Midterm 5 Solutions

If the terms have opposite s1 z and s2 z eigenvalues

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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...
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