1

Young diagram of shape $(a,b)$ has $a$ boxes in the 1st row, $b$ boxes in the second row.

Objective: decompose the following direct product of irreps, and then determine their dimensions given $SU(3)$

$$(3,1) \otimes (2,1).$$

I've determined this to be

$$(5,2) \oplus (3,1) \oplus (4,3) \oplus (2,2) \oplus (1) \oplus (4) \oplus (3,1).$$

I want to check that my dimensions are right (I'm having trouble with the inequivalence of irreps and their complex conjugates). I have

$$15 \otimes 8 = 42 \oplus 24^* \oplus 15_a \oplus 15_b \oplus 15_c \oplus 6^* \oplus 3$$

where the labels on the 15's denote inequivalent irreps and the *'s represent complex conjugates

Qmechanic
  • 220,844
liu111111119
  • 43

2 Answers2

0

Despite the whiff of doing your homework for you, and your having scrambled the order of the irreps so as to correspond their dimensionalities properly, yes, the answers are:

(2,1) ⊗ (3,1) = (5,2) ⊕ (3,1) ⊕ (4,3) ⊕ (2,2) ⊕ (1) ⊕ (4) ⊕ (3,1),

and, crucially, in their proper order, not your question's scrambled order, their dimensionalities unambiguously resulting from the hook rule and checkable by the arithmetic conservation of states (!) are

8 ⊗ 15 = 42 ⊕15 ⊕ 24$^*$⊕ 6$^*$⊕3 ⊕15 ⊕15 .

So, the only nontrivial equation is the first, resulting from distributing the 3 boxes, a,a,b, of the (2,1) onto (3,1). This *can only result in tableaux with 7,7-3=4, and 7-3-3=1 boxes, as you only chuck out 3-box towers--singlets.

This, then answers your question, about the conjugate reps: you cannot have a 6, so (2) with 2 boxes, or a 24, so (4,1) with 5 boxes. Moreover, note two of your 15s are of the original (3,1) type and the third one, the penultimate in this array, is a (4), instead. Analogously, you can never have a 15$^*$, so (3,2), as this has 5 boxes, an unachievable number in the Kronecker multiplication involved.

One might well wonder how you'd have lingering doubts and of what type they could be.

Cosmas Zachos
  • 67,623
0

An $SU(3)$ Young diagram with $a$ boxes in the 1st row and $b$ boxes in the second row corresponds to $SU(3)$ weights $(n,\bar{n})=(a\!-\!b,b)$.

To derive OP's $SU(3)$ fusion rules one can use a Wick-type theorem, explained in my Phys.SE answer here. The Wick-type calculus yields$^1$ $$\begin{align} {\bf 15}\otimes {\bf 8} ~=~&(2,1)\otimes (1,1)\cr\cr ~\cong~&:(2,1)\otimes (1,1): ~\oplus~:(\underline{2,1)\otimes (1},1): ~\oplus~:(2,\underline{1)\otimes (1,1}):\cr\cr ~\oplus~&:(\overline{2,1)\otimes (1,1}):~\oplus~:(\overline{\underline{2,1)\otimes (1},1}):\cr\cr ~\oplus~&:(2,\overline{1)\otimes (1},1): ~\oplus~:(\overline{2,\overline{1)\otimes (1},1}):\cr\cr ~=~&(3,2)~\oplus~(1,3)~\oplus~(4,0)~\oplus~(2,1)~\oplus~(0,2)~\oplus~(2,1)~\oplus~(1,0)\cr\cr ~=~&{\bf 42}~\oplus~\overline{\bf 24}~\oplus~{\bf 15}^{\prime}~\oplus~{\bf 15}~\oplus~\overline{\bf 6}~\oplus~{\bf 15}~\oplus~{\bf 3}. \end{align} $$

$^1$ The Wick-type theorem excludes a double antisymmetrization term

$$\begin{matrix} :(2,&1)\otimes (1,&1):\cr |&---| &\cr &|---&|\end{matrix} \quad~=~\quad (2,1)~=~{\bf 15}.$$

Qmechanic
  • 220,844