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. 41
( 42 .)



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i,j



In this appendix, let us ¬rst prove the following angular momen-
tum commutation relation (¯ = 1)
h


[Lij , Lkl ] = iδjl Lik + iδik Ljl ’ i δjk Lil ’ iδil Ljk (B3)


We make use of the commutation relations


[xi , pj ] = iδij (B4)


and

[xi , xj ] = 0 = [pi , pj ] (B5)


© 2001 by Chapman & Hall/CRC
to note that the right-hand-side of (B3)


= (xj pl ’ pl xj )(xi pk ’ xk pi ) + (xi pk ’ pk xi )(xj pl ’ xl pj )

’(xj pk ’ pk xj )(xi pl ’ xl pi ) ’ (xi pl ’ pl xi )(xj pk ’ xk pj )

= xj pl xi pk ’ pl xj xi pk ’ xj pl xk pi + pl xj xk pi + xi pk xj pl

’pk xi xj pl ’ xi pk xl pj + pk xi xl pj ’ xj pk xi pl + pk xj xi pl

+xj pk xl pi ’ pk xj xl pi ’ xi pl xj pk + pl xi xj pk + xi pl xk pj ’ pl xi xk pj

= xj pl (pk xi + iδik ) + xi pk (pl xj + iδjl ) ’ xj pk (pl xi + iδil )

’xi pl (pk xj + iδjk ) ’ xj pl (pi xk + iδik ) ’ xi pk (pj xl + iδjl )

+xj pk (pi xl + iδil ) + xi pl (pj xk + iδjk ) + pl xk (xj pi ’ pj xi )

+pk xl (xi pj ’ xj pi )

= xj pi (’pl xk + pk xl ) ’ xi pj (’pl xk + pk xl ) + (pk xl ’ pl xk )Lij

[using the de¬nition (B1)]

= (xj pi ’ xi pj )(pk xl ’ pl xk ) + (pk xl ’ pl xk )Lij

= Lji (xl pk ’ xk pl ) + (xl pk ’ xk pl )Lij

= Lij Lkl ’ Lkl Lij .
= Left-hand-side of (B3).


We next write


L2 f = L2 f = L12 L12 f = (x1 p2 ’ x2 p1 )(x1 p2 ’ x2 p1 )f
1 12

(B6)
‚ ‚ ‚ ‚2
= ’ x1 ’ x2 x1 ’ x2 f
‚x2 ‚x1 ‚x2 ‚x1


© 2001 by Chapman & Hall/CRC
for an arbitrary function f . Now

‚ ‚
x1 ’ x2 f
‚x2 ‚x1
D’1
1 ‚x2 ‚x1 ‚f
= x1 ’ x2
h2 ‚θj ‚θj ‚θj
j
j=0

D’1
x2 ‚ x2 ‚f
1
=
h2 ‚θj x1 ‚θj
j
j=0

D’1
x2 ‚ ‚f
1
= (tan θ1 )
h2 ‚θj ‚θj
j
j=0


x2 2 ‚f
1
= sec θ1
h2 ‚θ1
1

r2 cos2 θ1 sin2 θ2 . . . sin2 θD’1 2 ‚f
= sec θ1
r2 sin2 θ2 . . . sin2 θD’1 ‚θ1

‚f
= (B7)
‚θ1
Therefore
‚2
L2 =’ 2 (B8)
1
‚θ1
Next
j’1
L2 = Lij Lij = L2 + L13 L13 + L23 L23 (B9)
2 1
j=2,3 i=1

where
2
‚ ‚
L2 f 2
= (x1 p3 ’ x3 p1 ) f = ’ x1 ’ x3 f (B10)
13
‚x3 ‚x1

2
‚ ‚
L2 f 2
= (x2 p3 ’ x3 p2 ) f = ’ x2 ’ x3 f (B11)
23
‚x3 ‚x2


© 2001 by Chapman & Hall/CRC
Consider

‚ ‚
x2 ’ x3 f
‚x3 ‚x2
D’1
1 ‚x3 ‚x2 ‚f
= x2 ’ x3
h2 ‚θj ‚θj ‚θj
j
j=0

D’1
1 2‚ x3 ‚f
= x
h2 2 ‚θj x2 ‚θj
j
j=0

D’1
x2 ‚ ‚f
2
= (cot θ2 cosecθ1 )
h2 ‚θj ‚θj
j
j=0


x2 ‚ x2 ‚
‚f ‚f
2
+2
= (cot θ2 cosecθ1 ) (cot θ2 cosecθ1 )
h2 ‚θ1 ‚θ1 h2 ‚θ2 ‚θ2
1 2

r2 sin2 θ1 sin2 θ2 . . . sin2 θD’1 ‚f
=’ cot θ2 cosecθ1 cot θ1
r2 sin2 θ2 . . . sin2 θD’1 ‚θ1

r2 sin2 θ1 . . . sin2 θD’1 ‚f
— cosec2 θ2 cosecθ1
’2 2
r sin θ3 . . . sin2 θD’1 ‚θ2

‚f ‚f
= ’ cos θ1 cot θ2 ’ sin θ1
‚θ1 ‚θ2
(B12)
Therefore

2
‚ ‚
x2 ’ x3 f
‚x3 ‚x2

‚ ‚ ‚f ‚f
= cos θ1 cot θ2 + sin θ1 cos θ1 cot θ2 + sin θ1
‚θ1 ‚θ2 ‚θ1 ‚θ2
2f
‚f ‚
= ’ sin θ1 cos θ1 cot2 θ2 + cos2 θ1 cot2 θ2 2
‚θ1 ‚θ1
‚2f
‚f
2
+ cos θ1 cot θ2 + sin θ1 cos θ1 cot θ2
‚θ2 ‚θ1 ‚θ2


© 2001 by Chapman & Hall/CRC
‚f
’ sin θ1 cos θ1 cosec2 θ2
‚θ1
(B13)
‚2f ‚2f
+ sin2 θ1 2
+ sin θ1 cos θ1 cot θ2
‚θ1 ‚θ2 ‚θ2
Again

(x1 p3 ’ x3 p1 )

D’1
x2 ‚ x3 ‚f
1
=
h2 ‚θj x1 ‚θj
j
j=0


x2 ‚f x2 ‚f
1 1 2
= 2 ‚θ cot θ2 sec θ1 tan θ1 + h2 ‚θ (’cosec θ2 ) sec θ1
h1 1 2
2

r2 cos2 θ1 sin2 θ2 . . . sin2 θD’1 ‚f
= cot θ2 sec θ1 tan θ1
r2 sin2 θ2 . . . sin2 θD’1 ‚θ1

r2 cos2 θ1 sin2 θ2 sin2 θ3 . . . sin2 θD’1 ‚f
cosec2 θ2 sec θ1

r2 sin2 θ3 . . . sin2 θD’1 ‚θ2

‚f ‚f
= sin θ1 cot θ2 ’ cos θ1
‚θ1 ‚θ2

(x1 p3 ’ x3 p1 )2 f

‚ ‚ ‚ ‚
= sin θ1 cot θ2 ’ cos θ1 sin θ1 cot θ2 ’ cos θ1 f
‚θ1 ‚θ2 ‚θ1 ‚θ2

‚2f
‚f2 2 2
= sinθ1 cos θ1 cot θ2 + sin θ1 cot θ2 2
‚θ1 ‚θ1

‚2f
‚f
2
+ sin θ1 cos θ1 cosec θ2 ’ cos θ1 sin θ1 cot θ2
‚θ1 ‚θ1 ‚θ2
2 2
+ sin2 θ1 cot θ2 ‚θ2 ’ sin θ1 cos θ1 cot θ2 ‚θ1 ‚θ2 + cos2 θ1 ‚ f
‚f ‚f

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. 41
( 42 .)



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