Energy operator

Using the energy operator to the Schrödinger equation:

${\displaystyle i\hbar {\frac {\partial }{\partial t}}\Psi (\mathbf {r} ,\,t)={\hat {H}}\Psi (\mathbf {r} ,t)\,\!}$

can be obtained:

${\displaystyle {\begin{aligned}&{\hat {E}}\Psi (\mathbf {r} ,\,t)={\hat {H}}\Psi (\mathbf {r} ,\,t)\\\end{aligned}}\,\!}$

where i is the imaginary unit, ħ is the reduced Planck constant, and ${\displaystyle {\hat {H}}}$ is the Hamiltonian operator.

In a stationary state additionally occurs the time-independent Schrödinger equation:

${\displaystyle {\begin{aligned}&E\Psi (\mathbf {r} ,\,t)={\hat {H}}\Psi (\mathbf {r} ,\,t)\\\end{aligned}}\,\!}$

where E is an eigenvalue of energy.

The relativistic mass-energy relation:

${\displaystyle E^{2}=(pc)^{2}+(mc^{2})^{2}\,\!}$

where again E = total energy, p = total 3-momentum of the particle, m = invariant mass, and c = speed of light, can similarly yield the Klein–Gordon equation:

${\displaystyle {\begin{aligned}&{\hat {E}}^{2}=c^{2}{\hat {p}}^{2}+(mc^{2})^{2}\\&{\hat {E}}^{2}\Psi =c^{2}{\hat {p}}^{2}\Psi +(mc^{2})^{2}\Psi \\\end{aligned}}\,\!}$

that is:

${\displaystyle {\frac {\partial ^{2}\Psi }{\partial t^{2}}}=c^{2}\nabla ^{2}\Psi -\left({\frac {mc^{2}}{\hbar }}\right)^{2}\Psi \,\!}$