Lecture 06

Lecture Lecture 6: Angular Angular momentum: momentum: tensor operators and the Wigner-Eckart theorem (10/6/2005) We first need to finish the Wigner functions – the matrix elements of the rotation operators. Then the main goal for today is the Wigner-Eckart theorem that fixes the matrix elements of tensor operators with respect to the basis of spin eigenstates by “pure thought” based on rotational symmetry.

Additiona Additionall reading if you wish: Sakurai Sakurai ch. 3.10

Commutators of angular momentum with vectors We have checked the commutators ˆ i ,  ˆ ˆk [L ¯ ijk L L j ] = i¯  ih h but we have not yet discussed simpler commutators such as ˆ i , rˆ j ], [L

ˆ i , p [L  ˆ j ].

Why don’t we do it now? In the case of the position, ˆ i , rˆ j ] = [lmi rˆl pˆm , rˆ j ] = lmi rˆl [ pˆm , rˆ j ] = lmi rˆl ( ihδ  [L ¯ δ mj h mj ) =

−

−ih ¯ h

 =  i¯ ¯ ijl rˆl . rˆ  = i h h

lji l

In the case of the momentum, it’s an almost identical calculation: ˆ i , p [L  ˆ j ] = [lmi rˆl pˆm, p  ˆ j ] =  lmi pˆm[ˆrl , p  ˆ j ] =   lmi pˆm (ihδ  ¯ δ  jl ) =  i¯ ¯  jmi pˆm  = i  =  i¯ ¯ ijm pˆm . h  ih h h h In all three cases - the commutators of  L of  L i  with L j  as well as r as  r j  as well as p as  p j (more generally V  generally  V  j j ), we may write the result as ˆ i ,  ˆ [L ¯ ijk  ˆ V  j ] = i¯  ih h V k ,

  =    V  L, x,    P ,   S,...

    – or more generWhat What is the reason reason for this genera generall rule? rule? It’s It’s becaus becausee L   ally J    - “generates” all rotations of the three-dimensional space; the rotation operators are some functions (exponentials) of the angular momentum as discusse discussed d in the previous previous lecture. lecture. The commutato commutators rs of the angular momentum with these vectors are therefore determined by the transformation rules 1

for these vectors under the rotations. And all vectors transform in the same vector-like way, which explains the general commutator above. A similar argument also implies that the angular momentum commutes with all scalars   i.e.  quantities that are invariant under rotations (i.e. those that do not changed when you rotate the axes). The vectors and scalars are  just two examples of a more general class of objects called the  tensors .

Tensors What is a tensor? Well, it is approximately speaking a collection of numbers parameterized by several vector-like indices; in this sense they generalize vectors. What is an important feature of a vector? If you change your axes from (x,y,z ) to (x , y  , z  ), the components of a vector V i  should transform as V i = Rij V  j



 j =1,2,3

where Rij   is the matrix encoding the rotation. Position vectors, momenta, spin, angular momentum itself, and other vectors always transform in this ˆ i ,  ˆ way; the previous commutator [ L V  j ] may be thought of as an infinitesimal version of the transformation rule involving Rij   above. What is the third simplest example of a tensor after a scalar and a vector? It is T ij = U i V  j      an outer product (i.e. tensor product) of two vectors U, V . Sometimes we use the word “outer” to distinguish the product from the “inner product”      where the indices of  U, V   are summed over i.e. connected “inside”; in the outer product, they remain free indices “outside”. How does T ij   transform under the rotations? T ij = U i V  j  =



k=1,2,3

Rik U k



R jl V l  =

l=1,2,3

R

ik

R jl U k V l

k,l

where the summation always goes over the (three) values of the indices. Sometimes we use the “Einstein summation convention” in which the sign may be omitted for all indices that appear twice, and a summation is automatically understood. You can see that U k V l  at the end of the previous equation is nothing else than T kl   again, and therefore you may write the transformation rule for T ij as

 

T ij = Rik R jl T kl 2

where the Einstein summation rule was adopted. Our special form of  T ij ,  ,   namely U i V  j , only depended on six components of vectors U V . But you may now imagine that a general tensor T ij  is a collection of 3 3 = 9 independent components, and the last relation between T ij and T kl   defines the general transformation rule for the “tensor”.

×

In other words, tensors are objects defined by their property that they transform just like outer products of vectors. For T ij , the components may be written in a 3 3 table, namely a matrix. It has nine components but it is natural to divide the components into two parts: 1 1 S ij = (T ij  + T  ji ), Aij = (T ij T  ji ). 2 2 These are called the symmetric part – note that  S ij = S  ji  – and the antisymmetric part – note that Aij = A ji . The symmetric part has six independent components (11,12,13,22,23,33) while the antisymmetric part has the remaining three independent components (12,13,23). The full tensor may be written as T ij = S ij  + Aij

×

−

−

as you can easily check. The three pieces of information contained in  Aij may be mapped into a vector W k  (actually, Aij  is a pseudovector or an axial vector because it behaves differently under parity – mirror reflections of space): 1 W k = ijk Aij 2 We already discussed the vectors and A ij  is not too different; W k  calculated from it will transform much like the other vectors, and the commutator ˆ i , W  ˆ j ] will follow the usual rules. In fact, if  T ij = U i V  j , then simply W    = [L   V    . U  The six numbers encoded in  S ij  are more original. Actually, they can be further separated to two parts (which means that the original tensor T ij is naturally divided to  three  parts):  1 sij = s ij δ ij skk , t = s kk 3 which are called the traceless part of the tensor (note that sii  = 0) and the trace t. The inverse relation is, of course,  1 sij = s ij  + tδ ij . 3 Aij =  ijk W k

⇔

 ×

−

3

Note that sij  now contains five independent components (six minus the single trace that must vanish). The original tensor  T ij  was divided into three parts and they do not mix with each other; we call them “irreducible representations”. What does it mean that they don’t mix? It means that if you split T ij  into its symmetric traceless part, the trace, and the antisymmetric part, each of the three contributions will only depend on the initial value of the 9 same   part of the tensor before you transformed it. In other words, the 9 matrix M ij,kl  = R ik R jl

×

whose indices are really “double indices” ij or kl may be conjugated in such a way that it becomes block diagonal with 5 5, 3 3, 1 1 blocks. Each of these blocks transforms one of the three “irreducible” parts of the tensor to a rotated version of itself.

×

×

×

Tensor notation and decompositions Which mathematical notation do we have in order to express the fact that the two-index objects originally obtained from outer products of two vectors decompose into these three “decoupled” pieces? Well, it’s

3

⊗ 3 =  5 ⊕ 3 ⊕ 1.

Both sides are formally vector spaces. The dimension of the vector space is 3 3 = 5 + 3 + 1. Checking that these dimensions work is generally a necessary but not a sufficient condition for the “boldface” decomposition to be true. The left hand side is the space of tensors T ij   while the right hand side is the direct sum of the spaces of all vectors sij , Aij , and t  (the latter being one-dimensional). In other words, every general tensor T ij in 3 3  may be written as a linear combination of three terms sij , aij , and t (times δ ij /3) that belong to the spaces 5,  3 , 1, respectively. Actually, the equation above is isomorphic to our rule for adding the spins of two spin-one particles:

×

⊗

(spin 1) (add) (spin 1) = (spin 2) (or) (spin 1) (or) (spin 0) We can obtain the previous equation by replacing the symbols (spin j) by the vector spaces  2j + 1  where the numbers label the right dimension; replacing (add) by the tensor product; and replacing (or) by the direct sum. 4

Now you can easily conjecture that all (2 j+1)-dimensional representations may be expressed as a special kind of tensors with j indices – and the cases with j = 0, 1, 2 above are just three examples. And you are right. These tensors with (2 j + 1) components are called spherical tensors . If you want to know how the allowed values of the j-index tensors may be written as various (anti)symmetrized products of copies of the vectors U i , do the following:

•   write the spherical harmonic Y  (θ, φ) as Y  (ˆn) •  realize that indeed, it can be written as an  l-th order polynomial in nˆ •   substitute U   for nˆ  into these (2l + 1) different polynomials •  alternatively, the previous three steps may be written by replacing z  x y cos θ →  → U  , sin θ cos φ →  → U  , sin θ sin φ →  → U  . r r r lm

i

lm

i

z

x

y

•  consider the most general linear combination of such  Y 

lm

 ) for a fixed (U 

l but all values of  m

In mathematical terms, ( ) ( ) =q    (k) = D qqk (R)T q k T q(k) = Y lm =k (U ) transforms as T q 



where the D’s are the Wigner functions – the matrix elements of the rotation operator – for spin k.

Commutators of angular momentum and tensors ˆz and J  ˆ±  on the states Aaron has shown you the derivation of the action of  J   jm .

| 

J z  jm = m¯h  jm ,

| 

| 

 

¯  j( j + 1) J ±  jm = h

| 

− m(m + 1)| j, m ± 1.

We say that the states  jm “transform in a particular way under the angular momentum (i.e. the rotations). What does it mean to say that an operator T q(k) transforms in the same way as the states  jm ? It means to postulate the commutators that otherwise reproduce the structure of the previous formula (with j = k and m = q ):

| 

| 

(k)

(k )

[J z , T q ] = q h ¯ T q ,

(k)

 

[J ±, T q ] = ¯h k(k + 1) 5

(k) q±1

− q (q  + 1)T 

.

You can view the validity of these commutation rules to be a definition of  spherical tensors T q(k). It is always possible to obtain a spherical tensor of  rank k from spherical tensors U q(k ) , V q ( k ) of ranks k 1 and k 2   assuming that 1

2

1

k

2

∈ {|k − k |, |k − k | + 1, . . . k  + k } 1

2

1

2

1

2

much like for the angular momentum simply by taking a sum of   products  of  U, V   over the z -projections of the spin with the same Clebsch-Gordan coefficients as we used for the states: T q(k) =

c q1

q2

k1 k2 k q1 q2 q

U q(k ) V q(k 1

1

2

)

2

It is then guaranteed that T q(k) is a spherical tensor of rank k i.e. it transforms correctly. Using the known transformation properties of spherical tensors U, V , the proof of the statement about T   mimicks the structure of similar proofs about the states, but we won’t reproduce it here.

The Wigner-Eckart theorem We are just approaching one of the punch lines of this lecture. The WignerEckart theorem says that the dependence of matrix elements of arbitrary tensor operators on the magnetic quantum numbers m, m (the third polarization of the spin) is completely determined by the CG coefficients: 





 jk j  mqm

(k )

α , j , m |T  |α,j,m = c q





(k)

α , j√ ||T  ||α, j 2 j + 1

Here, α , α refer to all non-angular-momentum quantum numbers. The doublebarred matrix element is called the reduced matrix element and is defined by this equation. The equation is however non-trivial because it says that a unique reduced matrix element exists for all values of  m , m. The theorem thus implies that the m , m  quantum numbers are purely geometrical. Also, we may easily derive that most of the matrix elements vanish, including those for m =  m + q  or j  > j + k or j  <  j k ,



|− |

simply because the corresponding CG coefficient equals zero.

6

A proof  To prove the theorem, we may decide to perform the following steps:

•  find a recursive relation between CG coefficients with different values of  m 1 , m2 , m

•  show that the matrix elements satisfy the same recursion relation; this allows one to prove that the theorem holds for all values in the set by a generalized mathematical induction

•   we also need a “beginning” step of the mathematical induction.

But this will be trivial because the overall normalization of the reduced matrix elements can be chosen in such a way that the theorem holds for these “initial” values of the quantum numbers automatically.

Start with J ˆ±  j1 j2 jm

|



 ˆ1± + J  ˆ2± )  j1 j2 jm = (J   ˆ1± + J  ˆ2± ) = (J   j1 m1  j2 m2  j1 m1  j2 m2  j1 j2 jm

|   | |  | |    ⇒  j( j + 1) − m(m ± 1)| j(m ± 1) =     j ( j  + 1) − m (m ± 1)| j (m ± 1)| j m  = m1 m2

1

1

1

1

1

1

2

2

m1 m2

 

+  j2 ( j2  + 1)

− m (m ± 1)| j m | j (m ± 1) 2

2

1

1

2

2



Multiply this by  j1 m1  j2 m2  to get

 

| |    =  j( j + 1) − m(m ± 1)c      =  j ( j  + 1) − m (m ± 1)δ   j1 j2 j m1 m2 (m±1)

1

1

1

1

+ = +

   j ( j  + 1) − m (m ± 1)δ      j ( j  + 1) − m (m ± 1)c 2

2

1

1

 j2 ( j2  + 1)

δ 

m1 (m±1) m2 m2

m1 m2

δ 

2

2





 j1 j2 j (m1 ∓1)m2 m

 +





 j1 j2 j m1 (m2 ∓1)m

.

1

m1 m1 m 2(m2 ±1)

1

− m (m ± 1)c 2

2



c jm j m j 1 2 1

2

m

Look at the equality between the first and the last form of the expression above and erase (in your mind) the primes from m1 , m2   on both sides to 7

simplify your life. At any rate, this relation allows you to generate all CG coefficients with the value of  m being m 1 (and by induction, for all values of  m) in terms of the CG coefficients with the value of  m  being m. Now we can perform a related calculation of some matrix elements.

±

 ˆ , T  ]|αjm  = α j   m |[J  = h ¯ k(k + 1) − q (q  ± 1)α j m |T  |αjm =   = h ¯  j ( j + 1) − m (m ∓ 1)α j (m ∓ 1)|T  |αjm  −   − h¯  j( j + 1) − m(m ± 1)α j m |T  |αj(m ± 1)  



±

(k)

q

 







(k) q±1



 



(k )



q

 



(k)

q

Comparing this relation with the relation for the CG coefficients, we see that the coefficient coincide up to a simple renaming j1 k, m1 q ,    j j,m m , j2 j, m2 m and the sign flip (a permutation of the three terms). The matrix elements of  T q(k) are therefore proportional to the CG coefficients which is the content of the Wigner-Eckart theorem.

 →

 →

 →

→

 →

→

Applications of the theorem Scalar operators A scalar operator S   cannot change the angular momentum of a state. By this statement we mean that the matrix elements are only nonzero if the initial and final spin states are identical. It’s because  j 0 j  m0m

c

 

= δ  jj δ mm 



 



⇒ α j m |S |αjm = δ 

δ 

 jj  mm

α√   j ||S ||αj  2 j + 1

Electromagnetic transitions By electromagnetic transitions, we mean absorption or emission of a photon. Most of these transitions occur because of the electric dipole moment that is proportional to the operator r  (the relative position of the electron compared to the nucleus, for example). The probability amplitudes are thus derived from the matrix elements α j  m ˆr αjm .



||



Looking at the Wigner-Eckart theorem and the CG coefficients, this is only nonzero if ∆ j = 1 or 0. Moreover, even the transition from  j  = 0 to j  = 0 with ∆ j = 0 is forbidden.

±

8