Rotation group R (3) Ling-Fong Li. (Institute) Continuous Group & SU (2) 1 / 46

Transcription

1 Rotation group R (3) Ling-Fong Li (Institute) Continuous Group & SU (2) / 46

2 In nite group : group with in nite number of elements. Label group elements by real parameters (group parameters) A (α, α 2, α n ), α, α 2, α n group parameters Continuous group : group parameters continuous Compact group: group parameters vary over compact domain. For example, 3-dim rotational group is compact because angles of rotation, varies over compact interval [, 2π] Lorentz group is not compact because, β = v varies over non-compact interval, [, ). c Example: SO (2), 2 2 real orthogonal matrices with det = is an one-parameter continuous group. Choose parameter to be the angle of rotation φ, φ 2π.and is a compact group. Denote group element by, R (φ) The group multiplication is R (φ ) R (φ 2 ) = R (φ + φ 2 ) Visualize group elements as points on the unit circle and are labeled by the angle. This is called group parameter space. Group Integration Group invariant measure In nite group, rearrangement theorem, n j= f (A j ) = n j= f (A j B ) = n j= f (BA j ), B, A j 2 G (Institute) Continuous Group & SU (2) 2 / 46

3 plays essential role in representation theory. For continuous group, sum over group elements =) integration in group parameter space, Z Z da W (α, α n ) d α d α n where W (α, α n ) is a measure (or weight function). De ne group integration (or choose a measure W ) such that rearrangement theorem holds. This is called the group invariant integration (measure). The measure W (α, α n ) should be chosen such that Z Z u (A) da = Z u (AB ) da = u (BA) da where Z Z u (A) da u (α, α n ) W (α, α n ) d α d α n and u (A) is some arbitray function of group elements. Use notation,!α!β A = A (α, α n ) = A, B = A (β, β n ) = A Then!β!α!γ BA = A A = A (Institute) Continuous Group & SU (2) 3 / 46

4 where! γ =! γ!β,! α are some functions determined by the group multiplication. Write the integrations as, da = W d (BA) = W!α d n α = W (α, α n ) d α d α n!γ d n γ = W (γ, γ n ) d γ d γ n Thus group invariant measure should have the property da = d (BA), or W!α!γ d n α = W d n γ Note that left (or right) multiplication by a xed group element, say B, is a - mapping of G onto itself. Thus giving a set of group elements in some region V of the parameter space, under the left (or right) multiplication by B, these elements will move to other region V. Since the total number of element in V is the same as those in V, we get, ρv = ρ V, (Institute) Continuous Group & SU (2) 4 / 46

5 where ρ (ρ ) is the density of elements at V (V ). Thus if we take the measure W (α, α n ) to be the density of the group elements at! α, then W!α!γ d n α = W d n γ, where! γ =!!β,! γ α!α i.e. W is a group invariant measure.!α To get the density of elements W consider an in nitesmal volume element V = d α d α n in the neighborhood of the origin(identity) i.e. I = A (). Under left multiplication by B, they move to V, W!α!γ V = W V Or!α W!γ = V W V Thus ratio of weight functions are ratio of the volume elements. We will normalize the density! such that W =, i.e. density is at origin. Setting! α =!, we get W!α!!γ!β = W =, and W = W (Institute) Continuous Group & SU (2) 5 / 46

6 Then!β W = V V!β Recall that the change in volume elements under the transformation induced by B = A is given by the Jacobian of the change of the variables! α!!!β,! γ α. Thus we get V = d α d α n = (α, α n ) (γ, γ n )!α d γ! d γ n = (α, α n ) = (γ, γ n )!α V! = Thus the invariant group measure is given by W!β = (γ, γ n ) (α, α n )!α! = Thus to nd the group invariant measure, need to know the change of group parameters under the multiplication of group element. (Institute) Continuous Group & SU (2) 6 / 46

7 SO (2) group. The group multiplication is given by R (α) R (β) = R (γ), with γ = α + β where α, β, γ are angles of rotations. The group invariant measure is W (β) = γ = α i.e.group elements of SO (2) are uniformly populated along the unit circle =) the points per unit length is same everywhere. This explains the feature W (β) = in this case. The group integration is Z 2π d α Since this is an Abelian group, all irreps are -dimensional and of the form e imα, m = integer The great orthogonality theorem is of the form, Z 2π d α e imα e im α = δmm 2π If we had chosen any other measure f (α) 6= constant, then we would not have the great orthogonality theorem. (Institute) Continuous Group & SU (2) 7 / 46

8 SU (2) group- 2 2 unitary matrice with determinant=, UU = U U =, det U = Write unitary matrix U as a U = c b d then The conditions imply that U = det U d b c a, U a c = b d U = U, and det U = a = d, c = b, jaj 2 + jbj 2 = Most general 2 2 unitary matrix is Parametrize a and b in terms of real variables, a b U = b a, with jaj 2 + jbj 2 = a = u + iu 2, b = u 3 + iu 4, u i real (Institute) Continuous Group & SU (2) 8 / 46

9 then we have u 2 + u u u 2 4 = Thus group elements of SU (2) are now on the surface of a sphere in 4-dimensional space. This suggests the invariant measure as Z Z dr = du du 4 δ u 2 + u u u 2 4 It is intuitively clear that multiplication by an SU (2) element corresponds to a rotation on the surface of sphere and this measure is invariant. To show this, we need to prove that Z Z dr f (R ) = drf (R ), where R = SR and f is some arbitrary function. Using the parametrizations, R = R = we get from the relation R = SR, u + iu2 u3 iu4 s + is u3 iu4 u iu2, S = 2 s 3 is 4, s 3 is 4 s is 2 u + iu 2 u 3 iu 4, u 3 iu 4 u iu 2 8 >< >: u = s u s 2 u 2 s 3 u 3 s 4 u 4 u 2 = s 2u + s u 2 s 4 u 3 + s 3 u 4 u 3 = s 3u + s 4 u 2 + s u 3 s 2 u 4 u 4 = s 4u s 3 u 2 + s 2 u 3 + s u 4 (Institute) Continuous Group & SU (2) 9 / 46

10 The Jacobian of the transformation is then J = (u, u 2, u 3, u 4 ) (u, u 2, u 3, u 4 ) = s s 2 s 3 s 4 s 2 s s 4 s 3 s 3 s 4 s s 2 s 4 s 3 s 2 s It is easy to see that J = because it is the determinant of an orthogonal matrix using the fact that s 2 + s s s 4 2 =. Then we have du du4δ u 2 + u2 2 + u3 2 + u4 2 = (u, u 2, u 3, u 4 ) (u, u 2, u 3, u 4 ) du du 4 δ u 2 + u2 2 + u3 2 + u4 2 = du du 4 δ u 2 + u2 2 + u3 2 + u4 2 (Institute) Continuous Group & SU (2) / 46

11 Rotation Group O (3) Homomorphism to SU (2) Group Elements of rotation group O (3) will be denoted by P! n (θ), the operator which rotates the system by angle θ about the axis! n. Write P! n (θ) = i θ! J! n h A then! J will be called the generator of rotation. we can also parametrize the rotation operator in terms of the familiar Euler rotations, R (α, β, γ) = P z (α) P y (β) P z (γ) = exp ijz h α exp ijy h β exp ijz h γ, where α, γ 2π, β π. We will now show that O (3) is homomorphic to SU (2) group. For any vector! r = (x, y, z) de ne a 2 2 hermitian matrix by h =! σ! r = z x iy x + iy z where! σ are the Pauli matrices. It is easy to see that h has the properties Tr (h) =, det h = x 2 + y 2 + z 2 = r 2 (Institute) Continuous Group & SU (2) / 46

12 Let U any 2 2 unitary matrix U and de ne a new matrix h by h = UhU () Then h is also hermitan, traceless and has the same determinant as h, h = h, Tr h =, det h = det h If we expand h in terms of Pauli matrices, h =! σ! r then the relation between! r and! r is just a 3-dimensional rotation becuse of the determinants, det h = x 2 + y 2 + z 2 = det h = x 2 + y 2 + z 2 i.e. relation between! r and! r is a linear transformation and can be written as r i = R ij r j where R is an orthogonal matrix. This establishes the correspondence between SU (2) matrix and 3-dimensional rotation. Note that from Eq() we see that U and U, give the same h and hence the same rotation. So the correspondence U! R is a homomorphism rather than isomorphism. (Institute) Continuous Group & SU (2) 2 / 46

13 Rotation about z-axis Suppose U is diagonal. Then the general form is given by, e i α/2 U = e i α/2 and h = z x iy x + iy z = UhU = z (x iy ) e i α (x + iy ) e i α z This gives the relation 8 < : x = cos αx + sin αy y = sin αx + cos αy z = z which is clearly a rotation around z z-axis. axis. Thus a diagonal U corresponds to rotation about (Institute) Continuous Group & SU (2) 3 / 46

14 Rotation about y-axis If U is real B U cos β 2 sin β 2 C A sin β 2 cos β 2 Then h = = cos β 2 sin β 2 C A z x iy x + iy z sin β cos β 2 2 z cos β + x sin β x cos β iy z sin β iy + x cos β z sin β z cos β x sin β cos β 2 sin β 2 sin β 2 cos β 2 C A and 8 < x = cos βx sin βz y = y : z = sin βx + cos βz i.e. a rotation about y axis. Rotation in terms of Euler angles The 2 2 unitary matrix corresponding to rotation by Euler angles, is then (Institute) Continuous Group & SU (2) 4 / 46

15 U (α, β, γ) = P z (α) P y (β) P z (γ) (2) e i α/2 B cos β sin β = e i 2 2 C e i γ/2 sin β cos β A e i γ/2 (3) 2 2 B e i(α+γ)/2 cos β e i(α γ)/2 sin β 2 2 C a b e i(α+γ)/2 sin β e i(α+γ)/2 cos β A = b a, 2 2 where a = e i(α+γ)/2 cos β 2 write b = e i(α γ)/2 sin β 2 a = u + iu 2 = e i(α+γ)/2 cos β 2, b = u 3 + iu 4 = e i(α γ)/2 sin β 2 the group invariant integration can be converted to the integration over Euler angles by computing the Jacobian. Z Z du du 2 du 3 = d αd βd γj (Institute) Continuous Group & SU (2) 5 / 46

16 First write Z dr = = Z Z du du 4 δ u 2 + u u u 2 4 du du 2 du 3 2 ju 4 j [δ (u 4 a) + δ (u 4 + a)] where q a = u 2 + u2 2 + u2 3 The Jacobian is then of the form J = (u, u 2, u 3 ) (α, β, δ) 2 3 = sin (α + γ) /2 cos β/2 cos (γ + γ) /2 sin β/2 sin (α + γ) /2 cos β/2 4 cos (α + γ) /2 cos β/2 sin (α + γ) /2 sin β/2 cos (α + γ) /2 cos β/2 5 8 sin (α γ) /2 sin β/2 cos (α γ) /2 cos β/2 sin (α γ) /2 sin β/2 = 8 2 sin (α γ) /2 cos β/2 sin 2 β/2 The integration measure in terms of Euler angles is of the form, Z Z dr = du du 4 δ u 2 + u2 2 + u3 2 + u4 2 = Z 2π Z π Z 2π d α sin βd β d γ 6 The factor is an overall normalization factor and is usually neglected for conveience. 6 (Institute) Continuous Group & SU (2) 6 / 46

17 Irreducible Representation of SU (2) The 2 2 matrice of SU (2) can be viewed as a rotations in the complex 2 dimensional space C 2. We can use the induced transformation on functions of these 2 dimensional coordinates to generate other representations. ξ,η basis vector for 2 2 matrice of SU (2) a b P U (ξ, η) = (ξ, η) U = (ξ, η) b a = (aξ + b η, bξ + a η) i.e. ξ = P U ξ = aξ + b η η = P U η = bξ + a η Consider the action of P U on the mononomial ξ λ η µ, with λ + µ = 2j, where j is an integer or half integer. As an example, take j =. There are 3 di erent monomials, Then under the induced transformation ξ 2, ξη, η 2 ξ 2 = (aξ + b η) 2 = a 2 ξ 2 + 2ab ξη + b 2 η 2 ξ η = (aξ + b η) ( bξ + a η) = abξ 2 + (aa bb ) ξη + a b η 2 (Institute) Continuous Group & SU (2) 7 / 46

18 η 2 = ( bξ + a η) 2 = b 2 ξ 2 2a bξη + a 2 η 2 Write this in matrix notation ξ 2 p 2 ξ η η 2 p 2 C A p a 2 2ab p b 2 2ab (aa bb p p ) 2a b b 2 2a b a 2 A ξ 2 p 2 ξη η 2 p 2 C A We have inserted the factor of α, β, γ p 2 in order to get a unitary matrix. In terms of Euler angles D () B 2 ( + cos β) e i(α+γ) p sin βe i α 2 2 ( cos β) e i(α γ) p sin βe i γ cos β p2 sin βe i γ 2 2 ( cos β) e i(α γ) p sin βe i α 2 2 ( + cos β) e i(α+γ) C A As discussed before these matrice for di erent values of a, b will form the representation matrices of the SU (2) group. For case β = γ =, we have rotation about z axis, D () (α) e i α e i α A (Institute) Continuous Group & SU (2) 8 / 46

19 However, in Cartesian coordinates the rotation around x y z A cos α sin α sin α cos α axis is, x y z A Rotation matrix D () (α) corresponding to basis (x + iy, z, x Cartesian coordinates by a unitary transformation, Generalize this to arbitrary x + iy z A i x y A x iy i z φ j m = n jm (ξ) j+m (η) j m, m = j, j +,, j iy ), spherical basis, related to the (Institute) Continuous Group & SU (2) 9 / 46

20 where n jm is a normalization factor to be determined later. Since the transformation from (ξ, η) to ξ, η is linear and homogeneous the transform of φ j m will have same j but di erent m. More explicitly, P U φ j m = = n jm (aξ + b η) j+m ( bξ + a η) j m n jm j m j+m s= r = (j + m)! r! (j + m r )! (j m)! s! (j m s)! ( b) s (a ) j m s a r (b ) j+m r ξ r ξ s (η) j+m r (η) j m s De ne m by Then P U φ j m = r,m s = j + m r, j m j njm n jm " (j + m)! (j m)! ( ) j+m r r! (j + m r )! (j + m r )! (r m m )! # a r (b ) j+m r (b) j+m r (a ) r m m φ j m In terms of Euler angles, a = e i(α+γ)/2 cos β 2, b = e i(α γ)/2 sin β 2 (Institute) Continuous Group & SU (2) 2 / 46

21 and we can write where P U φ j m = φ j m m D j m (α, β, γ) m D j m m (α, β, γ) = e imα d j m m (β) e imγ (4) with d j m m (β) = (j + m)! (j m)!n jm ( ) n jm j+m k k! k (j + m k)! (j + m k)! (k m m )! cos β 2k m m sin β 2j 2k+m+m 2 2 (5) Note that sum over k covers all thos values for which the argument of the factorial functions are positive. (Institute) Continuous Group & SU (2) 2 / 46

22 Properties of D j m (α, β, γ) m D j m (α, β, γ) s form (2j + ) dim rep of SU (2), because group induced transformation m always generates a representation of the group as discussed in Note 2. 2 D j m m (,, ) = δ mm, the identiy matrix. To see this we set β = in Eq(5) and the non-zero term is where 2k = 2j + m + m. This implies that j + m k = 2 m m, j + m k = 2 m m and the positivity of the arguments of the factorial functions gives m = m and d j m m () = δ mm. 3 For matrix D j m (α, β, γ) to be unitary, we require m D j m m (α, β, γ) = D j m m ( γ, β, α) ) d j m m ( β) = d j mm (β) It is straightforward to show that this xes the constant n jm to be q n jm = (j + m)! (j m)! (Institute) Continuous Group & SU (2) 22 / 46

23 and the d nction is then d j m m (β) = ( ) k p (j + m)! (j m)! (j + m )! (j m )! k! k (j + m k)! (j + m k)! (k m m )! cos β 2k m m sin β 2j 2k+m+m 2 2 (6) (7) 4 For each j, the rep D j m (α, β, γ) is irreducible. This can seen as follows. Suppose 9 a m matrix M such that MD j (α, β, γ) = D j (α, β, γ) M (8) Consider following cases: α 6=, β = γ = From Eq(6), only non-zero term is 2j 2k + m + m = which gives Then d j m m () = δ mm D j m m (α,, ) = e imα δ mm (Institute) Continuous Group & SU (2) 23 / 46

24 and Eq(8) implies that e imα e im α M mm = Thus M mm = if m 6= m, i.e. M is diagonal. 2 α 6=, β 6=, γ 6= Since M is diagonal, Eq(8) gives M mm D j m m = Dj m m M m m (no sum) But for arbitrary (α, β, γ), D j m m is not zero. Thus M mm = M m m, or M is a multiple of identity. Schur s lemma implies that D j m m (α, β, γ) is irreducible. 5 If we replace β by β + 2π, we see that from Eq(2) that the 2 2 matrix U has the property, U! U For the representation matrice this implies that d j (β + 2π) = ( ) 2j d j (β), or D j ( U ) = ( ) 2j D j (U ) Thus for j =integer D j ( U ) = D j (U ) single valued representation j = half integer D j ( U ) = D j (U ) double valued representation (Institute) Continuous Group & SU (2) 24 / 46

25 6 Representation of generators J i, i =, 2, 3 Recall that ijz R (α, β, γ) = exp h α exp Take β = γ =, α, we get R (α, β, γ) ' ijy h β exp ij z h α ijz h γ From we get D j m m (α,, ) ' im α δ mm D j m m (J z ) = hmδ mm Similarly, take α = γ = π 2, and β small, we can get D (J x ) α = γ = and β small, we can get D (J y ) In particular, for j = /2, we have D /2 (j x ) = h 2, D /2 (j y ) = h i 2 i, D /2 (j z ) = h 2 which are the Pauli matrices up to h 2. (Institute) Continuous Group & SU (2) 25 / 46

26 7 Great Orthogonality Theorem reads Z 2π Z π Z R π 2π d α sin βd β d γd j k m (α, β, γ) D m n n (α, β, γ) = δ jk δ mn δ m n d α R π sin βd β R π d γ (2j + ) Using D j m (α, β, γ) given in Eq(4) we can integrate over angles α and β to reduce the m integral to orthogonality relation on d j m m (β), Z π sin β d βd j m m (β) d k m m (β) = 2δ jk (2j + ) 8 Characters of irreps Since all rotations of same angle are in the same class, choose rotation about z-axis to compute the trace χ (j) (θ) = Tr hd i h i j! n, θ = Tr D j (θ,, ) = j e imθ = m= j sin j + θ 2 sin θ 2 Note that χ (j) (θ) χ (j ) (θ) = 2 cos jθ (Institute) Continuous Group & SU (2) 26 / 46

27 Theorem: There are no irreps of SU (2) group other than D j (α, β, γ). Proof: Suppose D is another irrep with characterχ (θ), not contained in D j. Then χ (θ) = χ ( θ) since they are in the same class. Then from orhogonality theorem χ (θ)? χ (j) (θ) for all j. =) h i χ (θ)? χ (j) (θ) χ (j ) (θ) = 2 cos jθ From the property of Fourier series, cos jθ, 2j =, 2, 3, form a complete set of even functions in the range < θ < π. Thus χ (θ) =. for all θ. Basis Functions Suppose we de ne for j = l =integer r D (l) 4π m (α, β, γ) = 2l + Y l m (β, γ) then we have D (l) m (α, β, γ) = ( )m r 4π 2l + Y m l (β, α) (Institute) Continuous Group & SU (2) 27 / 46

28 Suppose we have the relation for multiplication of group elements R (α 2, β 2, γ 2 ) R (α, β, γ ) = R (α, β, γ) The corresponding representation matrices will satisfy Setting m =, the relation is Dm l m (α, β, γ) = Dm l m" (α 2, β 2, γ 2 ) Dm"m l (α, β, γ ) m" Y m l (β, γ) = Yl m" (β 2, γ 2 ) Dm"m l (α, β, γ ) m" Thus functions fyl m (β, γ)g form the basis for irrep D l (α, β, γ). We can rewrite this in a more familiar spherical angles notation as, P R Yl m (θ, φ) = Yl m θ, φ = Yl m" (θ, φ) Dm"m l (α, β, γ) (9) m" where cos θ = cos θ cos β + sin θ sin φ cos (φ γ) Example : (Institute) Continuous Group & SU (2) 28 / 46

29 l = Y (θ, φ) = Y (θ, φ) = r 3 8π sin θe i φ = r 3 8π cos θ = r 3 z 8π r, r 3 8π (x + iy ), r 2 l = 2 Y (θ, φ) = r 3 8π sin θe i φ = r 3 (x iy ) 8π r Y2 2 (θ, φ) = r 5 4 2π sin2 θe 2i φ x 2 y 2 + 2ixy r 2 (x + iy )2 r 2 r 5 2π sin θ cos θe i φ z (x + iy ) r 2 Y2 (θ, φ) = 2 Y2 (θ, φ) = r 5 4 2π 3 cos2 θ 2z 2 x 2 y 2 r 2 2z 2 (x + iy ) (x iy ) r 2 Y2 (θ, φ) = r 5 2 2π sin θ cos θe i φ z (x iy ) r 2 Y2 2 (θ, φ) = r 5 4 2π sin2 θe 2i φ x 2 y 2 2ixy r 2 (x iy )2 r 2 (Institute) Continuous Group & SU (2) 29 / 46

30 3 For the special case of m = we have and from Eq(9) we get the addition theorem, r 2l + Yl (θ, φ) = 4π P l (cos θ) P l cos θ 4π = 2l + m Yl m (θ, φ) Yl m (β, γ) where cos θ = cos θ cos β + sin θ sin β cos (φ γ) (Institute) Continuous Group & SU (2) 3 / 46

31 Product Representations Addition of Angular Momentum Suppose D (j ) and D (j 2) are 2 irreps of SU (2). How to reduce the product representation D (j ) D (j 2) D (j j 2 ), i.e. write D (j j 2 ) as sum of irreps?. Using the characters of irreps (take j > j 2 ), χ (j j 2 ) (θ) = χ (j) (θ) χ (j2) (θ) sin j + 2 = = = 8 < sin 2 θ : 2 sin θ 2 sin θ sin j 2 + θ 2 = cos (j j 2 ) θ cos (j + j 2 + ) θ sin 2 θ 2 sin 2 θ 2 [cos (j j 2 ) θ cos (j j 2 + ) θ] + [cos (j j 2 + ) θ cos (j j 2 + 2) θ] + [cos (j + j 2 ) θ cos (j + j 2 + ) θ] j j θ + sin j j = ; θ + sin (j + j 2 ) θ the character of D (j ) D (j 2) has the decomposition, χ (j j 2 ) (θ) = j +j 2 J=jj j 2 j χ (J) (θ) (Institute) Continuous Group & SU (2) 3 / 46

32 which implies the reduction, D (j j 2 ) = D (j ) D (j 2) = j +j 2 J=jj j 2 j D (J) () We can relate this to the addition of angular momenta. Let f m j (ˆr ) be the basis for D (j) irrep and f m 2 j 2 (ˆr ) be the basis for D (j2). Then the product f m j (ˆr ) f m 2 j 2 (ˆr ) are the basis for D (j) D (j2), i.e. i h P R hf m j f m 2 j 2 = P R f m j i h P R f m 2 i j 2 m f m j D (j ) For in nitesmal rotation around, say z-axis, (setting h = ) m m P R = e i αjz ' ( i αj z ), α m 2 f m 2 j 2 D (j 2) m 2 m 2 A Let! J acting on f m j and leaving f m 2 j 2 alone and! J 2 acting on f m 2 j 2 and leaving f m j alone. Then we can write i P R hf m j f m 2 j 2 = h ( i αj z ) f m j i h = ( i αj z ) f m j f m 2 j 2 ( i αj z ) f m 2 j 2 i ' [ i α (J z + J 2z )] f m j f m 2 j 2 (Institute) Continuous Group & SU (2) 32 / 46

33 where J z = J z + J 2z We can extend this to other components to write! J =! J +! J 2 which is just the total angular momentum. =) in nitesmal rotation of product of the basis can be written in terms of total angular momenta. From Eq() we see that decomposition of the product of irreps is related to the additon of angular momenta. (Institute) Continuous Group & SU (2) 33 / 46

34 Clebsch-Gordon Coe cients In the decomposition of product representations D (j j 2 ) = D (j ) D (j 2) = D (jj j 2 j) + D (jj j 2 j+) + + D (j +j 2 ) =)representation matrices D (j j 2 ) can be changed into a direct sum of irreps D j. In terms of matrix elements we have D (j j 2 ) m m 2 ;m = D (j ) m 2 m m D (j 2) m 2 m2 On the other hand, D (jj j 2 j) D J = B D (jj j 2 j+)... D (j +j 2 ) C A J M ;JM = δ JJ DMM J where J, J label the boxes and M, M labels the row and colums within each box. Let A be the unitary matrix which transform D (j ) D (j 2) into, Writing out in terms of matrix elements, D (j ) D (j 2) = A A (Institute) Continuous Group & SU (2) 34 / 46

35 h D (j j 2 ) m m 2 ;m = A i m 2 m m 2 ;J M J M ;JM [A] JM ;m m 2 () New notation for the matrix elements of the similarity transformation, [A] JM ;m m 2 hjm jj m j 2 m 2 i These are called Clebsch-Gordon coe cients. Since A is unitary, AA : =, and A A = we have j mj 2 m2jj M J M jj m j 2 m 2 = δm m δ m 2 m 2 J M m m 2 hjm jj m j 2 m 2 i j m j 2 m 2 jj M = δ JJ δ MM Note that unitary matrix A is not uniquely de ned. Let B be a unitary matrix which commutes with, B = B, or = B B Then D = A A = (BA) BA Thus if A block diagonalizes D, so does A = BA. Since is of the form D (J ) = D (J 2) (Institute) Continuous Group & SU (2) 35 / D C A

36 A JM ;m m 2 = e i θ J [A] JM ;m m 2 From Schur s lemma, B must be of the form, B = c I c 2 I 2... I C A, where c i = e i θr Or B JM ;J M = δ JJ δ MM e i θ j and Thus similarity transformation is de ned up to J dependent phases. Condon-Shortly convention : Choose hjjjj j j 2 J j i to be real and positive, then it turns out that all Clebsch-Gordon coe cients are real. In terms of Clebsch-Gordon coe cients Eq ) becomes D (j ) m m D (j 2) m 2 m2 = D J M M JM jj mj 2 m 2 hjm jj m j 2 m 2 i J,M,M (Institute) Continuous Group & SU (2) 36 / 46

37 Theorem: Let ψ m j (ˆr ) be the basis for D (j) irrep and ψ m 2 j 2 (ˆr ) be the basis for D (j2). Then are basisi for D J. Proof: P R φ J M = hjm jj m j 2 m 2 i m m 2 = hjm jj m j 2 m 2 i m m 2 = M "m m 2 D J M "M φ J M = hjm jj m j 2 m 2 i ψ m j ψ m 2 j 2 m m 2 P R ψ m j J,M,M D J JM "jj m j 2 m 2 P R ψ m 2 j 2 = hjm jj m j 2 m 2 i D (j ) m m D (j 2) m m m 2 m ψ m j 2 ψ m 2 j 2 2 M M " J M jj mj 2 m2 J M m jj m j 2 m 2 ψ ψ m j ψ m 2 j 2 Thus φ J M does transform according to the representation D J. As a consequence if! j 3 =! j +! j 2 then ψ m j ψ m 2 j 2 ψ m 3 j 3 hj m j 2 m 2 jj 3, m 3 i hj 3, m 3 j 3 m 3 ji m m 2 m 3 is invariant under SU (2) transformations. j ψ m 2 j 2 (Institute) Continuous Group & SU (2) 37 / 46

38 Rotation group and Quantum Mechanics In quantum mechanics, implement symmetry transformations by unitary operator U on the states jψi jψi! ψ = U jψi, for all states so that φ jψ = hφjψi At the same time the operators change as A! A = UAU so that φ ja jψ = hφjajψi If A = A, or [U, A] =, we say that A is invariant under the transformation U. In particular, if Hamiltonian H is invariant under the symmetry transformation U [U, H ] = (2) Suppose, then [U, H ] =, [U 2, H ] = [U U 2, H ] = [U, H ] U 2 + U [U 2, H ] = (Institute) Continuous Group & SU (2) 38 / 46

39 Collection of all such operators form a group. Suppose jψi is an eigenstate of H with energy E, Then from Eq(??) we see that H jψi = E jψi HU jψi = UH jψi = E (U jψi) which means that U jψi is also an eigenstate of H with same energy E. If we run the operator U through the whole group G, we get U jψi, U 2 jψi, U 3 jψi, If we select a linear independent set out of these, then we have degeneracy of energy levels.on the other hands, if we consider the eigenfunction ψ (x ) of time independent Schrodinger equation, H ψ (x ) = E ψ (x ) we can use induced tranformation on this eignefunctions to get P R ψ (x ) = ψ R x, P R2 ψ (x ) = ψ R 2 x, As discussed before, if we select a linearly independent set out of these transformed functions, they will form a basis of irrep. Since they are also eigenfunctions of Hamiltonian with same E, we get the result, degenercy = dim of irrep (Institute) Continuous Group & SU (2) 39 / 46

40 !x For example, if choose ψ = z, then the basis for the transformed functions will be, x, y, z. and they form the basis of l = irrep of rotational group. In the case of hydrogen atom, the Hamiltonian is of the form H = p2 2m e 2 4πεr and is invariant under rotation [R (α, β, γ), H ] = where R is an arbitrary 3-dimensional rotation. As a consequence, the degeneracy of the energy levels is 2l +. Another important result is that for matrix elements between states belong to di erent irrep are zero, D φ m j jφm j E δ jj δ mm This follows from the great orthogonal theorem as follows. We can perform a rotation on this matrix element to get D φ m j jφm j E D = P R φ m j jp R φ m j E D E = D j l,l l m (R ) D j lm (R ) φ l j jφl j, In order to use great orthogonality theorem we sum over all group elements, D φ m j jφm j E Z Z dr = D dr P R φ m j jp R φ m j E Z = l,l, D E dr D j l m (R ) D j lm (R ) φ l j jφl j (Institute) Continuous Group & SU (2) 4 / 46

41 Then from we get the result Z drd j l m (R ) D j lm (R ) δ jj δ mm δ ll D φ m j jφm j E δ jj δ mm (Institute) Continuous Group & SU (2) 4 / 46

42 Wigner-Eckart Theorem In quantum mechanics, need to compute matrix elements of certain operators. We can use the representation theory to simplify the computations. The strategy is to write the operator and wave functions in terms of irrep of group and use the rep theory. Tensor operators Suppose D (α) is an irrep of group G and dimension of D (α) is d α. A set of operators T (α) i, i =, 2, d α, transforming under the symmetry group G as P R T (α) i P R = j T (α) j D (α) ji (R ) is said to be irreducible tensor operators corresponding to D (α) irreps. For the case of SO (3) (or SU (2) group), if the operators Tj m, m = j, j +, j satisfy the relation, P R (α, β, γ) Ti m PR (α, β, γ) = Ti m D j m m (α, β, γ) m then we say T m j are tensor operators of rank. Remarks : are irreducible tensors of rank j in SO (3). For example, T () x + iy, T () z, T () x iy (Institute) Continuous Group & SU (2) 42 / 46

43 If T m j are irreducible tensor of rank j and T m 2 j 2 are irreducible tensor of rank j 2 then hjmjj m j 2 m 2 i T m j T m 2 j 2 m,m 2 are irreducible tensor of rank j. In particular, since hjjmj mi = ( )j m p 2j + we see that the combination Tj m Sj m ( ) m m is an invariant operator under rotations. Here S m j is another tensor operator of rank j. 2 If T m j are irreducible tensor operator of rank j, so is This follows from the fact that h i Sj m = ( ) m Tj m D j m m (α, β, γ) = ( )m m D j m (α, β, γ), m (Institute) Continuous Group & SU (2) 43 / 46

44 3 Combining remarks () and (2) we get that Tj m m Tj m is a SO (3) invariant operator. Theorem (Wigner Eckart) If φ m j are basis for D j, φ m j are basis for D j and T q k are irreducible tensor operator of rank k, then E D E Dφ φ m j jt q k jφm j = j m jkqjm j jjt k jjφ j p 2j + E where Dφ j jjt k jjφ j are the reduced matrix elements and are independent of m, m, q. Proof: Since T q k D φ m j are irreducible tensor operator, we can write jt q k jφm j E D = P R φ m j jp R T q k P R jp R φ m j = D D j l m (R ) D j lm (R ) l,l q, E φ l q j jtk jφl j E D k q q (R ) = D j l m (R ) DM J M (R ) JM jkq jl hjm jkqjmi J,M,M l,l q, D E φ l q j jtk jφl j (Institute) Continuous Group & SU (2) 44 / 46

45 Summing over all group elements in SU (2), we get Z D E dr φ m j jt q k jφm j = = D Z φ m j dr jt q k jφm j E Z J,M,M l,l q, dr D j l m (R ) D J M M (R ) JM jkq jl hjm jkqjmi D E we have used the fact φ m j jt q k jφm j is independent of R. Using great orthogonality theorem Z Z drd j l m (R ) DM J M (R ) = δ Jj δ l M δ m M dr D E φ l q j jtk jφl j we get D φ m j jt q k jφm j E = j m jkqjm j l jkq jl D l,l q E Dφ = j m jkqjm j jjt k jjφ j p 2j + φ l q j jtk jφl j E where Dφ j jjt k jjφ j E = p 2j + l,l q j l jkq jl D is the reduced matrix element and is independent of m, m, q. Remarks: φ l q j jtk jφl j (Institute) Continuous Group & SU (2) 45 / 46 E

46 D E The m, m, q dependence in the matrix element φ m j jt q k jφm j are all contained in hj m jkqjmi which are universal and independent of details of φ m j, φ m j and T q k. Then in the ratios of matrix elements we have D E φ m 2 j jt q k jφm j D E = hj m 2 jkqjm i φ m 4 j jt q k jφm 3 hj m 4 jkqjm 3 i j Thus, we can calculate only one such matrix element explicitly and use Clebsch-Gordon coe cients to obatin other matrix elements with same j, j and k. 2 Since Clebsch-Gordon coe cients hj m jkqjmi has the property that it vanishes unless,jk jj j j + k, and m = q + m, the matrix elements will satify the same selection rules independent of the nature of the wavefunctions φ m j, φ m j or the operator T q k. D Example: Dipole matrix elements combinations, r + = φ m j j! r jφ m j E,! r = (x, y, z). Note that the linear p 2 (x + iy ), r = z, r = p 2 (x iy ) are the basis for the irrep D mm. Thus we have the selection rules: matrix elements vanish unless j = j, but if j = then j = and m = m or m (Institute) Continuous Group & SU (2) 46 / 46