rotations.nb: 11/3/04::13:48:13 1
Rotations and Angular momentum
Á 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 ' s 5. Tensor operators.
Á 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 vx numbers, e.g. v = v . The transpose of a vector is vT = v , v , v . There is an inner product defined between two ML y ]\ x y z M ] M vz ] + / T M T ] vectors by u ÿ v =Nv ÿ^u = uv cos f, 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 vx v' = R v and v' , v' , v' = v , v , v RT ML y ]\ ML y ]\ x y z x y z M ] M ] M v'z ] M vz ] + / + / M ]]] M ]]] N ^ N ^ To preserve the inner product, it is requird that RT ÿ R = 1
u'ÿ v' = uRT ÿ Rv = u1v = u ÿ v
As an example, a rotation by f around the z-axis (or in the xy-plane) is given by rotations.nb: 11/3/04::13:48:13 2
cos f sin f 0 R f = sin f cos f 0 z ML ]\ M ] + / M 0 0 1 ] M ]]] N ^ 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.
z
y x
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
cos f 0 sin f 1 0 0 R f = 0 1 0 and R f = 0 cos f sin f y ML ]\ x ML ]\ M ] M ] + / M sin f 0 cos f ] + / M 0 sin f cos f ] M ]]] M ]]] N ^ N ^ The sign of sin f 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 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅN N 1 independent rotations. + 2 /
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 by è x ' = R x = R x ? ? ? è where a distinction has been made between the operator R, 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. è ? R r ≈ n` = r ≈ R ÿ n` + ? ?/ ? ? Direction 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. ? rotations.nb: 11/3/04::13:48:13 3
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. T The constraint that the matrix is orthogonal, Ri j R j 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 3 ä3 unit matrix acts as an identity element for the group. 1 R = R 1 = R inverse: Each element has an inverse R 1 = RT , RT R = RRT = 1
associativity: 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, Det R = 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 ML ]\ M ] M 0 0 1 ] M ] N ^ 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. rotations.nb: 11/3/04::13:48:13 4
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. + / rotations.nb: 11/3/04::13:48:13 5
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
1 d 0 0 i 0 R d = d 1 0 or J = i 0 0 z ML ]\ z ML ]\ M ] M ] + / M 0 0 1 ] M 0 0 0 ] M ]]] M ]]] N ^ N ^ Similarly, for this representation
0 0 0 0 0 i J = 0 0 i , J = 0 0 0 x ML ]\ y ML ]\ M ] M ] M 0 i 0 ] M i 0 0 ] M ]]] M ]]] N ^ N ^ 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. rotations.nb: 11/3/04::13:48:13 6
R f = Lim P Rz f n nض i + / - + s /1 n = Lim 1 i Jz f n nض = e i Jz f+ + s //
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 + /
Á 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 ¶? ¹¹¶ ¹¶ ¹¶ used for all of these applications, if a sufficiently detailed definition is suppli ? ed 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 eigenstate
x' = 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 + / ` i ` D n, f = e ÅÅÅÅ Jÿn f + / 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. + / rotations.nb: 11/3/04::13:48:13 7
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 2 2 diagonalized. For SO(3) or SU(2) this would be J and one other component of J , typically taken to be Jz . J 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 2 2 2 the quadratic operator C = i Oi where the Oi are the infinitesimal generators of the group. J is such a quadratic casimir operator. In groups with more than one casimir operator, they all commute w½ ith 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,
2 2 2 2 J , Jz = Jx + Jy + Jz , Jz 2 2 # ' = #Jx + Jy , Jz '
= #Jx Jx, Jz + 'Jx, Jz Jx + Jy Jy, Jz + Jy, Jz Jy
= i# Jx Jy '+ J#y Jx +'i Jy Jx +# Jx Jy ' # ' = 0 + / + /
ü Labeling of states
2 Since J , Jz can be diagonalized simultaneously, we can specifiy the states by j, m , where m is the eignevalue when 2 Jz operates on the state and the operation of J yields a j ?
Jz j, m = m j, m 2 J « j, m? = a j« j, m? « ? « ? 2 where the eigenvalue of J 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 J 2 commutes with all the generators of the group, it also commutes with any function of those generators, and in particular J 2, D = 0. As with other commutation relations, this implies that the rotation operators don't change the eigenvalue of # J 2 ',
2 2 J D jm = DJ jm = Da j jm = a j D jm + ?/ ? ? + ?/ rotations.nb: 11/3/04::13:48:13 8
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' ?
j ? D jm = S Dmm' jm' m' ? ? j where the exact value of the coefficients Dmm' depends on the parameters describing the rotation.
Á 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 =+ cos /q 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 ÿ L 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 ÅÅÅÅ∑xÅÅÅ x b . For Lz the analagous relation is ; ? ; « ? ` ∑ ` n Lz b = i ÅÅÅÅ∑fÅÅÅ n b ; ? ∑ ; « ?∑ ` = i x ÅÅÅÅÅÅÅ∑y y ÅÅÅÅÅÅÅ∑x n b ` = xx p, y xy px n0 ;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 ; ? + / ; « ? rotations.nb: 11/3/04::13:48:13 9
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 sin f ÅÅÅÅ∑qÅÅÅ cot q cos f ÅÅÅÅ∑fÅÅÅ , ∑ ∑ 0 Ly Ø i cos f ÅÅÅÅÅÅÅ∑q cot q sin f ÅÅÅÅÅÅÅ∑f , ∑ 0 Lz Ø i ÅÅÅÅÅÅÅ∑f
These can be combined to yield the Casimir operator L2
2 L2 Ø 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅ1 ÅÅÅÅÅÅÅÅÅ∑ + ÅÅÅÅÅÅÅÅÅÅ1 ÅÅÅÅÅÅÅ∑ sin q ÅÅÅÅÅÅÅ∑ sin 2 q ∑2f sin q ∑q ∑q , 0 2 It is not a coincidence that L 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.
2 ü Eigenstates of L , Lz
2 Eigenstates of angular momentum can be chosen to be simultaneous eigenvalues of J and Jz . For orbital angular 2 2 momentum we expect eigenstates of L and Lz . Suppose, therefore, that b is an eigenstate of Lz and L with eigenvalues m and l l + 1 respectively. Then labeling the state by lm gives ? + / ? ` ∑ ` ` n Lz lm = i ÅÅÅÅ∑fÅÅÅ n lm = m n lm ; ? ; « ? ; « ? This equation is independent of q, so it is reasonable to separate variables in both the coordinates and the states
n` = z, f = z ä f
lm? = l m?ä m ? ? ? ? ? where m -kets exist in the f -space, and l m -kets exist in the z or q -space. The notation l m suggests 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 l m f m ; « ? ; « ? ; « ? The differential equation can then be rewritten by dividing through by z l m , ; « ? ∑ i ÅÅÅÅ∑fÅÅÅ f m = m f m ; « ? ; « ? and solved
f m = ÅÅÅÅÅÅÅÅ1ÅÅÅÅÅÅ eimf 2 p ; « ? r where the normalization is chosen so that rotations.nb: 11/3/04::13:48:13 10
m m' = m 1 m' ; « ? = ;m „ f ? f f m' = ; „ f +¼ m f ? ;f /m ' ? 2 p = ¼ „; f ÅÅÅÅÅÅÅÅ«1 ? e;i m'« m f? 0 2 p + / = d¼mm'
2 Similarly, in the z-space, z l m is a function only of q. In the differential equation for L one can substitute the solution for Lz ; « ?
2 n` L2 lm = 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅ1 ÅÅÅÅÅÅÅÅÅ∑ + ÅÅÅÅÅÅÅÅÅÅ1 ÅÅÅÅÅÅÅ∑ sin q ÅÅÅÅÅÅÅ∑ n` lm sin 2 q ∑2f sin q ∑q ∑q ; ? , 0 ; « ? or
2 z L2 l = 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ m + ÅÅÅÅÅÅÅÅÅÅÅ1 ÅÅÅÅÅÅÅ∑ sin f ÅÅÅÅÅÅÅ∑ z l m sin 2 f sin f ∑q ∑q m ; ? , 0 ; « ? which has for solutions the associated Legendre polynomials.
m z l m = clm Pl z ; « ? + / The normalization is given by the requirement
ml l m = 1 ; « ? = „ z ml z z l m * m 2 = ¼ „ z c;lm «clm? ;Pl« ?z = c2 ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅl++m !+ // ¼lm 2 l+1 +l m/! + / or, with a conventional choice of phase, c = m ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 l+1 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅl m ! lm 2 +l+m/! + / + / Combining the two forms, gives the spherical harmonics
n` lm = Y q, f = m ÅÅÅÅÅÅÅÅÅÅÅÅÅ2 l+1 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅl m ! eimf Pm cos q lm 4 p +l+m/! l ; « ? + / + / + / + / 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/' ¼ ? ; so that ` ` ` yb q, f = n „ W' n' n' b ` ` ` + /= ;„W +¼' n n' ?n;' b/ ? ` ` ` = ¼ sin q;' „« q' „? f; ' d«n,?n' n' b = y¼ b q, f + / ; « ? + / To make the last step valid, the d-function needs to be defined as rotations.nb: 11/3/04::13:48:13 11
` ` 1 d n, n' = ÅÅÅÅÅÅÅÅsin qÅÅÅ d q' q d f' f = d cos q' cos q d f' f = d z' z d f' f + / + / + / + / + / + / + / Next consider the lm states ? 1 = S lm lm lm ? ; lm l'm' = m m' ml l' m' = dmm' ml l' m' = dmm' dll' ; « ? ; « ? ; « ? ; « ? The projection operators onto the l-subspace are defined by
l Pl = S l m ml m= l ? ; For a given m, one may also define a sum over l
¶ Pm = S l m ml l=m ? ;
In fact, each of the operators Pm = 1 since in the q-space any of the sets lm q 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 z, f; lm« lm? f lm = S ; „W«' z,?f; lm« ?lm z', f' z', f' f lm ¼ ; « ? ; « ? ; « ? im' f' = S „ z' „f' z, f lm ml z' m f' f q' e lm 1¼ ; « ? ; « ? ; « ? + / i m' m f' = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ S „ z' z, f lm ml z' f q' „ f' e 2 p lm + / r ¼ ; « ? ; « ? + / ¼ = 2 p S „ z' z, f lm ml z' f q' dmm' r lm ¼ ; « ? ; « ? + / = 2 p S „ z' z, f lm' m'l z' f q' r l ¼ ; « ? ; « ? + / = 2 p S „ z' z l m' m'l z' f q' f m' r l ¼ ; « ? ; « ? + / ; « ? im' f = „ z' S z l m' m'l z' f q' e l à ; « ? ; « ? + / Now, the last line requires that
f q = „ z' S z l m' m'l z' f q' l + / Ã ; « ? ; « ? + / ¶ which is only valid for an arbitrary f if S z l m' m'l z' = z z' , which in turn requires that Pm' = S l m' m'l = 1. l l=m' ? ; Since the derivation is independent of m', it follows that ; « ? ; « ? Pm; =« 1? for any m.
ü Some identities (in progress)
ü Spin (in progress) rotations.nb: 11/3/04::13:48:13 12
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
2 J , J≤ = 0. # ' The commutator with Jz is given by
Jz, J≤ = Jz, Jx ≤ i Jy
# ' = #Jz, Jx ≤ i J'z, Jy
= #i Jy ≤ i' i J# x '
= ≤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, « ? 2 0 = J , J+ j, m 2 2 # = 'J« J+ ? J+ J j, m 2 = +J a j J+ j/, «m ? + / « ? 2 J+ j, m = 0 or J+ j, m is an eigenstate of J 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 +# = J'z J+ / «J+ Jz ? J+ j, m = +Jz m 1 J+ j,/m« ? + / « ? rotations.nb: 11/3/04::13:48:13 13
It follows that either
J+ j, m = 0 or + J+ « 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 or J « 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 J 2 , 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
2 2 2 2 J Jz = Jx + Jy 1 = ÅÅÅÅ2 J+ J + J J+ 1 † † = ÅÅÅÅ2 +J+ J+ + J+ J+/ , 0 where at the end, both terms are positive definite since
† + * + + 2 j, m J+ J+ j, m = j, m + 1 c jm c jm j, m + 1 = c jm < ¬ ¬ @ ; « + / « ? « « It follows that
2 2 2 2 J Jz j, m = a j m j, m > 0 or a j m > 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 2 2 += Jx + Jy/ + i Jx, Jy/ 2 2 = J Jz J#z '
so, when operating on the highest state of the representation
2 2 J J+ j, mmax = J Jz Jz j, mmax 2 « = a j ? mmax+ 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 14
2 2 J+ J j, mmin = J Jz + Jz j, mmin ? 2 ? « = a j m+min + 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 = 2 mmax = n, or
n mmax = ÅÅ2ÅÅ
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 J 2« j, m? = j «j + 1? j, m « ? + / « ? ü Other matrix elements
From above
† + * + j 2 j, m J+ J+ j, m = j, m + 1 c jm c jm j, m + 1 = c+m ; « « ? ; « + / « ? « « † 2 2 whereas, we can also use the relation J J+ = J+ J+ = J Jz Jz
† 2 2 j, m J+ J+ j, m = j, m J Jz Jz j, m < ¬ = j ¬j + 1 @ m; m +«1 « ? = +j m / j + m+ + 1 / + / + / or
+ c jm = j m j + m + 1 r + / + / Similarly
c jm = j + m j m + 1 r + / + / ü Explicit matrix form
It is often convenient to show the angular momentum operators in explicit matrix form. The form of the matrix depends 2 1 3 on the representation. For example, the figure shows matrices corresponding to J for the j = ÅÅ2ÅÅ , 1, ÅÅ2ÅÅ representations
15 ÅÅÅÅ4ÅÅÅ 0 0 0 2 0 0 ÅÅ3ÅÅ 0 L 0 ÅÅÅÅÅÅÅ15 0 0 \ 2 2 4 2 2 2 2 M 4 ] J j= ÅÅÅÅ1 = , J j=1 = L 0 2 0 \, J j=ÅÅÅÅ3 = M ] 2 L 0 ÅÅÅÅ3 \ M ] 2 M 0 0 ÅÅÅÅÅÅÅ15 0 ] M 4 ] M 0 0 2 ] M 4 ] + / M ] + / M ] + / M ] M ] M 0 0 0 ÅÅÅÅÅÅÅ15 ] N ^ M 4 ] N ^ M ] N ^ rotations.nb: 11/3/04::13:48:13 15
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 jm states. Since + / J 2 is a Casimir operator, it is diagonal and proportional to the identity matrix in 2 each representation, but with a different eigenvalue. ? Jz can be simultaneously diagonalized with J , but has different 1 3 eigenvalues for each state. The matrix form for Jz for the j = ÅÅ2ÅÅ , 1, ÅÅ2ÅÅ representations is
3 ÅÅ2ÅÅ 0 0 0 1 0 0 ÅÅ1ÅÅ 0 L 0 ÅÅÅÅ1 0 0 \ 2 M 2 ] Jz j= ÅÅÅÅ1 = , Jz j=1 = L 0 0 0 \, Jz j= ÅÅÅÅ3 = M ] 2 L 0 ÅÅÅÅ1 \ M ] 2 M 0 0 ÅÅÅÅ1 0 ] M 2 ] M 0 0 1 ] M 2 ] + / M ] + / M ] + / M ] M ]] M 0 0 0 ÅÅÅÅ3 ] N ^ M 2 ] N ^ M ] N ^ 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 1 3 j = ÅÅ2ÅÅ , 1, ÅÅ2ÅÅ representations is
0 3 0 0 0 2 0 0 1 0r 0 2 0 r ML ]\ J+ 1 = , J+ j=1 = L \, J+ 3 = M ] j= ÅÅÅÅ2 M 0 0 2 ] j= ÅÅÅÅ2 M ] L 0 0 \ M ] M 0 0 0 3 ] + / M ] + / M 0 0r 0 ] + / M ] M ] M r ] N ^ M ] M 0 0 0 0 ] N ^ M ] 1 N i ^ J is similar, but on the lower off-diagonal. Jx = ÅÅ2ÅÅ J+ + J , Jy = ÅÅ2ÅÅ J+ J are determined by adding the matrix components, entry by entry. + / + /
Á 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 16
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 2 J2 by j2 m2 . Where ji describes the eigenstate with respect to the casimir operator Ji and mi is the eigenvalue of « ? Ji z . 2 2 The two operators commute with each other « ? J1, J2 = 0, so a full set of commuting opertors is J1 , J2 , 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 2 operators J and Jz
2 2 2 J = J1 + J2 + 2 J1 ÿ J2 Jz = J1 z + J2 z
There are many cases where the Hamiltonian includes a term of the form J1 ÿ J2 , for example the spin-orbit coupling is proportional to L ÿ S . In this case the relation
1 2 2 2 J1 ÿ J2 = ÅÅ2ÅÅ J J1 + J2 + / is useful. It is also possible to write J1 ÿ J2 in terms of raising and lowering operators
1 J1 ÿ J2 = J1 z J2 z + ÅÅ2ÅÅ J1+ J2 + J1 J2+ + /
ü two alternative basis sets for the j1, j2 representation
2 The last relation makes it clear that because J contains pieces that raise and lower the states of J1 z and J2 z , eigenstates 2 j1 m1 j2 m2 , are not generally eigenstates of J . 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 2 2 scheme one chooses 4 commuting operators J1 , J2 , J1 z, and J2 z . In this case the states are labeled by j1 m1 j2 m2 . ? 2 2 2 Alternatively, in the J, Jz scheme a different set of four operators J , J1 , J2 , and Jz define the states. It is 2 2 2 straightforward to show that J , Ji = Jz, Ji = 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 17
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 = S j1 j2 m1 m2 j1 j2 m1 m2 j1 j2 jm m1 m2
? j1 j2 ? ; « ? = S c jm,m1 m2 j1 j2 m1 m2 m1 m2 ? where 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
j1 + j2 + 1 m m ¥ D j Nm = 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 m Ø 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 18
j= j1+ j2 N = S 2 j + 1 j= j1 j2
i=2 j2 = S 2 D j + i + 1 i=0 + / i=2 j2 = 2 D j + 1 2 j2 + 1 + 2 S i i=0 + / + / = 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 = j2 ? ? Proceeding with the application of J
J j1, j2, j = j1 + j2, m = j1 + j2 = J1 + J2 j1, j2, m1 = j1, m2 = j2 c j1+ j2, j1+ j2 j1, j2, j = j1 + j2, m =?j1 + +j2 1 =/ ? cj1, 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 r highest m-state of the + j =/j+1 + j2 representation,/
c c j1, j1 j2, j2 j1, j2, j = j1 + j2, m = j1 + j2 1 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1, j2, m1 = j1 1, m2 = j2 + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1, j2, m1 = j1, m2 = j2 1 j1+ j2, j1+ j2 j1+ j2, j1+ j2 ? ? ? 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, j1 j1 c = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ and similarly j1+ j2, j1+ j2 1; j1 1, j2 c j + j , j + j j1+ j2 1 2 1 2 c j1 j2 j2, j2 j2 c = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ j1+ j2, j1+ j2 1; j1, j2 1 c j + j , j + j j1+ j2 1 2 1 2
In response to a question in class, if j1 5 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
j1 j2 2 S c jm;m1,m m1 m1
The 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 = c c j2, j2 j1, j1 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1, j2, m1 = j1 1,?m2 = j2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅc j1, j2, m1 = j1, m2 = j2 1 j1+ j2, j1+ j2 j1+ j2, j1+ j2 ? ? rotations.nb: 11/3/04::13:48:13 19
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 th combinations of these which are orthogonal and normalized, which can be labeled by index a =« S? ai i . Then an n i state can be added to the a-basis which will be orthogonal to the others by construction ? ?
i1 i2 …in 1 2 n 1 an = e ai1 ai2 …ain 1 in ? ? ü Example of two spin 1/2 reps
1 As an explicit example of angular momentum addition, consider the case of adding two spins of s1 = s2 = ÅÅ2ÅÅ . Since s1 and s2 are specified I'll adopt the shorter notation where they are suppressed.
sm sm = m1 m2 m1 m2 ? ? In the m1 m2 -basis there are four states
1 1 1 1 1 1 1 1 ÅÅ2ÅÅ , ÅÅ2ÅÅ , ÅÅ2ÅÅ , ÅÅ2ÅÅ , ÅÅ2ÅÅ , ÅÅ2ÅÅ , and ÅÅ2ÅÅ , ÅÅ2ÅÅ « @ « @ « @ « @ 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
1 1 11 sm = ÅÅÅÅ ÅÅÅÅ 2 2 m1 m2 ? @
Acting with S = S1 + S2 on both sides gives the algebraic
1 1 S 11 sm = S1 + S2 ÅÅÅÅ , ÅÅÅÅ 2 2 m1 m2 ? 1 1 @ 1 1 + 1 1 / 1 1 c11 10 sm = c ÅÅÅÅ ÅÅÅÅ ÅÅÅÅ2 , ÅÅÅÅ2 + c ÅÅÅÅ ÅÅÅÅ ÅÅÅÅ2 , ÅÅÅÅ2 2 2 m1 m2 2 2 m m ? @ A 1 2 1 1 1 1 2 10 sm = ÅÅÅÅ2 , ÅÅÅÅ2 + ÅÅÅÅ2 , ÅÅÅÅ2 m1 m2 m1 m2 r ? @ @ 1 1 1 1 1 1 10 sm = ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ 2 2 2 m m 2 2 2 ? r A 1 2 r Am1 m2 1 1 ÅÅÅÅ2 ÅÅÅÅ2 10 sm = S c10;m m m1, m2 m 1 2 ? 1 B 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 r m-state c j j = 2 j . In the last line, the sum over + / +m-states defines the Clebsch-Gordon coefficients. Evidently/ r 1 1 1 1 ÅÅÅÅ2 ÅÅÅÅ2 1 ÅÅÅÅ2 ÅÅÅÅ2 1 c 1 1 = ÅÅÅÅ and c 1 1 = ÅÅÅÅ 10; ÅÅÅÅ , ÅÅÅÅ 2 10; ÅÅÅÅ ,ÅÅÅÅ 2 2 2 2 2
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 10 sm state, it remains to determine the orthogonal m = 0 linear combination, which in this case is ? rotations.nb: 11/3/04::13:48:13 20
1 1 1 1 1 1 00 sm = ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ 2 2 2 m1 m2 2 2 2 m1 m2 ? r ? r ? This result aslo defines the CG coefficients
1 1 1 1 ÅÅÅÅ2 ÅÅÅÅ2 1 ÅÅÅÅ2 ÅÅÅÅ2 1 c 1 1 = ÅÅÅÅ and c 1 1 = ÅÅÅÅ 00; ÅÅÅÅ , ÅÅÅÅ 2 10; ÅÅÅÅ , ÅÅÅÅ 2 2 2 2 2 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.
1 1 1 1 1 1 S 10 sm = S1 + S2 ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ + S1 + S2 ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ 2 2 2 m1 m2 2 2 2 m1 m2 r r ? + 1 / 1 ? 1 + 1 / 1 1? c10 1, 1 sm = ÅÅÅÅÅÅÅÅÅÅ 0 + c 1 1 ÅÅÅÅ , ÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅ c 1 1 ÅÅÅÅ , ÅÅÅÅ + 0 2 ÅÅÅÅ2 ÅÅÅÅ2 2 2 m1 m2 2 ÅÅÅÅ2 ÅÅÅÅ2 2 2 m1 m2 r r ? 1 , 1 1 ?1 0 1 1 , ? 0 2 1, 1 sm = ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ + ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ 2 2 2 m1 m2 2 2 2 m1 m2 r r r 1, 1 =? ÅÅÅÅ1 , ÅÅÅÅ1 ? ? sm 2 2 m1 m2 ? ? which makes sense, since the lowest m-state should also be defined uniquely.
As a check, applying S to the 00 sm state
1 1 ?1 1 1 1 S 00 sm = S1 + S2 ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ S1 + S2 ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅ , ÅÅÅÅ 2 2 2 m1 m2 2 2 2 m1 m2 r r 1? + 1/ 1 ? 1 + 1 / 1 ? 0 = ÅÅÅÅÅÅÅÅÅÅ 0 + c 1 1 ÅÅÅÅ , ÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ c 1 1 ÅÅÅÅ , ÅÅÅÅ + 0 2 ÅÅÅÅ2 ÅÅÅÅ2 2 2 m1 m2 2 ÅÅÅÅ2 ÅÅÅÅ2 2 2 m1 m2 r r 0 = ÅÅÅÅÅÅÅÅÅÅ1 , ÅÅÅÅ1 , ÅÅÅÅ1 ÅÅÅÅÅÅÅÅÅÅ1? 0ÅÅÅÅ1 , ÅÅÅÅ1 , = 0 ? 0 2 2 2 m1 m2 2 2 2 m1 m2 r ? r ? yields the null state on both sides.
2 2 2 2 2 1 Another check is to apply S = S1 + S2 + 2 S1 ÿ S2 = S1 + S2 + 2 S1 z S2 z + ÅÅ2ÅÅ S1+ S2 + S1 S2+ to all four states. For the triplet states S2 = 2, so + /
2 2 1 1 3 3 1 1 1 1 1 1 S1 + S2 + 2 S1 z S2 z + S1+ S2 + S1 S2+ ÅÅ2ÅÅ ÅÅ2ÅÅ = ÅÅ4ÅÅ + ÅÅ4ÅÅ + 2 ÅÅ2ÅÅ ÅÅ2ÅÅ + 0 + 0 ÅÅ2ÅÅ ÅÅ2ÅÅ = 2 ÅÅ2ÅÅ ÅÅ2ÅÅ 2 2 1 @ 1 3 3 1 1 @ 1 @ 1 1 1 +S1 + S2 + 2 S1 z S2 z + +S1+ S2 + S1 S2+// ÅÅÅÅ2 ÅÅÅÅ2 + = ÅÅÅÅ4 + ÅÅÅÅ4 + 2 ÅÅÅÅ2+ ÅÅÅÅ2// + 0 + 0 ÅÅÅÅ2 ÅÅÅÅ2 = 2 ÅÅÅÅ2 ÅÅÅÅ2 + + // @ + + / + / + // @ @ and for the m = 0state
2 2 1 1 1 1 S1 + S2 + 2 S1 z S2 z + S1+ S2 + S1 S2+ ÅÅ2ÅÅ , ÅÅ2ÅÅ + ÅÅ2ÅÅ , ÅÅ2ÅÅ = 3 3 1 1 1 , 1 @1 1 @0 1 1 1 1 + ÅÅÅÅ4 + ÅÅÅÅ4 2 ÅÅÅÅ2+ ÅÅÅÅ2 + 1 + 0 ÅÅÅÅ2 /,/ ÅÅÅÅ2 + ÅÅÅÅ2 , ÅÅÅÅ2 = 2 ÅÅÅÅ2 , ÅÅÅÅ2 + ÅÅÅÅ2 , ÅÅÅÅ2 + + // , @ @0 , @ @0 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 3 3 1 1 ÅÅ4ÅÅ + ÅÅ4ÅÅ 2 ÅÅ2ÅÅ ÅÅ2ÅÅ 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 21
Á 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 ¹¹¶ 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 a ¶ ¶ ¶ Xi?= ¼„ x x ? x;i x« ? ¶ ¶ ¶ ¼ ? ; 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 22
At this point it is important to distinguish the effects of rotations in three distinct scenarios. First, one may rotate the states a Ø 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 c; « ? ombination of the operators ¼ ? 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
Xi Ø 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 ar¼ ? < « ¼ ?e 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' ¶ ¶ ¶ ¶ ¶ ¶ ¼ = ?„ x'; a x' ¼ xi ' x ' a? ; ¶ ¶ ¶ = ¼ „ x ;a x ?xi x; a ? ¶ ¶ ¶ = ¼Xi ; ? ; ?
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 x Ø x' is performed. The result of these considerations is that the effect of a rotation on¼ ¼ the position operator can be described either by a transformation of the basis states,
1 Xi ' = „ x x' xi x' = RXi R ¶ ¶ ¶ ¼ ? ; or by a transformation of the operators
Xi ' = „ x x' Ri j x j x' = Ri j X j ¶ ¶ ¶ ¼ ? ; 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 1 on different repesentations, so one may use the notation Xi ' = DXi D instead. Further, although convenient to use the position basis for pedagogical purposes, any complete set of basis kets would do, so one may replace „ x Ø S . In jm addition, since rotations do not mix states from different representations, the D-operators can be chosen to be finite ¼ 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 ? ? ? R1 has 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 1 relation can also be written in the inverse form a = R2 a ', and so ? ?
1 ? ? R2 R1 a = R2 R1 R2 a ' ? ? rotations.nb: 11/3/04::13:48:13 23
Comparing the two results, one finds that the rotations themselves transform as
1 R1 ' = R2 R1 R2
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 unchange; « « ? d 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' 1 ; « « ?= b; R«1 R«' R?1 a
; « 1 « ? one finds R = R1 R' R1 or
1 R' = R1 RR1
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
+ / 1 Ry' b = Rz a Ry b Rz a + / + / + / + / Defining the combined rotation R a, b = Rz a Ry b
+ / 1 + / + / Rz' g Ry' b Rz a = Rz' g Rz a Ry b Rz a Rz a + / =+ R/z' g+ R/z a R+y /b + / + / + / + /
= Rz'+g/ R a+ ,/b + / 1 = R a+ ,/b +Rz g/ R a, b R a, b = R+a, b/ Rz+g/ + / + / = R+z a R/y b+ R/ z 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 1 ie nÿJ operators. To do this consider the relation R' = R1 RR1 in the case where R is infinitesimal, R = e 1 ie n ÿ J . 1 1 On the left, one has R' = 1 ie n ÿ J ' with Ji ' = Ri j J j , and on the right R1 RR1 = R1 1 ie n ÿ J R1 . Equating the two sides, one finds the same relation that was obtained for the position vector + / rotations.nb: 11/3/04::13:48:13 24
1 Ji ' = R1 i j J j = D1 Ji D1 + / 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 11 component of J1 '. On
the left, J1 '11 is made up of the 11 components of J1, J2, and J3 weighted by R11 , R12 , and R13 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 as¹¹¶ ¹¶ ¹¹¶
1 Vi Ø V 'i = Ri j V j = RV i R
It is useful to consider the infinitesimal form of this relation. On the left,
k k Ri j V j = 1 ie n ÿ J i j V j = di j ie n J i j V j + / + / k where J i j = ieki j . On the right, + / 1 k k RV i R = 1 ie n ÿ J Vi 1 ie n ÿ J = Vi ie n J , 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
k J , 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. ¹¹¶ ¹¶
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≤ .
1 Define coordinates xq with q taking the values of q = 1, 0, 1, by x0 = z, x≤ = ÅÅÅÅÅÅÅÅÅÅ x ≤ iy . I don't have a name for 2 r 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 25
ÅÅÅÅ1ÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅi ÅÅ 0 2 2 r r U = ML 0 0 1 ]\ M ] M 1 i ] M ÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅ 0 ] M 2 2 ] M r r ] N ^ ifnÿJ The next step is to re-examine the defining relation for rotations in cartesian coordinates x'i = Ri j x j = e i j x j . è è 1 First, the new coordinates can be used to rewrite the generator: n ÿ J = n q J q , where J ≤ = ÅÅÅÅÅÅÅÅÅÅ J≤ is defined to account 2 + / r 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,
0 0 S1 è 1 J ≤ = ÅÅÅÅÅÅÅÅÅÅ 0 0 i 2 ML ]\ r M ] M ≤1 i 0 ] M ]]] N ^ Second, the new coordinates can be used as the object of the rotations
ifnÿJ x'q = Rqq' xq' = e qq' xq' + / è Again there is the choice of writing n ÿ J as ni Ji or n q J q . In either case, though, J must be written in the form for † è è † operating on xq instead of xi . Following the usual rules for transforming operators J 'i = UJi U or J 'q = UJ q U . The J ' have the same form as when operating on the j = 1 representation, for example,
1 0 0 0 1 0 l=1 è 1 l=1 J '0 = J = 0 0 0 , J '+ = ÅÅÅÅÅÅÅÅÅÅ J+ = 0 0 1 z+ / ML ]\ 2 + / ML ]\ M ] r M ] M 0 0 1 ] M 0 0 0 ] M ] M ] (5/11/04 - I am not sure I don't have some "typos" N ^ still in this section)N ^
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 1 V≤ = ÅÅÅÅÅÅÅÅÅÅ Vx ≤ iVy 2 r + / 1 1 1 Later this object will be identified as a rank-1 spherical tensor Tq where q = 1, 0, 1 , and T0 = V0 , T≤1 = V≤ . + / Expressing the effect of the rotation on Vq by its effect on the eigenstates
1 m èm RV q R = 1 ie n ÿ J Vq 1 ie n ÿ J = Vq ie n J , Vq + / + / $ ( where m has been used to distinguish the generator from the spherical vector component. Meanwhile, rotating the operator components yields
m m Rqq' Vq' = 1 ie n ÿ J qq' Vq' = dqq' ie n J qq' Vq' + / + / equating the two results, for each m one has, rotations.nb: 11/3/04::13:48:13 26
èm èm J , Vq = J qq' Vq'
$ ( èm On the left, the exact form of J and Vq depend on the representation upon which the operators act. On the right, Vq is èm defined relative to the representation but J q q' is always the j = 1representation since it is acting on a vector operator. èm Explicit forms for J are
1 0 0 0 1 0 0 0 0 è0 è+ è J = 0 0 0 , J = 0 0 1 , and J = 1 0 0 ML ]\ ML ]\ ML ]\ M ] M ] M ] M 0 0 1 ] M 0 0 0 ] M 0 1 0 ] M ]]] M ]]] M ]]] N ^ N ^ N ^ As an example, suppose Vq is the angular momentum operator itself, and that the equation is applied to the j = 2 èm èm representation. Then, the Vq are 5ä5matrices, and the J in the commutator is also 5ä5, but the J qq' on the right is 3ä3 to mix the components of the vector operator.
ü Cartesian tensors
i j One can form more complicated tensors. For example, T = Xi X j is a rank-2 tensor. A rank- n tensor has n-indicies, T i1 i2 …in , and transforms under rotations as
T i1 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 and V 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
S V Q 1 1 S O = di j Oi j , Oi = ei jk O jk , and Oi j = ÅÅÅÅ2 Oi j + O ji ÅÅÅÅ3 O + / These three tensors are invariant under rotations. In terms of labels, the reduction of UV 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 = UV can be understood in terms of anglar momentum addition
1 ä 1 = 0 + 1 + 2 rotations.nb: 11/3/04::13:48:13 27
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
m m Rqq' Vq' = 1 ie n ÿ J qq' Vq' = dqq' ie n J qq' Vq' + / + / k k k Correspondingly, a spherical tensor of rank- k is T with 2 k + 1 components Tq . Tq transforms as + / k k k k m k m k T Ø T ' = Rqq' T = 1 ie n ÿ J T = dqq' ie n J qq' T q q q' qq' q' + / q' + / + / where the label k has been added to J m 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 T k with components k Tq , 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
T k = a Tk a q a q ; ? ; « « ? Under rotations,
k k 1 Tq Ø RTq' R
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.
k k k Tq Ø Dqq' Tq'
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 J m . k 1 Thus, for an infinitesimal rotation, the form RTq' R becomes
k 1 k m èm k RTq R Ø Tq ie n J , Tq $ ( k k k m k m k k m Similarly D T Ø T ien J qq' T , which requires knowledge of J qq' . These operators are known from the qq' q' q + / q' + / action of Jz, J≤ on angular momentum states in different / j-representations,
k 0 k ≤ ≤ J qq' = dqq' q and J qq' = c dq,q'≤1 + / + / kq
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 28
k k Jz, Tq = qTq k ≤ k #J≤, Tq' = ckq Tq≤1 # ' ≤ S where ckq = k q k ≤ q + 1 , as for when the angular momentum generators act on states j = k, m = q . r + / + / ? 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 U k1 , V k2 , with
k1 k2 k1 k2 components Uq1 , Vq2 . Then the product Uq1 Vq2 can be expressed as a linear combination of objects with tensor properties
k1 k2 k1 k2 k Uq Vq = S ckq,q q Tq (Eq. 1) 1 2 kq 1 2
k k There are 2 k1 + 1 2 k2 + 1 possible products that can be formed from the components of U 1 and V 2 . Under the set of rotations, the various components of + / + / U k1 transform into each other. Similarly for V k2 and the T k . It follows that for k k k1 k2 each T all components Tq are included in the set of all possible Uq1 Vq2 products.
The next step is to consider the effect of Jz
k1 k2 k1 k2 k1 k2 Jz, Uq1 Vq2 = Uq1 Jz, Vq2 + Jz, Uq1 Vq2
k1 k2 # = q'1 + q2 #Uq1 Vq2' # ' + / 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 . kmax k1 k2 The operator Tkmax = Uk1 Vk2 is the unique q = kmax operator.
Following the development of angular momentum addition, the next step is recognize that there will be two operators
k1 k2 k1 k2 kmax with q = kmax 1 built out of linear combinations of Uq1 1 Vq2 and Uq1 Vq2 1 . One will belong to T and can be built with J . The other will be an orthogonal linear combination, and must be the highest q-component of an operator k 1 T max . The operation with J on the left side of Eq. 1 above gives
k1 k2 k1 k2 k1 k2 J , Uk1 Vk2 = Uk1 J , Vk2 + J , Uk1 Vk2 # = c ' U k1 # V k2 + c' # U k1 V k2' k1 k1 k1 1 k2 k2 k2 k1 k2 1
while on the right-hand side
J , T k1+k2 = c T k1+k2 k1+k2 k1+k2,k1+k2 k1+k2 1 # ' or
c c Tk1+k2 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅk1 k1 U k1 V k2 + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅk2 k2 U k1 V k2 k1+k2 1 c k1 1 k2 c k1 k2 1 k1+k2,k1+k2 k1+k2,k1+k2 rotations.nb: 11/3/04::13:48:13 29
The second linear combination will be orthogonal to this one, and will become the head component of the T k1+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 k1 k2 k1 k2 in the product of U V . Since the construction technique is the same, the coefficients ckq,q1 q2 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 T k is a rank- k tensor operator, then T 'k = f r T k 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 tens+ / + / or properties from the è k normalization. Accordingly, one can uniquely define a normalized rank- k tensor operator T , and then an arbitrary tensor operator Tk can be written as
è k Tk = f r T + / For a given application f r may depend on k , but it does not depend on q, so all components of T k scale in the same way + /
k è k Tq = f r Tq + / 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 notation ?
nlm = nl r lm W ? ? ? where 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 k of states. It seems reasonable to guess that operating with Tq 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 30
k Indeed, consider the set of states that can result from operation by any component q of Tq 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 k combinations of Tq jm will be exactly the representations present in the combination of two representions k, j. ? To see this explicity, consider the product,
k èk j Tq jm = S cqm, j' m' j'm' 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 T k is normalized.
Operating with Jz , on the right
èk j èk j Jz S cqm, j' m' j'm' = S m' cqm, j' m' j'm' j' m' j' m' ? ? and on the left
k k k Jz Tq jm = Jz, Tq + Tq Jz jm k k =? qT+#q + Tq ' m jm / ? k = +q + m Tq jm/ ? èk j =+ S q +/m c qm,?j' m' j'm' j' m' + / ? It follows that m' = q + m term by term, and the sum is restricted to just a sum over different j'.
k èk j Tq jm = S cqm, j' m'=q+m j'm' = q + m j' ? ?
Similar 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 k uniquely by Tk j j = j + k, j + k . Similarly, there will be two states produced with m' = k + j 1, by the k k combinations Tk 1? j j or Tk 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
k J Tk j j = J j + k, j + k = c j+k, j+k j + k, j + k 1 ? ? ? and on the left
k k k J Tk j j = J , Tk + Tk J j j k k =? ckk+#Tk 1 j' j + c j j T/k j,?j 1 ? ? Combining the results, rotations.nb: 11/3/04::13:48:14 31
c j j k ckk k j + k, j + k 1 = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ T j, j 1 + ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ T j j c j+k, j+k k c j+k, j+k k 1 ? ? ? 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
k k j Tq jm = dm',q+m S cqm, j' m' j'm' j' ? ? 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
k k j k n' j' m' Tq n jm = cqm, j' m' n' j' T 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
k è k Tq = f r Tq + / and so
k è k n' j' m' Tq n jm = n' j' m' f r T q n jm è k ; =« j«' m' T? ü Some examples m selection rules j selection rules 10 10 10=0 rotations.nb: 11/3/04::13:48:14 32 ü stuff