# Finding helicity eigenstates

Question: Give the mode expansion of the $$A_i$$ in terms of plane wave $$$$\epsilon^{\pm}_i(p)e^{-ip \cdot x} \ \ \ \ \ \text{ and } \ \ \ \ \ \epsilon^{*\pm}_i(p)e^{-ip \cdot x}$$$$ where the polarisation vectors for the helicity eigenstates satisfy $$$$\epsilon_{ijk}\hat{p}_j \epsilon_k^{\pm}(p) = \pm i \epsilon_i^{\pm}(p) \ \ \ \ \ \text{ with } \ \ \ \ \ \hat{p}_i \equiv \frac{p_i}{|p|}.$$$$ Find the helicity eigenstates for $$\hat{p} = \hat{e}_z$$.

Solution I know that the mode expansion is given by $$A_i = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2E_{\mathbf{p}}}} \sum_{\lambda = \pm} \left[ \epsilon_i^{\lambda}(\mathbf{p})a_{\mathbf{p}}^{\lambda}e^{-ip\cdot x} + \epsilon_i^{*\lambda}(\mathbf{p})a_{\mathbf{p}}^{\lambda \dagger}e^{+ip\cdot x} \right],$$ but I have no idea how to determine the helicity eigenstates.

The Helicity operator is given by $$$$\hat{\Lambda}=\int \mathrm{d}^{3} k\left(\hat{a}_{k+}^{\dagger} \hat{a}_{k+}-\hat{a}_{k-}^{\dagger} \hat{a}_{k-}\right)$$$$
where we have \begin{aligned} \hat{a}_{k+} &=\frac{1}{\sqrt{2}}\left(\hat{a}_{k 1}-\mathrm{i} \hat{a}_{k 2}\right) \\ \hat{a}_{k-} &=\frac{1}{\sqrt{2}}\left(\hat{a}_{k 1}+\mathrm{i} \hat{a}_{k 2}\right) \end{aligned} for the usual creation and annihilation operators $$\hat{a}_{k\lambda}^{\dagger}, \hat{a}_{k\lambda}$$. The operators $$\hat{a}_{k -}^{\dagger},\hat{a}_{k +}^{\dagger}, \hat{a}_{k-},\hat{a}_{k,+}$$
$$$$\left[\hat{a}_{k^{\prime}+}, \hat{a}_{k+}^{\dagger}\right]=\left[\hat{a}_{k^{\prime}-}, \hat{a}_{k-}^{\dagger}\right]=\delta^{3}\left(k-k^{\prime}\right)$$$$
Then an helicity eigenstate $$|k,+\rangle$$ for example,is given by $$\hat{a}_{k,+}^{\dagger} |0\rangle$$