Rotations and Angular momentum

� Intro

The material here may be found in

Sakurai Chap 3: 1-3, (5-6), 7, (9-10)Merzbacher Chap 11, 17.

Chapter 11 of Merzbacher concentrates on orbital angular momentum. Sakurai, and Ch 17 of Merzbacher focus on angular momentum in relation to the group of rotations. Just as linear momentum is related to the translation group,

angular momentum operators are generators of rotations. The goal is to present the basics in 5 lectures focusing on

1. J as the generator of rotations.2. Representations of SO�3�3. Addition of angular momentum

4. Orbital angular momentum and Ylm ' s5. Tensor operators.

� Rotations & SO(3)

ü Rotations of vectors

Begin with a discussion of rotations applied to a 3-dimensional real vectorspace. The vectors are described by three real

numbers, e.g. v =���������� vx

vy

vz

���������� . The transpose of a vector is vT = �vx, vy, vz� . There is an inner product defined between two

vectors by uT ÿ v = vT ÿu = uv cosf , where f is the angle between the two vectors. Under a rotation the inner product between any two vectors is preserved, i.e. the length of any vector and the angle between any two vectors doesn't change. A rotation can be described by a 3ä3real orthogonal matrix R which operates on a vector by the usual rules of

matrix multiplication

���������� v'x

v'yv'z

���������� = R

���������� vx

vy

vz

���������� and �v'x, v'y, v'z� = �vx, vy, vz� RT

To preserve the inner product, it is requird that RT ÿR= 1

u' ÿ v' = uRT ÿRv

= u1v

= u ÿv

As an example, a rotation by f around the z-axis (or in the xy-plane) is given by

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Rz�f� = ���������� cosf -sinf 0

sinf cosf 0

0 0 1

����������

The sign conventions are appropriate for a right handed coordinate system: put the thumb of right hand along z-axis,

extend fingers along x-axis, and curl fingers in direction of y-axis.

x

y

z

The direction of rotation for f is counter-clockwise when looking down from the +zdirection, i.e. rotate the x-axis into the y-axis. Similarly the rotations around the x and y axes are

Ry�f� = ���������� cosf 0 sinf

0 1 0

-sinf 0 cosf

���������� and Rx�f� = �

��������� 1 0 0

0 cosf -sinf

0 sinf cosf

����������

The sign of sinf in Ry is related to the handed-ness of the coordinate system and the sense of rotation. For 3-dimensions, it is equivalent to talk about rotations around the z-axis, or rotations in the xy-plane. In any other number

of dimensions, the correct language is to talk about rotations in the xi x j -plane, where xi defines one of the coordinate directions of the vector space. Thus, while for 3-dimensions there are 3 independent rotations, in N -dimensions, there will be N�N-1�ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ2 independent rotations.

ü Direction kets

To make the correspondence to quantum states, just as a translation was defined by its action on position eigenkets, a rotation around the origin also can act on position eigenkets byx ' = R

è x = Rx

where a distinction has been made between the operator Rè

, which acts on the state, and the rotation matrix R which acts on the coordinates. Since the rotations don't change the length of the vector, it is possible to define spherical coordinates, r, q, f , and spherical position kets, x Ø �r�≈ �n�, where r determines the radial position, and n indicates

the direction from the origin. The rotations act only on the n degrees of freedom.

Rè �r≈ n� = r≈ Rÿ nDirection kets will be used more extensively in the discussion of orbital angular momentum and spherical harmonics,

but for now they are useful for illustrating the set of rotations. The set of all direction kets �n� can be visualized by the surface of a sphere, and the rotations are the set of all possible ways to reorient that sphere.

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ü Orthognal group SO(3)

The set of all possible rotations form a group. Consider the four properties: closure, identity, inverse and associativity.

Using the picture of rotations as reorientations of a sphere, one can construct visualizations to illustrate each property.

With greater mathematical rigor, the set of all possible rotations form the group SO(3), where O Ø orthogonal, 3 Ø 3 dimensions, and S Ø special, which in this case means the matrix has a determinant of 1. The rotations are described by three continuous, but bounded, parameters. From the matrix point of view, a 3ä3matrix has nine degrees of freedom.

The constraint that the matrix is orthogonal, Ri jT Rj k = di k yields 6 conditions, i.e. three for i = k and three for i ∫ k.

The properties of a group are obeyed:

closure: For any two orthogonal matrices R1 and R2 , the product R3 = R1 R2 , is also orthogonal. The combination of two rotations is also a rotation.identity: The 3ä3 unit matrix acts as an identity element for the group. 1 R = R1 = R

inverse: Each element has an inverse R-1 = RT , RT R= R RT = 1associativity: R1�R2 R3� = �R1 R2� R3

Unlike the Translation group, SO(3) is not abelian, i.e. in general R1 R2 ∫ R2 R1 .

The significance of the S-condition, DetR= 1, is that reflections are not included in the group, i.e. for three dimensions one cannot turn a right-handed object into a left-handed object by doing a rotation. If we allowed reflections, e.g.

���������� -1 0 0

0 -1 0

0 0 -1

����������

Then, the group would be O(3) instead of SO(3). O(3) is called "disconnected" since not all elements of the group can be reached by a succession of infinitesimal transformations. SO(3) is connected.

The rotation matrices R are just one "representation" of the group SO(3). For two different representations, there has to be a 1 1 mapping of the elements of one representation to the other. The mapping has to preserve the combination law.

Consider two representations R and S. Label a rotation by a subscript which represents the three parameters to define a rotation, and identify Ra ¨ Sa , If R3 = R1 R2 , then we must have S3 = S1 S2 to preserve the combinatin law.

ü Full set of rotations: �n` , f�

There are two common methods for parameterizing rotations. The first is to choose an axis for rotation and then perform

a rotation by an angle between 0 and p. The axis of rotation can be chosen anywhere on the sphere. Why not 0¨ 2 p ? Then rotations with poles on opposite sides of the sphere would be redundant. An explicit form for R�n, f� will be given after developing the language of infinitesimal rotations.

Draw your own picture showing the rotation of a sphere around an off �

axis pole. The sphere represents the set of states, n �,

i.e. the set of direction kets. The rotation reorients the sphere.

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ü Euler angles

The second parameterization is to give Euler angles. In this method one describes where the "north pole" moves to

under a rotation, and the orientation of the sphere after the pole has been moved. The location of the pole is determined by first choosing a longitude by rotating around the z-axis, then a latitude by rotating around the new y-axis. Finally, the orientation of the sphere is given by a final rotation around the new z-axis. Pictorially,

Draw more pictures, showing the sequence of rotations to move the pole,

and then reorient the sphere around the new pole.

In the Euler parameterization the range of angles is

a = �0, 2 p�, b = �0, p�, g = �0, 2 p�and an arbitrary rotation is given by

R�a, b, g� = Rz'�g� Ry'�b� Rz�a�Note that the z' and y' rotations are not defined with respect to the original coordinate axes, but rather with respect to

where those axes have moved with the reorientation of the sphere. Later it will be shown that

Rz'�g� Ry'�b� Rz�a� = Rz�a� Ry�b� Rz�g�where the order has been reversed, but now all rotations are conveniently defined around the axes of the original coordinate system.

ü Equivalency of the two parameterizations

The two parameterizations may not seem equivalent, but they are, as can be seen by a pictorial mapping of �n, f� to �a, b, g� . Observe that there are two ways to produce the same set of Euler angles consistent with the restriction of f to �0, p� .

This picture didn' �t make it into the classroom presentation. it ' �s a bit of work

The two techniques have different uses. Euler angles tend to be more useful for building up actual rotation matrices in a calculation. This is because Rz and Ry are generally fairly easy to construct for a representation, and the matrix multiplication is straightforward. The �n, f� notation has advantages in some analytic manipulations, as we will see below.

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ü J as the generator of infinitesimal rotations.

In analogy to the discussion of translations and time evolution, it is useful to build up the finite rotations from

generators of infinitesimal rotations. Recognizing that we are eventually interested in a quantum mechanical formulation, it is useful to develop this formalism in a way that realizes the rotations as unitary operations. For example, an infinitesimal rotation around the z-axis is given by

Rz�d� = 1- i d Jz

where Jz is the generator of infinitesimal rotations around the z-axis. Since R is unitary (note: orthogonal matrices are unitary), J must be Hermitian. In the present case, to leading order in d

Rz�d� = ���������� 1 -d 0

d 1 0

0 0 1

���������� or Jz =

���������� 0 -i 0

i 0 0

0 0 0

����������

Similarly, for this representation

Jx =

���������� 0 0 0

0 0 -i

0 i 0

���������� , Jy =

���������� 0 0 i

0 0 0

-i 0 0

����������

Note that since this is still a classical discussion I haven't put in any factors of �.

ü Commutation relations

The generators obey the commutation relations

Ji, J j� = i ei jk Jk

where ei jk = 1, if �i j k � is an even permutation of�x y z�= -1, if �i j k� is an odd permutation of�x y z�= 0, if any two of�i j k � are equal

as a useful aside ei jk elmk = dil d jm - dim d jl .

The commutation relations are a property of the group, not just a particular representation. The collection of all the commutator relations for the generators is sometimes called the algebra of the generators of the group, or just the algebra of the group.

ü Finite rotations

For rotations around a particular axis, it should be clear that we can build up an arbitrary rotation by a sequence of infinitesimal rotations, similar to the procedure for building up a finte translation as the limiting product of a large number of infinitesimals.

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R�f� = Limnض

�Pi

Rz�f �n��= Lim

nض�1- i Jz �f � n��n

= e-i Jz f

It is now possible to give a form for an arbitrary rotation in the �n, f� parameterization. The infinitesimal rotation

around the n-axis is given by

R�n, d� = 1- i d n ÿ J

There is no concern about which component of J is operated on first, since the effects of commutation amongst the different genreators shows up at second order in the infinitesimal d. The finite rotation is then given by

R�n, f� = e-i f nÿJ

� Representations of SO(3)

ü relation between D and R - notational

Rotations can act to change a wide variety of objects, e.g. classical position vectors, position eigenstates x� , operators

such as X����, P

�� or L

��, angular momentum states lm etc. In principle, the notation R to denote a rotation operator can be

used for all of these applications, if a sufficiently detailed definition is supplied for the case at hand. In practice, a common convention is to use R when the object in question has the properties of a position vector, but to use a notation

D when operating on angular momentum states or objects with similar characteristics. For example, to operate on a classical vector use

x'i = Ri j x j

or to operate on a position eigenstatex' = R x = Ri j x j

In contrast, to perform a rotation on a state a which is an angular momentum state a = lm one would write� a'� = D�R� � lm�where D�R� is an operator that depends on orbital angular momentum l . One would still write D in the exponential form

D �n, f� = e-iÅÅÅÅ� Jÿnf

but the form of the generators would be specific to the set of states, or representation, which is the object of the rotation. The set of possible representations is quite large. To simplify the discussion as much as possible, one defines the irreducible representations of SO�3� which is the subject of the next section.

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ü Irreducible representations

ü Casimir operators and maximal set of commuting operators.

The first item of business is to determine a maximal set of commuting operators that can be simultaneaously diagonalized. For SO(3) or SU(2) this would be J2 and one other component of J , typically taken to be Jz. J2 is an example of what is known as a Casimir operator. Casimir operators commute with all operators within the algebra of the

group. Other groups may have more than one Casimir. The number of Casimir operators is equal to the "rank" of the group. SO(4) and SU(3) for example have two Casimir operators and are rank 2. Generally, the most useful Casimir is the quadratic operator C2 =�i Oi

2 where the Oi are the infinitesimal generators of the group. J2 is such a quadratic casimir operator. In groups with more than one casimir operator, they all commute with each other.

In addition to the casimir operators, one can choose a set of operators equal in number to the rank from the group algebra that also commute with each other. For example in SU(3) one can find two generators that are diagonal. So in general, the dimension of a maximal set of commuting operators with which to define the representations of a group is

twice the rank.

For completeness,

J2, Jz� = Jx2 + Jy

2 + Jz2, Jz�

= Jx2 + Jy

2, Jz�= Jx Jx, Jz� + Jx, Jz� Jx + Jy Jy, Jz� + Jy, Jz� Jy

= -i�Jx Jy + Jy Jx� + i�Jy Jx + Jx Jy�= 0

ü Labeling of states

Since J2, Jz can be diagonalized simultaneously, we can specifiy the states by j, m , where m is the eignevalue when Jz operates on the state and the operation of J2 yields a j

Jz � j, m� =m � j, m�J2 � j, m� = a j � j, m�where the eigenvalue of J2 is not yet determined. Note that the labeling of the states is rather arbitrary. In the case of Jz,

it is convenient to use the eigenvalue directly. We will also take the states j, m to be normalized.

ü j -representations as irreducible representations

As J2 commutes with all the generators of the group, it also commutes with any function of those generators, and in particular J2, D� = 0. As with other commutation relations, this implies that the rotation operators don't change the

eigenvalue of J2 ,

J2�D jm� = DJ2 jm = Da j jm = a j�D jm�

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On the other hand Jz, D� ∫ 0, and so the rotations mix states of different m-value but not of different j -value. In this case we say that all the states of a given j -value, taken together, form a representation of the group. The dimension of

the representation is equal to the number of distinct basis states which may be chosen for the same j -value. In the case of j -representations we say that the representation is irreducible, i.e. it is not possible to break the representation down into two subspaces that don't mix under the action of rotations. Thus, the result of performing a rotation on a state jm

is given by a linear combination of all states jm'

D jm = Sm' Dmm'

j jm'where the exact value of the coefficients Dmm'

j depends on the parameters describing the rotation.

� Orbital angular momentum

This section develops orbital angular momentum operators in a manner analagous to the development of momentum as

the generator of translations.

ü Direction kets

Begin with the direction kets n , or q, f . The angular behavior of a particle in a given state, say b , is given by yb�q, f� = �n � b . This is in direct analogy to the spatial wave function yb�x� = �x � b . Instead of q, f one can use

z= cosq as the polar coordinate, and states z, f , so that yb�z, f� = �n � b .

ü Rotations

The rotation acting on the direction ket gives n' = D�a, b, g� n = R�a, b, g� ÿ n , where here R is reserved to denote the rotation of vectors, D to denote the transformation of states, and the rotation is labeled by the Euler angles. This is

analagous to x' = T�D� x = x+ D .

ü Infinitesimal Rotations

The generators of R were defined above as 3ä3 matrices Jx,y,z. Let the corresponding genertors to operate on states be labeled Lx,y,z. For example, they act on direction kets by n Ø n' � �1- idn ÿL� n .

ü L as a differential operator

For spatial translations, the differential form of the momentum operator was uncovered by showing that �x� p b = -i� ∑ÅÅÅÅÅÅÅ∑x �x � b . For Lz the analagous relation is �n� Lz b = -i� ∑ÅÅÅÅÅÅÅ∑f �n � b= -i��x ∑ÅÅÅÅÅÅÅ∑y - y ∑ÅÅÅÅÅÅÅ∑x � �n � b= �xx py - xy px� �n � b

More generally, a rotation around the k-axis (or in the i j -plane) is given by�n� Lk b = ei jk �xi p j - x j pi� �n � b

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A little effort shows that the algebra of the generators L is Li, L j� = i ei jk Lk

The generators can be written explicitly in terms of angular variables in the usual coordinates

Lx Ø -i��-sinf ∑ÅÅÅÅÅÅÅ∑q - cotq cosf ∑ÅÅÅÅÅÅÅ∑f �Ly Ø -i��cosf ∑ÅÅÅÅÅÅÅ

∑q- cotq sinf ∑ÅÅÅÅÅÅÅ

∑f�

LzØ -i� ∑ÅÅÅÅÅÅÅ∑f

These can be combined to yield the Casimir operator L2

L2 Ø -�2� 1ÅÅÅÅÅÅÅÅÅÅÅÅÅsin2 q

∑2ÅÅÅÅÅÅÅÅÅ∑2f +

1ÅÅÅÅÅÅÅÅÅÅsinq ∑ÅÅÅÅÅÅÅ∑q sinq ∑ÅÅÅÅÅÅÅ∑q �

It is not a coincidence that L2 and Lz are the differential operators that arise from performing the separation of

variables. Although I haven't proven it myself, I suspect strongly that there is an equivalence between the ability to separate variables and the ability to find a maximal set of commuting variables.

ü Eigenstates of L2, Lz

Eigenstates of angular momentum can be chosen to be simultaneous eigenvalues of J2 and Jz. For orbital angular

momentum we expect eigenstates of L2 and Lz. Suppose, therefore, that b is an eigenstate of Lz and L2 with eigenvalues m and l�l + 1� respectively. Then labeling the state by lmgives�n� Lz lm = -i� ∑ÅÅÅÅÅÅÅ

∑f �n � lm =m �n � lm

This equation is independent of q , so it is reasonable to separate variables in both the coordinates and the statesn = z, f = zä flm = lmä mwhere m-kets exist in the f-space, and lm-kets exist in the �z� or q-space. The notation lmsuggests that for each value of m, there are a set of states labeled by l . The allowed range of l is l ¥ m� . The amplitude �n � lm separates as

well�n � lm = �z � lm �f � mThe differential equation can then be rewritten by dividing through by �z � lm,

-i� ∑ÅÅÅÅÅÅÅ∑f �f � m =m �f � m

and solved�f � m = 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ����������2 p eimf

where the normalization is chosen so that

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�m � m' = �m� 1 m'= �m� �� „f f �f�� m'= � „f �m � f �f � m'= �0

2 p„f 1ÅÅÅÅÅÅÅÅ2 p e

i�m'-m� f= dmm'

Similarly, in the z-space, �z � lm is a function only of q . In the differential equation for L2 one can substitute the solution for Lz�n� L2 lm = -�2� 1ÅÅÅÅÅÅÅÅÅÅÅÅÅ

sin2 q ∑

2ÅÅÅÅÅÅÅÅÅ∑2f

+ 1ÅÅÅÅÅÅÅÅÅÅsinq ∑ÅÅÅÅÅÅÅ∑q sinq ∑ÅÅÅÅÅÅÅ

∑q� �n � lm

or�z� L2 lm = -�2� -m2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅsin2 f

+ 1ÅÅÅÅÅÅÅÅÅÅÅsinf ∑ÅÅÅÅÅÅÅ∑q sinf ∑ÅÅÅÅÅÅÅ∑q � �z � lm

which has for solutions the associated Legendre polynomials.�z � lm = clm Plm�z�

The normalization is given by the requirement�ml � lm = 1

= � „z �ml � z �z � lm

= � „z clm clm* �Pl

m�z��2

= clm2 2ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 l+1

�l+�m��!ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ�l-�m��!or, with a conventional choice of phase, clm = �-�m ���������������������������2 l+1ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 �l-�m��!ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ�l+�m��!Combining the two forms, gives the spherical harmonics

�n � lm = Ylm�q, f� = �-�m ���������������������������2 l+1ÅÅÅÅÅÅÅÅÅÅÅÅÅ4 p �l-�m��!ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ�l+�m��! eimf Pl

m�cosq�ü Orthogonality and completeness

There are a number of orthogonality and completeness relations regarding the states lm and the direction-kets n . Start with �n � n' = d�n, n'�1= � „W ' n' �n'�so that

yb�q, f� = �n� �� „W ' n' �n'�� b= � „W ' �n � n' �n' � b= � sinq ' „ q ' „f ' d�n, n'� �n' � b= yb�q, f�

To make the last step valid, the d-function needs to be defined as

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d�n, n'� = 1ÅÅÅÅÅÅÅÅÅÅÅsinq d�q '- q� d�f '- f� = d�cosq '- cosq� d�f '- f� = d�z'- z� d�f '- f�Next consider the lm states

1=Slm �lm� �lm��lm � l 'm' = �m � m' �ml � l 'm' = dmm' �ml � l 'm' = dmm' dl l '

The projection operators onto the l -subspace are defined by

Pl = Sm=-l

l

�l�m �ml �For a given m, one may also define a sum over l

Pm = Sl=m

¶

�l�m �ml �In fact, each of the operators Pm = 1 since in the q-space any of the sets �lmq� form a complete set. This can be seen

by considering an angular wavefunction with a defined value of m=m',

y = f �q� eim' f = �z, f � f = S

lm �z, f � lm �lm � f

= Slm � „W ' �z, f � lm �lm � z', f ' �z', f ' � f

= Slm � „z' „f ' �z, f � lm �ml � z' �m � f ' f �q '� eim' f'

= 1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ����������2 p Slm � „z' �z, f � lm �ml � z' f �q '� � „f ' ei �m'-m� f'

=��������

2 p Slm � „z' �z, f � lm �ml � z' f �q '� dmm'

=��������

2 p Sl � „z' �z, f � lm' �m'l � z' f �q '�

=��������

2 p Sl � „z' �z � lm' �m'l � z' f �q '� �f � m'

= � „z' Sl �z � lm' �m'l � z' f �q '� eim' f

.

Now, the last line requires that

f �q� = � „z' Sl �z � lm' �m'l � z' f �q '�

which is only valid for an arbitrary f if Sl �z � lm' �m'l � z' = �z � z' , which in turn requires that Pm' = S

l=m'

¶

�l�m' �m'l � = 1.

Since the derivation is independent of m', it follows that Pm = 1 for any m.

ü Some identities (in progress)

ü Spin (in progress)

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ü Eigenvalue problem using the algebra of the generators for the group

We have already discussed the eigenvalue problem in terms of orbital angular momentum. That discussion was couched in the language of solutions to partial differential equations with boundary conditions. In this discussion, we'll derive the eigenstate spectrum algebraically. This discussion is given in Merzbacher, chapter 11.

ü Raising and lowering (ladder) operators J±

It is useful to define the raising and lowering operators

J≤ = Jx ≤ i Jy

In general, after choosing a maximal set of commuting operators, it is possible to define linear combinations of the remaining generators which act as pairs of ladder operators. In the case of SO(3) there is just one such pair. Since they are linear combinations of generators, one has

J2, J≤� = 0.

The commutator with Jz is given by

Jz, J≤� = Jz, Jx ≤ i Jy�= Jz, Jx� ≤ i Jz, Jy�= i Jy ≤ i �-i Jx�= ≤J≤

And,

J+, J-� = Jx + i Jy, Jx - i Jy�= i� Jy, Jx�- Jx, Jy��= 2 Jz

Similar relations hold for other groups.

ü Effect of ladder operators on states

First, one can show that operating on a state � j, m� with a ladder operator does not change the value of j . For example,

0= J2, J+� � j, m�= �J2 J+ - J+ J2� � j, m�= �J2 - a j� J+ � j, m�

J+ � j, m� = 0 or J+ � j, m� is an eigenstate of J2 with eigenvalue a j .

Next, observe that when acting on a state � j, m� ladder operators act to raise or lower the value of m. For example, consider the operation of J+

0= � Jz, J+�- J+� � j, m�= �Jz J+ - J+ Jz- J+� � j, m�= �Jz-m- 1� J+ � j, m�

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It follows that either

J+ � j, m� = 0 orJ+ � j, m� = c jm

+ � j, m+ 1�i.e. either J+ annihilates the state, or J+ creates an eigenstate with the same value of j , but with an eigenvalue for Jz which is increased by 1. Similarly, J- lowers the eigenvalue of Jz

J- � j, m� = 0 orJ- � j, m� = c jm

- � j, m- 1�The coefficients c jm

+ and c jm- are as yet undetermined.

From this discussion one may see that starting from a particular state � j, m� one can generate a sequence of states with the same eigenvalue of J2 , and values of m that differ by integers.

ü Representations are finite dimensional

The next step is to show that the j -representations are finite dimensional. Consider

J2 - Jz2 = Jx

2 + Jy2

= 1ÅÅÅÅ2 �J+ J- + J- J+�= 1ÅÅÅÅ2 �J+ J+† + J+

† J+�where at the end, both terms are positive definite since

j, m ! J+† J+ ! j, m" = � j, m+ 1 � �c jm

+ �* c jm+ � j, m+ 1 = � c jm

+ �2It follows that

�J2 - Jz2� � j, m� = �a j -m2� � j, m� > 0 or �a j -m2� > 0. Since a j does not change through out the sequence, it

follows that for each j -representation m has both an upper bound mmax and a lower bound mmin .

ü Determination of a j

Consider

J- J+ = �Jx - i Jy� �Jx + i Jy�= Jx

2 + Jy2 + i Jx, Jy�

= J2 - Jz2 - Jz

so, when operating on the highest state of the representation

J- J+ � j, mmax� = �J2 - Jz2 - Jz� � j, mmax�

= �a j -mmax2 -mmax� � j, mmax�

= 0

since J+ � j, mmax� = 0. It follows that a j =mmax�mmax+ 1� . Similarly, operating with

rotations.nb: 11/3/04::13:48:13 13

J+ J- � j, mmin� = �J2 - Jz2 + Jz� � j, mmin�

= �a j -mmin2 +mmin� � j, mmin�

= 0

since J- � j, mmin� = 0. It follows that a j =mmin�mmin - 1� . There is also the constraint that mmax=mmin + n, where n is

some non-negative integer. These relations for a j , mmax, mmin can only be satisfied if mmin = -mmax. This, in turn, implies that mmax-mmin = 2mmax= n, or

mmax=nÅÅÅÅ2

This gives the desired result. The maximum value of m is integer or half-integer. Conventionally we label this quantum number j . The representations of SU(2) are labeled by j . They have dimension 2 j + 1. The states are labeled by a

second quantum number m, which runs from - j to j . The eigenvalue equations are then

Jz � j, m� =m � j, m�J2 � j, m� = j� j + 1� � j, m�ü Other matrix elements

From above� j, m � J+† J+ � j, m = � j, m+ 1 � �c jm

+ �* c jm+ � j, m+ 1 = � c+m

j �2whereas, we can also use the relation J- J+ = J+

† J+ = J2 - Jz2 - Jz

j, m ! J+† J+ ! j, m" = � j, m � J2 - Jz

2 - Jz � j, m= j� j + 1� -m�m+ 1�= � j -m� � j +m+ 1�

or

c jm+ =

����������������������������������������� j -m� � j +m+ 1�Similarly

c jm- =

����������������������������������������� j +m� � j -m+ 1�ü Explicit matrix form

It is often convenient to show the angular momentum operators in explicit matrix form. The form of the matrix depends

on the representation. For example, the figure shows matrices corresponding to J2 for the j = 1ÅÅÅÅ2 , 1, 3ÅÅÅÅ2 representations

�J2� j= 1ÅÅÅÅ2= �

2 ������ 3ÅÅÅÅ4 0

0 3ÅÅÅÅ4

������ , �J2� j=1 = �2

���������� 2 0 0

0 2 0

0 0 2

���������� , �J2� j= 3ÅÅÅÅ2

= �2

������������������

15ÅÅÅÅÅÅÅ4 0 0 0

0 15ÅÅÅÅÅÅÅ4 0 0

0 0 15ÅÅÅÅÅÅÅ4 0

0 0 0 15ÅÅÅÅÅÅÅ4

������������������

rotations.nb: 11/3/04::13:48:13 14

The matrices operate on a �2 j + 1�-dimensional state vector, which defines a state of the j -representation as a linear combination of the jmstates. Since J2 is a Casimir operator, it is diagonal and proportional to the identity matrix in

each representation, but with a different eigenvalue. Jz can be simultaneously diagonalized with J2 , but has different eigenvalues for each state. The matrix form for Jz for the j = 1ÅÅÅÅ2 , 1, 3ÅÅÅÅ2 representations is

�Jz� j= 1ÅÅÅÅ2= �

������ 1ÅÅÅÅ2 0

0 - 1ÅÅÅÅ2

������ , �Jz� j=1 = �

���������� 1 0 0

0 0 0

0 0 -1

���������� , �Jz� j= 3ÅÅÅÅ2

= �

������������������

3ÅÅÅÅ2 0 0 0

0 1ÅÅÅÅ2 0 0

0 0 - 1ÅÅÅÅ2 0

0 0 0 - 3ÅÅÅÅ2

������������������

The other generators, chosen to be either Jx, Jy or J+, J- , are not diagonal in the basis where Jz is diagonal, but the matrix entries are easily determined by the c+, c- coefficients given above. For example, the matrix form for J+ for the

j = 1ÅÅÅÅ2 , 1, 3ÅÅÅÅ2 representations is

�J+� j= 1ÅÅÅÅ2= �

���� 0 1

0 0���� , �J+� j=1 = �

�����������

0�����

2 0

0 0�����

2

0 0 0

����������� , �J+� j= 3ÅÅÅÅ2

= �

�����������������

0�����

3 0 0

0 0 2 0

0 0 0�����

3

0 0 0 0

�����������������

J- is similar, but on the lower off-diagonal. Jx =1ÅÅÅÅ2 �J+ + J-� , Jy = -

iÅÅÅÅ2 �J+ - J-� are determined by adding the matrix components, entry by entry.

� Angular momentum addition

Suppose one has two particles and it is known that one has angular momentum j1 and the second has angular momentum j2 . It is a common queston to ask, "What are the possible angular momentum states for the combined

system. Or it may be that one knows the orbital angular momentum and spin of a particle, but it is needed to know the total angular momentum. A third case is that one knows the spin of two particles, but wants to know the total spin for the system. All three cases are mathematically equivalent.

Perhaps even more important than forming a state from two angular momentum degrees of freedom, is determining the

resulting angular momentum when operating on a state with a "spherical-tensor" operator, such as occurs when calculating the rates and selection rules for radiative transitions. Tensor operators will be addressed in the following section, but the algebraic concepts are essentially identicle to those developed for the addition of angular momentum

presented here.

The notes below consider the general case of adding two angular momentum degrees of freedom, described by the operators J1 and J2 .

rotations.nb: 11/3/04::13:48:13 15

ü Product of two spaces

As usual, when there are two degrees of freedom, one can describe the full set of states for a system as the direct

product of the states for the two subspaces. In this case, the states of the J1 operator are given by � j1 m1� and those of J2 by � j2 m2� . Where j i describes the eigenstate with respect to the casimir operator Ji

2 and mi is the eigenvalue of Ji z . The two operators commute with each other J1, J2� = 0, so a full set of commuting opertors is �J1

2, J22, J1 z, J2 z� . A

state for the full system is then given by the direct product

� j1 m1 j2 m2� = � j1 m1�≈ � j2 m2�If it is known that the first particle has j1 and the second j2 , then there are n1 = 2 j1 + 1 possible states for the first, n2 = 2 j2 + 1 possible states for the second, and the full system has n = n1 n2 = �2 j1 + 1� �2 j2 + 1� possible states, with varying values of m1 and m2 .

ü total J

It often happens that one is interested in the total angular momentum. The total angular momentum operator is given by

J = J1 + J2 .

where the vector nature of the operators has been emphasized. The states of total angular momentum are defined by the operators J2 and Jz

J2 = J12 + J2

2 + 2 J1 ÿ J2

Jz = J1 z+ J2 z

There are many cases where the Hamiltonian includes a term of the form J1 ÿJ2 , for example the spin-orbit coupling is

proportional to L ÿS. In this case the relation

J1 ÿJ2 =1ÅÅÅÅ2 �J2 - J1

2 + J22�

is useful. It is also possible to write J1 ÿ J2 in terms of raising and lowering operators

J1 ÿJ2 = J1 z J2 z+1ÅÅÅÅ2 �J1+ J2- + J1- J2+�

ü two alternative basis sets for the j1, j2 representation

The last relation makes it clear that because J2contains pieces that raise and lower the states of J1 z and J2 z, eigenstates j1 m1 j2 m2 , are not generally eigenstates of J2 . Such states are, however, eigenstates of Jz. The effect of �J1+ J2- + J1- J2+� is to leave the total m=m1 +m2 unchanged, since if m1 is raised m2 is lowered, and vice versa.

It appears that there are two ways one can describe the angular momentum states of the system. First, in the J1 z J2 z scheme one chooses 4 commuting operators J1

2, J22, J1 z, andJ2 z. In this case the states are labeled by j1 m1 j2 m2 .

Alternatively, in the J, Jz scheme a different set of four operators J2, J12, J2

2, andJz define the states. It is

straightforward to show that J2, Ji2� = Jz, Ji

2� = 0 for i = 1, 2. In this case the states are labeled by j m j1 j2 .

rotations.nb: 11/3/04::13:48:13 16

The values of j1 and j2 are common to both schemes. Accordingly, if the values of j1 and j2 are well defined, it is common to list them first or to omit them entirely. In the J1 z J2 z scheme one has, j1 j2 m1 m2 , or uses the short hand m1, m2 . Corespondingly, in the J Jz scheme one has j1 j2 j m or j m .

ü definition of clebsch gordon coefficients

Since j1 and j2 are valid in either representation, it follows that the n= �2 j1 + 1� �2 j2 + 1� states present in the m1, m2 scheme must be rearranged into the same number of states in the jm scheme, although the allowed values of

j and m are still undetermined. It follows that the � j1 j2 jm� states can be written as linear combinations of j1 j2 m1 m2 states,

j1 j2 jm = Sm1 m2

j1 j2 m1 m2 � j1 j2 m1 m2 � j1 j2 jm= S

m1 m2

c jm,m1 m2

j1 j2 j1 j2 m1 m2where the last line defines the Clebsch-Gordon coefficients.

ü Reduce product of j1, j2 to irreducible representations of J

The next step is to determine which representations of total angular momentum are found in the product of j1≈ j2 . The key to this is to consider a) the multiplicity of states with a particular value of m, and b) to realize that the representations of j must be complete, implying that all states from m= - j to j must be present.

Without any loss of generality, take j1 ¥ j2 and define D j = j1 - j2 . Then the multiplicity of states with eigenvalue m is given by

Nm =j1 + j2 + 1- � m � � m � ¥ D j

2 j2 + 1 � m � § D j

The highest m-state has m= j1 + j2 . The increase in multiplicity with decreasing m corresponds to the different ways in which J1- and J2- can be applied to decrease the total m. The increase stops when m has been decreased by 2 j2 + 1

steps, since the number of times J2- can be applied is limited. The multiplicity holds steady until a similar constraint applies to J1- at which point the multiplicity decreases until m= -� j1 + j2� . The multiplicity is symmetric under mØ -m.

The highest m-state is unique. Since the representations must be complete, the highest m-state must be accompanied by

a set of states j m for j = j1 + j2 with �2 j + 1� values of m, - j <m< j , included. Specifically, there are two states with m= j1 + j2 - 1, and one of these belongs to the j = j1 + j2 representation. The other heads a representation with j = j1 + j2 - 1. Similarly, there are three states with m= j1 + j2 - 2, one of which heads a new representation with

j = j1 + j2 - 2.

All told this procedure results in complete representations with j1 + j2 ¥ j ¥ j1 - j2 . The total number of states in these representations is

rotations.nb: 11/3/04::13:48:13 17

N = Sj= j1+ j2

j= j1- j2 2 j + 1

= Si=2 j2

i=0 2 �D j + i� + 1

= �2 D j + 1� �2 j2 + 1� + 2 Si=2 j2

i=0 i

= �2 D j + 1� �2 j2 + 1� + �2 j2� �2 j2 + 1�= �2 � j1 - j2� + 1+ 2 j2� �2 j2 + 1�= �2 j1 + 1� �2 j2 + 1�

which is equal to the number of states in the product representation j1 j2 m1 m2 .

ü Procedure for calculating CG coeffs

It remains to find the Clebsch-Gordon coefficients. This is done iteratively by repeated use of the operator

J- = J1- + J2- . The process begins with the observation that there is only one state with maximal total m, and it belongs to the representation j = j1 + j2 ,

j1, j2, j = j1 + j2, m= j1 + j2 = j1, j2, m1 = j1, m2 = j2Proceeding with the application of J-

J- j1, j2, j = j1 + j2, m= j1 + j2 = �J1- + J2-� j1, j2, m1 = j1, m2 = j2c j1+ j2, j1+ j2- j1, j2, j = j1 + j2, m= j1 + j2 - 1 =

c j1, j1- j1, j2, m1 = j1 - 1, m2 = j2 + c j2, j2

- j1, j2, m1 = j1, m2 = j2 - 1

where the c jm- =

����������������������������������������� j +m� � j -m+ 1� were defined above. Dividing through by c j1+ j2, j1+ j2- , we obtain the second

highest m-state of the j = j1 + j2 representation, j1, j2, j = j1 + j2, m= j1 + j2 - 1 =cj1, j1-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1+ j2, j1+ j2- j1, j2, m1 = j1 - 1, m2 = j2 + cj2, j2

-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅcj1+ j2, j1+ j2- j1, j2, m1 = j1, m2 = j2 - 1

Comparing to the definition of the CG coefficient, for the case j = j1 + j2, m= j1 + j2 - 1, m1 = j1 - 1, m2 = j2

c j1+ j2, j1+ j2-1; j1-1, j2j1 j2 =

cj1, j1-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1+ j2, j1+ j2- =�������������j1ÅÅÅÅÅÅÅÅÅÅÅÅÅÅj1+ j2

and similarly

c j1+ j2, j1+ j2-1; j1, j2-1j1 j2 =

cj2, j2-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1+ j2, j1+ j2- =�������������j2ÅÅÅÅÅÅÅÅÅÅÅÅÅÅj1+ j2

In response to a question in class, if j1 � j2 the j = jmax representation is predominately composed of the larger of the j1 and j2 representations. At least in this case it is clear that the CG coefficients are normalized so that

Sm1

c jm;m1,m-m1

j1 j2 �2The other m= j1 + j2 - 1 state, which plays the role of the head of the j = j1 + j2 - 1 representation is given by j1, j2, j = j1 + j2 - 1, m= j1 + j2 - 1 =

cj2, j2-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1+ j2, j1+ j2- j1, j2, m1 = j1 - 1, m2 = j2 - cj1, j1

-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅcj1+ j2, j1+ j2- j1, j2, m1 = j1, m2 = j2 - 1

rotations.nb: 11/3/04::13:48:13 18

and is explicitly orthogonal to the j = j1 + j2 state.

This procedure can now be repeated applying J- to the two states above to fill out the j = j1 + j2 and j = j1 + j2 - 1

representations. A third state may be constructed, orthogonal to the first two, which will head the j = j1 + j2 - 2 representation. It may appear a bit awkward to create the new orthogonal state, but one can use the following construction. Suppose one has an n-dimensional vector space spanned by an orthonormal basis � i� , and n- 1linear combinations of these which are orthogonal and normalized, which can be labeled by index a = S

i ai i . Then an nth

state can be added to the a-basis which will be orthogonal to the others by constructionan = ei1 i2 …in ai11 ai2

2 …ain-1n-1 in

ü Example of two spin 1/2 reps

As an explicit example of angular momentum addition, consider the case of adding two spins of s1 = s2 =1ÅÅÅÅ2 . Since s1

and s2 are specified I'll adopt the shorter notation where they are suppressed.smsm= m1 m2m1 m2

In the m1 m2-basis there are four states � 1ÅÅÅÅ2 , 1ÅÅÅÅ2 �, � 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 �, � - 1ÅÅÅÅ2 , 1ÅÅÅÅ2 �, and � - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 �In the s-basis there are representations extending from smax = s1 + s2 to smin = s1 - s2 . In this case, there are just two representations smax= 1 and smin = 0. The highest m-state has m= 1 and belongs to the highest s-representation s= 1

�11�sm= � 1ÅÅÅÅ2 1ÅÅÅÅ2 �

m1 m2

Acting with S- = S1- +S2- on both sides gives the algebraic

S- �11�sm= �S1- +S2-� � 1ÅÅÅÅ2 , 1ÅÅÅÅ2 �m1 m2

c11- �10�sm= c 1ÅÅÅÅ2

1ÅÅÅÅ2- �- 1ÅÅÅÅ2 , 1ÅÅÅÅ2 �

m1 m2+ c 1ÅÅÅÅ2

1ÅÅÅÅ2- � 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 �

m1 m2�����2 �10�sm= �- 1ÅÅÅÅ2 , 1ÅÅÅÅ2 �

m1 m2+ � 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 �

m1 m2�10�sm=1ÅÅÅÅÅÅÅÅÅÅ������2 �- 1ÅÅÅÅ2 , 1ÅÅÅÅ2 �

m1 m2

+ 1ÅÅÅÅÅÅÅÅÅÅ������2 � 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 �

m1 m2�10�sm= Sm1

c10;m1 m2

1ÅÅÅÅ2 1ÅÅÅÅ2 �m1, m2�

where in the third line, the result csm- =

����������������������������������������s+m� �s-m+ 1� has been used. In general, it is noted that for the highest m-state c j j

- =��������

2 j . In the last line, the sum over m-states defines the Clebsch-Gordon coefficients. Evidently

c10; 1ÅÅÅÅ2 ,- 1ÅÅÅÅ2

1ÅÅÅÅ2 1ÅÅÅÅ2 =�������1ÅÅÅÅ2 and c

10;- 1ÅÅÅÅ2 , 1ÅÅÅÅ2

1ÅÅÅÅ2 1ÅÅÅÅ2 =�������1ÅÅÅÅ2

This result could have been written down from the general result above for the CG coefficients of the m= j1 + j2 - 1

state of the j = j1 + j2 representation.

Having determined the 10sm state, it remains to determine the orthogonal m= 0 linear combination, which in this case is

rotations.nb: 11/3/04::13:48:13 19

00sm=1ÅÅÅÅÅÅÅÅÅÅ������2 - 1ÅÅÅÅ2 , 1ÅÅÅÅ2

m1 m2- 1ÅÅÅÅÅÅÅÅÅÅ������

2 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2

This result aslo defines the CG coefficients

c00; 1ÅÅÅÅ2 ,- 1ÅÅÅÅ2

1ÅÅÅÅ2 1ÅÅÅÅ2 = -�������1ÅÅÅÅ2 and c

10;- 1ÅÅÅÅ2 , 1ÅÅÅÅ2

1ÅÅÅÅ2 1ÅÅÅÅ2 =�������1ÅÅÅÅ2

Note that there is an arbitrariness to which of these CG coefficients gets the minus sign.

The next step is to apply S- again to determine the m= jmax- 2 states.

S- 10sm= �S1- +S2-� 1ÅÅÅÅÅÅÅÅÅÅ������2 - 1ÅÅÅÅ2 , 1ÅÅÅÅ2

m1 m2+ �S1- +S2-� 1ÅÅÅÅÅÅÅÅÅÅ������

2 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2

c10- 1, -1sm=

1ÅÅÅÅÅÅÅÅÅÅ������2 �0+ c 1ÅÅÅÅ2

1ÅÅÅÅ2- - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2� + 1ÅÅÅÅÅÅÅÅÅÅ������

2 �c 1ÅÅÅÅ2

1ÅÅÅÅ2- - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2+ 0������

2 1, -1sm=1ÅÅÅÅÅÅÅÅÅÅ������2 - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2+ 1ÅÅÅÅÅÅÅÅÅÅ������

2 - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m21, -1sm= - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 m1 m2

which makes sense, since the lowest m-state should also be defined uniquely.

As a check, applying S- to the 00sm state

S- 00sm= �S1- +S2-� 1ÅÅÅÅÅÅÅÅÅÅ������2 - 1ÅÅÅÅ2 , 1ÅÅÅÅ2

m1 m2- �S1- +S2-� 1ÅÅÅÅÅÅÅÅÅÅ������

2 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2

0= 1ÅÅÅÅÅÅÅÅÅÅ������2 �0+ c 1ÅÅÅÅ2

1ÅÅÅÅ2- - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2�- 1ÅÅÅÅÅÅÅÅÅÅ������

2 �c 1ÅÅÅÅ2

1ÅÅÅÅ2- - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2+ 0�

0= 1ÅÅÅÅÅÅÅÅÅÅ������2 - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2- 1ÅÅÅÅÅÅÅÅÅÅ������

2 - 1ÅÅÅÅ2 , - 1ÅÅÅÅ2

m1 m2= 0

yields the null state on both sides.

Another check is to apply S2 = S12 + S2

2 + 2 S1 ÿS2 = S12 + S2

2 + 2 S1 z S2 z+1ÅÅÅÅ2 �S1+ S2- +S1- S2+� to all four states. For the

triplet states S2 = 2, so

�S12 + S2

2 + 2 S1 z S2 z+ �S1+ S2- +S1- S2+�� � 1ÅÅÅÅ2 1ÅÅÅÅ2 � = � 3ÅÅÅÅ4 +

3ÅÅÅÅ4 + 2 1ÅÅÅÅ2 1ÅÅÅÅ2 + �0+ 0�� � 1ÅÅÅÅ2

1ÅÅÅÅ2 � = 2 � 1ÅÅÅÅ2 1ÅÅÅÅ2 ��S1

2 + S22 + 2 S1 z S2 z+ �S1+ S2- +S1- S2+�� �- 1ÅÅÅÅ2 -

1ÅÅÅÅ2 � = � 3ÅÅÅÅ4 +3ÅÅÅÅ4 + 2 �- 1ÅÅÅÅ2 � �- 1ÅÅÅÅ2 � + �0+ 0�� �- 1ÅÅÅÅ2 -

1ÅÅÅÅ2 � = 2 �- 1ÅÅÅÅ2 -1ÅÅÅÅ2 �

and for the m= 0state�S12 + S2

2 + 2 S1 z S2 z+ �S1+ S2- +S1- S2+�� � 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 � + �- 1ÅÅÅÅ2 , 1ÅÅÅÅ2 � =� 3ÅÅÅÅ4 +3ÅÅÅÅ4 - 2 1ÅÅÅÅ2

1ÅÅÅÅ2 + �1+ 0�� � 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 �+ �- 1ÅÅÅÅ2 , 1ÅÅÅÅ2 � = 2 � 1ÅÅÅÅ2 , - 1ÅÅÅÅ2 �+ �- 1ÅÅÅÅ2 , 1ÅÅÅÅ2 �where the term �S1+ S2- +S1- S2+� acts to exchange the two spins, which yields (+1) for a symetric state.

For the singlet state, S2 = 0. Comparing to the 10 calculation, only the exchange term changes sign, which results in � 3ÅÅÅÅ4 +3ÅÅÅÅ4 - 2 1ÅÅÅÅ2

1ÅÅÅÅ2 - �1+ 0�� = 0, so everything checks.

A few comments: Comparing to the discussion of identicle particles, the S= 0 state is antisymmetric and is a singlet, whereas the S= 1states are symmetric and form a triplet. In general, when adding angular momentum for the case j1 = j2 , the jmax representation is a symmetric under exchange of m1 ¨ m2 , the j = jmax- 1 representaion will be

anti-symmetric, and the lower j representations alternate accordingly.

rotations.nb: 11/3/04::13:48:13 20

� Tensor operators

The discussion so far has focussed on the effect of rotations on states. In a position basis, the states are transformed in an obvious way: the coordinates describing the position ket are transformed. However, since position kets are not

typically eigenstates of the Hamiltonian, it is typical to consider the eigenstates of angular momentum, described by the quantum numbers j, m. Under rotations the quantum number j does not change, but m does. One may say that the set of states with the same j constitutes a representation consisting of �2 j + 1� components. The components are mixed up by

rotations, but the representations are not.

As usual in quantum mechanics, transformations may affect operators as well as states. For rotations, the operators may be charactrized either in a coordinate representation or in terms of spherical tensor operators which are conveniently described in the same language used for angular momentum states. A spherical tensor operator of rank k consists of �2 k+ 1� components, each identified by an azimuthal number q. The relation between identifiers for the states and operators is k¨ j and q¨m. The use of k, q is purely conventional to cue the association with an operator as opposed to a state. One may think of acting with a spherical tensor operator on a state as "adding" angular momentum to the

state, and the algebra for this angular momentum addition is similar to that for combining the angular momentum content of two states, as described above. Similarly, spherical tensor operators may be combined to form new operators with different angular momentum content.

The material below proceeds in two steps. The first discusses properties of rotations on simple operators: a) a general

discussion of the effect of rotations on operators, b) an explicit application to the behavior of rotations on themselves and c) specializing to the effect of rotations on the generators of rotations Ji leading to d) a more general discussion of vector operators under rotations, and e) isolating that discussion to the effect of infinitesimal rotations on a vector

operator. Then the second step begins with f) a generalization of vector operators to a definition of tensor operators, g) their transformations under infinitesimal rotations, h) their addition properties, i) the result of operating on states with tensor opertors, and culminating with j) a discussion of the Wigner-Eckart theorem.

ü General remarks

To focus on the effects of rotations, consider the position operator X����= �X, Y, Z� and the set of position kets x� , which

may be used as a basis set to specify both the state a of a system and the operators acting on those states. In terms of the position kets, one may expand

a = � „ x� x� �x� � aXi = � „ x� x� xi �x��

In this form the position operator is diagonal, and has been written in component form. The expectation for Xi in a particular state is

�Xi = �a � Xi � a

rotations.nb: 11/3/04::13:48:13 21

At this point it is important to distinguish the effects of rotations in three distinct scenarios. First, one may rotate the states aØ a ' = R a . This does not involve changing the position basis kets, but it does imply a transformation of

the expansion coefficients �x� � a . Second, one may rotate the position operators Xi Ø X 'i = Ri j X j = � „ x� x� Ri j x j �x�� . Again, this does not involve a change of the basis kets, merely a different linear combination of the operators �X, Y, Z� . The new operators are still diagonal, but with different eigenvalues. If one leaves the states unchanged, but rotates the

operators, then the expectation values change as a classical position vector would change under rotations �XiØ Ri j �X j . Third, one may change coordinate systems. In this case, the state of the system a is unchanged, but the basis kets for describing the system are changed: � „ x� x� �x� � Ø � „ x� x� ' �x� ' � , where x� ' = R x� . Note that the

integration is still over all position kets. In addition, the position operators are also transformed so that, for example, the x-direction transforms to the x-direction in the new basis. A change of viewer should not change the expectation value for an observable, and indeed

Xi = � „ x� x� xi �x�� Ø � „ x� x� ' Ri j x j �x� '�= � „ x� ' �a x� ' xi ' �x� '� a= � „ x� �a x� xi �x�� a= Xi

where in the second line one uses the equivalance of the integration volumes � „ x� = � „ x� ', and in the third line a change of variables from xØ x' is performed. The result of these considerations is that the effect of a rotation on the position operator can be described either by a transformation of the basis states,

Xi ' = � „ x� x� ' xi �x� '� = RXi R-1

or by a transformation of the operators

Xi ' = � „ x� x� ' Ri j x j �x� '� = Ri j X j

Since the symbol R is used in different fashions (either to act on the basis states or in the space of operators) there is

some danger of confusion! Fortunately, the D-notation is established to describe more generally how rotations operate on different repesentations, so one may use the notation Xi ' = DXi D-1 instead. Further, although convenient to use the position basis for pedagogical purposes, any complete set of basis kets would do, so one may replace � „ xØ S

jm. In

addition, since rotations do not mix states from different representations, the D-operators can be chosen to be finite

dimensional, so as to apply to a single irreducible representation at a time.

ü The effect of rotations on rotations

Suppose one has a state a and two rotations R1 and R2 . Define the state a1 = R1 a , and the state a2 = R2 a1 = R2 R1 a . In this picture, R2 acts to rotate the state a1 . Alternatively, one could view the action of R2

to be a rotation on the system where R1has operated, i.e.

R2�R1 a� = R1 ' a '

where R1 ' defines the action of the rotation R1 as viewed from the coordinate system of R2 , and a ' = R2 a . The latter relation can also be written in the inverse form a = R2

-1 a ', and so

R2 R1 a = R2 R1 R2-1 a '

rotations.nb: 11/3/04::13:48:13 22

Comparing the two results, one finds that the rotations themselves transform as

R1 ' = R2 R1 R2-1

ü Alternative

Consider the effects of a rotation R1 on the matrix element �b � R � a . In the picture where the transformation is considered to be a change of coordinate system, the matrix element should be left unchanged by the transformation. Using 's to denote the transformed states and operators, and denoting the transforming rotation by R1 �b � R � a fl �b' � R' � a'

= �b � R1-1 R' R1 � a

one finds R= R1-1 R' R1 or

R' = R1 RR1-1

as above.

ü Application to Euler angles.

In the discussion of Euler angles, it was stated that

Rz'�g� Ry'�b� Rz�a� = Rz�a� Ry�b� Rz�g�where the 's indicated a rotation around the body axis, for example Ry' is the rotation around the y-axis as seen after the rotation by Rz�a� . Given the transformation properties of rotations, one may write

Ry'�b� = Rz�a� Ry�b� Rz-1�a�

Defining the combined rotation R�a, b� = Rz�a� Ry�b�Rz'�g� Ry'�b� Rz�a� = Rz'�g� Rz�a� Ry�b� Rz

-1�a� Rz�a�= Rz'�g� Rz�a� Ry�b�= Rz'�g� R�a, b�= R�a, b� Rz�g� R-1�a, b� R�a, b�= R�a, b� Rz�g�= Rz�a� Ry�b� Rz�g�

as advertised.

ü The effect of rotations on the generators of rotation

The generators of rotations Ji constitute a three component vector, with the property that n ÿ J acts as the generator for

rotations around an arbitrary axis n. It seems reasonable that these should transform in a manner similar to the position operators. To do this consider the relation R' = R1 RR1

-1 in the case where R is infinitesimal, R= e-ie nÿJ � 1- ie n ÿ J . On the left, one has R' = 1- ie n ÿ J ' with Ji ' = Ri j J j , and on the right R1 RR1

-1 = R1�1- ie n ÿ J� R1-1 . Equating the two

sides, one finds the same relation that was obtained for the position vector

rotations.nb: 11/3/04::13:48:13 23

Ji ' = �R1�i j J j = D1 Ji D1-1

where on the left-hand side R mixes the different operators, and on the right-hand side the D notation has been adopted to emphasize the transformation of the basis kets used to describe the vector space.

To see the difference in the two determinations of Ji ', consider a particular component, say, the 11component of J1 '. On the left, J1 '11 is made up of the 11 components of J1, J2, andJ3 weighted by R11, R12, andR13 respectively. On the

right, J1 '11 is made from a linear combination of all 9 components of J1 , but the components of J2 , and J3 are not used.

ü Vector operators

The behavior of X���� and J

�� can be generalized to define a vector operator V

��� as a set of three operators Vi which under a

rotation transform as

Vi Ø V 'i = Ri j V j = RVi R-1

It is useful to consider the infinitesimal form of this relation. On the left,

Ri j V j = �1- ie n ÿ J�i j V j = �di j - ie nk Jki j � V j

where �Jk�i j = -ieki j . On the right,

RVi R-1 = �1- ie n ÿ J� Vi�1- ie n ÿ J� = Vi - ie nk Jk, Vi�Note that in this equation J operates on the components of an individual Vi as opposed to mixing the compenents of V . Equating the two, the requirement for V to be a vector operator is that the components of V obey the commutator relations

Jk, Vi� = -ieki j V j

The general vector operator can be defined by commutation relations in analogy to the angular momentum commutation relations.

This can be checked explicitly for the case of orbital angular momentum acting on vector operators X���� and P

��.

ü Cartesian vectors vs "spherical vector"

One may recall that the description of angular momentum states is simplified by introducing the raising and lower operators J≤ , and then choosing J0, J≤ (as opposed to Jx, Jy, Jz) to be the basis for the space of angular momentum

generators. A similar redefinition can be made for vector operators, but before proceeding, it is useful to introduce a transformation of cartesian coordinates which parallels the transformation from Jx, Jy, Jz to J0, J≤ .

Define coordinates xq with q taking the values of q = 1, 0, -1, by x0 = z, x≤ =1ÅÅÅÅÅÅÅÅÅÅ������2 �x≤ i y� . I don't have a name for

these coordinates. In analogy to the spherical tensors to be defined below, I'm tempted to call these coordinates

"spherical" coordinates, but that name is already taken. Spherical cartesian? Ugh. In any event, the dot product of two vectors can be rewritten as u ÿ v= ui vi = uq v-q . The new coordinates can be expressed as a transformation of the usual cartesian coordinates, xq = U xi , where the transformation matrix is

rotations.nb: 11/3/04::13:48:13 24

U =

��������������

1ÅÅÅÅÅÅÅÅÅÅ������2

- iÅÅÅÅÅÅÅÅÅÅ������2

0

0 0 11ÅÅÅÅÅÅÅÅÅÅ������2

iÅÅÅÅÅÅÅÅÅÅ������2

0

��������������

The next step is to re-examine the defining relation for rotations in cartesian coordinates x'i = Ri j x j = �e-ifnÿJ�i j x j . First, the new coordinates can be used to rewrite the generator: n ÿ J = n-q J

èq , where J

è≤ =

1ÅÅÅÅÅÅÅÅÅÅ������2 J≤ is defined to account

for the difference in definition between the new coordinates and the conventional definition of J≤ . As written, the

rotation is operating on the usual cartesian coordinates, so the generators must be written in that basis as well, for example,

Jè≤ =

1ÅÅÅÅÅÅÅÅÅÅ������2

���������� 0 0 �1

0 0 -i

≤1 i 0

����������

Second, the new coordinates can be used as the object of the rotations

x'q = Rqq' xq' = �e-ifnÿJ�qq' xq'

Again there is the choice of writing n ÿ J as ni Ji or n-q Jè

q . In either case, though, J must be written in the form for operating on xq instead of xi . Following the usual rules for transforming operators J 'i = U Ji U† or J

è'q = U J

èq U† . The

J ' have the same form as when operating on the j = 1 representation, for example,

J '0 = Jz�l=1� =

���������� -1 0 0

0 0 0

0 0 1

���������� , Jè'+ =

1ÅÅÅÅÅÅÅÅÅÅ������2 J+

�l=1� =���������� 0 1 0

0 0 1

0 0 0

����������

(5/11/04 - I am not sure I don't have some "typos" still in this section)

ü Spherical vector

The definition of a spherical vector operator can now be given in terms of the components of a cartesian vector operator. Specifically, if Vi is a vector operator, then the spherical vector with components V0, V≤can be defined by

V0 = Vz

V≤ =1ÅÅÅÅÅÅÅÅÅÅ������2 �Vx ≤ iVy�

Later this object will be identified as a rank-1 spherical tensor Tq1 where q = �1, 0,-1� , and T0

1 = V0 , T≤11 = V≤ .

Expressing the effect of the rotation on Vq by its effect on the eigenstates

RVq R-1 = �1- ie n ÿ J� Vq�1- ie n ÿ J� = Vq - ie n-m#Jèm, Vq$where m has been used to distinguish the generator from the spherical vector component. Meanwhile, rotating the

operator components yields

Rqq' Vq' = �1- ie n ÿ J�qq' Vq' = �dqq' - ie n-m Jmqq'� Vq'

equating the two results, for each m one has,

rotations.nb: 11/3/04::13:48:13 25

#Jèm, Vq$ = Jè

qq'm Vq'

On the left, the exact form of Jèm

and Vq depend on the representation upon which the operators act. On the right, Vq is defined relative to the representation but J

èq q'm

is always the j = 1representation since it is acting on a vector operator.

Explicit forms for Jèm

are

Jè0=

���������� -1 0 0

0 0 0

0 0 1

���������� , Jè+=

���������� 0 1 0

0 0 1

0 0 0

���������� , and J

è-=

���������� 0 0 0

1 0 0

0 1 0

����������

As an example, suppose Vq is the angular momentum operator itself, and that the equation is applied to the j = 2 representation. Then, the Vq are 5ä5matrices, and the J

èmin the commutator is also 5ä5, but the J

èqq'm

on the right is 3ä3

to mix the components of the vector operator.

ü Cartesian tensors

One can form more complicated tensors. For example, Ti j = Xi X j is a rank-2 tensor. A rank-n tensor has n-indicies,

Ti1 i2 …in , and transforms under rotations as

Ti1 i2 …in Ø T ' i1 i2 …in = Ri1 j1 Ri2 j2 …Rin jn T j1 j2 … jn

This rank-n tensor has 3n components. It may be reducible, since one might be able to find subsets of the components of T which are invariant under rotation.

ü Break cartesian into irreducible tensors

For example, one can form a 9-component rank-2 tensor Oi j from two vectors Ui andV j

Oi j = Ui V j

but this tensor can be reduced to three irreducible tensors OS, OV, OQ , known as scalar, vector and quadrupole tensors. Their explicit construction is

OS = di j Oi j , OiV = ei jk O jk , and Oi j

Q = 1ÅÅÅÅ2 �Oi j +O ji � - 1ÅÅÅÅ3 OS

These three tensors are invariant under rotations. In terms of labels, the reduction of U V can be expressed as

V ä V = S+V +Q

S, V, Q have 1,3, and 5 independent components respectively, so one may express the reduction as

3 ä 3 = 1+ 3+ 5

Lastly, one may identify S, V, Q as operators associated with irreducible representtions of angular momentum

corresponding to j = 0, 1, 2, respectively. Then the break down of O= U V can be understood in terms of anglar momentum addition�1� ä �1� = �0� + �1� + �2�

rotations.nb: 11/3/04::13:48:13 26

where the labels are j -values. This provides the rational for generalizing to spherical tensors.

ü Spherical tensors

The spherical vector was defined through its properties under rotations. For an infinitesimal rotation

Rqq' Vq' = �1- ie n ÿ J�qq' Vq' = �dqq' - ie n-m Jmqq'� Vq'

Correspondingly, a spherical tensor of rank-k is Tk with �2 k+ 1� components Tqk . Tq

k transforms as

Tqk Ø T 'q

k = Rqq' Tq'k = �1- ie n ÿ J�qq' Tq'

k = �dqq' - ie n-m J�k� mqq'� Tq'

k

where the label �k� has been added to Jm to make explicit that the representation of J must match the rank of the tensor. Although the term rank-k is used, it should be understood that this corresponds to an irreducible representation-k of angular momenetum

ü Spherical tensor operators

Motivated by the definition of a sperical vector operator, one can define a spherical tensor operator Tk with components Tq

k , by the behavior of the operators when transformed by a rotation. The expectation value of an operator when the

system is in a particular state is given by

�Tqka= �a � Tq

k � aUnder rotations,

Tqk Ø RTq'

k R-1

A reasonable definition of a spherical tensor operator is that the expectation value transforms under rotations in the same way that a classical tensor would. This should be true for any state a and therefore should be a property of the

operator, i.e.

Tqk Ø Dqq'

k Tq'k

ü Definition of tensor operators by their commutation properties

By considering infinitesimal rotations one can define the tensor operators in terms of commutation properties with Jm. Thus, for an infinitesimal rotation, the form RTq'

k R-1 becomes

RTqk R-1 Ø Tq

k - ie n-m#Jèm, Tqk$

Similarly Dqq'k Tq'

k Ø Tqk - ien-m J�k� m

qq'� Tq'k , which requires knowledge of J�k� m

qq' . These operators are known from the

action of Jz, J≤ on angular momentum states in different j -representations,

J�k� 0qq' = dqq' q and J�k�≤

qq' = ckq≤ dq,q'≤1

The definition of a tensor operator can then be specified in terms of how Jz, J≤ act on a tensor operator through the commutation relations

rotations.nb: 11/3/04::13:48:13 27

Jz, Tqk� = qTq

k J≤, Tqk� = ckq

≤ Tq≤1k

where ckq≤ =

���������������������������������������k� q� �k≤ q+ 1� , as for when the angular momentum generators act on states j = k, m= q .

ü Combining tensor operators

Just as angular momentum states can be combined via the procedure for angular momentum addition, spherical tensor operators can be combined with a similar addition formula. Consider two normalized tensor operators Uk1, Vk2 , with components Uq1

k1, Vq2k2 . Then the product Uq1

k1 Vq2k2 can be expressed as a linear combination of objects with tensor

properties

Uq1k1 Vq2

k2 = Skq ckq,q1 q2

k1 k2 Tqk (Eq. 1)

There are �2 k1 + 1� �2 k2 + 1� possible products that can be formed from the components of Uk1 andVk2 . Under the set of rotations, the various components of Uk1 transform into each other. Similarly for Vk2 and the Tk . It follows that for

each Tk all components Tqk are included in the set of all possible Uq1

k1 Vq2k2 products.

The next step is to consider the effect of Jz

Jz, Uq1k1 Vq2

k2� = Uq1k1 Jz, Vq2

k2� + Jz, Uq1k1� Vq2

k2

= �q1 + q2� Uq1k1 Vq2

k2

or, in the sum over k, q, it is required that q = q1 + q2 . This in turn implies that there is a maximum value for q, namely

qmax= k1 + k2 . Considering that the tensor operators must be complete, k also has a maximum value, kmax= k1 + k2 . The operator Tkmax

kmax = Uk1

k1 Vk2

k2 is the unique q= kmaxoperator.

Following the development of angular momentum addition, the next step is recognize that there will be two operators

with q = kmax- 1 built out of linear combinations of Uq1-1k1 Vq2

k2 and Uq1k1 Vq2-1

k2 . One will belong to Tkmax and can be built with J- . The other will be an orthogonal linear combination, and must be the highest q-component of an operator

Tkmax-1 . The operation with J- on the left side of Eq. 1 above gives

J-, Uk1

k1 Vk2

k2� = Uk1

k1 J-, Vk2

k2� + J-, Uk1

k1� Vk2

k2

= ck1 k1- Uk1-1

k1 Vk2

k2 + ck2 k2- Uk1

k1 Vk2-1k2

while on the right-hand side

J-, Tk1+k2

k1+k2� = ck1+k2,k1+k2- Tk1+k2-1

k1+k2

or

Tk1+k2-1k1+k2 =

ck1 k1-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅck1+k2,k1+k2- Uk1-1

k1 Vk2

k2 +ck2 k2-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅck1+k2,k1+k2- Uk1

k1 Vk2-1k2

rotations.nb: 11/3/04::13:48:13 28

The second linear combination will be orthogonal to this one, and will become the head component of the Tk1+k2-1 operator. Following the same procedures used to fnd the irreducible representation content in the product of two

representations j1 j2 , one can proceed to repetitively apply J- to fill out the list of components for the tensor operators in the product of Uk1 Vk2 . Since the construction technique is the same, the coefficients ckq,q1 q2

k1 k2 are in fact exactly

Clebsch-Gordon coefficients.

ü Normalization of tensor operators

The definition of tensor operators has appropriately concentrated on the behavior under rotations. As such, changing the

normalization or even multiplying by an arbitrary scalar function of r will not affect the properties under rotations. Accordingly, if Tk is a rank-k tensor operator, then T 'k = f �r� Tk is also a rank-k tensor operator. This follows directly from the observation that R f�r� = f �r� for any rotation R. There is nothing particularly wrong with having a large

number of tensor operators of the same rank, but it seems simpler to separate the tensor properties from the

normalization. Accordingly, one can uniquely define a normalized rank-k tensor operator Tèk

, and then an arbitrary

tensor operator Tk can be written as

Tk = f �r� Tèk

For a given application f �r� may depend on k, but it does not depend on q, so all components of Tk scale in the same

way

Tqk = f �r� Tèq

k

One may think of this as being equivalent to the separation of variables that allows one to write a wave function in the form

Ynlm = Rnl�r� Ylm�q, f�States are typically written in the form nlm where the separation of variables is not explicit, but it would be valid to use a notationnlm = nlr lmWwhere the subscripts indicate that the two terms depend on radial and angular degrees of freedom.

The separation above defines the concept of a normalized tensor operator, but doesn't say what that normalization should be. Common sense suggests that it should correspond to the normalization of states jm so that when combining operators and states the result will be well defined normalized states. This is the subject of the Wigner-Eckart theorem.

ü Acting on states with tensor operators

We have seen that the tensor operators can be combined using the same rules as apply for angular momentum addition of states. It seems reasonable to guess that operating with Tq

k on a state jm will result in another example of angular

momentum addition, i.e. operating on representation j with an operator Tk will produce states in representations j ', with k- j� < j ' < k+ j present in the product.

rotations.nb: 11/3/04::13:48:13 29

Indeed, consider the set of states that can result from operation by any component q of Tqk on any state m of jm . Since

there are q components and m states, there are n = qm linearly independent outcomes. Further, one can consider the set of all rotations on the result. The rotations will ensure that if one state of a representation is present in the result, then

all states of the representation will be present. On the other hand, the rotations will not increase the number of possible combinations of operators and initial states. Simple state counting suggests that the representations present in all combinations of Tq

k jmwill be exactly the representations present in the combination of two representions k, j .

To see this explicity, consider the product,

Tqk jm = S

j' m' cèqm, j' m'

k j j 'm'the coefficients cè are written in the same notation as Clebsch-Gordon coefficients, but they have not yet been shown to be identicle. There are two considerations: first whether or not the ratio of any two coefficients cè is the same as for the

ratio of CG coefficients, and second whether or not the magnitude of the coefficients matches. The second is a question of normalization, and it will be assumed that in the sense of the previous subsection, the the operator Tk is normalized.

Operating with Jz, on the right

Jz Sj' m' cèqm, j' m'

k j j 'm' = Sj' m' m' cèqm, j' m'

k j j 'm'and on the left

Jz Tqk jm = � Jz, Tq

k� + Tqk Jz� jm

= �qTqk + Tq

k m� jm= �q+m� Tq

k jm= S

j' m'�q+m� cèqm, j' m'

k j j 'm'It follows that m' = q+m term by term, and the sum is restricted to just a sum over different j '.

Tqk jm = S

j' cèqm, j' m'=q+m

k j j 'm' = q+mSimilar to the addition of states, the highest m'-state in the operator-state product is given m'max= k+ j , and the necessity of complete representations implies that the highest representation is j 'max= k+ j . This state is produced

uniquely by Tkk j j = j + k, j + k . Similarly, there will be two states produced with m' = k+ j - 1, by the

combinations Tk-1k j j or Tk

k j j - 1 . One linear combination of these states will be a member of the j 'max representation and the other will head a representation with j ' = j 'max- 1. The exact linear combinations are determined

by operating with J- on the right

J- Tkk j j = J- j + k, j + k = c j+k, j+k

- j + k, j + k- 1and on the left

J- Tkk j j = � J-, Tk

k� + Tkk J-� j j

= ckk- Tk-1

k j j + c j j- Tk

k j, j - 1Combining the results,

rotations.nb: 11/3/04::13:48:13 30

j + k, j + k- 1 = cj j-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅcj+k, j+k- Tk

k j, j - 1 + ckk-

ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅcj+k, j+k- Tk-1

k j j Once again, this process is replicating the arithmetic of angular momentum addition of states, with the same Clebsch-Gordon coefficients. The ~ over the c-coefficients in

Tqk jm = dm',q+m S

j' cqm, j' m'

k j j 'm'has been removed.

ü Wigner-Eckhardt theorem

Having demonstrated the equivalance to angular momentum addition, the W-E theorem can be stated. The matrix element for a tensor operator, acting between two states of known angular momentum is given by�n' j ' m' � Tq

k � n jm = cqm, j' m'k j �n' j ' �� Tk �� n j

where the c-coefficients are Clebsch-Gordon coefficients that know about the angular momentum variables but are independent of the normalization or radial degrees of freedom. The "double-barred" or "reduced" matrix element

contains information about the radial dependence of the operator and the states, and/or any other degrees of freedom which are not part of the angular momentum discussion. In the notation of normalization discussion above, an arbitrary tensor operator can be expressed as the product of a scalar function and a normalized operator

Tqk = f �r� Tèq

k

and so

�n' j ' m' � Tqk � n jm = n' j ' m' ! f �r� Tèq

k ! n jm"= j ' m' � T

èqk � jm"

W �n' j ' � f �r� � n jr

= cqm, j' m'k j �n' j ' � f �r� � n jr

where in the second line the product notation nlm = nlr lmW has been used. The Clebsch-Gordon coefficient comes

from the angular matrix element and the rotation properties of the operator. What is typically called the reduced matrix element is identified as the radial matrix element, including any normalization which affects the overall strength of the operator.

ü Some examples

m selection rulesj selection rules10 10 10=0

rotations.nb: 11/3/04::13:48:14 31

ü stuff

rotations.nb: 11/3/04::13:48:14 32