Young Tableaux

We discussed Young tableaux in connection with the irreducible represen-tations of the symmetric groups. We will now see that they are useful for dealing with irreducible representations of Lie groups. We will begin by dis-cussing this for SU(3), but the real advantage is that it generalizes to SU(n).

12.1 Raising the indices

The crucial observation is that the 3 representation is an antisymmetric com-bination of two 3s, so we really do not need the second fundamental sentation to construct higher representations. We can write an arbitrary repre-sentation as a tensor product of 3s with appropriate symmetry.

In

fact, as we have seen, Young tableaux correspond to irreducible representations of the permutation group, and the connection with the irreducible representations of SU(3) (and SU(N), as we will see later) is that the irreducible representa-tions of SU(3) transform irreducibly under permutation of the labels of the indices.

Consider a general representation, (

n, m).

It is a tensor (in the old lan-guage) with components

Ai.1···i!'

Jl···Jm (12.1)

separately symmetric in upper and lower indices, and traceless. We can raise all the lower indices with E tensors to get

ail···inklli···kmlm

= €ilk1£1 ••. Ejmkmlm Ai.1···i"!' Jl···Jm

(12.2) Clearly, it is antisymmetric in each pair, ki

++

^.e.i. and symmetric in the ex-change of pairs ki,.e.i

++

^kjlj.

178 DOI: 10.1201/9780429499210-13

12.1. RAISING THE INDICES 179 Now for each such tensor, we can associate a Young tableau:

(12.3)

What we would like to do is to find a rule that associates with the Young tableau the specific symmetry of the tensor (12.2). We can do that by think-ing about the highest weight of the representation, ( n,

m).

Because the low-ering operators preserve the symmetry, if we find the symmetry of the tensor components describing the highest weight, all the states will have that sym-metry. The highest weight is associated with the components in which all the is are 1, and all the k,

.e

pairs are 1,3. All of these can be obtained by antisymmetrizing the k,

.e

pairs from the component in which all the ks are 1, and all the fs are 3. But this one component is symmetric under arbitrary permutations of the is and ks, and separately symmetric under permutations of the .es. Thus we will obtain a tensor with the right symmetry if we start with an arbitrary tensor with n +2m components, and first symmetrize all the is and ks, and separately the .es, and then antisymmetrize in every k,

.e

pair.

In the Young tableau language, this is very easy to state. We first symmetrize in the components in the rows, then antisymmetrize in the components in the columns. The result is symmetric in the is and in the k,

.e

pairs as (12.2) must be. But it also has a property that is the analog of tracelessness. Because we have raised the indices with ^ES, the condition of tracelessness becomes

(12.4) This vanishes for a tensor with the symmetry properties just described be-cause of the symmetrization of the components in the rows.

Thus a Young tableau like the one above is a rule for symmetrizing a tensor to project out a specific irreducible representation. For example, if

aj1]2k1 is a general tensor with three upper indices, but no special symmetry property, the Young diagram

produces the tensor

~ ~

aj1]2k1

+

^aj2]1k1

-ak1]2]1 _ a]2k1]1

which transforms according to the (1,1) (or adjoint) representation.

(12.5)

(12.6)

180 CHAPTER 12. YOUNG TABLEAUX

We can generalize the concept to Young tableau with more rows. The general rule is the same. Put indices in the boxes. Symmetrize in the indices in the rows. Then antisymmetrize in the indices in the columns.

In SU(3), the tensors corresponding to Young tableaux with more than three boxes in any column vanish because no tensor can be completely an-tisymmetric in four or more indices which take on only three values. Any column with three boxes corresponds to a factor of ^€ in the three indices

-€ijk =

(12.7) So tableaux of the form

II ^II~:

^{I (12.8)}

describe the same representation as (12.3).

12.2 Clebsch-Gordan decomposition

We can now give, without proof, an algorithm for the Clebsch-Gordan decom-position of a tensor product. To decompose the tensor product of irreducible representations a and f3 corresponding to tableaux

A

and

B,

you build onto

A

using the boxes of

B

in the following way. Begin by putting as in the top row of

B

and bs in the second row. Take the boxes from the top row of

B

and put them on A, building to the right and/or down, to form legal tableau (that is collections of boxes in which the numbers of boxes in the rows are not increasing as you go down, and the numbers of boxes in the columns are not increasing as you go the right), with no two as in the same column. Then take the second row and add the boxes to each of the resulting tableau to form legal tableaux with one further condition. Reading along the rows from right to left from the top row down to the bottom row, the number of as must be greater than or equal to the number of bs. This avoids double counting. The tableaux produced by this construction correspond to the irreducible representations in a®/3.

Examples:

12.2. CLEBSCH-GORDAN DECOMPOSITION ¹⁸¹

D

^Q9

^{@] = [li]}

^EB

[ij

^(12.9)

3®3=6EB3

(12.10)

3®3=8EB1 It is less trivial to do this one the other way.

(12.11)

The tableaux that are crossed out do not satisfy the constraint that the number of

as

is greater than or equal to the number of

bs.

Needless to say, it is easier to do it the other way, because you have fewer boxes to move around.

Sometimes the tableaux that are produced in the first stage are useless, because there is no possible second state. Here's an example:

(12.12)

Of course we would actually never do it this way. This is the decomposition of 3 Q9 3 =

6

EB 3. This is just the complex conjugate of 3 ^Q9 3 = 6 EB 3 which we have already done, and which was much easier because it involved fewer boxes. That's a useful lesson.

Im Dokument Lie Algebras in Particle Physics (Seite 197-200)