Until recently, I had the impression that any representation $R_1 \otimes R_2$ for spins $J_1$ and $J_2$ is reducible, for example, into $(2 \min{(J_1,J_2)}+1)$ multiplets.

$$ J_1 \otimes J_2 = (J_1 + J_2)\oplus(J_1+J_2 -1)\oplus\cdots\oplus|J_1-J_2| \,. \tag1$$

For example, I would naively think that $$1 \otimes \frac1{2} = \frac3{2}\oplus \frac1{2}, \tag2$$

or that

$$ \color{red}{3 \otimes 3} = 6 \oplus 5 \oplus 4 \oplus 3 \oplus 2 \oplus 1 \oplus 0 \,. \tag3$$

I believe this is known as the Clebsch-Gordan decomposition.

Anyway, I recently learned (by word of mouth) that a tensor product of representations can be decomposed into a tensor sum of irreducible representations using a neat trick called the Littlewood-Richardson Rule (using Young Tableau).

However, using this different method ("trick"), the results obtained no longer coincide. For example, here we see that,

$$ \color{red}{3 \otimes 3} = 6 \oplus \overline3 \,. \tag4$$

What is the deal with the overline on $3$? Is it perhaps a shorthand for $5 \oplus 4 \oplus 3 \oplus 2 \oplus 1 \oplus 0 \,?$

Over here, on the other hand,

$$ \color{red}{3 \otimes \overline3} = 8 \oplus 1 \,. \tag5$$

I figured from the above posts that it has something to do with the symmetry/antisymmetry of certain subspaces of the algebra I am trying to find a represention for. Apart from this, most of the other arguments remain incomprehensible to me. I would rather I read an introductory book on the material.

I searched extensively over several classic textbooks on quantum field theory to find information on this decomposition rule. But I failed to discover anything useful.

So here is my question:

$\color{red}{\text{Why is there a discrepancy between (3) and (4)? }}$ In particular, is there any connection between Clebsch-Gordan decomposition and whatever is done with Young Tableau?


3 Answers 3


It seems that your (4) refers to the decomposition of $su(3)$ irreps whereas your (3) applies to the decomposition of $su(2)$ irreps.

The overline refers to the $su(3)$ irrep conjugate to $3$ ($\bar{3}$would be the antiquark irrep). Whereas all irreps of $su(2)$ are self-conjugate so in $su(2)$ there is no difference between $3$ and $\bar 3$, conjugate irreps in $su(3)$ are different - so $\bar 3$ and $3$ are different Indeed in $su(3)$ the tensor product $3\otimes 3=6+\bar 3$, where $6$ is the symmetric part of the tensor product and $\bar 3$ is the antisymmetric part.

Note that HEP people has the sad habit of referring to irreps by their dimension: whereas $3$ and $\bar 3$ are distinct, they have the same dimension hence the bar. In the more mathematical literature, where irreps are distinguished by Dynkin labels, $3$ is $(1,0)$ and $\bar 3$ is $(0,1)$, while $6$ is actually $(2,0)$. Thus: $3\otimes 3\to 6\oplus \bar 3$ is written $(1,0)\otimes (1,0)\to (2,0)\oplus (0,1)$.

The Littlewood Richardson rule applies for $su(3)$ and $su(2)$ decomposition however, whereas the $su(2)$ irrep $j=3/2$ would be represented by a Young diagram with $3$ boxes, the $su(3)$ irrep $3$ is represented by a Young diagram with a single box, and $\bar 3$ by a Young diagram with two boxes in a single column of two rows.

If you prefer the notation in terms of partitions, then $3$ corresponds to the partition $\{1\}$ but $\bar 3$ to the partition $\{1,1\}$. $su(2)$ irreps correspond to partitions with a single row of length $2j$.

Details of applications of the Littlewood-Richardson rules can be found in very many places (see here and here for a start)


The dimension of the $su(3)$ irrep $(\lambda,\mu)$ (in Dynkin notation) is $$ \hbox{dim}(\lambda,\mu)=\frac{1}{2}(\lambda+1)(\mu+1)(\lambda+\mu+2) $$ so with this $$ \hbox{dim}(1,0)=\frac{1}{2} 2\times 1 \times 3=3 $$ and so forth. The difficulty of course is that there might be more than one pair of $(\lambda,\mu)$ which produce the same dimension: indeed $(\mu,\lambda)$ has the same dimension as $(\lambda,\mu)$. In this case, one of the irrep is assigned a $*$ as $(\lambda,\mu)$ and $(\mu,\lambda)$ are conjugate: the eigenvalues of the Cartan elements in one are the negatives of the eigenvalues in the other. It is possible there are more than two irreps with the same dimension, in which case a double $**$ is used. I do not know of a systematic convention to assign $*$ or $**$ but the standard go-to text for this is the review by Slansky.

The Dynkin labels $(\lambda,\mu)$ are the eigenvalues of the Cartan elements on the so-called highest weight state.

The Lie algebra $u(n)$ is spanned by operators $C_{ij}$ with commutation relations $[C_{ij},C_{km}]=C_{im}\delta_{jk}-C_{kj}\delta_{im}$. This is realized by matrices $$ C_{ij}=\cases{ 1& at row $i$ column $j$\\ 0& otherwise} \qquad \hbox{for} i\ne j. \tag{1} $$ The Cartan elements are diagonal with $$ h_k=\cases{1& at row $k$ and column $k$\\ -1 & at row $k+1$ and column $k+1$\\ 0& otherwise}\, . \tag{2} $$ Note that these differ from the standard diagonal Gell-Mann matrices. With this, $C_{ij}$ with $i<j$ is a raising operator and the highest weight state of a representation is the unique state killed by all raising operators, the eigenvalues of $h_k$ are its weight and the Dynkin labels are just these weights, which can be shown to be non-negative integers. With this convention the Dynkin label for $su(2)$ irreps should be $2j$ (which is necessarily an integer).

This definition also makes it obvious that the $N$-dimensional vector $(1,0,\ldots)$ is a highest weight for the $su(N)$ irrep $(1,0,\ldots)$. The other $N-1$ vectors span the $N$-dimensional space and the matrices resulting from (1) and (2) are used to define the fundamental representation of $su(N)$.

  • $\begingroup$ Thank you for your answer. Please allow me to ask a few followup questions. 1. How do you see that $3$ is $(1,0)$ and $6$ is $(2,0)?$ I can go the other way around, calculating dimension using Young tableau. However, there could be, in principle, several tableau with the same dimension. 2. Moreover, I read that $(1,0, \dots, 0)$ of any $su(N)$ corresponds to the fundamental representation (which I intuitively know as the representation where matrices are represented as themselves). How do you see that? I mean how do you translate ordinary intuition to Dynkin weights? Thank you very much! $\endgroup$ Jul 5, 2017 at 6:08
  • $\begingroup$ @NanashiNoGombe Added some material which I hope answers your question. $\endgroup$ Jul 5, 2017 at 13:00

I think equation (2) refers to spin $j_1=3$ and $j_2=3$ whose respective dimension is $2j+1=7$.
Now $7\times 7=49=1+3+5+7+9+11+13$, this is the sum rule for squares of integers: $7\times 7$ is the sum of $7$ odd numbers starting from $1$ spaced by $2$.

So the left hand side of (4) corresponds to the direct sum of seven spins, $j=0,1,2,3,4,5,6$
and the right hand side to the direct product of the two spins $j_1=3$ and $j_2=3$.


A simple sanity check about your discrepancies. The dimension of a tensor product representation is equal to the product of the dimensions of the factors, whereas the dimension of a direct sum is the sum of the dimensions.

So the $3 \otimes 3$ has dimension $9$. On the other hand, your right hand side would have dimension $21$, which is clearly non-sensical.

  • $\begingroup$ Are you aware that a spin $j$ representation is $(2j+1)$-dimensional? $\endgroup$ Oct 13, 2017 at 8:49

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