# How can unpolarised light be produced from 2 beams with orthogonal polarisation?

Say I have 2 beams of equal intensity with orthogonal polarisation. Is there a way, using linear polarizers, half and quarter waveplates , to combine those two beams in such a way that they produce an unpolarized light? And if it is possible, what would the intensity of resultant beam be?

The first thing I thought about was to use matrices and vectors to represent the polarised light, so we have initially two beams: $$E_{||} = ( 1, 0 ) \\E_{\bot } = (0,1)$$ First, I realised that passing those through a quarter wave plate will give me circularly polarised light. I though I could realise how to make unpolarized light by looking at the matrix algebra, but then I realised there isn't a way to represent unpolarized light with those vectors and matrices (I think what I am using here are called Jones matrices). So I guess I need to think about the underlying physics to realise what is going on. Okay, well I know that unpolarized light has to have some random phase difference between axis, but because I have no idea how to represent this mathematically I am struggling to come up with a combination of filters and waveplates that could make such a light.

EDIT: Okay, it seems there is no way to make unpolarised light out of two completely polarised coherent beams. But what if the original beams are produced by a lamp, which emits not at a perfect frequncy, but has a very small bandwidth (due to doppler shift, because light is emitted by fast travelling particles in a hot gas). Is there a way to make unpolarized light?

EDIT 2: I guess here the actual set up would be important. So the lamp emits light with a certain bandwidth. I guess that means that the frequency of light emitted is random, but close to some value. Then the light is split into two beams by a polarising beam splitter. So in my understanding, light emitted at time $t$ is not coherent with light emitted at some time $t+\delta t$. But because two beams are produced from the same original beam by a beam splitter, that two beams at any certain time are cohherent.

• You got a lovely answer, already, but for measurements you can also use a polarized beam and a rotator. By repeating the measurement for several different rotations and numerical averaging you should get the same answer, at least for linear system behavior. After all, linear superpositions are linear superpositions, no matter how they are being "calculated". You can let nature's analog computer do them for you, or you can have them done in your CPU with IEEE floats. :-) May 8, 2016 at 22:20
• If you look in the optics catalogs you will not find a "depolarizer", is they don't exist. About the best you can do is to put the beams through a diffuser, which randomizes the phases & directions of the beam. May 8, 2016 at 22:43

You need to mix them incoherently. If the beams are from different sources, then you're pretty much there, as they won't stay in phase.

You can't represent incoherent light with Jones vectors (2 components), but you can do it with density matrices (this approach has a huge importance in quantum mechanics, where incoherent mixtures are most important). So, 2×2 matrices, formed as time averages of tensor product of the Jones vector with itself. The meaning of the matrix is the correlation of components: diagonal terms are the energy fluxes (remember - it's amplitude squared), and the off-diagonal terms are cross-correlations between polarizations. There are no cross-correlations if they are incoherently mixed (unpolarized) so incoherent light has diagonal matrices. Anything between totally polarized and totally unpolarized light is likewise describable by this. Trace of the matrix is the total energy flux.

Consider this $x$ and $y$ polarizations: $E_x=(1,0)$ and $E_y=(0,1)$. Associated matrix: $$D_x^{ij}=E_x^i E_x^j=\begin{bmatrix}1&0\\0 & 0\end{bmatrix}$$ $$D_y^{ij}=E_y^i E_y^j=\begin{bmatrix}0&0\\0 & 1\end{bmatrix}$$

Coherent mixture: $E=E_x+E_y=(1,1)$ (diagonal polarization). The matrix for this: $$D^{ij}=E^i E^j=\begin{bmatrix}1&1\\1 & 1\end{bmatrix}$$

Incoherent mixture: $$D_x^{ij}+D_y^{ij}=\begin{bmatrix}1&0\\0 & 1\end{bmatrix}$$

So, to reiterate... mixing correlated beams (meaning it's from the same source, usually the same beam passed through polarizers, optically active materials, anisotropic plates and so on) creates polarized states. Polarized states are those for which the matrix can be split into a "square" of a single vector. Mixing uncorrelated beams produces a state that can't be written as a "square" of a single vector, and thus can't be reduced to the simple formalism. But you can always work with density matrices.

Formalism for working with matrices is similar to that for vectors, but you need to apply all the transforms on both sides.

How to get from physics to math? Imagine a single beam:

$$\vec{E}(t)=y(t)\vec{E}$$ That is, amplitude times polarization. Now consider two such beams. Compute the energy flux, which is a time average of the square. Mark averages as $\langle\rangle$. Consider also multiplying this by the jones matrix before calculating the square and integrating. So, we have this expression for the flux (ignore the constant factors of speed of light and stuff like that):

$$j=\langle(A^{ij}(y_1(t){E}_1^j+y_2(t){E}_2^j)^2\rangle$$ $$=\langle A^{ik}A^{ij}(y_1(t){E}_1^j+y_2(t){E}_2^j)(y_1(t){E}_1^k+y_2(t){E}_2^k)\rangle$$ $$=A^{ik}A^{ij}\langle(y_1^2(t){E}_1^j{E}_1^k+y_2^2(t){E}_2^j{E}_2^k+y_1 (t)y_2(t){E}_1^j{E}_2^k+y_2 (t)y_1(t){E}_2^j{E}_1^k\rangle=A^{ik}A^{ij}D^{jk}=Tr(ADA^T)$$ Now you know also why we need to preserve the whole matrix, not just the trace. Whatever Jones matrices you applied to the initial mixture of light, you need all four components of the density matrix to compute the product before taking the trace.

Now observe the averaged part. If $y_1$ and $y_2$ are incoherent, then the long term average of their product will give zero. In that case, you get only the diagonal terms. If they are at least partially coherent, the off diagonal terms will be nonzero. Remember that autocorrelations and cross-correlations drop off with time delay even for coherent beam, so if you use a beam splitter and rejoin the beams with time delay, you'll lose correlations. That's actually how you measure time coherence of the beam.

Anyway, we just reproduced the whole reasoning of how to deal with coherent/incoherent/partially coherent polarized light. All we needed is the fact that locally, amplitudes are additive, but when you time-average the intensity (flux, proportional to the amplitude squared), the result depends on how the beams are correlated (you may have more than 2, this was just an example). The density matrix formalism emerged out automatically, and their meaning too.

No beam is completely coherent with itself on long term (only a perfect sine wave would do that). There's always drop in coherence when you delay the beam with itself. That's what autocorrelation function is all about. Unless you have some funny light source with atypical correlation profile, you'll have exponentially decaying autocorrelation function with the characteristic time proportional to the inverse frequency bandwidth of the light (basically, the uncertainty principle). Beams from unrelated sources are never coherent at all. Think of two cars with blinkers on: unless they are powered by the same signal, they'll never blink in synchrony for long (on average, they'll be in opposite phase for as much time as in phase, which would make the cross-correlation - the average product - equal to zero).

Heads up: if you're dealing with complex amplitudes (required if you want to handle circular polarizations), then one of the terms in the product must be complex conjugated and matrix transpose becomes hermitian conjugate.

• So, if my source is emitting light within a certain bandwidt of frequencies ( a discharge lamp which produces a range of frequencies because of Doppler shift due particles of the gas inside the lamp moving very fast with different velocities), can I use that to create an incoherent mixture? May 8, 2016 at 22:02
• Every beam is time-coherent with itself at zero time delay, but auto-correlations drop (more quickly for higher bandwidth, usually within characteristic time, proportional to the inverse of frequency bandwidth). So you have to split the beam, apply sufficient time delay between them, rotate the polarization, and add rejoin them. Either that, or put something in the beam path that will scatter the beam multiple times, like a diffuse glass. That's bound to reduce the correlations and make the phases more random. May 8, 2016 at 22:10
• Okay, I think I get why I need to delay one of the beams, but why would I need to rotate the polarization of beam after delaying it? Wouldn't delaying one them, then combining them together make in unpolarized light already? May 8, 2016 at 22:15
• Well, if you start with a single polarized beam, split it in two, delay them, and add them together... you'll still only have a single polarization! Mixing [1,0;0,0] with [1,0;0,0] won't magically produce two ones on the diagonal. You still need equal amounts of x and y polarization, but you have to make them incoherent to avoid x and y moving in synchrony. May 8, 2016 at 22:18
• No, if they're already orthogonal you don't have to rotate them. May 10, 2016 at 6:37

I expect the answer is no because unpolarized is a 'random' mixture of all sorts of different polarizations mixed together. Much easier to make polarized from unpolarized with a filter. Perhaps in theory possible to make something that looks like unpolarized from polarized but requiring lots and lots of beam splitters wave plates etc. to make up the complicated mix of different polarizations normally present in unpolarized light.