# Linear and circular polarization in quantization of the EM field

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 $$\mathbf A(\mathbf x,t)=\sum_{\mathbf k, \lambda}\left[A_{\mathbf k,\lambda}e^{i(\mathbf k\cdot\mathbf x-\omega_k t)}\hat{\mathbf e}_{\mathbf k,\lambda} + A_{\mathbf k,\lambda}^*e^{-i(\mathbf k\cdot\mathbf x-\omega_k t)}\hat{\mathbf e}_{\mathbf k,\lambda}^*\right]$$ where $\omega_k=|\mathbf k|c$ and $\lambda=\pm$. The unit vectors $\hat{\mathbf e}_{\mathbf k\pm}$ are the circular polarization defined as $$\hat{\mathbf e}_{\mathbf k\pm}=\mp\frac{1}{\sqrt2}\big(\hat{\mathbf e}_{\mathbf k}^{(1)}\pm i\hat{\mathbf e}_{\mathbf k}^{(2)}\big)$$ 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 $$\hat{\mathbf e}_{\mathbf k\lambda}^{*}\times\hat{\mathbf e}_{\pm\mathbf k\lambda'}=\pm i\lambda\delta_{\lambda\lambda'}\hat{\mathbf k},\qquad(*)$$ 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?

• You need Sakurai's statement "where $\hat{\mathbf e}_{\mathbf k}^{(1)}$ and $\hat{\mathbf e}_{\mathbf k}^{(2)}$ are the linear unit vectors perpendicular to $\mathbf k$" but you also need to know that $\hat{\mathbf e}_{\mathbf k}^{(1)}$ and $\hat{\mathbf e}_{\mathbf k}^{(2)}$ are orthonormal. I think Sakurai has forgotten to tell you this. If you don't have orthogonality, you still get a relationship like (*) (because $\hat{\mathbf e}_{\mathbf k}^{(1)}$ and $\hat{\mathbf e}_{\mathbf k}^{(2)}$ define a plane orthogonal to $\hat{\mathbf{k}}$) but there is a nonunity scale factor encoding ... – WetSavannaAnimal Dec 29 '14 at 9:39
• .. the angle between $\hat{\mathbf e}_{\mathbf k}^{(1)}$ and $\hat{\mathbf e}_{\mathbf k}^{(2)}$ – WetSavannaAnimal Dec 29 '14 at 9:40
• I doubt the relation (*), i.e. the $\delta _{\lambda \lambda '}$. The vector product between any two different vectors in the plane perpendicular to $\hat k$ is along $\hat k$, and is non-zero. – Sofia Dec 29 '14 at 10:37
• which version of Sakurai are you using? On my edition chapter 7 is about Diffusion theory, and there is no chapter about the quantization of EM field – glS Dec 29 '14 at 11:29
• @glance The second edition, Chapter 7 Section 6 – Rui Chao Dec 30 '14 at 12:07

They are any pair of orthonormal vectors normal to the wavevector. It is just implicitly assumed that you have a scheme for defining what they are (I can see that this can cause problems on a first reading): you need to define a suitable $\hat{\mathbf{e}}^{(1)}_{\mathbf{-k}},\,\hat{\mathbf{e}}^{(2)}_{\mathbf{-k}}$ for each point $\mathbf{k}$ in momentum (wavevector) space. If you like, they are two possible polarisation directions.
There is a great deal of freedom in defining them. For example, you might take $\hat{\mathbf{e}}^{(1)}_{\mathbf{-k}}$ to be a unit vector in the same direction as $\mathbf{k}\times\hat{\mathbf{z}}$ when $\mathbf{k}$ is not parallel to $\hat{\mathbf{z}}$ and make some choice, say $\hat{\mathbf{e}}^{(1)}_{\mathbf{-k}} = \mathbf{\hat{x}}$ when $\mathbf{k}$ is parallel to $\hat{\mathbf{z}}$. Having chosen $\hat{\mathbf{e}}^{(1)}_{\mathbf{-k}}$, you then take $\hat{\mathbf{e}}^{(2)}_{\mathbf{-k}}=\mathbf{\hat{k}}\times\hat{\mathbf{e}}^{(1)}_{\mathbf{-k}}$ so that the triplet $\hat{\mathbf{e}}^{(1)}_{\mathbf{-k}}$, $\hat{\mathbf{e}}^{(2)}_{\mathbf{-k}}$, $\mathbf{\hat{k}}$ in that order forms a right handed basis.
The neat thing about the circular polarisation eigenvectors $\hat{\mathbf e}_{\mathbf k\pm}=\mp\frac{1}{\sqrt2}\big(\hat{\mathbf e}_{\mathbf k}^{(1)}\pm i\hat{\mathbf e}_{\mathbf k}^{(2)}\big)$ is that they are rotated through angle $\pm\phi$ about the wavevector simply by multiplication by the phase term $e^{i\,\phi}$: you don't need to call up messy $2\times2$ rotation matrices.
Therefore, if $\hat{\mathbf e}_{\mathbf k\pm}$ thought of as a function of $\mathbf{k}$ are a valid system of circular polarisation vectors, then so is the system $e^{i\,\phi(\mathbf{k})}\,\hat{\mathbf e}_{\mathbf k\pm}$, where $\phi(\mathbf{k})$ is a function of wavevector: it is the first system rotated by the angle $\phi(\mathbf{k})$ at wavevector $\mathbf{k}$.