Ensayo

CHAPTER 7

1. (a) The system under consideration has rotational degrees of freedom, allowing it to rotate about two orthogonal axes perpendicular to the rigid rod connecting the two masses. If wedefine the z axis as represented by the rod, then the Hamiltonian has the form



where I is the moment of inertia of the dumbbell.

(b) Since there are no rotations about the z axis, theeigenvalue of Lz is zero, so that the eigenvalues of the Hamiltonian are



with = 0,1,2,3,…

(c) To get the energy spectrum we need an expression for the moment of inertia. We use the fact thatwhere the reduced mass is given by




If we express the separation a in Angstroms, we get



The energy difference between the ground state and the first excited state iswhich leads to the numerical result



2. We use the connection to write


Next we have




and



We may use Eq. (7-46) to obtain the form for
3. We use to calculate. We may now proceed



and on the r.h.s. we insert



4. Again we use to work out


We calculate



and



which is easily obtained from the preceding result byinterchanging m1 and m2.

The remaining two terms yield



The remaining calculation is simple, since



5. The Hamiltonian may be written as


whose eigenvalues are


where –l ≤ m≤ l.
(b) The plot is given on the right.
(c) The spectrum in the limit that I1 >> I3 is just ,
with m = 0,1,2,…l. The m = 0 eigenvalue is nondegenerate, while the other ones are doubly degenerate(corresponding to the negative values of m).

6. We will use the lowering operator acting on Y44. Since we are not interested in the normalization, we will not carry the factor.7. Consider the H given. The angular momentum eigenstates are eigenstates of the Hamiltonian, and the eigenvalues are



with . Thus for every value of there will be (2 +1)...