Angular Momentum for beginners

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Physics 125

Course Notes

Angular Momentum

040429 F. Porter

1 Introduction

This note constitutes a discussion of angular momentum in quantum mechan-ics. Several results are obtained, important to understanding how to solveproblems involving rotational symmetries in quantum mechanics. We will of-ten give only partial proofs to the theorems, with the intent that the readercomplete them as necessary. In the hopes that this will prove to be a usefulreference, the discussion is rather more extensive than usually encountered

in quantum mechanics textbooks.

2 Rotations: Conventions and Parameteriza-

tions

A rotation by angle about an axis e (passing through the origin in R3)is denoted by Re(). Well denote an abstract rotation simply as R. It isconsidered to be positive if it is performed in a clockwise sense as we lookalonge. As with other transformations, our convention is that we think of a

rotation as a transformation of the state of a physical system, and not as achange of coordinate system (sometimes referred to as the active view). IfRe() acts on a configuration of points (system) we obtain a new, rotated,configuration: If x is a point of the old configuration, and x is its imageunder the rotation, then:

x= Re()x. (1)

That is,R is a linear transformation described by a 33 real matrix, relativeto a basis (e1, e2, e3).

Geometrically, we may observe that:

Re() = R1e (), (2)I = Re(0) =Re(2n), n= integer, (3)

Re()Re() = Re(+

), (4)

Re(+ 2n) = Re(), n= integer, (5)

Re = Re(). (6)1


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A product of the form Re()Re() means first do Re(), then do Re()

to the result. All of these identities may be understood in terms of ma-trix identities, in addition to geometrically. Note further that the set of all

rotations about a fixed axis eforms a one-parameter abelian group.It is useful to include in our discussion of rotations the notion also ofreflections: Well denote the space reflection with respect to the origin by P,for parity:

Px= x. (7)Reflection in a plane through the origin is called a mirroring. Lete be a unitnormal to the mirror plane. Then

Mex= x 2e(e x), (8)since the component of the vector in the plane remains the same, and the

normal component is reversed.Theorem: 1.

[P, Re()] = 0. (9)

[P, Me] = 0. (10)

(The proof of this is trivial, since P =I.)2.

P Me= MeP =Re(). (11)

3. P,Me, andRe() are involutions, that is:

P2 = I. (12)

M2e

= I. (13)

[Re()]2 = I. (14)

Theorem: Let Re() be a rotation and let e, e be two unit vectors per-

pendicular to unit vector e such that e is obtained from e accordingto:

e= Re(/2)e. (15)

Then

Re() =Me

Me

=Re

()Re

(). (16)Hence, every rotation is a product of two mirrorings, and also a productof two rotations by .

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M

M

x

R

e'

e''e

a+b=a

b

e'

e''

/2

/2/2

e( )

x

Figure 1: Proof of the theorem that a rotation about e by angleis equivalent

to the product of two mirrorings or rotations by .

Proof: We make a graphical proof, referring to Fig. 1.

Theorem: Consider the spherical triangle in Fig. 2.

1. We haveRe3(23)Re2(22)Re1(21) =I , (17)

where the unit vectors

{ei

}are as labelled in the figure.

2. Hence, the product of two rotations is a rotation:

Re2(22)Re1(21) =Re3(23). (18)

The set of all rotations is a group, where group multiplication isapplication of successive rotations.

Proof: Use the figure and label M1 the mirror plane spanned by e2, e3, etc.Then we have:

Re1(21) = M3M2

Re2(22) = M1M3

Re3(23) = M2M1. (19)

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1

2

3

e

e

e

3

2

1

.

.

.

Figure 2: Illustration for theorem. Spherical triangle vertices are defined asthe intersections of unit vectors e1, e2, e3 on the surface of the unit sphere.

Thus,

Re3(23)Re2(22)Re1(21) = (M2M1)(M1M3)(M3M2) =I . (20)

The combination of two rotations may thus be expressed as a problemin spherical trigonometry.

As a corollary to this theorem, we have the following generalization ofthe addition formula for tangents:

Theorem: IfRe() =Re()Re(), and defining:

= e tan /2

= etan /2

= etan /2, (21)

then

=+ +

1 . (22)This will be left as an exercise for the reader to prove.

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Theorem: The most general mapping x x of R3 into itself, such thatthe origin is mapped into the origin, and such that all distances arepreserved, is a linear, real orthogonal transformation Q:

x= Qx, where QTQ= I , and Q=Q. (23)

Hence,x y= x y points x, yR3. (24)

For such a mapping, either:

1. det(Q) = 1,Q is called a properorthogonal transformation, andis in fact a rotation. In this case,

x y= (x y) points x, yR3. (25)

or,

2. det(Q) =1, Q is called an improper orthogonal transforma-tion, and is the product of a reflection (parity) and a rotation. Inthis case,

x y=(x y) points x, yR3. (26)

The set of all orthogonal transformations on three dimensions forms agroup (denotedO(3)), and the set of all proper orthogonal transforma-tions forms a subgroup (O+(3) or SO(3) ofO(3)), in 1 : 1 correspon-dence with, hence a representation of, the set of all rotations.

Proof of this theorem will be left to the reader.

3 Some Useful Representations of Rotations

Theorem: We have the following representations of rotations (u is a unitvector):

Ru()x= uu

x + [x

uu

x]cos + u

x sin , (27)

andRu() =e

uJJJ =I+ (u JJJ)2(1 cos ) + u JJJsin , (28)

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whereJJJ = (J1, J2, J3) with:

J1=

0 0 00 0

1

0 1 0 , J2=

0 0 10 0 0

1 0 0 , J3 =

0 1 01 0 00 0 0

.(29)

Proof: The first relation may be seen by geometric inspection: It is a de-composition of the rotated vector into components along the axis ofrotation, and the two orthogonal directions perpendicular to the axisof rotation.

The second relation may be demonstrated by noticing that Jix= eix,where e1, e2, e3 are the three basis unit vectors. Thus,

(u JJJ)x= u x, (30)and

(u JJJ)2x= u (u x) = u(u x) x. (31)Further,

(u JJJ)2n+m = ()n(u JJJ)m, n= 1, 2, . . . ; m= 1, 2. (32)

The second relation then follows from the first.

Note that

Tr [Ru()] = TrI+ (u JJJ)2(1 cos )

. (33)

With

J21 = 0 0 00 1 0

0 0 1

, J22 =

1 0 00 0 0

0 0 1

, J23 =

1 0 00 1 0

0 0 0

, (34)

we haveTr [Ru()] = 3 2(1 cos ) = 1 + 2 cos . (35)

This is in agreement with the eigenvalues ofRu() being 1, ei, ei.

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Theorem: (Euler parameterization) Let R O+(3). Then R can berepresented in the form:

R= R(, , ) =Re3()R

e2()R

e3(), (36)

where the Euler angles , , can be restricted to the ranges:

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We note that e3 does not depend on since this first rotation is aboute3 itself. The polar angles ofe

3 are given precisely by and. Hence

and are uniquely determined (within the specified ranges, and

up to the ambiguous case mentioned in the theorem) by e3 = Re3,which can be specified to any desired orientation. The angle is thendetermined uniquely by the orientation of the pair (e1, e

2) in the plane

perpendicular to e3.

We note that the rotation group [O+(3)] is a group of infinite order (or, isan infinite group, for short). There are also an infinite number of subgroupsofO+(3), including both finite and infinite subgroups. Some of the importantfinite subgroups may be classified as:

1. The Dihedral groups,Dn, corresponding to the proper symmetries ofan n-gonal prism. For example,D6

O+(3) is the group of rotations

which leaves a hexagonal prism invariant. This is a group of order 12,generated by rotations Re3(2/6) and Re2().

2. The symmetry groups of the regular solids:

(a) The symmetry group of the tetrahedron.

(b) The symmetry group of the octahedron, or its dual (replacevertices by faces, faces by vertices) the cube.

(c) The symmetry group of the icosahedron, or its dual, the dodeca-hedron.

We note that the tetrahedron is self-dual.

An example of an infinite subgroup ofO+(3) isD, the set of all rotationswhich leaves a circular disk invariant, that is, including all rotations aboutthe normal to the disk, and rotations by about any axis in the plane of thedisk.

4 Special Unitary Groups

The set of all n

n unitary matrices forms a group (under normal matrix

multiplication), denoted by U(n). U(n) includes as a subgroup, the set ofall nn unitary matrices with determinant equal to 1 (unimodular, orspecial. This subgroup is denoted by S U(n), for Special Unitary group.

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e

e

3

1

Figure 4: A hexagonal prism, to illustrate the group D6.

The group of 2

2 unimodular unitary matrices, SU(2), has a special

connection withO+(3), which is very important in quantum mechanics. Con-sider the real vector space of all 2 2 traceless hermitian matrices, whichwe denote by V3. The 3 refers to the fact that this is a three-dimensionalvector space (even though it consists of 2 2 matrices). Hence, it can be putinto 1 : 1 correspondence with Euclidean 3-space,R3. We may make this cor-respondence an isometry by introducing a positive-definite symmetric scalarproduct onV3:

(X, Y) =1

2Tr(XY), X, YV3. (38)

Let u be any matrix in SU(2): u1 = u and det(u) = 1. Consider the

mapping: XX= uX u. (39)We demonstrate that this is a linear mapping ofV3 into itself: IfX is her-

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mitian, so is X. IfX is traceless then so is X:

Tr(X) = Tr(uXu) = Tr(Xuu) = Tr(X). (40)

This mapping ofV3into itself also preserves the norms, and hence, the scalarproducts:

(X, X) = 1

2Tr(XX)

= 1

2Tr(uXuuXu)

= (X, X). (41)

The mapping is therefore a rotation acting on V3 and we find that to everyelement ofS U(2) there corresponds a rotation.

Let us make this notion of a connection more explicit, by picking anorthonormal basis (1, 2, 3) ofV3, in the form of the Pauli matrices:

1 =

0 11 0

, 2=

0 ii 0

, 3 =

1 00 1

. (42)

Note that the Pauli matrices form an orthonormal basis:

1

2Tr() =. (43)

We have the products:

12 = i3, 23=i1, 31= i2. (44)

Different Pauli matrices anti-commute:

{, } + = 2I (45)

The commutation relations are:

[, ] = 2i. (46)

Any element ofV3 may be written in the form:

X=x =3i=1

xii, (47)

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where xR3. This establishes a 1 : 1 correspondence between elements ofV3 andR

3. We note that

1

2Tr [(a )(b )] =a b, (48)and

(a )(b ) =3i=1

aibi2i +

3i=1

j=i

aibjij

= (a b)I+ i(a b) . (49)

Finally, we may see that the mapping is isometric:

(X, Y) =1

2Tr(XY) =

1

2Tr [(x

)(y

)] =x

y. (50)

Lets investigateSU(2) further, and see the relevance ofV3: Every unitarymatrix can be expressed as the exponential of a skew-hermitian (A =A)matrix. If the unitary matrix is also unimodular, then the skew-hermitianmatrix can be selected to be traceless. Hence, everyu SU(2) is of the formu= eiH, where H= H and Tr(H) = 0. For every real unit vector e andevery real , we define ue()SU(2) by:

u() = exp i

2e

. (51)

Any element ofSU(2) can be expressed in this form, since every tracelesshermitian matrix is a (real) linear combination of the Pauli matrices.

Now let us relate ue() to the rotation Re():

Theorem: Let xR3, andX=x . Let

u() = exp i

2e

, (52)

and letue()Xu

e

() =X= x . (53)Then

x= Re()x. (54)

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Proof: Note that

ue() = exp i

2e

= Icos

2 i(e )sin

2. (55)

This may be demonstrated by using the identity (a )(b ) = (ab)I+ (a b) , and letting a= b = e to get (e )2 =I, and usingthis to sum the exponential series.

Thus,

x = ue()Xue()=

Icos

2 i(e )sin

2

(x )

Icos

2+ i(e )sin

2

= x cos22

+ (e )(x )(e )sin22

+ [i(e )(x ) + i(x )(e )] sin2

cos2

. (56)

But

(e )(x ) = (e x)I+ i(e x) (57)(x )(e ) = (e x)I i(e x) (58)

(e )(x )(e ) = (e x)(e ) + i [(e x) ] (e )= (e x)(e ) + i2 [(e x) e] = 2(e x)(e ) x , (59)

where we have made use of the identity (C B) A = B(A C) C(A B) to obtain (e x) e= x e(e x). Hence,

x =

cos2

2x + sin2

2[2(e x)e x]

+i sin

2cos

2[2i(e x) ] . (60)

Equating coefficients of we obtain:

x = x cos2

2

+ [2(e

x)e

x]sin2

2

+ 2(e

x)sin

2

cos

2= (e x)e + [x (e x)e]cos + (e x)sin = Re()x. (61)

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Thus, we have shown that to every rotation Re() corresponds at leastone u SU(2), and also to every element of SU(2) there corresponds arotation. We may restate the theorem just proved in the alternative form:

uXu = u(x )u = x (uu)= x = [Re()x] = x

R1

e ()

. (62)

But x is arbitrary, souu = R1

e (), (63)

or,u1u= Re(). (64)

More explicitly, this means:

u1iu=3j=1

Re()ijj . (65)

There remains the question of uniqueness: Suppose u1 SU(2) andu2SU(2) are such that

u1Xu1= u2Xu

2, XV3. (66)

Then u12 u1 commutes with every X V3 and therefore this matrix mustbe a multiple of the identity (left for the reader to prove). Since it is uni-tary and unimodular, it must equal I orI. Thus, there is a two-to-onecorrespondence between SU(2) and O

+

(3): To every rotation Re() corre-sponds the pair ue() andue() = ue(+ 2). Such a mapping ofSU(2)ontoO+(3) is called a homomorphism (alternatively called an unfaithfulrepresentation).

We make this correspondence precise in the following:

Theorem: 1. There is a two-to-one correspondence between SU(2) andO+(3) under the mapping:

u R(u), where Rij(u) =12

Tr(uiuj), (67)

and the rotation Re() corresponds to the pair:

Re() {ue(), ue() =ue(+ 2)}. (68)

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2. In particular, the pair of elements{I, I} S U(2) maps to IO+(3).

3. This mapping is a homomorphism:u

R(u) is a representation

ofS U(2), such that

R(uu) =R(u)R(u), u, uSU(2). (69)That is, the multiplication table is preserved under the map-ping.

Proof: 1. We have

u1iu=3j=1

Re()ijj. (70)

Multiply by k and take the trace:

Tr(u1iuk) = Tr

3j=1

Re()ijjk

, (71)

or

Tr(uiuk) =3j=1

Re()ijTr(jk). (72)

But 12Tr(jk) =jk , hence

Rik(u) = Re()ik =1

2Tr(uiuk). (73)

Proof of the remaining statements is left to the reader.

A couple of comments may be helpful here:

1. Why did we restrictuto be unimodular? That is, why are we consider-ing SU(2), and notU(2). In fact, we could have considered U(2), butthe larger group only adds unnecessary complication. All U(2) adds ismultiplication by an overall phase factor, and this has no effect in thetransformation:

XX= uXu. (74)This would enlarge the two-to-one mapping to infinity-to-one, appar-ently without achieving anything of interest. So, we keep things assimple as we can make them.

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2. Having said that, can we make things even simpler? That is, can weimpose additional restrictions to eliminate the double-valuedness inthe above theorem? The answer is no S U(2) has no subgroup which

is isomorphic withO+

3.

5 Lie Groups: O+(3) and SU(2)

Def: An abstract n-dimensional Lie algebra is an n-dimensional vectorspaceVon which is defined the notion of a product of two vectors ()with the properties (x,y,z V,ca complex number):

1. Closure: x y V.2. Distributivity:

x (y+ z) = x y+ x z (75)(y+ z) x = y x + z x. (76)

3. Associativity with respect to multiplication by a complex number:

(cx) y= c(x y). (77)

4. Anti-commutativity:x y=y x (78)

5. Non-associative (Jacobi identity):x (y z) + z (x y) + y (z x) = 0 (79)

We are especially interested here in Lie algebras realized in terms of matri-ces (in fact, every finite-dimensional Lie algebra has a faithful representationin terms of finite-dimensional matrices):

Def: A Lie algebra of matrices is a vector space M of matrices which isclosed under the operation of forming the commutator:

[M, M] = MM

MM

M,

M, M

M. (80)

Thus, the Lie product is the commutator: M M = [M, M]. Thevector space may be over the real or complex fields.

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withMthere is ann-dimensionalLie groupGof matrices:Gis the matrixgroup generated by all matrices of the form eX, where X M. We seethat O+(3) and SU(2) are Lie groups of this kind in fact, every element

of either of these groups corresponds to an exponential of an element of theappropriate Lie algrebra.1

6 Continuity Structure

As a finite dimensional vector space,M has a continuity structure in theusual sense (i.e., it is a topological space with the usual topology). Thisinduces a continuity structure (topology) onG (for O+(3) and S U(2), thereis nothing mysterious about this, but well keep our discussion a bit moregeneral for a while). G is an n-dimensional manifold (a topological spacesuch that every point has a neighborhood which can be mapped homeo-morphically onto n-dimensional Euclidean space). The structure ofG (itsmultiplication table) in some neighborhood of the identity is uniquely deter-mined by the structure of the Lie algebraM. This statement follows fromthe Campbell-Baker-Hausdorff theorem for matrices: If matrices X, Y aresufficiently small, theneXeY =eZ, where Z is a matrix in the Lie algebragenerated by matrices X and Y. That is, Z is a series of repeated commu-tators of the matrices X and Y. Thus, we have the notion that the localstructure ofGis determined solely by the structure ofMas a Lie algebra.

We saw that the Lie algebras ofO+(3) and S U(2) are isomorphic, hencethe groupO+(3) is locally isomorphic with S U(2). Note, on the other hand,

that the properties

(u J)3 =(u J) and (u J)2n+m = ()n(u J)m, (87)

for all positive integers n, m, are not shared by the Pauli matrices, whichinstead satisfy:

(u )3 =u . (88)Such algebraic properties are outside the realm of Lie algebras (the productsbeing taken are not Lie products). We also see that (as with O+(3) and

1This latter fact is not a general feature of Lie groups: To say that G is generated

by matrices of the form eX means that G is the intersection of all matrix groups whichcontain all matrices eX where X M. An element ofGmay not be of the form eX.

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SU(2)) it is possible for two Lie algebras to have the same local structure,while not being globally isomorphic.

A theorem describing this general situation is the following:

Theorem: (and definition) To every Lie algebraM corresponds a uniquesimply-connected Lie group, called the Universal Covering Group,defined byM. Denote this group byGU. Every other Lie group witha Lie algebra isomorphic withMis then isomorphic with the quotientgroup ofGU relative to some discrete (central all elements whichmap to the identity) subgroup ofGU iself. If the other group is simplyconnected, it is isomorphic withGU itself.

We apply this to rotations: The group SU(2) is the universal coveringgroup defined by the Lie algebra of the rotation group, hence SU(2) takes

on special significance. We note that SU(2) can be parameterized as thesurface of a unit sphere in four dimensions, hence is simply connected (allclosed loops may be continuously collapsed to a point). On the other hand,O+(3) is isomorphic with the quotient group S U(2)/I(2), whereI(2) is theinversion group in two dimensions:

I(2) =

1 00 1

,1 0

0 1

. (89)

7 The Haar Integral

We shall find it desirable to have the ability to perform an invariant inte-gral on the manifolds O+(3) and SU(2), which in some sense assigns anequal weight to every element of the group. The goal is to find a way ofdemocratically averaging over the elements of a group. For a finite group,the correspondence is to a sum over the group elements, with the same weightfor each element.

For the rotation group, let us denote the desired volume element byd(R). We must find an expression ford(R) in terms of the parameterization,for some parameterization of O+(3). For example, we consider the Eulerangle parameterization. Recall, in terms of Euler angles the representationof a rotation as:

R= R(, , ) =Re3()Re2()Re3(). (90)

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We will argue that the appropriate volume element must be of the form:

d(R) = K d sin dd, K >0. (91)

The argument for this form is as follows: We have a 1 : 1 correspondencebetween elements ofO+(3) and orientations of a rigid body (such as a spherewith a dot at the north pole, centered at the origin; let IO+(3) correspondto the orientation with the north pole on the +e3axis, and the meridian alongthe +e2 axis, say). We want to find a way to average over all positions of thesphere, with each orientation receiving the same weight. This correspondsto a uniform averaging over the sphere of the location of the north pole.

Now notice that if R(, , ) acts on the reference position, we obtainan orientation where the north pole has polar angles (, ). Thus, the (, )dependence ofd(R) must bed sin d. For fixed (, ), the angledescribesa rotation of the sphere about the north-south axis the invariant integralmust correspond to a uniform averaging over this angle. Hence, we intuitivelyarrive at the above form for d(R). The constant K >0 is arbitrary; we pickit for convenience. We shall choose K so that the integral over the entiregroup is one:

1 =O+(3)

d(R) = 1

82

20

d 0

sin d 20

d. (92)

We can thus evaluate the integral of a (suitably behaved) function f(R) =f(, , ) over O+(3):

f(R) =O+(3)

f(R)d(R) = 182 20

d 0

sin d 20

df(, , ). (93)

The overbar notation is intended to suggest an average.What about the invariant integral over SU(2)? Given the answer for

O+(3), we can obtain the result for S U(2) using the connection between thetwo groups. First, parameterize SU(2) by the Euler angles:

u(, , ) = exp i

23

exp

i

22

exp

i

23

, (94)

with

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uniquely, up to a set of measure 0 (when = 0, ). The integral of functiong(u) onS U(2) is thus:

g(u) =SU(2) g(u)d(u) =

1

162 20 d

0 sin d

40 dg [u(, , )] , (96)

with the volume element normalized to give unit total volume:

SU(2)

d(u) = 1. (97)

A more precise mathematical treatment is possible, making use of mea-sure theory; well mention some highlights here. The goal is to define ameasure(S) for suitable subsets ofO+(3) (orSU(2)) such that ifR0 is anyelement ofO+(3), then:

(SR0) =(S), where S R0 = {RR0|RS} . (98)

Intuitively, we think of the following picture: Smay be some region inO+(3),andSR0 is the image ofSunder the mapping RRR0. The idea then, isthat the regionsSandSR0should have the same volume for allRO+(3).Associated with such a measure we have an integral.

R

S

SR

0

0

Figure 5: Set Smapping to setSR0, under the rotation R0.

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It may be shown2 that a measure with the desired property exists and isunique, up to a factor, for any finite-dimensional Lie group. Such a measureis called a Haar measure. The actual construction of such a measure must

deal with coordinate system issues. For example, there may not be a goodglobal coordinate system on the group, forcing the consideration of localcoordinate systems.

Fortunately, we have already used our intuition to obtain the measure(volume element) for the Euler angle parameterization, and a rigorous treat-ment would show it to be correct. The volume element in other parameter-izations may be found from this one by suitable Jacobian calculations. Forexample, if we parameterize O+(3) by:

Re() =eJJJ, (99)

where e, and|| , then the volume element (normalized again tounit total volume of the group) is:

d(R) = 1

421 cos

2 d3(), (100)

whered3() is an ordinary volume element on R3. Thus, the group-averagedvalue off(R) is:

O+(3)

f(R)d(R) = 1

42

O+(3)

1 cos 2

d3()f

eJJJ

(101)

= 1

424

de 2

0

(1

cos )dfeJJJ . (102)Alternatively, we may substitute 1 cos = 2 sin2 2 . ForSU(2) we have thecorresponding result:

d(u) = 1

42desin

2

2d, 02. (103)

We state without pro0f that the Haar measure is both left- and right-invariant. That is, (S) =(SR0) =(R0S) for all R0 O+(3) and for allmeasurable sets S O+(3). This is to be hoped for on physical grounds.The invariant integral is known as the Haar integral, or its particular real-

ization for the rotation group as the Hurwitz integral.2This may be shown by construction, starting with a small neighborhood of the identity,

and using the desired property to transfer the right volume element everywhere.

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8 Unitary Representations ofSU(2)(and O+(3))

A unitary representation ofS U(2) is a mapping

u SU(2) U(u)U(n) such that U(u)U(u) = U(uu), u, uSU(2).(104)

That is, we represent the elements of SU(2) by unitary matrices (notnecessarily 2 2), such that the multiplication table is preserved, eitherhomomorphically or isomorphically. We are very interested in such mappings,because they permit the study of systems of arbitrary angular momenta,as well as providing the framework for adding angular momenta, and forstudying the angular symmetry properties of operators. We note that, forevery unitary representationR T(R) ofO+(3) there corresponds a unitaryrepresentation ofSU(2): u U(u) =T[R(u)]. Thus, we focus our attentiononS U(2), without losing any generality.

For a physical theory, it seems reasonable to to demand some sort ofcontinuity structure. That is, whenever two rotations are near each other,the representations for them must also be close.

Def: A unitary representationU(u) is calledweakly continuousif, for anytwo vectors , , and any u:

limuu

| [U(u) U(u)] = 0. (105)

In this case, we write:

w-limuuU(u

) = U(u), (106)

and refer to it as the weak-limit.

Def: A unitary representation U(u) is called strongly continuous if, forany vectorand any u:

limuu

[U(u) U(u)] = 0. (107)

In this case, we write:

s-limuuU(u) = U(u), (108)

and refer to it as the strong-limit.

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Strong continuity implies weak continuity, since:

|| [U(u) U(u)] | [U(u) U(u)] (109)

We henceforth (i.e.

, until experiment contradicts us) adopt these notions ofcontinuity as physical requirements.An important concept in representation theory is that of (ir)reducibility:

Def: A unitary representation U(u) is called irreducible if no subspace ofthe Hilbert space is mapped into itself by every U(u). Otherwise, therepresentation is said to be reducible.

Irreducible representations are discussed so frequently that the jargon ir-rep has emerged as a common substitute for the somewhat lengthy irre-ducible representation.

Lemma: A unitary representation U(u) is irreducible if and only if every

bounded operatorQ which commutes with every U(u) is a multiple ofthe identity.

Proof of this will be left as an exercise. Now for one of our key theorems:

Theorem: Ifu U(u) is a strongly continuous irreducible representation ofSU(2) on a Hilbert spaceH, thenHhas a finite number of dimensions.

Proof: The proof consists of showing that we can place a finite upper boundon the number of mutually orthogonal vectors inH: LetEbe any one-dimensional projection operator, and , be any two vectors inH.Consider the integral:

B(, ) =SU(2)

d(u)|U(u)EU(u1). (110)This integral exists, since the integrand is continuous and bounded,because U(u)EU(u1) is a one-dimensional projection, hence of norm1.

Now

|B(, )| =SU(2)

d(u)|U(u)EU(u1)

SU(2)

d(u)||U(u)EU(u1)| (111)

SU(2)

d(u)U(u)EU(u1) (112) , (113)

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where we have made use of the Schwarz inequality and of the factSU(2) d(u) = 1.

B(, ) is linear in , anti-linear in , and hence defines a bounded

operatorB0 such that:

B(, ) =|B0. (114)

Letu0SU(2). Then

|U(u0)B0U(u10 ) =SU(2)

d(u)|U(u0u)EU((u0u)1)(115)

=SU(2)

d(u)|U(u)EU(u1) (116)= |B0, (117)

where the second line follows from the invariance of the Haar integral.Since and are arbitrary vectors, we thus have;

U(u0)B0 = B0U(u0), u0SU(2). (118)

SinceB0commutes with every element of an irreducible representation,it must be a multiple of the identity, B0 = pI.

SU(2)d(u)U(u)EU(u1) =pI. (119)

Now let{n|n = 1, 2, . . . , N } be a set of N orthonormal vectors,n|m= nm, and take E=|11|. Then,

1|SU(2)

d(u)U(u)EU(u1)|1 = p1|I|1 =p, (120)

=SU(2)

d(u)1|U(u)|11|U(u1)|1

=SU(2)

d(u)|1|U(u)|1|2 >0. (121)

Note that the integral cannot be zero, since the integrand is a contin-

uous non-negative definite function ofu, and is equal to one for u = I.

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Thus, we have:

pN =N

n=1

n

|pIn

(122)

=Nn=1

SU(2)

d(u)n|U(u)EU(u1)n (123)

=Nn=1

SU(2)

d(u)n|U(u)|11|U(u1)n (124)

=Nn=1

SU(2)

d(u)U(u)1|nn|U(u)1 (125)

=SU(2)

d(u)U(u)1|N

n=1|nn|U(u)1 (126)

SU(2)

d(u)U(u)1|I|U(u)1 (127)

SU(2)

d(u)U(u)12 = 1. (128)(129)

That is, pN 1. But p >0, so N


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Note that we can think of equivalence as just a basis transformation. Thedesired theorem is:

Theorem: Any finite-dimensional (continuous) representationu

W(u) of

SU(2) is equivalent to a (continuous) unitary representation u U(u)ofS U(2).

Proof: We prove this theorem by constructing the required similarity trans-formation: Define matrix

P =SU(2)

d(u)W(u)W(u). (131)

This matrix is positive definite and Hermitian, since the integrand is.Thus Phas a unique positive-definite Hermitian square rootS:

P =P> 0 =

P =S=S > 0. (132)

Now, let u0, u SU(2). We have,W(u0)W(u)W(u) =

W(uu0)W(uu0)

W(u10 ). (133)

From the invariance of the Haar integral, we find:

W(u0)P =SU(2)

d(u)W(uu0)W(uu0)

W(u10 ) (134)

= P W(u10 ), u0SU(2). (135)Now define, for all u SU(2),

U(u) =SW(u)S1. (136)

The mapping uU(u) defines a continuous representation ofS U(2),and furthermore:

U(u)U(u) =SW(u)S1

SW(u)S1

=

S1

W(u)SSW(u)S1

=

S1

P W(u1)W(u)S1

=

S1

P S1

= S1

SSS1

= I. (137)

That is,U(u) is a unitary representation, equivalent to W(u).

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We have laid the fundamental groundwork: It is sufficient to determine allunitary finite-dimensional irreducible representations ofS U(2).

This brings us to some important tool theorems for working in group

representaion theory.Theorem: Let u D(u) and u D(u) be two inequivalent irreducible

representations ofSU(2). Then the matrix elements ofD(u) and D(u)satisfy:

SU(2)d(u)Dmn(u)D

rs(u) = 0. (138)

Proof: Note that the theorem can be thought of as a sort of orthogonalityproperty between matrix elements of inequivalent representations. LetVbe theN-dimensional carrier space of the representation D(u), andletVbe theN-dimensional carrier space of the representation D(u).

LetA be any N N matrix. Define another N N matrix, A0 by:A0

SU(2)

d(u)D(u1)AD(u). (139)

Consider (in the sceond line, we use the invariance of the Haar integralunder the substitution u uu0):

D(u0)A0 =SU(2)

d(u)D(u0u1)AD(u)

=SU(2)

d(u)D(u1)AD(uu0)

=SU(2)

d(u)D(u1)AD(u)D(u0)

= A0D(u0), u0SU(2). (140)

Now define N N matrix B andN N matrix B by:BA0A0, BA0A0. (141)

Then we have:

D(u)B = D(u)A0A0

= A0D(u)A0

= A0A0D

(u)

= BD(u), u SU(2). (142)

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Similarly,D(u)B= B D(u), u SU(2). (143)

Thus, B , an operator onV, commutes with all elements of irreducible

representationD, and is therefore a multiple of the identity operatoronV : B=bI. Likewise, B= bI onV.

If A0 = 0, this can be possible only if N = N, and A0 is non-singular. But if A0 is non-singular, then D

and D are equivalent,since D(u)A0 = A0D, u SU(2). But this contradicts the as-sumption in the theorem, hence A0 = 0. To complete the proof, selectAnr = 1 for any desired n, r and set all of the other elements equal tozero.

Next, we quote the corresponding orthonormality theorem among ele-

ments of the same irreducible representation:Theorem: Let u D(u) be a (continuous) irreducible representation of

SU(2) on a carrier space of dimension d. ThenSU(2)

d(u)Dmn(u1)Drs(u) = msnr/d. (144)

Proof: The proof of this theorem is similar to the preceding theorem. LetAbe an arbitrary d d matrix, and define

A0 SU(2)

d(u)D(u1)AD(u). (145)

As before, we may show that

D(u)A0= A0D(u), (146)

and hence A0 =aI is a multiple of the identity. We take the trace tofind the multiple:

a = 1

dTr

SU(2)

d(u)D(u1)AD(u)

(147)

=

1

dSU(2) d(u)Tr

D(u

1

)AD(u)

(148)

= 1

dTrA. (149)

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This yields the result

SU(2) d(u)D(u1)AD(u) =

Tr(A)

d I. (150)

Again, select A with any desired element equal to one, and all otherelements equal to 0, to finish the proof.

We consider now the set of all irreducible representations ofSU(2). Moreprecisely, we do not distinguish between equivalent representations, so this setis the union of all equivalence classes of irreducible representations. Use thesymbolj to label an equivalence class, i.e.,j is an index, taking on values inan index set in 1:1 correspondence with the set of all equivalence classes. WedenoteD(u) = Dj(u) to indicate that a particular irreducible representationuD(u) belongs to equivalence class j. Two representations Dj(u) andDj

(u) are inequivalent if j= j . Let dj be the dimension associated withequivalence class j. With this new notation, we may restate our above twotheorems in the form:

SU(2)d(u)Djmn(u

1)Dj

rs =1

djj msnr. (151)

This is an important theorem in representation theory, and is sometimesreferred to as the General Orthogonality Relation.

For much of what we need, we can deal with simpler objects than the fullrepresentation matrices. In particular, the traces are very useful invariantsunder similarity transformations. So, we define:

Def: The character (u) of a finite-dimensional representation uD(u)ofS U(2) is the function on S U(2):

(u) = Tr[D(u)] . (152)

We immediately remark that the characters of two equivalent representationsare identical, since

TrSD(u)S1

= Tr[D(u)] . (153)

In fact, we shall see that the representation is completely determined by thecharacters, up to similarity transformations.

Let j(u) denote the character of irreducible representation Dj(u). Theindex j uniquely determines j(u). We may summarize some importantproperties of characters in a theorem:

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Theorem: 1. For any finite-dimensional representationu D(u) ofSU(2):(u0uu

10 ) =(u), u, u0SU(2). (154)

2.(u) =(u) = (u1) =(u), u SU(2). (155)

3. For the irreducible representationsu Dj(u) ofS U(2):SU(2)

d(u)j(u0u1)Dj

(u) = 1

djjj D

j(u0) (156)

SU(2)

d(u)j(u0u1)j(u) =

1

djjj j(u0) (157)

SU(2) d(u)j(u1)j(u) = SU(2) d(u)

j(u)j(u) =jj . (158)

Proof: (Selected portions)

1.

(u0uu10 ) = Tr

D(u0uu

10 )

= TrD(u0)D(u)D(u

10 )

= (u). (159)

2.

(u1

) = TrD(u

1

)

= TrD(u)

1= Tr

(SU(u)S1)1

, where U is a unitary representation,

= TrS1U(u)S

= Tr

U(u)

= (u). (160)

The property (u) =(u) holds forS U(2), but not more gener-ally [e.g., it doesnt hold for SU(3)]. It holds forSU(2) becausethe replacement u u gives an equivalent representation forSU(2). Let us demonstrate this. Consider the parameterization:

u= ue() = cos

2I i sin

2e . (161)

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Now form the complex conjugate, and make the following similar-ity transformation:

2u12 = 2

cos

2I+ i sin

2e 12

= cos

2I+ i sin

2[e1212 e2222+ e3232]

= cos

2I i sin

2e

= u. (162)

We thus see that u and u are equivalent representations forSU(2). Now, for representation D (noting that i2SU(2)):

D(u) =D(i2)D(u)D1

(i2), (163)

and hence, (u) =(u).

3. We start with the general orthogonality relation, and use

j(u0u1) =

m,n

Djnm(u0)Djmn(u

1), (164)

to obtainSU(2)

d(u)j(u0u1)Dj

rs(u) =SU(2)

d(u)

m,nDjnm(u0)D

jmn(u

1)Dj

rs(u)

=m,n

Djnm(u0)1dj

jj nrms

= 1

djjj D

jrs(u0). (165)

Now take the trace of both sides of this, as a matrix equation:

SU(2)

d(u)j(u0u1)j(u) =

1

djjj j(u0). (166)

Let u0 = I. The character of the identity is just dj, hence we

obtain our last two relations.

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9 Reduction of Representations

Theorem: Every continuous finite dimensional representation u D(u) of

SU(2) iscompletely reducible

,i.e.

, it is the direct sum:D(u) =

r

Dr(u) (167)

of a finite number of irrepsDr(u). Themultiplicitiesmj (the numberof irrepsDr which belong to the equivalence class of irreps characterizedby index j ) are unique, and they are given by:

mj =SU(2)

d(u)j(u1)(u), (168)

where (u) = Tr[D(u)]. Two continuous finite dimensional represen-

tationsD(u) and D(u) are equivalent if and only if their characters(u) and(u) are identical as functions onS U(2).

Proof: It is sufficient to consider the case where D(u) is unitary and re-ducible. In this case, there exists a proper subspace of the carrierspace ofD(u), with projectionE, which is mapped into itself byD(u):

ED(u)E= D(u)E. (169)

Take the hermitian conjugate, and relabelu u1:ED(u1)E

= D(u1)E (170)ED(u1)E = ED(u1) (171)

ED(u)E = ED(u). (172)

Hence, ED(u) =D(u)E, for all elements uin S U(2).

Now, let E =I E be the projection onto the subspace orthogonalto E. Then:

D(u) = (E+ E)D(u)(E+ E) (173)

= ED(u)E+ ED(u)E (174)

(since, e.g., ED(u)E = D(u)EE = D(u)E(I E) = 0). Thisformula describes a reduction ofD(u). IfD(u) restricted to subspace

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E (or E) is reducible, we repeat the process until we have only ir-reps remaining. The finite dimensionality of the carrier space ofD(u)implies that there are a finite number of steps to this procedure.

Thus, we obtain a set of projections Er such that

I =r

Er (175)

ErEs = rsEr (176)

D(u) =r

ErD(u)Er, (177)

where D(u) restricted to any of subspaces Er is irreducible:

D(u) =

rDr(u). (178)

The multiplicity follows from

SU(2)

d(u)j(u1)j(u) = jj . (179)

Thus,

SU(2)

d(u)j(u1)(u) =

SU(2)

j(u1)Tr

r

Dr(u)

(180)

= the number of terms in the sum

withD

r

=D

j

(181)= mj . (182)

Finally, suppose D(u) and D are equivalent. We have already shownthat the characters must be identical. Suppose, on the other hand, thatD(u) and D are inequivalent. In this case, the characters cannot beidentical, or this would violate our other relations above, as the readeris invited to demonstrate.

10 The Clebsch-Gordan Series

We are ready to embark on solving the problem of the addition of angularmomenta. LetD(u) be a representation ofSU(2) on carrier space V, and

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let D(u) be a representation ofSU(2) on carrier space V, and assume V

andV are finite dimensional. LetV =V V denote the tensor productofV andV .

The representationsD and Dinduce a representation onV in a naturalway. Define the representationD(u) onVin terms of its action on any Vof the form = as follows:

D(u)( ) = [D(u)] [D(u)] . (183)Denote this representation as D(u) =D (u) D(u) and call it the tensorproductofD andD . The matrixD(u) is the Kronecker product ofD (u)andD (u). For the characters, we clearly have

(u) = (u)(u). (184)

We can extend this tensor product definition to the product of any finitenumber of representations.

The tensor product of two irreps, Dj

(u) and Dj

(u), is in general notirreducible. We know however, that it is completely reducible, hence a directsum of irreps:

Dj

(u) Dj(u) = j

CjjjDj(u). (185)

This is called a Clebsch-Gordan series. The Cjjj coefficients are some-times referred to as Clebsch-Gordan coefficients, although we tend to use thatname for a different set of coefficients. These coefficients must, of course, benon-negative integers. We have the corresponding identity:

j(u)j(u) =j

Cjjjj(u). (186)

We now come to the important theorems on combining angular momentain quantum mechanics:

Theorem:

1. There exists only a countably infinite number of inequivalent ir-reps ofSU(2). For every positive integer (2j +1), j= 0, 1

2, 1, 3

2, . . .,

there exists precisely one irrepDj(u) (up to similarity transforma-

tions) of dimension (2j+ 1). As 2j runs through all non-negativeintegers, the representations Dj(u) exhaust the set of all equiva-lence classes.

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2. The character of irrepDj(u) is:

j[ueee()] =sin 2j+1

2

sin1

2

, (187)

andj(I) =dj = 2j+ 1. (188)

3. The representation Dj(u) occurs precisely once in the reductionof the 2j-fold tensor product (u)2j , and we have:

SU(2)

d(u)j(u) [Tr(u)]2j = 1. (189)

Proof: We take fromSU(2) d(u)j(u)j(u) = jj the suggestion that the

characters of the irreps are a complete set of orthonormal functionson a Hilbert space of class-functions ofSU(2) (A class-function is afunction which takes the same value for every element in a conjugateclass).

Consider the following function on S U(2):

(u) 1 18

Tr(u I)(u I)

=

1

2

1 +

1

2Tr(u)

, (190)

which satisfies the conditions 1 > (u) 0 for u= I, and u(I) = 1(e.g., noting that ueee() = cos

2

I + a traceless piece). Hence, we have

the lemma: Iff(u) is a continuous function onS U(2), then

limn

SU(2) d(u)f(u) [(u)]

nSU(2) d(u) [(u)]

n =f(I). (191)

The intuition behind this lemma is that, as n , [(u)]n becomesincreasingly peaked about u = I.

Thus, ifDj(u) is any irrep ofSU(2), then there exists an integernsuchthat:

SU(2) d(u)j(u1) [Tr(u)]n = 0. (192)

Therefore, the irrepDj(u) occurs in the reduction of the tensor product(u)n.

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Next, we apply the Gram-Schmidt process to the infinite sequence{[Tr(u)]n |n= 0, 1, 2, . . .} of linearly independent class functions on SU(2)to obtain the orthonormal sequence{Bn(u)|n= 0, 1, 2, . . .} of classfunctions:SU(2)

d(u)Bn(u)Bm(u) = 1

20

d sin2

2Bn[ue3()] Bm[ue3()] =nm,

(193)where we have used the measure d(u) = 1

42deeesin

2 2

d and the factthat, since Bn is a class function, it has the same value for a rotationby angle about any axis.

Now, writen() =Bn[ue3()] =Bn[ueee()] . (194)

Noting that

[Tr(u)]n =

2cos

2

n=

nm=0

n

m

ei(

n

2m), (195)

we may obtain the result

n() =sin [(n + 1)/2]

sin /2 =Bn[ueee()] =B

n[ueee()] , (196)

by adopting suitable phase conventions for the Bs. Furthermore,

SU(2) d(u)Bn(u) [Tr(u)]

m = 0 ifn > m,1 ifn = m.

(197)

We need to prove now that the functions Bn(u) are characters of theirreps. We shall prove this by induction onn. First,B0(u) = 1;B0(u) isthe character of the trivial one-dimensional identity representation:D(u) = 1. Assume now that for some integer n0 0 the functionsBn(u) for n = 0, 1, . . . , n0 are all characters of irreps. Consider thereduction of the representation (u)n0+1:

[Tr(u)]n0+1 =n0n=0

Nn0,nBn(u) +jJn0

cn0,jj(u), (198)

where Nn0,n andcn0,j are integers0, and the cn0,j sum is over irrepsDj such that the characters are not in the set{Bn(u)|n= 0, 1, . . . , n0}(j runs over a finite subset of the Jn0 index set).

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With the above fact that:

SU(2)d(u)Bn(u) [Tr(u)]

n0+1 =

0 ifn > n0+ 1,1 ifn = n0+ 1,

(199)

we have,Bn0+1(u) =

jJn0

cn0,jj(u). (200)

Squaring both sides, and averaging over SU(2) yields

1 =jJn0

(cn0,j)2. (201)

But the cn0,j are integers, so there is only one term, with cn0,j = 1.Thus, Bn0+1(u) is a character of an irrep, and we see that there are a

countably infinite number of (inequivalent) irreducible representations.Let us obtain the dimensionality of the irreducible representation. TheBn(u), n = 0, 1, 2, . . . correspond to characters of irreps. Well labelthese irredusible representations Dj(u) according to the 1 : 1 mapping2j= n. Then

dj = j(I) =B2j(I) =2j(0) (202)

= lim0

sin [(2j+ 1)/2]

sin /2 (203)

= 2j+ 1. (204)

Next, we consider the reduction of the tensor product of two irreduciblerepresentations ofSU(2). This gives our rule for combining angular momen-tum. That is, it tells us what angular momenta may be present in a systemcomposed of two components with angular momenta. For example, it maybe applied to the combination of a spin and an orbital angular momentum.

Theorem: Let j1 and j2 index two irreps of SU(2). Then the Clebsch-Gordan series for the tensor product of these representations is:

Dj1

(u) Dj2

(u) =

j1+j2j=|j1j2|D

j

(u). (205)

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Equivalently,

j1(u)j2(u) =j1+j2

j=|j1j2|

j(u) = j

Cj1j2jj(u), (206)

where

Cj1j2j =SU(2)

d(u)j1(u)j2(u)j(u) (207)

=

1 iffj1+j2+j is an integer, and a triangle canbe formed with sides j1, j2, j,

0 otherwise.

(208)

The proof of this theorem is straightforward, by considering

Cj1j2j = 1

2

0d sin

(j1+

12

)

sin

(j2+12

) jm=j

eim, (209)

etc., as the reader is encouraged to carry out.We have found all the irreps ofSU(2). Which are also irreps ofO+(3)?

This is the subject of the next theorem:

Theorem: If 2j is an odd integer, then Dj(u) is a faithful representationofSU(2), and hence is not a representation ofO+(3). If 2j >0 is aneven integer (and hence j >0 is an integer), then Reee()Dj[ueee()]is a faithful representation of O+(3). Except for the trivial identityrepresentation, all irreps ofO+(3) are of this form.

The proof of this theorem is left to the reader.We will not concern ourselves especially much with issues of constructing

a proper Hilbert space, such as completeness, here. Instead, well concentrateon making the connection between S U(2) representation theory and angularmomentum in quantum mechanics a bit more concrete. We thus introducethe quantum mechanical angular momentum operators.

Theorem: Letu U(u) be a (strongly-)continuous unitary representationofSU(2) on Hilbert space

H. Then there exists a set of 3 self-adjoint

operatorsJk, k= 1, 2, 3 such that

U[ueee()] = exp(ieee JJJ). (210)

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To keep things simple, well consider now U(u) = D(u), where D(u) isa finite-dimensional representation the appropriate extension to thegeneral case may be demonstrated, but takes some care, and well omit

it here.1. The functionD[ueee()] is (for eeefixed), an infinitely differentiable

function of. Define the matricesJ(eee) by:

J(eee)i

D [ueee()]

=0

. (211)

Also, letJ1 = J(eee1),J2= J(eee2),J3 = J(eee3). Then

J(eee) = eee JJJ=3

k=1(eee eeek)Jk. (212)

For any unit vector eeeand any , we have

D [ueee()] = exp(ieee JJJ). (213)The matricesJk satisfy:

[Jk, J] =ikmJm. (214)

The matricesiJk, k= 1, 2, 3 form a basis for a representation ofthe Lie algebra ofO+(3) under the correspondence:

Jk iJk, k= 1, 2, 3. (215)The matricesJk are hermitian if and only ifD(u) is unitary.

2. The matricesJk, k = 1, 2, 3 form an irreducible set if and only ifthe representation D(u) is irreducible. If D(u) = Dj(u) is irre-ducible, then for any eee, the eigenvalues of eee JJJare the numbersm=j, j+ 1, . . . , j 1, j, and each eigenvalue has multiplicityone. Furthermore:

JJJ2 =J21 + J22 + J

23 =j(j+ 1). (216)

3. For any representationD(u) we have

D(u)JJJ2D(u1) = JJJ2; [Jk, JJJ2] = 0, k= 1, 2, 3. (217)

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Proof: (Partial) Consider first the case whereD(u) = Dj(u) is an irrep. LetMj ={m|m=j, . . . , j}, let eeebe a fixed unit vector, and define theoperatorsFm(eee) (withmMj) by:

Fm(eee) 14

40

deimDj[ueee()]. (218)

Multiply this defining equation by Dj[ueee()]:

Dj [ueee()]Fm(eee) =

1

4

40

deimDj[ueee(+ )] (219)

= 1

4

40

deim()Dj[ueee()] (220)

= e(im)Fm(eee). (221)

Thus, eitherFm(eee) = 0, or e(im) is an eigenvalue ofDj[ueee()]. But

j [ueee()] =j

n=jein, (222)

and hence,

Tr[Fm(eee)] =

1 ifmMj ,0 otherwise.

(223)

Therefore,Fm(eee)= 0.We see that the

{Fm(eee)

}form a set of 2j+1 independent one-dimensional

projection operators, and we can represent the Dj(u) by:

Dj[ueee()] =j

m=jeimFm(eee). (224)

From this, we obtain:

J(eee) i

Dj [ueee()]

=0

=j

m=jmFm(eee), (225)

and Dj[ueee()] = exp [iJ(eee)] , (226)which is an entire function of for fixed eee.

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Since every finite-dimensional continuous representation D(u) is a di-rect sum of a finite number of irreps, this result holds for any suchrepresentation:

D[ueee()] = exp [iJ(eee)] . (227)Let www be a unit vector with components wk, k = 1, 2, 3. Considersmall rotations about www:

uwww() =ueee1(w1)ueee2(w2)ueee3(w3)ueee(,www)[(, www)], (228)

where(, www) =O(2) for small . Thus,

exp[iJ(www)] =eiw1J1eiw2J2eiw3J3eiJ(eee). (229)

Expanding the exponentials and equating coefficients of terms linear in

yields the result:

J(www) =w1J1+ w2J2+ w3J3 = www JJJ. (230)

To obtain the commutation relations, consider

ueee1()ueee2()u1eee1() (231)

=

cos

2I i sin

21

cos

2I i sin

22

(232)

cos2

I+ i sin

21 (233)

= cos

2I i sin

2(cos 2+ sin 3) (234)

=ueee2cos +eee3sin (). (235)

Thus,

exp(iJ1) exp(iJ2)exp(iJ1) = exp [i(J2cos + J3sin )] .(236)

Expanding the exponentials, we have:

n=0

=0

m=0

n+mJn1 J2Jm1

n!!m! (i)n+im =

r=0

(i)rr(J2cos +J3sin )r1r!

.

(237)

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We equate coefficients of the same powers of, . In particular, theterms of order yield the result:

[J1, J

2] =iJ

3. (238)

We thus also have, for example:

[J1, JJJ2] = [J1, J

21 + J

22 + J

23 ] (239)

= J1J2J2 J2J2J1+ J1J3J3 J3J3J1 (240)= (iJ3+ J2J1)J2 J2(iJ3+ J1J2) + (241)

(iJ2+ J3J1)J3 J3(iJ2+ J1J3) (242)= 0. (243)

As a consequence, we also have:

D(u)JJJ2D(u1) = eieeeJJJJJJ2eieeeJJJ = JJJ2. (244)

In particular, this is true for an irrep:

Dj(u)JJJ2Dj(u1) = JJJ2. (245)

Therefore JJJ2 is a multiple of the identity (often referred to as a casimiroperator).

Let us determine the multiple. Take the trace:

Tr(JJJ2) = 3Tr(J23 ) (246)

= 3Tr

Dj [ueee3()]

2=0

(247)

= 3Tr

lim0

0

1

Dj (ueee3(+ )) Dj (ueee3())

(248)

Dj (ueee3(+

)) Dj (ueee3())

=0

(249)

=

3Tr lim0

0

1

Dj (ueee3( + )) (250)

Dj (ueee3()) Dj (ueee3()) + Dj (ueee3(0))

(251)

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= 3Tr

2

2Dj [ueee3()]

=0

(252)

=

3 2

2j [u

eee3()]

=0

. (253)

There are different ways to evaluate this. We could insert

j [ueee3()] =sin(j+ 12)

sin 12

, (254)

or we could use

j [ueee3()] =j

m=jeim. (255)

In either case, we find:

Tr(JJJ2) =j(j+ 1)(2j+ 1). (256)

Since dj = 2j+ 1, this gives the familiar result:

JJJ2 =j(j+ 1)I. (257)

Finally, well compute the eigenvalues of eee JJJ= J(eee) =jm=jmFm(eee).We showed earlier that the Fm(eee) are one-dimensional projection oper-ators, and it thus may readily be demonstrated that

J(eee)Fm(eee) =mFm(eee). (258)

Hence, the desired eigenvalues are m= {j, j+ 1, . . . , j 1, j}.

11 Standard Conventions

Lets briefly summarize where we are. Well pick some standard conven-tions towards building explicit representations for rotations in quantum me-chanics.

The rotation group in quantum mechanics is postulated to be SU(2),with elements

u= ueee() = ei

2eee

, (259)describing a rotation by angle about vector eee, in the clockwise senseas viewed along eee.

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O+(3) is a two to one homomorphism ofS U(2):

Rmn(u) =1

2Tr(umun). (260)

To every representation of SU(2) there corresponds a representationof the Lie algrebra ofSU(2) given by the real linear span of the threematricesiJk, k= 1, 2, 3, where

[Jm, Jn] =imnpJp. (261)

The vector operator JJJ is interpreted as angular momentum. Its squareis invariant under rotations.

The matrix group SU(2) is a representation of the abstract group

SU(2), and this representation is denoted D12 (u) = u. For this rep-

resentation, Jk= 12

k.

Every finite dimensional representation of SU(2) is equivalent to aunitary representation, and every unitary irreducible representation ofSU(2) is finite dimensional. Therefore, the generating operators, Jk,can always be chosen to be hermitian.

Let 2j be a non-negative integer. To every 2j there corresponds aunique irrep by unitary transformations on a 2j +1-dimensional carrierspace, which we denote

Dj =Dj(u). (262)

These representations are constructed according to conventions whichwe take to define the standard representations:The matricesJk are hermitian and constructed according to the follow-ing: Let|j, m, m=j, j+ 1, . . . , j 1, j be a complete orthonormalbasis in the carrier space such that:

JJJ2|j, m = j(j+ 1)|j, m (263)J3|j, m = m|j, m (264)

J+|j, m

= (j m)(j+ m+ 1)|j, m+ 1 (265)

J|j, m =

(j+ m)(j m+ 1)|j, m 1, (266)

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whereJJ1 iJ2. (267)

According to convention, matrices J1 andJ3 are real, and matrix J2 is

pure imaginary. The matrix

Dj [ueee()] = exp(ieee JJJ), (268)describes a rotation by about unit vector eee.

Ifj = 12

-integer, then Dj(u) are faithful representations ofSU(2). Ifjis an integer, thenDj(u) are representations ofO+(3) (and are faithfulif j > 0). Also, if j is an integer, then the representation Dj(u) issimilar to a representation by real matrices:

Rmn(u) = 3

j(j+ 1)(2j+ 1) TrDj(u)JmD

j

(u)Jn

. (269)

In the standard basis, the matrix elements ofDj(u) are denoted:Djm1m2(u) =j, m1|Dj(u)|j, m2, (270)

and thus,

Dj(u)|j, m =j

m=jDjmm(u)|j, m. (271)

Let

|

be an element in the carrier space ofDj, and let

|

=Dj(u)

|

be the result of applying rotation Dj(u) to|. We may expand thesevectors in the basis:

| = m

m|j, m (272)

| = m

m|j, m. (273)

Thenm=

m

Djmm(u)m. (274)

Since matrices uand Dj(u) are unitary,Djm1m2(u

1) =Djm1m2(u) =Djm2m1(u). (275)

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The representation Dj is the symmetrized (2j)-fold tensor product ofthe representation D

12 = SU(2) with itself. For the standard repre-

sentation, this is expressed by an explicit formula for matrix elements

Djm1m2(u) as polynomials of degree 2jin the matrix elements ofSU(2):

Define the quantities Djm1m2(u) form1, m2=j, . . . , j by:|u|2j = (1u111+ 1u122+ 2u211+ 2u222)2j (276)

= (2j)!m1,m2

(277)

j+m11 jm12

j+m21

jm22

(j+ m1)!(j m1)!(j+ m2)!(j m2)!Djm1m2(u).

We defer to later the demonstration that the matrix elementsDjm1m2(u)so defined are identical with the earlier definition for the standard rep-

resentation. A consequence of this formula (which the reader is encour-aged to demonstrate) is that, in the standard representation,

Dj(u) = Dj(u) (278)

Dj(uT) = DjT(u). (279)

Also, noting that u= 2u2, we obtain

Dj(u) = exp(iJ2)Dj(u) exp(iJ2), (280)making explicit our earlier statement that the congugate representationwas equivalent in SU(2) [We remark that this property does not hold

forS U(n), ifn >2].

We can also describe the standard representation in terms of an ac-tion of the rotation group on homogeneous polynomials of degree 2jof complex variables x and y. We define, for each j = 0, 12 , 1, . . ., andm= j, . . . , j the polynomial:

Pjm(x, y) xj+myjm

(j+ m)!(j m)!; P001. (281)

We also define Pjm 0 if m / {j, . . . , j}. In addition, define thedifferential operators Jk, k= 1, 2, 3, J+, J:

J3 = 1

2(xx yy) (282)

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J1 = 1

2(xy+ yx) =

1

2(J++ J) (283)

J2 = i

2(yx

xy) =

i

2(J

J+) (284)

J+ = xy =J1+ iJ2 (285)

J = yx = J1 iJ2 (286)These definitions give

JJJ2 =1

4

(xx yy)2 + 2(xx+ yy)

. (287)

We let these operators act on our polynomials:

J3Pjm(x, y) = 1

2(xx yy)

xj+myjm

(j+ m)!(j m)!

(288)

= 1

2[j+ m (j m)]Pjm(x, y) (289)

= mPjm(x, y). (290)

Similarly,

J+Pjm(x, y) = (xy) xj+myjm(j+ m)!(j m)!

(291)

= (j m) xj+m+1yjm1

(j+ m)!(j m)!(292)

= (j m)(j+ m + 1)!(j m 1)!

(j+ m)!(j m)! Pj,m+1(x, y)

=

(j m)(j+ m + 1)Pj,m+1(x, y). (293)Likewise,

JPjm(x, y) =

(j+ m)(j m+ 1)Pj,m1(x, y), (294)and

JJJ2Pjm(x, y) = j(j+ 1)Pjm(x, y). (295)

We see that the actions of these differential operators on the monomials,Pjm, are according to the standard representation of the Lie algebra of

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the rotation group (that is, we compare with the actions of the standardrepresentation for JJJon orthonormal basis|j, m).Thus, regarding Pjm(x, y) as our basis, a rotation corresponds to:

Dj(u)Pjm(x, y) =m

Djmm(u)Pjm(x, y). (296)

Now,

D12 (u)P1

2m(x, y) =

m

D12

mm(u)P12m(x, y) (297)

=m

umm(u)P12m(x, y). (298)

Or,

uP12m(x, y) = u12mP12

12 (x, y) + u 12mP12 12 (x, y). (299)

With P1212

(x, y) =x, and P12 12

(x, y) = y , we thus have (using normal

matrix indices now on u)

uP1212

(x, y) = u11x + u21y, (300)

uP12 12

(x, y) = u12x + u22y. (301)

Hence,

Dj(u)Pjm(x, y) = Pjm(u11x+ u21y, u12x+ u22y) (302)

=m

Djmm(u)Pjm(x, y). (303)

Any homogeneous polynomial of degree 2j in (x, y) can be written as aunique linear combination of the monomials Pjm(x, y). Therefore, theset of all such polynomials forms a vector space of dimension 2j+ 1,and carries the standard representationDj of the rotation group if theaction of the group elements on the basis vectorsPjm is as above. Notethat

Pjm(x, y)Pjm

(x, y) =

j+mx jmy x

j+myjm

(j+ m)!(j m)!(j+ m)!(j m)!

= mm . (304)

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Apply this to

Pjm(u11x+ u21y, u12x+ u22y) =m

D12

mm(u)P12m(x, y) : (305)

Pjm(x, y)m

D12

mm(u)P12m(x, y) = D

jmm(u). (306)

Hence,

Djmm(u) = Pjm(x, y)Pjm(u11x + u21y, u12x + u22y), (307)

and we see that Djmm(u) is a homogeneous polynomial of degree 2j inthe matrix elements ofu.

Now,

j

m=j

Pjm(x1, y1)Pjm(x2, y2) =j

m=j

(x1x2)j+m(y1+ y2)

jm

(j+ m)!(j m)! . (308)

Using the binomial theorem, we can write:

(x1x2+ y1y2)2j

(2j)! =

jm=j

(x1x2)j+m(y1+ y2)

jm

(j+ m)!(j m)! . (309)

Thus,j

m=jPjm(x1, y1)Pjm(x2, y2) =

(x1x2+ y1y2)2j

(2j)! . (310)

One final step remains to get our asserted equation defining the Dj(u)standard representation in terms ofu:

m1,m2

j+m11 jm12

j+m21

jm22

(j+ m1)!(j m1)!(j+ m2)!(j m2)!Djm1m2(u) (311)

=m1,m2

Pjm1(1, 2)Djm1m2

(u)Pjm2(1, 2) (312)

=m2

Pjm2(u111+ u212, u121+ u222)Pjm2(1, 2) (313)

= 1

(2j)!(1u111+ 1u122+ 2u211+ 2u222)

2j . (314)

The step in obtaining Eqn. 313 follows from Eqn. 303, or it can be

demonstrated by an explicit computation. Thus, we have now demon-strated our earlier formula, Eqn. 277, for the standard representationforDj.

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12 Special Cases

We have obtained a general expression for the rotation matrices for an irrep.

Let us consider some special cases, and derive some more directly usefulformulas for the matrix elements of the rotation matrices.

1. Consider, in the standard representation, a rotation by angle aboutthe coordinate axes. Let 1= exp(i1/2) =i1. Using

Djmm(u) = Pjm(x, y)Pjm(u11x + u21y, u12x + u22y), (315)

we find:

Djmm(1) = Pjm(x, y)Pjm(iy, ix), (316)= Pjm(x, y)()jPjm(x, y), (317)

= eij

mm

. (318)Hence,

exp(iJ1)|j, m= eij |j, m. (319)Likewise, we define

2 = exp(i2/2) =i2, (320)3 = exp(i3/2) =i3, (321)

which have the properties:

12 = 21= 3, (322)23 = 32= 1, (323)31 = 13= 2, (324)

and hence,Dj(2) =D

j(3)Dj(1). (325)

In the standard representation, we already know that

exp(iJ3)|j, m =eim|j, m. (326)Therefore,

exp(

iJ2)|j, m

= exp(

iJ3) exp(

iJ1)

|j, m

(327)

= exp(iJ3) exp(ij)|j, m (328)= exp(i(j m))|j, m. (329)

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2. Consider the parameterization by Euler angles, , :

u= eJ3eJ2eJ3 , (330)

(hereJk= i2k) or,Dj(u) =Dj(, , ) =eiJ3eiJ2eiJ3, (331)

where it is sufficient (for all elements ofS U(2)) to choose the range ofparameters:

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Thedj functions are homogeneous polynomials of degree 2jin cos(/2)and sin(/2). Note that slightly different conventions from those hereare sometimes used for the big-Dand little-d functions.

The Djm1m2(u) functions form a complete and essentially orthonormalbasis of the space of square integrable functions on S U(2):

SU(2)

d(u)Djm1m2(u)Dj

m1m2(u) =

jj m1m1m2m22j+ 1

. (344)

In terms of the Euler angles, d(u) = 1162 d sin dd, and

jj

2j+ 1 =

1

162

20

d 0

sin d 40

d (345)

ei(m1+m2)ei(m1+m2)djm1m2()dj

m1m2() (346)

= 12

0sin ddjm1m2()d

jm1m2

(). (347)

3. Spherical harmonics and Legendre polynomials: TheYm functions arespecial cases of the Dj . Hence we may define:

Ym(, )

2 + 1

4 Dm0(,, 0), (348)

where is an integer, and m {, + 1, . . . , }.This is equivalent to the standard definition, where the Yms are con-

structed to transform under rotations like |m basis vectors, and where

Y0(, 0) =

2 + 1

4 P(cos ). (349)

According to our definition,

Y0(, 0) =

2 + 1

4 D00(0, , 0) =

2 + 1

4 d00(). (350)

Thus, we need to show that d00() = P(cos ). This may be done bycomparing the generating function for the Legendre polynomials:

=0

tP(x) = 1/

1 2tx+ t2, (351)

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with the generating function for d00(), obtained from considering thegenerating function for the characters of the irreps ofS U(2):

j j(u)z

2j

= 1/(1 2 z+ z2

), (352)

where= Tr(u/2). We havent discussed this generating function, andwe leave it to the reader to pursue this demonstration further.

The other aspect of our assertion of equivalence requiring proof con-cerns the behavior of our spherical harmonics under rotations. WritingYm(eee) =Ym(, ), where, are the polar angles of unit vector eee, wecan express our definition in the form:

Ym[R(u)eee3] = 2 + 14

D0m(u1), (353)

since,

Ym[R(u)eee3] = Ym(eee) =Ym(, ) (354)

=

2 + 1

4 Dm0(,, 0) (355)

=

2 + 1

4 j, m|D(u)|j, 0 (356)

= 2 + 14

D(u)(j, 0)||j, m

(357)

=

2 + 1

4 j, 0|D(u)|j, m. (358)

To anyu0SU(2) corresponds R(uo) on any functionf(eee) on the unitsphere as follows:

R(uo)f(eee) =f

R1(u0)eee

. (359)

Thus,

R(uo)Ym(eee) = YmR1(u0)eee

(360)

= YmR1(u0)R(u)eee3

(361)

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= YmR(u10 u)eee3

(362)

=

2 + 1

4 D0m(u

1u0) (363)

=

2 + 1

4

m

D0m(u1)Dmm(u0) (364)

=m

Dmm(u0)Ym(eee). (365)

This shows that the Ym transform under rotations according to the|, m basis vectors.We immediately have the following properties:

(a) If JJJ= ixxx is the angular momentum operator, thenJ3Ym(eee) = mYm(eee) (366)JJJ2Ym(eee) = ( + 1)Ym(eee). (367)

(b) From thDD orthogonality relation, we further have: 20

d 0

d sin Ym(, )Ym(, ) =mm. (368)

The Ym(, ) form a complete orthonormal set in the space ofsquare-integrable functions on the unit sphere.

We give a proof of completeness here: Iff(, ) is square-integrable,and if 2

0d 0

d sin f(, )Ym(, ) = 0,(, m), (369)

we must show that this means that the integral of|f|2 vanishes.This follows from the completeness of Dj(u) on SU(2): Extendthe domain of definition offtoF(u) =F(, , ) =f(, ),u SU(2). Then,

SU(2)

d(u)F(u)Djmm(u) =

0 ifm= 0, by assumption0 ifm= 0 since F(u) is independent

of, and40 de

im = 0.

(370)Hence,

SU(2) d(u)|F(u)|2 = 0 by the completeness ofDj(u), and

therefore

(4) d|f|2 = 0.

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(c) We recall thatY0(, 0) =

2+14

P(cos ). With

R(uo)Ym(eee) =Ym R1(u0)eee=

m

Dmm(u0)Ym(eee), (371)

we have, form = 0,

R(uo)Y0(eee) = Y0

R1(u0)eee

=m

Dm0(u0)Ym(eee) (372)

=

4

2 + 1

m

Ym(eee)Ym(eee), (373)

where we have defined R(u0)eee3= eee. But

eee

eee= eee

[R(u0)eee3 = [R

1(u0)eee]

eee3, (374)

and

Y0[R1(u0)eee] = function of [R1(u0)eee] only, (375)

=

2 + 1

4 P(cos ), (376)

where cos = eee eee. Thus, we have the addition theorem forspherical harmonics:

P(eee eee) = 4

2 + 1m Ym(eee)Ym(eee). (377)

(d) In momentum space, we can therefore write

(3)(ppp qqq) = (p q)4pq

=0

(2 + 1)P(cos ) (378)

= (p q)

4pq

=0

m=

Ym(p, p)Ym(q, q),(379)

wherep= |ppp|, and is the angle between pppand qqq.(e) Let us see now how we may compute the dj() functions. Recall

Djmm(u) =Pjm(x, y)Pjm(u11x + u21y, u12x + u22y). (380)

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With

u(0, , 0) = ei

22 = cos

2I i sin

22 (381)

=

cos 2 sin 2sin

2 cos

2

, (382)

we have:

djmm() = Djmm(0, , 0) (383)

=Pjm(x, y)Pjm(x cos

2+ y sin

2, x sin

2+ y cos

2) (384)

=j+mx

jmy (x cos

2

+ y sin 2

)j+m

(x sin 2

+ y cos 2

)jm

(j+ m)!(j m)!(j+ m)!(j m)!

.(385)

Thus, we have an explicit means to compute the little-dfunctions.An alternate equation, which is left to the reader to derive (againusing the Pjm functions), is

djmm() = 1

2

(j+ m)!(j m)!(j+ m)!(j m)! (386)

20

dei(mm)(cos

2+ ei sin

2)j+m

(cos

2 ei sin

2)jm

In tabulating the little-dfunctions it is standard to use the label-

ing:

dj =

djjj . . . djj,j

... . . .

...djj,j . . . d

jj,j

. (387)

For example, we find:

d12 () =

cos 2 sin 2sin 2 cos

2

, (388)

and

d1() =

1

2

(1 + cos )

1

2sin 1

2

(1

cos )12sin cos 1

2sin

12

(1 cos ) 12sin 1

2(1 + cos )

. (389)

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13 Clebsch-Gordan (Vector Addition) Coef-ficients

We consider the tensor (or Kronecker) product of two irreps, and its reductionaccording to the Clebsch-Gordan series:

Dj1(u) Dj2(u) =j1+j2

j=|j1j2|Dj(u). (390)

The carrier space of the representationDj1Dj2 is of dimension (2j1+1)(2j2+1). It is the tensor product of the carrier spaces for the representations Dj1

andDj2 . Corresponding to the above reduction of the kronecker product, wehave a decomposition of the carrier space into orthogonal subspaces carrying

the representations Dj

.For each j we can select a standard basis system:

|j1j2;j, m=m1,m2

C(j1j2j; m1m2m)|j1, m1;j2, m2, (391)

where the latter ket is just the tensor product of the standard basis vectors|j1, m1 and |j2, m2. The coefficientsCare the vector-addition coefficients,or Wigner coefficients, or, more commonly now, Clebsch-Gordan (CG)coefficients. These coefficients must be selected so that the unit vectors|j1j2;j, m transform under rotations according to the standard representa-tion.

We notice that the CG coefficients relate two systems of orthonormalbasis vectors. Hence, they are matrix elements of a unitary matrix with rowslabeled by the (m1, m2) pair, and columns labeled by the (j, m) pair. Thus,we have the orthonormality relations:

m1,m2

C(j1j2j; m1m2m)C(j1j2j ; m1m2m) = jj mm , (392)

j,m

C(j1j2j; m1m2m)C(j1j2j; m1m2m) = m1m1m2m2 . (393)

The inverse basis transformation is:

|j1, m1;j2, m2 =j,m

C(j1j2j; m1m2m)|j1j2;j, m. (394)

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We wish to learn more about the CG coefficients, including how to com-pute them in general. Towards accomplishing this, evaluate the matrix ele-ments of the Clebsch-Gordan series, with

Dj1m1m1

(u) =j1, m1|Dj1(u)|j1, m1, (395)

etc. Thus we obtain the explicit reduction formula for the matrices of thestandard representations:

Dj1m1m1

(u)Dj2m2m2

(u) = j1, m1;j2, m2|j

Dj(u)|j1, m1;j2, m2 (396)

=j

m

C(j1j2j; m1m

2m

)j1j2;j, m|Dj(u)m

C(j1j2j; m1m2m

)|j1j2;j, m

=

j,m,mC(j1j2j; m

1m

2m

)C(j1j2j; m1m2m

)Djmm(u). (397)

Next, we take this equation, multiply by Djmm(u), integrate overSU(2),and use orthogonality to obtain:

C(j1j2j; m1m

2m

)C(j1j2j; m1m2m

) = (2j+1)SU(2)

d(u)Djmm(u)Dj1m1m1

(u)Dj2m2m2

(u).

(398)Thus, the CG coefficients for the standard representations are determined bythe matrices in the standard representations, except for a phase factor whichdepends on (j1j2j) [but not on (m1m2m)].

The coefficients vanish unless:

1. j {|j1j2|, |j1j2|+1, . . . , j1 +j2} (from the Clebsch-Gordan series).2. m= m1+ m2 (as will be seen anon).

3. m {j, j +1, . . . , j},m1 {j1, j1 +1, . . . , j1},m2 {j2, j2 +1, . . . , j2} (by convention).

Consider the Euler angle parameterization for the matrix elements, giv-ing:

C(j1j2j; m1m

2m

)C(j1j2j; m1m2m

)

= (2j+ 1) 1162

20

d 40

d 0

sin d (399)

exp {i [(m+ m1+ m2)+ (m+ m1+ m2)]} djmm()dj1m1m1

()dj2m2m2

().

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We see that this is zero unless m =m1+m2 andm

=m1+ m2, verifying

the above assertion. Hence, we have:

C(j1j2j; m1m2m)C(j1j2j; m1m2m) (400)=

(2j+ 1)

2

0

sin ddjmm()dj1m1m1

()dj2m2m2

().

Now, to put things in a more symmetric form, letm3=m,m3 = m,and use

dj3m3,m

3() = ()m3m3 dj3m

3,m

3(), (401)

to obtain the result:

2

2j+ 1()m3m3 C(j1j2j3; m1m2, m3)C(j1j2j3; m1m2, m3) (402)

= 0

sin ddj1m1m1

()dj2m2m2

()dj3m3m3

().

The d-functions are real. Thus, we can choose our arbitrary phase inC forgiven j1j2j3 so that at least one non-zero coefficient is real. Then, accordingto this formula, all coefficients for given j1j2j3 will be real. We thereforeadopt the convention that all CG coefficients are real.

The right-hand side of Eqn. 402 is highly symmetric. Thus, it is usefulto define the 3-j symbol (Wigner):

j1 j2 j3

m1 m2 m3

()j1j2m3C(j1j2j3; m1m2, m3)

2j3+ 1 . (403)

In terms of these 3-j symbols we have:

j1 j2 j3m1 m

2 m

3

j1 j2 j3m1 m

2 m

3

(404)

=1

2

0

sin d()dj1m1m1

()dj2m2m2

()dj3m3m3

(),

form 1+ m2+ m

3= m

1+ m

2+ m

3 = 0.

According to the symmetry of the right hand side, we see, for example,that the square of the 3-j symbol is invariant under permutations of the

columns. Furthermore, since

djmm( ) = ()jm

djmm(), (405)

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we find that j1 j2 j3m1 m2 m3

= ()j1+j2+j3

j1 j2 j3

m1

m2

m3

. (406)

It is interesting to consider the question of which irreps occur in thesymmetric, and in the anti-symmetric, tensor products ofDj with itself:

Theorem:

Dj(u) Dj(u)

s

=[j]n=0

D2j2n(u), (407)

Dj(u) Dj(u)

a

=

[j 12]

n=0

D2j2n1(u), (408)

where s denotes the symmetric tensor product, a denotes the anti-symmetric tensor product, and [j] means the greatest integer which isnot larger than j .

Proof: We prove this by considering the characters. Let S(u) [Dj(u) Dj(u)]s,andA(u)[Dj(u) Dj(u)]a. Then, by definition,

S(m1m2)(m

1m2) =

1

2

Djm

1m1

Djm2m2

+ Djm1m2

Djm2m1

, (409)

A(m1m2)(m

1m2) =

1

2

Djm

1m1

Djm2m2 Djm

1m2

Djm2m1

. (410)

(411)Taking the traces means to set m1 =m

1 andm

2 =m

2, and sum over

m1 andm2. This yields

TrS(u) = 1

2[2j(u) + j(u

2)] (412)

TrA(u) = 1

2[2j(u) j(u2)]. (413)

We need to evaluatej(u2). Ifuis a rotation by, thenu2 is a rotation

by 2, hence,

j(u2) =

jm=j

e2im (414)

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=2j

k=2jeik

2j1k=2j+1

eik +2j2

k=2j+2eik . . . (415)

=

2jn=0

()n2jn

k=2j+n eik (416)

=2jn=0

()n2jn(u). (417)

Next, consider 2j(u). Since

j(u)j(u) =j+j

j=|jj|j(u), (418)

we have

2j(u) =2jk=0

k(u) =2jn=0

2jn(u). (419)

Thus,

TrS(u) = 1

2

2jn=0

[2jn(u) + ()n2jn(u)] (420)

=[j]n=0

2j2n(u). (421)

Similarly, we obtain:

TrA(u) =

[j 12]

n=0

2j2n1(u). (422)

This completes the proof.

This theorem implies an important symmetry relation for the CG coeffi-cients when two j s are equal. From Eqn. 398, we obtain

C(j1j1j; m1m2m)C(j1j1j; m1m2m) = (2j+1)SU(2) d(u)D

j

m

m

(u)D

j1

m

1m

1 (u)D

j1

m

2m

2 (u).(423)

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But

Dj1m1m1

(u)Dj1m2m2

(u) = S(m1m2)(m

1m2)(u) + A(m

1m2)(m

1m2)(u), (424)

and the integral of this with Dj(u) picks the Dj piece of this quantity,which must be either m1 m2 symmetric or anti-symmetric, according tothe theorem we have just proven. If symmetric, then

C(j1j1j; m2m

1m

) = C(j1j1j; m1m2m

), (425)

and if anti-symmetric, then

C(j1j1j; m2m

1m

) =C(j1j1j; m1m2m). (426)Lets try a simple example: Let j1 = j2 =

12

. That is, we wish to

combine two spin-1/2 systems (with zero orbital angular momentum). Fromthe theorem:

D

12 D 12

s

=[1/2]n=0

D12n =D1, (427)

D

12 D 12

a

=[0]n=0

D12n1 =D0. (428)

Hence, the spin-1 combination is symmetric, with basis

|j= 1, m= 1; m1 =

1

2

, m2 = 1

2, (429)

12

|1, 0; 12 , 12 + |1, 0; 12 , 12

, (430)

|1, 1; 12 , 12, (431)and the spin-0 combination is antisymmetric:

12

|0, 0; 1

2, 1

2 |0, 0; 1

2, 12

. (432)

The generalization of this example is that the symmetric combinationsarej = 2j1, 2j1

2, . . ., and the antisymmetric combinations are j = 2j1

1, 2j1 3, . . .Therefore,C(j1j1j; m

2m

1m

) = ()2j1+jC(j1j1j; m1m2m). (433)

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Also, j j Jm1 m2 M

= ()2j+J

j j Jm2 m1 M

, (434)

as well as the corresponding column permutations, e.g., j J jm1 M m2

= ()2j+J

j J Jm2 M m1

, (435)

We adopt a standard construction of the 3-j and CG coefficients: First,they are selected to be real. Second, for any triplet j1j2j the 3-j symbolsare then uniquely determined, except for an overall sign which depends on

j1j2j only. By convention, we pick (this is a convention when j1, j2, j3 are alldifferent, otherwise it is required):

j1 j2 j3m1 m2 m3

= j2 j3 j1

m2 m3 m1

= ()j1+j2+j3 j2 j1 j3

m2 m1 m3

.(436)

That is, the 3-j symbol is chosen to be symmetric under cyclic permutationof the columns, and either symmetric or anti-symmetric, depending on j1+

j2+j3, under anti-cyclic permutations.Sometimes the symmetry properties are all we need, e.g., to determine

whether some process is permitted or not by angular momentum conserva-tion. However, we often need to know the CG (or 3-j) coefficients themselves.These are tabulated in many places. We can also compute them ourselves,and we now develop a general formula for doing this. We can take the fol-

lowing as the defining relation for the 3-j symbols, i.e., it can be shown tobe consistent with all of the above constraints:

G({k}; {x}, {y}) (x1y2 x2y1)2k3(x2y3 x3y2)2k1(x3y1 x1y3)2k2

(2k3)!(2k1)!(2k2)!(j1+j2+j3+ 1)!

m1m2m3

j1 j2 j3m1 m2 m3

Pj1m1(x1, y1)Pj2m2(x2, y2)Pj3m3(x3, y3), (437)

where 2k1, 2k2, 2k3 are non-negative integers given by:

2k3= j1+j2

j3; 2k1= j2+j3

j1; 2k2= j3+j1

j2. (438)

Well skip the proof of this consistency here. The interested reader maywish to look at T.Regge,Il Nuovo CimentoX (1958) 296; V. Bargmann, Rev.

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Mod. Phys. 34 (1962) 829. Since Pjm(x, y)Pjm(x, y) = mm , we obtainthe explicit formula:

j1 j2 j3m1 m2 m3

= Pj1m1(x1 , y1)Pj2m2(x2 , y2)Pj3m3(x3 , y3)G({k}; {x}, {y}).

(439)For example consider the special case in which j3 = j1+ j2 (and m3 =

(m1+ m2)). In this case, k3= 0,k1 = j2, andk2= j1. Thus, j1 j2 j1+j2m1 m2 m1 m2

(440)

= j1+m1x1

j1m1y1

j2+m2x2

j2m2y2

j1+j2m1m2x3

j1+j2+m1+m2y3

(j1+ m1)!(j1 m1)!(j2+ m2)!(j2 m2)!(j1+j2 m1 m2)!(j1+j2+ m1+ m2)!

(x2y3 x3y2)2j2(x3y1 x1y3)2j1

(2j2)!(2j1)!(2j1+ 2j2+ 1)!(441)

= ()j1+j2+m1m2 (2j1)!(2j2)!(j1+j2 m1 m2)!(j1+j2+ m1+ m2)!

(2j1+ 2j2+ 1)!(j1+ m1)!(j1 m1)!(j2+ m2)!(j2 m2)! . (442)

Forj1= j3 = j ,j2= 0, we findj 0 jm 0 m

=

()j+m2j+ 1

. (443)

We may easily derive the corresponding formulas for the CG coefficients.In constructing tables of coefficients, much computation can be saved by

using symmetry relations and orthogonality of states. For example, we havereally already computed the table for the 1

2 1

2= 1 0 case:

C(1212j; m1m2m)

j 1 1 0 1

m1 m2 m 1 0 0 -1

12

12 1

12 12 1/

2 1/

2

12 12 1/2 1/21

2 1

2 1

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14 Wigner-Eckart Theorem

Consider a complex vector spaceHop of operators onH which is closed

under U(u)QU1

(u), where Q Hop, and U(u) is a continuous unitaryrepresentation ofSU(2). We denote the action of an element of the groupSU(2) on Q Hopby

U(u)Q= U(u)QU1(u). (444)

Corresponding to this action of the group elements, we have the actionof the Lie algebra ofS U(2):

JkQ= [Jk, Q]. (445)

We have obtained this result by noting that, picking a rotation about the kaxis, withU(u) = exp(iJk),

U(u)QU1(u) = (1 iJk)Q(1 + iJk) + O(2) (446)= Q i[Jk, Q] + O(2), (447)

and comparing with

U(u)Q= exp(iJk)Q= Q iJkQ+ O(2). (448)

We may also compute the commutator:

[Jk, J]Q = Jk[J, Q] J[Jk, Q] (449)= [Jk, [J, Q]] [J, [Jk, Q]] (450)= [[Jk, J], Q] (451)

= [ikmJm, Q]. (452)

Hence,[Jk, J] =ikmJm. (453)

Def: A set of operatorsQ(j, m), m=j, j + 1, . . . , jconsists of the 2j + 1components of a spherical tensor of rank j if:

1. J3Q(j, m) = [J3, Q(j, m)] =mQ(j, m).

2. J+Q(j, m) = [J+, Q(j, m)] =

(j m)(j+ m + 1)Q(j, m+ 1).

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3. JQ(j, m) = [J, Q(j, m)] =

(j+ m)(j m + 1)Q(j, m 1).Equivalently,

JJJ2Q(j, m) =3k=1

(Jk)2Q(j, m) (454)

=3k=1

[Jk, [Jk, Q(j, m)]] (455)

= j(j+ 1)Q(j, m), (456)

andU(u)Q(j, m)U1(u) =

m

Djmm(u)Q(j, m). (457)

That is, the set of operators Q(j, m) forms a spherical tensor if they

transform under rotations like the basis vectors|jm in the standardrepresentation.For such an object, we conclude that the matrix elements ofQ(j, m) must

depend on the m values in a particular way (letting k denote any otherquantum numbers describing our state in the situation):

(jm)(k)|Q(j, m)|(j m)(k) (458)= j m|(|jm|j m)j, k||Qj||j, k (459)=A(j, j, j)C(jj j ; mmm)j , k||Qj||j , k,(460)

where a common convention lets

A(j, j , j ) = ()j+jj/

2j+ 1. (461)

The symbolj , k||Qj||j , kis called the reduced matrix element of thetensorQj . Eqn. 460 is the statement of the Wigner-Eckart theorem.

Let us try some examples:

1. Scalar operator: In the case of a scalar operator, there is only onecomponent:

Q(j, m) =Q(0, 0). (462)

The Wigner-Eckart theorem reads

(j m)(k)|Q(0, 0)|(jm)(k) (463)=

()jj2j+ 1

C(0j j; 0mm)j, k||Q0||j , k.(464)

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But

C(0j j ; 0mm) = j m|(|00|jm) (465)= jjmm , (466)

and hence,

(j m)(k)|Q(0, 0)|(jm)(k) =jjmm j, k||Q0||j , k

2j + 1 . (467)

The presence of the kronecker deltas tells us that a scalar operatorcannot change the angular momentum of a system, i.e., the matrixelement of th

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