3 A scattering event

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.

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Figure 1.3.1 The alter­na­tives that con­tribute to the prob­a­bility with which two par­ti­cles scatter elas­ti­cally at right angles. The lines do not rep­re­sent tra­jec­to­ries; they merely indi­cate pos­sible iden­ti­ties between the incoming and out­going particles.

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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.

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