Midterm 5 Solutions

Calculating the energy eigenvalues printed by

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 0 Aê2 0 0 0 0 00 00 0 0 -1 2 0 02 0 0 —w + 20 2 -1 0 0 0 -2 0 0 01 00 0 0 0 0 0 — w0 - HA ê 2L A 0 A -— w0 - HA ê 2L 0 0 0 Aê2 which agrees with our result from Method 1. ü Calculating the energy eigenvalues Printed by Mathematica for Students 0 0 0 1 3 4 MidtermExam5Solutions.nb Calculating the energy eigenvalues ` We see from the “outside” block - which is diagonal - that H has eigenvalue A with multiplicity 2, corresponding to the eigenstates +z, +z\ and -z, -z\. What remains is to diagonalize the “inside” block, as detB A 2 — w0 - -l A -— w0 - A BIl + A2 M 2 A 2 -l F=0 - —2 w2 F - A2 = 0 0 l2 + l A - J 3 A2 4 + —2 w2 N = 0 0 l± = 1 B- A 2 A l± = - 2 ± A2 + 4 J ± 3 A2 4 + —2 w2 N F 0 IA2 + —2 w2 M 0 A Therefore, the four energy eigenvalues are E œ : 2 , A , 2 A l+ = - 2 + A IA2 + —2 w2 M , l- = - 2 0 IA2 + —2 w2 M > 0 ü Part b We have the total spin operator ` ` ` Stot = S1 + S2 Thus ` 2 S1 ` 2 S1 `2 `2 `` `2 Stot = S1 + 2 S1 ÿ S2 + S2 ` `2 `2 `2 ÿ S2 = Stot - S1 - S2 ` `2 3 —2 ` ÿ S2 = Stot - 2 1ཽ ` ` ¬༼ S1 and S2 commute `2 ` ¬༼ Si = —2 ji H ji + 1L 1ཽ = so for w0 = 0, our Hamiltonian is given by ` H= A —2 `2 KStot - 3 —2 2 ` 1ཽO `2 Therefore, the energy eigenstates when w0 = 0 are the same as the eigenstates of Stot . ü Part c ` ü Calculating the energy eigenstates of H There are again two ways of doing this. ü Printed by Mathematica for Students 3 —2 4 ` 1 1ཽ for spin- 2 MidtermExam5Solutions.nb 5 Method 1 ` See lecture note 5. From the matrix representation and eigenvalues of H from (a) for w0 = 0 ` H Hw0 = 0L Ø ` ` Sz Äȩ Sz basis Aê2 0 0 0 0 -A ê 2 A 0 0 A -A ê 2 0 0 0 0 Aê2 A A A , 2, 2 with eigenvalues E œ 9 2 , - 3A = 2 for which we solve for the eigenvectors as Aê2 0 0 0 0 -A ê 2 A 0 0 A -A ê 2 0 0 0 0 Aê2 a b c d Aê2 0 0 0 0 -A ê 2 A 0 0 A -A ê 2 0 0 0 0 Aê2 a b c d = A 2 fl a=a -b + 2 c = b 2b - c = c d=d fl a b c d a = -3 a -b + 2 c = -3 b 2 b - c = -3 c d = -3 d flb=c flb=c and =- 3A 2 a b c d fl fl fl fl a b b d...
View Full Document

Ask a homework question - tutors are online