S.K. Saikin
October 1, 2009
Lectures 7-8
7 Angular Momentum I
Content:
• Definition of Orbital Angular Momentum.
• Commutation Rules.
• Matrix Representation of Angular Momentum.
7.1
Definition
Let us use the same definition as in classical mechanics
L̄ = r̄ × p̄,
(1)
where r̄ and p̄ are the position and the momentum operators respectively. The vector product (1)
can be written as
¯
¯
¯ êx êy êz ¯
¯
¯
L̄ = ¯¯ x y z ¯¯ .
(2)
¯ px py pz ¯
¯ The Cartesian
In the spatial basis we need to replace the momentum operator as p̄ = −i~∇.
components of the orbital angular momentum are
µ
¶
∂
∂
Lx = −i~ y ·
−z·
,
∂z
∂y
µ
¶
∂
∂
Ly = −i~ z ·
−x·
,
(3)
∂x
∂z
µ
¶
∂
∂
Lz = −i~ x ·
−y·
.
∂y
∂x
Exercise: How it can be written in the momentum basis?
We should introduce one more operator
L2 = L2x + L2y + L2z .
(4)
The angular momentum defined by Eq. (1) has units of ~. Sometimes it is easier to operate with
a unitless angular momentum assuming ~ = 1. This is only a mathematical trick. A physical
quantity should have units.
1
7.2 Commutation Rules
Commutators of an orbital angular momentum operator L̄ with a position operator r̄, a momentum
operator p̄ and itself may be written in a compact form
[Lj , Ak ] = i~²jkl Al ,
(5)
where j, k, l are for the Cartesian components of the vectors, and the antisymmetric tensor ²jkl is
defined as

j, k, l
cyclic
 1
−1
j, k, l anticyclic
²jkl =

0
otherwise
Example 1:
[Lz , x] = [xpy − ypx , x] = xpy x − ypx x − xxpy + xypx = x[py , x] + y[x, px ] = i~y.
(6)
Example 2:
[Lz , py ] = [xpy − ypx , py ] =
(7)
xpy py − ypx py − py xpy + py ypx = [py , y]px + [x, py ]py = −i~px .
One more useful commutator:
[L2 , Li ] = 0.
(8)
The vector of the angular momentum operator doesn’t commute with itself, but it commutes with
its absolute value. We can define a basis |ψi, where two of the operators L2 and Li , i = x, y, z are
simultaneously diagonal. Usually, the basis states are chosen to diagonalize Lz . In this case Lx
and Ly are non-diagonal.
7.3
Matrix representation of the angular momentum operator
¯ It satisfies all the
1. Let us introduce operator a mode general operator of angular momentum J.
commutation rules given above, but not Eq. 1. For the sake of simplicity we assume that ~ = 1.
2. We also define the rising J+ and the lowering J− operators as
J± = Jx ± Jy .
(9)
J−† = J+ .
(10)
These operators are non-Hermitian
2
Components of the angular momentum operator in terms of J± are
1
(J+ + J− ),
2
−i
=
(J+ − J− ).
2
Jx =
Jy
(11)
And the square of angular momentum is
1
J 2 = Jz2 + (J+ J− + J− J+ ).
2
(12)
3. Commutation rules and useful relations:
[J 2 , J± ] = 0,
(13)
[Jz , J± ] = ±J± ,
(14)
[J+ , J− ] = 2Jz .
(15)
Combining Eq. (14) with Eq. (12) we get
J+ J− = J 2 − Jz2 + Jz ,
J− J+ = J 2 − Jz2 − Jz .
(16)
(17)
Operators J± are non-diagonal in the chosen basis. However, J± J∓ are diagonal.
4. Let us define the basis states |µ, νi that satisfy to two eigenvalue problems
J 2 |µ, νi = ν|µ, νi,
Jz |µ, νi = µ|µ, νi.
(18)
(19)
We will construct the matrices of the operators J 2 , Jz , and J± in this basis. Because J± commutes
with J 2 we get
J 2 J± |µ, νi = J± J 2 |µ, νi = νJ± |µ, νi
(20)
The state J± |µ, νi is an eigenstate of the operator J 2 with the same eigenvalue. Similarly, using the
commutation relations for Jz and J± we get that the state J± |µ, νi is an eigenstate of the operator
Jz with the eigenvalue µ ± 1 respectively. Thus, we can write that
J± |µ, νi = γµ,ν |µ ± 1, νi,
3
(21)
where γµ,ν is a complex coefficient, which may depend on the eigenvalues of J 2 and Jz .
5. Let us impose the condition that the norm of the eigenvector is not negative
hµ, ν|µ, νi ≥ 0.
(22)
It has been shown above that if we apply the operator J+ to the state |µ, νi the resulting state should
be proportional to |µ + 1, νi. Its norm also should be not negative. This condition can be written
as
hµ, ν|J− J+ |µ, νi ≥ 0.
(23)
We can substitute the J− J+ operator with Eq. 17 and obtain the relation for eigenvalues of J 2 and
Jz
ν − µ(µ − 1) ≥ 0,
which set up the upper boundary on the value of µ
r
1
1
µmax ≤ − +
+ ν.
2
4
(24)
(25)
Let assume that µmax = j, then ν = j(j + 1). Actually, from the the norm inequality we can
get only a weaker condition ν ≥ j(j + 1). To get the equality we need to use the definition of J 2 ,
¯ Another way to get ν is using the relation
see Eq. (4) and matrices of all the components of J.
hµmax , ν|J− J+ |µmax , νi = 0.
(26)
This is true because J+ |µmax , νi = 0. Then, we can substitute J− J+ with Eq. (17) and get ν =
j(j + 1). Applying J+ to the state with a maximal value of µ we should get 0. But we can apply
J− . If we apply it k-times the final state will be proportional to |j − k, νi. This state also should
satisfy the condition of Eq. 22. In the way similar to the discussed above we can get that
kmax = 2j,
(27)
where kmax is a maximal number of J− operators that we can apply to the state |j − k, νi.
Excercise: Derive the lower boundary for µ.
In Eq. (27) km ax is an integer number. Therefore, j can be
1
3
j = 0, , 1, , ...
2
2
(28)
m = −j, −j + 1, ..., j − 1, j
(29)
and for each j the value of m can be
4
Now, for each j we can construct matrices of J 2 and Jz .
hm0 , j 0 |Jz |m, ji = mδjj 0 δmm0
(30)
hm0 , j 0 |J 2 |m, ji = j(j + 1)δjj 0 δmm0
(31)
and
For example for j = 3/2


3/2 0
0
0
 0 1/2
0
0 

Jz = 
 0
0 −1/2
0 
0
0
0
−3/2
(32)
and

1

3
0
J2 = 

0
4
0
0
1
0
0
0
0
1
0

0
0 

0 
1
(33)
For the orbital angular momentum operator the the values of m can be integer only. This can be
shown using a spatial representation of L̄. Half-integer values may appear when we study spin.
For the operators J± we know that
J+ |m, ji = λm,j |m + 1, ji
(34)
hm + 1, j|J+ |m, ji = λm,j
(35)
or
In a most general case the coefficient λm,j can be a complex number. If we take a Hermitian
conjugate of the last equation we get
J− |m + 1, ji = λ∗m,j |m, ji
(36)
Now, we can calculate matrix elements in the following way
J− J+ |m, ji = |λm,j |2 |m, ji = (J 2 − Jz2 − Jz )|m, ji,
(37)
or
λm,j =
p
j(j + 1) − m(m + 1) =
5
p
(j − m)(j + m + 1)
(38)
The matrix elements of the operators J± are defined by the following relations
p
hm0 , j 0 |J+ |m, ji = (j − m)(j + m + 1)δj 0 j δm0 m+1
(39)
and
hm0 , j 0 |J− |m, ji =
p
(j − m + 1)(j + m)δj 0 j δm0 m−1 .
(40)
For example, for j = 3/2

0

0
J+ = J−† = 
 0
0
√
3
0
0
0

0 0
2 √0 
.
0
3 
0 0
(41)
At this point we can come back and prove that the eigenvalue of the operator J 2 is j(j + 1). We
just need to write it in terms of the components of the angular momentum.
Exercise 1: Show that ν = j(j + 1) (j = 3/2) using the matrix form for J+ , J− , Jz , and J 2 .
Exercise 2: Write matrices Jx,y,z ,, J± , J 2 for j = 2.
7.4 Spin
Now, it is easy to introduce a spin S̄ as an intrinsic angular momentum of a particle. It possesses
all the properties of J. Its basis states may be written as |ms , si. Because the spin of a particle does
not depend on its orbital motion for a given type of particles s = const. For electrons, protons and
neutrons s = 1/2.
References
[1] W. H. Louisell, Quantum statistical properties of radiation. (Wiley, New York, 1990), Part
II, Chapters 2.6.
[2] J. J. Sakurai, Modern Quantum Mechanics (Addison-Wesley, New York, 1994), Chapters
3.5-3.6.
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