For analytic solution when $V = 0$, we have: $$ f = f_0 exp(i (k_x x+k_y y - \omega t)) $$ where $\omega = \alpha (k_x^2 + k_y^2)$.
set $f(m,n,t)$, $$ i \frac {f(m,n,t+1)-f(m,n,t)} {\Delta t} = -\alpha \left( \frac {f(m+1,n,t)-2f(m,n,t)+f(m-1,n,t)} {\Delta x^2} + \frac {f(m,n+1,t)-2f(m,n,t)+f(m,n-1,t)} {\Delta y^2} \right)- V(m,n,t) f(m,n,t) $$ when $V(m,n,t)=0$, we have $$ i \frac {f(m,n,t)exp(-i\omega\Delta t)-f(m,n,t)} {\Delta t} = -\alpha \left( \frac {f(m,n,t)exp(i k_x \Delta x)-2f(m,n,t)+f(m,n,t)exp(-i k_x \Delta x)} {\Delta x^2} + \frac {f(m,n,t)exp(i k_y \Delta y)-2f(m,n,t)+f(m,n,t)exp(-i k_y \Delta y)} {\Delta y^2} \right) $$
$$ i \frac {exp(-i\omega\Delta t)-1} {\Delta t} = -\alpha \left( \frac {exp(i k_x \Delta x)-2+exp(-i k_x \Delta x)} {\Delta x^2} + \frac {exp(i k_y \Delta y)-2+exp(-i k_y \Delta y)} {\Delta y^2} \right) $$$$ exp(-i \omega \Delta t) = 1+i\alpha \Delta t \left( \frac {sin^2(\frac{k_x \Delta x}{2})} {(\frac{\Delta x}{2})^2} + \frac {sin^2(\frac{k_y \Delta y}{2})} {(\frac{\Delta y}{2})^2} \right) $$