tube".  The coefficient $\sigma$ in the $V(r)$ I wrote above is commonly known as the "string tension".]  In very rough terms, string breaking can be expected once there is enough energy in the system to kick a quark out to a large enough distance $r$ for $\sigma r \subset V$ to be a few hundred MeV---larger than the masses of the two final-state mesons.
Of course, when such few-hundred-MeV energies are at play, we shouldn't expect few-MeV up and down quarks to be non-relativistic.  This is why the "Cornell potential" $V(r)$ I wrote above is most useful for heavier charm and bottom quarks, which are massive enough to make a non-relativistic potential description a reasonable approximation for some purposes.  This is another big reason to consider this potential merely "effective".  In lattice field theory calculations we generally consider the "static potential" defined in the limit where the quarks are infinitely massive (and hence are unable to move with any finite kinetic energy).
