Say we are representing the polarisation of the electric field as a two-element complex Jones’ vector where the elements describe the horizontal and vertical linear polarisation components. We can describe the action of a horizontal polariser by the matrix
\[\begin{equation*}\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}\end{equation*}\]
First, let’s make sure this makes sense. Horizontally polarised light should pass straight through and vertically polarised light should be blocked:
\[\begin{equation*}\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
=
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}
\;\;\;\;
\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}
\begin{pmatrix}
0 \\
1 \\
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
\end{pmatrix}\end{equation*}\]
Diagonally polarised light will let through at half the intensity:
\[\begin{equation*}\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}
\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}
=\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 \\
0 \\
\end{pmatrix}\end{equation*}\]
and
\(\left(1\left/\sqrt{2}\right.\right)^2=1/2\) as expected.
To extend this description to a linear polariser at an arbitrary angle \(\theta\), we imagine first rotating the light by \(-\theta\), passing through a horizontal polariser, then rotating back by \(\theta\). Each of these operations can be represented by a matrix, rotation of a 2D vector by \(\theta\) anti-clockwise is given by
\[\begin{equation*}R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}.\end{equation*}\]
(see
rotation-matrices). So the entire sequence, ordered right to left is
\[\begin{equation*}P_{\theta }=R(\theta )
\begin{pmatrix}
1 & 0 \\
0 & 0 \\
\end{pmatrix}
R(-\theta )=
\begin{pmatrix}
\cos ^2\theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^2\theta \\
\end{pmatrix}\end{equation*}\]
In particular horizontal, vertical, diagonal and anti-diagonal polarisers have the form:
\[\begin{align*}P_{H} &= P_{0} = \begin{pmatrix} 1&0\\0&0 \end{pmatrix}\;\; &
P_{V} &= P_{\pi/2} = \begin{pmatrix} 1&0\\0&0 \end{pmatrix} \\
P_{D} &= P_{\pi/4} = \frac{1}{2}\begin{pmatrix} 1&1\\1&1 \end{pmatrix}\;\; &
P_{A} &= P_{-\pi/2} = \frac{1}{2}\begin{pmatrix} 1&-1\\-1&1 \end{pmatrix}\end{align*}\]
1 Malus’ Law
Now say we have horizontally polarised light and it passes through a linear polariser set at angle \(\theta\), the result is:
\[P_{\theta }\left(
\begin{array}{c}
1 \\
0 \\
\end{array}
\right)=\left(
\begin{array}{c}
\cos ^2\theta \\
\cos\theta \sin\theta \\
\end{array}
\right)\]
If we look at the intensity of the transmitted light, we recover Malus’ Law:
\[\left(
\begin{array}{c}
\cos ^2\theta \\
\cos \theta \sin \theta \\
\end{array}
\right)^{\dagger }\left(
\begin{array}{c}
\cos ^2\theta \\
\cos \theta \sin \theta \\
\end{array}
\right)=\cos ^2\theta \]
2 Crossed Polarisers
Similarly, if we cross two linear polarisers none of the light gets through
\[P_{\theta +\pi /2}P_{\theta }=\left(
\begin{array}{cc}
\sin ^2\theta & -\cos \theta \sin \theta \\
-\cos \theta \sin \theta & \cos ^2\theta \\
\end{array}
\right)\left(
\begin{array}{cc}
\cos ^2\theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^2\theta \\
\end{array}
\right)=\left(
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
\end{array}
\right)\]
but then if we insert a polariser in-between the crossed ones orientated at a diagonal to both, we suddenly start transmitting light again. That is, the operation becomes
\[P_{\theta +\pi /2}P_{\theta +\pi /4}P_{\theta }=\left(
\begin{array}{cc}
-\frac{1}{4} \sin (2 \theta ) & -\frac{1}{2} \sin ^2(\theta ) \\
\frac{\cos ^2(\theta )}{2} & \frac{1}{4} \sin (2 \theta ) \\
\end{array}
\right).\]
and it’s easy to see that this would not produce a zero matrix.
3 As projectors
An alternative way of viewing polarisers is to consider linear polarised light orientated at an angle \(\theta\):
\[|\theta\rangle =R(\theta )\left(
\begin{array}{c}
1 \\
0 \\
\end{array}
\right)=\left(
\begin{array}{c}
\cos \theta \\
\sin \theta \\
\end{array}
\right),\]
where we have written the polarisation vector using the
Dirac braket notation. Now, if we form the following outer-product
\(|\theta \rangle\langle \theta| \), something that is known as a projector (yes I’m sneakily teaching you quantum mechanics), then the result is exactly the polariser orientated at angle
\(\theta\),
\(P_\theta\):
\[|\theta \rangle\langle \theta| =\left(
\begin{array}{c}
\cos \theta \\
\sin \theta \\
\end{array}
\right)^{\dagger }\left(
\begin{array}{c}
\cos \theta \\
\sin \theta \\
\end{array}
\right)=\left(
\begin{array}{cc}
\cos ^2\theta & \cos \theta \sin \theta \\
\cos \theta \sin \theta & \sin ^2\theta \\
\end{array}
\right)\]
Physically constructing polarisers for circular polarisation is difficult. It turns out it’s much easier to use a quarter-wave plate followed by a linear polariser.