I am going through the "Quantization of the EM field" in Chapter 7 of Sakurai's Modern Quantum Mechanics, which basically goes like:
The vector potential satisfies wave function $\nabla^2\mathbf A-\frac{1}{c}\frac{\partial^2\mathbf A}{\partial t^2}=0$ and the Coulomb gauge $\nabla \mathbf A=0$. The general solution for $\mathbf A$ is \begin{equation} \mathbf A(\mathbf x,t)=\sum_{\mathbf k, \lambda}\left[A_{\mathbf k,\lambda}e^{i(\mathbf k\cdot\mathbf x-\omega_k t)}\hat{\mathbf e}_{\mathbf k,\lambda} + A_{\mathbf k,\lambda}^*e^{-i(\mathbf k\cdot\mathbf x-\omega_k t)}\hat{\mathbf e}_{\mathbf k,\lambda}^*\right] \end{equation} where $\omega_k=|\mathbf k|c$ and $\lambda=\pm$. The unit vectors $\hat{\mathbf e}_{\mathbf k\pm}$ are the circular polarization defined as \begin{equation} \hat{\mathbf e}_{\mathbf k\pm}=\mp\frac{1}{\sqrt2}\big(\hat{\mathbf e}_{\mathbf k}^{(1)}\pm i\hat{\mathbf e}_{\mathbf k}^{(2)}\big) \end{equation} where $\hat{\mathbf e}_{\mathbf k}^{(1)}$ and $\hat{\mathbf e}_{\mathbf k}^{(2)}$ are the linear unit vectors perpendicular to $\mathbf k$. Then the author says that, with these definition it is easy to show \begin{equation} \hat{\mathbf e}_{\mathbf k\lambda}^{*}\times\hat{\mathbf e}_{\pm\mathbf k\lambda'}=\pm i\lambda\delta_{\lambda\lambda'}\hat{\mathbf k},\qquad(*) \end{equation} where $\hat{\mathbf k}$ is a unit vector in the direction of $\mathbf k$.
I know how to prove $\hat{\mathbf e}^*_{\mathbf k\lambda}\times\hat{\mathbf e}_{\mathbf k\lambda'}=i\lambda\delta_{\lambda\lambda'}\hat{\mathbf k}$. The question is that, to prove the second part of $(*)$, it seems that we have to define what is $\hat{\mathbf e}_{-\mathbf k}^{(1)}$ and $\hat{\mathbf e}_{-\mathbf k}^{(2)}$. But what is a proper definition of these?