?

E2=E2E2=m2a4126hm2a436h2=m2a490h2 -7. Consider a free particle constrained to move over the rectangular region 0xa, 0yb. The energy eigenfunctions of this system are n,,(x,y)=(ab4)1/2sinanxxsinbnyynx=1,2,3,ny=1,2,3, The Hamiltonian operator for this system is H^=2m2(x22+y22) Show that if the system is in one of its eigenstates, then E2=E2E2=0