2.4Amplitudes: arrows that add

Follows §2-6 of Peres. About 18 minutes. You need: Lesson 2.3. Complex numbers are introduced here from scratch.

Probabilities do not add when paths cannot be told apart. Yet the results are perfectly regular: a half here, nothing there. Something must be adding. This lesson finds out what, and in doing so sets up the one calculation on which all of quantum theory runs.

The hint from waves

Interference is old news for waves. When two ripples cross on a pond, their heights add. Where crest meets crest the water rises twice as high; where crest meets trough it stays flat. The energy of a wave goes as the square of its height, so two equal ripples that meet crest to crest carry, at that spot, not twice the energy of one but four times, and where they meet crest to trough, none.

The brightness of light is its energy, and in terms of photons brightness is how many arrive, which is to say, the probability that each one arrives. So the hint is this. Probability behaves like the energy of a wave. Behind each probability there should be a quantity that behaves like the height: something that adds when paths combine, and is squared at the very end to give the probability. It is called an amplitude.

One number is not enough to describe the height of a wave, though. Two ripples of the same size can arrive in step, or out of step, or anything between, and what they add up to depends on which. So an amplitude needs two pieces of information: a size, and a phase that says where in its cycle it is. A thing with a size and an angle is an arrow.

Arrows, and how to do arithmetic with them

From here on, an amplitude is an arrow drawn on a flat sheet. It has a length and it points at some angle, measured anticlockwise from “due east”. Only two operations are needed.

Adding. Put the tail of the second arrow at the tip of the first. The sum runs from the start to the finish. Two arrows pointing the same way give one of their combined length. Two equal arrows pointing opposite ways give nothing.

Multiplying. Multiply the lengths and add the angles. So multiplying by an arrow of length 1 at 90° leaves the length alone and turns the other arrow a quarter-turn anticlockwise.

Figure 1. Arrows on a plane. Under “Add”, drag the two arrows and compare the two bottom bars: the sum of the squares, and the square of the sum. Under “Multiply”, watch the lengths multiply and the angles add.

Mathematicians call these arrows complex numbers, and write them in a way that makes the arithmetic mechanical. An arrow that reaches $a$ to the east and $b$ to the north is written $a + ib$ . The symbol $i$ stands for the arrow of length 1 pointing due north, at 90°.

Now apply the multiplication rule to $i \times i$ : the lengths give $1\times1 = 1$ , the angles give $90^\circ + 90^\circ = 180^\circ$ . That is the arrow of length 1 pointing due west, which is the ordinary number $-1$ . So

i^2 = -1 .

That equation has a reputation for mystery. Here it only says that two quarter-turns make a half-turn. Everything else follows by ordinary algebra, treating $i$ as a symbol and replacing $i^2$ by $-1$ wherever it turns up.

Refresher: the notation for complex numbers you will need

For $z = a + ib$ , the number $a$ is called the real part and $b$ the imaginary part. Its length is written $|z|$ , and by Pythagoras $|z|^2 = a^2 + b^2$ .

The mirror image of $z$ in the east–west line is called its conjugate and written with a bar: $\bar z = a - ib$ . It has the same length and the opposite angle. Multiplying a number by its own conjugate gives its length squared: $\bar z z = a^2 + b^2 = |z|^2$ .

An arrow of length 1 at angle $\varphi$ is written $e^{i\varphi}$ , which is shorthand for $\cos\varphi + i\sin\varphi$ . So $e^{i\cdot 0} = 1$ , $e^{i\cdot 90^\circ} = i$ , $e^{i\cdot 180^\circ} = -1$ . An arrow of length $r$ at angle $\varphi$ is $r e^{i\varphi}$ , and the multiplication rule becomes $r_1e^{i\varphi_1}\cdot r_2e^{i\varphi_2} = r_1r_2\,e^{i(\varphi_1+\varphi_2)}$ .

The three rules

The proposal, borrowed from the theory of waves and then tested against experiment, is this.

Every step from a preparation to an outcome has an amplitude, an arrow. The probability of the step is the square of the arrow’s length.
For several steps in succession along one path, multiply the amplitudes.
When several paths lead to the same outcome and nothing records which was taken, add the amplitudes of the paths, and square the length of the total. When something does record which was taken, square each path separately and add the results.

F. Composition of amplitudes. The phases of the amplitudes can be chosen so that, when several paths lead from the initial state to the final outcome and the process leaves no trace of which was taken, the amplitude for the outcome is the sum of the amplitudes of the paths.

Rule 2 costs nothing: squared lengths multiply when arrows multiply, so the probability of a path is still the product of the probabilities of its steps, as it should be. Rule 1 is a definition. All the physics is in rule 3.

The three magnets again

Try the rules on the last lesson’s experiment. From an outcome of the vertical magnet ( $+$ or $-$ ) to an outcome of the horizontal one ( $L$ or $R$ ) every probability is $\tfrac12$ , so every amplitude has length $1/\sqrt2$ . Suppose their angles are such that three of them point east and one points west:

\begin{array}{c|cc} & \text{from } + & \text{from } - \\ \hline \text{to } L & \tfrac{1}{\sqrt2} & \tfrac{1}{\sqrt2} \\[4pt] \text{to } R & \tfrac{1}{\sqrt2} & -\tfrac{1}{\sqrt2} \end{array}

and the same table serves for the return trip from $L$ or $R$ back to $+$ or $-$ . Now switch the middle magnet off, so that the two routes cannot be told apart, and add:

\begin{aligned} + \to + &: \quad \overbrace{\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2}}^{\text{via } L} + \overbrace{\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2}}^{\text{via } R} = \tfrac12 + \tfrac12 = 1, && \text{probability } 1^2 = 1 \\[6pt] + \to - &: \quad \overbrace{\tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2}}^{\text{via } L} + \overbrace{\tfrac{1}{\sqrt2}\cdot\left(-\tfrac{1}{\sqrt2}\right)}^{\text{via } R} = \tfrac12 - \tfrac12 = 0, && \text{probability } 0 \end{aligned}

Every atom that left the first magnet by $+$ leaves the third by $+$ , as repeatability demands. With the middle magnet on, the routes are recorded, each is squared on its own, and you get $\tfrac14 + \tfrac14 = \tfrac12$ for both lines: the eight equal clusters. The whole difference between the two experiments is one minus sign.

An interferometer

The cleanest place to watch amplitudes at work is an instrument built to do nothing else. A half-silvered mirror gives a photon two routes; ordinary mirrors bring the routes together at a second half-silvered mirror; two detectors watch its two exits. A slab of glass in one route delays the light there and so turns that route’s arrow through an angle $\varphi$ of your choosing.

Figure 2. One photon at a time through two routes. The arrow diagram shows the two amplitudes arriving at detector D1 and their sum. Work through the three challenges under the figure.

Here is the sum for detector D1. Each half-silvered mirror contributes a factor of length $1/\sqrt2$ , so each route arrives with an arrow of length $\tfrac12$ . One of them has been turned through $\varphi$ :

\text{amplitude at D1} = \tfrac12 + \tfrac12 e^{i\varphi}, \qquad \text{probability} = \left|\tfrac12 + \tfrac12 e^{i\varphi}\right|^2 = \cos^2\!\frac{\varphi}{2} .

At $\varphi = 0$ the arrows line up and every photon goes to D1. At 180° they cancel and none does. In between, the detector’s share runs smoothly from all to nothing. And when the routes are marked, so that the photon’s path is on record, rule 3 says to square first: $\tfrac14 + \tfrac14 = \tfrac12$ , whatever the glass does.

What the amplitudes must satisfy

The minus sign in the table above was not a lucky guess. Repeatability pins the amplitudes down. Write $C_{\mu m}$ for the amplitude to go from outcome $m$ of one test to outcome $\mu$ of another, and $\Gamma_{n\mu}$ for the amplitude of the return step. With nothing recording $\mu$ , an atom that starts at $m$ must end at $n = m$ and nowhere else:

\sum_\mu \Gamma_{n\mu}\,C_{\mu m} \;=\; \begin{cases} 1 & \text{if } n = m \\ 0 & \text{otherwise.} \end{cases}

Reciprocity (rule D) says that the return step has the same probability as the outward one, so $\Gamma_{m\mu}$ and $C_{\mu m}$ have the same length. The natural way to satisfy both conditions is to make the return amplitude the mirror image of the outward one, $\Gamma_{m\mu} = \bar C_{\mu m}$ . Then the condition reads

\sum_\mu \bar C_{\mu n}\,C_{\mu m} \;=\; \begin{cases} 1 & \text{if } n = m \\ 0 & \text{otherwise.} \end{cases}

A table of complex numbers with this property is called unitary. In words: take any two different columns, multiply them entry by entry (conjugating one), and the products sum to zero; do the same with a column and itself and you get 1. Check it on the table above: the two columns give $\tfrac12 - \tfrac12 = 0$ , and each with itself gives $\tfrac12 + \tfrac12 = 1$ .

If the entries were ordinary real numbers, that property would describe a rotation: the table that converts the coordinates of a point from one set of perpendicular axes to another. So amplitudes behave like the coordinates of something, and changing from one test to another behaves like turning the axes. That is the thread Part III picks up.

Amplitudes come first

One last point, which changes how the whole subject is laid out. We arrived at amplitudes as a device for explaining probabilities. It is tempting to keep thinking of the probabilities as the real thing and the amplitudes as bookkeeping. That gets it the wrong way round.

Suppose there are three tests, with tables of amplitudes $C$ from the first to the second, $\Gamma$ from the second to the third, and $\mathcal C$ from the first directly to the third. If the middle test is left unrecorded, rule 3 gives

\mathcal C_{rm} \;=\; \sum_\mu \Gamma_{r\mu}\,C_{\mu m} ,

the amplitudes of each step multiplied and summed over the hidden middle. Square the lengths and you have a relation between three tables of probabilities that can each be measured separately. But it cannot be written in terms of the probabilities alone: it needs the angles. The amplitudes are the fundamental objects, and probabilities are what you get from them at the last step. From here on the theory is built on amplitudes.

This lesson follows §2-6. The book also works out how far the probabilities determine the angles: completely for two outcomes, by drawing a triangle for three, and not uniquely for four or more, which is one more reason to treat the amplitudes as primary.

What to keep

An amplitude is an arrow, a complex number. Arrows add tip to tail; they multiply by multiplying lengths and adding angles. $i$ is a quarter-turn, and $i^2 = -1$ .
Probability is the squared length of an amplitude.
Along one path, multiply the amplitudes of the steps.
Rule F: for paths that cannot be told apart, add the amplitudes and then square. For paths that can, square and then add.
Repeatability and reciprocity make the table of amplitudes between two tests unitary: its columns are mutually perpendicular and of length 1.

Check yourself

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

1Two paths lead to a detector. Each has an amplitude of length $\tfrac12$ , and the two arrows point the same way. Nothing records which path was taken. What is the probability of a click?

2The same two paths, but now one arrow has been turned through 90° relative to the other. What is the probability of a click?

3What is i × i, and why?

4A path has two steps, with amplitudes $0.6$ and $0.5\,i$ . What is the probability of following the whole path?

5In the interferometer the glass turns one route’s arrow through $\varphi = 60^\circ$ , and the routes are not marked. What is the probability that a photon reaches D1?

6When should you add the probabilities of the paths rather than their amplitudes?