# Convention for defining wave polarisation

Suppose I have a plane wave with wavevector $$\mathbf{k}$$ and polarization $$\mathbf{\hat{e}}$$. Is there a convention or some logical way of defining the polarization given the wavevector? I know that the polarization vector must be perpendicular to the wavevector, so the only free parameter to specify is the angle of rotation about the wavevector. What is confusing me is in the case where $$\mathbf{k}$$ is completely general and in some arbitrary direction, how do you specify the polarization vector and what is the angle of rotation relative to? Is there a vector equation to specify $$\mathbf{\hat{e}}$$ given an arbitrary $$\mathbf{k}$$ and perhaps angle of rotation?

• What you have described is the constraint that $\mathbf k \cdot\mathbf{\hat{e}} = 0$. This equation describes the plane of possible polarization vectors. – user111476 May 19 at 1:26
• What you have said about the angle of rotation being a free parameter is not correct, or in the very least it is not worded very well. Perhaps you mean the orientation of the polarisation vector within the plane perpendicular to the wave vector, but in any case, one cannot define the orientation of the polarisation other than by performing an actual physical measurement (for a completely general wave in which the only information we have is the wavevector). – MC2k May 19 at 1:35
• @Alex Yes so supposing that I am free to choose the polarization vector in the plane of possible polarization vectors, how do I practically go about specifying it, mathematically speaking? Sorry I realize my question wasn't worded completely clearly. – PaperBee May 19 at 2:32
• @PaperBee You need to be more specific, the polarization vector is not unique. In principle, any unit vector in the plane specified by $\mathbf{k}\cdot \mathbf{\hat{e}}$ is valid. One way is to fix the $x$ and $y$ axes to lie in this plane (i.e. $\mathbf k = k \mathbf{\hat{z}}$) and then parameterize the polarization vector in terms of an angle from the $x$-axis $\theta$, and a relative phase $\phi$, i.e. $\mathbf{\hat{e}} = \begin{pmatrix} \cos \theta \\ e^{i \phi }\sin \theta \\ 0 \end{pmatrix}$. You may be interested in looking at the so called "Stokes parameters". – user111476 May 19 at 3:10
• @Alex So what I am asking is how do I mathematically generalize your example if $\mathbf{k}$ is not in the $\mathbf{\hat{z}}$, but rather it is a vector with $\mathbf{\hat{x}}$, $\mathbf{\hat{y}}$, and $\mathbf{\hat{z}}$ components? – PaperBee May 19 at 3:37

As you already mentioned, the polarization is perpendicular to the wavevector, this is expressed mathematically as $$\mathbf k \cdot \mathbf{\hat e} = 0$$ and specifies a plane that your polarization vector can lie in. To start, you can assume that $$\mathbf{k} = k\mathbf{\hat z}$$ so that the polarization lies in the $$x,y$$ plane. Then on can parameterize the polarization vector in terms of an angle from the $$x$$-axis $$θ$$, and a relative phase $$ϕ$$, i.e. $$\mathbf{\hat e} = \cos \theta \mathbf{\hat x}+ e^{i \phi} \sin \theta\mathbf{\hat y}.$$ Given some aribitrary wavevector in a pre-determined basis $$\mathbf k = k_x \mathbf{\hat x}+k_y \mathbf{\hat y} + k_z \mathbf{\hat z}$$, you can always find a rotation matrix $$R$$ such that $$R \mathbf k = k \mathbf{\hat z}$$ where $$k = \sqrt{k_x^2+k_y^2+k_z^2}$$.

• Thanks this has been really helpful, any tips on how I can go about finding $R$? – PaperBee May 19 at 5:13

We have, for a free space plane wave, that

$$\hat{\boldsymbol{e}} \cdot \hat{\boldsymbol{k}} = 0$$.

This tells us that $$\hat{\boldsymbol{e}}$$ lies in the plane perpendicular to $$\hat{\boldsymbol{k}}$$. You are correct that in a fully general situation there is nothing else which determines the direction for $$\hat{\boldsymbol{e}}$$ within this plane.

However, in practice, we can make some conventional choice.

# gravity

Often a laser beam is propagating horizontal to the Earth's surface, for example it is being shined parallel to an optical table which includes lenses, mirrors, etc. to control the laser beam. In this case the gravity direction, the direction perpendicular to the table, indicates a preferred direction. Light with a polarization vector parallel to gravity is called vertically polarized and the other polarization (parallel to the table) is called horizontally polarized.

If the laser is traveling in the gravity direction You'll need a different convention. Perhaps North-South or East-West polarized could work.

## Relative to a Surface

The other important convention is to define polarization relative to a surface with which the light is interacting, like a mirror. For concreteness lets imagine a mirror lying on a table which we call the $$xy$$ plane. Let's imagine a laser beam is approaching the mirror from above and at angle. Suppose the wave vector $$\hat{\boldsymbol{k}}$$ is contained in the $$xz$$ plane. That is, $$\hat{\boldsymbol{k}}$$ has a component in the $$z$$ direction (moving downwards) and a component in the $$x$$ direction, parallel to the table. We can define a vector $$\hat{\boldsymbol{n}}$$ which is normal to the surface of the mirror. In this case the surface normal vector points in the $$z$$ direction.

As long as the beam is not parallel with the surface normal the surface normal $$\hat{\boldsymbol{n}}$$ and $$\hat{\boldsymbol{k}}$$ span a 2D plane called the plane of incidence. The beam travels within this plane towards the mirror before reflection and away from the mirror after reflection. In this case the plane of incidence is the $$xz$$ plane.

We can then consider two conventional polarization with respect to this plane. For one polarization the polarization vector is normal to this plane, that is, it has no component in the plane of incidence. This polarization is called $$s$$-polarized light. Apparently $$s$$ stands for senkrecht which is German for perpendicular, indicating that this light is perpendicular to the plane of incidence.

The orthogonal polarization lies completely in the plane of incidence (but still perpendicular to $$\hat{\boldsymbol{k}}$$) and is called $$p$$-polarized light. I guess $$p$$ stands for parallel indicating that the light is parallel to the plane of incidence.

It is useful to define polarizations with respect to the surface normal because it turns out that certain optical surfaces will have different behaviors depending on whether the incident light is s or p-polarized.

If the light is propagating parallel to the surface normal then there is no plane of incidence and both polarizations are parallel to the surface itself so there is again no a-priori way to indicate one polarization versus another.

## No Reference?

If there is truly no other reference in the problem then it is enough to simply say the beam propagates in the $$\hat{\boldsymbol{k}}$$ direction and has two polarizations $$\hat{\boldsymbol{e}}_1$$ and $$\hat{\boldsymbol{e}}_2$$ which are orthogonal to $$\hat{\boldsymbol{k}}$$ and each other. If there are no other directions in the problem then you might define your coordinate system so that $$\hat{\boldsymbol{k}}$$ points in the $$z$$ direction then you can take $$x$$ and $$y$$ to be the directions for the two polarizations. This shouldn't be a problem if there are no other references in the problem.

• Thanks for your reply. So the consensus that I am getting is that I need to rotate $\mathbf{\hat{k}}$ to be in a direction that is easy to work with, eg $\mathbf{\hat{z}}$, then specify $\mathbf{\hat{e}}$ before finally applying the inverse rotation matrix. – PaperBee May 19 at 23:36
• It's not EXACTLY clear what you're asking. Are you trying to perform some analytical or numerical calculation or are you just trying to get a conceptual understanding of something? Perhaps you could clarify this in your question. – jgerber May 19 at 23:42
• It is for a numerical application, to be implemented in code, so I require a precise mathematical approach allowing me to choose/specify $\mathbf{\hat{e}}$ given some arbitrary $\mathbf{k}$. – PaperBee May 20 at 1:26

I would proceed as follows

\begin{align} \hat{k} =& k_x\hat{x} + k_y\hat{y} + k_z\hat{z}\\ \hat{e} =& e_x\hat{x} + e_y\hat{y} + e_z\hat{z} \end{align}

\begin{align} \hat{e}\cdot\hat{e} =& e_x^2+e_y^2+e_z^2 = 1 \\ \hat{k}\cdot\hat{e} =& k_xe_x+k_y e_y+k_z e_z = 0 \end{align}

We are always free to arbitrarily choose $$e_y=0$$ for one of the polarizations

So

\begin{align} e_z=&\sqrt{1-e_x^2}=-\frac{k_x}{k_z}e_x\\ \end{align}

and thus

$$e_x^2 = \frac{1}{\frac{k_x^2}{k_z^2}-1} \\$$

So

\begin{align} e_x =& \frac{1}{\sqrt{\frac{k_x^2}{k_z^2}-1}}\\ e_y =& 0\\ e_z =& \frac{-\frac{k_x}{k_z}}{\sqrt{\frac{k_x^2}{k_z^2}-1}} \end{align}

From which $$\hat{e}$$ can be written down. The second polarization is found by computing

$$\hat{e}' = \hat{e}\times\hat{k}$$

I guess you have to watch out for a few special cases like $$k_z=0$$ or $$k_x=k_z$$.

All in all I think @Alex's solution is a little more elegant. Google rotation matrices.