Midterm 3 Solutions

Midterm 3 Solutions - Intermediate QMI Midterm 3 Solutions...

This preview shows pages 1–4. Sign up to view the full content.

Intermediate QMI Midterm 3 Solutions Problem 1 Given the raising and lowering operators S ` ± = S ` x ± Â S ` y and that S ` ± = S ` ° . ü Part a We have B S ` z , S ` ± F = B S ` z , S ` x ± Â S ` y F = B S ` z , S ` x F = Â S ` y ± Â B S ` z , S ` y F = -Â S ` x = ± S ` x + Â S ` y = ± J S ` x ± Â S ` y N therefore B S ` z , S ` ± F = ± S ` ± ü Part b We have S ` z B S ` ± j , m ^F = :B S ` z , S ` ± F + S ` ± S ` z > j , m ^ ¬ B S ` z , S ` ± F = ± S ` ± S ` z j , m ^ = m j , m ^ = B ± S ` ± + m S ` ± F j , m ^ S ` z B S ` ± j , m ^F = H m ± 1 L S ` ± j , m ^ Therefore, m £ = m + 1 for S ` + j , m ^ and m £ = m - 1 for S ` - j , m ^ . ü Part c We have Z j , m S ` ° S ` ± j , m ^ = Z j , m S ` ± S ` ± j , m ^ = S ` ± j , m ^ 2 Computing another way, we take S ` ° S ` ± = J S ` x ° Â S ` y N J S ` x ± Â S ` y N = S ` x 2 ± Â S ` x S ` y ° Â S ` y S ` x + S ` y 2 = J S ` x 2 + S ` y 2 N = S ` 2 - S ` z 2 ± Â B S ` x , S ` y F Â S ` z Printed by Mathematica for Students

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
S ` ° S ` ± = S ` 2 - S ` z 2 ° — S ` z Using S ` 2 j , m _ = 2 j H j + 1 L j , m _ S ` z j , m ^ = m j , m ^ gives S ` ° S ` ± j , m ^ = K S ` 2 - S ` z 2 ° — S ` z O j , m _ = A 2 j H j + 1 L - 2 m 2 ° — 2 m E j , m _ Z j , m S ` ° S ` ± j , m ^ = 2 @ j H j + 1 L - m H m ± 1 LD S ` ± j , m ^ 2 = 2 @ j H j + 1 L - m H m ± 1 LD Using the result from (b) therefore gives S ` ± j , m ^ = j H j + 1 L - m H m ± 1 L j , m ± 1 _ ü Part d We have from (c) Z j , m £ S ` ± j , m ^ = j H j + 1 L - m H m ± 1 L d m £ m ± 1 this gives the representations of S ` ± for a spin-1 particle in the 8 j , m \< basis S ` + Ø 8 j , m \< basis 0 2 0 0 0 2 0 0 0 and S ` - Ø 8 j , m \< basis 0 0 0 2 0 0 0 2 0 Thus S ` x = 1 2 J S ` + + S ` - N Ø 8 j , m \< basis 2 0 1 0 1 0 1 0 1 0 and S ` y = - Â 2 J S ` + - S ` - N Ø 8 j , m \< basis 2 0 0 Â 0 0 Â 0 Finally S ` z Ø 8 j , m \< basis 1 0 0 0 0 0 0 0 - 1 since S ` z is diagonal in this basis. 2 MidtermExam3Solutions.nb Printed by Mathematica for Students
ü Statistics Mean = 5.32 Stddev = 2.38 Problem 2 (Problem 3.2 on page 106) We have S ` n = S ` ÿ n Ø S ` z basis 2 cos q sin q cos f - Â sin q sin f sin q cos f + Â sin q sin f - cos q = 2 cos q -Â f sin q

This preview has intentionally blurred sections. Sign up to view the full version.

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

{[ snackBarMessage ]}

Page1 / 9

Midterm 3 Solutions - Intermediate QMI Midterm 3 Solutions...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online