Solutions to practice problem set 3

1. Use Taylor's theorem for dealing with any small quantities:

   

   provided x-a is small. In this case, a=0:

   

2. or
3. 1. The field splits the states according to the value of . There are four different values of ( ) so there are four states with different energies in a magnetic field.
   2. The energy shift caused by the field is proportional to . Thus the difference in energy between two states depends only on (and not on the actual values of the spins) so only one line appears.
4. 1. Instead of electrons pairing up into orbitals, up to four electrons could share an orbital because particles with a spin of 3/2 can have four different values of . Thus, there would be 16 main groups.
   2. Particles with integer spin are bosons and are not subject to an exclusion principle. Accordingly, all of the electrons could sit in the same orbital, regardless of how many of them there were. There would be no periodicity as we know it: Every electron would be a valence electron or, to state the matter another way, each element would be in its own ``group''.
5. If is a solution of Schrödinger's equation, . Then we have

   

6. The trial wavefunction is
   

   > phi := x -> x*(x-L)+c2*x^2*(x-L)^2;

   

   
   

   For the particle in a box, . Therefore is simply
   

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

   

   
   

   is
   

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

   

   > Evar := pKp/ip;

   

   
   

   Once we have the variational energy, we minimize it to find the best value of the variational parameter:
   

   > s1:=solve(diff(Evar,c2),c2);

   

   
   

   There are two solutions, only one of which can be right. Recall that we are looking for the value of which minimizes the variational energy. Substitute the two solutions back into Evar:
   

   > evalf(subs(c2=s1[1],Evar));

   

   > evalf(subs(c2=s1[2],Evar));

   

   > c2:=simplify(s1[2]);

   

   > hbar:=h/(2*Pi);

   

   > simplify(Evar);

   

   > evalf(");

   

   
   

   The exact value of the ground-state energy is or . This is very close to the value obtained with our simple variational wavefunction but, in accordance with the variational theorem, slightly lower.

   To normalize the wavefunction, replace by , where A is a normalization factor. We want . This gives , but we have already computed this inner product:
   

   > A := simplify(1/sqrt(ip));