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I am trying to understand the topic in the title but i find some difficulties.

For example, i understand that $(\frac{1}{2},0)\otimes(\frac{1}{2},0)=(1,0)\oplus(0,0)$ which is a consequence of Clebsch-Gordan decomposition and the scalar representation is given by the antisymmetric product. The representation (1, 0) can be represented by an antisymmetric, self-dual second rank tensor.

However, if i consider the following:$(\frac{1}{2},\frac{1}{2})=(\frac{1}{2},0)\otimes(0,\frac{1}{2})$, i don't understand how to decompose it. I know that represents a four vector field, with a temporal scalar component (spin 0) and vector component( spin 1) but i don't really understand why.

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[quote][...]However, if i consider the following:(1/2,1/2)=(1/2,0)⊗(0,1/2), i don't understand how to decompose it. I know that represents a four vector field, with a temporal scalar component (spin 0) and vector component (spin 1) but i don't really understand why. [/quote]

There is nothing to decompose with respect to the full $\mbox{SL}(2,\mathbb{C})$ because the spinor tensor with one undotted index and one dotted index is irreducible. Actually, you should write it backwards and apply the same rules as you already did for the 1-form representations:

$$\left(\frac{1}{2},0\right)\otimes \left(0,\frac{1}{2}\right) = \left(\frac{1}{2},\frac{1}{2}\right) $$, and this because $\frac{1}{2}+0 = \frac{1}{2}-0 = \frac{1}{2}$ for both terms.

Now here is something that addresses your question. $\mbox{SU}(2)$ is a proper subgroup of $\mbox{SL}(2,\mathbb{C})$. So one may ask if the representation $\left(\frac12,\frac12\right) $, which as I said is irreducible with respect to $\mbox{SL}(2,\mathbb{C})$, is perhaps reducible with respect to $\mbox{SU}(2)$. The answer is positive. The general formula below enter image description here becomes

$$\left(\frac12,\frac12\right)|_{\mbox{SU}(2)} \equiv D^0 \oplus D^1 $$

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