Lecture 2: Angular momentum and rotation
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Angular momentum of a composite system
Let J1 and J2 be two angular momentum operators. One might imagine them to be:
• the orbital angular momentum and spin of a particle;
• the angular momentum of two spin-less particles;
• the spin of two particles in bound state.
and a-priori we do not know the relative orientation of the two vectors. And indeed, this orientation can only within a
small range of quantized values.
Classical angular-momentum addition works by simply adding components. In quantum mechanics, we cannot know
the three components simultaneously, but we can know the magnitude-squared, J 2 (the Casimir operator) and one component; Jz being the conventional choice.
The total angular momentum operator and z-component have the eigenvalues:
J12 | j1 , m1 i =
j1 ( j1 + 1)~ | j1 , m1 i
| j2 , m2 i =
j2 ( j2 + 1)~ | j2 , m2 i
J22
(1)
J1z | j1 , m1 i
= m1 ~ | j 1 , m1 i
J2z | j2 , m2 i =
m2 ~ | j 2 , m2 i
though let’s drop the ~ from now on. Consider combined operators:
J2
= (J1x + J2x )2 + (J1y + J2y )2 + (J1z + J2z )2
Jz
=
acting on ψ( j1 , j2 , m1 , m2 )
J1z + J2z
= | j1 , m1 i | j2 , m2 i
ψ is an eigenfunction of Jz with eigenvalue m = m1 + m2 but it is not generally an eigenfunction of J 2 . However, linear
combinations of ψ can produce eigenfunctions Ψ( j, m) of J 2 with eigenvalues j( j + 1), with | j1 − j2 | < j < | j1 + j2 |.
Ψ( j, m) =
j1
X
j2
X
C( j, j1 , j2 , m1 , m2 ) ψ( j1 , j2 , m1 , m2 )
(2)
m1 =− j1 m2 =− j2
The coefficients C( j, j1 , j2 , m1 , m2 ) are the Clebsch-Gordan1 coefficients. They give the probability (=C2 ) that we measure, from a combined system of | j1 , m1 i | j2 , m2 i, a combined total angular momentum of j( j + 1).
1 Clebsch
and Gordan were two 19th century mathematicians who identified these coefficients in the development of Lie algebra.
1
Let’s derive these coefficients by first noting that the angular momentum operators obey commutation relation:
h
i
Ji , J j = i jk i Jk
where i jk
h
i
J 2 , Ji
=
+1 (xyz) −1 (zyx)
=
0 (i = x, y, z)
And define raising / lowering operators
(3)
0 (otherwise)
(4)
J± = J x ± iJy
(5)
[Jz , J± ] = ±J±
(6)
which, from eq. 3, satisfies
Using this commutation relation:
Jz J− | j, mi =
(J− Jz − J− ) | j, mi
=
J− (Jz − 1) | j, mi
=
(m − 1) J− | j, mi
Similarly m + 1 is the eigenvalue of Jz applied to the “raised” state: (J+ | j, mi)
As the eigenvalue of the raised state drops out so simply, we know there can only be a constant term in front of the
“raised” eigenstate. We write,
J± | j, mi = C± ( j, m) | j, m ± 1i
(7)
with the boundary condition that C± is required to be 0 for J− | j, −j i and J+ | j, +j i.
To derive these constants we note that J+ is complex-conjugate of J−
h j, m | J− | j, m + 1i
= C− ( j, m + 1)h j, m | j, mi = C− ( j, m + 1)
=
h j, m | J+∗ | j, m + 1i
(8)
and from bra-ket algebra (noting J+∗ ≡ J+† ) reverse eq. 8
h j, m + 1 | J+ | j, mi∗
=
C+∗ ( j, m)h j, m + 1 | j, m + 1i∗ = C+∗ ( j, m)
so the lowering coefficient of the m + 1 state is the same as the raising coefficient of the m state:
C− ( j, m + 1) = C+∗ ( j, m) = C
(9)
C is real. Thus applying both operators successively:
J− J+ | j, mi = C2 | j, mi
(10)
where the double operator can be broken down to:
J− J+
=
J x2 + Jy2 + i(J x Jy − Jy J x )
=
J x2 + Jy2 + Jz2 − Jz2 + i[J x , Jy ]
=
J 2 − Jz2 + Jz
=
J 2 − Jz (Jz − 1)
(11)
from where we can easily identify the eigenvalues.
C+ ( j, m) =
C∗+ ( j, m)
C2
=
C− ( j, m)
=
= C− ( j, m + 1) =
2
j( j + 1) − m(m − 1)
p
j( j + 1) − m(m − 1)
p
j( j + 1) − m(m + 1)
(12)
An example application and calculation
Consider two particles | j1 , m1 i and | j2 , m2 i forming a combined state | j, mi where j1 = 1 and j2 = 1/2. Evidently, j can
be 1/2 or 3/2. Two key points:
• Coefficients should normalise to 1
• When raising (lowering) a state to an m higher (lower) than J (−J), state = 0
3/2 states:
Evidently the max. (3/2) and min. (-3/2) m states are:
3 3
, i =
2 2
3 3
| ,− i =
2 2
1 1
, i
2 2
1 1
| 1, −1i | , − i
2 2
|
| 1, 1i |
Using definition:
C± ( j, m) =
1 1
J− | , i =
2 2
J− | 1, 1i =
J− | 1, 0i =
J− | 1, −1i =
j( j + 1) − m(m ± 1)
1 1
1 | ,− i
2 2
√
2 | 1, 0i
√
2 | 1, −1i
p
0
Now operating on the combined state
3 3
, i =
2 2
√ 3 1
3| , i =
2 2
3 1
| , i =
2 2
1 1
, i
2 2
√
1 1
1 1
2 | 1, 0i | , i+ | 1, 1i | , − i
2 2
2 2
r
r
1 1
1 −1
2
1
| 1, 0i | , i +
| 1, 1i | ,
i
3
2 2
3
2 2
J− |
J− | 1, 1i |
In analogous way:
3 1
| ,− i =
2 2
r
2
1 1
| 1, 0i | , − i +
3
2 2
r
1
1 1
| 1, −1i | , i
3
2 2
1/2 states:
|
1 1
, i =
2 2
a | 1, 1i |
1 1
1 1
, − i + b | 1, 0i | , i
2 2
2 2
with a2 + b2 = 1. Now apply J+
J+ |
1 1
, i =
2 2
=
a+
√
2b = 0 →
a=
q
2
3
and b = −
q
0
a | 1, 1i |
1
3
3
1 1
1 1
, i + b | 1, 1i | , i
2 2
2 2
Our Clebsch-Gordan coefficients for j1 = 1 and j2 =
1⊗
1
2
m1
+1
+1
0
0
−1
−1
And for a j1 =
1
2
1
2
⊗
m1
+ 12
+ 21
− 12
− 12
3
2
3
2
j
m
m2
+ 12
− 12
+ 12
− 12
+ 12
− 12
j
m
m2
+ 12
− 12
+ 12
− 12
1
3
2
1
2
1
2
1
2
√
√1/3
2/3
√
√2/3
− 1/3
1
2
1
2
− 12
3
2
− 12
√
√1/3
− 2/3
√
√2/3
1/3
3
2
− 32
1
1
2
and j2 =
1
2
combination
1
1
1
0
0
0
1
−1
1
√
√1/2
1/2
√
√1/2
− 1/2
1
In the exam, Clebsch-Gordan coefficient tables are provided, though you should still be able to derive them if asked.
They are also recording in most of the recommended texts and in the PDG:
http://pdg.lbl.gov >
4
“Reviews, Tables, Plots” >
“Mathematical Tools”
Rotations
Consider a small rotation , about the y-axis: ~x 0 = Ry ()~x
 
 x0
y0
 
z0
=

 cos 

− sin 1
 
sin   x
  
 y
cos z
And its inverse: ~x = R−1
x0
y ()~
 
 x
 
y
z
=

cos 

sin 1
 
− sin   x0
  
 y0
cos z0
Impose invariance:
ψ 0 (~x 0 ) = ψ(~x) = ψ(R−1
x 0)
y ()~
4
Without losing generality take a specific point ~x 0 = ~a and find a relation between ψ 0 (~a) and ψ(~a):
ψ 0 (~a) =
ψ(R−1
a)
y ()~
=
ψ(a x cos − az sin , ay , az cos + a x sin )
=
ψ(a x − az , ay , az + a x )
∂ψ(~a)
∂ψ(~a)
ψ(~a) + a x
− az
+ ...
∂z
∂x
"
!#
∂
∂
ψ(~a) 1 + a x − az
∂z
∂x
i
1 − Jy ψ(~a)
~ }
| {z
≈
≈
≈
as → 0
Taylor expansion
"
#
∂
∂
as : Jy = −i~ z − x
∂x
∂z
Ur ()
Conservation ↔ Invariance
Now consider the time variation of Ur :
"
#
"
#
d
d
d
dUr
hφ(t) | Ur | ϕ(t)i =
hφ(t) | Ur | ϕ(t)i + hφ(t) |
| ϕ(t)i + hφ(t) | Ur
| ϕ(t)i
dt
dt
dt
∂t
dUr
i
= hφ(t) |
| ϕ(t)i + hφ(t) | Ur H − HUr | ϕ(t)i
dt
~
Hence, Ur is invariant if Jy (the tricky part of Ur ) is independent of time and commutes with the Hamiltonian.
i.e. the eigenvalues of Ur () are constant. Angular momentum is conserved due to the requirement that the
wavefunction be invariant under rotation.
Rotation is a Lie group and so any rotation can be expressed as the successive application of the infinitesimal rotation:
Ur (β)
i~
i β n
lim 1 −
Jy
n→∞
~n
i
= e− ~ βJy
=
(13)
or for some, general 3D rotation: Ur (~θ) = e− ~ θ· J .
ˆ
In the language of operators, the angular momentum operator Jy is said to be the generator of rotations about the y
axis.
Euler angles
Generic rotation can be parameterised by three Euler angles defining three successive rotations. These could be:
1. rotation by angle γ about the z-axis.
2. rotation by angle β about the intermediate y-axis.
3. rotation by angle α about the final z0-axis.
This is inconvenient as these definitions use two changes bases. However, and importantly, a general 3D rotation can be
achieved using just the original basis (“intrinsic rotations”):
1. rotation by angle α about the original z-axis.
2. rotation by angle β about the original y-axis.
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3. rotation by angle γ about the original z-axis.
Thus, a general rotation of a wavefunction can be represented by D-matrices:
j
(αβγ)
Dm0,m
= h j, m0 | e−i/~αJz e−i/~βJy e−i/~γJz | j, mi
j
(β)
= e−i/~(αm+γm) dm0,m
(14)
where the change of the z-projection (m → m0 ) only occurs during the rotation around ŷ. The two rotations around the
original ẑ leave the quantum numbers unchanged.
j
Rotation matrix: dm0,m
(β)
Although the y-projection is unchanged, the z-direction has changed so the quantum number m is not the same — it is
now projected onto a new z 0 -axis. A state | j, mi transforms under a rotation β about the y-axis into a linear combination
on the (2 j + 1) | j, m 0 i states.
X
j
e−(i/~)βJy | j, mi =
dm0m
(β) | j, m 0 i
(15)
m0
Which we apply to the final state h j, m0 | to examine the rotation from m to m 0 projection.
dmj 0 m (β)
=
h j, m0 | e−(i/~)βJy | j, mi
(16)
So, lets calculate the matrix the specific case of j = 1. There are 3 projection (−1, 0, 1) so there are 3 × 3 elements of dmj 0 m
in this case. I calculate a couple here, the rest can be done as an exercise.
1. From inspecting the expansion of e−iβJy (with ~ = 1):
e−iβJy = 1 − iβJy −
1 2 2
i
1
β Jy + β3 Jy3 + β4 Jy4 + ...
2!
3!
4!
2. Look seperately for solutions of Jy2n+1 | j, mi and Jy2n | j, mi
p
3. Recall the raising/lowering operators J± = J x ±iJy and the Clebsch-Gordon coefficients C±jm = j( j + 1) − m(m ± 1)
J+ − J−
=
Jy
=
so, Jy | 1, 1i =
=
J x + iJy − J x + iJy
i
− [J+ − J− ]
2
i
√
i h 11 − C+· 0 − C−11 | 1, 0i with C−11 = 2
2
i
√ | 1, 0i
2
6
4. Operate again with Jy
Jy2 | 1, 1i =
=
=
i
√ Jy | 1, 0i
2
1
√ (C+10 | 1, 1i − C−10 | 1, −1i)
2 2
√
1
(| 1, 1i − | 1, −1i)
as C+10 = C−10 = 2
2
5. And again:
Jy3 | 1, 1i =
=
=
=
=
1
Jy (| 1, 1i− | 1, −1i)
2
1 −i
−i
( [J+ − J− ] | 1, 1i − [J+ − J− ] | 1, −1i)
2 2
2
√
i √
( 2 | 1, 0i + 2 | 1, 0i)
4
i
√ | 1, 0i
2
Jy | 1, 1i
6. Note the cyclical pattern and conclude:
Jy2n+1 | 1, 1i =
Jy2n | 1, 1i =
i
√ | 1, 0i
2
1
(| 1, 1i − | 1, −1i)
2
h1, 1 | Jy2n+1 | 1, 1i =
0
| 1, 1i =
0
h1, −1 |
h1, 0 |
7. Which then leads to, for each specific h j, m0 | state:
Jy2n+1
Jy2n+1
| 1, 1i =
h1, 1 | Jy2n | 1, 1i
h1, −1 | Jy2n | 1, 1i
=
8. With these relations the
2
1
2
= − 21
h1, 0 | Jy2n | 1, 1i =
dmj 0 m
√i
0
coefficients are calculated as:
1
d11
= h1, 1 | e−iβJy | 1, 1i =
=
=
=
1
d−11
= h1, −1 | e−iβJy | 1, 1i =
=
1
1
i
3
(−iβJ
h1, 1 | 1 + (−iβJ
(−iβJy )2 + (−iβJy )4 + ... ... | 1, 1i
y) +
y) +
2!
3!
4!
1 β2
i β4
1−
+
− ... ...
2! 2
4! 2
1
1
1
(1 + (1 − β2 + β4 − ...))
2
2!
4!
1
(1 + cos β)
2
1
1
1
(1 − (1 − β2 + β4 − ...))
2
2!
4!
1
(1 − cos β)
2
7
h1, −1 || 1, 1i = 0 of course
Example: e−e+ → µ−µ+
The incoming left-handed electron annihilates with the right-handed positron. In the electromagnetic process, a γ is
exchanged with the outgoing µ-pair. Final state particles must have opposite helicity in conserving j ≡ 1.
A11 = A(e− L e+R → µ− L µ+R ) ∝
A1−1 = A(e− L e+R → µ−R µ+ L ) ∝
dσ
d cos θ
1
(1 + cos θ)
2
1
= (1 − cos θ)
2
1
d1,1
=
1
d1,−1
=
|A11 |2 + |A1−1 |2
∝
1 + cos2 θ
The two amplitudes should be of equal intensity due to parity conservation with an electromagnetic interaction. See the
following reproduction of e+ e− → µ+µ− data from the TASSO experiment at DESY (c. 1980).
BTW: the asymmetry is evidence for the off-shell influence of the parity-violating Z0 interfering with the γ exchange.
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