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.


1 Answer 1


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 $


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.