Spin of the sress-energy tensor in d=2

[navocane]

If we suggest a generic quantum field theory and assume that the theory is Poincare-Invariant (i.e. the corresponding Ward-Identities are satisfied) than the stress energy tensor in two dimensions can be written in terms of complex coordinates $$z,\bar{z}$$as
$$T^{zz}(z,\bar{z})=T^{00}-T^{11}-2iT^{10}$$
$$T^{\bar{z}\bar{z}}(z,\bar{z})=T^{00}-T^{11}+2iT^{10}$$
$$T^{z\bar{z}}(z,\bar{z})=T^{\bar{z}z}(z,\bar{z})=T^{00}+T^{11}\equiv -\Theta(z,\bar{z})$$
My question is how to find the Spin of the components $$T^{zz},T^{\bar{z}\bar{z}},\Theta$$. The authors of the paper I'm studying claim
$$ST^{zz}=2, ST^{\bar{z}\bar{z}}=-2 , S\Theta=0$$ but i don't see how . Can anyone help?

 

[azntoon]

The spin of a tensor is determined by the transformation properties of its components under rotations. In two dimensions, rotations are generated by the angular momentum operator, so the spin of the components can be determined by looking at the action of the angular momentum operator on them. For example, since T^{zz} transforms as a tensor of rank two, it will have spin 2 under rotations. This can be seen by considering the action of the angular momentum operator on T^{zz}:L_zT^{zz}=i\partial_zT^{zz}-i\partial_{\bar z}T^{zz}This equation implies that T^{zz} transforms as a vector with spin 2 under rotations. Similarly, T^{\bar z \bar z} will have spin -2 under rotations, since it is a tensor of rank two and has the opposite transformation properties of T^{zz}. Finally, since the trace of a tensor is invariant under rotations, $\Theta$ will have spin 0.