Klein-Gordon equationKleinGordonEquation2008-03-16 15:21:392008-03-16 15:21:39Definition% this is the default PlanetPhysics preamble. as your knowledge
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% define commands hereThe Klein-Gordon equation is an equation of mathematical physics that describes spin-0 particles. It is given by:
\[
(\Box + \left(\frac{m}{\hbar c}\right)^2) \psi = 0
\]
Here the $\Box$ symbol refers to the wave operator, or D'Alembertian, and $\psi$ is the wavefunction of a particle.
It is a Lorentz invariant expression.
\subsection{Derivation}
Like the Dirac equation, the Klein-Gordon equation is derived from the relativistic expression for total energy:
\[
E^2 = m^2c^4 + p^2c^2
\]
Instead of taking the square root (as Dirac did), we keep the equation in squared form and replace the momentum and energy with their operator equivalents, $E = i \hbar \partial_t$, $p = -i \hbar \nabla$. This gives (in disembodied operator form)
\[
-\hbar^2 \frac{\partial^2}{\partial t^2} = m^2 c^4 - \hbar^2 c^2 \nabla^2
\]
Rearranging:
\[
\hbar^2(c^2 \nabla^2 - \frac{\partial^2}{\partial t^2}) + m^2 c^4 = 0
\]
Dividing both sides by $\hbar^2 c^2$:
\[
(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}) + \frac{m^2 c^2}{\hbar^2} = 0
\]
Identifying the expression in brackets as the D'Alembertian and right-multiplying the whole expression by $\psi$ , we obtain the Klein-Gordon equation:
\[
(\Box + \left(\frac{m}{\hbar c}\right)^2) \psi = 0
\]