Equação de Klein-Gordon

$(◻+\frac{m^2{c}^2}{{\hslash }^2})\psi (x,t)=0$

onde

$(◻=\frac{{\partial }^2}{c^2\partial t^2}-{\mathrm{\nabla }}^2)$

então

$\frac{{\partial }^2\psi }{\partial t^2}=c^2{\mathrm{\nabla }}^2\psi -\frac{m^2{c}^4}{{\hslash }^2}\psi$

$\frac{{\partial }^2\psi }{\partial t^2}=c^2\frac{{\partial }^2\psi }{\partial x^2}-\frac{m^2{c}^4}{{\hslash }^2}\psi$ (em uma dimensão)

no método das diferenças finitas:

$\frac{{\partial }^2\psi (x,t)}{\partial t^2}\approx \frac{\psi (x,t+\mathrm{\Delta }t)-2\psi (x,t)+\psi (x,t-\mathrm{\Delta }t)}{(\mathrm{\Delta }t{)}^2}$

$\frac{{\partial }^2\psi (x,t)}{\partial x^2}\approx \frac{\psi (x+\mathrm{\Delta }x,t)-2\psi (x,t)+\psi (x,-\mathrm{\Delta },t)}{(\mathrm{\Delta }x{)}^2}$

ou seja:

$\frac{\psi (x,t+\mathrm{\Delta }t)-2\psi (x,t)+\psi (x,t-\mathrm{\Delta }t)}{(\mathrm{\Delta }t{)}^2}=c^2\frac{\psi (x+\mathrm{\Delta }x,t)-2\psi (x,t)+\psi (x,-\mathrm{\Delta },t)}{(\mathrm{\Delta }x{)}^2}-\frac{m^2{c}^4}{{\hslash }^2}\psi$