From page 30, David J Logan textbook, applied PDE textbook.
Schrodinger PDE with zero potential (Logan p. 30)
Solve \[ I \hbar f_t = - \frac {\hbar ^2}{2 m} f_{xx} \] With boundary conditions \begin {align*} f(0,t) &= 0\\ f(L,0) &=0 \end {align*}
Mathematica ✓
ClearAll["Global`*"]; pde = I*h*D[f[x, t], t] == -((h^2*D[f[x, t], {x, 2}])/(2*m)); bc = {f[0, t] == 0, f[L, t] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, f[x, t], {x, t}, Assumptions -> L > 0], 60*10]]; sol = sol /. K[1] -> n;
\[\left \{\left \{f(x,t)\to \underset {n=1}{\overset {\infty }{\sum }}e^{-\frac {i h n^2 \pi ^2 t}{2 L^2 m}} c_n \sin \left (\frac {n \pi x}{L}\right )\right \}\right \}\]
Maple ✓
restart; interface(showassumed=0); pde :=I*h*diff(f(x,t),t)=-h^2/(2*m)*diff(f(x,t),x$2); bc:=f(0,t)=0,f(L,t)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],f(x,t)) assuming L>0),output='realtime'));
\[f \left ( x,t \right ) =\sum _{n=1}^{\infty }{\it \_C1} \left ( n \right ) \sin \left ( {\frac {n\pi \,x}{L}} \right ) {{\rm e}^{{\frac {-{\frac {i}{2}}h{\pi }^{2}{n}^{2}t}{m{L}^{2}}}}}\]
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