# Choosing the polarization vectors, Quantization of EM field

In Sakurai, Modern Quantum Mechanics 3rd edition, page 465, the book writes:

The set of solutions to $$\nabla^2\boldsymbol{A}-\frac{1}{c^2}\frac{\partial^2\boldsymbol{A}}{\partial t^2}=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(7.164)$$ are naturally written as $$\boldsymbol{A}(x,t)=\boldsymbol{A}(\boldsymbol{k})e^{\pm i\boldsymbol{k}\cdot\boldsymbol{x}}e^{\pm i\omega t}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(7.165)$$ where $$\omega=\omega_k=|\boldsymbol{k}|c$$. The Coulomb gauge condition implies $$\boldsymbol{k}\cdot\boldsymbol{A}(\boldsymbol{k})=0.$$ The book then says we can pick two unit vectors $$\hat{\boldsymbol{e}}^{(1)}_\boldsymbol{k}$$ and $$\hat{\boldsymbol{e}}_{\boldsymbol{k}}^{(2)}$$ perpendicular to $$\boldsymbol{k}$$ which can be used to decompose $$\boldsymbol{A}(\boldsymbol{x},t)$$. My question is: how are we picking $$\hat{\boldsymbol{e}}^{(1)}_\boldsymbol{k}$$ and $$\hat{\boldsymbol{e}}_{\boldsymbol{k}}^{(2)}$$? Because I believe for topological reasons you can't pick them continuously. Also what to do with $$\boldsymbol{k}=0$$?

Also why is Sakurai using circular polarization not linear polarization as in Ryder's QFT for example. What advantage does circular polarization offer?

I think you can define the polarization vectors $$\mathbf e_{\mathbf k}^{(1)}, \mathbf e_{\mathbf k}^{(2)}$$ for the given $$\mathbf k$$ in any way, provided their dot product is zero. There is no need for them to be a continuous function of $$\hat{\mathbf{~k}}$$, at least any discontinuity seems fine if the field is expressed as the Fourier series (sum over discrete wavevectors), not as the Fourier integral... but I'm not sure on this.

For example, one can work with spherical coordinates and define those two polarization vectors based on the vectors $$\mathbf e_\phi, \mathbf e_\theta$$ associated with the direction of $$\mathbf k$$. This fails on the poles, but, we can complete the definition there by an arbitrary prescription such as $$\mathbf e_{\mathbf k}^{(1)} = (1,0,0)$$ and $$\mathbf e_{\mathbf k}^{(2)}=(0,1,0)$$. Of course, this is ugly but there does not seem to be any mathematical problem with this.

Also why is Sakurai using circular polarization not linear polarization as in Ryder's QFT for example. What advantage does circular polarization offer?

It's just a different choice of basis for expansion of the same field in the same space. Different bases can be advantageous in different circumstances.

If the field has an important component that is really right-polarized (say, the light is coming off a crystal that turns a light beam with linear polarization into a light beam with right circular polarization), then its expansion coefficients in the linear polarization basis $$\mathbf e_{\mathbf k}^{(1)}, \mathbf e_{\mathbf k}^{(2)}$$ will be varying with time, but in the circular polarization basis, the expansion coefficients will be constant. This may be useful in calculations, in a similar way in which we remove trivial time dependence (factor of $$e^{i\omega t}$$) when using the phasor method to find stationary solutions of linear differential equations in AC circuits. Or like in QT when removing time-dependence from central quantities when going from the Schrodinger picture to the interaction picture, or when using the so-called rotating-wave approximation.

The two bases are related to each other, in this way: $$\mathbf e_{\mathbf k}^{(R)} = \mathbf e_{\mathbf k}^{(1)} + i \mathbf e_{\mathbf k}^{(2)},$$

$$\mathbf e_{\mathbf k}^{(L)} = \mathbf e_{\mathbf k}^{(1)} - i \mathbf e_{\mathbf k}^{(2)}.$$