Midterm 3 Solutions

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

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

View Full Document Right Arrow Icon
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
Background image of page 1

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

View Full Document Right Arrow Icon
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
Background image of page 2
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
Background image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
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 Right Arrow Icon
Ask a homework question - tutors are online