string theory - Decomposition of the gravitino into helicity $pm frac{3}{2}$ and $pm frac{1}{2}$ components

I'm reading this book on string theory. When they decompose two dimensional gravitino (formula 7.16)

$$\chi_\alpha = \frac{1}{2}\rho^\beta \rho_\alpha \chi_\beta + \frac{1}{2}\rho_\alpha \rho^\gamma \chi_\gamma,$$ they interpret summands as helicity $\pm \frac{3}{2}$ and $\pm \frac{1}{2}$ components. Here $\rho^\alpha$ are generator of the $1+1$ Clifford algebra. I have the following questions about this decomposition

1. How helicity operator is mathematically implemented in this case?


2. Why the above claim is correct?



3. Is it the same thing as decomposition of tensor product of vector and spinor representations of the Clifford algebra into two irreducible representations of "spin $\frac{3}{2}$ and $\frac{1}{2}$"?

Answer

Here the object $\chi_\alpha$ has an explicit 2D vector index, as well as an implicit 2D spinor index. There for it is in the $\textbf{1}\otimes\frac{\textbf{1}}{\textbf{2}} =\frac{\textbf{1}}{\textbf{2}}\oplus \frac{\textbf{3}}{\textbf{2}}$ representation of the $SO(1,1)$ group.

Now the question is how do we isolate the $\frac{\textbf{1}}{\textbf{2}}$ representation in the sum, from the $\frac{\textbf{3}}{\textbf{2}}$ representation? we will succeed if we can form an object out of $\chi_\alpha$ which is in the $\frac{\textbf{1}}{\textbf{2}}$ representation of $SO(1,1)$ and remove that object from the original $\chi_\alpha$.

You can easily see that $\rho^\alpha \chi_\alpha \equiv \xi$ does the job elegantly, because the vector index is contracted and all is left is the spinor index. So $\rho^\alpha \chi_\alpha\equiv \xi$ extracts the spin half component of the gravitino, in much the same way as the trace of a tensor extracts the scalar component of a tensor.

Now we want to form an object $\psi_\alpha$ out of $\chi_\alpha$, which satisfies $\rho^\alpha\psi_\alpha =0$, because as we understand now such object would have the spin half component removed! you can see using simple gamma matrix identity that $\psi_\alpha \equiv \frac{1}{2} \rho^\beta \rho_\alpha \chi_\beta = \chi_\alpha - \frac{1}{2}\rho_\alpha \xi$, does the job and satisfies the condition. Which means it is the way to project the spin three halves component out of $\chi_\alpha$.

So schematically $\chi = \xi\oplus\psi$ is the consequence of $\textbf{1}\otimes\frac{\textbf{1}}{\textbf{2}} =\frac{\textbf{1}}{\textbf{2}}\oplus \frac{\textbf{3}}{\textbf{2}}$

Note: in two dimensions the helicity operator is $\propto \rho^0\rho^1$ just like the $\gamma^5$ in four dimensions

Note: this is the exactly same procedure to understand the Rarita-Schwinger condition on spin $3/2$ field in four dimensions