Circular Polarization States for Light,
and quarter-wave Plates

We are given a monochromatic, polarized light wave travelling in a single, constant direction, which we call x. We may describe it by its electric field vector:

$${\vec E} = {\vec E}_m\,\sin(kx-\omega t)\;,$$ (1)

where ${\vec E}_m$ is the (vector) amplitude, which must be perpendicular to the x-direction.
Given that the wave moves with speed $v \equiv c/n$, with n the index of refraction of the material through which it moves, the (single) color of the light is determined by any of various different parameters that we traditionally use to describe it:

$$\begin{array}{rl} \hbox{the frequency,~~}\nu \;,& \; \hbox{th... ...;,&\; \hbox{the wavenumber,~~} k = 2\pi/\lambda\;. \end{array}$$ (2)

Note also that when the light wave is travelling through some material, with the index of refraction different from 1, then the frequencies are unchanged, while the wavelength and wavenumber do change.

Various polarization states of the wave are characterized by the way the amplitude vector for the electric field depends on time:

• When ${\vec E}_m$ is constant in time it always points in the same direction; we refer to this case as plane polarization.
• When the magnitude, $\vert{\vec E}_m\vert$ , is constant, but the direction rotates with time, we refer to that as circular polarization.
• When not even the magnitude is constant, we refer to that as elliptical polarization.

The elliptical case is clearly the most general case; however, it is more complicated than we need to consider here. Furthermore, your text does a reasonable job of describing the details of plane polarized light. Therefore, the purpose of these extra notes is to consider the case of circular polarization. In principle there are two sorts of circular polarization; the one where the electric vector rotates to the right and the one where it rotates to the left, referred to as left-circular and right-circular polarizations. However, while it is definitely true that there are these two kinds (or possibilities), there is considerable disagreement about how one labels which one is which; therefore, we will not do so, but simply refer to both as circularly-polarized light, remembering, occasionally, that there are indeed two kinds. It is true that an individual photon created by a single, individual atom always has an electric field that is circularly polarized. However, this is an uncommon way to be looking at light waves, since these waves usually involve very many photons--trillions at the very least--so that the polarization state of any light wave we are observing will be some sort of an average over very many, differently-oriented states of the individual photons; therefore, it is usually expected that any incoming light wave is unpolarized unless something has been done to it to select that state in a certain way.

We want now to spend a little time discussing methods to select the polarization state of an outgoing light wave.

Plane Polarization by Using Polarizers

As we know from the textbook, there are particular materials, known as polarizers, that selectively absorb that light which has its electric field in some particular direction, while passing light directly onward when its electric field is polarized in the perpendicular direction. This perpendicular direction your textbook refers to as the polarizing direction, or axis. When a light wave strikes a polarizer with an electric field vector that is in neither of these two (perpendicular) directions, we can easily visualize what happens by making a standard vector decomposition of that electric field into its components relative to a basis for (2-dimensional) vectors that is just taken along the two perpendicular directions defined by the polarizer. So, if the incoming electric field vector, ${\vec E}_m$, makes an angle $\theta$ relative to the polarizing direction, we may decompose that vector into its component along the polarizing direction, which will have value $\vert{\vec E}_m\vert\,\cos\theta$, and its component perpendicular to the polarizing direction, which would then have value $\vert{\vec E}_m\vert\,\sin\theta$. This second portion would be absorbed so that only the light along the polarizing direction would exit the material. The light exiting this polarizer would then be plane polarized, and have less intensity than when it entered since a portion of its energy had been absorbed. See the Law of Malus in the text, which basically has two parts:

• if the incoming light is plane polarized, at angle $\theta$ relative to the polarizing direction, then the ratio of the intensity of the outgoing light to that of the incoming light is given by $I = I_0\,\cos^2\theta$.
• if the incoming light is unpolarized, then we should treat it as a large, random assembly of plane polarized incoming beams, which says we should average over all possible angles. The average of $\cos^2\theta$ is one half, so that the final intensity is given by $I = \frac{1}{2}\,I_0$.

Birefringent materials

The previous process, of the absorption of one of the vector components of the light while passing the other component directly through, could be described as changing the speed of the two components, so that one has speed zero and the other has its speed unchanged.
This mode of description suggests that we could consider rather more sophisticated materials, where the light is also split into two (or more) components, and these two components pass through the material with different speeds, but neither of them being zero.

These more general materials are referred to as birefringent, or doubly refracting, materials. Typically such materials are crystals, so that they can easily have a special direction within them, called the optic axis, and light moving through this crystal which has its polarization vector (or a component of that vector) along the optic axis moves with a different speed than light with its polarization vector perpendicular to the optic axis. To describe this, we must associate with the crystal two different indices of refraction, n_o, the index for ``ordinary'' rays, and n<sub>e</sub>, the index for ``extraordinary" rays. (As an amusing fact, it is sometimes true that extraordinary rays travel faster than ordinary ones, as in calcite and dolomite, while in other materials the extraordinary rays travel slower than the ordinary ones, as in quartz and ice.) [It should also be pointed out that there are in the world trirefringent materials, which need three different indices of refraction; topaz is an example. However, we shall not discuss them here.] By allowing a light ray to be incident on a birefringent material we may manipulate its polarization state. In the general case the incoming light will have a polarization vector that has non-zero components both along the optic axis and perpendicular to it. These two components will then move through the crystal with two different speeds, at two different directions--because Snell's Law for refraction depends on the speed--and therefore through two different distances. In the general case this will cause two different beams of light to exit the crystal where only one entered; our standard example is a calcite crystal. Both those beams will of course be plane polarized, whether or not the incoming light was polarized. A somewhat special case, but a very interesting one and a common one, is obtained when the crystal has its optic axis in one of its faces, i.e., in the plane of its surface. (One could also cut the crystal to cause this if it didn't happen naturally.) The scenario described above is still the valid one for an arbitrary incoming light ray; however, if we now consider the special case of light incident at zero angle on the surface, i.e, normally incident, then Snell's Law will tell us that the light ray will go straight on through, i.e., the angle of incidence is 0, so the angle of refraction is also 0, independent of the speed of the wave inside the crystal. Therefore the two beams will exit at the same place, and will be recombined into a single light ray again. However, since the two beams required different times to make the transit through the crystal the two beams being recombined--upon exit--will have begun their journey at different times; i.e., they will be components of two different incoming waves, and therefore will have different phases. By changing the thickness of the crystal, we can manipulate this phase difference when the two are recombined. In general crystals which have been cut this way are referred to as wave plates. In the next section we will describe how this feature may be used to create, for instance, circularly-polarized light from plane-polarized light, or vice versa.

quarter-wave Plates and Circular Polarization

If the birefringent material is of thickness T, then the time required to pass through it is simply t = T/v = n(T/c). Therefore if the two components have indices of refraction n_o and n_e, the difference in time required for them to pass through the crystal is $\Delta t = \vert n_e-n_o\vert(T/c)$. To convert a time difference into a phase difference, for the two components about to be combined, we multiply by the angular frequency, $\omega$, which is the same for both waves:

$$\Delta\phi = \omega\,\Delta t = \omega (T/c)\vert n_e-n_o\vert = 2\pi\,T\vert\lambda_e^{-1}-\lambda_o^{-1}\vert\;.$$ (3)

We now suppose that our crystal has incoming, plane-polarized light ray normally incident upon it. As well, label the direction of the optic axis by $\hat o$ and the perpendicular direction in the crystal surface by $\hat e$. Then the statement that the incoming polarization direction is at an angle $\theta$ relative to the optic axis of the crystal allows us to divide the incoming beam up into two components, as it enters the crystal, at time t and at location x:

$${\vec E}_{\rm in} = {\vec E}_m\,\sin(kx - \omega t) = \vert{... ...)\left\{\cos\theta\,{\hat o} + \sin\theta\,{\hat e}\right\}\;.$$ (4)

Notice that the direction of the electric field is independent of time, and that the portion inside the braces has magnitude 1. The intensity associated with this electric field is the time-average of its square. Since the average over a period of the square of a sine function is one half, the intensity is proportional to $\frac12 {\vec E}_m^2$. When the light exits the crystal, having travelled a distance T, the two components will be re-combined into a total electric field vector again, being now at time t' and location x'. However, one component has travelled longer than the other, and therefore has a different phase, i.e., started earlier:

$${\vec E}_{\rm out} = \vert{\vec E}_m\vert\,\cos\theta\,\sin(... ...rt\,\sin\theta\,\sin(kx'-\omega t' + \Delta\phi)\,{\hat e}\;.$$ (5)

In general this is quite a complicated-appearing equation, and gives us light with some variety of elliptical polarization. However, in the following special case we can use this phenomenon to create circularly-polarized light. We must first require that the phase difference be one quarter of an entire cycle, i.e., $\pi/2$ radians, or (of course) five quarters, or nine quarters, etc. Secondly, we must require that the initial angle be 45 degrees. Since the cosine of 45 degrees is the same as the sine, the magnitude of the two components is therefore the same, and we have

$$\begin{array}{rcl} {\vec E}_{\rm out} & = & \frac{1}{\sqrt{2}... ...'){\hat o} + \cos(kx'-\omega t'){\hat e}\right\}\;. \end{array}$$ (6)

The quantity in the braces is now rotating as time progresses, but always has the same length, i.e., just 1. This is our requirement for circular polarization! Moreover we notice that the time-average of the intensity is still $\frac12 {\vec E}_m^2$, the same as it was when it entered. We may summarize by saying that if the phase difference of the two waves is one quarter of an entire cycle, or an entire wavelength, then this crystal may be used to create circularly-polarized waves from plane polarized ones. Therefore, they are customarily referred to as quarter-wave plates. The thickness will of course depend on the difference in the indices of refraction. We may determine an explicit formula easily enough from our formula above for the phase difference, which we now require to be $\pi/2$, where we also now look at the case when $\lambda_e < \lambda_o$:

$$\pi/2 = \Delta\phi = 2\pi T[\frac{1}{\lambda_e} - \frac{1}{\... ...ft\{\frac{1}{\lambda_e} - \frac{1}{\lambda_o}\right\}^{-1}\;.$$ (7)

We see that a quarter-wave plate satisfies that requirement for only one particular wavelength, although ``nearby" wavelengths should not be too different. However, if, for instance, we have a quarter-wave plate for violet light, at $\lambda =$400 nm, then it would be a half-wave plate for light at 800 nm, so that red light, being ``almost" that long, would have its plane of polarization changed instead of becoming circularly polarized. To better understand the last statement, let us retreat to Eq. (5) and consider the case where the phase difference is half a wave, i.e., $\Delta\phi = \pi$. In that case, again at an angle of 45 degrees, the outgoing wave looks like

$$\begin{array}{rcl} {\vec E}_{\rm out} & = & \frac{1}{\sqrt{2}... ...x'-\omega t')\left\{{\hat o} - {\hat e}\right\} \;. \end{array}$$ (8)

This time the quantity in the braces is again constant, and therefore plane polarized. However, for our 45-degree incidence case, it entered as ${\hat o} + {\hat e}$, and is now exiting as ${\hat o} - {\hat e}$. These two directions are perpendicular to one another, so that the plane of polarization is still constant, but has been rotated by 90 degrees relative to its original direction.