Chapter 4

Momento Angular

Contenido: Definicion de momento angular (MA). Deduccion de las matricesde Pauli. Acople de dos MA. Coeficientes de Clebsch-Gordan. Ejemplo deacople: spin-spin, spin-orbital. Sımbolos: 3j de Wigner. Acople de tres MA.Sımbolo de Wigner 6j. Acople de cuatro MA. Sımbolo de Wigner 9j. Ejemlode acople de cuatro momentos angulares: acoples jj y LS de dos partıculas.Cambio de acoples. Ejemplo de cambio de acople en la funcion de onda de dospartıculas. Diagramatica. Ejercicios.

Descripcion: Los nucleones (protones y neutrones) tienen momento angularde espın intrınseco y de momento angular orbital, de modo que se hace nece-sario manejar el algebra del acople de momentos angulares. Los coeficientesde Clebsh-Gordan y su forma mas simetrica, los sımbolos 3j estan relacionadoscon el acople de dos momentos angulares. El acople de tres o cuatro momen-tos angulares da diferentes opciones de acoplar pares de momento angular. Latransformacion de una de las opciones a otra se realiza a traves de los coeficientes6j o 9j, respectivamente.

Credit: The notes of this chapter were built from the following Refs.: [3, 5,9, 8]

4.1 Angular-momentum operators

The nucleons have an intrinsic spin and (in general) some orbital angular mo-mentum. The angular momentum of a nucleus is built from the coupling ofthese angular momenta. For this reason it is important to study the algebraof angular momentum, which is done in this chaper. The notion of angularmomentum can be related to abstract rotations of state vectors in a abstractHilbert space.

The angular momentum is an operator with three components which (thecomponents) satisfy the following commutation rules (this commutation haveto be postulated in order to go from the classical concept of angular momentum

22

over the quantum mechanics one)

[J1, J2] = J1J2 − J2J1 = i~J3 (4.1)

[J2, J3] = i~J1 (4.2)

[J3, J1] = i~J2 (4.3)

these and the other commutation are more generally expreseed using the anti-symmetric three-dimensional Levi-Civita permutation symbol ǫijk

[Ji, Jj ] = i~∑

k

ǫijkJk (4.4)

with

ǫijk =

0 if two of the indices are the same1 if the indices are different in cyclic permutations(c.p.):123, 231, 312−1 if the indices are different witn other differen than the above c.p.

(4.5)We have used the numeration 1, 2, 3 to represent in more compact form theconmuation rules of the component of the angular momentum. These indexesmean to represent the cartesian x, y, z components.

The orthonormalized eigenstates of the operator J2 are labelled by the quan-tum numbers (j,m)

J2|jm〉 = (J2x + J2

y + J2z )|jm〉 = j(j + 1)~2|jm〉 (4.6)

Jz|jm〉 = m~|jm〉 (4.7)

where〈jm|j′m′〉 = δjj′δmm′ (4.8)

with m = −j,−j + 1, · · · , j − 1, j.The coordinate representation of the abstract angular momentum eigenstate

|jm〉 for a single particle are the spherical harmonics

〈θφ|jm〉 = Yjm(θ, φ) (4.9)

For a given value j (wich gives the length of the angular momentum vectorJ), and m (which gives the projection on the z-axis of a Cartesian coordinatesystem) the other states belonging to the multiplet of j can be obtained by usingthe raising and lowering (or ladder) operators

J+ = Jx + iJy (4.10)

J− = Jx − iJy (4.11)

which satisfy the following commutation relations

[J+, J−] = 2~Jz (4.12)

[J±, Jz] = ∓~J± (4.13)

The ladder operators J± commute with J2 then they do not change theangular momentum but they change the value of m,

Jz(J±|jm〉) = m~(J±|jm〉) + [Jz, J±]|jm〉 (4.14)

= (m± 1)(J±|jm〉) (4.15)

23

Using the Condon-Shortley phase convention (which takes the coefficient α±

in J±|jm〉 = ~α±|jm± 1〉 real and positive) we have

J±|jm〉 = ~

√

j(j + 1)−m(m± 1)|jm± 1〉 (4.16)

= ~

√

(j ±m+ 1)(j ∓m)|jm± 1〉 (4.17)

The angular momenta components can be also combine in another convenientway to give origin to the so called spherical components Jµ of the angularmomentum

J+1 = − 1√2(Jx + iJy) = − 1√

2J+ (4.18)

J0 = Jz (4.19)

J−1 =1√2(Jx − iJy) =

1√2J− (4.20)

with the following commutation relations

[Jz, Jµ] = µ~Jµ (4.21)

[J±, Jµ] = ~

√

(1∓ µ)(2 ± µ)Jµ±1 (4.22)

In a system consisting of several particles the operator

Jµ =∑

i

(J i)µ (4.23)

is the µth component of the total angular momentum if (J i)µ is the µth com-ponent of the angular momentum of particle labelled by i.

For example, for a two particle system the z-component of the total angularmomentum (we will see later what we mean by total angular momentum) is

jz = (j1)z + (j2)z (4.24)

Spin Matrices For the case j = 12 the angular momentum J = S with | 12m〉,

m = ± 12 its eigenvectors. The matrix representation can be obtained from the

relations

Sz |jm〉 = ~m|jm〉 (4.25)

S±|jm〉 = ~

√

j(j + 1)−m(m± 1)|jm± 1〉 (4.26)

Sx =S+ + S−

2(4.27)

Sy =S+ − S−

2i(4.28)

for j = 12 and m = ± 1

2 .

24

First we calculate

Sz

∣

∣

∣

∣

1

2,±1

2

⟩

= ±~

2

∣

∣

∣

∣

1

2,±1

2

⟩

S+

∣

∣

∣

∣

1

2,±1

2

⟩

= ~

√

1

2

(

1

2+ 1

)

−(

±1

2

)[(

±1

2

)

+ 1

] ∣

∣

∣

∣

1

2,

(

±1

2

)

+ 1

⟩

S−

∣

∣

∣

∣

1

2,±1

2

⟩

= ~

√

1

2

(

1

2+ 1

)

−(

±1

2

)[(

±1

2

)

− 1

]∣

∣

∣

∣

1

2,

(

±1

2

)

− 1

⟩

(4.29)

Then,

S+

∣

∣

∣

∣

1

2,±1

2

⟩

= ~

√

3

4−(

±1

2

)[(

±1

2

)

+ 1

] ∣

∣

∣

∣

1

2,

(

±1

2

)

+ 1

⟩

S+

∣

∣

∣

∣

1

2,1

2

⟩

= ~

√

3

4−(

1

2

)[(

1

2

)

+ 1

] ∣

∣

∣

∣

1

2,

(

1

2

)

+ 1

⟩

= 0

S+

∣

∣

∣

∣

1

2,−1

2

⟩

= ~

√

3

4−(

−1

2

)[(

−1

2

)

+ 1

]∣

∣

∣

∣

1

2,

(

−1

2

)

+ 1

⟩

= ~

∣

∣

∣

∣

1

2,1

2

⟩

Then,

S+

∣

∣

∣

∣

1

2,1

2

⟩

= 0 (4.30)

S+

∣

∣

∣

∣

1

2,−1

2

⟩

= ~

∣

∣

∣

∣

1

2,1

2

⟩

(4.31)

A similar calculation would give

S−

∣

∣

∣

∣

1

2,1

2

⟩

= ~

∣

∣

∣

∣

1

2,−1

2

⟩

(4.32)

S−

∣

∣

∣

∣

1

2,−1

2

⟩

= 0 (4.33)

Acomodando las matrices de modo que

• la 1ra columna representa el ket∣

∣

12 ,

12

⟩

• la 2da columna representa el ket∣

∣

12 ,− 1

2

⟩

• la 1ra fila representa el bra⟨

12 ,

12

∣

∣

• la 2ra fila representa el bra⟨

12 ,− 1

2

∣

∣

tendremos la siguiente representacion para Sz y S±

Sz =

(

~

2 00 −~

2

)

(4.34)

S+ =

(

0 ~

0 0

)

(4.35)

25

S− =

(

0 0~ 0

)

(4.36)

Sx =S+ + S−

2(4.37)

=

(

0 ~

2~

2 0

)

(4.38)

Sy =S+ − S−

2i(4.39)

=1

2i

(

0 ~

−~ 0

)

(4.40)

= −i(

0 ~

2

−~

2 0

)

=

(

0 −i~2i~2 0

)

(4.41)

Matrices de Pauli: definamos las matrices de Pauli como

S =~

2σ (4.42)

entonces

σz =

(

1 00 −1

)

(4.43)

σx =

(

0 1−1 0

)

(4.44)

σy =

(

0 −ii 0

)

(4.45)

Introduccion al isospin: Hacer el Ejercicio 5.

4.2 Coupling of two angular momenta

Let J1 and J2 be two commuting angular momentum vectors. They couldcorrespond to the angular momenta of two different nucleons or they couldcorrespond to the intrinsic spin and the orbital angular momenta of a singlenucleon

[J1,J2] = 0 (4.46)

[J1k, J2l] = 0 for all k, l = x, y, z (4.47)

Each of the angular momenta has its eigenstates,

J2k|jkmk〉 = jk(jk + 1)~2|jkmk〉 k = 1, 2 (4.48)

Jkz |jkmk〉 = mk~|jkmk〉 (4.49)

The sum J = J1+J2 is an angular momentum since it satisfies the properties

[Ji, Jj ] = i~∑

k

ǫijkJk (4.50)

26

(jm)

j_1 j_2

Figure 4.1: Coupling of two angular momenta.

The eigenvectors of the total angular momentum J are

|j1m1j2m2〉 = |j1m1〉|j2m2〉 (4.51)

They are orthonormalized

〈j1m1j2m2|j′1m′

1j′

2m′

2〉 = δj1j′1δm1m′

1δj2j′2δm2m′

2(4.52)

The vectors |j1m1j2m2〉 are eigenstates of the operators

{J21, J1z ,J

22, J2z} (4.53)

These four operators form a complete set of commuting operators,

J2k|j1m1j2m2〉 = jk(jk + 1)~2|j1m1|j2m2〉 (4.54)

Jkz |j1m1j2m2〉 = mk~|j1m1j2m2〉 (4.55)

Then, the vectors |j1m1j2m2〉 form a complete set called uncouple basis.The angular momenta J1 and J2 can be coupled to a single angular mo-

mentum J in such a way that the following set of operators form an alternativeset of commuting operators

{J21,J

22,J , Jz} (4.56)

with |J1 −J2| ≤ J ≤ J1 +J2, i.e. J1, J2 and J must form a triangle, see Fig.4.1

The simultaneous eigenvectors of these four operators are |j1j2jm〉,J2

k|j1j2jm〉 = jk(jk + 1)~2|j1j2jm〉 (4.57)

J2|j1j2jm〉 = j(j + 1)~2|j1j2jm〉 (4.58)

Jz |j1m1jm〉 = m~|j1j2jm〉 (4.59)

(4.60)

with〈j1j2jm|j′1j′2j′m′〉 = δj1j′1δj2j′2δjj′δmm′ (4.61)

The complete set of states |j1j2jm〉 is called couple basis.Comment: The uncouple basis contains two angular momentum vectors and

their projections on the z-axis while the couple basis has three angular mo-mentum vectors and the projection of the coupled angular momentum. Thisimplies that coupling two angular momenta leaves only their ’lengths’ well de-fined while their z projections become fuzzy, i.e. they do not have any sharpquantum number associated to them.

27

4.2.1 Clebsch-Gordan coefficients

Both, the uncouple and coupled sets form a basis (for fixes j1 and j2), i.e.

I =∑

m1m2

|j1m1j2m2〉〈j1m1j2m2| (4.62)

and

I =∑

jm

|j1j2jm〉〈j1j2jm| (4.63)

then we can change from one to the other,

|j1j2jm〉 =∑

m1m2

|j1m1j2m2〉〈j1m1j2m2|j1j2jm〉 (4.64)

or with the follwing notation

|j1j2jm〉 =∑

m1m2

〈j1m1j2m2|jm〉|j1m1j2m2〉 (4.65)

where we have introduced the Clebsh-Gordan coefficient 〈j1m1j2m2|jm〉.Because the norm of the states is preserved, this lineal transformation is unitary.

The following are the basic properties of angular momentum coupling andClebsch-Gordan coefficients:

i) The projection quantum numbers have to fulfil the addtion law 〈j1m1j2m2|jm〉 =0 unless m1 +m2 = m.

ii) The coupled angular momenta have to fulfill the triangular condition |j1 −j2| ≤ j ≤ j1 + j2, denoted as ∆(j1j2j).

iii) The allowed values of the total angular momentum comes from the relationj1 + j2 + j =integer, i.e. j = |j1 − j2|, |j1 − j2|+1, · · · , j1 + j2 − 1, j1 + j2,with j1 and j2 integer or half-integer.

iv) The Clebsch-Gordan coefficients are chosen to be real, and so that 〈j1j1j2j2|j1+j2j1 + j2〉 = +1, 〈j1m1j2 − j2|jm〉 ≥ 0.

These conditions fix the phases of all Clebsch-Gordan coefficients.Clebsh-Gordan coefficients are orthogonal (demostrate it)

∑

m1m2

〈j1m1j2m2|jm〉〈j1m1j2m2|j′m′〉 = δjj′δmm′ (4.66)

and complete

∑

jm

〈j1m1j2m2|jm〉〈j1m′

1j2m′

2|jm〉 = δm1m′

1δm2m′

2(4.67)

The uncouple states |j1m1j2m2〉 in term of the coupled basis are give by(demostrar: multiplicar por

∑

jm〈j1m′1j2m

′2|jm〉 y usar 4.67)

|j1m1j2m2〉 =∑

jm

〈j1m1j2m2|jm〉|j1j2jm〉 (4.68)

28

Racah [1] showned that the Clebsch-Gordan coefficients can be written inthe form of a finite series,

〈j1m1j2m2|j3m3〉 = j3

√

(j1 + j2 − j3)!(j1 − j2 + j3)!(−j1 + j2 + j3)!

(j1 + j2 + j3 + 1)!√

√

√

√

3∏

i=1

(ji +mi)!(ji −mi)!

∑

k

(−)k

k!(j1 + j2 − j3 − k)!(j1 −m1 − k)!(j2 +m2 − k)!

1

(j3 − j2 +m1 + k)!(j3 − j1 −m2 + k)!(4.69)

Some symmetry properties of the Clebsch-Gordan coefficients (from Ref.[5, 9]):

• 〈j1m1j2m2|jm〉 = (−)j1+j2−j〈j2m2j1m1|jm〉• 〈j1m1j2m2|jm〉 = (−)j1+j2−j〈j1 −m1j2 −m2|j −m〉• 〈j10j20|j0〉 = 0 unless j1 + j2 + j =even.

• 〈j1m1j2m2|jm〉 = (−)j1−m1j

j2〈j1m1j −m|j2 −m2〉

• 〈jmj −m|00〉 = (−)j−m

j

• 〈jm00|j0〉 = 1

where j =√2j + 1

4.2.2 Spin-Spin coupling

• Desarrollar en la clase.

• Dar el desarrollo formal del singlete y el triplete

• Dar la forma exlicita del singlete para mostrar que la combinacion anti-simetrica de los espines.

χSMS (σ1σ2) =[

χ1/2(σ1)χ1/2(σ2)]

SMS(4.70)

4.2.3 Single particle wave function

The single particle wave function is writen as

φslmlms(rσ) = R(r) Ylml

(r) χsms(σ) (4.71)

The couple single particle wave function reads

ψsljmj(rσ) = R(r) Yljmj

(rσ) (4.72)

with (dar los detalles en la clase)

Yljmj(rσ) = [Yl(r)χs(σ)]jm (4.73)

where [. . . ]jm means couple to jm. Definir el rango de j

29

Utilidad autovalor |jm〉: Calcular la accion de l.s sobre Yljmj(rσ).

4.2.4 The Wigner 3j symbol

The phase factor which appears in the firs two previous relations can be donemore symmetric by defining the so-called 3j symbols,

(

j1 j2 j3m1 m2 m3

)

=(−)j1−j2−m3

j3〈j1m1j2m2|j3 −m3〉 (4.74)

The Clebsch-Gordan in terms of the 3j reads

〈j1m1j2m2|j3m3〉 = (−)j2−j1−m3 j3

(

j1 j2 j3m1 m2 −m3

)

(4.75)

The following are its basic symmetric properties (from Ref. [2, 9])

•(

j1 j2 j3m1 m2 m3

)

=

(

j2 j3 j1m2 m3 m1

)

=

(

j3 j1 j2m3 m1 m2

)

= (−)j1+j2+j3

(

j1 j3 j2m1 m3 m2

)

•(

j1 j2 j3−m1 −m2 −m3

)

= (−)j1+j2+j3

(

j1 j2 j3m1 m2 m3

)

•(

j1 j2 j3m1 m2 m3

)

= 0 unless

{

∆(j1j2j3)m1 +m2 +m3 = 0

•(

j1 j2 0m1 m2 0

)

= (−)j1−m1

j1δj1j2δm1−m2

•(

j1 j2 j30 0 0

)

= 0 unless j1 + j2 + j3 =even.

where ∆(j1j2j3) denotes the triangular condition |j1 − j2| ≤ j3 ≤ j1 + j2

4.3 Coupling of three angular momenta. TheWigner 6j symbol

The coupling of three commuting angular momentum vectors J1, J2 and J3

so that J = J1 + J2 + J3 is the total angular momentum can be done in thefollowing three different ways

• J12 = J1 + J2, J = J12 + J3

• J23 = J2 + J3, J = J23 + J1

• J13 = J1 + J3, J = J13 + J2

with the following properties

1. The values of the quamtum number j (corresponding to J) do not dependon the coupling order

2. The states corresponding to different coupling schemes are no the same

30

(jm)

j_1

j_2j_3

j_12

Figure 4.2: Coupling of three angular momenta.

3. The number of linearly independent states in all three coupling is the sameand each of the set form a complete set

I =∑

J12

|j1j2(j12)j3; jm〉〈j1j2(j12)j3; jm| (4.76)

I =∑

J23

|j1j2j3(j23); jm〉〈j1j2j3(j23); jm| (4.77)

I =∑

J13

|j1j3(j13)j2; jm〉〈j1j3(j13)j2; jm| (4.78)

As for the case of two angular momenta, one can change from one basis tothe other, for example the change for the basis |j1j2(j12)j3; jm〉 to the basis|j1j2j3(j23); jm〉 is given by

|j1j2j3(j23); jm〉 =∑

J12

|j1j2(j12)j3; jm〉〈j1j2(j12)j3; jm|j1j2j3(j23); jm〉

=∑

J12

(−)j1+j2+j3+j j12j23

{

j1 j2 j12j3 j j23

}

|j1j2(j12)j3; jm〉

where the six j enclosed in the braces {· · · } is called the 6j symbol. It is pro-portional to the overlap of two state vectors related to two different couplingschemes of three angular momenta. They are members of the unitary transfor-mation since the norm of the states are preserved. 6j symbols are related to atransformation betweenn two basis sets where all the states have a good totalangular momentum.

The following explicit expression of the 6j symbol can be obtained from isdefinition in the basis transformation

{

j1 j2 j12j3 j j23

}

=∑

m1m2m3

∑

m12m23m

(−)j3+j+j23−m3−m−m23

(

j1 j2 j12m1 m2 m12

)(

j1 j j23m1 −m m23

)

(

j3 j2 j23m3 m2 −m23

)(

j3 j j13−m3 m m12

)

(4.79)

31

Symmetry:

1. The angular momenta involved in a 6j symbol have to satisfy certaintriangular conditions given by (analysis theses conditions with the aboveexpression)

{

j1 j2 j3l1 l2 l3

}

= 0 unless

∆(j1j2j3)∆(l1l2j3)∆(l1j2l3)∆(j1l2l3)

(4.80)

2. Exchange of any two columns leave the value of the symbol unchanged:

{

j1 j2 j3l1 l2 l3

}

=

{

j2 j1 j3l2 l1 l3

}

=

{

j3 j1 j2l3 l1 l2

}

= · · · (4.81)

3. Exchange of the upper and lowe arguments in each of any two columnsleave the value of the symbol unchanged:

{

j1 j2 j3l1 l2 l3

}

=

{

l1 l2 j3j1 j2 l3

}

=

{

j1 l2 l3l1 j2 j3

}

= · · · (4.82)

Altogether these exchanges create 24 symmetry transformations.An alternative notation to the 6j symbol are the so-calle Racah symbol of

Racah W coefficient,

W (j1j2l2l1; j3l3) = (−)j1+j2+l2+l1

{

j1 j2 j3l1 l2 l3

}

(4.83)

If one of the angular momentum is zero the 6j reads

{

j1 j2 j30 j′3 j′2

}

=(−)j1+j2+j3

j2j3δj2j′2δj3j′3∆(j1j2j3) (4.84)

4.4 Coupling of three angular momenta. The

Wigner 9j symbol

Let J1, J2, J3 and J4 be four commuting angular momentum vectors, withJ = J1 + J2 + J3 + J4 their vector sum.

Let us consider two of the many different way to combine them,

• J12 = J1 + J2, J34 = J3 + J4, J = J12 + J34

• J13 = J1 + J3, J24 = J2 + J4, J = J13 + J24

the have the following properties

1. The values of the quamtum number j (corresponding to J) do not dependon the coupling order

2. The states corresponding to different coupling schemes are no the same

32

(jm)

j_1

j_2 j_3

j_4j_12

j_34

Figure 4.3: Coupling of four angular momenta.

3. The number of linearly independent states in all three coupling is the sameand each of the set form a complete set

I =∑

j12j34

|j1j2(j12)j3j4(j34); jm〉〈j1j2(j12)j3j4(j34); jm| (4.85)

I =∑

j13j24

|j1j3(j13)j2j4(j24); jm〉〈j1j3(j13)j2j4(j24); jm| (4.86)

4.4.1 Ejemplo: Acople de dos partıculas

Cada partıcula tiene un momento angular orbital y un momento angular debidoal spin:

• j1 → l1 con |l1ml1〉 → Yl1ml1

• j2 → s1 con |s1ms1〉 → χs1ms1

• j3 → l2 con |l2ml2〉 → Yl2ml2

• j4 → s2 con |s2ms2〉 → χs2ms2

Acople jj:

• j1 = l1 + s1

• j2 = l2 + s2

• J = j1 + j2

Luego, las bases acoplada para las particulas 1 y 2 antes de acoplarlas entreellas resulta

• |j1m1〉 → [Yl1χs1 ]j1m1= Yj1m1

• |j2m2〉 → [Yl2χs2 ]j2m2= Yj2m2

La base de las dos particulas acopladas resulta

• |JM〉 → [Yj1Yj2 ]JM explicitar esta expresion utilizando los Clebsch-Gordancorrespondientes

• En termino de la notacion de la presentacion general serıa |JM〉 → |l1s1(j1)l2s2(j2); JM〉

• Adaptar la figura 4.3 para este caso.

33

Acople LS:

• L = l1 + l2

• S = s1 + s2

• J = L+ S

Luego, las bases acoplada para las particulas 1 y 2 antes de acoplarlas entreellas resulta

• |LML〉 → [Yl1Yl2 ]LML= YLML

• |SMs〉 → [χs1χs2 ]j2m2= χSMS

La base de las dos particulas acopladas resulta

• |JM〉 → [YLχS ]JM explicitar esta expresion utilizando los Clebsch-Gordancorrespondientes

• En termino de la notacion de la presentacion general serıa |JM〉 → |l1l2(L)s1s2(S); JM〉

• Adaptar la figura 4.3 para este caso.

4.4.2 Change of coupling

A change from one basis to the other can be accomplished by a linear unitarytransformation,

|j1j3(j13)j2j4(j24); jm〉 =∑

J12J34

|j1j2(j12)j3j4(j34); jm〉

〈j1j2(j12)j3j4(j34); jm|j1j3(j13)j2j4(j24); jm〉

=∑

J12J34

j12j34j13j24

j1 j2 j12j3 j4 j34j13 j24 j

|j1j2(j12)j3j4(j34); jm〉

where the nine j enclosed in the braces {· · · } is called the 9j symbol.From the above equation we can get the following explicit expression for the

9j symbol

j1 j2 j12j3 j4 j34j13 j24 j

=∑

m1m2m3m4

∑

m12m34m13m24m(

j1 j2 j12m1 m2 m12

)(

j3 j4 j34m3 m4 m34

)

(

j13 j24 jm13 m24 m

)(

j1 j3 j13m1 m3 m13

)

(

j2 j4 j24m2 m4 m24

)(

j12 j34 jm12 m34 m

)

(4.87)

34

The following relation link the 9j symbol with the 6j symbol when the totalangular momentum is nill

j1 j2 j12j3 j4 j34j13 j24 0

=δj12j34

j12

δj13j24

j13(−)j12+j13+j2+j3

{

j1 j2 j12j4 j3 j13

}

(4.88)

Symmetry:

1. From the 9j symbol in terms of the 3j ones, we observe that there are sixtriangular conditions. The triangular relations imply that the 9j symbolis zero unless the triangular rule is valid for all of its comluns and rows

j1 j2 j12j3 j4 j34j13 j24 j

= 0 unless

∆(j1j2j12)∆(j3j4j34)∆(j13j24j)∆(j1j3j13)∆(j2j4j24)∆(j12j34j)

(4.89)

2. The value does not change when its columns and rows are interchanged:

j1 j2 j12j3 j4 j34j13 j24 j

=

j1 j3 j13j2 j4 j24j12 j34 j

(4.90)

3. The exchange of any two columns or any two rows changes the symbolonly by a phase factor:

j2 j1 j12j4 j3 j34j24 j13 j

= (−)σ

j1 j2 j12j3 j4 j34j13 j24 j

= (−)σ

j4 j3 j34j2 j1 j12j24 j13 j

= · · ·

(4.91)with σ = j1 + j2 + j12 + j3 + j4 + j34 + j13 + j24 + j

Altogether these exchanges create 24 symmetry transformations.

4.4.3 Relation between the jj-coupling andthe LS-coupling

From the general expresion for recoupling ([2] pag. 125) we gets,

Ψ[n1l1s1(j1),n2l2s2(j2);JM ](x1x2) =∑

LS

√

(2j1 + 1)(2j2 + 1)(2L+ 1)(2S + 1)

l1 l2 Ls1 s2 Sj1 j2 J

Ψ[n1l1n2l2(L),s1s2(S);JM ](x1x2)

35

from deShalit-Talmi pag. 516

l1 l2 Ls1 s2 Sj1 j2 J

=

s1 s2 Sj1 j2 Jl1 l2 L

=

j1 j2 Jl1 l2 Ls1 s2 S

(4.92)

We are interested in the single part of the two-particle wave function S = 0(from deShalit-Talmi pag 517 )

j1 j2 Jl1 l2 L1/2 1/2 0

=(−)−j2−J−l1−1/2

J ˆ1/2

{

j1 j2 Jl1 l2 1/2

}

=(−)−J−1/2+j1+l2

√2J

W (j1j2l1l2, J1/2) (4.93)

Then, the two particle wave function reads,

ΨS=0[n1l1s1(j1),n2l2s2(j2);JM ](x1x2) = (−)J−l2−j1+1/2 j1j2√

2W (j1l1j2l2, 1/2J)

Ψ[n1l1n2l2(J),s1s2(0);JM ](x1x2) (4.94)

Working out the radial part we get,

Ψ[n1l1n2l2(J),s1s2(0);JM ](x1x2) = S [ϕl1(r1)ϕl2(r2)]LM χ00(σ1σ2) (4.95)

where S is the symmetric normalization operator

S [ϕl1(r1)ϕl2(r2)]LM =[ϕl1(r1)ϕl2(r2)]LM + [ϕl1(r2)ϕl2(r1)]LM

√

2(1 + δl1l2)(4.96)

We already demostrated that the single part of the spin is antisymmetric.Here we shows that the radial part is symmetric. This also shows that thetwo-particle wave function is antisymmetric.

4.5 Graphical methods in angular momentum

One may use graphical method for the calculation containing sums of products ofClebsch-Gordan coefficients or Wigner 3j-coefficients. This goal is accomplishedby making a correspondence between diagrams and algebraic formulae whereeach term of the formula is represented by a component of an appropiate graph.The following diagramatic rules are from Ref. [8]:

Rule 1: The Wigner 3j symbol is represented by a node with three lines joinedto it, see Fig. 4.97

(

j1 j2 j3m1 m2 m3

)

=

j_1 m_1

j_3 m_3

j_2 m_2

+ =−

j_1 m_1

j_2 m_2

j_3 m_3

(4.97)

The anti-clockwise orientation is denoted by a + sign while a clockwise orien-tation is denoted by a − sign. A rotate diagram represent the same 3j-symbolas the original one.

36

Rule 2: The anti-symmetric symbol or ’metric tensor’

(

j1m1m2

)

= (−)j1+m1δm1,−m2

is denoted by a line with an arrow on it, see Fig. 4.98

δj1,j2

(

j1m1m2

)

=j_1 m_1 j_2 m_2

(4.98)

Rule 3: A line with no arrow represents the expression δj1j2δm1m2, see Fig.

4.99

δj1j2δm1m2=

j_1 m_1 j_2 m_2 (4.99)

Rule 4: By joining the above three diagrams, more complicated diagramasmay be constructed. Two lines representing the same total angular momentumcan be joined. This implies that the z components of the two angular momentashould be set equal and summed over. In this case the z can be omited fromthe diagram.

Internal lines: lines which hoin nodes are called internal lines.External lines: external lines have one end connected to a node and one endfree.Closed diagrams: closed diagrams have no external lines.

Example: Construction of the graph representing of the left hand side of thefirst orthogonality relation for 3j symbols below:

∑

m1m2

(

j1 j2 j3m1 m2 m3

)(

j1 j2 j′3m1 m2 m′

3

)

=1

j23δj3j′3δm3m′

3(4.100)

(i) The left hand side 3j symbol is represented by the left figure in 4.97 withpositive node; let us it called left diagram.(ii) The right hand side 3j symbol can be represented by the right figure in 4.97with the replacement j3m3 → j′3m

′3, with negative node; let us it called right

diagram. This is a convenient election for the orientation because it will allowus to joined the legs wiht the same j than the left diagram.(iii) We rotate the left diagram untill the leg j3 is horizontal pointing to the left.This implies that the leg with j2 will be upp and the leg with j1 will be down.(iv) We rotate the right diagram untill the leg j′3 is horizontal pointing to theright, with j2 in the upper left leg ang j1 in the lower left leg.(v) Using the summation

∑

m2we join the left and right legs with j2 forming a

line without arrow with label j2.(vi) We repeated the process in (v) with the summation

∑

m1and the legs with

j1 forming a line without arrow with label j1.Then, we get the identity of Fig. 4.101

j_3 j’_3

j_2

j_1

+ −

= 12j3+1δj3j′3δm3m′

3

(4.101)

Rules for transforming graphs Calculation using the graphical techniqueimplies that after the algebraic equation have been given in graphical way with

37

the above rules, one have to transform the graph looking for components of thegraph wichn may be identified with standard invariant functions such as theRacah W-functions, the 6j symbol or the 9j symbol. Then the transformedgraph is reconverted to an algebrraic formula.

About arrow.

The transformation of the graph involve adding or removing arrows or deformingthe diagram to get it into some standard form. These operation are given bythe following rules:

A. A line with two oppositely directe arrows is equivalent to a line with noarrows.

B. A line corresponding to an angular momentum j with two arrows in thesame direction is equivalent to a line with no arrows times a factor (−)2j .

C. If an arrow on a line with angular momentum j is reversed the graph mustbe multiplied by a factor (−)2j .

D. Three arrows may be added at a node, one to each line joined to the node,without changing the value of the graph provided the arrows are directtedeither all away from or all towrads, the node.

E. The direction of all arrows and the signs of all nodes may be changed simul-taneously in a closed diagram without altering the value of the diagram.

Normal form of graph: The number and orientation of arrows in a graphmay be changed as it is needed following the above rules A-D. It is said the graphis in its normal form if there is exactly one arrow on every internal line. Onlythose diagrams which can be put into normal form represent formulae arisingfrom coupling of angular momenta. The normal form of a graph is not unique asthe directions of arrows may be changed in many different ways without alteringthe value of the graph (rule E).

Example: Example 1 in pag. 122 of Ref. [8].About derformation.

A graph may be deformed in any way withou altering its value under the fol-lowing conditions:

i. The direction of any arrow relative to the nodes it conects is not changed.

ii. The sign of a node is changed if the cyclic order of the angular momenta atthe node is reversed.

Example: Example 2 in pag. 123 of Ref. [8].

4.6 List of some coefficients

It follows of some Clebsch-Gordan and three-j coefficients.

38

〈jmjm′|00〉 =(−)j−m

jδm,−m′ (4.102)

(

j1 j2 0m1 m2 0

)

=(−)j1−m1

j1δj1j2δm1,m2

(4.103)

(

j1 j2 j30 0 0

)

= 0 unless j1 + j2 + j3 = even (4.104)

〈j1m1j2m2|j3m3〉 = (−)j2−j1−m3 j3

(

j1 j2 j3m1 m2 −m3

)

(4.105)

(

j j 1m −m− 1 1

)

= (−)j−m

√

2(j −m)(j +m+ 1)

2j(2j + 1)(2j + 2)(4.106)

(

j j 1m −m 0

)

= (−)j−m m√

j(j + 1)(2j + 1)(4.107)

4.7 Exercises

Ex. 1. Write the eigenvalue equations for ℓ2 and ℓz. Give

the posible values of the z component of the orbital angular-momentum

. Description: Let ℓ be the orbital angular-momentum operator

ℓ =1

~r × p (4.108)

obeys the commutation relations 4.1. Their eigenfunctions are the spherical har-monic Ylm(θ, φ). The eigenvectors Ylm describe a state with angular momentuml and z component m.

Ex. 2. Write the eigenvalue equations for s2 and sz. Give

the posible values of the z component of the spin operator

Description: The nucleon spin operators operator s = 1/2 satisfying the com-mutation relations 4.1. Writen in terms of the Pauli matrices they are

sµ =1

2σµ (4.109)

with components

σx =

(

0 11 0

)

(4.110)

σy =

(

0 −ii 0

)

(4.111)

σz =

(

1 00 −1

)

(4.112)

39

The simultaneously eigenvectors of the spin operators s2 and sz are,

χ1/2 =

(

10

)

(4.113)

χ−1/2 =

(

01

)

(4.114)

Ex. 3. Calculates the Clebsch-Gordan coefficient 〈JM |JM10〉This coefficient is going to be use in the Tensor chapter.

Solution: From table we get the following value for the three-j coefficient

(

J J 1M −M 0

)

= (−)J−M M√

J(J + 1)(2J + 1)(4.115)

Using the relation between the three-j coefficient and the Clebsch-Gordanone

(

j1 j2 j2m1 m2 m3

)

=(−)j1−j2−m3

√2j3 + 1

〈j1m1j2m2|j3 −m3〉 (4.116)

We procedu as follows: let us take j1 = J , m1 =M , j2 = 1, m2 = 0, j3 = J ,−m3 =M we get

〈JM |JM10〉 =√2J + 1

(

J 1 JM 0 −M

)

(−)−J+1+M

=√2J + 1(−)J+1+J

(

J J 1M −M 0

)

(−)−J+1+M

=√2J + 1(−)J+M

(

J J 1M −M 0

)

=√2J + 1(−)J+M (−)J−M M

√

J(J + 1)(2J + 1)

=M

√

J(J + 1)(4.117)

Ex. 4. Consider the total angular momentum j = ℓ + s

(i) Give the eigenvectors which are simultaneously eigenvectors of {ℓ2, ℓz, s2, sz}.(ii) Give the eigenvectors which are simultaneously eigenvectors of {ℓ2, s2, j2, jz}.(iii) For the eigenvectors which are simultaneously eigenvectors of {ℓ2, s2, j2, jz}calculates the action of ℓ · s on them.

Ex. 5. Calculates the raising and lowering operators t+ andt−. Calculates their action on the eigenvector of t.

Description: Let us considerer the proton and the neutron as two differentmanifestation of the same particle: the nucleon. The two different states of the

isotopic spin t = 12σ (with σ the Pauli matrices) are written as π =

(

01

)

and

40

ν =

(

10

)

, with t3ν = 12ν and t3π = − 1

2π.

Solution:

t3 =1

2

(

1 00 −1

)

(4.118)

t+ =

(

0 10 0

)

(4.119)

t− =

(

0 01 0

)

(4.120)

t−ν = π (4.121)

t−π = 0 (4.122)

t+ν = 0 (4.123)

t+π = ν (4.124)

Ex. 6. Consider two particles, each one with orbital an-gular momentum ℓ and intrinsic spin s = 1/2. Couple thefour angular momenta to a total angular momentum J inthe following two orders

(i) L = ℓ1 + ℓ2, S = s1 + s2, J = L+ S. Use diagramatic representation.(ii) j1 = ℓ1 + s1, j2 = ℓ2 + s2, J = j1 + j2. Use diagramatic representation.

Ex. 7. Find the relation between the above two coupling

Use diagramatic representation like in Fig. 4.1.

Ex. 8. Find the coefficient k in the following relation.[

Ylχ1/2

]

jm= k

[

χ1/2Yl]

jm(4.125)

Ex. 9. Find the graphical representation for the product ofthe 3j symbols and the righ hand side of equation (4.126)(second orthogonality relation for the 3j symbol).

∑

j3

(2j3 + 1)

(

j1 j2 j3m1 m2 m3

)(

j1 j2 j3m′

1 m′2 m3

)

= δm1m′

1δm2m′

2(4.126)

Solution:

(

j1 j2 j3m1 m2 m3

)(

j1 j2 j3m′

1 m′2 m3

)

=

+ −

j_3

j_2 m’_2

j_1 m’_1j_1 m_1

j_2 m_2 (4.127)

41

δm1m′

1δm2m′

2=

j_1 m’_1j_1 m_1

j_2 m’_2j_2 m_2(4.128)

42

Bibliography

[1] G. Racah, Phys. Rev. 62, 438, 1942.

[2] A. de-Shalit and I. Talmi, Nuclear Shell Theory, New York, 1963.

[3] G. F. Bertsch. The practitioner’s shell model. North-Holland PublishingCompany - Amsterdam. 1972.

[4] Libro DeShalit de 1974 (completar informacion)

[5] R. D. Lawson. Theory of the nuclear shell model. Clarendon Press - Oxford.1980.

[6] M. Shaw and A. Williart. Fısica Nuclear: problemas resueltos. Alianza.Universidad Textos. 1996.

[7] D. Lunney, J. M. Pearson and C. Thibault, Rev. Mod. Phys. 75, 1021,2003.

[8] D. M. Brink and G. R. Satchler. Angular Momentum. Third Edition. Ox-ford Science Publications. 2002.

[9] J. Suhonen. From nucleons to nucleus. Springer-Verlag - Berlin. 2007.

[10] B. R. Martin, Nuclear and Particle Physics. Wiley. John Wiley & Sons Ltd.2009.

[11] M. Wang, G. Audi, A. H. Wapstra, F. G. Kondev, M. Mac Cormick, X.Xu, and B. Pfeiffer, Chinese Phys. C 36, 1603, 2012.

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