Why does a short lifetime of a nucleon excitation imply a broad mass distribution? In this Coursera lecture by Mercedes Paniccia on inelastic scattering of electrons off hadrons, at 1:45 she says that

At moderate $q^2$, inelastic processes produce excited states of the nucleon, also called resonances, such as $\Delta^+$, which have the same quantum numbers as the proton. These resonances have an extremely short lifetime and therefore a broad mass distribution. 

I assume that the broad mass distribution means that the invariant mass of the produced resonance can assume values over a broad range. Why does the resonance lifetime being short imply that its mass distribution is broad?
 A: A typical propagator in QFT at tree level looks like $(p^2-M^2)^{-1}$. Naively, this tells you that if the four-momentum of your virtual particle becomes on shell, $p^2=M^2$, then the amplitude for this process diverges. However, if you take into account loops in your scattering processes, you can arrive at the famous Breit-Wigner distribution, which states the the propagator gets corrected to
$$\frac{1}{p^2-M^2+iM\Gamma},$$
where $\Gamma$ is the total decay rate of your virtual particle. Squaring this gives you cross section which will look like
$$\sigma\sim\frac{1}{(p^2-M^2)^2+M^2\Gamma^2}.$$
Thus, our original divergent peak at $p^2=M^2$ has been broadened. The full-width the distribution of $\sigma$ against $\sqrt{p^2}$ at half-maximum is given by $\Gamma$. Thus, if the decay rate is very high (the lifetime is very short), then the width is very wide, and the distribution of the cross section against $\sqrt{p^2}$ has a very broad peak (equivalently, a very broad mass distribution).

A: By the generalization of Heisenberg's uncertainty principle, $\Delta E \Delta t \geq \frac{\hbar}{2}$ and so if the lifetime is very short and well defined, the energy distribution is badly defined (the masses are broad). 
