If a charged pion decays via a charged current $W$, the $W$ boson is much heavier than the pions. How is this not a problem? Could the processes be off-shell in such a drastic manner?
It is tempting consider the $Z$ or $W$ as virtual particles, however if we were to consider the Drell Yan process, weak interaction only contributes to the scattering cross section at a centre of mass energy above 60 GeV. It seems that there is actually no essential difference, except in the first case we are considering a bound state and in the later case we are dealing with a scattering process.
Best Answer
Of course charged pion decays involve the exchange of virtual Ws, like all low-energy weak interactions. As you may have learned in your introductory HEP course, charged pions are the longest-lived hadrons, since, being the lightest, cannot decay strongly, but only weakly, so their decay is slow: 2.6 $\cdot 10^{-8}$s.
In natural units, this amounts to $$ \Gamma_{\pi^{\pm}}\sim 2.5~\cdot 10^{-15}MeV, $$ as compared to a routine strong decay, $$ \Gamma_\rho \sim 150 MeV. $$
Since the W involved is necessarily virtual, its propagator amounts to a 4th power of $M_W$ in the denominator of the width, so, veeery crudely, from dimensional analysis, $$ \Gamma_{\pi^{\pm}}\propto \frac{ m_\mu^2(K.E.)^3}{M_W^4} $$ within a few orders of magnitude of the above experimental width! (Have used 4MeV for the kinetic energy of the muon.)
So the comparison to Drell-Yann is inapposite: you never see the suppressed low-energy Drell-Yan process. But the pion has no other way to decay, beyond this freakishly small width channel, suppressed by 17 orders of magnitude w.r.t. its fellow hadrons' strong decays!