# Can the energy-time uncertainty principle be applied to rest energy?

In Introduction to Quantum Mechanics (3rd edition) by David J.Griffiths, when deriving the energy-time uncertainty principle，$$\Delta$$E is defined as $$\sigma _H$$, the standard deviation of H. And in an example following the derivation (Example 3.7), the $$\Delta$$ particle is introduced. The problem says that the $$\Delta$$ particle lasts about $$\frac{1}{10^{23}}$$ s before spontaneously disintegrating, and the uncertainty of its rest energy ($$\text{mc}^2$$) is about 120 MeV. Therefore, the product $$\Delta E \Delta t=\frac{6 \text{Me} V s}{10^{22}}>\frac{\hbar }{2}$$ satisfies the energy-time uncertainty principle. Since when H acts on a wave function, it gives the sum of kinetic energy and potential energy, and $$\Delta$$E is defined as $$\sigma _H$$, I think $$\Delta$$E should be the uncertainty of kinetic energy and potential energy. Can $$\Delta$$E be the rest energy of a particle?

$$H=\int \hbar \omega_{\vec{p}} N_{\vec {p}}d^3p$$
$$N$$ is the particle number density operator.
For single particle states, the eigenvalue is $$\hbar\omega_{p}=\hbar\sqrt {p^2+m^2}=\hbar(m+\frac{p^2}{2m}+..)$$ , which includes the rest energy.