# $(A,B)$-Representation of Lorentz Group: Coefficient functions of fields

I have a question regarding the construction of general causal fields in Weinberg's book on quantum field theory.

In his conventions a field that transforms according to the irreducible $$(A,B)$$ representation of the Lorentz group is given by (eq. 5.7.1) $$$$\psi_{ab}=(2\pi)^{-3/2}\sum_{\sigma}\int d^3p\left[\kappa a(\boldsymbol{p},\sigma)e^{ip\cdot x}u_{ab}(\boldsymbol{p},\sigma)+\lambda a^{c\dagger}(\boldsymbol{p},\sigma)e^{-ip\cdot x}v_{ab}(\boldsymbol{p},\sigma)\right]\, .\tag{5.7.1}$$$$ Here, $$a$$ and $$a^{\dagger}$$ are the usual creation and annihilation operators, $$u_{ab}$$ and $$v_{ab}$$ are coefficients carrying an irreducible representation of the Lorentz group, and $$\kappa$$ and $$\lambda$$ are coefficients.

The zero-momentum coefficients $$u_{ab}(0,\sigma)$$ have to fulfill the conditions $$$$\sum_{\bar{\sigma}}u_{\bar{a}\bar{b}}(0,\bar{\sigma})\boldsymbol{J}^{(j)}_{\bar{\sigma}\sigma}=\sum_{ab}\mathcal{J}_{\bar{a}\bar{b},ab}u_{ab}(0,\sigma)\tag{5.7.1a}$$$$ $$$$-\sum_{\bar{\sigma}}v_{\bar{a}\bar{b}}(0,\bar{\sigma})\boldsymbol{J}^{(j)*}_{\bar{\sigma}\sigma}=\sum_{ab}\mathcal{J}_{\bar{a}\bar{b},ab}v_{ab}(0,\sigma),\tag{5.7.1b}$$$$ where $$\boldsymbol{J}^{(j)}_{\bar{\sigma}\sigma}$$ are the angular momentum matrices in the$$j$$- representations of the rotation group, and $$\mathcal{J}_{\bar{a}\bar{b},ab}v_{ab}(0,\sigma)$$ are the angular momentum matrices in the $$(A,B)$$ representation of the Lorentz-group.

Weinberg shows that $$u_{ab}(0,\sigma)$$ is given by $$$$u_{ab}(0,\sigma)=(2m)^{-1/2}C_{AB}(j\sigma;ab)\, ,\tag{5.7.4}$$$$ where $$C_{AB}(j\sigma;ab)$$ is the Clebsch-Gordan coefficient and the normalization was chosen for convenience. However, when I try to calculate the coefficient $$u_{ab}$$ in the $$(1/2,1/2)$$ representation and want to relate them to the $$u^{\mu}$$ obtained when working directly in the vector representation of the Lorentz group I cannot reproduce them. , where $$$$u^{\mu}(0,\sigma=0)=(2m)^{-1/2}\begin{pmatrix}0\\0\\0\\1\end{pmatrix}\qquad u^{\mu}(0,\sigma=1)=-\frac{1}{\sqrt{2}}(2m)^{-1/2}\begin{pmatrix}0\\1\\+i\\0\end{pmatrix}$$$$ $$$$u^{\mu}(0,\sigma=-1)=\frac{1}{\sqrt{2}}(2m)^{-1/2}\begin{pmatrix}0\\1\\-i\\0\end{pmatrix}\, .$$$$ What is the procedure of translating from $$(A,B)$$ to a mixture of Lorentz indices and spinor indices in more general cases, such as in the Rarita-Schwinger field?

The relations between the four dimensional vector indices and the $$(a,b)$$ indices can be found in this way. First let me define the four dimensional Pauli matrices $$\sigma^\mu_{ab}$$. They have an $$a$$ type index, a $$b$$ type index and a $$\mu$$ index (so they interpolate between the $$(1/2,1/2)$$ and the vector). \begin{align} \sigma^0 &= \left(\begin{array}{cc} -1 & 0\\ 0 & -1\end{array}\right)\,,& \sigma^1 &= \left(\begin{array}{cc} 0 & 1\\ 1 & 0\end{array}\right)\,,\\ \sigma^2 &= \left(\begin{array}{cc} 0 & -i\\ i & 0\end{array}\right)\,,& \sigma^3 &= \left(\begin{array}{cc} 1 & 0\\ 0 & -1\end{array}\right)\,. \end{align} Let me define also the conjugate matrices as $$\bar{\sigma}^{\mu ba} = ( \sigma^0, -\sigma^{1,2,3})$$. A Lorentz vector can be converted into bispinor notation $$\mathrm{v}_{ab} = \sigma^{\mu}_{ab} v_\mu\,,\quad v^\mu = -\mbox{\frac{1}{2}}\,\mathrm{v}_{ab} \bar{\sigma}^{\mu ba}\,,\quad\det(\mathrm{v}_{ab}) = -v^2\,.$$ Irreducible representation of the form $$(A,B)$$ are tensor with $$2A$$ $$a$$ type indices and $$2B$$ $$b$$ type indices and the $$a$$ indices are symmetrized among themselves, as well as the $$b$$ indices. So something like $$\mathrm{v}_{(a_1\ldots a_A)(b_1\ldots b_B)}\,.$$ Since $$a$$ and $$b$$ take values $$1,2$$ antisymmetrization is equivalent to contraction with $$\epsilon_{a_1a_2}$$ or $$\epsilon_{b_1b_2}$$ thus we don't have irreducible representations with indices antisymmetrized.
If $$A=B$$, in order to obtain a Lorentz tensor from it just contract with Pauli matrices $$v^{\mu_1\ldots \mu_{A}} = \left(-\frac{1}{2}\right)^A \bar{\sigma}^{\mu_1 a_1b_1}\cdots \bar{\sigma}^{\mu_A a_Ab_A}\mathrm{v}_{(a_1\ldots a_A)(b_1\ldots b_A)}\;.$$ If instead $$A\neq B$$ we can contract as many indices as possible with Pauli matrices leaving $$|A-B|$$ $$a$$ or $$b$$ indices uncontracted. For the Rarita-Schwinger $$(A,B) = (1,1/2)\oplus(1/2,1)$$. We have then $$\psi^\mu_{a_2} = \bar{\sigma}^{\mu a_1 b_1} \Psi_{(a_1 a_2) b_1}\,,\qquad \psi^\mu_{b_2} = \bar{\sigma}^{\mu a_1 b_1} \Psi_{a_1(b_1 b_2)}\,.$$ This will lead to expression which are totally symmetric in the $$\mu$$ indices. We know that there are also representations with the indices being antisymmetrized. For those we can define another two objects. $$\sigma^{\mu\nu\phantom{a_1}a_2}_{\phantom{\mu\nu}a_1} = \frac{1}{4}\left(\sigma^{\mu}_{a_1b}\bar{\sigma}^{\nu b a_2} - \sigma^{\nu}_{a_1b}\bar{\sigma}^{\mu b a_2}\right)\;,\qquad \bar{\sigma}^{\mu\nu b_1}_{\phantom{\mu\nu b_1}b_2} = \frac{1}{4}\left(\bar{\sigma}^{\mu b_1a}\sigma^{\nu}_{a b_2} - \bar{\sigma}^{\nu b_1a}\sigma^{\mu}_{a b_2} \right)\;.$$ I am not aware of a general procedure to reduce every $$(A,B)$$ tensor to $$\mu$$ tensors, but I can make you an example for the field strength representation $$(A,B) = (1,0)\oplus(0,1)$$. Calling $$F^{\mu\nu}$$ the self-dual component and $$\tilde{F}^{\mu\nu}$$ the anti-self-dual component we have $$F^{\mu\nu} = \sigma^{\mu\nu a_1 a_2} \mathrm{F}_{(a_1 a_2)}\,,\quad \tilde{F}^{\mu\nu} = \bar{\sigma}^{\mu\nu b_1 b_2} \mathrm{F}_{(b_1 b_2)}\,.$$ The $$a$$ and $$b$$ indices are raised and lowered with the two dimensional $$\epsilon$$ tensor. Let me remind also the definition for self-dual and anti-self-dual: $$\varepsilon^{\mu\nu\rho\lambda}F_{\rho\lambda} = -2 i F^{\mu\nu}\,,\qquad \varepsilon^{\mu\nu\rho\lambda}\tilde{F}_{\rho\lambda} = 2 i \tilde{F}^{\mu\nu}\;.$$ This notation is a bit heavy to take in, but it's very precise. The standard convention is spelled out in one of the appendices of Wess & Bagger's Supersymmetry and Supergravity. In their conventions $$a$$ indices are denoted $$\alpha,\beta,\ldots$$ and $$b$$ indices as $$\dot{\alpha},\dot{\beta}\ldots$$.