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I want to understand why in $\mathcal N=2$ SUSY representation (from Wess & Bagger book on SUSY and SUGRA, the second table on page 14):

$$Q_\alpha^A Q^B_\beta |1\rangle=(1)^4 \oplus 0 \oplus 2,\quad \text{with } A,B=1,2 \text{ and $|1\rangle$ is the Clifford vacuum}.$$

My wrong (miss a spin-1) method is: First symmetrize the two spin indices and antisymmetrize the A and B indices (since the generators Q anticommute) which gives $(1)^{(AB)}=1$, then antisymmetrize the two spin-indices and symmetrize A and B which gives $(0)^{\{AB\}}=0^3$, now I have $$(1\oplus 0^3)\otimes 1= 1^3\oplus 2+0.$$ (I am not sure about the $1^3$ part which I think must be instead $1^4$ but I do not know why!?)

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Not sure where you're missing the 1 from, but

$$0 \otimes 1=1$$

and

$$ 1\otimes 1 = 0 \oplus 1 \oplus 2$$

give the result in the book.

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  • $\begingroup$ How did you find the last one? ($1 \otimes 1 = (2)_{symmetric} \oplus (0)_{antisymmetric} $ ?) $\endgroup$ – Mohamed Vall Dec 21 '16 at 12:06
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    $\begingroup$ Usual SU(2) representation direct product decomposition is $i \otimes j = (i+j) \oplus \dots \oplus |i-j|$. $\endgroup$ – user121664 Dec 21 '16 at 12:37
  • $\begingroup$ As for why what you're doing isn't correct, well, symmetric and antisymmetric in what indices? Spin 1 are 3d vectors, so antisymmetric has 3 components and is spin 1, while symmetric decomposes into trace and traceless, which give 0 and 2. $\endgroup$ – user121664 Dec 21 '16 at 12:40
  • $\begingroup$ I agree. when I did the Dynkin diagram I found $3\otimes 3 =1+3+5$ which exactly $0+1+2$. Thanks again $\endgroup$ – Mohamed Vall Dec 21 '16 at 12:45

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