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:

$-\hbar ^{2}{\frac {\partial ^{2}\psi ({\mathbf {r}},t)}{\partial t^{2}}}=(-\hbar ^{2}c^{2}\nabla ^{2}+m^{2}c^{4})\psi ({\mathbf {r}},t)\qquad \qquad \qquad \qquad \qquad (1)$

Klein-Gordon can also be written as the following:

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

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

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

Potentials couple to the Klein-Gordon equation as four-vectors. A good example of this is the potential four-vector:

Failed to parse (unknown function "\math"): {\displaystyle \Phi ^{\mu} = ( \phi ; \bold{A} )<\math> Coupling this to the momentum four-vector, <math>p^{\mu} = ( \frac{E}{c} ; \bold{p} )<\math> where <math>E \rightarrow i\hbar \frac{\partial}{\partial t}} and ${\mathbf {p}}\rightarrow -i\hbar {\mathbf {\nabla }}$ in the quantum limit, we find the conjugate momentum four vector:

$P^{\mu }=p^{\mu }-{\frac {e}{c}}\Phi ^{\mu }=({\frac {E}{c}}-{\frac {e}{c}}\phi ;{\mathbf {p}}-{\frac {e}{c}}{\mathbf {A}})$

Squaring the conjugate momentum four vector and multiplying by $c^{2}$, we obtain the Klein-Gordon equation in an electromagnetic field:

Failed to parse (syntax error): {\displaystyle c^2 P^{\mu}P_{\mu} = m^2 c^4 = (E – e\phi)^2 – (c\bold{p} – e\bold{A})^2}

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

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 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=\pm c\sqrt{\bold p^2+m^2c^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 (1) 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 \bold \psi^{*}} from the left, 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 -\frac{\hbar^2}{c^2}\psi^{*} \frac{\partial ^2\psi(\bold r, t)}{\partial t^2}=\psi^{*}(-\hbar^2\nabla^2+m^2c^2)\psi(\bold r, t)\qquad \qquad \qquad (3)}

Multiplying the complex conjugate form of (1) 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 \bold \psi} from the left, 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 -\frac{\hbar^2}{c^2}\psi \frac{\partial ^2\psi^{*}(\bold r, t)}{\partial t^2}=\psi(-\hbar^2\nabla^2+m^2c^2)\psi^{*}(\bold r, t)\qquad \qquad \qquad (4)}

Subtracting (4) from (3), 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 -\frac{\hbar^2}{c^2}\left( \psi^{*} \frac{\partial ^2\psi}{\partial t^2}-\psi \frac{\partial ^2\psi^{*}}{\partial t^2}\right) =\hbar^2\left( \psi\nabla^2\psi^{*}-\psi^{*}\nabla^2\psi\right) }

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

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

this give us the continuity equation:

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 \frac {\partial \rho}{\partial t}+\nabla \bold j = 0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad(5)}

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 \rho = \frac {i\hbar}{2mc^2}\left( \psi^{*}\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^{*}}{\partial t}\right) \qquad \qquad \qquad \qquad \qquad \; \; \; \;(6)}

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 j = \frac {\hbar}{2mi}(\psi^{*}\nabla\psi-\psi\nabla\psi^{*})\qquad \qquad \qquad \qquad \qquad \qquad \qquad \; \;(7)}

From (5) we can see that the integral of the density 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 \rho} over all space is conserved. However, 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\rho} is not positively definite. Therefore, we can neither interpret 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 \rho} as the particle probability density nor can we interpret 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 j} as the particle current. The appropriate interpretation are charge density 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 e\rho(\bold r,t)} and electric current 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 e\bold j(\bold r, t)} since charge density and electric current can be either positive or negative.

Nonrelativistic limit

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

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=[(pc)^2+(mc^2)^2]^{1/2}=mc^2\left[ 1+(\frac{p}{mc})^2\right] ^{1/2}\approx mc^2\left( 1+\frac{1}{2}\left( \frac{p}{mc}\right) ^2\right) =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}

In the nonrelativistic limit the first term is considered negligibly small. As a result, for free particles in this limit we get back the Schrodinger equation:

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}