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See A. Zee, QFT in a nutshell, Appendix B, eq. (24) (p. 469 in first edition with a typo $55^*\to50^*$, cf. Zee errata; p. 530 in second edition.)

Where does the $50^*$ in $SU(5)$: $$10\otimes10= 5^*\oplus 45^*\oplus 50^*\tag{24}$$ come from?

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  • $\begingroup$ It is the symmetric traceless tensor. $\endgroup$ – Neuneck Mar 23 '14 at 20:47
  • $\begingroup$ Yes @Neuneck, that was my thought too but it did not answer my original question which turned out to be a typo in the Zee book. $\endgroup$ – Your Majesty Mar 24 '14 at 8:46
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Using the Littlewood-Richardson (LR) rules for Young tableaux, one may show that

$$ \begin{array}{c} [~~]\cr [~~] \end{array} \quad\otimes\quad \begin{array}{c} [a]\cr [b] \end{array} \quad=\quad\begin{array}{c} [~~]\cr [~~] \cr [a]\cr [b]\end{array} \quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~]&\cr [b] \end{array} \quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~]&[b] \end{array} \quad\oplus\quad\underbrace{\begin{array}{rcl} [~~]&[a]&[b]\cr [~~] \end{array}}_{\text{forbidden}} $$

The last extension is forbidden because of the LR rule that When reading the extension in arabic order, one must demand that the number of $a$'s $\geq$ the number of $b$'s $\geq$ the number of $c$'s$\geq\ldots $ at any point while reading, cf. Refs. 1 and 2.

For $su(N)$ irreps, the corresponding dimensions read

$$\frac{(N-1)N}{2}\otimes\frac{(N-1)N}{2} $$ $$=\frac{(N-3)(N-2)(N-1)N}{4!} \oplus\frac{(N-2)(N-1)N(N+1)}{2\cdot 4} \oplus\frac{(N-1)N^2(N+1)}{2\cdot 3\cdot 2} .$$

For $N=5$, this becomes

$${\bf 10}\otimes {\bf 10} = \bar{\bf 5} \oplus \bar{\bf 45} \oplus \bar{\bf 50}.$$

References:

  1. H. Georgi, Lie Algebras in Particle Physics, 1999, Section 13.2.

  2. J.J. Sakurai, Modern Quantum Mechanics, 1994 (1st edition), Section 6.5.

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  • $\begingroup$ Why the conjugate? $\endgroup$ – Mitor Jan 12 '15 at 1:12
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    $\begingroup$ @Mitor: Good question. In general, if two irreps have the same dimensions, additional work is needed to identify the relevant irreps. We leave it to the reader to give the full argument in the case at hand. Hint: Since the two ${\bf 10}$'s correspond to antisymmetric $5\times 5$ complex matrices $A^{ij}$ and $B^{kl}$, one may guess that the $\bar{\bf 5}$ comes from contraction with a Levi-Civita symbol $C_m=A^{ij}B^{kl}\epsilon_{ijklm}$. And so forth. $\endgroup$ – Qmechanic Jan 12 '15 at 13:36
  • $\begingroup$ Note that the second edition (2011) of Sakurai's Modern Quantum Mechanics does not contain the relevant section. In the preface, the following can be found: "Chapter 7 has two new sections that contain a significant expansion of the existing material on identical particles. (The section on Young tableaux has been removed.) " $\endgroup$ – Danu Apr 8 '15 at 22:59
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This notation is typical of the terrible habit of high energy theorists to label irreps by their dimension, and some educated guess is required to figure out what is $50$ and what is $50^*$.

I will label representations of su(5) by their highest weight (or Dynkin labels), i.e. by the 4-dimensional vector of non-negative integers $\Lambda=(\lambda_1,\lambda_2,\lambda_3,\lambda_4)$. In this way I think $\Lambda^*=(\lambda_4,\lambda_3,\lambda_2,\lambda_1)$.

Now, according to table 28 of the authoritative review by Slansky (R. Slansky, Group theory for unified model building, Physics Reports Volume 79, Issue 1, December 1981, Pages 1–128 and available here: http://cds.cern.ch/record/134739/files/198109187.pdf), the irrep 10 is actually $(0100)$ in terms of Dynkin label.

The key point is that the tensor product $(0100)\otimes(0100)$ will certainly contain upon decomposition the irrep with Dynkin label $(0200)$ since the highest weight state of $(0200)$ will be the direct product of the highest weights in each copy of the $(0100)\sim 10$.

The irrep $(0200)$ is not listed in Table 28 but instead its conjugate $(0020)$ is identified as $50$. Thus, in all likelyhood $(0200)\leftrightarrow 50^*$.

Why Slansky (and others) would have chosen to label $(0020)\sim 50$ (rather than $(0200)$) is not known to me, and would seem a matter of convention rather than mathematical merit.

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  • $\begingroup$ For reference: Table 28 is on (pdf) page 198 of the pdf you linked to (195 according to the page numbers). $\endgroup$ – Quantum spaghettification Oct 16 '17 at 17:40

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