This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Physics 216 Solutions to Problem Set 3 Spring 2010 1. A system of three unperturbed states consisting of a degenerate pair of states of energy E 1 and a nondegenerate state of energy E 2 is subsequently perturbed, and is represented by the Hamiltonian matrix: H = E 1 a E 1 b a b E 2 , (1) where E 2 > E 1 . The quantities a and b are to be regarded as perturbations that are of the same order but small compared with E 2 E 1 . We shall write write the Hamiltonian matrix as H = H (0) + H (1) , where H (0 ) = E 1 E 1 E 2 and H (1) = a b a b . for the unperturbed Hamiltonian and the perturbation, respectively. The unper turbed eigenvalues of H (0) are E 1 (doubly degenerate) and E 2 . The unperturbed eigenstates are eigenvectors of H (0) , which can be chosen as: vextendsingle vextendsingle 1 )big = 1 , vextendsingle vextendsingle 2 )big = 1 , vextendsingle vextendsingle 3 )big = 1 . Note that there is no first order correction to the eigenvalues due to the perturbation, since ( 3  H (1)  3 ) = 0 and the 2 2 matrix whose elements are ( i  H (1)  j ) = 0 , i, j = 1 , 2 . Thus, we must employ secondorder perturbation theory to find shifts in the unper turbed eigenvalues. (a) Use second order nondegenerate perturbation theory to calculate the per turbed eigenvalues. Is this procedure correct? Suppose we apply nondegenerate perturbation theory (which is not the correct procedure given that the unperturbed Hamiltonian matrix possesses degenerate eigen values). Using, E (2) n = summationdisplay m negationslash = n ( n  H (1)  m ) 2 E n E m , (2) 1 we compute the following matrix elements: ( 1  H (1)  2 ) = ( 2  H (1)  1 ) = 0 , ( 1  H (1)  3 ) = ( 3  H (1)  1 ) = a, ( 2  H (1)  3 ) = ( 3  H (1)  2 ) = b. Thus, eq. (2) yields: E (2) 1 =  a  2 E 1 E 2 , E (2) 2 =  b  2 E 1 E 2 , E (2) 3 =  a  2 +  b  2 E 1 E 2 . Since  3 ) is a nondegenerate eigenstate, it follows that E (2) 3 is the correct second order energy shift corresponding to this eigenstate. However, as already noted, the procedure employed above is not correct when applied to the degenerate subspace. (b) Use second order degenerate perturbation theory to calculate the perturbed eigenvalues. The correct procedure is to use secondorder degenerate perturbation theory. First, we must compute the 2 2 matrix, W ji = summationdisplay k negationslash = j ( j  H (1)  k )( k  H (1)  i ) E j E k , j, i = 1 , 2 , where the indices j and i run over the degenerate states. We have already computed W 11 and W 22 in part (a). Thus, we only need to compute W 12 = W 21 = ab E 1 E 2 ....
View
Full
Document
 Spring '09
 Physics

Click to edit the document details