1.3Superposition is not a mixture

In the last lesson the crystal was thick enough to pull its two beams right apart, one polarised horizontally and one vertically. Make the crystal thinner and the beams overlap. The question for this lesson is simple to ask: what kind of photons are in the overlap?

Two beams in the same place

Start with the wave picture, where the answer is not in doubt. Where the beams overlap, the electric field is the sum of two swings: one side to side, of size $E_x$, and one up and down, of size $E_y$. Both swing at the same frequency. But they need not swing in step. Light travels at slightly different speeds along the two routes through the crystal, so one swing comes out a fraction of a cycle behind the other. That lag is called the phase difference, written $\delta$, and it depends on how thick the crystal is.

Add a side-to-side swing to an up-and-down swing and the tip of the field traces out a shape. Which shape depends on $\delta$.

So there are three cases, and names for them.

• $\delta = 0$: the two swings rise and fall together and the field runs back and forth along a slanted line. This is ordinary linear polarisation, at an angle set by the ratio of $E_y$ to $E_x$. With equal sizes it is at 45°, which we shall call diagonal. A lag of 180° gives the line slanting the other way, anti-diagonal.
• $\delta = \pm 90^\circ$ with equal sizes: the field keeps a constant size and turns steadily, like the hand of a clock. This is circular polarisation. It comes in two kinds, right-handed and left-handed, according to which way it turns.
• Anything else: an ellipse. Elliptical polarisation is the general case, and the other two are its extremes.

Opticians make these to order. A slice of crystal cut to give exactly a quarter-cycle lag is called a quarter-wave plate, and it turns diagonal light into circular light and back again.

So which photons are in the overlap?

Now the photon picture. The beam that went into the crystal was made of photons polarised at some angle. Inside, the horizontal route and the vertical route are both open. Where they come back together, there are two things you might say about the photons.

You might say: each photon took one route or the other, so the overlap contains some horizontal photons and some vertical photons, shuffled together. Call that a mixture.

Or you might say: every photon in the overlap is the same, and its polarisation is the one the wave picture gives, the line or circle or ellipse of Figure 1. Call that a superposition of horizontal and vertical.

There is one case where there can be no argument. Shrink the crystal to nothing. Then there is no lag, $\delta = 0$, and also no crystal: nothing has been done to the light at all. What comes out must be exactly what went in, photons polarised along the original direction. Not a shuffle of horizontal and vertical ones.

You should not accept that kind of argument without an experiment to back it, and there is one. Take equal beams with no lag, so that the wave picture says “diagonal”, and send the light into a second, thick crystal turned through 45°. That crystal sorts diagonal from anti-diagonal.

This is the distinction the whole subject turns on, so here it is once more in a table. Both kinds of light split evenly when tested for horizontal against vertical. They part company when asked a different question.

A photon in a superposition of horizontal and vertical is not secretly one or the other. It is in a single, perfectly definite state of polarisation, as definite as horizontal or vertical themselves. It is just a state for which the horizontal-or-vertical question has no certain answer.

Name the source

Here are four sources. One really does toss a coin for each photon. The other three produce superpositions with different lags. You have three analysers. Work out which source you have been given.

The horizontal-against-vertical analyser gave half and half every time, whatever the source. It is blind to the lag $\delta$. The other two analysers see nothing but the lag: diagonal, anti-diagonal and circular light differ only in $\delta$, and they give completely different counts.

That invisible-then-decisive quantity, a phase, is the first sign of what is coming. In Part II it turns into the reason probabilities fail to add, and in Part III into the reason quantum theory is written with complex numbers.

Every polarisation has an opposite

One more fact from the optics bench, which later lessons lean on. Horizontal and vertical are a pair: a crystal can be set up so that horizontal photons always go up and vertical ones always go down. Diagonal and anti-diagonal are another such pair. So are right and left circular: put a quarter-wave plate in front of the crystal and the arrangement sorts those two with certainty instead.

The same is true of any ellipse. For each polarisation there is exactly one other, its opposite, such that a suitable arrangement of plates and a crystal sends the first kind always up and the second always down. (The opposite of an ellipse is the ellipse of the same shape turned through a right angle and traced the other way round.) Two polarisations related like this are called orthogonal.

So each orthogonal pair is a question that can be put to a photon, with two possible answers. There are infinitely many such questions. A photon that is certain to give one answer to one of them will, in general, give uncertain answers to all the others.

This lesson follows §1-3 of the book. Exercises 1.3 to 1.6 there ask you to design the optics for arbitrary elliptical polarisations; exercise 1.7 is the subject of the next lesson.