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) \begin{equation} \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} \end{equation} 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 \begin{equation} \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} \end{equation} \begin{equation} -\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} \end{equation} 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 \begin{equation} u_{ab}(0,\sigma)=(2m)^{-1/2}C_{AB}(j\sigma;ab)\, ,\tag{5.7.4} \end{equation} 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 \begin{equation} 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} \end{equation} \begin{equation} u^{\mu}(0,\sigma=-1)=\frac{1}{\sqrt{2}}(2m)^{-1/2}\begin{pmatrix}0\\1\\-i\\0\end{pmatrix}\, . \end{equation} 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?


1 Answer 1


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 $\dot{\alpha},\dot{\beta}\ldots$.


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