Lorentz group representation of spin 2 particles

Question

I am beginner in particle physics and I have the following understanding of Lorentz group. Correct me if I am wrong. We know that Lorentz algebra is characterised by the eigenvalues $m(m + 1)$ and $n(n+ 1)$ of the square of the operators $$ \vec{N} = \frac{1}{2} (\vec{J} + i \vec{K}) \\ \vec{N^{\dagger}}= \frac{1}{2} (\vec{J} - i \vec{K}) , $$ where $\vec{J}$ is the generator of rotation and $\vec{K}$ the generator of boosts. Then the pair $(m,n)$ gives different representations of the Lorentz group.

My problem is:

I am reading the book of Pierre Ramond on field theory, Page 22. I don’t understand why $ (0,0) \oplus (1,1) $ is the symmetric part and $ (0,2) \oplus (2,0) $ (I think in the book it’s mistakenly written as $ (0,1) \oplus (1,0) $) is the antisymmetric part in the above equation 1.4.64? Here the symbols have usual meaning.

Answer 1

Note that

$$(a,b) \otimes (c,d) = (a \otimes c, b \otimes d) \tag{1}$$

so

$$(1/2,1/2) \otimes (1/2,1/2) = (1/2\otimes 1/2, 1/2\otimes 1/2)$$ $$ = (1 \oplus 0, 1/2\otimes 1/2)$$ $$ = (1, 1/2\otimes 1/2) \oplus (0, 1/2\otimes 1/2)$$ $$ = (1, 1 \oplus 0) \oplus (0, 1 \oplus 0)$$ $$ = (1, 1 ) \oplus (1, 0) \oplus (0, 1 )\oplus (0,0)$$ $$ = \left[ (0, 0) \oplus (1,1) \right]\oplus \left[ (1,0) \oplus (0,1)\right]$$

It's correct and not a typo. Typically when using product representations (your example is Rarita-Schwinger without the Dirac part), you'll find a spin background that leads to extra degrees of freedom, unless you impose some conditions.

$(1/2,1/2)$ can be interpreted as the coupling of two spinors with each other $(1/2,0) \otimes (0,1/2) = (1/2,1/2)$. As thus you (might) already know that they can couple to a triplet and a spinor. In terms of dof this is $2 \otimes 2 = 3 \oplus 1$, in terms of spin this is $1/2 \otimes 1/2 = 1 \oplus 0$.

Your "raw" spin-2 field thus has the following spin contents:

$$(1 \oplus 0) \otimes (1 \oplus 0) = 1\otimes 1 \oplus 0 \otimes 1 \oplus 1 \otimes 0 \oplus 0\otimes 0$$ $$ = \underbrace{2 \oplus 1 \oplus 0}_{(1,1)} \oplus \underbrace{1 \oplus 1}_{(1,0)\oplus(0,1)} \oplus \underbrace{0}_{(0,0)}$$

So your $(1,1)$ tensor needs $5+3+1 = 9$ degrees of freedom.

It has two Lorentz indices $T_{\mu\nu}$ and thus 16 components in total. Making it symmetric means there are only $10$ independend components. Making it traceless fixes another component, thus the required $9$.

From this you can follow, that in order to have the needed $9$ dof, your tensor must be traceless and symmetric.

Your $(1,0) \oplus (0,1)$ tensor has $3+3=6$ degrees of freedom. If you again start with a 2-index tensor $T_{\mu\nu}$ with 16 entries, antisymmetry $T_{\mu\nu}= -T_{\nu\mu}$ eliminates the diagonal (-4 indices) and poses a condition on the remaining 12 (-6 indices) leaving you with a total of 6 independent entries. The electro-magnetic fieldstrength tensor $F_{\mu\nu}$ is an example for a $(1,0)\oplus(0,1)$ tensor.

Answer 2

This isn't a complete answer, but

1. Just by counting dimensions, $(0,1)\oplus (1,0)$ is correct. $(1/2,1/2)$ has dimension 4. So $(1/2,1/2)\otimes (1/2,1/2)$ dimension 16. In the decomposition, $(0,0)$ has dimension $1\times 1=1$, $(1,1)$ has dimension $3\times 3=9$, $(0,1)$ has dimension $1\times 3=3$ and $(1,0)$ has dimension $3\times 1=3$. $1+9+3+3=16$ so it checks out.
2. About Weyl spinors, those are the $(1/2,0)$ and $(0,1/2)$ representations. One is 'left-handed' and the other 'right-handed' (not sure which is which.) The representations are on $\Bbb{C}^2$ given by the formulas $$\begin{split} A\circ_1 v = Av\\ A\circ_2 v = \overline{A}v \end{split}$$ where $v\in\Bbb{C}^2$, $A$ is either in $\mathfrak{sl}(2,\Bbb{C})$ or $SL(2,\Bbb{C})$ and $\overline{A}$ denotes complex conjugation of the matrix elements. Note that deciding which one is $(1/2,0)$ and which one is $(0,1/2)$ is basis-dependent and a matter of convention. The map $A\mapsto \overline{A}$ is an outer automorphism of $\mathfrak{sl}(2,\Bbb{C})$ as a real Lie algebra that flips the signs of $J_1,K_2,J_3$ and has the effect of swapping $(p,q)$ and $(q,p)$.