Midterm 3 Solutions

# In this basis printed by mathematica for students 2 0

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Unformatted text preview: 0 0 2 0 0 2 0 and ` S- Ø8 j, m\&lt; basis — 0 0 0 2 0 0 0 0 2 0 Thus ` Sx = 1 2 ` ` JS+ + S- N Ø8 j, m\&lt; basis — 2 010 1 0 1 and 010 ` ` Â` Sy = - 2 JS+ - S- N Ø8 j, m\&lt; basis Finally 10 0 ` Sz Ø8 j, m\&lt; basis — 0 0 0 0 0 -1 ` since Sz is diagonal in this basis. Printed by Mathematica for Students — 2 0 -Â 0 Â 0 -Â 0Â0 MidtermExam3Solutions.nb ü Statistics Mean = 5.32 Stddev = 2.38 Problem 2 (Problem 3.2 on page 106) We have ` ` Sn = S ÿ n The eigenvalues — detB 2 — 2 Ø ` Sz basis cos q sin q cos f - Â sin q sin f sin q cos f + Â sin q sin f -cos q — 2 l± are solutions to cos q - l ‰-Â f sin q ‰Â f sin q -cos q - l F=0 Il2 - cos2 qM - sin2 q = 0 l2 - 1 = 0 l± = ± 1 Thus we solve — 2 cos q ‰-Â f sin q Âf ‰ sin q -cos q a b = — 2 cosHqL a + ‰-Â f sin HqL b = l± a ‰Â f sinHqL a - cosHqL b = l± b l± a b a= ‰-Â f sin HqL b l± - cos q b= ‰Â f sinHqL a l± + cos q ﬂ Using 1 - cos q = 2 sin2 sin q = 2 sin q 2 cos q 2 q 2 1 + cos q = 2 cos2 q 2 Printed by Mathematica for Students Gives a+ = ‰-Â f sin HqL b+ 1 - cos q = ‰-Â f cosHqê2L b+ sinHqê2L a+ ‰-Â f cos q = — 2 cos q ‰-Â f sin q ‰Â f sin q -cos q 3 a= cosHqL a + ‰-Â f sin HqL b = l± a ﬂ ‰Â f sinHqL a - cosHqL b = l± b 4 b= MidtermExam3Solutions.nb ‰ sin HqL b l± - cos q ‰Â f sinHqL a l± + cos q Using 1 - cos q = 2 sin2 sin q = 2 sin q 2 cos q 2 q 2 1 + cos q = 2 cos2 q 2 Gives ‰-Â f sin HqL b+ 1 - cos q a+ = ‰Â f sinHqL a+ 1 + cos q b+ = ‰-Â f cosHqê2L b+ sinHqê2L = = ‰Â f sinHqê2L a+ cosHqê2L a+ b+ ﬂ = ‰-Â f cos sin q 2 q 2 for the l = l+ = 1 case. We see that these constraints are actually the same, as we expect. For the l = l- = -1 case ‰-Â f sin HqL b-1 - cos q a- = ‰Â f sinHqL a-1 + cos q b- = ==- ‰-Â f sinHqê2L bcosHqê2L ab- ﬂ ‰Â f cosHqê2L asinHqê2L =- ‰-Â f sin c...
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