( ◻ + m 2 c 2 ℏ 2 ) ψ ( x , t ) = 0 {\displaystyle \left(\Box +{\frac {m^{2}c^{2}}{\hbar ^{2}}}\right)\psi (x,t)=0}
onde
( ◻ = ∂ 2 c 2 ∂ t 2 − ∇ 2 ) {\displaystyle \left(\Box ={\frac {\partial ^{2}}{c^{2}\partial t^{2}}}-\nabla ^{2}\right)}
então
∂ 2 ψ ∂ t 2 = c 2 ∇ 2 ψ − m 2 c 4 ℏ 2 ψ {\displaystyle {\frac {\partial ^{2}\psi }{\partial t^{2}}}=c^{2}\nabla ^{2}\psi -{\frac {m^{2}c^{4}}{\hbar ^{2}}}\psi }
= ∂ 2 ψ ∂ t 2 = c 2 ∂ 2 ∂ x 2 ψ − m 2 c 4 ℏ 2 ψ {\displaystyle ={\frac {\partial ^{2}\psi }{\partial t^{2}}}=c^{2}{\frac {\partial ^{2}}{\partial x^{2}}}\psi -{\frac {m^{2}c^{4}}{\hbar ^{2}}}\psi }
no método das diferenças finitas:
∂ 2 ψ ( x , t ) ∂ t 2 ≈ ψ ( x , t + Δ t ) − 2 ψ ( x , t ) + ψ ( x , t − Δ t ) ( Δ t ) 2 {\displaystyle {\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}}}}
∂ 2 ψ ( x , t ) ∂ x 2 ≈ ψ ( x + Δ x , t ) − 2 ψ ( x , t ) + ψ ( x , − Δ , t ) ( Δ x ) 2 {\displaystyle {\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}}}}
