The Clebsch-Gordan Theorem: Proving the Symmetric Spinor Tensors as IRR of SU(2)

[sshaep]

Clebsch-Gordan Theorem??

symmetric spinor tensors are IRR of SU(2), i.e., $$T_{\undergroup{\alpha_1\cdots\alpha_r}}$$

The Clebsch-Gordan theorem says,

$${\{j_1\}}\otimes{\{j_2\}}={\{j_1+j_2\}}\oplus{\{j_1+j_2-1\}}\oplus\cdots\oplus{\{|j_1-j_2|\}}$$.

Can I prove this theorem by symmetrizing the tensor product,

$$T_{\alpha_1\cdots\alpha_{2j_1}}\otimes T_{\beta_1\cdots\beta_{2j_2}}$$=(express sum of fully symmetric tensors) ??

 

[lbrits]

My suspicion is yes, but it will be hard. Smells like something that you'd do with Young tableaux.

Edit: I think the CG theorem involves both symmetric and antisymmetric tensors. Do you know how to prove the theorem in the usual way (highest weight procedure)?

 

[StatusX]

The completely symmetric part of the tensor product will give you the higest spin representation. After that, I would guess you contract indices using the invariant symbol:

$$\epsilon^{\alpha \beta} = \left( \begin{array}{cc} 0 &1 \\ -1 & 0 \end{array} \right)$$

In other words, to couple the rank n representation T with the rank m represenation S, define:

$$A_{\alpha_1 ... \alpha_{n+m-2}} = \epsilon^{\alpha \beta} T_{\alpha \alpha_1 ... \alpha_{n-1}} S_{\beta \alpha_n ... \alpha_{n+m-2}$$

$$B_{\alpha_1 ... \alpha_{n+m-4}} = \epsilon^{\alpha \beta} \epsilon^{\gamma \delta }T_{\alpha \gamma \alpha_1 ... \alpha_{n-2}} S_{\beta \delta \alpha_{n-1} ... \alpha_{n+m-4}$$

and so on, where you also symmetrize over the $\alpha_i$.