# Young Diagram SO(3) Doublet

1. Jun 5, 2012

### alexvas

Hello all!

Clebsch-Gordon Coefficients tell us that in an SO(3) representation 3 x 2 = 4 + 2 (x/+ is the tensor product/sum). In practicing my Young Diagrams I tried to re-create this calcuation, but can't seem to figure out how to draw a doublet. I would appreciate some advice! (The triplet is, of course, easy - but the quartet is also a challenge)

I am taking an introduction to Particle Physics class so am not an expert - answers that are clear and simple are much appreciated!

Thanks!

2. Jun 5, 2012

### fzero

Presumably you are using Young tableaux where the boxes represent vector indices in the $\mathbf{3}$. In that case, you cannot represent a spinor representation directly. For SO(3), it's clear that you can use the SU(2) Young tableaux to compute products of SO(3) spinor representations. There the basic block is the fundamental $\mathbf{2}$ of SU(2), which is of course the spinor of SO(3).

For SO(N) in general, I seem to recall that there is a similar way of using a spinor, rather than vector, representation as the basic block, but there are more complicated rules for counting that I don't think I've ever completely worked out.

Last edited: Jun 6, 2012
3. Jun 5, 2012

### alexvas

Hmmmm, I have never used the Young Tableau to compute products of SU(2) spinors - I've only used it for SO(3) and SO(2). I've been told that the algebras of SO(3) and SU(2) are the same - but I presume that Young diagrams would have to be used differently? A singlet in SO(3) is three vertically-stacked boxes, in SO(2) it's two. What would a singlet be in SU(2)?

4. Jun 6, 2012

### fzero

The SU(2) Young tableaux are a bit boring. If you stack two boxes, you get the singlet, so the nontrivial irreps correspond to n boxes in a row and have dimension n+1. The dimension counting is similar, but you don't have to remove the traces like you do for SO.

5. Jun 6, 2012

### alexvas

"You don't have to remove the tracers"

Do you mean the "hooks", Ie the number by which you divide the product of the numbers in the boxes? In this case, wouldn't two boxes on top of each other (in SU(2)) be a doublet rather than a singlet (a two in the top box and a one in the bottom box, multiplied)? Moreover, how would you get a triplet, since every product will be even?

6. Jun 6, 2012

### fzero

By remove the traces, I mean for SO(N), the irreducible symmetric representations are traceless. So for SO(3), $\mathbf{3\otimes 3 = 1\oplus 3_a \oplus 5_s}$, and so on.

When you stack boxes on top of one another, you antisymmetrize the index, so in SU(2), you get the singlet. To compute the dimension, you have rules for generating the numerator and denominator by labeling the boxes. The numerator comes from

[2]
[1]

but the denominator has the same labels

[2]
[1]

so the dimension is 1.

The symmetric product of $\mathbf{2}$s is the triplet, corresponding to two boxes in the same row. Here we get

[2][3]

and

[2][1]

so 6/2 = 3.

In SO(3), the symmetric product of two $\mathbf{3}$s gives

[3][4]

and

[2][1]

so 12/2 = 6, but this is actually a reducible representation, since we haven't removed the trace part. When we account for this, we get $\mathbf{1\oplus 5}$. You might have learned different rules for SO(N), but you should get the same answer. Similarly, you have to find that the completely symmetric product of three $\mathbf{3}$s is $\mathbf{1\oplus 3\oplus 7}$.

7. Jun 7, 2012

### samalkhaiat

When dealing with the fundamental and adjoint representations, it is more useful to use tensor methods. I will below explain the methods for $SU(2)$ and $SO(3)$.
$SU(2)$: Here the fundamental representation (i.e., the smallest non-trivial space on which the elements of $SU(2)$ act) is spanned by 2-component vector $X_{a}, a = 1,2$. We often write $[n]$ for the n-dimensional (irreducible) representation space spanned by objects having n independent components. Now, let us consider the tensor product $X_{a}Y_{b}$ which represents $[2] \otimes [2]$. We write
$$X_{a}Y_{b} = \frac{1}{2}(X_{a}Y_{b} + X_{b}Y_{a}) + \frac{1}{2}(X_{a}Y_{b} - X_{b}Y_{a}). \ \ (1)$$
Since the index a and b take values in the set {1,2}, the first object on the right-hand-side has only 3 independent elements; $\{X_{1}Y_{1}, (1/2)(X_{1}Y_{2}+X_{2}Y_{1}),X_{2}Y_{2}\}$. We think of it either as 3-component vector $V_{i}, i = 1,2,3$ or as $2 \times 2$ symmetric matrix $G_{ab}=G_{ba}$. Both span the same representation space $[3]$ of $SU(2)$. The 2nd object in Eq(1) has only one independent element; $\{(1/2)(X_{1}Y_{2}-X_{2}Y_{1})\} = (1/2)\epsilon^{ab}X_{a}Y_{b}$. Thus, it belongs to the trivial (invariant) representation space $[1]$; notice that it is proportional to the invariant $SU(2)$ “metric”. So, we can rewrite Eq(1) as
$$X_{a}Y_{b} = G_{ab} + (1/2) \epsilon_{ab} \epsilon^{cd} X_{c}Y_{d}.$$
This shows that
$$[2]\otimes [2] = [3] \oplus [1].$$
Next, let us decompose the tensor product $[3] \otimes [2]$. For this we need to break the tensor $G_{ab}X_{c}$ into symmetric and anti-symmetric parts. So, we write
$$G_{ab}X_{c} = \frac{1}{3}(G_{ab}X_{c} + G_{ca}X_{b} + G_{bc}X_{a}) + \frac{1}{3}(G_{ab}X_{c}-G_{bc}X_{a}) + \frac{1}{3}(G_{ab}X_{c}-G_{ca}X_{b}). \ \ (2)$$
So what do we have here? The 1st object on the right-hand-side is a rank-3 totally symmetric tensor $T_{(abc)}$ in 2-dimension. Such tensor has only 4 independent components; $\{T_{(111)},T_{(112)},T_{(122)},T_{(222)}\}$. We can think of this as a vector spanning 4-dimensional vector space $[4]$.
The 2nd + 3rd objects on the RHS of Eq(2) can be written as
$$X_{abc}\equiv \epsilon_{ac}Z_{b} + \epsilon_{cb}Z_{a},$$
where
$$Z_{a} = G_{ab}\epsilon^{bc}X_{c}.$$
Now, using the identity
$$\epsilon_{ab}Z_{c} + \epsilon_{ca}Z_{b} + \epsilon_{bc}Z_{a} = 0,$$
we finally find
$$G_{ab}X_{c} = T_{(abc)} + \epsilon_{ab}Z_{c},$$
which proves
$$[3] \otimes [2] = [4] \oplus [2].$$
$[SO(3)]$: I will give you the result of decomposing the tensor product $[3] \otimes [3]$ and leave you to fill in the details to conclude that
$$[3] \otimes [3] = [5] \oplus [3] \oplus [1],$$
follows from
$$U^{i}V^{j} = \frac{1}{2}(g^{ij} - \frac{1}{3}\delta^{ij}g) + \frac{1}{2}A^{ij} + \frac{1}{6}\delta^{ij}g,$$
where $g^{ij} = g^{ji}, A^{ij} = -A^{ji}$ and $g = \delta_{ij}g^{ij}$.
Question for you: Why didn’t we subtract a trace from the symmetric $SU(2)$ tensor $G_{ab}$?

Sam

Last edited: Jun 7, 2012
8. Jun 7, 2012

### fzero

The SO(N) metric is $\delta_{ij}$, so there's no difference between co- and contra-variant vectors. Arbitrary tensor representations are therefore reducible by contractions. For SU(2), we can use $\epsilon_{ab}$ as a metric, while SU(N>2) has complex representations and a Hermitian metric. In those cases, the tensor representations are irreducible. For SU(2), the metric trace of a symmetric tensor actually vanishes.

9. Jun 8, 2012

### samalkhaiat

Ok, except that the question meant for the original poster.

Sam