Tensor Operators and the Wigner Eckart Theorem

November 11, 2009 Tensor Operators and the Wigner Eckart Theorem Vector operator The ket α transforms under rotation to α D(R) α. The expectation value of a vector operator in the rotated system is related to the expectation value in the original system as α V i α α D V i D α R ij α V j α With D(R) e ī h J ˆnθ, and R ij an orthogonal X rotation matrix. Define a vector operator as an object that transforms according to In the case of an infinitesimal rotation D V i D R ij V j. RV V + δθˆn V. The effect of an infinitesimal rotation of V about J ˆn by δθ gives us (1 + ī h J nδθ)v(1 ī J ˆnδθ) V + δθˆn V h ī [J ˆn, V] ˆn V h Then i h [J i, V j ]δθ j ɛ ijk V k [V j, J i ] i hɛ ijk V k [V i, J k ] i hɛ ijk V j [V i, J j ] i hɛ ijk V j [V x, J z ] i hv y 1

or more generally [V i, J j ] i hɛ ijk V k We can take that last as the definition of a vector operator. A rotation of a vector operator is accomplished by computing D VD It is convenient if we can write the operator in a basis of angular momentum eigenstates since we know how to write D in that basis. So we might write the position operator as x r 8π 2 (Y 1 1 + Y1 1 ) y r 8π 2i (Y 1 1 Y 1 4π z r Y 1 0 1 ) Y 0 1 Y ±1 1 z 4π r 4π x ± iy 2r where r x 2 + y 2 + z 2. Some examples of vector operators are momentum, position, and angular momentum. If we write our vector using spherical harmonics as a basis then our definition of a vector operator reads where m ±1, 0. D V m D D 1 mm V m Cartesian tensor operator We can form a rank 2 cartesian tensor T by taking the product of two cartesian vector operators U and V. T i,j V i U j Then the transformation property follows as D T ij D R ik V k R jl V l R ik R jl V k U l There are 9 components of this second rank cartesian tensor. And the components do not transform irreducibly. In particular we can write T as T ij δ ij V U + ( V iu j V j U i ) + ( V iu j + V j U i 2 2 2 V U δ ij )

Under rotations, the first term has a single component and is invariant, the second term has components and transforms like a vector, and the third term is a symmetric, traceless tensor with 5 independent terms. Meanwhile, suppose that T x i p j where x and p are position and momentum operators. The tensor will include a scalar x p a vector x p and a symmetric traceless tensor 1(x 2 ip j + x j p i ) 1 x p. We could write the operator in a spherical basis using spherical harmonics where we let and then Y ±1 1 (V ) Y ±1 1 x ± iy 8π r 8π sin θe±iφ, Y 0 1 1 z 1 4π r 4π cos θ 1 1 8π (V x ± iv y ) V 8π sin θe±iφ, Y1 0 (V ) 4π V z V 4π cos θ So we have a recipe for translating a cartesian vector into a spherical l 1 tensor. Then we can combine the l 1 tensors using Clebsch Gordon coefficients pretty much the same as Y m l T k q q 1,q 2 kql 1 l 2 l 1 q 1 l 2 q 2 E l 1 q1 F l 2 q 2 m 1,m 2 l, ml 1, l 2 l 1, m 1, l 2, m 2 Y m 1 l 1 Y m 2 l 2 And how does a spherical harmonic transform under rotations? First let s start with an example. Let s suppose that we have a cartesian vector operator V (V x, V y, V z ). We can also write it as V ±1 1 2 (V x ± iv y ), V 0 V z. Let s consider an infinitesimal rotation of the vector about the z-axis. We will rotate the cartesian version using an orthogonal matrix (SO()) and the spherical version using a representation of SU(2). The SO() rotation of the cartesian vector is here 1 ɛ 0 V x V x ɛv y V R z (ɛ)v ɛ 1 0 V y ɛv x + V y (1) 0 0 1 V z V z The SU(2) rotation of the spherical vector is V D z (ɛ)v (1 ī h ɛj z) V 1 V 0 V 1 1 iɛ 0 0 0 1 0 0 0 1 + iɛ V 1 V 0 V 1 V 1 (1 iɛ) V 0 V 1 (1 + iɛ)

Converting V spherical to V cartesian we get V cart V 1 + V 1 i(v 1 V 1) V 0 V 1 + V 1 iɛ(v 1 V 1 ) i(v 1 V 1 iɛ(v 1 + V 1 )) V 0 V x + ɛv y V y ɛv x (2) V z Comparing Equation 1 with 2 we see that the SU(2) representation of the rotation by ɛ is D( ɛ) when the SO() representation is R z (ɛ). We see this again as follows ˆn D(R) ˆn D(R 1 ) l, m m l, m D l m m(r 1 ) ˆn D(R 1 ) l, m ˆn l, m Dm l m(r 1 ) m Yl m (ˆn ) Yl m (ˆn)D m l m(r 1 ) m We can turn Yl m (ˆn ) into a more general vector operator V by setting V ± V Y 1 ±1 and V 0 V Y1 0. D (R)Y m l (V)D(R) m Y m l (V)D l m m(r 1 ) Yl m (V)D l mm (R) m Now we define a spherical tensor as an object that transforms according to D T k q D m D m q(r 1 )T k m m D k qm (R)T k m On the left we rotate the state by R and then measure by taking the expectation value. On the right we first measure the expectation value of each component of the operator and then rotate those expectation values by R 1. Then an infinitesimal rotation gives us (1 + ī h J ˆnθ)T k q (1 ī h J nθ) m km (1 + ī h J ˆn) kq T k m T k q + ī h m km J ˆnθ kq T k m [J ˆnθ, T k q ] m km J ˆnθ kq T k m 4

If ˆn ẑ then or if ˆn 1 2 (ˆn x ± iˆn y ), [J z, T k q ] q ht k q () [J ±, T k q ] h m k(k + 1) q(q ± 1)km kq ± 1 T k m h k(k + 1) q(q ± 1)T k q±1 Selection Rule We show that α, j, m T k q α, j, m 0, unless m q + m (4) Using Equation we have 0 α, j, m [J z, Tq k ] hqtq k α, j, m h α, j, m Tq k (m m q α, j, m h α, j, m Tq k α, j, m (m m q) so if (m m q) 0 then the expectation value of T k q does. If T k q us a vector operator (k 1), then the matrix element α, j, m Tq k α, j, m is zero unless m m ±1, 0 and j j 1, 0. Tensor operators transform under rotations the same as angular momentum eigenkets. From Equation 4 we see that T k q jm j, q + m. The effect of the operator is to add q units of angular momentum. Combining Rotation matrices We can use the Clebsch Gordan matrix to combine rotation matrices. Let s go back to the general transformation from the m 1, m 2 basis to the j, m basis. j, m m 1,m 2 j 1, m 1, j 2, m 2 j 1, j 2, m 1, m 2 j, m 5

We can rotate the pieces by applying the rotation operator and we have or we could write D(R) j, m D 1 (R)D 2 (R) j 1, m 1, j 2, m 2 j 1, m 1, j 2, m 2 j, m D(R) j, m D 1 (R)D 2 (R) j 1, m 1 j 2, m 2 j 1, m 1, j 2, m 2 j, m Then multiply from the left by j, m and we get D j m,m m 1,m 2 Clebsch-Gordan Series m 1,m 2 If we invert the above we have j 1, m 1, j 2, m 2 j 1, j 2, m 1, m 2 j, m m 1,m 2 m 1,m 1 Dj 2 m 2,m 2 j 1, j 2, m 1, m 2 j, m j 1, m 1, j 2, m 2 j, m (5) j 1, m 1, j 2, m 2 j, m j, m j 1, m 1, j 2, m 2 j,m Then rotation and multiplication from the left by the dual gives m 1,m Dj 2 1 m 2,m j, m D j, m j 2 1, m 1, j 2, m 2 j, m j, m j 1, m 1, j 2, m 2 j,m,j,m The rotation does not change j so m 1,m Dj 2 1 m 2,m D j 2 m,m j 1, m 1, j 2, m 2 j, m j, m j 1, m 1, j 2, m 2 (6) j,m,m This is called the Clebsch Gordan series or the Kronecker or direct product of representations and we can write +j 2 +j 2 1... j 2 or +j 2 0 0 0 +j 2 1 0.... 00 00 α. 0 j 2 6

Integration over rotations We want to consider drd(r) α j,m dr D(R) j, m j, m α where Then dr 4π 0 dα 4π dγ π dβ sin β 4π 0 4π 0 2 dr D j mm δ m,0δ m,0δ j,0 1 This follows because integration over all angles averages over all directions and the average is zero except for the that state that has no direction, namely j m 0. Now we can use this result along with Equation 6 and write m 1 m Dj 2 1 m 2 m dr dr D j 2 m,m j 1, m 1, j 2, m 2 j, m j, m j 1, m 1, j 2, m 2 j,m,m j 1, m 1, j 2, m 2 0, 0 0, 0 j 1, m 1, j 2, m 2 We can guess at the relevant Clebsch Gordon coefficients. Write 0, 0, j 1, j 2 m 1,m 2 j 1, m 1, j 2, m 2 0, 0, j 1, j 2 j 1, m 1, j 2, m 2 It is clear that j 1 and j 2 must be equal and m 1 m 2 for the coefficients to be non zero. If for example j 1 j 2 2 then 0, 0, 2, 2 a 2, 2, 2, 2 +b 2, 1, 2, 1 +c 2, 0, 2, 0 +d 2, 1, 2, 1 +e 2, 2, 2, 2 No matter how we rotate the state there will be equal parts of all five components so it must be that all the coefficients are equal magnitude. Then for the normalization to be right 1 j, m, j, m 0, 0, j, j 2j + 1 and we can write m 1 m Dj 2 1 m 2 m dr 2 j 1, m 1, j 2, m 2 0, 0 0, 0 j 1, m 1, j 2, m 2 δ j 1,j 2 δ m1, m 2 δ m 1, m 2 2j + 1 7

Then using D(R 1 ) D (R) we can see that D j m,m ( 1)m m D j m, m and we can write m 1 m 1 D j 2 m 2 m 2 dr δ j 1,j 2 δ m1,m 2 δ m 1,m 2 2j + 1 (7) Integration of a triple product of rotation matrices Begin with the Clebsch Gordon series, Equation 6. m 1,m 1 Dj 2 m 2,m 2 j,m,m D j m,m j 1, m 1, j 2, m 2 j, m j, m j 1, m 1, j 2, m 2 and multiply both sides by D J MM and integrate over R and use Equation 7 dr m 1,m 1 Dj 2 m 2,m 2 DJ MM where we have used dr Meanwhile we have already shown that Substitution into 8 gives j,m,m D j m,m j 1, m 1, j 2, m 2 j, m j, m j 1, m 1, j 2, m 2 D J MM j 1, m 1, j 2, m 2 j, m j, m j 1, m 1, j 2, m 2 δ m,mδ m,m δ j,j 2j + 1 j,m,m j 1, m 1, j 2, m 2 J, M J, M 1 j 1, m 1, j 2, m 2 2J + 1 D j m,m (R) ( 1) m m D j m, m (R). 4π 4π Dm0(α, l β, γ) 2l + 1 Y lm(β, α) 2l + 1 Y lm(n) dωylm(ω)y l1,m 1 (Ω)Y l2,m 2 (Ω) (2l 1 + 1)(2l 2 + 1) l 1, m 1, l 2, m 2 LM l 1, 0, l 2, 0 L0 4π(2L + 1) (8) 8

Wigner Eckart Theorem Spherical tensor operators transform according to D T k q D q D k q,q T k q which implies that T k q q DD k q,q T k q D α, j, m Tq k α, j, m α, j, m D α, j, m 1 Dq,q k α, j, m1 Tq k α, j, m 2 α, j, m 2 D α, j, m q,m 1,m 2 Dmm j 1 Dq,q k α, j, m1 Tq k α, j, m 2 D j m 2 m q,m 1,m 2 Before integrating let s look at that more carefully. D T k q D q D k qq T k q jm D T k q D j m q D k qq jm T k q j m jm D jm 1 jm 1 Tq k j m 2 j m 2 D j m D k qq jm T q k j m m 1 m 2 q D j mm 1 jm 1 Tq k j m 2 D j m 2 m D k qq jm T q k j m m 1 m 2 q jm 1 T k q j m 2 D 1j q,m,m m 1 md k qq jm T k q j m D 1 j Dm j 1 md k qq jm T q k j m D j q,m,m m 2 m m m 2 Then integrate over R using Equation 8 and we get dr α, j, m 1 Tq k α, j, m 2 dr Dm j 1 md k qq αjm T q k α j m D j m 2 m q,m,m 1 2j + 1 9 q,m,m αjm T k q α j m jm 1 kqj m 2 kq j m jm

α, j, m1 T k q α, j, m 2 jm 1 k, q, j, m 2 2j + 1 With the sum over q, m and m we can rewrite that last equation as α, j, m1 T k q α, j, m 2 jm 1 k, q, j, m 2 2j + 1 αj T k α j q,m,m αjm T k q α j m kq j m jm All of the remaining dependence on m 2, q, and m 1 is in the Clebsch Gordon coefficient j, m 1 k, q, j, m 2. The so called reduced matrix element α, k T k α, j does not depend on m 1, m 2 or q. Perhaps the significance is more obvious if we write it as α, j, m1 T k q α, j, m 2 jm 1 k, q, j, m 2 2j + 1 c j,j (α, α ) c j (α) is a number that depends only on j, j and α. α represents aspects of the state that do not depend on orientation, like the radial dependence of a wave function. So for an initial and final state with j, j, α, α and a spherical tensor operator T k q, once we have computed α, j, m 1 Tq k α, j, m 2 for a particular m1, m 2 and q, we can with the help of the Clebsch Gordon coefficient on the right, determine the reduced matrix element c j,j α,α. Then we compute any of the others by multiplying by the appropriate Clebsch Gordon coefficient. Tensor operator recursion relationship α, j, m [J ±, T k q ] α, j, m h k(k + 1) q(q ± 1) α, j, m Tq±1 k α, j, m (j (j + 1) m (m 1) α, j, m 1 Tq k α, j, m j(j + 1) m(m ± 1) α, j, m Tk k α, j, m ± 1 h k(k + 1) q(q ± 1) α, j, m Tq±1 k α, j, m We see that this is the same recursion relationship as for the Clebsch Gordon coefficients. Therefore it is reasonable to conclude that the matrix elements are all proportional to Clebsch Gordon coefficients which is precisely what the Wigner Eckart 10

Theorem says, namely α, j, m Tq k α, j, m α j T k α, j jmkq j, m, j, k 2j + 1 Operator for E2 transitions An example of a tensor operator. The interaction of an electromagnetic field with a charged particle will correspond to a term in the Hamiltonian H int e 2mc p A If the fields are in the form of a plane wave then A ɛa 0 e k r ɛ is the polarization vector. If k 2πλ is small or λ is large compared to the extent of the wave function (an atom), the we can expand in powers of k and A ɛa 0 (1 + ik r 1 2 (k r)2 +... and H int ea 0 (ɛ p + i(ɛ p)(k r) +...) 2mc The second term is the E2 transition operator and it will have the form Implications E2 ij p i r j Matrix elements of a scalar operator T 0 0 are zero unless m m and j j α, j m α S α, j, m δ jj δ j S αj mm 2j + 1 For a vector operator m ±1, 0 and j ±1, 0 and there is no 0 0 transition. In fact that transition is forbidden for any rank except 0. Must be a higher order, two photon transition. 11

Example of Wigner Eckart symmetry Consider the dipole operator r. The matrix element f r i Rf(r)Y l f,m f (ry 1,m )R i (r)y li,m i d r Rf(r)rR i (r)r 2 dr Yl f,m f (Y 1,m )Y li,m i dω Rf(r)rR i (r)r 2 drl i, m i, 1, m l f, m f l i, 0, 1, 0 l f, 0 (2l i + 1)(2l + 1) 4π(2l f + 1) l i, m i, 1, m l f, m f l i, 0, 1, 0 l f, 0 (2l i + 1)(2l + 1) R 4π(2l f + 1) f(r)rr i (r)r 2 dr r 1 m l i, m i, 1, m l f, m f f, l f T 1 i, l i 2lf + 1) where r ± ry 1,±1 (θ, φ), r ry 1,0 (θ, φ) Projection theorem The theorem shows that an expectation value taken between states with the same j, that a rank one tensor can be written as λj where λ is some constant that is independent of the z-component of angular momentum of the initial and final states. α, jm V 1 q α, jm α, jm J V α, jm h 2 j(j + 1) First note that J V J 0 V 0 J + V J V + Then jm J q jm. α, j, m J V α, j, m m h α, jm V 0 α, j, m h j(j + 1) m(m 1) α, jm 1 V α, jm 2 j(j + 1) m(m + 1) α, jm + 1 V + α, jm 2 c jm α, j V αj where the Wigner-Eckart theorem is used in the last step. But J V is a scalar operator so its expectation value can have no m dependence. So c jm c j. We might have done the same exercise with V J and then we would get α, jm J 2 α, jm c j α, j J α, j 12

Meanwhile we could have written from the WE theorem or which implies α jm V q α, jm α, jm J q α, jm α, j V α, j α, j J αj α jm V q α, jm α, jm J q α, jm α, jm J V α, jm α, jm J J αjm α jm V q α, jm α, jm J V α, jm h 2 j(j + 1) α, jm J q α, jm J V is a scalar so its expectation value is independent of m. Therefore, the operator V λj where λ is independent of m and m. 1