Equação de Klein-Gordon

$\left(\Box +{\frac {m^{2}c^{2}}{\hbar ^{2}}}\right)\psi (x,t)=0$

onde

$\left(\Box ={\frac {\partial ^{2}}{c^{2}\partial t^{2}}}-\nabla ^{2}\right)$

então

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

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

no método das diferenças finitas:

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

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

ou seja:

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

isso nos leva a equação final:

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

chamarei $\alpha ={\frac {c\Delta t}{\Delta x}}$ e $\beta ={\frac {mc^{2}\Delta t}{\hbar }}$