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Proposition
The following recursive equations hold for k 0
β‰₯
L l,l k =ΞΆ(l k)l,l k 1 (4.77)
βˆ’| βˆ’ i βˆ’ | βˆ’ βˆ’ i
L l,l k 1 =ΞΆ(l k)l,l k (4.78)
+
| βˆ’ βˆ’ i βˆ’ | βˆ’ i
ΞΆ(l k)2 ΞΆ(l k+1)2 =2(l k) (4.79)
βˆ’ βˆ’ βˆ’ βˆ’
where the first equation is just a definition and need not be proved.
Proof
For the base case put k=0. Then the equations read
L l,l =ΞΆ(l)l,l 1 (4.80)
βˆ’| i | βˆ’ i
L l,l 1 =ΞΆ(l)l,l (4.81)
+
| βˆ’ i | i
ΞΆ(l)2 =2l (4.82)
The last equation follows from the definition (4.71) of the highest weight
state
L l,l =0,
+
| i
i.e. there is no coefficient corresponding to l+1, or ΞΆ(l+1)=0.
The state L l,l must be normalized, so that
βˆ’| i
l,l L +L l,l =ΞΆ(l)βˆ—ΞΆ(l) l,l 1l,l 1 =ΞΆ(l)βˆ—ΞΆ(l)=ΞΆ(l)2
h | βˆ’| i h βˆ’ | βˆ’ i
92
On the other hand, using the angular momentum algebra
l,l L L l,l = l,l L l,l =2l l,l L l,l =2l,
+ z z
h | βˆ’| i h | | i h | | i
so that
ΞΆ(l)2 =2l.
This proves (4.82).
Next performing some further algebra
1 1
L l,l 1 = L L l,l = [L ,L ]l,l =
+ + +
| βˆ’ i ΞΆ(l) βˆ’| i ΞΆ(l) βˆ’ | i
2 2l
L l,l = l,l =ΞΆ(l)l,l
z
ΞΆ(l) | i ΞΆ(l)| i | i
This proves (4.81).
For the induction step, assume that the equations (4.78) (4.79) are true for
a certain k. First we consider 4.79) for k+1. Calculate the norm of the state
L l,l k 1 using equation (4.77), which as pointed out, is just a definition,
βˆ’| βˆ’ βˆ’ i
we get
l,l k 1L L l,l k 1 =ΞΆ(l k 1) ΞΆ(l k 1) l,l k 2l,l k 2 =
+ βˆ—
h βˆ’ βˆ’ | βˆ’| βˆ’ βˆ’ i βˆ’ βˆ’ βˆ’ βˆ’ h βˆ’ βˆ’ | βˆ’ βˆ’ i
ΞΆ(l k 1)2.
βˆ’ βˆ’
On the other hand, using the angular momentum algebra and the induction
hypothesis
l,l k 1L L l,l k 1 = l,l k 1[L L ]+L L l,l k 1 =
+ + +
h βˆ’ βˆ’ | βˆ’| βˆ’ βˆ’ i h βˆ’ βˆ’ | βˆ’ βˆ’ | βˆ’ βˆ’ i
l,l k 12L z+L L l,l k 1 =2(l k 1)+ΞΆ(l k)βˆ—ΞΆ(l k)=
+
h βˆ’ βˆ’ | βˆ’ | βˆ’ βˆ’ i βˆ’ βˆ’ βˆ’ βˆ’
2(l k 1)+ΞΆ(l k)2.
βˆ’ βˆ’ βˆ’
Combining the last two equations yields
ΞΆ(l k 1)2 =2(l k 1)+ΞΆ(l k)2,
βˆ’ βˆ’ βˆ’ βˆ’ βˆ’
which is precisely (4.79) with k+1 instead of k.
Next we consider (4.78) for k + 1. Using the operator algebra and the
induction hypothesis yields
1
L l,l k 2 = L L l,l k 1 =
+ +
| βˆ’ βˆ’ i ΞΆ(l k 1) βˆ’| βˆ’ βˆ’ i
βˆ’ βˆ’
1
[L ,L ]+L L l,l k 1 =
+ +
ΞΆ(l k 1) βˆ’ βˆ’ | βˆ’ βˆ’ i
βˆ’ βˆ’ (cid:0) (cid:1)
1
2L +L L l,l k 1 =
z +
ΞΆ(l k 1) βˆ’ | βˆ’ βˆ’ i
βˆ’ βˆ’ (cid:0) (cid:1)
93
1
2(l k 1)+ΞΆ(l k)2 l,l k 1 =
ΞΆ(l k 1) βˆ’ βˆ’ βˆ’ | βˆ’ βˆ’ i
βˆ’ βˆ’ (cid:0) (cid:1)
ΞΆ(l k 1)2
βˆ’ βˆ’ l,l k 1 =ΞΆ(l k 1)l,l k 1 .
ΞΆ(l k 1) | βˆ’ βˆ’ i βˆ’ βˆ’ | βˆ’ βˆ’ i
βˆ’ βˆ’
The recursive equations (4.79) can be solved. The result is
ΞΆ(l k)= (k+1)(2l k). (4.83)
βˆ’ βˆ’