The Jones and Mueller Matrices

The Jones and Mueller Matrices

Generally speaking there are three types of devices that will transmit, but modify the polarization state of an incident wave. They are:

1. Polarizer: transmits a single polarization state and removes others
   
2. Phase retarder: Introduces a phase difference between wave components
   
3. Rotator: Rotates the direction of linearly polarized light

As discussed above, we will write these operations as matrix transforms such that

$\overset{\rightharpoonup }{E}'=\overleftrightarrow{M}\overset{\rightharpoonup }{E}$

where $\overleftrightarrow{M}$ is the Jones matrix, $\overset{\rightharpoonup }{E}$ is the input electric field vector, and $\overset{\rightharpoonup }{E}'$ is the modified electric field vector. The first element we describe is the linear polarizer. The Jones matrix for a linear polarizer aligned with the x-y axes is

$\overset{\rightharpoonup }{E}'=\left( \begin{array}{cc} p_x & 0 \\ 0 & p_y \\ \end{array} \right)\overset{\rightharpoonup }{E}$

where $0\leq p_j\leq 1$, $j=x,y$. Then, for complete transmission along the x-axis, $p_y=0$ and $p_x=1$, or

$\left( \begin{array}{c} E_x^{\prime } \\ E_y^{\prime } \\ \end{array} \right)=\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} \right)\left( \begin{array}{c} E_{0,x} \\ E_{0,y} \\ \end{array} \right)=\left( \begin{array}{c} E_{0,x} \\ 0 \\ \end{array} \right)$

and similarly for complete transmission along the y-axis, $p_x=0$ and $p_y=1$, or

$\left( \begin{array}{c} E_x^{\prime } \\ E_y^{\prime } \\ \end{array} \right)=\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right)\left( \begin{array}{c} E_{0,x} \\ E_{0,y} \\ \end{array} \right)=\left( \begin{array}{c} 0 \\ E_{0,y} \\ \end{array} \right)$

The phase retarder produces a total relative phase shift, $\varphi$, between the x- and y-components of the input wave. That means we would write

$\overset{\rightharpoonup }{E}'=\left( \begin{array}{cc} e^{i \varphi /2} & 0 \\ 0 & e^{-i \varphi /2} \\ \end{array} \right)\overset{\rightharpoonup }{E}$

or, just to be concrete

$\left( \begin{array}{c} E_x^{\prime } \\ E_y^{\prime } \\ \end{array} \right)=\left( \begin{array}{cc} e^{i \varphi /2} & 0 \\ 0 & e^{-i \varphi /2} \\ \end{array} \right)\left( \begin{array}{c} E_{0,x} \\ E_{0,y} \\ \end{array} \right)=\left( \begin{array}{c} E_{0,x}e^{i \varphi /2} \\ E_{0,y}e^{-i \varphi /2} \\ \end{array} \right)$

and

$\overset{\rightharpoonup }{E}'=E_{0,x}e^{i \varphi /2}\hat{i}+E_{0,y}e^{-i \varphi /2}\hat{j}=e^{i \varphi /2}\left(E_{0,x}\hat{i}+E_{0,y}e^{-i \varphi }\hat{j}\right)$

where the last equality explicitly shows that $E_y$ is phase advanced from $E_x$ by $\varphi$.

Quarter Wave Plate

The Jones matrix representing a quarter wave plate is given when $\varphi =\pi /2$ (remember, π/2 is a quarter of the full wave, which is 2π)

$\overleftrightarrow{M}_{\text{QWP}}=\left( \begin{array}{cc} e^{i \pi /4} & 0 \\ 0 & e^{-i \pi /4} \\ \end{array} \right)=e^{i \pi /4}\left( \begin{array}{cc} 1 & 0 \\ 0 & e^{-i \pi /2} \\ \end{array} \right)=e^{i \pi /4}\left( \begin{array}{cc} 1 & 0 \\ 0 & -i \\ \end{array} \right)$

or simply

$\overleftrightarrow{M}_{\text{QWP}}=e^{i \pi /4} \left( \begin{array}{cc} 1 & 0 \\ 0 & -i \\ \end{array} \right)$

Half Wave Plate

The Jones matrix representing a half wave plate is given when $\varphi=\pi$ (remember, π is a half of the full wave, which is 2π)

$\overleftrightarrow{M}_{\text{HWP}}=\left( \begin{array}{cc} e^{i \pi /2} & 0 \\ 0 & e^{-i \pi /2} \\ \end{array} \right)=e^{i \pi /2}\left( \begin{array}{cc} 1 & 0 \\ 0 & e^{-i \pi } \\ \end{array} \right)=i \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$

Rotated Wave Retarders

All of the matrices we have derived just now have been aligned with the x- and y-axes.

It turns out that if we want to rotate a vector counter-clockwise (when the axis about which the vector is being rotated is pointing toward you, $\hat{z}$ in these examples) you simply multiply the vector by the 2D rotation matrix:

$\overleftrightarrow{R}(\theta )=\left( \begin{array}{cc} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)$

see the figure below

For instance, if we started with a vector aligned with the x-axis and wanted to rotate it by that in the counter-clockwise direction, we would have

$\overset{\rightharpoonup }{v}'=\left( \begin{array}{cc} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)\overset{\rightharpoonup }{v}$

or, in this particular example

$\left( \begin{array}{c} \cos (\theta ) \\ \sin (\theta ) \\ \end{array} \right)=\left( \begin{array}{cc} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$

writing this in unit vector notation

$\overset{\rightharpoonup }{v}'=\cos (\theta )\hat{i}+\sin (\theta )\hat{j}$

which is simply a vector that points anywhere in the plane with magnitude 1.

Now, if we took that same vector and rotated it by NEGATIVE θ, we would get the figure on the left below here

In math, we would write

$\overset{\rightharpoonup }{v}'=\overleftrightarrow{R}(-\theta )\overset{\rightharpoonup }{v}=\left( \begin{array}{cc} \cos (\theta ) & \sin (\theta ) \\ -\sin (\theta ) & \cos (\theta ) \\ \end{array} \right)\overset{\rightharpoonup }{v}$

This turns out to be the same thing as rotating the AXES by a positive angle as you can see in the second figure on the right above. So if we want to rotate AXES we multiply by $\overleftrightarrow{R}(-\theta )$.

Now, if we want to 'rotate' a matrix so that the axes of the matrix are no longer aligned with 'x' and 'y', we write

$\overleftrightarrow{M}'=\overleftrightarrow{R}(\theta )\overleftrightarrow{M}\overleftrightarrow{R}(-\theta )$

So for instance, if we want to write the Jones matrix for a general phase retarder, which is

$\overleftrightarrow{M}_{\varphi }=\left( \begin{array}{cc} e^{i \varphi /2} & 0 \\ 0 & e^{-i \varphi /2} \\ \end{array} \right)$

in some rotated axes. We would write

$\overleftrightarrow{M}_{\varphi }^{\prime }=\overleftrightarrow{R}(\theta )\overleftrightarrow{M}_{\varphi }\overleftrightarrow{R}(-\theta )=\left( \begin{array}{cc} \cos (\theta ) & -\sin (\theta ) \\ \sin (\theta ) & \cos (\theta ) \\ \end{array} \right)\left( \begin{array}{cc} e^{i \varphi /2} & 0 \\ 0 & e^{-i \varphi /2} \\ \end{array} \right)\left( \begin{array}{cc} \cos (\theta ) & \sin (\theta ) \\ -\sin (\theta ) & \cos (\theta ) \\ \end{array} \right)$

multiplying these products out, we get

$\overleftrightarrow{M}_{\varphi }^{\prime }=\left( \begin{array}{cc} e^{-\frac{i \varphi }{2}} \left(\sin ^2(\theta )+e^{i \varphi } \cos ^2(\theta )\right) & i \sin (2 \theta ) \sin \left(\frac{\varphi }{2}\right) \\ i \sin (2 \theta ) \sin \left(\frac{\varphi }{2}\right) & e^{-\frac{i \varphi }{2}} \left(\cos ^2(\theta )+e^{i \varphi } \sin ^2(\theta )\right) \\ \end{array} \right)$

and simplified using Euler's form for sine, $\sin (\varphi /2)=\frac{1}{2i}\left(e^{i \varphi /2}-e^{-i \varphi /2}\right)$ and factoring out $e^{i \varphi /2}$, we have

$\overleftrightarrow{M}_{\varphi }^{\prime }=e^{i \varphi /2} \left( \begin{array}{cc} \cos ^2(\theta )+e^{-i \varphi } \sin ^2(\theta ) & \left(1-e^{-i \varphi }\right)\sin (2 \theta ) \frac{1}{2} \\ \left(1-e^{-i \varphi }\right)\sin (2 \theta ) \frac{1}{2} & e^{-i \varphi }\cos ^2(\theta )+\sin ^2(\theta ) \\ \end{array} \right)$

or, getting rid of the factors of 1/2 with a trig. identity Links to an external site.

$\overleftrightarrow{M}_{\varphi }^{\prime }=e^{i \varphi /2} \left( \begin{array}{cc} \cos ^2(\theta )+e^{-i \varphi } \sin ^2(\theta ) & \left(1-e^{-i \varphi }\right)\sin (\theta )\cos (\theta ) \\ \left(1-e^{-i \varphi }\right)\sin (\theta )\cos (\theta ) & e^{-i \varphi }\cos ^2(\theta )+\sin ^2(\theta ) \\ \end{array} \right)$

Again, we can use this to write the Jones matrix for a quarter wave plate i.e., when $\varphi =\pi /2$

$\overleftrightarrow{M}_{\text{QWP}}^{\prime }(\theta )=e^{i \pi /4} \left( \begin{array}{cc} \cos ^2(\theta )+e^{-i \pi /2} \sin ^2(\theta ) & \left(1-e^{-i \pi /2}\right)\sin (\theta )\cos (\theta ) \\ \left(1-e^{-i \pi /2}\right)\sin (\theta )\cos (\theta ) & e^{-i \pi /2}\cos ^2(\theta )+\sin ^2(\theta ) \\ \end{array} \right)$

simplified using $e^{-i \pi /2}=-i$

$\overleftrightarrow{M}_{\text{QWP}}^{\prime }(\theta )=e^{i \pi /4} \left( \begin{array}{cc} \cos ^2(\theta )-i \sin ^2(\theta ) & (1+i)\sin (\theta )\cos (\theta ) \\ (1+i)\sin (\theta )\cos (\theta ) & \sin ^2(\theta )-i \cos ^2(\theta ) \\ \end{array} \right)$

Let's test to make sure this returns what we got before when θ=0

$\overleftrightarrow{M}_{\text{QWP}}(0)=e^{i \pi /4} \left( \begin{array}{cc} 1 & 0 \\ 0 & e^{-i \pi /2} \\ \end{array} \right)=e^{i \pi /4} \left( \begin{array}{cc} 1 & 0 \\ 0 & -i \\ \end{array} \right)$

which is indeed what we derived before. And when $\theta =\pi /2$

$\overleftrightarrow{M}_{\text{QWP}}^{\prime }(\pi/2)=e^{i \pi /4} \left( \begin{array}{cc} -i & 0 \\ 0 & 1 \\ \end{array} \right)$

We get a quarter-wave plate with the fast axis horizontal, which is the original QWP rotated by 90^o. Then the half-wave plate is given when $\varphi =\pi$

$\overleftrightarrow{M}_{\text{HWP}}^{\prime }(\theta )=e^{i \pi /2} \left( \begin{array}{cc} \cos ^2(\theta )+e^{-i \pi } \sin ^2(\theta ) & \left(1-e^{-i \pi }\right)\sin (\theta )\cos (\theta ) \\ \left(1-e^{-i \pi }\right)\sin (\theta )\cos (\theta ) & e^{-i \pi }\cos ^2(\theta )+\sin ^2(\theta ) \\ \end{array} \right)$

simplified using $e^{-i \pi }=-1$

$\overleftrightarrow{M}_{\text{HWP}}^{\prime }(\theta )=e^{i \pi /2} \left( \begin{array}{cc} \cos ^2(\theta )- \sin ^2(\theta ) & 2\sin (\theta )\cos (\theta ) \\ 2\sin (\theta )\cos (\theta ) & \sin ^2(\theta )-\cos ^2(\theta ) \\ \end{array} \right)=e^{i \pi /2} \left( \begin{array}{cc} \cos (2\theta ) & \sin (2\theta ) \\ \sin (2\theta ) & -\cos (2\theta ) \\ \end{array} \right)$

and again, when $\theta =0$ and using $e^{i \pi /2}=i$, we get

$\overleftrightarrow{M}_{\text{HWP}}^{\prime }(0)=i \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$

More generally speaking, when we apply a HWP to an arbitrary input beam

$\overleftrightarrow{M}_{\text{HWP}}^{\prime }(\alpha )\overset{\rightharpoonup }{E}(\theta )=e^{i \pi /2} \left( \begin{array}{cc} \cos (2\alpha ) & \sin (2\alpha ) \\ \sin (2\alpha ) & -\cos (2\alpha ) \\ \end{array} \right)\left( \begin{array}{c} \cos (\theta ) \\ \sin (\theta ) \\ \end{array} \right)=e^{i \pi /2}\left( \begin{array}{c} \cos (2 \alpha -\theta ) \\ \sin (2 \alpha -\theta ) \\ \end{array} \right)$

we get a π phase shift of the y-component relative to the x-component of the wave. Also, note that for instance, when θ=0

$\overleftrightarrow{M}_{\text{HWP}}^{\prime }(\alpha )\overset{\rightharpoonup }{E}(0)=e^{i \pi /2}\left( \begin{array}{c} \cos (2 \alpha ) \\ \sin (2 \alpha ) \\ \end{array} \right)$

the output polarization angle is rotated by TWICE the angle that the fast axis of the HWP makes with the x-axis. So, if you rotate a HWP by 10^o, the output polarization (assuming it wasn't aligned with either the fast or slow axis) will be rotated by 20^o.

Finally, applying the same formalism to a polarizer with transmission axis aligned with the x-axis, we get

$\overleftrightarrow{M}_{\text{LP}}^{\prime }(\theta )=\left( \begin{array}{cc} \cos ^2(\theta ) p_x & \sin (\theta ) \cos (\theta ) p_x \\ \sin (\theta ) \cos (\theta ) p_x & \sin ^2(\theta ) p_x \\ \end{array} \right)$

Examples

Now it turns out that the real power of the Jones and Mueller (shown later) matrices is the fact that they can be used in series to model a complicated optical system.

Do you need to model an optical system composed of a linear polarizer with transmission axis +45^o to the x-axis, quarter-wave plate with fast axis on the y-axis, and ANOTHER quarter-wave plate with fast axis on the y-axis? If the incident wave is polarized along the x-axis, just multiply those three Jones matrices together in the following way

$\overset{\rightharpoonup }{E}_{\text{out}}=\frac{e^{i \pi /2}}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & -i \\ \end{array} \right)\left( \begin{array}{cc} 1 & 0 \\ 0 & -i \\ \end{array} \right)\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right)\overset{\rightharpoonup }{E}_{\text{in}}$

where $\overset{\rightharpoonup }{E}_{\text{in}}=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$, then we would have

$\overset{\rightharpoonup }{E}_{\text{out}}=\frac{e^{i \pi /2}}{2}\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)$

Let's break down what this systems is doing element by element. The incident wave is polarized along the x-axis:

$\overset{\rightharpoonup }{E}_{\text{in}}=\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$

or

$\overset{\rightharpoonup }{E}_{\text{in}}=E_{0,x}\hat{i}+0\hat{j}$

which is just a linearly polarized wave oscillating in the x-direction. The linear polarizer at +45^o to the x-axis (matrix) multiplied by this gives

$\overset{\rightharpoonup }{E}_{\text{out}}=\frac{1}{2}\left( \begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)=\frac{1}{2}\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)$

or (aside from scaling factors)

$\overset{\rightharpoonup }{E}_{\text{out}}=E_{0,x}\hat{i}+E_{0,y}\hat{j}$

which is just a linearly polarized wave oscillating at +45^o from the x-axis. So the action of a polarizer rotated +45^o from the x-axis on a wave polarized along the x-axis is to transmit half the power but now the electric field oscillates +45^o from the x-axis.

Next we'll tackle the action of the two quarter-wave plates with fast axis aligned with the x-axis:

$\overleftrightarrow{M}_{\text{QWP}}\overleftrightarrow{M}_{\text{QWP}}=\left( \begin{array}{cc} 1 & 0 \\ 0 & -i \\ \end{array} \right)\left( \begin{array}{cc} 1 & 0 \\ 0 & -i \\ \end{array} \right)=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$

which, seems oddly familiar. Remember that

$\overleftrightarrow{M}_{\text{HWP}}(0)=i \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$

Woah!! So two quarter wave plates act like a half-wave plate??! YES they do! So the action of two consecutive quarter-wave plates acting on a wave polarized +45^o to the x-axis should be to rotate the wave so it is now polarized -45^o to the x-axis

$\overset{\rightharpoonup }{E}_{\text{out}}=\overleftrightarrow{M}_{\text{QWP}}\overleftrightarrow{M}_{\text{QWP}}\overset{\rightharpoonup }{E}_{\text{in}}=\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)$

And there you have it

$\overset{\rightharpoonup }{E}_{\text{out}}=\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)$

or

$\overset{\rightharpoonup }{E}_{\text{out}}=E_{0,x}\hat{i}-E_{0,y}\hat{j}$

which is a wave polarized -45^o w.r.t. the x-axis.

A final note on the Jones matrices: Matrix multiplication is non-commutative, that means order matters. Furthermore, the matrices must be multiplied in 'reverse-order' so-to-speak. This just ensure that the input wave 'encounters' the optics in the correct order. For instance, in the example above, the incident beam went through a polarizer, and two quarter wave plates. The order of the matrices was

$\overleftrightarrow{M}_{\text{QWP}}\overleftrightarrow{M}_{\text{QWP}}\overleftrightarrow{M}_{\text{LP}}$

which when read aloud, is backwards from what first intuition would say. However, we multiply the incident electric field vector on the RIGHT to get the output field, so the order above is correct.

The following is a table containing Jones and Mueller matrices. While we did not derive any of the Mueller matrices, they operate on the Stokes vectors the same way the Jones matrices operate on the Jones vectors and have the advantage that they work for partially polarized light, unlike the Jones matrices. However, they do come at the cost of higher complexity.