quantum field theory - Why helicity is proportional to the spin of particle and has two values?

How can it be shown without using the little group formalism?

Let's have the Wigner's classification for the irreducible represetation of the Poincare group. For the massless case the eigenvalues of two Casimir operators of the group, the squares of Pauli-Lubanski operator and momentum operator, $\hat {W}_{\alpha}W^{\alpha}, \hat {P}_{\alpha}\hat {P}^{\alpha}$, is equal to zero.

Together with $\hat {W}_{\alpha}\hat {P}^{\alpha} = 0$ it leads to an expression $\hat {W}_{\alpha} = \hat {h}\hat {P}_{\alpha}$, where eigenvalues of $\hat {h}$ has dimension like angular momentum. It is called helicity. I want to get it "properties" without using small groups formalism (by the other words, not as Weinberg).

Answer

1. Construction of the helicity formula using 3-vector notation

The zero component of the pauli Lubanski vector

$W^0 = \epsilon^{0 ijk}J_{ij}p_k = \epsilon^{ijk}J_{ij}p_k $

The angular momentum genrerators

$ j^k = \epsilon^{ijk}J_{ij}$

Thus

$W^0 = j^k p_k = \vec{j}.\vec{p} $

The orbital angular momentum

$ \vec{l} = \vec{x} \times \vec{p}$

is orthogonal to the momentum:

$ \vec{l}.\vec{p} = 0$

And since the total angular momentum is the vector sum of the orbirtal and the spin angular momenta

$ \vec{j} = \vec{l} + \vec{\Sigma}$

Thus

$W^0 = j^k p_k = \vec{j}.\vec{p} = (\vec{j-l}).\vec{p} = \vec{\Sigma}.\vec{p} $

Now, since

$W^0 = \hat{h} p_0$

and for a massless particle

$ p_0 = p$

We obtain:

$ \hat{h} = \frac{\vec{\Sigma}.\vec{p}}{p_0} = \vec{\Sigma}.\hat{p}$

1. The helicity operator

$$\hat{h} = \Sigma.\hat{p}$$

where $\Sigma$ is the spin operator and $\hat{p}$ is the momentum unit vector is a projection along the axis $\hat{p}$ of a spin operator, thus one might expect it to have for a helicity $\lambda$ the eigenvalues $\lambda$, $\lambda-1$, ..., $-\lambda$.

However, the eigenvectors corresponding to all eigenvalues except $\pm \lambda$ are not physical, because they describe longitudinal polarizations which do not exist in free massless particles.

Here is an example of the massless spin-1 case (photon). In this case, we may choose the spin operators as:

$ \Sigma_x = \begin{bmatrix} 0 & 0& 0\\ 0 & 0& -i\\ 0 & i& 0 \end{bmatrix}$

$\Sigma_y = \begin{bmatrix} 0 & 0& i\\ 0 & 0& 0\\ -i & 0& 0 \end{bmatrix}$

$\Sigma_z = \begin{bmatrix} 0 & -i & 0\\ i & 0& 0\\ 0 & 0& 0 \end{bmatrix}$

The action of the Helicity operator on (say), the electric field in the momentum representation is:

$$\hat{h} \vec{E} = i\begin{bmatrix} 0 & -\hat{p}_z & \hat{p}_y\\ \hat{p}_z & 0& -\hat{p}_x\\ -\hat{p}_x & \hat{p}_x& 0 \end{bmatrix}\begin{bmatrix} E_x\\ E_y\\ E_z \end{bmatrix} = i \hat{p}\times \vec{E}$$

Thus:

$$\hat{h}^2 \vec{E} = - \hat{p} \times ( \hat{p}\times \vec{E}) = \vec{E} -\hat{p}(\hat{p}. \vec{E})$$

But, since for a free electromagnetic field:

$$\hat{p}. \vec{E} = 0$$

We get:

$$\hat{h}^2 = 1$$,

and the only admissible eigenvalues are $\pm 1$