Chapter 5. Angular Momentum
Review: If the potential has spherical symmetry, V(r) =V(r) , e.g., Coulomb potential
V(r) = Z 1Z 2e 2 / r , coordinate separation can be performed using the spherical coordinates
ψ(r) = R(r)Y (θ,φ) ,
(5.1)
where the angular wavefunctions Y (θ,φ) satisfy the angular equation:
2 ⎤
⎡ 1 ∂ ⎛
⎞
⎜⎜sin θ ∂ ⎟⎟ + 1 ∂ ⎥Y m (θ,φ) + l(l + 1)Y m (θ,φ) = 0 ,
⎢
⎟
l
⎢ sin θ ∂θ ⎜⎝
∂θ ⎟⎠ sin 2 θ ∂φ 2 ⎥⎥⎦ l
⎢⎣
(5.2)
while the radial wavefunctions R(r) satisfy the radial equation:
⎡  2 d ⎛ d ⎞ l(l + 1) 2
⎤
⎜⎜r 2 ⎟⎟ +
⎢−
⎥ R(r) = ER(r) .
+V(r)
⎟
2
⎢ 2mr 2 dr ⎜⎝ dr ⎟⎠
⎥
2mr
⎢⎣
⎥⎦
(5.3)
In last chapter we focused on radial wavefunctions, now we study the angular wavefunctions.
5.1 Spherical Harmonics and Orbital Angular Momentum
The solutions Yl m (θ,φ) to Eq. (5.2) are called spherical harmonics. Explicitly,
Yl m (θ,φ) = AlmPl m [cos(θ)]e imφ ,
Yl
−m
m
m
*
(θ,φ) = (−1) [Yl (θ,φ)] ,
m ≥0
(5.4)
m ≥0
with l = 0,1, 2, and m = 0,±1,±2,,±l . Here Pl m (x) is the associated Legendre Polynomial,
m
2 m/2
Pl (x) = (1 − x )
Pl
−m
l+m
dm
1
2 m/2 d
P (x) = l (1 − x )
(x 2 −1)l ,
m l
l+m
dx
2l!
dx
m
(x) = Pl (x),
m ≥0
(5.5)
m ≥0
where Pl (x) is the Legendre Polynomial,
Pl (x) =
1 dl 2
(x −1)l .
l
l
2 l ! dx
(5.6)
A summary of properties of Legendre polynomials and associated Legendre polynomials:
Pl (1) = 1;
⎧ 0
⎪
⎪
Pl m (±1) = ⎨
⎪
±1
⎪
⎪
⎩
Pl (−x) = (−1)l Pl (x);
Pl m (−x) = (−1)l−m Pl m (x)
56
(m ≠ 0)
(m = 0)
(5.7)
(5.8)
∫
1
−1
Pl ′(x)Pl (x)dx =
2
δ ;
2l + 1 l ′,l
∫
1
−1
Pl ′m (x)Pl m (x)dx =
2 (l + m)!
δ ,
2l + 1 (l −m)! l ′,l
m ≥ 0 (5.9)
Using the normalization conditions Eq. (5.9) and
∫
2π
0
′
e −im φe imφ dφ = 2πδm′,m ,
(5.10)
the normalization parameter Alm is determined (m ≥ 0) :
2l + 1 (l −m)!
,
4π (l + m)!
Alm = (−1)m
(5.11)
so that
∫
[Yl ′m (θ,φ)]*Yl m (θ,φ)dΩ ≡ ∫
2π
′
Ω
0
∫
π
0
′
[Yl ′m (θ,φ)]*Yl m (θ,φ)sin(θ)dθ dφ = δl ′,l δm′,m . (5.12)
For m = 0 , e imφ = 1 we obtain
2l + 1
P (cos θ) ,
4π l
Yl 0(θ,φ) =
Y00(θ,φ) =
1
P (cos θ) =
4π 0
1
.
4π
(5.13)
(5.14)
Now we can express eigenfunctions for central potential in both Cartesian and spherical coordinates; in particular, for a free particle V = 0 :
∞
l
ψk (r) = ∑ ∑ C lm jl (kr)Yl m (θ,φ) .
(5.15)
l=0 m=−l
One can employ (1) the normalization condition for spherical Bessel functions,
∞
π
2
∫0 jl (kr)jl (k ′r)r dr = 2k 2 δ(k − k ′) ;
(5.16)
(2) the addition theorem for spherical harmonics:
4π l
Pl (cos γ) =
[Yl m (θ′, φ′)]*Yl m (θ,φ) ,
∑
2l + 1 m=−l
(5.17)
where γ is the angle between two vectors r = (r, θ,φ) and r′ = (r ′, θ′, φ′) ; and
(3) the well-known expression for Legendre polynomials:
∞
e ik⋅r = ∑ (2l + 1)i l jl (kr)Pl (k̂ ⋅r̂) ,
l=0
to find the coefficients C lm . You will encounter it when studying quantum scattering.
57
(5.18)
5.2 Orbital Angular Momentum
Spherical harmonics Yl m (θ,φ) are eigenfunctions of operator L2 , with L the orbital angular
momentum operator from the correspondence principle:
L = r×p = r×(−i∇) .
(5.19)
In spherical coordinates,
∂
1 ∂
1
∂
r = rr̂; ∇ = r̂
+ θ̂
+ φ̂
. (5.20)
∂r
r ∂θ
r sin θ ∂φ
Then L in spherical coordinates:
⎛ 1 ∂
∂⎞
L = i ⎜⎜θ̂
− φ̂ ⎟⎟⎟.
⎜⎝ sin θ ∂φ
∂θ ⎟⎠
(5.21)
Specifically, Eq. (5.21) leads to
⎛
∂
∂ ⎞
Lx = +i ⎜⎜⎜sin(φ)
+ cos(φ)cot(θ) ⎟⎟⎟ ,
∂θ
∂φ ⎟⎠
⎝
⎛
∂
∂ ⎞
Ly = −i ⎜⎜cos(φ) − sin(φ)cot(θ) ⎟⎟⎟ ,
⎜⎝
∂θ
∂φ ⎟⎠
Lz = −i
(5.22)
(5.23)
∂
.
∂φ
(5.24)
⎡ 1 ∂ ⎛
∂ ⎞⎟
1 ∂ 2 ⎤⎥
⎜
⎢
⎟
.
L = L + L + L = − ⎢
⎜⎜sin θ ⎟ +
∂θ ⎟⎠ sin 2 θ ∂φ 2 ⎥⎥⎦
⎢⎣ sin θ ∂θ ⎝
2
2
x
2
y
2
z
2
(5.25)
Comparing Eq. (5.2) with Eq. (5.25), we find that Yl m (θ,φ) are eigenfunctions of L2 :
L2Yl m (θ,φ) = l(l + 1) 2Yl m (θ,φ) .
(5.26)
In addition, Eqs. (5.24) and (5.4) ⇒
LzYl m (θ,φ) = mYl m (θ,φ)
(5.27)
Alternative derivation: One can show that
L2 = r2p2 −(r ⋅ p)2 + ir ⋅ p ,
(5.28)
Then using Eq. (4.3), one obtains
⎡
⎡ 1 ∂ ⎛
1 ∂ ⎛⎜ 2 ∂ψ ⎞⎟⎤⎥
∂ψ ⎞⎟
1 ∂2 ψ
2 ⎢
⎜
⎟
⎟
L2ψ(r) =  2r 2 ⎢⎢−∇2ψ(r) + 2
r
=
−
sin(θ)
+
⎜
⎟
⎟
⎢ sin(θ) ∂θ ⎜⎜⎝
∂θ ⎟⎠ sin 2(θ) ∂φ 2
r ∂r ⎜⎝ ∂r ⎟⎠⎥⎥⎦
⎢⎣
⎢⎣
the same operator as expressed in Eq. (5.25).
58
⎤
⎥,
⎥
⎥⎦
(5.29)
The spherical harmonics Yl m (θ,φ) are the orbital angular momentum states in the spherical
coordinates representation:
Yl m (θ,φ) = n̂ l,m = θ,φ l,m ,
(5.30)
where the directional eigenket n̂ is determined by θ and φ . Eqs. (5.26) and (5.27) indicate
that
L2 l,m = l(l + 1) 2 l,m ,
(5.31)
Lz l,m = m l,m ,
(5.32)
i.e., l,m are the common eigenkets for operators L2 and Lz , and they form a complete finitedimensional orthonormal basis:
l ′, m ′ l,m = δl ′,l δm′,m ,
Eq. (5.12) ⇒
m
∑
lm ml = 1.
(5.33)
(5.34)
m=−l
5.2.1 Commutation Relations of Orbital Angular Momentum
From the fundamental commutator [X,P ] = i , it is straightforward to derive the fundamental
commutation relations of orbital angular momentum:
[Li , Lj ] = iijk Lk ,
(5.35)
[L2, Li ] = 0 .
(5.36)
Here i, j,k runs from 1 to 3, denoting x, y, z directions, respectively.  is the Levi-Civita tensor:
⎧
⎪
+1
⎪
⎪
⎪
ijk = ⎨ −1
⎪
⎪
⎪
⎪
⎩ 0
if (i, j,k) = (1, 2, 3) or (2, 3,1) or (3,1, 2),
if (i, j,k) = (1, 3, 2) or (3, 2,1) or (2,1, 3),
(5.37)
if i = j or j = k or k = i,
i.e., ijk is +1 if (i, j,k) is an even permutation of (1,2,3), −1 if (i, j,k) is an odd permutation of
(1,2,3), and 0 if any index is repeated. Alternatively,
ijk = (i − j)(j −k)(k −i) / 2.
Eq. (5.35) can be expressed explicitly as
59
(5.38)
[Lx ,Ly ] = iLz ;
[Ly ,Lz ] = iLx ;
[Lz ,Lx ] = iLy .
(5.39)
Let’s prove them from the fundamental commutator and the definition of orbital angular momentum [Eq. (5.19)]. We prove the first one, and the other two are similar.
[Lx ,Ly ] = [ypz − zpy , zpx − xpz ] = [ypz , zpx ]+[−zpy , − xpz ]
= ypx [pz ,z ]+ py x[z, pz ] = i(−ypx + xpy ) = iLz .
Using Eq. (5.36) [or Eq. (5.39)] we can prove Eq. (5.36):
[L2,Lz ] = [L2x ,Lz ]+[L2y ,Lz ]+[L2z ,Lz ] = [L2x ,Lz ]+[L2y ,Lz ],
[L2x ,Lz ] = Lx [Lx ,Lz ]+[Lx ,Lz ]Lx = −i(Lx Ly + Ly Lx ); similarly, [L2y ,Lz ] = i(Lx Ly + Ly Lx ).
Then [L2,Lz ] = 0. One can verify that [L2,Lx ] = [L2,Ly ] = 0.
5.2.2 Ladder Operators
One might ask what are Lx l,m and Ly l,m ? Obviously l,m are not eigenkets of Lx or Ly
(Why? If so, what fundamental principle is violated?) Recall in simple harmonic oscillator, how
did we find X n and P n , where n are eigenkets of H but not of X or P ?
Basically, we follow the same route using the ladder operators defined as
L± ≡ Lx ± iLy ,
(5.40)
which indicates that L+ = L†− since Lx and Ly are Hermitian. Then Lx , Ly and L2 can be expressed in terms of ladder operators:
1
1
Lx = (L+ + L− ); Ly = (L+ − L− )
2
2i
1
L2 = (L+L− + L−L+ ) + L2z .
2
(5.41)
Let’s work out the commutation relations. Eqs. (5.35) and (5.36) ⇒
[L+, L− ] = 2Lz ; [Lz , L± ] = ±L± ,
(5.42)
[L2,L± ] = 0 .
(5.43)
60
Next, we want to find L± l,m algebraically. In fact, we can determine L2 l,m and
Lz l,m algebraically from the basic commutation relations Eqs. (5.35) and (5.36) [together with
(5.42) and (5.43)] without invoking spherical harmonics – mathematically, these basic commutation relations uniquely define the operations of L2 and Li ( i = x,y,z;± ) on the common eigenkets, a,b , of L2 and Lz :
L2 a,b = a a,b ; Lz a,b = b a,b .
(
(5.44)
)
The commutation relation, Eq. (5.42) ⇒ Lz L− a,b = (L−Lz − L− ) a,b = (b − )L− a,b ⇒
L− a,b = C − a,b −  ; similarly, L+ a,b = C + a,b +  .
(5.45)
We can prove that a ≥ b 2 . Since L2 − L2z = (L+L− + L−L+ ) / 2 and
L+ = L†− ⇒ a,b L+L− a,b =|C − |2 ≥ 0 ; L− = L†+ ⇒ a,b L−L+ a,b =|C + |2 ≥ 0 ,
(
)
a −b 2 = a,b L2 − L2z a,b ≥ 0 ,
(5.46)
which suggests that there must exist a bmax for each value of a so that
L+ a,bmax = 0 .
(5.47)
Thus L−L+ a,bmax = 0 , and L−L+ = (Lx −iLy )(Lx + iLy ) = L2x + L2y + i[Lx ,Ly ] = L2 − L2z − Lz
(
)
2
⇒ L2 − L2z − Lz a,bmax = 0 ⇒ a −bmax
− bmax = 0 ⇒
a = bmax (bmax + ) .
(5.48)
L− a,bmin = 0 ⇒ a = bmin (bmin − ) .
(5.49)
Similarly,
Comparing Eqs. (5.48) and (5.49), one finds that bmax = −bmin and bmin ≤ b ≤ bmax .
Eq. (5.45) indicates that applying L+ to a,bmin n times one can reach a,bmax (subject to a
constant factor). Therefore, bmax = bmin + n , with n a non-negative integer, and
bmax = n / 2; bmin = −n / 2 .
61
(5.50)
Denote l = n / 2 [Note: here l can be either integer (orbital angular momentum) or half-integer
(spin or total angular momentum; spin can be integer as well)], then we obtain
a = l(l + 1) 2
(5.51)
from Eq. (5.48). Denote m = b , then m = −l,−l + 1,,l −1,l. Instead of eigenvalues a and b ,
we can use quantum numbers l and m to denote the common eigenkets of L2 and Lz by l,m .
( )
L+ l,m = C +lm l,m + 1 ⇒ l,m L− = l,m + 1 C +lm
(
*
⇒
)
|C +lm |2 = l,m L−L+ l,m = l,m L2 − L2z − Lz l,m
= l(l + 1) 2 −m 2 2 −m 2 = (l −m)(l + m + 1) 2 .
Therefore,
L+ l,m = (l −m)(l + m + 1)  l,m + 1 .
(5.52)
Similarly one can easily derive
L− l,m = (l + m)(l −m + 1)  l,m −1 .
(5.53)
Finally, a summary of the crucial conclusions for orbital (actually any) angular momentum:
[Li ,Lj ] = iεijk !Lk
[L2,Li ] = 0
⎫
⎪
⎪
⎬⇒
⎪
⎪
⎭
L2 l,m = l(l + 1) ! 2 l,m
Lz l,m = m! l,m
l ′, m ′ l,m = δl ′,l δm′,m
l
∑
L+ l,m = (l −m)(l + m + 1) ! l,m + 1
L− l,m = (l + m)(l −m + 1) ! l,m −1
lm ml = 1
m=−l
5.3 Quantum Rotation and Formal Theory of Angular Momentum
Angular momentum J is the generation operator for rotation operator D(R) , like momentum p
is the generation operator for displacement D(r) . Here we need to distinguish D(R) and R :
ψ
R
= D(R) ψ ,
62
(5.54)
where the ket state ψ changes to another ket state ψ
R
under a rotation Rn̂ (ϕ) represented by
a 3×3 orthogonal matrix, where the unit vector n̂ denotes direction of the rotation axis while ϕ
the rotation angle. For example,
⎡ cosφ −sin φ 0 ⎤
⎢
⎥
⎥
Rz (φ) = ⎢⎢ sin φ
cosφ 0 ⎥ .
⎢
⎥
0
1 ⎥⎦
⎢⎣ 0
(5.55)
Question: How many dimensions does D(R) have?
We can easily construct D(R) from the general infinitesimal translation:
U  = 1 −iG .
(5.56)
Here G is Hermitian while U is unitary (Not Hermitian in general). For example,
 → dx , G → p̂x /  ;  → dt , G → H /  . We thus define angular momentum operator J i for
the i-th axis (x, y or z) associated with an infinitesimal rotation of angle dφ about this axis:
 → dφ, G →
Ji

, Diˆ(dφ) = 1 −i
Ji

dφ .
(5.57)
In general, for an arbitrary rotation axis along n̂ direction:
⎛ J ⋅ n̂ ⎞⎟
⎟dφ .
Dn̂ (dφ) = 1 −i ⎜⎜
⎜⎝ ! ⎟⎟⎠
(5.58)
For a finite rotation angle φ :
N
⎡
⎡ ⎛ J ⋅ n̂ ⎞ ⎤
⎛ J ⋅ n̂ ⎞⎟ φ ⎤
⎟⎟ ⎥
⎟⎟ ⎥ = exp ⎢−i ⎜⎜
Dn̂ (φ) = limN →∞ ⎡⎢Dn̂ (φ / N )⎤⎥ = limN →∞ ⎢⎢1 −i ⎜⎜⎜
⎥
⎢ ⎜⎝  ⎟⎟⎠ φ ⎥ .
⎣
⎦
⎟

N
⎝
⎠
⎢⎣
⎥⎦
⎢⎣
⎥⎦
N
We know p̂x = −ih
(5.59)
d
(Question: How did we derive it by considering space displacedx
ment?). Similarly, if a state ψ can be represented in real space, then J z = −i
d
, which redφ
covers Eq. (5.24), i.e., it is the orbital angular momentum Lz . However, if a state ket ψ is associated with intrinsic degree of freedom, such as spin, then J cannot be written in terms of realspace coordinates explicitly. Question: how can we study J and rotation D(R) ?
63
5.3.1 The Fundamental Commutation Relations of Angular Momentum
Unlike px and py , J x and J y don’t commute, this is because two space displacements along different directions commute (Abelian) but two rotations about two different axes don’t (NonAbelian). Since for every rotation R in real space there exists one rotation D(R) in an appropriate
ket space, we postulate that D(R) has the same group properties as R :
D(R1 )D(R2 ) = D(R1R2 ) ,
(5.60)
D(R)D(R −1 ) = D(0) = 1 ⇒ D †(R) = D −1(R) = D(R −1 ) ,
(5.61)
D(R1 )D(R2 )D(R3 ) = [D(R1 )D(R2 )]D(R3 ) = D(R1 )[D(R2 )D(R3 )] .
(5.62)
Based on Eq. (5.55) and the similar matrix representations for Rx (φ) and Ry (φ) , one can
prove that for an infinitesimal rotation angle  :
Rx ()Ry ()− Ry ()Rx () = Rz (2 )− R(0) ,
(5.63)
(Clearly rotations around x and y axes don’t commute!) which suggests that
Dx ()Dy ()− Dy ()Dx () = Dz (2 )− D(0) .
(5.64)
Using the Taylor series (up to the second order) to Eq. (5.59), one obtains that
[J x , J y ] = iJ z .
(5.65)
Repeating this argument with rotations along other axes, the fundamental commutation relations of angular momentum are obtained:
[J i , J j ] = iijk J k .
(5.66)
This is exactly the same as Eq. (5.35) for orbital angular momentum L .
5.3.2 Matrix Representations of Angular Momentum and Rotation
The fundamental commutation relations [Eq. (5.66)] offer the foundation for investigating rotations for any quantum states. Similar to orbital angular momentum, from Eq. (5.66), one can derive all the following crucial expressions:
[J2,J i ] = 0 ,
(5.67)
[J +, J − ] = 2J z ; [J z , J ± ] = ±J ± ,
(5.68)
64
J2 j,m = j(j + 1)  2 j,m
J z j,m = m j,m
J + j,m = (j −m)(j + m + 1)  j,m + 1
(5.69)
J − j,m = (j + m)(j −m + 1)  j,m −1
Here J ± = J x ± iJ y and j,m (m = −j,−j + 1,, j −1, j) form a complete orthonormal basis
for a fixed j :
j ′, m ′ j,m = δj ′,j δm′,m ,
j
∑
(5.70)
j,m m, j = 1 .
(5.71)
m=−j
Employing Eqs. (5.66)-(5.71), one can easily construct the matrix representations for angular
momentum J i using the basis { j,m } with a fixed value of j , then these irreducible matrices
have (2j + 1)×(2j + 1) matrix elements of j, m ′ J i j,m . Furthermore, one can construct the
matrix representations for rotation Dn̂ (ϕ) , and the irreducible matrices also have
(2j + 1)×(2j + 1) matrix elements denoted by Dm(j′),m (R) :
⎛ J ⋅ n̂ ⎞⎟
Dm(j′),m (R) = j, m ′ exp ⎜⎜−i
ϕ⎟⎟ j,m .
⎜⎝
 ⎟⎠
(5.72)
Some properties of Dm(j′),m (R) :
D(R1 )D(R2 ) = D(R1R2 ) ⇒
∑D
m ′′
(j )
m ′,m ′′
(R1 )Dm(j′′),m (R2 ) = Dm(j′),m (R1R2 ) ,
(j )
D(R −1 ) = D †(R) ⇒ Dm(j′),m (R −1 ) = [Dm,
(R)]* ,
m′
j,m
R
≡ D(R) j,m = ∑ j, m ′ j, m ′ D(R) j,m = ∑ Dm(j′),m (R) j, m ′ .
m′
(5.73)
(5.74)
(5.75)
m′
Evaluation of Dm(j′),m (R) is non-trivial, and Euler rotations and Euler angles can simplify it
greatly. Any arbitrary rotation can be decomposed to rotations about z- and y-axes:
Rn̂ (ϕ) = Rz (α)Ry (β)Rz (γ) ⇒ Dn̂ (ϕ) ≡ D(α, β, γ) = Dz (α)Dy (β)Dz (γ) ,
where α , β and γ are Euler angles. Eq. (5.76) ⇒
65
(5.76)
Dm(j′),m (α, β, γ) = j, m ′ exp(−iαJ z / )exp(−iβJ y / )exp(−iγJ z / ) j,m
= exp[−i(m ′α + mγ)] j, m ′ exp(−iβJ y / ) j,m .
(5.77)
Thus we need only evaluate d (j )(β) for the rotation around y-axis, whose matrix elements are
dm(j′),m (β) = j, m ′ exp(−iβJ y / ) j,m .
(5.78)
5.4 Spin-Half Systems
We will first apply the general theory of angular momentum to study the simplest but crucial
case: the two-level spin-half system. In contrast to orbital angular momentum, spin has no counterparts in classical physics. It is an intrinsic property of a particle, like (rest) mass and charge,
which won’t change. The formal theory of spin requires quantum field theory (QFT), and here
the non-relativistic, phenomenological aspects of spin are discussed.
5.4.1 Eigenstates of Spin Operators and Pauli Matrices
Mathematically, spin-1/2 just a special case of angular momentum with s = j = 1 / 2 , J → S ,
and all the commutation relations and properties of angular momentum [Eqs. (5.66)-(5.71)] are
exactly the same.
Let’s begin with basis
{ s,m } :
↑ ≡ 1 ≡
⎡
⎤
1 1
, = ⎢ 1 ⎥;
⎢ 0 ⎥
2 2
⎣
⎦
↓ ≡ 2 ≡
⎡
⎤
1 1
,− = ⎢ 0 ⎥ .
⎢ 1 ⎥
2 2
⎣
⎦
(5.79)
Using this basis, spin operators can be written in terms of Pauli matrices (How to obtain them?):
Si =
⎡
⎤
⎡
⎤
⎡

0 ⎤⎥ .
σi ; σx = ⎢ 0 1 ⎥ , σy = ⎢ 0 −i ⎥ , σz = ⎢ 1
⎢ 1 0 ⎥
⎢ i
⎢ 0 −1 ⎥
2
0 ⎥⎦
⎣
⎦
⎣
⎣
⎦
(5.80)
It is straightforward to verify that
[σi , σj ] = 2iijk σk ; [σ 2, σi ] = 0 ,
(5.81)
which satisfy Eqs. (5.66) and (5.67). In addition,
{σ , σ } ≡ σ σ
i
j
i
j
+ σj σi = 2δi,j ; σx2 = σy2 = σz2 = 1 .
66
(5.82)
⎡
⎤
A ket state χ = a 1 +b 2 = ⎢ a ⎥ , and its corresponding bra state is
⎢ b ⎥
⎣
⎦
χ = 1 a * + 2 b* = ⎡⎢ a *, b* ⎤⎥ . Then the expectation value of Si for this state is
⎣
⎦
Si =
⎡
⎤
⎡ *
* ⎤
σi ⎢ a ⎥ .
a
,
b
⎢
⎥
⎦ ⎢ b ⎥
2⎣
⎣
⎦
(5.83)
Question: If Si is measured, what are outcomes and possibilities?
In order to answer this question, we need to find the eigenvalues and eigenstates for Si . Diagonalizing Pauli matrices [Eq. (5.80)], we obtain:
1
1 ⎡⎢ 1 ⎤⎥
,
2 ⎢⎣ ±1 ⎥⎦
(5.84)
1 ⎡⎢ 1
2 ⎢⎣ ±i
(5.85)
Sx χx± = ±

χ ;
2 x±
χx± =
Sy χy± = ±

χ ;
2 y±
χy± =

χ ;
2 z±
⎡
⎤
⎡
⎤
χz+ = 1 = ⎢ 1 ⎥ , χz− = 2 = ⎢ 0 ⎥ .
⎢ 0 ⎥
⎢ 1 ⎥
⎣
⎦
⎣
⎦
Sz χz± = ±
(1
2
1
2
(
)
± 2 =
)
1 ±i 2 =
⎤
⎥,
⎥
⎦
(5.86)
Next, if we measure spin along an arbitrary direction n̂ for the ket state χ , what are the
outcomes and possibilities? Similarly, we need to find the eigenvalues and eigenstates χn̂± for
operator S ⋅ n̂ =
 
σ ⋅ n̂ . Here the directional unit vector
2
n̂ = (nx , ny , nz ) = (sin θ cosφ, sin θ sin φ, cos θ) ,
⎡
−iφ

σ ⋅ n̂ = σx sin θ cosφ + σy sin θ sin φ + σz cos θ = ⎢⎢ iφ cos θ e sin θ
− cos θ
⎢⎣ e sin θ

Diagonalize the matrix σ ⋅ n̂ , eigenstates of S ⋅ n̂ are
χn̂+
⎡
cos(θ / 2)
= ⎢⎢ iφ
⎢⎣ e sin(θ / 2)
⎤
⎥,
⎥
⎥⎦
χn̂−
⎡
sin(θ / 2)
= ⎢⎢
iφ
⎢⎣ −e cos(θ / 2)
(5.87)
⎤
⎥.
⎥
⎥⎦
⎤
⎥ ; and S χ = ±  χ .
⎥
n̂
n̂±
2 n̂±
⎥⎦
(5.88)
(5.89)
You might verify that χx± , χy± and χz± are special cases for χn̂± . The possibilities of
finding ± / 2 are χn̂± χ
2
if spin is measured along direction n̂ .
67
5.4.2 Rotations About the z-axis in Spin-Half Systems
First, consider the rotation by an angle φ about the z-axis: χ
R
= Dz (φ) χ . Here
n
1
−iφ / 2) σzn .
(
n=0 n !
∞
Dz (φ) = exp (−iφSz /  ) = exp (−iφσz / 2) = ∑
(5.90)
This is very simple, since σz is diagonal:
⎡ n
⎤ ⎧
1
0 ⎥ ⎪⎪⎪ σz (for odd n)
⎢
σ =⎢
=⎨
n ⎥
⎢⎣ 0 (−1) ⎥⎦ ⎪⎪⎪⎩ 1 (for even n)
n
z
(5.91)
2
4
3
⎡
⎤
⎡ ⎛ ⎞
⎤
n
1
1 ⎛⎜ φ ⎞⎟
1 ⎛⎜ φ ⎞⎟
1 ⎜ φ ⎟ 1 ⎛⎜ φ ⎞⎟
⎢
⎥
⎢
⎥
n
Dz (φ) = ∑ (−iφ / 2) σz = ⎢1 − ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ −⎥ I −i ⎢ ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ + ⎥ σz
2! ⎝ 2 ⎟⎠
4! ⎝ 2 ⎟⎠
⎢
⎥
⎢ 1! ⎝ 2 ⎟⎠ 3! ⎝ 2 ⎟⎠
⎥
n=0 n !
⎣
⎦
⎣
⎦
= cos(φ / 2)I −i sin(φ / 2)σz .
∞
We thus obtain Dz (φ) in a 2×2 matrix:
⎡ −iφ/2
0
Dz (φ) = ⎢⎢ e
iφ/2
e
⎢⎣ 0
⎤
⎥.
⎥
⎥⎦
(5.92)
⎡
⎤
Let’s see if there is any interesting physics here. The ket χ = ⎢ a ⎥ ; after the rotation:
⎢ b ⎥
⎣
⎦
χ
R
⎡ −iφ/2
0
= Dz (φ) χ = ⎢⎢ e
iφ/2
e
⎢⎣ 0
⎤⎡
⎤ ⎡ −iφ/2a
⎥⎢ a ⎥ = ⎢ e
⎥ ⎢ b ⎥ ⎢ e iφ/2b
⎥⎦ ⎣
⎦ ⎢⎣
If a 2π rotation around z-axis is applied, the ket state becomes χ
2π
⎤
⎥.
⎥
⎥⎦
(5.93)
= − χ ; it goes back to it-
self only after a rotation of 4nπ (n = 1, 2,) . This is called Berry Phase of π for spin-half particles due to ultra-relativity. We’ll study this next semester.
The next question is how the expectation value (measureable!) changes due to rotation.
Before the rotation, Eq. (5.83) ⇒
Sx = χ Sx χ =
⎡ *
⎤
a , b* ⎥⎦ σx
2 ⎢⎣
⎡ a ⎤
⎢
⎥ = (a *b +b*a)  .
⎢ b ⎥
2
⎣
⎦
(5.94)
After the rotation, the expectation value depends on the rotation angle:
Sx
R
=R χ Sx χ
R
=
 ⎡ iφ/2 *
e a , e −iφ/2b*
2 ⎢⎣
⎤σ
⎥⎦ x
68
⎡ −iφ/2
a
⎢ e
⎢ e iφ/2b
⎢⎣
⎤
⎥ = (e iφ/2a *b +e −iφ/2b*a)  .
⎥
2
⎥⎦
(5.95)
Alternative Approaches
We might consider the change in operator after a rotation, instead of the change in states – recall
the Schrödinger picture and Heisenberg picture.
Sx
R
=R χ Sx χ
R
= χ SxR χ ,
(5.96)
where the rotated spin operator
iφS /
−iφS /
SxR ≡ D †(R)Sx D(R) = e z Sxe z .
(5.97)
There are a number of ways to show that for a rotation around the z-axis
SxR = Sx cosφ −Sy sin φ .
(
(5.98)
)
Approach 1: using the representation Sx = 1 2 + 2 1  / 2 .
 iφSz /
−iφS /
e
1 2+ 2 1 e z
2


= e iφ/2 1 2 e iφ/2 +e −iφ/2 2 1 e −iφ/2 = e iφ 1 2 +e −iφ 2 1
2
2


=
1 2 + 2 1 cosφ + i
1 2 − 2 1 sin φ
2
2
= Sx cosφ −Sy sin φ .
(
SxR =
)
(
)
(
)
(
(
)
)
Approach 2: Using the Baker-Hausdorff Lemma [Eq. (3.52)].
SxR = e
iφSz /
Sxe
−iφSz /
= Sx + λ[Sz ,Sx ]+
2
λ2 ⎡
⎤ + λ ⎡S , ⎡S ,[S ,S ]⎤ ⎤ +  , where
S
,[S
,S
]
2! ⎢⎣ z z x ⎥⎦ 2! ⎢⎣ z ⎢⎣ z z x ⎥⎦ ⎥⎦
λ = iφ /  . Because [Sz ,Sx ] = iSy , ⎡⎢Sz ,[Sz ,Sx ]⎤⎥ = −(i)2 Sx , 
⎣
⎦
⎡
⎤
⎡
⎤
(λi)2 (λi)4
(λi)3 (λi)5
SxR = Sx ⎢⎢1 −
+
+ ⎥⎥ + Sy ⎢⎢λi −
+
+ ⎥⎥
2!
4!
3!
5!
⎢⎣
⎥⎦
⎢⎣
⎥⎦
⎡
⎤
⎡
⎤
(−φ)2 (−φ)4
(−φ)3 (−φ)5
= Sx ⎢⎢1 −
+
+ ⎥⎥ + Sy ⎢⎢(−φ)−
+
+ ⎥⎥
2!
4!
3!
5!
⎢⎣
⎥⎦
⎢⎣
⎥⎦
= Sx cosφ −Sy sin φ .
Approach 3: Using the property of a vector operator.
In classical physics, a vector V transform as
3
Vi → Vi R = ∑ RijVj , for i = 1, 2, 3
j =1
69
(5.99)
under a rotation R . In quantum mechanics, the correspondence principle indicates that the expectation value of V under rotation transforms as
Vi
Eq. (5.100) suggests that
R
ψ Vi ψ
R
R
ψ Vi ψ
→ Vi
3
R
= ∑ Rij Vj .
(5.100)
j =1
3
R
= ∑ Rij ψ Vj ψ , while
j =1
= ψ D †(R)Vi D(R) ψ for an arbitrary state ψ ; thus we find that
3
Vi R ≡ D †(R)Vi D(R) = ∑ RijVj .
(5.101)
j =1
⎡ cosφ −sin φ 0 ⎤
⎢
⎥
⎢
⎥
Here the rotation R = Rz (φ) = ⎢ sin φ
cosφ 0 ⎥ , and Eq. (5.101) ⇒
⎢
⎥
0
1 ⎥⎦
⎢⎣ 0
⎡ S ⎤
⎢ x ⎥
⎢
⎥
R
Sx = ⎡⎢cosφ, − sin φ, 0⎤⎥ ⎢ Sy ⎥ = Sx cosφ −Sy sin φ . Then it is straightforward to show the ex⎣
⎦⎢
⎥
⎢ S ⎥
⎢⎣ z ⎥⎦
pectation value of Sx under rotation satisfying Eq. (5.95).
5.4.3 Rotations About n̂ in Spin-Half Systems
In general, evaluating matrix elements of D(R) [Eqs (5.77) and (5.78)] for an arbitrary rotation
is hard; however, for spin half s = j = 1 / 2 , we obtain Dn̂ (ϕ) readily using properties of Pauli
matrices.

Dn̂ (ϕ) = exp (−iϕS ⋅ n̂ /  ) = exp (−i σ ⋅ n̂ϕ / 2)
⎡
⎤
⎡ (ϕ / 2) 
⎤
(ϕ / 2)2 
(ϕ / 2)4 
(ϕ / 2)3 
2
4
⎢
⎥
⎢
= ⎢1 −
(σ ⋅ n̂) +
(σ ⋅ n̂) −⎥ −i ⎢
(σ ⋅ n̂)−
(σ ⋅ n̂)3 + ⎥⎥ .
2!
4!
3!
⎢⎣
⎥⎦
⎢⎣ 1!
⎥⎦
We need to use a very important property of Pauli matrices:



(σ ⋅ a)(σ ⋅ b) = a ⋅ b + i σ ⋅(a ×b) ,
(5.102)
where a and b are two vectors in three dimensions. Proving Eq. (5.102) requires employing the

properties of σ [Eqs. (5.80)-(5.82)], which are summarized as
70
3
σi σj = δi,j + i ∑ ijk σk .
(5.103)
σi† = σi ; det(σi ) = −1; Tr(σi ) = 0.
(5.104)
k=1
Other properties include
⎡
3
a3
a1 −ia 2 ⎤⎥


⎢
σ ⋅ a is a 2×2 matrix: σ ⋅ a = ∑ σja j = ⎢
⎥ , then
⎢ a1 + ia 2
−a 3 ⎥
j =1
⎣
⎦
3
3
3
3


1
(σ ⋅ a)(σ ⋅ b) = ∑ σja j ∑ σkbk = ∑ ∑ a jbk ⋅ ({σj , σk } +[σj , σk ])
2
j =1
k=1
j =1 k=1
3
3

= ∑ ∑ a jbk ⋅ (δj,k + ijkl σl ) = a ⋅ b + iσ ⋅(a ×b) .
j =1 k=1
In our case, a = b = n̂ , and they are real vectors, then

(σ ⋅ a)2 =| a |2
⎧ 1
(for even n)

 n ⎪
⎪
2
2
⇒ (σ ⋅ n̂) =| n̂ | = 1 ⇒ (σ ⋅ n̂) = ⎨ 
⎪
σ ⋅ n̂ (for odd n)
⎪
⎪
⎩
(5.105)
Now we can evaluate Dn̂ (ϕ) :
⎡
⎤
⎡ (ϕ / 2) (ϕ / 2)3
⎤

(ϕ / 2)2 (ϕ / 2)4
Dn̂ (ϕ) = I ⎢⎢1 −
+
−⎥⎥ −i(σ ⋅ n̂) ⎢⎢
−
+ ⎥⎥
2!
4!
3!
⎢⎣
⎥⎦
⎢⎣ 1!
⎥⎦

= I cos(ϕ / 2)−i(σ ⋅ n̂)sin(ϕ / 2)
Using n̂ = (nx ,ny ,nz ) = (sin θ cosφ, sin θ sin φ, cos θ) , we write Dn̂ (ϕ) explicitly,
⎡
⎢ cos ϕ −in sin ϕ
⎢
z

2
2
Dn̂ (ϕ) = I cos(ϕ / 2)−i(σ ⋅ n̂)sin(ϕ / 2) = ⎢
⎢
⎢ −(in −n )sin ϕ
x
y
⎢
2
⎣
ϕ
2
ϕ
ϕ
cos + inz sin
2
2
−(inx + ny )sin
⎤
⎥
⎥
⎥ (5.106)
⎥
⎥
⎥
⎦
How do we know the rotation operator [Eq. (5.106)] obtained is correct? Let’s examine if the
eigenstates of S ⋅ n̂ , χn̂± , can be obtained using Eq. (5.106). Physically, χn̂± can be obtained
from the basis state ↑
(or
χz+
) by two rotations: about the y-axis by an angle θ , followed by
a rotation about the z-axis by an angle φ . Mathematically,
71
χn̂+
⎡
φ
φ
⎢
⎡ 1 ⎤ ⎢ cos 2 −i sin 2
⎥=⎢
= Dz (φ)Dy (θ) ⎢
⎢ 0 ⎥ ⎢
⎣
⎦ ⎢
0
⎢
⎣
⎡ −iφ/2
⎤
e
cos(θ / 2) ⎥
⎢
=⎢
⎥,
⎢ e iφ/2 sin(θ / 2) ⎥
⎣
⎦
⎤⎡
⎥ ⎢ cos θ
0
⎥⎢
2
⎥⎢
⎥
⎢
φ
φ
θ
cos + i sin ⎥⎥ ⎢⎢ sin
2
2 ⎦⎣
2
θ
−sin
2
θ
cos
2
⎤
⎥
⎥⎡ 1 ⎤
⎥
⎥⎢
⎥⎢ 0 ⎥
⎦
⎥⎣
⎥
⎦
which is slightly different from Eq. (5.89) obtained by diagonalization. Question: is it all right?
An arbitrary rotation can also be described by three Euler angles, α , β and γ , and then the
rotation matrices are
⎡
⎢ cos β
⎢
2
d (1/2)(β) = exp (−iβJ y /  ) = ⎢
⎢
⎢ sin β
⎢
2
⎣
⎡ −iα/2
0
D (1/2)(α, β, γ) = Dz(1/2)(α)d (1/2)(β)Dz(1/2)(γ) = ⎢⎢ e
iα/2
e
⎢⎣ 0
⎡
⎢ e −i(α+γ)/2 cos β
⎢
2
=⎢
⎢
⎢ e i(α−γ)/2 sin β
⎢
2
⎣
β
2
β
e i(α+γ)/2 cos
2
−e −i(α−γ)/2 sin
β
2
β
cos
2
−sin
⎡
β
⎢
⎤ ⎢ cos
2
⎥⎢
⎥⎢
⎥⎦ ⎢ sin β
⎢
2
⎣
⎤
⎥
⎥
⎥,
⎥
⎥
⎥
⎦
β
2
β
cos
2
−sin
⎤
⎥
⎥
⎥ .
⎥
⎥
⎥
⎦
(5.107)
⎤
⎥
⎥ ⎡ e −iγ/2
0
⎥⎢
iγ/2
⎥⎢ 0
e
⎥ ⎢⎣
⎥
⎦
⎤
⎥
⎥
⎥⎦
(5.108)
5.5 Rotation Matrices with Higher j-Values
For j = 1 (It could be spin-one or l = 1 , or spin-orbital coupled total angular momentum), following exactly the same procedure, we can construct matrix representations for angular momentum operators and rotation operators using the basis set
{ j = 1,m } with m = −1, 0,1 .
Written explicitly in column matrices, the three basis eigenkets are
⎡ 1 ⎤
⎢
⎥
1 ≡ 1,1 = ⎢⎢ 0 ⎥⎥ ,
⎢ 0 ⎥
⎣
⎦
⎡ 0 ⎤
⎢
⎥
2 ≡ 1, 0 = ⎢ 1 ⎥ ,
⎢
⎥
⎢ 0 ⎥
⎣
⎦
⎡ 0 ⎤
⎢
⎥
3 ≡ 1,−1 = ⎢ 0 ⎥ .
⎢
⎥
⎢ 1 ⎥
⎣
⎦
Using the ladder operators J ± = J x ± iJ y and Eqs. (5.69), we obtain
72
(5.109)
⎡ 0 1 0
 ⎢⎢
Jx =
⎢ 1 0 1
2⎢ 0 1 0
⎣
⎤
⎡ 0 −i 0 ⎤
⎡ 1
⎥
⎢
⎥
⎢

⎥, J =
⎢
⎥
⎢ 0
;
obviously,
J
=

i
0
−i
⎥
y
z
⎢
⎥
⎢
2⎢
⎥
⎥
⎢ 0
0
i
0
⎦
⎣
⎦
⎣
0 0 ⎤⎥
0 0 ⎥⎥ .
0 −1 ⎥⎦
(5.110)
In general, evaluating the non-trivial rotation matrix d (j )(β) is difficult; however, for
j = 1 / 2 and j = 1 , it is easy. One can verify (your homework assignment) that for j = 1 ,
d (1)(β) = exp (−iβJ y /  ) = 1 −i sin(β)(J y /  ) −(1 − cos β)(J y /  ) .
2
Plugging Eq. (5.110) into Eq. (5.111), we find that
⎡
− sin(β) / 2
⎢ (1 + cos β) / 2
⎢
(1)
d (β) = ⎢ sin(β) / 2
cos β
⎢
⎢
sin(β) / 2
⎢ (1 − cos β) / 2
⎣
⎤
(1 − cos β) / 2 ⎥
⎥
− sin(β) / 2 ⎥⎥ .
⎥
(1 + cos β) / 2 ⎥
⎦
(5.111)
(5.112)
For an arbitrary j , the Wigner’s formula for dm(j′),m (β) is derived from the Schwinger’s oscillator model using the second quantization. See textbook on page 232.
ADDITION OF ANGULAR MOMENTA
Many problems in quantum mechanics require the addition of two or more angular momentum
variables. One example is for electron states in atoms, where the spin and orbital motion are separate angular momentum components. In classical mechanics, angular momentum is a vector,
and the addition of angular momentum is just the addition of vectors. In quantum mechanics,
each angular momentum variable is quantized eigenstates (Q: what experiments demonstrate
that?); therefore, adding these angular momenta is different than the addition of vectors.
The addition of angular momenta manifests many-particle quantum systems. The total angular momentum state (except for spin-orbital coupling of a particle) is an eigenstate of a manyparticle system, which can be constructed by combining single-particle eigenstates. We want to
examine which properties of the simple systems are retained by the composite system.
5.6 Direct (Tensor, Kronecker) product of vector space
Let’s begin with the math involved. Let U and V be vector spaces with bases ui (i = 1, 2, …,
m) and v j ( j = 1, 2, …, n), respectively. Consider an mn-dimensional vector space W whose
73
(
basis is in one-to-one correspondence with the pairs ui , v j
) or
ui ;v j or simply uiv j . The
space W thus defined is called the direct product of U and V , denoted by W =U ⊗V , with
the operation involving vectors and scalars
(
)
α u1v1 + β u2v 2 ≡ α u1 + β u2 , α v1 + β v 2 ,
(5.113)
and the inner product of W is defined as
uv u ′v ′ ≡ u u ′ v v ′ .
(5.114)
If operators  and B̂ operate on spaces of U and V , respectively, then the direct-product
operator T̂ = Â ⊗ B̂ works on space W ,
(
)
T̂ uv = (Â ⊗ B̂) uv = Â u , B̂ v ,
(5.115)
and its matrix elements
Tij,kl = (A ⊗ B)ij,kl ≡ uiv j  ⊗ B̂ ukvl
(5.116)
= ui  uk v j B̂ vl = Aik B jl .
5.7 Simple examples
5.7.1 Spin-orbital combined state
A realistic description of a spin-1/2 particle must include both the space degree of freedom and
the internal degree of freedom. Thus the associated Hilbert space is spanned by the direct product
of the infinite-dimensional position space spanned by eigenkets { r } and the two-dimensional
spin space spanned by { ± } .
Explicitly, the basis of the Hilbert space is
r;± = r ⊗ ± .
(5.117)
The (total) wave function of a spin-half particle is written as
ψ±(r) = r,± ψ .
⎛ ψ (r) ⎞⎟
⎜
⎟⎟ , or
Its two components are arranged in column matrix form, ⎜⎜ +
⎜⎜ ψ (r) ⎟⎟⎟
⎝ −
⎠
for spin-up and spin-down wave functions, respectively.
74
(5.118)
⎛ 1 ⎞⎟
⎛
⎜⎜
⎟⎟ ψ(r) and ⎜⎜ 0
⎜⎜⎝ 0 ⎟⎠
⎜⎜⎝ 1
⎞⎟
⎟⎟ ψ(r)
⎟⎠
The total angular momentum J = L + S = L ⊗ I + I ⊗ S , and the rotation operator
⎛ L ⋅ n ⎞⎟
⎛ S ⋅ n ⎞⎟
D (total)(R) = D (orb)(R) ⊗ D (spin)(R) = exp ⎜⎜−i
ϕ⎟⎟ ⊗ exp ⎜⎜⎜−i
ϕ⎟⎟ .
⎜⎝
 ⎟⎠
 ⎟⎠
⎝
(5.119)
Here L and S operate in the real and inner spaces, respectively. If ψ(r) corresponds to an orbital angular momentum state, then we can use nlm to denote it, and the product space will be
nlm ⊗ ± , which are the common eigenstates of L2 , Lz , S2 and Sz . However, we can also
construct the simultaneous engenstates for J2 , Jz , L2 and S2 , which will be studied later on.
5.7.2 Combining two spin-1/2 particles
Consider a system of two spin-half particles, and S1 and S2 are their spin operators. Its Hilbert
space is spanned by four vectors,
s1m1,s 2m2 = s1m1 ⊗ s 2m2 .
(5.120)
Since si = 1 / 2 and mi = ±1 / 2 with i = 1, 2 , we can use a more compact notation to represent
the product basis: + + , + − , − + and −− , where + − =
1 1 1 1
, ; ,− , and so
2 2 2 2
forth. Obviously, for the single-particle operators
Si2 s1m1,s 2m2 =  2si (si + 1) s1m1,s 2m2 ,
(5.121)
Si,z s1m1,s 2m2 = mi s1m1,s 2m2 .
(5.122)
Now we consider the two-particle system, and we want to study (1) all possible values for the
magnitude and the z-component of the system spin, and (2) the corresponding system states that
possess these values. Again, this is an eigenvalue/eigenstate problem.
In this case, the total (in contrast to individual) angular momentum operator is
S = S1 + S2 = S1 ⊗ I + I ⊗ S2 ,
(5.123)
which is the generator of rotation for the whole system, and it obeys the basic commutators
⎡S ,S ⎤ = iε S .
ijk k
⎣⎢ i j ⎦⎥
Here the problem is to find the eigenvalues {s, m} and eigenstates sm of S 2 and Sz .
75
(5.124)
First, considering Sz = S1,z + S 2,z , we obtain
Sz s1m1,s 2m2 = (m1 + m2 ) s1m1,s 2m2 = m s1m1,s 2m2 ,
(5.125)
where m = m1 + m2 . Then consider the operator S2 ,
S2 = S12 + S22 + 2S1 ⋅ S2
(5.126)
= S12 + S22 + 2S1,zS 2,z + S1,+S 2,− + S1,−S 2,+ .
By explicit evaluation, we obtain the representation matrix for S2 is
++ +−
⎡
⎢
⎢
2
 ⎢
⎢
⎢
⎢⎣
2
0
0
0
0
1
1
0
−+
0
1
1
0
−−
0
0
0
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
++
(5.127)
+−
−+
−−
Diagonalize the matrix above, and we find the eigenvalues [ s(s + 1) 2 ] are associated with
s = 1 and s = 0 , corresponding to the eigenstates: + + ,
2
+ − + − + , −− and
2
(
)
2
+ − − − + . They are also eigenstates of Sz with eigenvalues of , 0,−, 0 , respectively.
2
(
)
Now we can change the basis from s1m1,s 2m2 to sm :
2
+− + −+ ,
2
2
1,−1 = −− , 0, 0 =
+− − −+ .
2
1,1 = + + ,
1, 0 =
(
)
(
)
(5.128)
The three spin-1 states are called triplets, and the solitary spin-0 state is called the singlet. Under
the exchange of the two particles, the triplet states are symmetric, while the singlet state is antisymmetric. The basis change can be symbolically written as
1 1
⊗ = 1⊕0,
2 2
(5.129)
which means that the direct product of two spin-half Hilbert spaces is a direct sum of a spin-1
space and spin-0 space. (Q: is this product space reducible?) Using this basis set, the matrix rep-
76
resentations for both S 2 and Sz are diagonal, and the matrix representations for Sx and Sy are
obtained by using ladder operators S± .
Adding angular momentum is essentially the change of basis, from that diagonallizes
(S ,S ,S
2
1
2
2
1,z
)
(
)
,S 2,z to one that disagonalizes S 2,Sz ,S1,z ,S 2,z , i.e., to find out the coefficients ap-
pearing on the right-hand side of Eq. (5.128), which are Clebsch-Gordan coefficients. We will
discuss them in the following formal theory of angular momenta addition.
5.8 Formal theory of angular momentum addition
Consider two angular momentum operators J1 and J2 (for two subspaces), what are the eigenvalues and eigenstates for J2 and J z , where J = J1 + J2 = J1 ⊗ I + I ⊗ J2 . The total J satisfies
the basic angular momentum commutation relations,
⎡J ,J ⎤ = iε J .
ijk k
⎣⎢ i j ⎦⎥
(5.130)
and the associated rotation operator is similar to Eq. (5.119):
⎛ J ⋅ n ⎞⎟
⎛ J ⋅ n ⎞⎟
D(R) = D1(R) ⊗ D2(R) = exp ⎜⎜⎜−i 1 ϕ⎟⎟ ⊗ exp ⎜⎜⎜−i 2 ϕ⎟⎟ .
⎟⎟⎠
⎟⎟⎠
⎜⎝
⎜⎝


(5.131)
Similar to the previous section, the basis can be j1 j 2;m1m2 = j1m1 ⊗ j 2m2 , which are
simultaneous eigenkets for J12 , J22 , J 1,z and J 2,z :
J12 j1 j 2;m1m2 = j1(j1 + 1) 2 j1 j 2;m1m2 ,
(5.132)
J22 j1 j 2;m1m2 = j 2(j 2 + 1) 2 j1 j 2;m1m2 ,
(5.133)
J 1,z j1 j 2;m1m2 = m1 j1 j 2;m1m2 ,
(5.134)
J 2,z j1 j 2;m1m2 = m2 j1 j 2;m1m2 .
(5.135)
However, we want to find the combined system’s total angular momentum, thus we need the
simultaneous eigenkets for J12 , J22 , J2 and J z , j1 j 2; jm ,
J12 j1 j 2; jm = j1(j1 + 1) 2 j1 j 2; jm ,
(5.136)
J22 j1 j 2; jm = j 2(j 2 + 1) 2 j1 j 2; jm ,
(5.137)
77
J2 j1 j 2; jm = j(j + 1) 2 j1 j 2; jm ,
(5.138)
J z j1 j 2; jm = m j1 j 2; jm .
(5.139)
Let’s first look at the commutations among these operators: (1) J12 , J22 , J 1,z and J 2,z all mutually commute; (2) J12 , J22 , J2 and J z all mutually commute; (3) Question: How about
⎡ J2,J ⎤ and ⎡ J2,J ⎤ , are they also zero because ⎡ J2,J ⎤ = 0 ?
1,z ⎦⎥
2,z ⎦⎥
z ⎦⎥
⎢⎣
⎣⎢
⎣⎢
The problem we want to solve is to construct j1 j 2; jm from the basis set j1 j 2;m1m2 :
j1 j 2; jm = ∑ ∑ j1 j 2;m1m2
m1
m2
j1 j 2;m1m2 j1 j 2; jm .
(5.140)
Here are the Clebsch-Gordan (CG) coefficients, which unfortunately depend on 6 parameters,
and often written as j1 j 2m1m2 jm , ( j1 j 2m1m2 | jm ) , or C j j (jm;m1m2 ) and C jjmm ,j m , or using
1 2
1 1
2
2
Wigner’s 3-j symbol:
j1 −j 2 +m
j1 j 2;m1m2 j1 j 2; jm = (−1)
⎛ j
j2
⎜
2j + 1 ⎜⎜ 1
⎜⎜ m m
2
⎝ 1
⎞⎟
⎟⎟ .
⎟⎟
⎟⎠
j
−m
(5.141)
One can assembly these CG coefficients into a (2j1 + 1)(2j 2 + 1)×(2j1 + 1)(2j 2 + 1) orthogonal
(i.e., real and unitary) matrix C, which converts the basis from j1 j 2;m1m2 to j1 j 2; jm . The CG
coefficients are simply the matrix elements of C.
5.8.1 Properties of CG coefficients
(1) The triangle inequality:
j1 j 2m1m2 jm ≠ 0
only if j1 − j 2 ≤ j ≤ j1 + j 2 ,
(5.142)
which means that geometrically we must be able to form a triangle with three sides of j1 , j 2 and j .
This also ensures that the dimensions of two sets of basis are the same,
N=
j1 +j 2
∑
j =|j1 −j 2|
2j + 1 = (2j1 + 1)(2j 2 + 1) ,
(5.143)
suggesting that
j1 ⊗ j 2 = (| j1 − j 2 |) ⊕(| j1 − j 2 | +1) ⊕⊕(j1 + j 2 ) ,
(j )
D 1 (R) ⊗ D
(j 2 )
(R) = D
(|j1 −j 2|)
(R) ⊕ D
78
(|j1 −j 2|+1)
(R) ⊕ ⊕ D
(5.144)
(j1 +j 2 )
(R) .
(5.145)
(2) j1 j 2m1m2 jm ≠ 0 only if m = m1 + m2 . Combining Eq. (5.142), we get −j ≤ m ≤ j .
(3) They are real by convention (they could be set to complex numbers),
j1 j 2m1m2 jm = jm j1 j 2m1m2 .
(5.146)
Therefore, the CG matrix C is orthogonal, and the orthogonality conditions are
∑∑
j1 j 2m1m2 jm
j1 j 2m1′m2′ jm =δm ,m′ δm ,m′
(5.147)
∑∑
j1 j 2m1m2 jm
j1 j 2m1m2 j ′m ′ =δj, j ′δm,m′
(5.148)
j
m1
m
m2
1
1
2
2
As a special case, we set j ′ = j and m ′ = m = m1 + m2 , then
∑∑
m1
m2
j1 j 2m1m2 jm
2
=1
(5.149)
(4) The overall sign convention:
j1 j 2; j1, j − j1 j, j > 0 .
(5.150)
Here m1 = j1 , m2 = j − j1 and m = j , specifically.
(5) The symmetry of CG coefficients:
j +j 2 −j
j1 j 2m1m2 jm = (−1) 1
j1 j 2;−m1,−m2 j,−m .
(5.151)
5.8.2 Recursion relations for the CG coefficients
In order to obtain CG coefficients, we can use the recursion relations. With j1 , j 2 and j fixed,
the CG coefficients with different m1 and m2 are related each other recursively. We start with
J ± j1 j 2; jm = (J 1,± + J 2,± )∑ ∑ j1 j 2m1m2
m1
m2
j1 j 2m1m2 j1 j 2; jm .
(5.152)
Using the ladder operators, we obtain (with substitutions of m1 → m1′ and m2 → m2′ )
(j  m)(j ± m + 1) j1 j 2; j,m ± 1 = ∑ ∑ j1 j 2; m1′m2′ j1 j 2; jm ×
(
m1′
m2′
)
(j1  m1′)(j1 ± m1′ + 1) j1 j 2; m1′ ± 1, m2′ + (j 2  m2′ )(j 2 ± m2′ + 1) j1 j 2; m1′, m2′ ± 1 .
79
(5.153)
In order to get the recursion relation below, we multiply by j1 j 2;m1m2 on the left and use
orthonormality. On the right hand side, the first term is non-zero only when m1′ ± 1 = m1 and
m2′ = m2 , and the second term is non-zero only when m1′ = m1 and m2′ ± 1 = m2 .
(j  m)(j ± m + 1) j1 j 2;m1m2 j1 j 2; j,m ± 1 =
(j1  m1 + 1)(j1 ± m1 ) j1 j 2;m1  1,m2 j1 j 2; jm
(5.154)
+ (j 2  m2 + 1)(j 2 ± m2 ) j1 j 2;m1,m2  1 j1 j 2; jm .
Here we find that the nonvanishing condition for the CG coefficients become m1 + m2 = m ± 1 .
Question: is this in contradiction to CG coefficients property (2) on last page?
Using the recursion relation [Eq. (5.154)], orthogonality condition [Eq. (5.149)] and the
overall sign convention [Eq. (5.150)],
CG coefficients can be uniquely determined. The figure here shows the J +
and J − recursion relations, on the left
and right panels, respectively.
(a) (m1,m2 ) ⇔ (m1 −1,m2 ) and (m1,m2 −1) ,
Fig. 10.1: CG coefficients related by the recursion relations.
(b) (m1,m2 ) ⇔ (m1 + 1,m2 ) and (m1,m2 + 1) .
The recursive relations are convenient for computer programming; however, using the ladder
operators is also very handy in finding CG coefficients.
5.8.3 CG coefficients for addition of two spin-half states
Let’s first look at CG coefficients for the combined system of the two spin-1/2 particles:
jm
⎡
⎤
⎢ 1,1 ⎥ ⎡
⎢
⎥ ⎢
⎢ 1, 0 ⎥ ⎢
⎢
⎥ ⎢
⎢
⎥=⎢
⎢ 1,−1 ⎥ ⎢
⎢
⎥ ⎢
⎢
⎥ ⎢
0,
0
⎢
⎥ ⎣
⎣
⎦
m1m2
⎡
0 ⎤⎥ ⎢⎢
⎥⎢
0 1/ 2
1/ 2 0 ⎥ ⎢
⎥⎢
0
0
0
1 ⎥⎢
⎥⎢
0 1 / 2 −1 / 2 0 ⎥ ⎢
⎦⎢
⎣
1
0
0
80
⎤
++ ⎥
⎥
+ − ⎥⎥
⎥
−+ ⎥
⎥
⎥
−− ⎥
⎦
Notice that in this form, the columns contain NOT the components of vectors, but the basis vectors – here one column is NOT a vector in the Hilbert space, but a set of vectors forming a complete basis in the Hilbert space. We can revert the above relation,
m1m2
⎡
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
jm
⎤
++ ⎥ ⎡
⎥ ⎢
+ − ⎥⎥ ⎢⎢
⎥=⎢
−+ ⎥ ⎢
⎥ ⎢
⎥ ⎢
−− ⎥ ⎣
⎦
⎡
⎤
⎤ ⎢ 1,1 ⎥
⎥
⎥⎢
⎥
⎥⎢
0 1/ 2 0
1 / 2 ⎥ ⎢ 1, 0 ⎥
⎥
⎥⎢
0 1 / 2 0 −1 / 2 ⎥⎥ ⎢⎢ 1,−1 ⎥⎥
⎥
⎥⎢
0
0
1
0
⎦ ⎢ 0, 0 ⎥
⎣
⎦
1
0
0
0
From the first expression, we can obtain 1, 0 = 2 ( + − + − + ) , and so on, while the second
2
expression gives + − = 2 ( 1, 0 + 0, 0 ) , and so on.
2
Let’s re-derive the CG coefficients for spin-1/2 + spin-1/2 using ladder operators (or equivalently, using the recursion relation). From Eqs. (5.149) and (5.150), we obtain + + 1,1 = 1 ,
then 1,1 = + + = 1 + + + 0 + − + 0 − + + 0 −− . We’ll omit zero CG coefficients.
S− 1,1 = (S1,− + S 2,− ) + + ⇒
1 1 1 1
(1 + 1)(1 −1 + 1) 1, 0 = ( + )( − + 1) + − + − +
2 2 2 2
(
)
⇒ 1, 0 = 2 ( + − + − + ) . Operating S− on both sides we obtain 1,−1 = −− .
2
Now we need to construct 0, 0 . Here m1 + m2 = 0 , so 0, 0 = c1 + − +c2 − + . Question: how to determine c1 and c2 ? We obtain 0, 0 =
2
+ − − − + , and thus complete the
2
(
)
calculation of all these associated CG coefficients.
5.8.4 CG coefficients for orbital-spin addition
Let’s summarize the procedures for finding CG coefficients:
1) Obtain the top CG coefficient ( j = j1 + j 2 and m = j ), which is j1 j 2; j1, j 2 jj = 1 .
2) Using the recursion relations or ladder operators to obtain j1 j 2m1m2 jm fixed j .
81
3) Decreasing j by one ( j = j1 + j 2 −1 ), obtain the top coefficients ( m = j ) using orthogonality condition [Eq. (5.149)] and the normal conditions of j + 1, j j, j = 0 .
4) Using the recursion relations or ladder operators to obtain j1 j 2m1m2 jm fixed j .
5) Repeating (3) and (4) until j =| j1 − j 2 | , the very bottom of the CG coefficients.
6) One can also use the symmetry of CG coefficients [Eq. (5.151)] to halve the workload.
We consider the problem of adding the orbital- and spin-angular momenta of a single spinhalf particle (e.g., electron) to form the total angular momentum state. We have j1 = l , with
l = 0,1, 2, and m1 = −l,−l + 1,,l ; j 2 = 1 / 2 , and m2 = ±1 / 2 . Then j = l ± 1 / 2 (l > 0),
j = 1 / 2 (l = 0), and m = −j,−j + 1,, j .
Therefore we can suppress j1 j 2 in the notation CG coefficients, which are written as
m −m2 ,m2 j,m . Using the recursion relation Eq. (5.154), we consider j = l + 1 / 2 ,
1
1
1 1
1
(l + + m + 1)(l + −m) m − , l + ,m
2
2
2 2
2
1
1
1 1
1
= (l + m + )(l −m − ) m + , l + ,m + 1 .
2
2
2 2
2
(5.155)
Then we get the recursive relation for fixed j from m to m+1:
1
1 1
1
2 m + 1 , 1 l + 1 ,m + 1 ,
m − , l + ,m =
2 2
2
3
2 2
2
l +m +
2
l +m +
(5.156)
and from m to the maximum value of m = j = l + 1 / 2 ,
1
3
l +m +
1 1
1
2
2 m + 3 , 1 l + 1 ,m + 2 = 
m − , l + ,m =
2 2
2
3
5
2 2
2
l +m +
l +m +
2
2
l +m +
=
1
2 l + 1 , 1 l + 1 ,l + 1 .
2l + 1
2 2
2
2
l +m +
82
(5.157)
1 1
2 2
1
2
Since l + , l + ,l +
1
= 1 [Eqs. (5.149) and (5.150)], then
2
1 1
1
m − , l + ,m =
2 2
2
l +m +
2l + 1
1
2 .
(5.158)
Similarly,
1 1
1
m + ,− l + ,m =
2 2
2
1 1
1
m ± , l − ,m = ±
2 2
2
l −m +
2l + 1
1
2 ,
(5.159)
1
2.
(5.160)
l ±m +
2l + 1
Using these CG coefficients, we obtain the spin-angular wave functions
Ψlj =l±1/2,m ≡ θ,φ; ± jm
1
1
l m +
1
1
⎛ 1 ⎞⎟
⎛
⎞
m−
2 Y 2 (θ,φ)⎜⎜
2 Y m+ 2 (θ,φ)⎜⎜ 0 ⎟⎟
⎟
=±
+
⎟
l
l
⎜
⎜
⎜⎝ 0 ⎟⎠
⎜⎝ 1 ⎟⎟⎠
2l + 1
2l + 1
⎛
⎞⎟
⎜⎜
1 m−12
⎟
⎜ ± l ± m + Yl (θ,φ) ⎟⎟
⎟⎟
1 ⎜⎜
2
⎟⎟ .
=
⎜⎜
1
⎟⎟
2l + 1 ⎜⎜
1 m+ 2
⎜⎜
l  m + Yl (θ,φ) ⎟⎟⎟
⎟⎠
⎜⎝
2
l ±m +
(5.161)
They are simultaneous eigenfunctions of L2, S2, J2 and J z . They are also eigenfunctions of spin-
(
)
orbital coupling operator S ⋅ L = J2 − L2 − S2 / 2 , with eigenvalues
⎡
⎤ 2
⎢ j(j + 1)−l(l + 1)− 3 ⎥  .
⎢
4 ⎥⎦ 2
⎣
83
(5.162)