# Clebsch-Gordan Coefficients in Causal Fields

I am trying to understand Section 5.7 in Weinberg's field theory book. The task is to construct causal fields transforming according to the general $$(A,B)$$ representations of the proper orthochronous Lorentz group:

$$\psi_{ab}(x) = (2 \pi)^{-3/2}\sum_\sigma \int \mathrm{d}^3p \left[\kappa \, a(\mathbf{p},\sigma)\, \mathrm{e}^{\mathrm{i} p \cdot x} u_{ab}(\mathbf{p},\sigma) + \lambda a^{c \dagger}(\mathbf{p},\sigma)\mathrm{e}^{-\mathrm{i} p \cdot x} v_{ab}(\mathbf{p},\sigma)\right].$$

In particular I want to know how to calculate the $$u_{ab}(0,\sigma)$$'s and how they relate to polarization. Weinberg says that these are simply the Clebsch-Gordan coefficients:

$$u_{ab}(0,\sigma) = \frac{1}{\sqrt{2m}} C_{AB}(j \sigma ; a b)$$

For instance, in the case of a spin-1 massive boson we know that fields get dressed with polarization vectors: $$\epsilon^\mu(0,0) \sim (0,0,0,1)$$ $$\epsilon^\mu(0,-1) \sim (0,1,-\mathrm{i},0)$$ $$\epsilon^\mu(0,1) \sim (0,1,\mathrm{i},0)$$

I guess we can write these as bispinors via contraction with pauli matrices: $$u_{ab}(\sigma = \{0, \, 1, \, -1\}) = \left\{\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right),\left( \begin{array}{cc} 0 & 2 \\ 0 & 0 \\ \end{array} \right),\left( \begin{array}{cc} 0 & 0 \\ 2 & 0 \\ \end{array} \right)\right\}$$

I have three related questions:

1. Those bispinors are supposedly proportional to $$C_{A B}(j \sigma ; ab)$$ for some choice of $$A \, B, \, j.$$ What are the values of $$A, \, B, \, j$$?
2. How do we start with the notion of a "spin-1 massive boson" and deduce the aforementioned values?
3. What about a spin-2 massive boson? What are the $$C_{A B}(j \sigma ; ab)$$ and how do we extract the 5 polarization tensors from them?

There is a similar question ($(A,B)$-Representation of Lorentz Group: Coefficient functions of fields) but the scope of this one is purposefully far more limited since the root of my difficulty with the subject is that the presentation is very general and lacking in explicit calculation.

A massive spin $$1$$ boson is best represented by a vector field $$V^{\mu}$$, which transforms with respect to the $$(\frac{1}{2},\frac{1}{2})$$ representation of the Lorentz group, with the Lorentz constraint $$\partial_{\mu}V^{\mu}=0$$. The Clebsch-Gordan coefficients that are relevant here are $$C_{1/2,1/2}(1\sigma,ab)$$

The matrices you arrive at are

$$u_{ab}(\sigma=\{-1,0,1\})=\left\{\begin{pmatrix}1&0\\0&0\end{pmatrix},\begin{pmatrix}0&1/\sqrt2\\1/\sqrt2&0\end{pmatrix},\begin{pmatrix}0&0\\0&1\end{pmatrix}\right\}$$

There isn't an immediate relation between these and what you wrote down.

A massive spin $$2$$ boson is represented with a symmetric rank two tensor $$g_{\mu\nu}$$, which transforms with respect to the $$(1,1)\oplus(0,0)$$ representation of the Lorentz group, with analogous constraints which eliminate the two spin $$0$$ components and a spin $$1$$ component. The relevant Clebsch-Gordan coefficients are $$C_{1,1}(2\sigma,ab)$$

$$u_{ab}(\sigma=\{-2,-1,0,1,2\})=\left\{\begin{pmatrix}1&0&0\\0&0&0\\0&0&0 \end{pmatrix},\begin{pmatrix}0&1/\sqrt2&0\\1/\sqrt2&0&0\\0&0&0 \end{pmatrix},\begin{pmatrix}0&0&1/\sqrt6\\0&\sqrt{2/3}&0\\1/\sqrt6&0&0 \end{pmatrix},\begin{pmatrix}0&0&0\\0&0&1/\sqrt2\\0&1/\sqrt2&0 \end{pmatrix},\begin{pmatrix}0&0&0\\0&0&0\\0&0&1 \end{pmatrix}\right\}$$

Not a complete answer, but working out the spin 2 massive bosons from spin 1 massive bosons is a fun little exercise. I find it confusing to use tables of Clebsch Gordan coefficients, and find it easier to just redo them from scratch.

The spin 1 rep has three vectors, $$|1\rangle, |0\rangle, |-1\rangle$$, which in the rest frame $$k^\mu = (m,0,0,0)$$ correspond to to the polarization vectors $$\epsilon^\mu_1 = \tfrac{1}{\sqrt{2}}(0, 1, i, 0)$$, $$\epsilon^\mu_0 = (0, 0, 0, 1)$$, $$\epsilon^\mu_{-1} = \tfrac{1}{\sqrt{2}}(0, -1, i, 0)$$. Note that they all satisfy the constraint $$k^\mu \epsilon_\mu = 0$$.

Now, under the spin $$1$$ rep, we have \begin{align} J_\pm^{(1)} &= J_x^{(1)} \pm i J_y^{(1)} \\ J_-^{(1)} | 1 \rangle &= \sqrt{2} |0 \rangle \\ J_-^{(1)} | 0 \rangle &= \sqrt{2} |-1 \rangle \\ J_z^{(1)} | m \rangle &= m |m\rangle \end{align} If we tensor two spin $$1$$ reps together, we get $$1 \otimes 1 = 2 \oplus 1 \oplus 0$$. We only want the $$2$$ rep. We can deduce the Clebsch Gordan decomposition for ourselves in the following manner. We begin by making a state with $$m = 2$$ and then using the lowering operator repeatedly on the state. So, in our tensor product $$1 \otimes 1$$, the Lie algebra operators becomes $$J_i \equiv 1 \otimes J_i^{(1)} + J_i^{(1)} \otimes 1.$$ Now, the only option for the $$m=2$$ state is $$|2\rangle = |1 \rangle |1 \rangle$$ and you can check that $$J_z |2 \rangle = 2 |2 \rangle.$$ Then we compute the action of the lowering operator using our expression for $$J_\pm$$, making sure to define our lowered states with the proper normalization so that the states we write as $$|m\rangle$$ all have norm $$1$$. \begin{align} J_-|2\rangle &= \sqrt{2}( |1\rangle |0 \rangle+|0\rangle |1 \rangle) \equiv 2 |1\rangle \\ J_-|1\rangle &= 2 |0\rangle |0\rangle + |1\rangle |-1\rangle + |-1\rangle |1\rangle \equiv \sqrt{6} | 0 \rangle \\ J_-|0\rangle &= \frac{1}{\sqrt{3}}(3 |-1\rangle |0\rangle + 3 |0\rangle |-1\rangle ) \equiv \sqrt{6} |-1\rangle\\ J_-|-1\rangle &= 2 |-1\rangle |-1 \rangle \equiv 2 |-1\rangle. \end{align}

So, rephrasing our results, \begin{align} |2\rangle &= |1\rangle |1\rangle \\ |1\rangle &= \tfrac{1}{\sqrt{2}} (|1\rangle |0\rangle + |0\rangle |1\rangle )\\ |0\rangle &= \tfrac{1}{\sqrt{6}} ( 2 |0\rangle |0\rangle + |1\rangle |-1\rangle + |-1\rangle |1\rangle ) \\ |-1\rangle &= \tfrac{1}{\sqrt{2}} (|-1\rangle |0\rangle + |0\rangle |-1\rangle ) \\ |-2\rangle &= |-1\rangle |-1\rangle. \end{align}

Now, to get our spin 2 polarization tensors $$\epsilon^{\mu \nu}$$, we just have to combine our spin 1 polarization vectors $$\epsilon^\mu_{-1,0,+1}$$ in the manner exactly above. So, to give one example, $$\epsilon^{\mu\nu}_0 = \frac{1}{\sqrt{6}}(2 \epsilon_0^\mu \epsilon_0^\nu + \epsilon_1^\mu \epsilon_{-1}^\nu+ \epsilon_{-1}^\mu \epsilon_1^\nu ).$$ Notice that, by construction, these polarization tensors will satisfy $$k_\mu \epsilon^{\mu \nu} = k_\nu \epsilon^{\mu \nu} = 0$$. They also turn out to satisfy $$\epsilon^\mu_{\; \mu} = 0$$ and are symmetric. (Notice that we get a little taste of the slogan "gravity = E&M^2" from this method, even though we are working in the massive case.)

The results of taking these Kronecker products is \begin{align} \epsilon_2^{\mu \nu} &= \frac{1}{2} \begin{pmatrix} 0&0&0&0\\0&1&i&0\\0&i&-1&0\\0&0&0&0 \end{pmatrix} \\ \epsilon_1^{\mu \nu} &= \frac{1}{2}\begin{pmatrix} 0&0&0&0\\0&0&0&1\\0&0&0&i\\0&1&i&0 \end{pmatrix} \\ \epsilon_0^{\mu \nu} &= \begin{pmatrix} 0&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&2 \end{pmatrix} \\ \epsilon_{-1}^{\mu \nu} &= \frac{1}{2}\begin{pmatrix} 0&0&0&0\\0&0&0&-1\\0&0&0&i\\0&-1&i&0 \end{pmatrix} \\ \epsilon_{-2}^{\mu \nu} &= \frac{1}{2}\begin{pmatrix} 0&0&0&0\\0&1&-i&0\\0&-i&-1&0\\0&0&0&0 \end{pmatrix} \\ \end{align}

So, hopefully you can see that Clebsch Gordan coefficients are not so scary, and that you are always free to rederive them for yourself. Admittedly, we did not work from tensoring together four copies of the spin $$1/2$$ case, but rather used two copies of the spin $$1$$ case to ease our burden, but the principle is what is important.