Quantum theory

1.2One photon at a time

Follows §1-2 of Peres. About 15 minutes. You need: Sine and cosine.

The experiment in this lesson needs a lamp, a sheet of the plastic used in polarising sunglasses, and a clear crystal of calcite. It can be done on a classroom projector. It is also enough, once you accept that light comes in photons, to end the idea that the same cause always has the same effect.

Two spots of light

Light is a wave in the electric field, and the field swings from side to side, across the direction the light is travelling. The line along which it swings is called the polarisation. Light from a lamp is a jumble of every direction at once. A polariser tidies it up: it lets through only the part of the swing that lies along one line, its axis, so what comes out is polarised along that axis.

Calcite does something odder. It sends light two different ways depending on its polarisation: the horizontal part of the swing goes one way, the vertical part another. Look at a printed page through a lump of calcite and you see every letter twice. Put a screen behind it and you get two spots of light.

Now turn the polariser, so that the light reaching the crystal swings along a line tilted at an angle α\alpha from the horizontal. A tilted swing is part horizontal and part vertical. If the swing has size EE, the horizontal part has size Ecos⁡αE\cos\alpha and the vertical part Esin⁡αE\sin\alpha. The brightness of a beam goes as the square of the field, so the two spots have brightness

upper spot  ∝  cos⁡2α,lower spot  ∝  sin⁡2α.\text{upper spot} \;\propto\; \cos^2\alpha, \qquad \text{lower spot} \;\propto\; \sin^2\alpha .

Since cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1 whatever the angle, the two spots always share out all of the light between them. Nothing is lost in the crystal.

lamppolarisercalcite+75%−25%
upper spot, cos²α75.0%
lower spot, sin²α25.0%
polariser angle α30°
Figure 1. The bench in a bright beam. Drag the dial to turn the polariser. At 0° all the light goes to the upper spot, at 90° all of it goes to the lower one, and at 45° they are equal. The small disc above the polariser shows its axis face-on.
Refresher: the parts of a tilted arrow, and why brightness is a square

Draw an arrow of length 1 at an angle α\alpha above the horizontal. Its shadow on the horizontal axis has length cos⁡α\cos\alpha, and its shadow on the vertical axis has length sin⁡α\sin\alpha. These are its components. Pythagoras's theorem for the right-angled triangle they form says cos⁡2α+sin⁡2α=1\cos^2\alpha + \sin^2\alpha = 1.

The energy carried by any wave goes as the square of its size: a water wave twice as high carries four times the energy. So a beam whose field is cos⁡α\cos\alpha times the original is cos⁡2α\cos^2\alpha times as bright. At 30°, cos⁡30∘≈0.87\cos 30^\circ \approx 0.87, and the upper spot gets 0.872≈0.750.87^2 \approx 0.75 of the light.

So far this is nineteenth-century physics, and it is entirely satisfactory. The rule for the brightness of a beam that has passed a polariser, cos⁡2α\cos^2\alpha, has been in textbooks since 1809.

Now turn the lamp down

The previous lesson ended with the evidence that light arrives in lumps. Light of frequency ν\nu comes in photons, each carrying energy hνh\nu, and a detector either registers a whole one or registers nothing. So make the lamp so dim, and the detectors so quick, that photons can be counted one at a time, and ask the obvious question.

Predict first

One photon, polarised at 30°, reaches the crystal. In a bright beam three quarters of the light would go up and a quarter down. What do the two detectors record for this one photon?

lamppolarisercalcite+0−0
The printout appears here as the detectors click.
photons counted0
fraction that went +–
cos²α75.0%what a bright beam shows
running fraction, settling as photons arrive
polariser angle α30°
4
Figure 2. The same bench with the lamp turned right down. Each dot is one photon; the printout records which detector it reached. The dashed lines are only where a bright beam would go. Let it run, then send a thousand at once and watch the fraction settle on cos⁡2α\cos^2\alpha. Change the angle and it settles somewhere else.

Two things are on show here, and they pull in opposite directions. Over thousands of photons the fraction going up is cos⁡2α\cos^2\alpha, exactly the old brightness rule, as it must be: a bright beam is nothing but a great many photons. But the printout itself, the order of the pluses and minuses, has no pattern at all.

Can the next click be predicted?

Perhaps there is a pattern and you have not found it. The way to find out is to bet on it.

polarised at 30°+−
Call the first photon to begin. There are 20.
photon0 / 20
called correctly0
always saying plus expects15of 20, on average
Figure 3. The polariser is fixed at 30°, so each photon goes up with probability 0.75. Call each one before it arrives. The best anyone has managed is to say “plus” every time and be right three times in four.

Every photon in that run was prepared in the same way: same lamp, same polariser at the same angle, same pinhole. The crystal and the detectors did not change either. And still some went up and some went down. There is nothing we know how to control, or even to look at, that says which way the next one will go.

This is the break with classical physics, and it is a deeper one than the fuzziness of the last lesson. Classical physics rests on determinism: set up the same conditions exactly and you get the same result. Here the same conditions give different results. What is reproducible is not the outcome but the fraction of each outcome over many trials.

That fraction is what we shall mean by probability throughout the course: prepare the same thing again and again, and the probability of an outcome is the fraction of trials in which it happens, in the limit of very many trials. The theory we are building predicts those fractions and nothing else.

A fair objection: maybe each photon carries some hidden detail that decides its fate, and the lamp simply hands out that detail at random. For polarised photons on their own, such a story can be made to work. Later in the book (chapter 6, Bell’s theorem) it turns out that any story of this kind has to pay a price most physicists find worse than randomness. For now, take the randomness at face value.

A filter is also a preparation

There is one more thing to learn from this apparatus, and it will matter in every later lesson. Put two polarisers in a row with their axes at right angles. The first lets through light polarised along 0°. The second passes the fraction cos⁡290∘=0\cos^2 90^\circ = 0 of that. The pair is black.

Now slide a third polariser between them, at 45°. It seems this can only make things darker. It does the opposite: light comes through.

0°45°50%90°25%
light that gets through25.0%
each step turns by45°
and passes cos² of that50.0%
1
Figure 4. Light enters from the left, already polarised at 0°. The last polariser is fixed at 90°. Add polarisers in between, each turned a little further than the one before. With one in the middle a quarter of the light survives; with many, nearly all of it does, and it comes out turned through a right angle.

Think of it one photon at a time. A photon polarised at 0° meets the middle filter at 45° and passes with probability cos⁡245∘=12\cos^2 45^\circ = \tfrac12. If it passes, it is now a photon polarised at 45°: it behaves in every later experiment exactly like a photon that a lamp and a 45° polariser had prepared. The filter has not merely let it through; it has prepared it afresh. At the last filter, 45° away again, it passes with probability a half once more. Half of a half is a quarter.

With NN equal steps of 90∘/N90^\circ/N the fraction that survives every one is

[cos⁡2 ⁣(90∘N)]N,\left[\cos^2\!\left(\frac{90^\circ}{N}\right)\right]^{N},

which is 0.25 for two steps, 0.42 for three, 0.78 for ten and 0.97 for ninety. A photon asked a long series of questions, each only slightly different from the last, almost always says yes to all of them, and is led gently round to a polarisation that it would never have reached in one jump. Peres sets this as the first exercise in the book, with crystals in place of filters.

What to keep

  • In a bright beam, a polariser at angle α\alpha to the crystal splits the light into fractions cos⁡2α\cos^2\alpha and sin⁡2α\sin^2\alpha.
  • A photon never splits. Each one goes wholly up or wholly down.
  • Identically prepared photons do different things. Only the probability, cos⁡2α\cos^2\alpha, is fixed by the preparation.
  • Probability means the fraction of trials, over very many trials of the same preparation.
  • A photon that passes a filter leaves it polarised along that filter’s axis, whatever it was before.

Check yourself

5 questions. Each can be answered once, and the reasoning is shown either way. Work them out before you click: guessing teaches nothing.

1The polariser is set at 60° to the direction that sends photons up. Out of 10,000 photons, roughly how many reach the upper detector?

photons

2With the lamp turned down, the lower beam is the weaker one. In what sense is it weaker?

3The polariser is at 45°. Twenty photons have gone through and the last five all went up. What is the probability that the next one goes up?

4Two polarisers are crossed, at 0° and 90°, and pass nothing. A third is slipped between them at 30° to the first. What fraction of the light leaving the first polariser now gets out of the last?

5Which sentence states what this experiment shows?