Which outgoing particle is identical with which incoming particle?

If no particles are created or annihilated in the process, then these are the only possibilities:

The lines connecting the incoming particles to the outgoing ones indicate possible identities; they are not meant to suggest that the particles move along continuous paths.

We can only find out which of these two possibilities actually happened if there is an actual event or state of affairs from the answer can be inferred. In other words, something must indicate whether S is the same particle as W (and thus N the same as E) or S is the same particle as E (and thus N the same as W). If this is the case, then Rule A applies:

p = |A₁|² + |A₂|²,

where A₁ and A₂ are the amplitudes associated with the two alternatives. p gives the probability with which we find one of the outgoing particles moving eastward and the other moving westward, given that one of the incoming particles was moving northward and the other southward. This result, obtained using Rule A, agrees with standard probability theory, according to which 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.

Evidently, p₁ = |A₁|² and p₂ = |A₂|².

What if there is no actual event or state of affairs from which one could learn which of the two possibilities took place? (This rules out, among other things, that the incoming particles are of different types.)

In this case Rule B applies:

p = |A₁ + A₂|².

If there are no preferred directions (owing to external fields, particle spins, and such), then the mirror symmetry of the alternatives gives us the right to assume that the probabilities |A₁|² and |A₂|² are equal. Hence the two amplitudes can differ at most by a phase factor: A₂ = e^iφ A₁. If we write A₁ as |A|e^iα, then A₂ = |A| e^i(α+φ).

What can we say about the value of φ? If we interchange the outgoing particles, the first alternative becomes the second, and A₁ changes to e^iφA₁ — it acquires a phase factor. The same happens if we interchange the incoming particles. An exchange of two particles thus causes a phase shift by φ, and so a twofold exchange causes a phase shift by 2φ. [Remember that (e^iφ)² = e^i2φ] But a twofold exchange restores the orginal situation. This means that e^i2φ equals 1, and that e^iφ is either +1 or -1.

Nature makes use of both possibilities. Every known particle is either a boson or a fermion. For bosons we have that A₂ = A₁, whereas for fermions we have that A₂ = -A₁. Thus for bosons Rule B gives us

p = |2A₁|² = 4|A|²,

whereas for fermions it gives us

p = |A₁ - A₁|² = 0.

In other words, for bosons the probability of scattering at right angles is larger if Rule B applies than it is if Rule A applies (in fact, twice as large), whereas for bosons it is smaller (in fact, zero). In the first case we say that the alternatives interfere constructively, in the second we say that they interfere destructively.

This has (almost) nothing to do with the classical phenomenon of wave interference. All that is meant by this phraseology is that in one case Rule B gives a larger probability than Rule A, and in the other case it gives a smaller probability.

Now comes our first million-dollar question. Suppose that the conditions stipulated by Rule B are satisfied; there is no actual event or state of affairs from which one could learn which alternative took place; nothing indicates which outgoing particle is identical with which incoming particle. Can we nevertheless assume that one of the alternatives took place? Does the question "Which outgoing particle is identical with which incoming particle?" nevertheless have an answer?

The answer, in both cases, is NO.

Imagine a large number of scattering events with one northbound and one southbound incoming particle and one eastbound and one westbound outgoing particle. If either alternative actually takes place in each event, the first alternative happens in a fraction n₁/n of all n = n₁ + n₂ events and the second happens in a fraction n₂/n of all events. In the limit in which n tends to infinity (n→∞), the fraction n₁/n tends to |A₁|² and the fraction n₂/n tends to  |A₂|². This means that the probability for scattering at right angles is the sum of absolute squares |A₁|² + |A₂|² rather than the absolute square |A₁ + A₂|² of a sum. Since the assumption that either alternative actually takes place leads to a wrong conclusion, it is wrong.