<<

. 40
( 42 .)



>>

. 


L2
1
‚ ‚ 

’ k’1
L2 k’1 
=’ sin θk 

k
sin2 θk
k’1
θk ‚θk ‚θk 
sin 


. 

. 
. 




L2 
1 ‚ ‚
sinD’2 θD’1 ‚θD’1 ’
L2
D’1 =
D’2

sin2 θD’1
sinD’2 θD’1 ‚θD’1
(A21)
Therefore, from (A13), we get
L2
1
‚ D’1 ‚
’ D’1
∇2 = D’1 r (A22)
D
r2
r ‚r ‚r
From (A21) it is clear that since θ1 , θ2 , . . . θD’1 are independent, the
operators L2 , L2 . . . L2
D’1 mutually commute. Hence, they have a
1 2
common eigenfunction Y (θ1 , θ2 , . . . , θD’1 ). Let us write

L2 Y (θ1 , θ2 , . . . , θD’1 ) = »k Y (θ1 , θ2 . . . θD’1 ) (A23)
k

where »k is the eigenvalue of L2 . Since the potential function is
k
independent of t, Y (θ1 , θ2 , . . . , θD’1 ) can be expressed as
D’1
Y (θ1 , θ2 , . . . , θD’1 ) = ˜k (θk ) (A24)
k=1



© 2001 by Chapman & Hall/CRC
Then we get from (A23)
Y
L2 ˜1 (θ1 ) = »1 Y (A25)
˜1 (θ1 ) 1
where we note that L2 is dependent on θ1 only. In other words
1

L2 ˜1 (θ1 ) = »1 ˜1 (θ1 ) (A26)
1

Similarly from L2 Y = »2 Y we get
2

L2 ˜1 (θ1 )˜2 (θ2 ) = »2 ˜1 (θ1 )˜2 (θ2 ) (A27)
2

where L2 is dependent on θ1 and θ2 only.
2
Using the explicit form of L2 we have
2

L2 1 ‚ ‚
1
’ ’ sin θ2 ˜1 (θ1 )˜2 (θ2 )
sin2 θ2 sin θ2 ‚θ2 ‚θ2

= »2 ˜1 (θ1 )˜2 (θ2 ) (A28)
or
˜2 (θ2 ) 2 ˜1 (θ1 ) ‚ ‚
L1 ˜1 (θ1 ) ’ sin θ2 ˜2 (θ2 )
2 sin θ2 ‚θ2 ‚θ2
sin θ2
= »2 ˜1 (θ1 )˜2 (θ2 ) (A29)
or
˜2 (θ2 ) ˜1 (θ1 ) ‚ ‚
»1 ˜1 (θ1 ) ’ sin θ2 ˜2 (θ2 )
sin2 θ2 sin θ2 ‚θ2 ‚θ2
= »2 ˜1 (θ1 )˜2 (θ2 ) (A30)
or
‚2
»1 ˜2 (θ2 ) ‚
’ 2 ˜2 (θ2 ) ’ cot θ2 ˜2 (θ2 ) = »2 ˜2 (θ2 ) (A31)
2 ‚θ2
‚θ2
sin θ2
implying

‚2 ‚ »1
’ 2 + cot θ2 ‚θ ’ ˜2 (θ2 ) = »2 ˜2 (θ2 ) (A32)
sin2 θ2
‚θ2 2

Let us suppose that

‚2 ‚ »k’1
’ + (k ’ 1) cot θk ’ ˜k (θk ) = »k ˜k (θk ) (A33)
2
sin2 θk
‚θk
‚θk


© 2001 by Chapman & Hall/CRC
Looking at the eigenvalue equation

L2 Y = »k+1 Y (A34)
k+1

which on expansion becomes

L2
1 ‚ ‚
k k
’ sin θk+1 ’ Y = »k+1 Y
2
k
sin θk+1 ‚θk+1 ‚θk+1 sin θk+1
(A35)
can also be expressed as

Y 1 ‚ ‚
sink θk+1
’ ˜k+1 (θk+1 )
˜k+1 (θk+1 ) sink θk+1 ‚θk+1 ‚θk+1

»k Y
+ = »k+1 Y
sin2 θk+1
or
‚2 ‚ »k
’ + k cot θk+1 ’ ˜k+1 (θk+1 )
2 ‚θk+1 sin2 θk+1
‚θk+1
= »k+1 ˜k+1 (θk+1 ) (A36)
The above shows that if (A33) holds for k, it also holds for k + 1
as well. Now, we have clearly shown in (A32), that it holds for
k = 2. Therefore, by the principle of induction, (A33) holds ∀k =
2, 3, . . . D ’ 1.
Let us write

‚2 ‚ »k’1
L2 (»k’1 ) =’ + (k ’ 1) cot θk ’ (A37)
k 2
sin2 θk
‚θk
‚θk

Hence we have

L2 ˜1 (θ1 ) = »1 ˜1 (θ1 ) (A38)
1

L2 (»k’1 )˜k (θk ) = »k ˜k (θk ), k = 2, 3, . . . , D ’ 1 (A39)
k

We now turn to the eigenvalues »k . For k = 2
2
»1 = l1 (A40)

»2 = l2 (l2 + 1) (A41)


© 2001 by Chapman & Hall/CRC
where
l2 = 0, 1, 2, . . . (A42)
l1 = ’l2 , ’l2 + 1, . . . l2 ’ 1, l2 . (A43)
Let us assume

»k’1 = lk’1 (lk’1 + k ’ 2) (A44)
where lk’1 is an integer. Setting

L+ (lk’1 ) = ’ lk’1 cot θk
k
‚θk

L’ (lk’1 ) = ’ ’ (lk’1 + k ’ 2) cot θk (A45)
k
‚θk
it follows by induction

»k = lk (lk + k ’ 1) (A46)

We ¬nally have from (A24) and (A39)

L2 YlD’1 ,lD’2 ,...,l2 ,l1 (θ1 , θ2 , . . . , θD’1 )
D’1

= lD’1 (lD’1+D’2 )YlD’1 ,lD’2 ,...,l2 ,l1 (θ1 , . . . , θD’1 ) (A47)
where YlD’1 ,lD’2 ,...,l2 ,l1 are the generalised spherical harmonics and

lD’1 = 0, 1, 2, . . .
LD’2 = 0, 1, 2, . . . , lD’1 ,
(A48)
.
.
.
l1 = ’l2 , ’l2 + 1, . . . , l2 ’ 1, l2

Substituting in (A1)

ψ(r) = R(r)YlD’1 ,lD’2 ,...l1 (θ1 , θ2 , . . . , θD’1 ) (A49)
and using (A22) and (A47), we obtain the radial part of the
Schroedinger equation as

¯ 2 d2
h D’1 d l(l + D ’ 2)
’ + + R(r) + V (r)R(r) = ER(r)
2m dr2 r2
r dr
(A50)


© 2001 by Chapman & Hall/CRC
To eliminate the ¬rst derivative, we make the substitution

R(r) = r(1’N )/2 u(r)

so that (A21) reduces to

¯ 2 d2 u ±l
h
’ ’ 2 u(r) + V (r)u(r) = Eu(r) (A51)
2m dr2 r

where ±l = 1 (D ’ 1)(D ’ 3) + l(l + D ’ 2).
4




© 2001 by Chapman & Hall/CRC
Appendix B

Derivation of the Results (A19) and (A21)

Consider the angular momentum components de¬ned by


Lij = ’Lji = xi pj ’ xj pi , i = 1, 2, . . . , j ’ 1 j = 2, . . . D (B1)


We also set


L2 = Lij Lij i = 1, 2, . . . j ’ 1 j = 2, 3, . . . , k + 1 (B2)
k

<<

. 40
( 42 .)



>>