Optics - Faraday Rotator using waveplates I'm trying to replicate the effect of a 45° Faraday rotator using a series of waveplates instead.
I've encountered some difficulties using the jones matrix notation, the main book I'm using is: "Polarized Light" by Goldstein.
By viewing the problem on the Poincaré sphere I think the soltion is a lambda/4 waveplate + lambda/4 waveplate rotated by 45°.
Unfortunately, the mathematical results via Jones calculus don't support my idea.
 A: One crucial difference between a waveplate and a Faraday rotator is that the former is reciprocal and the latter is not reciprocal. So you positively cannot fully realise a Faraday rotator with waveplates. So, depending on what exactly it is about the $45^o$ rotator you are trying to realise, your sought behaviour may or may not be realisable. The unidirectional behaviour of a Faraday rotator is realisable with waveplates, the full bidirectional behaviour is not.
Case 1: One-Directional Behaviour Only Important
If, however, you seek the Faraday Rotator's rotation in one direction only, then you are in luck. A Faraday Rotator has the following one-pass $2\times2$ Jones matrix:
$$F(\theta) = \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)$$
i.e. it is a pure rotation matrix. The diagonalisation of this rotation matrix shows that:
$$F(\theta) = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&i\\i&1\end{array}\right)\left(\begin{array}{cc}e^{i\frac{\theta}{2}}&0\\0&e^{-i\frac{\theta}{2}}\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1&-i\\-i&1\end{array}\right)$$
i.e. the Faraday rotator imparts different pure delays to the circular polarisation states
A waveplate, on the other hand, imparts differing delays to the linear polarisation states; its Jones matrix in this notation is therefore:
$$W(\theta) = \left(\begin{array}{cc}e^{i\frac{\theta}{2}}&0\\0&e^{-i\frac{\theta}{2}}\end{array}\right)$$
Now a junction between two waveplates such that their fast axes are at $45^o$ to one another corresponds to a Jones matrix of $F\left(\pm\frac{\pi}{4}\right)$, depending on the sense of rotation between the two fast exis. For a general angle $\theta$ the Jones matrix is $F(\theta)$, but you must be aware that you have rotated your co-ordinate axes to keep aligned to the waveplate's axes. So, you must always compensate for the total co-ordinate rotation made in any concatenation of optical devices with the inverse rotation at the end to transform back to your originalco-ordinates. Now, you can directly check (hope you've got Mathematica handy) that:
$$F(\theta) =  W(\pi)\, F\left(\frac{\theta}{2}\right)\, W(\pi)\, F\left(-\frac{\theta}{2}\right)\tag{1}$$
$$F(\theta) =  W\left(\frac{\pi}{2}\right) F\left(-\frac{\pi}{4}\right) W\left(-2\,\theta\right) F\left(\frac{\pi}{4}\right) W\left(-\frac{\pi}{2}\right)\tag{2}$$
(1) represents two waveplates rotated relative to one another so that there is an angle $\theta/2$ between their axes. Note that the first $F\left(\frac{\theta}{2}\right)$ rotates our co-ordinate axes as well so that they stay aligned with the second waveplate's axes. So we must rotate our co-ordinates back to keep aligned with the system input co-ordinate axes. This is the reason for the $F\left(-\frac{\theta}{2}\right)$ on the end of the product in (1). This is a neat method because you need only two waveplates and you can make a variable rotation between them to implement any rotation you want. 
The second method, embodied by (2), calls for two quarter waveplates and a waveplate of delay $2\,\theta$ between fast and slow axis. We arrange them thus:


*

*The $2\,\theta$-waveplate is sandwiched between the two quarter waveplates;

*The fast axis of the $2\,\theta$ waveplate is at angle $\pi/4$ ($45^o$) to that of the first eighth waveplate;

*The slow axis of the second quarter waveplate is aligned to the fast axis of the first.
Case 2: Full Realisation with Correct Relationship between Forwards and Backwards Pass Behaviour
If you are trying to realise the full relationship between waves passing through the Faraday rotator in both directions, then you are doomed, owing to the Faraday rotator's non-reciprocity.
If you shine linearly polarised, monochromatic light through a waveplate at a perfect reflector, the polarisation rotation undergone by the light on coming back through the waveplate from the reflector is in the same sense as it was going towards the reflector. That is, if the polarisation rotates clockwise when looking along the direction of the wavevector (or Poynting vector, thus propagation direction) going one way through the waveplate, it will rotate in the same sense relative to the wavevector coming back. The nett effect is that the plane of polarisation in the reflected light will be the same as that for the incident light.
If however you swap the waveplate for the Faraday rotator, the polarisation rotation sense is reversed with the propagation direction. The result is that the plane of rotation of the reflected light differs from that of the incident light, and the angle of rotation is twice that undergone in a single pass. By choosing the length of the nonreciprocal magneto-optical material such rotation angle on a single pass is $45^o$, the reflected light is in the orthogonal polarisation state to the incident light if the former is linearly polarised. Thus, by aligning a linear polariser with the plane of incident light polarisation, all the incident light is passed but the reflected light is blocked. This principle is the grounding of the Faraday effect isolator, widely used to quell reflexions from optical networks and thus stabilise lasers, which are highly cantankerous if reflected light coherent with its output gets back into the laser cavity - what this baleful situation means is that there is now a partial cavity of very long length owing to the network reflexion which varies wildly with vibration and temperature.
Reciprocity, specifically Lorentz Reciprocity (see the "Reciprocity (electromagnetism) for the monochromatic electromagnetic field in a system follows whenever the operator $\Sigma$ relating the current density $\vec{J}$ and the electric field $\vec{E}$ in the materials making up the system is complex-symmetric, i.e. $\vec{J} = \Sigma\,\vec{E}$ where $\Sigma^T = \Sigma$ - naturally, this is the same as Hermitian if and only if the operators are all real. This condition is not fulfilled in magneto-optic materials with nondiagonal permitivity tensors in the presence of a biasing magnetic field, and this breaking of the reciprocity condition is the grounding of the Faraday rotator.
When dealing with two-directional systems such as discussed above, the raw Jones matrix notation needs to be beefed up a little as I discuss in my footnote. 

Footnote: Jones Matrix Notation for Two-Directional Two Ports
Indeed, instead of the $2\times 2$ Jones matrix describing the transformation between input and output polarisation state of a two port device, we need to write down the $4\times 4$ complete scattering matrix $S$ that defines the linear homogeneous relationship between the incident light polarisation state and scattered light polarisation state for each of the two ports; relative to linear polarisation states we have:
$$b = S\,a;\quad a=\left(\begin{array}{c}a_{1,\parallel}\\a_1,\perp\\a_{2,\parallel}\\a_2,\perp\end{array}\right);\quad b =\left(\begin{array}{c}b_{1,\parallel}\\b_1,\perp\\b_{2,\parallel}\\b_2,\perp\end{array}\right)$$ 
where $a_{1,\parallel},\,a_1,\perp$ and $a_{2,\parallel},\,a_2,\perp$ are the two complex amplitudes of the two linear orthogonal polarisation states for ports 1 and 2, respectively, and the $b$ quantities are the analogous scattered complex amplitudes. In this notation, Lorentz reciprocity is equivalent to:
$$S=S^T$$
A: You need three wave plates to make the rotator work. Add another QWP perpendicular to the first and it should work.
A: Probably it is a little bit late and you already know the answer but if not here is a short answer:
A rotator R(tita) is realizable as a sequence of two half wave plates with their fast axes making an angle tita with one another.
A: The Faraday Rotator consists of establishing a large uniform magnetic field that surrounds the optical beam.  These devices were not practical until advanced magnetic materials, such as Neodymium, were available in large sizes.
With this arrangement, it is easy to see that in one direction, the beam will traveling "with" the Magnetic field, and in the other direction, the beam will traveling "against" the field.  When traveling with the field, the polarization is rotated clockwise, and against the field, the polarization rotates counterclockwise.  Therefore, it is common to place a Faraday isolator designed to rotate the beam 45 degrees, and have it surrounded by two linear polarizers whose polarization axis are 45 degrees apart. In this way, any back reflections that enter the second of the two linear polarizers (in the opposite direction) will be rotated "in the same angular direction" as it was during its initial traversing the of the rotator, because the magnetic field direction has reversed with respect to the beam.  Therefore, the polarization axis will be rotated another 45 degrees, and the back reflections will be extinguished by the lead polarizer
