3.2The overlap rule

Every probability met so far, the $\cos^2\alpha$ of the crystal, the tables for pairs of magnets, the $\cos^2(\varphi/2)$ of the interferometer, is a case of one formula. This lesson states it, shows what it implies about how much can be known at once, and uses it to track down an unknown state.

The rule

First a piece of vocabulary. A test for the state $\mathsf v$ is a test that a system prepared in $\mathsf v$ always passes, and a system prepared in any state orthogonal to $\mathsf v$ always fails. A polariser with its axis horizontal is a test for the state “horizontal”.

H. The quantum expectation rule. Let $\mathsf u$ and $\mathsf v$ be normalised vectors. The probability that a system prepared in state $\mathsf u$ passes a test for state $\mathsf v$ is $|\ip{\mathsf u}{\mathsf v}|^2$.

The inner product measures how far one vector lies along the other, and the probability is the square of its size. That is all. Three checks against what you already know:

• For linear polarisations at angles $\alpha$ and $\beta$, the vectors are $(\cos\alpha, \sin\alpha)$ and $(\cos\beta, \sin\beta)$. Their inner product is $\cos\alpha\cos\beta + \sin\alpha\sin\beta = \cos(\alpha - \beta)$, and the probability is $\cos^2$ of the angle between them. That is lesson 1.2.
• Swapping $\mathsf u$ and $\mathsf v$ conjugates the inner product and leaves its size alone. So the probability is the same in both directions. Reciprocity, rule D, which we had to take on trust in lesson 2.2, is now a one-line consequence.
• The Schwarz inequality guarantees that $|\ip{\mathsf u}{\mathsf v}|^2$ lies between 0 and 1.

Same, different, and in between

The rule sorts pairs of states into three kinds.

Parallel. If $\mathsf v$ is $\mathsf u$ multiplied by a number, the probability is 1 in every test either can pass. They are the same state. In particular, multiplying a vector by $e^{i\theta}$, which turns all its components through the same angle, changes nothing that can be observed: $|\ip{\mathsf u}{e^{i\theta}\mathsf v}| = |\ip{\mathsf u}{\mathsf v}|$. So vectors and states are not in one-to-one correspondence. A state is a whole family of vectors differing by an overall factor, and we normally pick a normalised one to stand for the family.

Orthogonal. If $\ip{\mathsf u}{\mathsf v} = 0$, a system prepared in one never passes the test for the other. These are states that can be told apart with certainty, by a single test. They are what “different” means in the everyday sense.

Neither. Most pairs of states are in between, like polarisations at 0° and 30°. Such states are neither the same nor wholly different. No single test can tell them apart for certain, which is the fact lesson 1.4 was built on.

One caution about phases. An overall phase is invisible, but a relative phase is everything. The states $\mathsf u + \mathsf v$ and $\mathsf u + e^{i\theta}\mathsf v$ are different states. With $\mathsf u$ horizontal and $\mathsf v$ vertical, the first is diagonal light and the second, for $\theta = 90^\circ$, is circular.

The rule also lets the superposition principle be strengthened.

G*. The strong superposition principle. Every orthonormal basis represents a complete test that can be carried out.

Rule G said every vector is a possible preparation. This says every set of axes is a possible test. It is what entitles us to speak of “a test for the state $\mathsf v$” for an arbitrary $\mathsf v$: complete $\mathsf v$ to a set of axes, and the test exists.

The sphere of polarisations

The figure drew states as points on a sphere, and it is worth saying why that works. A polarisation vector has two complex components, four real numbers. Normalisation removes one and the invisible overall phase removes another. Two are left, and two numbers locate a point on the surface of a sphere, like latitude and longitude.

On this sphere horizontal and vertical sit at opposite ends of one axis, diagonal and anti-diagonal on a second, right and left circular on the third. Orthogonal states are always at opposite points. And the rule takes a pleasantly simple form: if $\Theta$ is the angle between two points, as seen from the centre,

Points 90° apart on the sphere, such as horizontal and diagonal, or horizontal and right circular, give a probability of exactly a half.

You cannot be sure of everything

Take the three tests whose outcomes lie along the three axes of the sphere: horizontal against vertical, diagonal against anti-diagonal, right against left. A photon that is certain to give one answer in any one of them is a coin toss in the other two. Tests related like that are called complementary, or mutually unbiased. They are as different from one another as tests can be.

What about states that are not on an axis? The figure lets you explore.

Write $s_1$ for the probability of “horizontal” minus that of “vertical”, a number that is $\pm1$ when the outcome is certain and 0 when it is a coin toss. Define $s_2$ and $s_3$ in the same way for the diagonal and circular tests. These three numbers are the coordinates of the state’s point on the sphere, so for every pure state

There is a fixed budget of certainty and it has to be shared among the three tests. Spend all of it on one ($s_1 = \pm1$) and nothing is left for the others. This is an uncertainty relation, the simplest there is, and it is worth being clear about what it says. It does not say that testing one thing disturbs another. It says something about states: there is no preparation at all for which both the horizontal test and the diagonal test have sure outcomes. No such vector exists.

Finding an unknown state

Lesson 1.4 showed that a single photon cannot be made to reveal its state. A supply of identically prepared photons can, and the three numbers above say how: measure them.

Send a large batch through the horizontal-against-vertical analyser; the fraction going each way gives $s_1$. A second batch through the diagonal analyser gives $s_2$, and a third through the circular one gives $s_3$. Those are the coordinates of the point. In terms of the vector, with components $\alpha$ and $\beta$: the first test gives their sizes, $|\alpha|^2$ and $|\beta|^2$, and the other two give the angle between them.

The general count goes like this. A pure state of a system with $N$ outcomes has $N$ complex components. Normalisation and the overall phase leave $2(N-1)$ real numbers to find. One complete test on many copies supplies $N - 1$ independent probabilities. So at least two tests are needed, and for polarisation ($N = 2$) two tests give two numbers and all but settle it; the third decides between the two points that remain, mirror images of each other in the plane of the first two axes.

Notice what kind of knowledge this is. We did not learn anything about any one photon: each was tested once and used up. What we learned is a fact about the source, that is, about the preparation. That is what a state has been all along.

This lesson follows §3-3 to §3-5. The book states two further rules here, I and J, which demand that actual apparatus, obeying the ordinary laws of physics, can be built to behave as rule H says. Whether that is possible is the “measurement problem”, taken up in its final chapter.