Quantum Mechanics II angular Momentum II : Addition of Angular Momentum Clebsch-Gordan Coefficients

2

m

m

m
Calculation of Clebsch-Gordan Coefficients

In the two limiting cases, i.e., for m₁ = j₁, m₂ = j₂; m = j₁ + j₂ and m₁ = −j₁, m₂ =
−j₂; m = −j₁ − j₂, the CG coefficients are 1. For the other cases we need to obtain

recursion relations. Note that



Σ
J_± | j₁, j₂; j, m⟩ = (J_1,± + J_2,±) | j₁, j₂; j, m⟩
= ⟨j₁, j₂; m₁^j , m^j₂ | j₁, j₂; j, m⟩(J_1,± + J_2,±) | j₁, j₂; m^j₁, m^j₂⟩
m^′₁,m^′₂

√ ^Σ
Applying the ladder operator on both sides


(j ∓ m)(j ± m + 1) | j₁, j₂; j, m ± 1⟩ = ⟨j₁, j₂; m₁^j , m₂^j
m^′₁,m^′₂

| j₁, j₂; j, m⟩

^√(j₁ ∓ m^j₁)(j₁ ± m^j₁ + 1) | j₁, j₂; m^j₁ ± 1, m^j₂⟩


^√(j₂ ∓ m^j₂)(j₂ ± m^j₂ + 1) | j₁, j₂; m^j₁, m^j₂ ± 1⟩ (16)

√√
Multiply both sides of the above equation by ⟨j₁, j₂; m₁, m₂ | and use orthogonality con- dition. The term on the left gives a CG coefficient. The first term on the right is non-zero only if m₁ = m^j₁ ± 1 and m₂ = m^j₂ while the second term is non-zero if m₁ = m^j₁ and m₂ = m₂^j ±1. Using these on the right, we get a relationship between three CG coefficients.

√
(j ∓ m)(j ± m₁ + 1)⟨j₁, j₂; m₁, m₂ | j₁, j₂; j, m ± 1⟩ = (j₁ ∓ m₁ + 1)(j₁ ± m₁)⟨j₁, j₂; m₁ ∓ 1, m₂ | j₁, j₂; j, m⟩
+ (j₂ ∓ m₂ + 1)(j₂ ± m₂)⟨j₁, j₂; m₁, m₂ ∓ 1 | j₁, j₂; j, m⟩ (17) Note that the left there are CG coefficients which require m₁ + m₂ = m ± 1.
3.1 Examples of CG Coefficient Calculation

Example 1 - Adding two spin 1/2 angular momenta :

We will denote the spins by
sˆ₁ and
sˆ₂ and total spin by
S^ˆ. This could, for instance,

represent the case of the spin states of hydrogen atom wherein the electron and the proton
each have spin 1/2. Clearly, the spin angular moments of two different particles commute. The total spin S can take values from s₁ − s₂ to s₁ + s₂, i.e. 0 and 1. Corresponding to s = 0, there is only one state with m = 0 while for s = 1, there are three states with m = −1, 0 and +1. The former is called a singlet while the latter trio are known as triplets. (s₁ = 1/2 has two states and s₂ = 1/2 also has two states, so that there are

2 × 2 = 4 states). We note that the state |
1 1
₂ , ₂ ; 1, 1⟩ in the representation | s₁, s₂, s, m⟩

is uniquely obtained from the state |
1 1 1
, ; ,
2 2 2
1
₂ ⟩ of | s₁, s₂, m₁, m₂⟩ representation.

1
| ₂ ,
1
₂ ; 1, 1⟩ =|
1 1 1 1

⟩
, ; ,
2 2 2 2

Apply J₋ on the left and (J₁₋ + J₂₋) on the right and use the formula J_|j, m⟩ = (j + m)(j − m + 1)

J₋ |
^√2 |

1 1 1 1
⟩

1 1		1 1 1 1
, 2 2	; 1, 1⟩ = (J1− + J2−) |	, ; , 2 2 2 2
1 1 ,
2 2	; 1, 0⟩ =| 2 , 2 ; −2 , 2 ⟩+ | 2 , 2 ; 2

1 1 1
,


1
−₂ ⟩

Thus the Clebsch-Gordon coefficients are

⟨1/2, 1/2; 1/2, 1/2 | 1/2, 1/2, 1, 1⟩ = 1

1
⟨1/2, 1/2; 1/2, −1/2 | 1/2, 1/2, 1, 0⟩ = ^√₂
1
⟨1/2, 1/2; −1/2, 1/2 | 1/2, 1/2, 1, 0⟩ = ^√₂
The coefficient for s = 1.m = −1 need not be calculated as it is obtained by symmetry to be also equal to 1. For s = 0, there is only state with m = 0. This state is orthogonal to the state | 1/2, 1/2, 1, 0⟩ and is given by
1
| 1/2, 1/2, 0, 0⟩ = ^√₂ (| 1/2, 1/2, 1/2, −1/2⟩− | 1/2, 1/2, −1/2, 1/2⟩)

Thus

1
⟨1/2, 1/2; 1/2, −1/2 | 1/2, 1/2, 1, 0⟩ = ^√₂

1
⟨1/2, 1/2; −1/2, 1/2 | 1/2, 1/2, 1, 0⟩ = −^√₂

It is conventional to indicate the state with m = 1/2 as | ↑⟩ or as α. The state with
m = −1/2 is then indicated as | ↓⟩ or as β. The triplet states are then s = 1 :
m = +1 :| ↑↑⟩
1
m = 0 : ^√₂ [| ↑↓⟩+ | ↓↑⟩)
m = −1 :| ↓↓⟩
It may be noted that the triplets are symmetric with respect to the two spins. The singlet state s = 0 is given by
1
^√₂ [| ↑↓⟩− | ↓↑⟩)
which is antisymmetric under exchange of spins.
When the magnitudes of the two angular momenta are different, it is conventional to choose the one with larger magnitude as the first spin and the other one as the second. This convention has some repercussion in fixing the signs of Clebsch-Gordan coefficients, which we will talk about in the next example.
Example 2: A Hydrogen atom is known to be in the state n = 3, l = 2. Find all the
values of the total angular momentum j and for each j, the eigenvalue of J² operator.
1 1 5 3

We have l = 2, s = 1/2. Thus allowed values of j are from 2 − ₂ to 2 + ₂ , i.e.
, . The
2 2

eigenvalue of J² being j(j + 1)k², we have, for j = 5/2 the eigenvalue is (35/4)k² and
(15/4)k² respectively.
Example 3: For two spins s = 1/2 each, show that ³ I + ¹ S^ˆ · S^ˆ is a projection operator
_{4 k2} 1 2

for S = 1 and ¹ I − ¹ S^ˆ
· S^ˆ
is a projection operator for S = 0

4

Solution:

_k2 1 2

Note ₃
I +
S^ˆ₁ · S^ˆ₂ = I +
1
[S² − S² − S²]

3

1
4 k² 4 2k^{2 1 2}
For S = 1, S² has eigenvalue 1(1 + 1)k² = 2k² and S² and S² have eigenvalue ³ k² each.
1 2 ₄
Thus we get for S = 1 this expression to have value 1 and for S = 0 it has value 0. Thus the operator projects S = 1 state.
Example 4: Adding j₁ = 1 with j₂ = 1/2
In this case, in the j₁, j₂; m₁, m₂ representation there are 3 × 2 = 6 states. The resultant

angular momentum which satisfies | j₁ − j₂ |≤ j ≤ j₁ + j₂ can take values
3
and
2
1
, the
2

former has 4 states and the latter 2, making a total of 6 states as before. We proceed as
follows. Start with the state having maximum j value and maximum possible m value

corresponding to this j. In this case it is j =
3
and m =
2
3
. This state is obtained
2

uniquely from j₁ = 1, m₁ = 1 and j₂ = 1/2, m₂ = 1/2. Thus
| 1, 1/2; 3/2, 3/2⟩ =| 1, 1/2; 1, 1/2⟩

√
We apply J₋ on the ket on the left and J₁₋ +J₂₋ to the ket on the right, using the formula
J₋ | j, m⟩ = (j + m)(j − m + 1) | j, m − 1⟩. We then get
^√3 | 1, 1/2; 3/2, 1/2⟩ = ^√2 | 1, 1/2; 0, 1/2⟩+ | 1, 1/2; 1, −1/2⟩ (18)
From this it follows that

3
⟨1, 1/2; 3/2, 1/2 | 1, 1/2; 0, 1/2⟩ = r²

3
⟨1, 1/2; 3/2, 1/2 | 1, 1/2; 1, −1/2⟩ = r¹
The remaining coefficients belonging to the negative values of m are obtained by symme- try. We now consider the j = 1/2 state with = ±1/2. Once again, we will only compute m = 1/2 term. Note that the state | 1, 1/2; 1/2, 1/2⟩ is orthogonal to the state with the same m but j = 3/2. Thus this state is orthogonal to

3

3
| 1, 1/2; 3/2, 1/2⟩ = r² | 1, 1/2; 0, 1/2⟩ + r¹ | 1, 1/2; 1, −1/2⟩

The state which is orthogonal to the state on the right is obviously

3

3
| 1, 1/2; 1/2, 1/2⟩ = ± r¹ | 1, 1/2; 0, 1/2⟩ − r² | 1, 1/2; 1, −1/2⟩!
Thus, there is an ambiguity on the overall phase. In order to fix the sign, we use what is known as Condon-Shortley sign convention according which the state with the highest m of the larger component of the angular momentum that is being added is assigned a positive sign. In case where j₁ = j₂, as was the case in the previous example,

the first mentioned one, i.e. j₁ will be taken to be the larger component for application of this convention. Thus in this case the state with j₁ = 1, m₁ = 1 will be taken to be positive. Thus we have
| 1, 1/2; 1/2, 1/2⟩ = r² | 1, 1/2; 1, −1/2⟩ − r¹ | 1, 1/2; 0, 1/2⟩
3 3
This gives two Clebsch-Gordan coefficients as follows

3
⟨1, 1/2; 1/2, 1/2 | 1, 1/2; 1, −1/2⟩ = r²

3
⟨1, 1/2; 1/2, 1/2 | 1, 1/2; 0, 1/2⟩ = −r¹