In a square, zero potential

Solve for \(f(x,y,t)\) \[ I f_t = - \frac {\hbar ^2}{2 m} \nabla ^2 f(x,y) \] With boundary conditions \begin {align*} f(0,y,t) &= 0\\ f(1,y,t) &=0 \\ f(x,1,t) &=0 \\ f(x,0,t) &=0 \end {align*}

And initial conditions \(f(x,y,0)=\sqrt {2} \left ( \sin (2\pi x) \sin (\pi y) + \sin (\pi x) \sin (2 \pi y) \right )\)

Mathematica ✓

ClearAll["Global`*"]; 
pde =  I*D[f[x, y, t], {t}] == -((hBar^2*Laplacian[f[x, y, t], {x, y}])/(2*m)); 
 initSum = f[x, y, 0] == Sqrt[2]*(Sin[2*Pi*x]*Sin[Pi*y] + Sin[Pi*x]*Sin[2*Pi*y]); 
 bcs = {f[0, y, t] == 0, f[1, y, t] == 0, f[x, 1, t] == 0, f[x, 0, t] == 0}; 
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bcs, initSum}, f[x, y, t], {x, y, t}], 60*10]];

\[\left \{\left \{f(x,y,t)\to 2 \sqrt {2} \sin (\pi x) \sin (\pi y) e^{-\frac {5 i \pi ^2 \text {hBar}^2 t}{2 m}} (\cos (\pi x)+\cos (\pi y))\right \}\right \}\]

Maple ✓

restart; 
interface(showassumed=0); 
pde :=  I* diff(f(x,y,t),t) = -hBar^2/(2*m) * (diff(f(x,y,t),x$2) +  diff(f(x,y,t),y$2)); 
ic  := f(x, y, 0) = sqrt(2)*(sin(2*Pi*x)*sin(Pi*y) + sin(Pi*x)*sin(2*Pi*y)); 
bc  := f(0, y, t) = 0, 
      f(1, y, t) = 0, 
      f(x, 1, t) = 0, 
      f(x, 0, t) = 0; 
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic,bc],f(x,y,t))),output='realtime'));

\[f \left ( x,y,t \right ) =\sqrt {2}\sin \left ( \pi \,x \right ) {{\rm e}^{{\frac {-5/2\,i{{\it hBar}}^{2}t{\pi }^{2}}{m}}}} \left ( 2\,\sin \left ( \pi \,y \right ) \cos \left ( \pi \,x \right ) +\sin \left ( 2\,\pi \,y \right ) \right ) \]

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2.3.2 In a square, zero potential