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?


1 Answer 1


In Quantum Field theory, the Hamiltonian isn't "kinetic energy plus potential energy". The Hamiltonian includes the rest energy.

The Hamiltonian of the free theory is:

$$H=\int \hbar \omega_{\vec{p}} N_{\vec {p}}d^3p$$

$N$ 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.

The inclusion of rest energy is important because kinetic energy isn't conserved in particle collisions, because the masses of products can differ from the masses of the incoming particles. Total energy is conserved.


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