Chemistry 3730 Fall 1997 Test 2 Solutions

1. 

2. 1. 

   2. If we neglect the last term, we have two independent hydrogenic problems. Thus treat the last term as a perturbation using a product of 1s hydrogenic wavefunctions as the solution of the known problem. Compute and add this quantity to (computed with Z=2, of course) to obtain an approximation to the energy of the ground state.
3. Start by defining the wavefunction and telling Maple that :
   

   > psi2p0 := (r,theta,phi) -> A*r*exp(-r/(2*a0))*cos(theta);

   

   > assume(a0>0);

   
   We want . Compute the inner product, which I will call N:
   

   > N := int(int(int(
     psi2p0(r,theta,phi)^2*r^2*sin(theta),r=0..infinity),
     theta=0..Pi),phi=0..2*Pi);

   

   
   

   If N=1, or, in somewhat simplified form,

   

4. 

   Using Taylor's theorem, we have

   

5. Define the potential energy and wavefunction and tell Maple that L>0:
   

   > V := x -> hbar^2/(m*L^3)*abs(x-L/2);

   

   > phi := x -> x*(L-x) + c*x^2*(L-x)^2;

   

   > assume(L>0);

   
   

   The variational energy is

   

   We will compute it in pieces. First :
   

   > Kip := int(phi(x)*(-hbar^2/(2*m))*diff(phi(x),x$2),x=0..L);

   

   
   

   Now :
   

   > Vip := int(phi(x)*V(x)*phi(x),x=0..L);

   

   
   

   Finally, we calculate :
   

   > ip:=int(phi(x)^2,x=0..L);

   

   > Evar:=(Kip+Vip)/ip;

   

   
   

   To find the best value of the variational parameter c, we minimize with respect to this parameter:
   

   > s1:=solve(diff(Evar,c)=0,c);

   

   
   

   The best solution is the one which gives the lowest energy:
   

   > evalf(subs(c=s1[1],Evar));

   

   > evalf(subs(c=s1[2],Evar));

   

   
   

   The first is obviously the superior solution so