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.

Continuity equation

Multiplying (9.2) by $\psi ^{*}$ from the left, we get:

$-{\frac {\hbar ^{2}}{c^{2}}}\psi ^{*}{\frac {\partial ^{2}\psi ({\mathbf {r}},t)}{\partial t^{2}}}=\psi ^{*}(-\hbar ^{2}\nabla ^{2}+m^{2}c^{2})\psi ({\mathbf {r}},t)\qquad \qquad \qquad (9.4)$

Multiplying the complex conjugate form of (9.2) by $\psi$ from the left, we get:

$-{\frac {\hbar ^{2}}{c^{2}}}\psi {\frac {\partial ^{2}\psi ^{*}({\mathbf {r}},t)}{\partial t^{2}}}=\psi (-\hbar ^{2}\nabla ^{2}+m^{2}c^{2})\psi ^{*}({\mathbf {r}},t)\qquad \qquad \qquad (9.5)$

Subtracting (9.5) from (9.4), we get:

$-{\frac {\hbar ^{2}}{c^{2}}}(\psi ^{*}{\frac {\partial ^{2}\psi }{\partial t^{2}}}-\psi {\frac {\partial ^{2}\psi ^{*}}{\partial t^{2}}})=\hbar ^{2}(\psi \nabla ^{2}\psi ^{*}-\psi ^{*}\nabla ^{2}\psi )$

$\Rightarrow -{\frac {\hbar ^{2}}{c^{2}}}{\frac {\partial }{\partial t}}(\psi ^{*}{\frac {\partial \psi }{\partial t}}-\psi {\frac {\partial \psi ^{*}}{\partial t}})+\hbar ^{2}\nabla (\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*})=0$

$\Rightarrow {\frac {\partial }{\partial t}}[{\frac {i\hbar }{2mc^{2}}}(\psi ^{*}{\frac {\partial \psi }{\partial t}}-\psi {\frac {\partial \psi ^{*}}{\partial t}})]+\nabla [{\frac {\hbar }{2mi}}(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*})]=0$

this give us the continuity equation:

${\frac {\partial \rho }{\partial t}}+\nabla {\mathbf {j}}=0\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (9.6)$

where $\rho ={\frac {i\hbar }{2mc^{2}}}(\psi ^{*}{\frac {\partial \psi }{\partial t}}-\psi {\frac {\partial \psi ^{*}}{\partial t}})\qquad \qquad \qquad \qquad \qquad \qquad (9.7)$

${\mathbf {j}}={\frac {\hbar }{2mi}}(\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*})\qquad \qquad \qquad \qquad \qquad \qquad \qquad (9.8)$

From (9.6) we can see that the integral of the density $\rho$ over all space is conserved. However, $\rho$ is not positively definite. Therefore, we can neither interpret $\rho$ as the particle probability density nor can we interpret ${\mathbf {j}}$ as the particle current. The appropriate interpretation are charge density for $e\rho ({\mathbf {r}},t)$ and electric current for $e{\mathbf {j}}({\mathbf {r}},t)$ since charge density and electric current can be either positive or negative.