2.2One test after another

A theory earns its keep by predicting something. For quantum tests the simplest thing to predict is how two tests in a row are related: given the outcome of the first, what are the odds for the second? The answer is a table of numbers, and the table turns out to obey two laws that no ordinary table of odds obeys.

Two magnets in a row

Use atoms of spin 1, for which a Stern–Gerlach magnet has three outcomes. Call them $+1$, $0$ and $-1$. The first magnet splits the beam in three. Behind it put a second magnet, tilted at some angle to the first, and big enough for all three beams to pass through. Each of the three splits in three again, and nine spots appear on the plate.

Each spot records two outcomes: where the atom left the first magnet, labelled $m$, and where it left the second, labelled $\mu$. (Peres uses Latin letters for the outcomes of one test and Greek for the other, so that you can always tell which test a label belongs to. We shall do the same.)

Let $I_m$ be the brightness of beam $m$ after the first magnet, and $I_{\mu m}$ the brightness of the spot it feeds. Their ratio is the fraction of the atoms in beam $m$ that the second magnet sends to $\mu$:

The first test is complete and repeatable, so each of its beams is a pure state. By rule A, the odds for the second test are then fixed. So the table $P_{\mu m}$ depends only on the two tests. The oven, and how many atoms it happens to put into each beam, do not come into it.

The columns add up, and so do the rows

Look at the sums printed round the edge of the table. Each column adds up to 1, and that is no surprise. A column is one beam from the first magnet, and every atom in it has to come out of the second magnet somewhere:

But each row adds up to 1 as well. There is no obvious reason why it should. A row collects the chances of reaching one particular spot from each of the three different beams; nothing says those should total anything in particular. For ordinary tables of odds they don’t.

The reason is rule C, the existence of a random mixture, and the argument is short. Feed the pair of magnets with atoms straight from the oven, a random mixture. The first magnet, being a complete test, gives each beam $m$ a share of $\tfrac1N$. Now regard the two magnets together as one piece of apparatus whose outcome is $\mu$. That is a complete test too, and it is being fed a random mixture, so each $\mu$ must also get a share of $\tfrac1N$. But the share of $\mu$ is made up of contributions from all the beams:

A table of non-negative numbers whose columns each sum to 1 is called stochastic. One whose rows do too is doubly stochastic. Every table linking two complete quantum tests is of this kind.

That is a strong restriction. A three-by-three table has nine entries, but once four of them are given the other five are forced.

Forwards is the same as backwards

Now exchange the two magnets, so that the tilted one comes first. This is a different experiment. Its table, call it $\Pi_{m\mu}$, gives the odds of outcome $m$ at the upright magnet for an atom that left the tilted one by $\mu$. When the experiment is done, the new table turns out to contain the same numbers as the old one:

D. Reciprocity. Let $\varphi$ and $\psi$ be two pure states. The probability that a system prepared in $\psi$ gives the outcome $\varphi$ in a complete test is equal to the probability that a system prepared in $\varphi$ gives the outcome $\psi$.

Nothing like this holds for everyday odds. Among playing cards, the chance that a king is a court card is 1; the chance that a court card is a king is $\tfrac13$. Turning a conditional probability round normally changes it. For pure quantum states it does not.

A good way to hold rule D in your head is to read the probability as a measure of likeness. A photon prepared diagonally passes the test for horizontal half the time: diagonal is half like horizontal. Rule D says that likeness is mutual. However alike $\psi$ is to $\varphi$, that is how alike $\varphi$ is to $\psi$.

For linear polarisations this is obvious from symmetry: the probability can only depend on the angle between the two directions, and that angle is the same whichever one you start from. But pairs of tests are not usually related by any symmetry, and at this stage rule D has to be accepted as an experimental fact. In Part III it will drop out of the mathematics in one line.

One consequence is worth stating separately. For pure states, working forwards and working backwards are the same calculation. Suppose a photon is known to have been prepared either diagonal or anti-diagonal, with nothing to choose between the two, and is then found to be horizontal. The odds that it had been prepared diagonal are exactly the odds that a diagonal photon is found horizontal. (It is the row sums that make this work: the two ways of arriving at “horizontal” have probabilities that already add up to 1.)

This lesson follows §2-4. The book notes that reciprocity holds even though each test is an irreversible process, and offers a comparison: a person carries on average a quarter of a grandparent’s genes, and the grandparent a quarter of the grandchild’s, though descent certainly runs one way only.