# Side-by-side quarter wave plates

1. Apr 26, 2013

### JesseM

Can anyone help out with the following question? Suppose we have an incident linearly polarized plane wave, polarized in the z-direction and propagating in the x-direction, and it passes through two quarter wave plates (their surfaces normal to the x-axis), and the left plate causes the fields emerging from the other side to rotate in the counter-clockwise direction as seen in this animation, the right plate causes the emerging field to rotate clockwise. What would be the relation between the angles of the E-field vectors emerging from the plates at any given moment? For example, at a moment when the E-field at the front of the left plate is pointing in the +y direction, would the E-field from the right plate also be parallel to the y-axis, and if so would it point in +y or -y at that moment?

Ideally I'd like to find a reference giving a general relation between the fields hitting one side of the plate and the fields emerging from the other as a function both of time and the thickness of the plate. (In order for a quarter-wave plate to function properly does there have to be some relation between its thickness and the wavelength of the wave hitting it, or can it theoretically be arbitrarily thin and still work perfectly? Will the same plate work equally well at producing circularly polarized waves for incoming linearly polarized wave of different wavelengths?) But a general relation isn't really necessary for the question I'm exploring (which is about what would happen in the classical analogue of the quantum experiment shown in the slides that accompany this article), if anyone knows how to answer the question above it'd be helpful.

2. Apr 26, 2013

### Andy Resnick

3. Apr 26, 2013

### JesseM

Not sure I understand the notation--if light polarized in the x direction (and propagating in the z direction) has normalized Jones vector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$, and that's multiplied by $e^{i(kz - \omega t)} = cos(kz - \omega t) + i sin(kz - \omega t)$, that means after we discard the imaginary component the x-component of the E field is proportional to $cos(kz - \omega t)$? (this equation seems to say that at x=0 and t=0 the field must be at its maximum, I guess the phase information was discarded when the Jones vector was normalized?) And if a right-hand circularly polarized beam has normalized Jones vector $\begin{pmatrix} 1 \\ -i \end{pmatrix}$ that means the complex amplitude of the x-component is $cos(kz - \omega t) + i sin(kz - \omega t)$ while the complex amplitude of the y-component is $-i cos(kz - \omega t) + sin(kz - \omega t)$, so when we discard the imaginary part the x-component is proportional to $cos(kz - \omega t)$ and the y-component is proportional to $sin(kz - \omega t)$? Does that mean if we pick a time and a place where $kz - \omega t = 0$, and the quarter wave plate that converts light polarized in the x direction to right-hand circularly polarized light is located at that position at that time, then the incoming linearly polarized beam is at maximum amplitude as it hits the plate, and the circularly polarized light coming out is also purely in the x-direction at that time and place, and also at maximum amplitude? If so, does that mean it's a general fact that if you have a quarter-wave plate oriented at the right way to convert incoming light linearly polarized in the x-direction to right-hand circularly polarized light, it'll always be true that at a moment where the incoming linearly polarized light is at maximum amplitude as it hits the plate, the circularly polarized light is purely in the x-direction as it comes out of the plate at that moment, and at maximum amplitude?

If I'm misunderstanding, then while a better understanding of Jones vectors/matrices would be nice, it's not really necessary for now--I'm just trying to do a numerical simulation, using the trick described in #2 http://docs.lumerical.com/en/fdtd/user_guide_planewave_circular_polarization.html [Broken] of combining the results of two simulations with linear polarized light on different axes to simulate circularly polarized light, so I just need to know the relation between the angles of the two E-field vectors coming out of side-by-side quarter wave plates, assuming the same linearly polarized plane wave (polarized in the z-direction, say) is hitting them both. At the moment the E-field vectors arbitrarily close to the plate on the front side (the side where the exiting beam is circularly polarized) are parallel (or antiparallel?), what is their direction? Do they line up when they are both in the z-direction too? And at this moment, what is the amplitude of the linearly polarized beam arbitrarily close to the plate on the other side, is it at its maximum?

Last edited by a moderator: May 6, 2017
4. May 6, 2013

### JesseM

OK, let me try to make this question a little more specific. Suppose the incoming wave, that hasn't gone through the plates yet, is traveling along the z-axis and is polarized at 45 degrees from the x axis and y axis, with the x-component described by $E_x = \frac{E_0}{\sqrt{2}} cos(\omega t - kz)$ and the y-component also described by $E_y = \frac{E_0}{\sqrt{2}} cos(\omega t - kz)$. From what I can gather from reading various sources on quarter wave plates such as this one, it seems as though they have a "fast axis" and a "slow axis" orthogonal to the fast one, and the effect of the quarter wave plate is to allow the component parallel to the fast axis to pass through the plate unchanged, while the component parallel to the slow axis gets a phase delay of pi/2, which would change a cosine into a sine. So given the incoming wave I described above, if the first plate had the fast axis parallel to the x-axis and the slow axis parallel to the y-axis, I think the wave emerging out of that plate should look like:

$E_x = \frac{E_0}{\sqrt{2}} \cos(\omega t - kz)$
$E_y = \frac{E_0}{\sqrt{2}} \sin(\omega t - kz)$

Which would be a circularly polarized wave whose polarization direction is rotating counter-clockwise (from the perspective of an observer watching the wave approaching along the z-axis). On the other hand, if the second plate had a fast axis parallel to the y-axis and the slow axis parallel to the y-axis, I think the wave emerging out of the second plate would look like:

$E_x = \frac{E_0}{\sqrt{2}} \sin(\omega t - kz)$
$E_y = \frac{E_0}{\sqrt{2}} \cos(\omega t - kz)$

Which would be a circularly polarized wave rotating clockwise (like the one shown in the image at the top of this wikipedia article). Can anyone confirm if this is right, or explain what I'm misunderstanding if it's wrong? It seems to match what is said on p. 276 of The Light Fantastic: A Modern Introduction to Classical and Quantum Optics, which can be seen on google books here. But what confuses me is that in the section on "Jones vectors and matrices" on p. 277-278 (also viewable on google books), they say that a quarter-wave plate can have the effect of converting an incoming wave like the one I described above into a left-circularly-polarized beam (which rotates counterclockwise) or a right-circularly-polarized beam (which rotates clockwise), depending on the whether the fast axis is along the x or y axis, and that if the original incoming wave is represented by the real component of $\begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} \exp(i \phi)$ (where $\phi = \omega t - kz$), then the outgoing left-circularly-polarized beam would be described by the real component of $\begin{pmatrix} 1 \\ -i \end{pmatrix} \exp(i \phi) / \sqrt{2}$ while the outgoing right-circularly-polarized beam would be described by the real component of $\begin{pmatrix} 1 \\ i \end{pmatrix} \exp(i \phi) / \sqrt{2}$ (equations 10.34 and 10.35 on page 278). But it seems to me that when you actually multiply this out and take the real component (keeping in mind that $\exp(i \phi) = \cos(\phi) + i \sin(\phi)$), then you find that the outgoing left-circularly-polarized/counter-clockwise beam looks like:

$E_x = \frac{E_0}{\sqrt{2}} \cos(\omega t - kz)$
$E_y = \frac{E_0}{\sqrt{2}} \sin(\omega t - kz)$

While the outgoing right-circularly-polarized/clockwise beam looks like:

$E_x = \frac{E_0}{\sqrt{2}} \cos(\omega t - kz)$
$E_y = \frac{E_0}{\sqrt{2}} - \sin(\omega t - kz)$

...which is different from what I had above. So, which is correct? What am I misunderstanding about the equations given in the textbook I linked to above?

edit: could it be that in Jones vector/matrix equations, we always assume that the top value of the vector refers to the fast axis and the bottom to the slow axis, rather than the top being the x-axis and the bottom being the y-axis? But that wouldn't seem to match the the chart here on the wikipedia article, where they give the Jones matrix for "quarter-wave plate with fast axis vertical" and "quarter-wave plate with fast axis horizontal", and aside from a constant factor the matrices are identical to the ones given on page 278 of the book, equation 10.33

edit 2: Eq. 4.7 on this page seems to show definitively that the second set of equations, were the x-component remained a cosine and the y-component turned into plus or minus sine, is the correct one, which means the first set I wrote down, where which became sine and cosine depended on whether the slow axis was parallel to x or y, was wrong. It still seems like that first set should follow from the equations on p. 276 of The Light Fantastic, though, so what am I misunderstanding about that page?

Last edited: May 6, 2013
5. May 7, 2013

### Andy Resnick

Oh my... sorry I didn't respond sooner. It's very straightforward, honest. Let's start with x-polarized light: $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. In terms of physics, it means we represent the polarization state of light as a vector, and we chose basis states x- and y-polarized light, because they are perpendicular and 'span the space' of polarization. Any polarization state can be represented in the Jones calculus as P = a$\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ + b$\begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

Ok, so we chose our initial state as $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Now, we propagate the light through two quarter-wave plates. In terms of the Jones calculus, our output polarization state P' = T * P, where T is the 2 x 2 'transfer matrix' of the optical system, which here is simply two quarter-wave plates, and let's call the transfer matrix of a quarter wave plate Q(θ) =$\begin{pmatrix} \cos2\theta & \sin2\theta \\ \sin2\theta & -\cos2\theta \end{pmatrix}$

where θ is the angle between the x-axis (the input polarization direction) and the 'fast axis' of the retarder. The transfer matrix is T = $Q_{2}(θ_{2})Q_{1}(θ_{1})$ (note the order of multiplication here), which is a giant messy 2x2 matrix. I'll cheat and write this as $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$

One thing you could do to simplify T is to set one or both of the fast axes to θ = 0, θ = 90, etc. I any case, P' = $\begin{pmatrix} a \\ c \end{pmatrix}$, which means the polarization vector has rotated, so now there are *both* x- and y-polarized light.

But you can let the angles vary in time as well, if you want to model rotating retarders. You can also vary the retardance by using a more general form, like:

\begin{pmatrix} e^{i\phi_x} \cos^2\theta+e^{i\phi_y} \sin^2\theta & (e^{i\phi_x}-e^{i\phi_y}) \cos\theta \sin\theta \\ (e^{i\phi_x}-e^{i\phi_y}) \cos\theta \sin\theta & e^{i\phi_x} \sin^2\theta+e^{i\phi_y} \cos^2\theta \end{pmatrix}

Now as for the e^(ikz+ iwt) business, that has basically been factored out- the expression $\begin{pmatrix} 1 \\ 0 \end{pmatrix}e^{i(kz-ωt)}$ just means the electric field points in the x-direction always, everywhere, and oscillates in time at a rate ω, and along the propagation direction with wavelength 2π/k.

Does this help?

Edit- I see you asked about circular polarization states- those are also basis states, so linear polarization states can be converted into circular polarization states- for example, the left-hand circular polarization state is constructed by adding the x- and y- states in *phase quadrature*- there is a factor e^(i*pi/2) = i present, which is what you wrote.

Last edited: May 7, 2013