# Spin representations of Lorentz group

In the context of Classical Field Theory, we know that irreducible representation are labelled by the values of the two Casimir operators of the Poincaré group: we can have massive fields $$P^2=m^2>0$$ with a certain spin $$s$$, $$W^2=m^2s(s+1)$$, or massless fields $$P^2=m^2=0$$ with a certain helicity $$h$$, $$W^\mu=h P^\mu$$.

Now, I want to obtain all the irreps of the Lorentz group (Poincaré representations are induced by these) for different values of $$m$$ and $$s,h$$ as projective representations, that is using the representation of the spin group which covers $$SO(1,3)$$. After some steps (see this) we understand that we are looking for complex representations of $$SL(2,\mathbb{C})\times SL(2,\mathbb{C})$$ (that is a tensors product of two copies of the same representation and they act on bispinor, a spinor and an anti-spinor) in order to obtain irreducible unitay representations of the Lorentz group for different types of fields (different values of mass and spin/helicity).

How do we do this?

For example:

1. I want a spin-0 representation ($$m>0$$ and $$s=0$$), so I expect that, since there is no spin, the spin representation is trivial $$R_0(I_{\mu\nu})=0$$ ($$I_{\mu\nu}$$ are the 6 generators of the Lorentz group) which act on a "trivial bispinor", which is just a 1-component real field (a scalar field). This is easy
2. I want a spin-$$1\over 2$$ representation ($$m>0$$ and $$s={1\over 2}$$), so I expect that, since a $$1\over 2$$-spinor has 2 components, the spin representation is a $$4\times 4$$ complex matrix $$R_{1\over 2}(I_{\mu\nu})=\sigma_{\mu\nu}$$ which act on a bispinor with 4 complex values (a spinor field). This is easy too, since we know that the (Weyl) representation of $$SL(2,\mathbb{C})\times SL(2,\mathbb{C})$$ in this case is just the defining representation of $$SL(2,\mathbb{C})$$ $$\times$$ the complex conjugate of the defining representation of $$SL(2,\mathbb{C})$$.
3. I want a spin-1 representation ($$m>0$$ and $$s=1$$), so, using the same arguments, I expect a bispinor with 6 components. Is this correct? How can I say that this is a four-vector complex field? Are they two different representation of the same field (like Weyl and Dirac representation for $$1\over 2$$-spin fields)?

Of course I know that it is a four-vector field, but I was wondering if there is an "easy way" to say that using a sort of heuristic argument like the one that I used in the previous examples.

I would suggest looking at the first chapter of Streater, R. F., & Wightman, A. S. (1989). PCT, Spin and Statistics, and All That. Princeton University Press. In there it is commented that all irreducible finite dimensional complex representations of $$\text{SL}(2,\mathbb{C})$$ are of the form $$D^{(j/2,k/2)}$$ $$\xi_{\alpha_1\cdots\alpha_j\dot{\beta}_1\cdots\dot{\beta}_k}\mapsto A_{\alpha_1}{}^{\rho_1}\cdots A_{\alpha_j}{}^{\rho_j}\bar{A}_{\dot{\beta}_1}{}^{\dot{\sigma}_1}\cdots\bar{A}_{\dot{\beta}_k}{}^{\dot{\sigma}_k}\xi_{\rho_1\cdots\rho_j\dot{\sigma}_1\cdots\dot{\sigma}_k},$$ where $$\xi_{\alpha_1\cdots\alpha_j\dot{\beta}_1\cdots\dot{\beta}_k}$$ is symmetric under permutations of the dotted and undotted indices separately.

1. The spin $$0$$ representation is $$D^{(0,0)}$$. In it the spinors have no indices $$\xi$$ and they transform trivially under $$\text{SL}(2,\mathbb{C})$$, so that they are scalars.
2. We have two spin 1/2 representations $$D^{(1/2,0)}$$ and $$D^{(0,1/2)}$$ corresponding to the Weyl spinors $$\xi_\alpha$$ and anti-Weyl (?) spinors $$\bar{\xi}_\dot{\alpha}$$. The Dirac field is then for example in $$D^{(1/2,0)}\oplus D^{(0,1/2)}$$ with Dirac spinors $$\psi_\mu=(\xi_\alpha,\bar{\xi}_\dot{\alpha})$$.
3. Spin 1 representations are $$D^{(1,0)}$$, $$D^{(1/2,1/2)}$$, and $$D^{(0,1)}$$. The middle one is the standard vector representation $$D^{(1/2,1/2)}=D^{(1/2,0)}\otimes D^{(0,1/2)}$$ consisting of spinors of the form $$\xi_{\alpha\dot{\beta}}$$. That they are 4-vectors can be seen under the standard isomorphism $$x_{\alpha\dot{\beta}}\mapsto x_\mu:=x_{\alpha\dot{\beta}}\sigma_\mu^{\dot{\beta}\alpha}$$, where $$\sigma_\mu=(1,\sigma_x,\sigma_y,\sigma_z)$$. Indeed, via this isomorphism one sees that four vectors carry a representation of $$\text{SL}(2,\mathbb{C})$$. To see this we first note that the inverse of this isomorphism is simply $$x_{\alpha\dot{\beta}}=x_\mu\sigma^\mu_{\alpha\dot{\beta}}$$, where $$\sigma^\mu=(1,\sigma_x,\sigma_y,\sigma_z)$$. The indexing in these sigma matrices is different than the one for the previous ones because they are suppose to represent the inverse matrix. It just so happens that the inverse of a Pauli matrix is itself. Then the representation on 4-vectors is $$x_\mu\mapsto A_\alpha{}^{\rho}\bar{A}_{\dot{\beta}}{}^{\dot{\sigma}}x_\nu\sigma^\nu_{\rho\dot{\sigma}}\sigma_\mu^{\dot{\beta}\alpha}$$, i.e the vector $$x_\mu$$ changes via the matrix $$\Lambda_\mu{}^\nu=\sigma_\mu^{\dot{\beta}\alpha}A_\alpha{}^{\rho}\sigma^\nu_{\rho\dot{\sigma}}\bar{A}_{\dot{\beta}}{}^{\dot{\sigma}}$$. One can show that this is in the connected component of the identity in the Lorentz transformations. More info on this can be found in Haag, R. (1996). Local Quantum Physics (2nd ed.). Springer Berlin Heidelberg. https://doi.org/10.1007/978-3-642-61458-3. However, the other spin 1 representations also appear. For example the representation $$D^{(1,0)}$$ is given by antisymmetric self-dual tensors $$B_{\mu\nu}$$. Using the fact that every 2-tensor $$F_{\mu\nu}$$ can be decomposed into a self-dual and an anti self-dual part, one can see that for example the Maxwell field strength tensor transforms in $$D^{(1,0)}\oplus D^{(0,1)}$$.
• I still have some truble with spin-1 representation. You say that spinors in $D^{(1/2,1/2)}$ are isomorphic to 4-vectors, but for a representation of the Lorentz group don't you need the tensor product of two representations? For example in the spin-$1/2$ representation, where you use the direct sum of two representations in order to obtain the non-irreducible representation that you need to represent a Lorentz transformation on a spin-$1/2$ field. I don't understand how do you do this in spin-1. Dec 2, 2020 at 9:38
• $D^{(1/2,1/2)}$ is indeed a tensor product of the representations $D^{(1/2,0}$ and $D^{(0,1/2)}$. I will add more details on how it gives rise to the vector representation of the Lorentz group. There is no connection of this (at least that I know of) with the procedure done to construct the Dirac representation. The representations $D^{(1/2,0)}$ and $D^{(0,1/2)}$ by themselves yield perfectly good spin 1/2 fields. They are usually bundled up into a Dirac field because mass terms usually mix them up. Dec 2, 2020 at 22:45