# 272: Polarized Light and Quantum Mechanics

- Page ID
- 141607

## The Linear Superposition

Unpolarized light consists of photons of all possible polarization angles. A photon polarized at an angle \( \theta\) with respect to the vertical can be written as a linear superposition of a vertically polarized photon, \( | v \rangle\), and a horizontally polarized photon, \( | h \rangle\). \( | v \rangle\) and \( | h \rangle\) are the polarization basis states.

\[ | \theta \rangle = | v \rangle \langle v | \theta \rangle + | h \rangle \langle h | \theta \rangle\]

From the figure above it can be seen that the projection of \( | \theta \rangle\) onto \( | v \rangle\) and \( | h \rangle\) are \( \cos \theta\) and \( \sin \theta\) respectively.

\[ | \theta \rangle = | v \rangle \cos \theta + | h \rangle \sin \theta\]

The probability that a photon polarized at an angle \( \theta\) will pass a vertical polarizer is

\[ | \langle | \theta \rangle | ^2 = \cos^2 \theta\]

By integrating this function over all possible angles we find that half of the incident light passes through a vertical polarizer. See the figure below.

\[ \frac{ \int_{0}^{2 \pi} \cos^2 \theta d \theta}{2 \pi} = 0.5\]

The photons that pass the vertical polarizer are now vertically polarized. That is they are eigenfunctions of that measurement operator.

The probability that a vertically polarized photon will pass a second filter that is vertically polarized is one.

\[ | \langle v | v \rangle |^2 = \cos^2 0^o = 1\]

The probability that a vertically polarized photon will pass a second filter that is horizontally polarized is zero.

\[ | \langle h | v \rangle |^2 = \cos^2 90^o = 0\]

The vertically polarized photon can be written as linear superposition of any other set of orthogonal basis states, for example \( |45^o \rangle\) and \( | -45^o \rangle\).

\[ |v \rangle = |45^o \rangle \langle 45^o | v \rangle + | -45^o \rangle \langle -45^o | v \rangle\]

\[ |v \rangle = |45^o \rangle \cos 45^o + | -45^o \rangle \cos -45^o\]

\[ |v \rangle = |45^o \rangle 0.707 + | -45^o \rangle 0.707 \]

Now if a 45^{o} polarizer is inserted in between the vertical and horizontal polarizers photons get through the horizontal polarizer that stopped them previously (see the last figure).

Here's the explanation. The probability that a vertically polarized photon will get through a polarizer oriented at an angle of 45^{o} is 0.5. See the figure below.

\[ | \langle 45^o | v \rangle |^2 = \cos^2 45^o = 0.5\]

Now the photon is in the state of \( |45^o \rangle\) which can be written as a linear superposition of \( |v \rangle\) and \( \h \rangle\).

\[ |45^o \rangle = |v \rangle \langle v | 45^o \rangle + | h \rangle \langle h | 45^o \rangle\]

\[ |45^o \rangle = |v \rangle \cos 45^o + | h \rangle \sin 45^o\]

\[ |45^o \rangle = |v \rangle 0.707 + | h \rangle 0.707 \]

Thus, the probability that this photon will pass the final horizontally oriented polarizer is

\[ | \langle H | 45^o \rangle |^2 = \sin^2 45^o = 0.5\]

See the figure below.

In this last figure the intensity of the light emerging from the final horizontal polarizing filter can be calculated compactly as

\[ \frac{ \int_{0}^{2 \pi} | \langle h | 45^o \rangle \langle 45^o | v \rangle \langle v | \theta \rangle|^2 d \theta}{2 \pi} = \frac{1}{8}\]

The term inside the integral is the probability that a photon with polarization θ will pass through the three filters. This expression is integrated over all values of θ.