# Fourier Optics - Why is polarization omitted?

I am wondering why in wave optics the polarisation is often omitted.

Examples:

1. In Fourier Optics, a wave at a point $$x \in \mathbb{R}^3$$ is given by a complex number $$U(x, y, z)$$ which describes amplitude and phase. But what about the polarisation direction?
2. The formulas for Fresnel and Fraunhofer diffraction are always written for scalar fields $$U(x, y, z)$$. It seems as if the polarisation direction does not make a difference when light is diffracted at an aperture. But is this really correct? If I try to imagine the electric field, I always come to the conclusion that the polarization of the light at the aperture should make a difference.

I am kind of confused about that and would be very grateful if somebody could help me to understand this correctly.

Kind Regards

If we assume implicitly, that the polarisation is constant, we are allowed to skip this detail in or description, because the result does not change. However, if we assume that different polarisation components are present, we have to address each component separately -- assuming that you consider two orthogonal components, which do not mix.

By the way, this is also true for the wavelength of the light -- there exists no light source which possesses only a single wavelength (i.e. frequency) component --, the concept of a perfectly plane wave, or the idea of a medium with a homogeneous index of refraction. These are conceptional descriptions.

Here an example: Suppose we'd like to describe a plane wave with "frequency" $$\omega$$, wave vector $$\vec k = \frac{2\pi}{\lambda} \, \vec e_z$$, and polarisation $$\vec p = \frac{1}{\sqrt{2}}(\vec e_x + \vec e_y)$$. We could either choose the cartesian coordinate system and write $$\vec E(\vec r, t) % = E_0 e^{-i(\omega t - \vec k\cdot \vec r)} \; \vec p = \frac{E_0}{\sqrt{2}} e^{-i(\omega t - k z)} \; \begin{pmatrix} 1\\ 1 \end{pmatrix}$$ or we could rotate the reference frame by 45° and use the vector basis $$\{\vec e_{p_{\parallel}}, \vec e_{p_{\perp}}\}$$, which leads to $$\vec E(\vec r, t) = E_0 e^{-i(\omega t - k z)} \; \begin{pmatrix} 1\\ 0 \end{pmatrix}$$ Now, if we assume that all electric fields are either parallel or anti-parallel to $$\vec e_{p_{\parallel}}$$, we know that we only have to consider the first component of the vector. Hence, by dropping the vector notation and implicitly only considering the first component of the vector, we are allowed to write $$E(\vec r, t) = E_0 e^{-i(\omega t - k z)}$$ Hence, the polarisation is no longer explicitly considered, but only implicitly.

• Thank you for your answer! Still I don't understand the theory completely behind the idea of wave optics without polarisation. It seems as if historically the wave nature of light was found much earlier than the effect of polarisation. But how can one imagine a wave without polarisation? I mean, even longitudinal waves have a polarisation, even though it's only the direction of propagation. Furthermore, which physical quantity does periodically change if it is not the electric/magnetic field? Sorry if my formulations are a bit squishy, but the character limit kind of confines me. Apr 5 '20 at 14:33
• I added a section. Apr 5 '20 at 14:55
• Could you expand your example a little further to include randomly polarized lights? Would that reduces to the scalar field case?
– WDC
Apr 5 '20 at 15:10
• This goes beyond the original question. Hence, I won't include that. However, a good description can be found e.g. in Goodman "Statistical Optics" book. Apr 5 '20 at 15:17
• If one likes to describe unpolarised light, one should consider a statistical mixture. For certain problems this leads to a scalar description, for others it won't. The description uses the concept of random variables. The reference I mentioned above is a good reference. Apr 5 '20 at 22:15