Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have read that polarized light is treated by Jones vectors and that to treat partially polarized light you have to use Stokes vectors and mueller matrices.

Nonetheless, the optics notes that my professor have given us have no mention of mueller calculus, and we have assigned exercises involving partially polarized light passing through polarizers, retarders... so I figured that perhaps the following is legitimate:

The Stokes parameters characterizing partially polarized light are the following:

$s_1=Vs_0\cos{2\alpha}$

$s_2=Vs_0\sin{2\alpha}\cos{\delta} $

$s_3=Vs_0\sin{2\alpha}\sin{\delta} $

from the Stokes vector $(s_0,s_1,s_2,s_3)$ we get $\alpha$ and $\delta$ and build a Jones vector using:

$|e\rangle=\left( \begin{array}{c} \cos{\alpha}\\ \sin{\alpha}e^{-i\delta} \end{array} \right) $

and from here we go on using jones matrices.

Is this doable? And if it is, why do people use mueller matrices if this can be done?

share|improve this question
1  
If you have completely polarized light, you work with pure $2$-dimensional states like $|e\rangle$ (a "Jones vector"), and you work with $2*2$ Jones Matrices representing transformations. However, with a partial polarized light, you have to work with a general $2*2$ density matrix. An apparatus like a Linear polarizer transforms a general density matrix too, the general density matrix has $4$ independent real parameters, and they correspond (disguised) to the $4$ Stokes parameters. so you have to use a $4*4$ Mueller matrix to represent the transformation acting on the $4$ Stokes parameters. –  Trimok Jan 2 at 20:56

1 Answer 1

up vote 1 down vote accepted

Your proposed method would work as long as you only pass light through linear optical components that do not change the light's degree of polarisaion or overall power, in which case you would be using the Jones calculus in disguise: you can keep the polarised and depolarised components separate.

But the method will not work in general. However, you can still use the Jones matrices to represent optical components, but you apply them in a new way.

Partial polarisation is a very hard thing to describe classically - it's almost the same (and as hard) as the classical discussion of partial coherence and one needs to have a thorough grasp of random processes to discuss it fully. Born and Wolf give a whole chapter to these concepts. But it is highly elegantly described in the quantum picture: partially polarised light is a statistical mixture of pure quantum states. I discuss both approaches in my answer here.

So now, you should first read up on the density matrix (see Wikipedia article of this name). The name "matrix" is a little misleading, because it is really a "state" (albeit a mixed one) written down as a $N\times N$ matrix (where $N$ is the dimensionality of the quantum states you are dealing with) and NOT a "transformation" or "operator" on states, as the name "matrix" would imply. It's written as a matrix because this is the most convenient way to get statistics out of it: the $n^{th}$ moment of a measurement by an observable $\hat{A}$ is computed as ${\rm Tr}(\rho \hat{A}^n)$ where $\rho$ is the density matrix representing the mixed state. So if the light state is a classical statistical mixture of polarisation states with $2\times 1$ Jones vectors $\vec{x}_1, \vec{x}_2, \cdots$ with the classical probabilities of each state being $p_1,p_2,\cdots$, then the density matrix is:

$$\rho = \sum\limits_j p_j \vec{x}_j \vec{x}_j^\dagger$$

(note the order: $\vec{x}_j \vec{x}_j^\dagger$ is a $2\times2$ projection matrix). Such matrices are readily seen to be Hermitian (i.e. $\rho = \rho^\dagger$)

So our $2\times 2$ mixed quantum light state is now represented as a general $2\times2$ Hermitian (i.e. $H = H^\dagger$) matrix:

$$\rho = \sum\limits_{j=0}^3 s_j \sigma_j$$

where $\sigma_0 = {\rm id}$ is the $2\times 2$ identity matrix and $\sigma_j$ are the Pauli spin matrices. The co-efficients $s_j$ are nothing but the Stokes vector. Any $2\times2$ Hermitian matrix can be written like this.

Now, if the light passes through a lossless component, so that its Jones matrix $U$ is unitary $U U^\dagger = U^\dagger U = {\rm id}$, then the density matrix becomes:

$$\rho^\prime = U \rho U^\dagger = s_0 {\rm id} + \sum\limits_{j=1}^3 s_j U \sigma_j U^\dagger$$

and the length of the "polarised" part of the light $(s_1, s_2, s_3)$ does not change. $s_0$ on the one hand and $(s_1, s_2, s_3)$ on the other stay separate and do not mix. The unitary matrix scrambles the $(s_1, s_2, s_3)$ but leaves their sum of squares constant and indeed, if we look only at the $(s_1, s_2, s_3)$ we are witnessing the group $SU(2)$ of unitary Jones matrices acting on the three dimensional Lie algebra $i\sigma_1, \sigma_2, \sigma_3)$ of $SU(2)$ through the Adjoint representation $SO(3)$ of $SU(2)$ - in everyday language we are seeing rotations of the Poincaré sphere.

However, if our optical component is not lossless, then the transformation $U$ is simply a general $2\times2$ Hermitian matrix and the $s_0$ and $(s_1, s_2, s_3)$ are mixed in a more general linear transformation. You can, if you like, still use your Jones matrices, but you must use them not acting on a state, but acting on the density matrix: i.e. instead of your pure state $x$ transforming like $x\mapsto U x$, your density matrix transforms by a so called spinor map $\rho\mapsto U \rho U^\dagger$.

Another way of doing this is simply to note that in the map $\rho\mapsto U \rho U^\dagger$, the four parameters $(s_0,s_1, s_2, s_3)$ defining the density matrix undergo linear transformations. So instead of spinor maps, we can use a $4\times4$ matrix to represent a general optical component. This of course is the Mueller matrix. For an optical component with general, nonunitary Jones matrix $U$, the corresponding elements of the Mueller matrix $M$ are:

$$M_{j\,k} = {\rm Tr}\left(\sigma_j^\dagger U \sigma_k U^\dagger\right)$$

The Mueller matrix acts on vectors in the linear space of $2\times2$ Hermitian matrices thought of as a vector space over $\mathbb{R}$. This space comes with an inner product for finding components of "vectors" the Killing form $\left<A,B\right> = {\rm Tr}(A^\dagger B) = {\rm Tr}(A B)$, which is how I wrote the expression above dowm. The Stokes vector is simply the density matrix living in this space but written as a $4\times 1$ real valued column and the Mueller matrix implements the linear spinor map on the rewritten density matrix.

More generally, the Mueller calculus is simply another way of calculating the transformations wrought on a density matrix for any finite dimensional quantum system by various operations, which can include unitary operators or Wigner-Friend kind conversion of pure states to mixed ones. Every $N$ dimensional quantum system implies an $N^2 \times N^2$ dimensional Mueller calculus when the density matrices are written as columns. Here the "basis vectors" are the matrices $\left|\left.x_j\right.\right>\left<\left.x_k\right.\right|$ where $x_j$ are the base quantum pure states. The $N^2 \times N^2$ Mueller matrix operates on the vector of co-efficients $\rho_{j,k}$ in the density matrix $\sum\limits_j\sum\limits_k \rho_{j,k}\left|\left.x_j\right.\right>\left<\left.x_k\right.\right|$.

Footnote: As Trimok has pointed out (thanks Trimok) the standard numbering of the Pauli matrices gives a reordering of the OP's Stokes parameters:

... with the OP conventions, you have the correspondence $s_1 \to s_z, s_2 \to s_x, s_3 \to s_y$ with $\rho = s_0\sigma_0 + s_x \sigma_x +s_y \sigma_y +s_z\sigma_z$

share|improve this answer
    
it's fucking awesome finding people willing to answer with these type of thoroughness. I'll read and I´ll tell –  silvrfück Jan 3 at 1:01
    
+1: Minor remark: with the OP conventions, you have the correspondence $s_1 \to s_z, s_2 \to s_x, s_3 \to s_y$ with $\rho = s_0\sigma_0 + s_x \sigma_x +s_y \sigma_y +s_z\sigma_z$ –  Trimok Jan 3 at 13:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.