Photons

In which I talk about photons and polarisation - an article I've long had in 'draft' form.

Quantum Cryptography (a background)

In this article I hope to illustrate some of the ideas behind the strange topic of Quantum Cryptography, though I won't be discussing cryptography itself, that comes later - just the necessary physics. First we must consider the nature of light (this can be generalised to any particle once we get all quantum mechanical, but let's stick with light for now).

Classically, light can be thought of as a wave. It's a transverse wave meaning that the 'oscillations' of the thing doing the waving are at right angles to the direction that the wave is travelling in. Another example of transverse waves are waves on the surface of water. These oscillations defined a 'plane' in which the waves are oscillating, and this plane can be oriented at any angle. Waves on the surface of water are vertically polarised. Though the plane of polarisation can be any angle, it is convenient to pick two planes which are at 90 degrees to each other. We can express any polarisation by talking about how much of each is present. Hence, we can talk of 'vertical' and 'horizontal' polarization. Here is an applet which demonstrates this.

You can see polaroid filters in action if you have a pair of polaroid glasses (often sold as 'anti-glare'). Find a light shining on a surface such as a desk. You don't want to be 'square on' to the surface, the light should be bouncing at an angle, 45 degrees is a good start. For the most obvious effect, don't use a mirror.

Look at the surface through your polaroid glasses, then rotate them 90 degrees, and keep looking. You should see the glare change in brightness. You will find that polaroid glasses are best at reducing glare from horizontal reflections when held normally. (See: Brewsters' Angle)

If you use your glasses for driving, you may find that you have trouble with the LCD screens on petrol pumps, this is because the LCD screen relies on polarising light!

If you take a polarised filter, this will ensure that all the light which passes through has the same polarisation. Classically, if a particular wave comes in with an amplitude of A, and a plane of polarisation at angle θ to the plane of polarisation, the amount of light which emerges has amplitude Acosθ. Suppose that we have two polaroid filters. Unpolarised light hits the first and emerges polarised. It emerges with amplitude, A (on average). This light hits the second filter. The two filters have an angle θ between their planes of polarisation - the amount of light which emerges is Acosθ. So, if the filters are aligned, the second filter has no effect. If it is turned 90 degrees, no light emerges (note, if it is turned 180 degrees, it has no effect - the sign of the amplitude doesn't matter, it's not 'negative light'!)

(Note that for real filters, there is a little scattering, so 90 degrees doesn't give total black, and zero degrees does give some reduction in intensity)

Imagine we have two filters, aligned at 90 degrees. No light emerges. This is because the cosine of 90 degrees is zero.

Now, insert a filter at 45 degrees between the two. What happens? More 'stuff' can only make the amount of light getting through smaller, right? The cunning reader will have assumed that I wouldn't ask the question if the answer were obvious. Some light emerges. In this circumstance, two filters allows through less light than three.

This counterintuitive result is easily explained. Imagine the second filter is at an angle of θ compared to the first.  The third is at 90 degrees. In other words, the angle from the second is (90-θ). From the first filter, we have light with amplitude Acosθ. This is then reduced by the third filter by cos(90-θ). The overall light intensity is now Acosθ.cos(90-θ) or Asinθcosθ, this reduces to A(sin2θ)/2. In other words, we get most light out when sin2θ=1, or when 2θ=90°, or when θ=45°

The newly inserted second filter is changing the polarisation of the light.

Take your time on polarisation, it's important that you understand the above if you're to comprehend subsequent articles. We'll put this aside for a while, though - the next step is to talk about photons.