Tensor Operators

A tensor operator is a set of operators that transforms under commutation with the generators of some Lie algebra like an irreducible representation of the algebra. In this chapter, we will define and discuss tensor operators for the SU(2) algebra discussed in chapter 3. A tensor operator transforming under the spin-s representation of SU(2) consists of a set of operators, Oj for f.= 1 to 2s+l (or -s to s), such that ·

(4.1) It is true, though we have not proved it, that every irreducible representa-tion is finite dimensional and equivalent to one of the representarepresenta-tions that we found with the highest weight construction. We can always choose all tensor operators for SU(2) to have this form.

4.1 Orbital angular momentum

Here is an example - a particle in a spherically symmetric potential. If the particle has no spin, then Ja is the orbital angular momentum operator,

(4.2) The position vector is related to a tensor operator because it transforms under the adjoint representation

(Ja, rb] = Eacd [rcPd, rb] = -i Eacd rc 8bd

=

^-i^{facb rc}

=

^rc^[J~j]cb

(4.3)

DOI: 10.1201/9780429499210-5

4.2. USING TENSOR OPERATORS

69

where

Jadj

is the adjoint representation, and we know from problem 3.C that this representation is equivalent to the standard spin 1 representation from the highest weight procedure.

4.2 Using tensor operators

Note that the transformation of the position operator in (4.3) does not have quite the right form, because the representation matrices

J;dj

are not the stan-dard form. The first step in using tensor operators is to choose the operator basis so that the conventional spin s representation appears in the commuta-tion relacommuta-tion ( 4.1 ). This is not absolutely necessary, but it makes things easier, as we will see. We will discuss this process in general, and then see how it works for ra.

Suppose that we are given a set of operators,

nx

^for^X ⁼ ¹^{to 2s+ 1}

that transforms according a representation D that is equivalent to the spin-s representation of SU(2):

(4.4) Since by assumption, D is equivalent to the spin-s representation, we can find a matrix

S

such that

or in terms of matrix elements

Then we define a new set of operators

Now 0£ satisfies

0£ = Dy

[S-

]ye

for/!= -s to s

[Ja,O£]

[Ja,

Dy] [S-¹

]ye

= nz[J~]zy

[S-

]ye

= Dz[S-¹]ze' [S]e'z' [J~]z'y [S-¹]ye

= 0£, [ J~]n

(4.5)

(4.6)

(4.7)

(4.8)

70 CHAPTER 4. TENSOR OPERATORS which is what we want. Notice that (4.8) is particularly simple for J3, because in our standard basis in which the indices£ label the J3 value, Jj, (or J~ for any s) is a diagonal matrix

[J~]£'£ = £

8u'

for£,£' =

-s

s.

^(4.9) Thus

[J3,

011 = o;,

^[J~]£'£

= £01.

^(4.10)

In practice, it is usually not necessary to find the matrix

S

explicitly. If we can find any linear combination of the

nx

which has a definite value of

J

3 (that means that it is proportional to its commutator with

J

3), we can take that to be a component of

0

^{8 ,}and then build up all the other

0

⁸components by applying raising and lowering operators.

For the position operator it is easiest to start by finding the operator r0 .

Since [J3, r3] = 0, we know that r3 ^hasJ3 = 0 and therefore that r3 <X ro.

Thus we can take

ro = r3 (4.11)

Then the commutation relations for the spin 1 raising and lowering operators give the rest

(J±,ro]= r±l

=F(n ± ir2}/..J2

^(4.12)

4.3 The Wigner-Eckart theorem

The interesting thing about tensor operators is how the product

01li,

^m,a) transforms.

Ja 01li, ^{m, a)}

[Ja, 0£]

jj, m, a)

+ 01 ^Ja

^jj,^{m, a)} ^(4.13)

0£,

jj,

m, a) [J:]£'

^£

+ 0£jj, m', a) [ Ji]m'm

This is the transformation law for a tensor product of spin

s

and spin j,

s

® j.

Because we are using the standard basis for the states and operators in which J3 is diagonal, this is particularly simple for the generator J3, for which (4.13) becomes

J301lj,m,a)

(£+m)01jj,m,a)

^(4.14) The

J

3 value of the product of a tensor operator with a state is just the sum of the

h

values of the operator and the state.

4.3. THE WIGNER-ECKART THEOREM 71 The remarkable thing about this is that the product of the tensor operator and the ket behaves under the algebra just like the tensor product of two kets. Thus we can decompose it into irreducible representations in exactly the same way, using the highest weight procedure. That is, we note that

li,

^j,^a)^with^J³ ⁼ ^j

+

^sis the highest weight state. We can lower it to construct the rest of the spin j

+

^srepresentation. Then we can find the linear combination of J3 = j

+

s - 1 states that is the highest weight of the spin j

+

^s^- 1 representation, and lower it to get the entire representation, and so on. In this way, we find explicit representations for the states of the irreducible components of the tensor product in terms of linear combinations of the 0£

jj,

a).

You probably know, and have shown explicitly in problem 3.A, that in this decomposition, each representation from j

+

s to

li - sl

appears exactly once. We can write the result of the highest weight analysis as follows:

2:0£ jj,M -f,a) (s,j,f,M -£I J,M) = kJ IJ,M) (4.15) e

Here

I

^J,

^M)

is a normalized state that transforms like the J3 = M compo-nent of the spin J representation and kJ is an unknown constant for each J (but does not depend on M). The coefficients

(s,

£,

M -

£ I

^J,

^M)

^are

determined by the highest weight construction, and can be evaluated from the tensor product of kets, where all the normalizations are known and the constants kJ are equal to 1: difference is the factor of kJ in (4.15).

We can invert ( 4.15) and express the original product states as linear com-binations of the states with definite total spin J.

O£lj,m,a)=

j+s 2: (J,l+mls,j,l,m)kJIJ,l+m)

J=li-sl (4.17)

1This is probably obvious, but as we will emphasize below, the operators are different because we do not have a scalar product for them.

CHAPTER4. TENSOROPERATORS

The coefficients ( J, M

I

^s,^j,^{f, M} ^-

^£)

are thus entirely determined by the algebra, up to some choices of the phases of the states. Once we have a convention for fixing these phases, we can make tables of these coefficients once and for all, and be done with it. The notation (J, f + mls, j, f, m) just means the coefficient of IJ, f + m) in the product Is,£) lj, m). These are called Clebsch-Gordan coefficients.

The Clebsch-Gordan coefficients are all group theory. The physics comes in when we reexpress the IJ, f + m) in terms of the Hilbert space basis states

I

^{J, f}

+ m, (3)

-kJIJ,f+m) =

L

kapiJ,f+m,(3)

,B ^(4.18)

We have absorbed the unknown coefficients kJ into the equally unknown co-efficients ka,B· These depend on a, j,

as

^and^s,because the original products do, and on (3 and J, of course. But they do not depend at all on form. We only need to know the coefficients for one value of£+ m. The ka,B are called reduced matrix elements and denoted

ka,B

=

(J, f31

as

^{lj, a)} ^(4.19)

Putting all this together, we get the Wigner-Eckart theorem for matrix ele-ments of tensor operators:

(J,m',f31

a1 ^lj,m,a)

_(4.20)

=

8m',Hm (J,f

+

mls,j,f,m) · (J,(31

as ^lj,a)

If we know any non-zero matrix element of a tensor operator between states of some given J, (3 and j, a, we can compute all the others using the algebra.

This sounds pretty amazing, but all that is really going on is that we can use the raising and lowering operators to go up and down within representations using pure group theory. Thus by clever use of the raising and lowering oper-ators, we can compute any matrix element from another. The Wigner-Eckart theorem just expresses this formally.

4.4. EXAMPLE 73 First, since

r·o

= r3,

(1/2,1/2,aJro J1/2,1/2,P) =A

^(4.23)

Then we know from (4.12) that

(4.24) Thus

(1/2, 1/2, aJ r1 11/2, -1/2, P)

(1/2,1/2,aJ J2(-r+1 +r_l) J1/2,-1/2,P}

1 (4.25)

J2(1/2, 1/2, al

1 ^r

+1 11/2,-1/2, m

Now we could plug this into the formula, and you could find the Clebsch-Gordan coefficients in a table. But I'll be honest with you. I can never re-member what the definitions in the formula are long enough to use it. In-stead, I try to understand what the formula means, and I suggest that you do the same. We could also just use what we have already done, decomposing

1/2

1

into irreducible representations. For example, we know from the highest weight construction that

J3/2, 3/2) =

+111/2, 1/2, P)

^(4.26) is a 3/2,3/2 state because it is the highest weight state that we can get as a product of an

re

operator acting on an

11/2, m)

state. Then we can get the corresponding

13/2, 1/2)

state in the same representation by acting with the lowering operator

J-13/2, 1/2)

If J-13/2, 3/2)

If ^ro ^11/2,1/2, ^m +If ^r +1 11/2,-1/2, m

^(4.27)

But we know that this 3/2 state has zero matrix element with any spin-112 state, and thus

0 (1/2, 1/2, aJ3/2, 1/2)

= {f(l/2,1/2,aJroJ1/2,1/2,P) +Jf(1/2, 1/2, al ^r +111/2, -1/2,

fJ)

(4.28)

74 Although we did not need it here, we can also conclude that

11/2, 1/2) =If

^ro

11/2,1/2, a)- If

+111/2, -1/2, a)

(4.29)

(4.30)

(4.31) is a 112,112 state. This statement is actually a little subtle, and shows the power of the algebra. When we did this analysis for the tensor product of j = 1 and j=112 states, we used the fact that the

11/2, 1/2)

must be orthogonal to the

13/2, 1/2)

states to find the form of the

11/2, 1/2)

state. We cannot do this here, because we do not know from the symmetry alone how to determine the norms of the states

(4.32) However, we know from the analysis with the states and the fact that the transformation of these objects is analogous that

J+

11/2, 1/2) = ⁰

^(4.33)

Thus it is a 112,112 state because it is the highest weight state in the represen-tation. We will return to this issue later.

There are several ways of approaching such questions. Here is another way. Consider the matrix elements

(1/2,m,al

l1/2,m',{3)

^(4.34) The Wigner-Eckart theorem implies that these matrix elements are all propor-tional to a single parameter, the ka/3· Furthermore, this result is a consequence of the algebra alone. Any operator that has the same commutation relations with Ja will have matrix elements proportional to ra. But Ja itself has the same commutation relations. Thus the matrix elements of ^r^aare proportional to those of Ja. This is only helpful if the matrix elements of Ja are not zero (if they are all zero, the Wigner-Eckart theorem is trivially satisfied). In this case, they are not (at least if a = {3)