PHYS 212A: Homework 5
December 8, 2013
3.21
a
We define Li = ijk xj pk . Writing x and p in terms of creation and annihilation operators gives,
Li = ijk i~(a†j + aj )(a†k − ak ) = ijk
i~
(aj a†k − a†j ak ) = ijk i~aj a†k
2
(1)
The other result can be derived similarly.
b
The states ketqlm are defined as eigenstates of the Lz and L2 operators. Here since N = 2q + l =
nx + ny + nz = 1, we have three possible kets |100i , |010i , |001i. It is helpful to consider what
happens to each state under the angular momentum operators.
Lz |100i = i~(ax a†y − ay a†x ) |100i = i~ |010i
(2)
Lz |010i = −i~ |100i
(3)
Lz |001i = 0
(4)
Under L2 any of the 3 states will have the same result, L2 |001i = ~2 [1(1 + 1)] |001i, this shows
|010iqlm = |001in . From inspection, we can determine that |01 ± 1iqlm = √12 (|100in ± i |010in )
c
Now we have 6 possible states to consider, and find that
1
|020iqlm = √ (|200i + |020i + |002i)
3
1
(5)
d
Following the same procedure as above, we find
Lz |200i = i~(|110i)
(6)
Lz |020i = −i~(|110i)
√
Lz |110i = i~ 2(|020i − |200i)
(7)
Lz |101i = i~(|011i)
(9)
(8)
Lz |011i = −i~(|101i)
(10)
(11)
By inspection, we can determine the unnormalized m = ±1 states are |101i ± i |011i. Similarly, for
m = ±2 states are √12 (|200i − |020i) ± i |110i
3.22
a
For x = 0 we have
g(x, t) = 1/(1 − t) = 1 + t + t2 + ... =
X
Ln (0)
tn
n!
(12)
n
|
= n! The other result follows similarly.
Comparing powers of t, we can see Ln (0) = ∂ g(x,t)
∂ tn t=0
b
∂g
−t
=
g
∂x
1−t
(13)
Inserting the series expression gives
(t − 1)
X
L0n (x)
X
tn X 0
tn
tn
−
Ln (x) = t
Ln (x)
n!
n!
n!
(14)
Rearranging indices to collect like powers of t, we find
nL0n−1 (x) − L0n (x) = nLn+1 (x)
(15)
∂g
1−t−x
=
g
∂t
(1 − t)2
(16)
c
(1 − 2t + t2 )
X
Ln (x)
X
tn−1
tn
= (1 − t − x)
Ln (x)
(n − 1)!
n!
Again grouping powers of t, we find the sought after expression.
2
(17)
d
This is most easily proved using the generator
−tx
−tx
−tx
−tx
−tx
e 1−t
e 1−t x
e 1−t tx
e 1−t t(1 − x) e 1−t t2 x
+
+
t[
−
−
] = 0 (18)
xg + (1 − x)g + ng = 0 −
(1 − t)2
(1 − t)3
(1 − t)2 (1 − t)2 (1 − t)3
00
0
3.15
3.24
3.26
3
4
5
6