Klein-Gordon equation

How to construct

Starting from the relativistic connection between energy and momentum:

$E^{2}={\mathbf {p}}^{2}c^{2}+m^{2}c^{4}$

Substituting $E\rightarrow i\hbar {\frac {\partial }{\partial t}}$ and ${\mathbf {p}}\rightarrow -i\hbar \nabla$, we get Klein-Gordon equation for free particles as follows:

$-\hbar ^{2}{\frac {\partial ^{2}\psi ({\mathbf {r}},t)}{\partial t^{2}}}=(-\hbar ^{2}c^{2}\nabla ^{2}+m^{2}c^{4})\psi ({\mathbf {r}},t)\qquad \qquad \qquad \qquad \qquad (9.2)$

Klein-Gordon can also be written as the following:

$(\square -K^{2})\psi ({\mathbf {r}},t)=0\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (9.3)$

where $\square =\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}$ is d'Alembert operator and $K={\frac {mc}{\hbar }}$.

Equation (9.3) looks like a classical wave equation with an extra term $K^{2}$.

For a charged particle couple with electromagnetic field, Klein-Gordon equation is as follows:

$[i\hbar {\frac {\partial }{\partial t}}-e\phi ({\mathbf {r}},t)]^{2}\psi ({\mathbf {r}},t)=([-i\hbar \nabla -{\frac {e}{c}}{\mathbf {A}}({\mathbf {r}},t)]^{2}c^{2}+m^{2}c^{4})\psi ({\mathbf {r}},t)$

Klein-Gordon is second order in time. Therefore, to see how the states of a system evolve in time we need to know both $\psi ({\mathbf {r}},t)$ and ${\frac {\partial \psi ({\mathbf {r}},t)}{\partial t}}$ at a certain time. While in nonrelativistic quantum mechanics, we only need $\psi ({\mathbf {r}},t)$

Also because the Klein-Gordon equation is second order in time, it has the solutions $\psi ({\mathbf {r}},t)=e^{i({\mathbf {p}}{\mathbf {r}}-Et)/\hbar }$ with either sign of energy $E=\pm c{\sqrt {{\mathbf {p}}^{2}+m^{2}c^{2}}}$. The negative energy solution of Klein-Gordon equation has a strange property that the energy decreases as the magnitude of the momentum increases. We will see that the negative energy solutions of Klein-Gordon equation describe antiparticles, while the positive energy solutions describe particles.