5.73 Lecture #23
23 - 1
Angular Momentum Matrix Elements
LAST TIME:
* all [, ]=0 Commutation Rules needed to block diagonalize

p 2  L2
H= r +
+ V (r) in nLM L basis set
2
2µ  2µr

* εijk Levi-Civita antisymmetric tensor — useful properties
* Commutation Rule DEFINITIONS of Angular Momentum and
“Vector” Operators [Li , L j ] = ih∑ ε ijk L k
k
[Li , Vj ] = ih∑ ε ijk Vk
k
Classification of operators: universality of matrix elements.
TODAY: Obtain all angular momentum matrix elements from the commutation
rule definition of an angular momentum, without ever looking at a
differential operator or a wavefuncton. Possibilities for phase
inconsistencies. [Similar derivation for angular parts of matrix
elements of all spherical tensor operators, Tq( k ).]
1.
Define Components of Angular Momentum using a Commutation Rule.
2.
Define eigenbasis for J2 and Jz
3.
show λ ≥ µ 2
4
raising and lowering operators (like a†, a and x~ ± i p )
|λµ⟩
~
J±|λµ⟩ gives eigenfunction of Jz belonging to µ ± h eigenvalue and
eigenfunction of J2 belonging to λ eigenvalue
5.
Must be at least one µMAX and one µMIN such that
J–(J+|λµMAX⟩) = 0
J+(J–|λµMIN⟩) = 0
n
n n
This leads to µmax = h  , λ = h2  + 1 .
2
2 2
6.
Obtain all matrix elements of Jx, Jy, J±, but there remains a phase
ambiguity
7.
Standard phase choice: “Condon and Shortley”
revised 4 November, 2002
5.73 Lecture #23
[
23 - 2
]
1. Commutation Rule J i , J j = ih∑ ε ijk J k
k
This is a general definition of angular momentum (call it J, L, S, anything!).
Each angular momentum generates a state space.
2. eigenfunctions of J2 and Jz exist
J 2 λµ = λ λµ
(Hermitian operators. Guaranteed
by symmetrization.)
J z λµ = µ λµ
but what are the values of λ,µ?
J2 and Jz are Hermitian, therefore λ,µ are real
3. find upper and lower bounds for µ in terms of λ : λ ≥ µ2
λµ ( J 2 − J 2z ) λµ = λ − µ 2
Want to show that this is ≥ 0.
but J 2 = J 2x + J 2y + J 2z
J 2 − J 2z = J 2x + J 2y
λ − µ 2 = λµ J 2x + J 2y λµ
completeness
λ − µ2 =
∑
λ ′,µ ′
[ λµ J λ ′µ′
x
λ ′µ ′ J x λµ
+ λµ J y λ ′µ ′ λ ′µ ′ J y λµ
λ ′µ ′ J x λµ = λµ J x λ ′µ ′ *
Hermitian:
λ − µ2 =
∑
λ ′,µ ′
]

 λµ J x λ ′µ ′
2
+ λµ J y λ ′µ ′
2

 ≥ 0
Thus λ − µ 2 ≥ 0 and λ ≥ µ 2 ≥ 0
and µ MAX ≤ λ1/ 2 , µ MIN ≥ −λ1/ 2
revised 4 November, 2002
5.73 Lecture #23
23 - 3
4. Raising/Lowering Operators
(not Hermitian:
] = [J , J ] ± i[J , J ]
[J z , J ±
z
x
z
( just like a,a† )
J†+ = J − )
J ± ≡ J x ± iJ y
y
[
= ihJ y ± i( −ihJ x ) = ± h J x ± iJ y
= ± hJ ±
]
right multiply by λµ
J z J ± = J ± J z ± hJ ±
J z ( J ± λµ ) = J ± ( J z λµ ) ± hJ ± λµ
= J ± µ λµ ± hJ ± λµ
= (µ ± h)( J ± λµ
(J
±
)
λµ ) is an eigenfunction of J z belonging to eigenvalue µ ± h.
Thus J ± “raises” or “lowers” J z eigenvalue in steps of h.
[
]
Similar exercise for J 2 , J ± to get effect of J ± on eigenvalue of J 2
[J
2
] [
] [
]
, J ± = J2 , J x ± i J2 , J y = 0
(
)
J 2 ( J ± λµ ) = J ± J 2 λµ = λ( J ± λµ
(We already know that [J
)
(J ± λµ ) belongs to same eigenvalue of J 2 as
2
] )
, Ji = 0
λµ
J ± has no effect on λ.
revised 4 November, 2002
5.73 Lecture #23
23 - 4
*
*
upper and lower bounds on µ are ±λ1/2
J± raises/lowers µ by steps of h
*
Since J x =
1
1
(J + + J − ) and J y = (J + − J − ),
2
2i
The only nonzero matrix elements of Ji in the |λµ⟩ basis set are those where
∆µ = 0, ±h and ∆ λ = 0. As for derivation of Harmonic Oscillator matrix
elements, we are not assured that all µ differ in steps of h. Divide basis states
into sets related by integer steps of h in µ.
2
5. For each set, there are µ MIN and µ MAX : λ ≥ µ
Thus, for each set
but
(
J + λµ MAX = 0
J − λµ MIN = 0
)(
)
J − J + = J x − iJ y J x + iJ y = J 2x + J 2y + iJ x J y − iJ y J x
[
= J 2x + J 2y + i J x , J y
= J 2x + J 2y + i (ihJ z )
]
= J 2x + J 2y − hJ z
but
J 2x + J 2y = J 2 − J 2z , thus
J − J + = J 2 − J 2z − hJ z
(
= (λ − µ
)
0 = J − J + λµMAX = J 2 − J 2z − hJ z λµMAX
2
MAX
)
− hµMAX λµMAX
λ = µ 2MAX + hµ MAX
Similarly for µ MIN
J + J − λµ MIN = 0
revised 4 November, 2002
5.73 Lecture #23
23 - 5
J + J − = J 2 − J 2z + hJ z
λ = µ 2MIN − hµ MIN
subtract 2 equations for λ
0 = µ 2MAX − µ 2MIN + h(µ MAX + µ MIN )
0 = (µ MAX + µ MIN )(µ MAX − µ MIN + h)
Thus µ MAX = −µ MIN
now factor
this equation
OR µ MAX = µ MIN − h
(impossible)
Thus for each set of λµ , µ goes from µ MAX to µ MIN in steps of h
µ MAX = µ MIN + nh = – µ MAX + nh
n
h
2
µ MAX =
Thus µ is either integer or half integer or both!
Thus there will at worst be only two non-communicating sets of λµ⟩ because if µ
were both integer and 1/2-integer, each would form a set of µ-values, the members of
which would be separated in steps of h.
Now, to specify allowed values of λ:
2
λ=µ
2
MAX
+ hµ MAX
let
nn 
n 
n 
h + h h = h2
=
+1
2 
2 
22 
n
≡j
2
µ MAX = hj
µ MIN = − hj
λ = h j ( j + 1)
2
!
j either integer or
half integer or both
revised 4 November, 2002
5.73 Lecture #23
23 - 6
rename our basis states
J 2 jm = h2 j ( j + 1) jm
J z jm = hm jm
valid for all operators that satisfy [A i , A j ] = ih Σ ε ijk A k
k
OK to define an ama basis set for any angular momentum operator defined as above.
6. Jx, Jy, J± matrix elements
recall page 23-3, but in new notation
(J ± raises / lowers m by 1)
jm ± 1 = N ± J ± jm
normalization factor (to be determined below)
1 = jm ± 1 jm ± 1 = ( N± J ± jm
)† ( N± J ±
jm
)
N±† = N±*
J†± = J m
1 = N±
2
!
jm J m J ± jm
J m J ± = ( J xmiJ y )( J x ± iJ y ) = J + J ± i[ J x , J y ]
2
x
2
y
use this to evaluate matrix
elements of JmJ±
= J 2 − J 2z ± i(ihJ z ) = J 2 − J 2z mhJ z
= J 2 − J z ( J z ± h)
1 = N±
N± =
2
1
[h
2
j ( j + 1) − h2 ( m( m ± 1))]
[ j( j + 1) − m(m ± 1)]
h
−1 / 2
e − iδ ±
arbitrary phase factor
from taking square root
1/ 2
J ± jm = h[ j ( j + 1) − m( m ± 1)] jm ± 1 e − iδ ±
Usual phase choice is δ ± = 0 for all j, m:
the “Condon and Shortley” phase choice
(sometimes δ ± = ± π 2
– so be careful)
revised 4 November, 2002
5.73 Lecture #23
23 - 7
std. phase choice: δ ± = 0
1/ 2
j ′m ′ J ± jm = hδ j ′jδ m ′m ±1[ j ( j + 1) − m( m ± 1)]
(or hδ
since J x =
( )
jj ′δ m ′m ±1[ j ( j + 1) − m m ′ ]
1/ 2
1
(J + + J − )
2
j ′m ′ J x jm =
h
2
)
remember matrix
elements of x and
p in harmonic
oscillator basis set?
{
δ j ′j δ m ′m +1[ j ( j + 1) − m( m + 1)]
1/ 2
1/ 2
+δ m ′m −1[ j ( j + 1) − m( m − 1)]
Jy =
1
(J + − J − )
2i
h
two sign
surprises
}
{
j ′m ′ J y jm = −i δ j ′j δ m ′m +1[ j ( j + 1) − m( m + 1)]
2
1/ 2
1/ 2
−δ m ′m −1[ j ( j + 1) − m( m − 1)]
}
This phase choice leaves all matrix elements of J 2 , J x and J ± real and positive,
but those of J y imaginary
[if use δ
±
= + π 2 , this gives J y real and J x , J ± imaginary]
Summary
j′ m ′ J 2 jm = δ j ′ jδ m ′ m h2 j( j + 1)
r
jm J jm = k√hm
r
h
1/2
jm ± 1 J jm = √
i m ij√ j( j + 1) − m ( m ± 1)
2
1
1
i√J x + √j J y = i√( J + + J − ) + √j ( J + − J − )
2
2i
1
1
i − ij√ + J − √
= J+ √
i + ij√
2
2
(
) [
(
]
)
(
)
revised 4 November, 2002