Heisenberg uncertainty principle and minimum energy In an exercise, given the average lifetime $\tau$ of a particle, the author estimates the minimum energy using the uncertainty principle formula : $\Delta E \Delta t \geq \hbar/2$, assuming $\Delta t = \tau$ and interpreting $\Delta E$ as the minimum energy.
Shouldn't $\Delta E$ and $\Delta t$ be uncertainties and not minimum or average values ?
 A: If we model particle decay as a Poisson process, then the probability density function for decay is an exponential distribution:
$$f(t;\lambda)=Ce^{-\lambda t},\text{where } t\in[0,\infty).$$
The normalization factor $C$ can be found by integrating over the time domain and requiring that the total probability be equal to $1$.
$$1=\int_0^{\infty}f(t;\lambda)\,dt=\int_0^{\infty}Ce^{-\lambda t}\,dt=\dfrac{C}{\lambda},$$
hence, $C=\lambda$.
The mean lifetime $\tau=\langle t\rangle$ under this distribution is
$$\langle t\rangle=\int_0^{\infty}t\,f(t;\lambda)\,dt=\int_0^{\infty}t\,\lambda e^{-\lambda t}\,dt=\frac{1}{\lambda},$$
so we can replace $\lambda$ with $\frac{1}{\tau}$ in our formulas. Then you can verify that the second moment of distribution is
$$\langle t^2\rangle=\int_0^{\infty}t^2\,f(t;\tau^{-1})\,dt=2\tau^2,$$
and so standard deviation is:
$$\left(\Delta t\right)^2=\langle t^2\rangle-\langle t\rangle^2=2\tau^2-\tau^2=\tau^2\\
\implies\Delta t=\tau.$$
As for the energy uncertainty, if the energy radiated by the decay is some value $E$, then energy that has been released from decay at time $t$ is either zero (if the decay hasn't happened yet) or $E$ (if the decay has already happened). So the spread of possible values is just $E$. If I ask you to guess a number between 0 and 100, you probably won't guess right but you will at least have a maximum uncertainty to within 100. So it's not unreasonable in this case to estimate the minimum uncertainty of energy here as simply the minimum energy itself. It'll least give the correct order of magnitude.
