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Question: Give the mode expansion of the $A_i$ in terms of plane wave \begin{equation} \epsilon^{\pm}_i(p)e^{-ip \cdot x} \ \ \ \ \ \text{ and } \ \ \ \ \ \epsilon^{*\pm}_i(p)e^{-ip \cdot x} \end{equation} where the polarisation vectors for the helicity eigenstates satisfy \begin{equation} \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|}. \end{equation} 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.

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The Helicity operator is given by \begin{equation} \hat{\Lambda}=\int \mathrm{d}^{3} k\left(\hat{a}_{k+}^{\dagger} \hat{a}_{k+}-\hat{a}_{k-}^{\dagger} \hat{a}_{k-}\right) \end{equation}

where we have \begin{equation} \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} \end{equation} for the usual creation and annihilation operators $\hat{a}_{k\lambda}^{\dagger}, \hat{a}_{k\lambda}$. The operators $\hat{a}_{k -}^{\dagger},\hat{a}_{k +}^{\dagger}, \hat{a}_{k-},\hat{a}_{k,+}$
\begin{equation} \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) \end{equation}

Then an helicity eigenstate $ |k,+\rangle $ for example,is given by $\hat{a}_{k,+}^{\dagger} |0\rangle $

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