Klein-Gordon equation

How to construct

Starting from the relativistic connection between energy and momentum:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E^2=\bold p^2c^2+m^2c^4}

Substituting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E \rightarrow i\hbar \frac{\partial}{\partial t}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold p \rightarrow -i\hbar \nabla} , we get Klein-Gordon equation for free particles as follows:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\hbar^2 \frac{\partial ^2\psi(\bold r, t)}{\partial t^2}=(-\hbar^2c^2\nabla^2+m^2c^4)\psi(\bold r, t)\qquad \qquad \qquad \qquad \qquad (9.2)}

Klein-Gordon can also be written as the following:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\square-K^2)\psi(\bold r, t)=0\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\;\;(9.3)}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \square=\nabla^2-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}} is d'Alembert operator and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K=\frac{mc}{\hbar}} .

Equation (9.3) looks like a classical wave equation with an extra term Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K^2} .

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [i\hbar \frac {\partial}{\partial t}-e\phi(\bold r, t)]^2\psi(\bold r, t)=([-i\hbar\nabla-\frac{e}{c}\bold A(\bold r, t)]^2c^2+m^2c^4)\psi(\bold 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 ${\mathbf {\psi }}^{*}$ 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 ${\mathbf {\psi }}$ 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 ${\mathbf {\rho }}$ over all space is conserved. However, ${\mathbf {\rho }}$ is not positively definite. Therefore, we can neither interpret ${\mathbf {\rho }}$ 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.

Nonrelativistic limit

In nonrelativistic limit when $v\ll c$ or $p\ll mc$, we have:

$E=[(pc)^{2}+(mc^{2})^{2}]^{1/2}=mc^{2}[1+({\frac {p}{mc}})^{2}]^{1/2}\approx mc^{2}(1+{\frac {1}{2}}({\frac {p}{mc}})^{2})=mc^{2}+{\frac {p^{2}}{2m}}$

So, the relativistic energy is different from classical energy by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle mc^2} , therefore, we can expect that if we write the solution of Klein-Gordon equation as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi e^{-imc^2t/\hbar}} and substitute it into Klein-Gordon equation, we will get Schrodinger equation for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \psi} .

Indeed, doing so we get:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\hbar ^2\frac {\partial ^2 \psi}{\partial t^2}e^{-imc^2t/\hbar}+2\frac {\partial \psi}{\partial t}imc^2 \hbar e^{-imc^2t/\hbar}+\psi m^2c^4 e^{-imc^2t/\hbar}=-\hbar ^2 c^2\nabla ^2 \psi e^{-imc^2t/\hbar}+m^2c^4 \psi e^{-imc^2t/\hbar}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow -\frac {\hbar ^2}{2mc^2} \frac{\partial ^2 \psi}{\partial t^2}+i\hbar \frac {\partial \psi}{\partial t}=-\frac {\hbar ^2}{2m}\nabla ^2 \psi}

The first term can be neglected in nonrelativistic limit, we get Schrodinger equation (for free particles:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar \frac {\partial \psi}{\partial t}=-\frac {\hbar ^2}{2m} \nabla ^2 \psi}