Midterm 5 Solutions

# If the terms have opposite s1 z and s2 z eigenvalues

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1+ S2- + S1- S2+ N -z, -z^ = 0 `` `` `` `` JS1+ S2- + S1- S2+ N +z, -z^ = —2 -z, +z^ JS1+ S2- + S1- S2+ N -z, +z^ = —2 +z, -z^ ` ` we see that if the terms have the same S1 z and S2 z eigenvalues (product is positive), then they are also an eigenstate of `` `` ` ` JS1+ S2- + S1- S2+ N with eigenvalue zero. If the terms have opposite S1 z and S2 z eigenvalues (product is negatve), then they are `` `` switched, so the only linear combinations of such terms that are also eigenstates of JS1+ S2- + S1- S2+ N are either symmetric or `2 antisymmetric. Therefore, the eigenstates of Stot are + z, + z\ 1 2 1 2 with eigenvalue —2 2 + 3 —2 2 = 2 —2 B + z, - z\ + -z, +z\F with eigenvalue —2 - —2 2 + 3 —2 2 B + z, - z\ - -z, +z\F with eigenvalue - —2 - —2 2 + - z, - z\ with eigenvalue —2 2 + 3 —2 2 = 2 —2 3 —2 2 =0 = 2 —2 This method has the advantage of picking the basis of the degenerate `2 ues of Stot for free. A -eigenspace 2 for us, and we get the corresponding eigenval- `2 ` ü Characterizing by eigenvalues of Stot and Stot, z `2 `2 One may find the eigenvalues of Stot using something similar to Method 2 above, using the matrix representation of Stot in the `2 ` ` ` ` Sz Äȩ Sz basis, or by noting that eigenvectors of Stot are also the eigenvectors of H with respective eigenvalues l ` 2 , lH related by Stot ` lH = A —2 Kl ` 2 Stot 3 —2 O 2 ` ` ` so we move straight to applying Stot, z = S1 z + S2 z , as ` — Stot, z +z, +z\ = I 2 + ` Stot, z : ` Stot, z 1 2 B + z, - z\ - + z, + z\ = — + z, + z\ — B + z, - z\ + 2 1 — M 2 -z, +z\F> = I 2 -z, +z\F = ` — Stot, z -z, -z\ = -I 2 + — M 2 1 2 — — 1 M: 2 2 I2 - — MB 2 B + z, - z\ + + z, - z\ - -z, +z\F> = 0 -z, +z\F = 0 - z , - z \ = -— - z , - z \ `2 ` Therefore, in terms of the total spin degrees of freedom Stot and Stot, z , we have the energy eigenstates when w0 = 0 1, 1\ = + z, + z\ 1, 0\ = 1 1, -1\ = 2 B + z, - z\ + - z, - z\ -z, +z\F 1 Printed 0y M= 0, b\ athematica f+z,tudents B or S -z\ 2 -z, +z\F Stot, z 2 B + z, - z\ - -z, +z\F = ` — Stot, z -z, -z\ = -I 2 + — M 2 2 I2 - 2 MB +z, -z\ - -z, +z\F = 0 - z , - z \ = -— - z , - z \ MidtermExam5Solutions.nb `2 ` Therefore, in terms of the total spin de...
View Full Document

Ask a homework question - tutors are online