A warming-up exercise, to begin with. Imagine two billiard balls coasting toward each other with equal speeds, then colliding. Since initially they move in opposite directions, they will also do so after the collision, and their speeds will again be equal. If the collision is perfectly head-on, each ball simply reverses its direction of motion. But suppose that the collision is uncontrollably somewhat off-center, so that each ball veers from its original direction of motion by some angle α. We cannot then predict whether the balls will scatter at (say) right angles (90° plus/minus a small angle), but we can estimate the probability with which they will do so.
There are two ways in which the balls can scatter at right angles, each with its own probability. If we call the incoming balls N and S (suggesting that they move northward and southward, respectively), we may call the outgoing balls E and W, and we may write p1 = p(N→E,S→W) and p2 = p(N→W,S→W). If either ball is as likely to be scattered eastward as westward, the two probabilities are equal, and the probability with which the balls will scatter at right angles is the sum
p = p(N→E,S→W) + p(N→W,S→E).
This is part of standard probability theory: an event that can come to pass in exactly two ways, with respective probabilities p1 and p2, comes to pass with probability p = p1 + p2.
Now we come to the real McCoy. Instead of billiard balls we use particles. We again assume that initially the two objects move in (more or less) opposite directions with (more or less) equal speeds, so that the same will be true after the… scattering event. (Where particles are concerned, the word “collision” invokes the wrong images.) We also assume that the scattering is elastic. This means that no particles are created or annihilated during the event. If the incoming particles (and thus the outgoing ones) are of different types, then it is possible to learn which outgoing particle is identical with which incoming particle, and Rule A applies: first square the magnitudes of the amplitudes A1 = A(N→E,S→W) and A2 = A(N→W,S→E) associated with the alternatives, then add the results:
p = |A(N→E,S→W)|2 + |A(N→W,S→E)|2.
This is again the sum of two probabilities, p1 = |A1|2 and p2 = |A2|2. If there are no preferred directions (due to external forces, particle spins, and such) then the two probabilities are equal, so that |A2| = |A1|. Still nothing to write home about.
But now suppose that the conditions stipulated by Rule B are met: there is no actual event or state of affairs from which one could learn which of the alternatives took place. (This rules out, among other things, that the incoming particles are of different types.) In this case we must first add the amplitudes and then square the magnitude of the result:
p = |A(N→E,S→W) + A(N→W,S→E)|2.
Previously we were adding real numbers; now we are adding complex numbers. When we were adding the squares of the magnitudes |A1| and |A2|, the phases of the amplitudes A1 and A2 were irrelevant. Now we need to think about the angle between the two complex numbers (visualized as arrows) or (what comes to the same) about the difference between their phases.
Take a look at Fig. 1.3.1 above. Observe that exchanging either the incoming or the outgoing particles is tantamount to exchanging the two alternatives and, correspondingly, the two amplitudes, so that A2 takes the place of A1 and vice versa. Since the two amplitudes have the same magnitude, there is a complex number c of unit magnitude such that A2 = A1 c. In other words, multiplication by c = [1:β] represents an exchange of the incoming or outgoing particles.
If the incoming or outgoing particles are exchanged twice, then (i) A1 gets multiplied by c2 and (ii) the original situation is restored. Thus A1 = A1 c2, whence it follows that c2 = [1:2β] = 1. This means that 2β must be equal to an integral multiple of 360°, and this leaves us with two possibilities: β = 0°, in which case A2 = A1, or β = 180°, in which case A2 = −A1.
Nature makes use of both possibilities. Every particle is either a boson or a fermion. Interchanging two indistinguishable bosons leaves the amplitudes unchanged, whereas interchanging two indistinguishable fermions is accompanied by a change of sign. Writing A for A1, we thus have
pb = |A + A|2 = |2A|2 = 4 |A|2
in the case of bosons, and
pf = |A — A|2 = 0
in the case of fermions. Compare this with the result we obtained for distinguishable particles:
p =|A|2 + |A|2 = 2 |A|2.
Indistinguishable bosons are thus twice as likely to scatter at right angles as particles that carry “identity tags” of some sort, whereas the probability with which indistinguishable fermions scatter at right angles is zero, rather than some positive number.
Now suppose that an initial measurement indicated that two indistinguishable particles are headed northward and southward, respectively, and that the next (relevant) thing that can be deduced from an actual event or state of affairs is that two particles are headed eastward and westward, respectively. Rule B therefore applies. Can we nevertheless assume that what actually happened is either of the alternatives? If we can, then the following questions have answers: Is E the same as N (in which case W is the same as S)? Is E the same as S (in which case W is the same as N)? Which incoming particle is identical with which outgoing particle?
Suppose that these questions have answers. If what actually happens during the scattering event corresponds to either of the alternatives illustrated in Fig. 1.3.1, then the probability with which the two particles scatter at right angles is the sum of the probabilities of the alternatives:
p = p1 + p2x with xp1 = |A|2x and xp2 = |A|2.
In reality we have that
pb = |A + A|2 = |2A|2 = 4 |A|2x andx pf = |A — A|2 = 0.
We have arrived at a contradiction, and this means we have made a wrong assumption. What actually happens is neither of the alternatives. Those questions have no answers.
Then what is it that actually happens? We will return to this question.