As you will remember, every particle is either a boson or a fermion. Let us use the symbol |a> for the state of a boson, and let us keep firmly in mind that a “state” in quantum mechanics is a probability algorithm; it serves to assign probabilities to possible outcomes of measurements — in this case any measurement to which the boson may be subjected.
As said, quantum states (and thus the possible outcomes of measurements that may be made) are determined by the actual outcomes of measurements that have been made. The symbol |a> accordingly plays a double role: it represents (i) the outcome of a measurement and (ii) the algorithm to be used for assigning probabilities to the possible outcomes of whichever measurement is made next.
If we want to know the probability with which a boson initially described by |a> is later found in the state |b>, we need to know the amplitude associated with this possibility. For this we shall use the symbol <b|a>.
Now suppose that a boson has been found “in” the state |a>: it is correctly and completely described in terms of the probabilities that |a> serves to assign. Also suppose that another boson has been found “in” the state |b>: it is correctly and completely described in terms of the probabilities that |b> serves to assign. (We shall further assume that |a> and |b> are possible outcomes of the same measurement so that, technically speaking, they are orthogonal.) What symbol shall we use for the state of the composite system made up of the two bosons?
If we use something like |a,b>, we introduce into our notation a distinction that does not correspond to anything in the actual world. In addition to the physically warranted distinction between the boson described by |a> and the boson described by |b>, there is then the physically unwarranted difference between the “left” boson and the “right” boson. To eliminate this physically unwarranted difference, we use the symmetric symbol
(|a,b> + |b,a>)/√2.
The division by √2 ensures that this two-boson state is normalized: the probabilities it assigns to the possible outcomes of any measurement to which the two-boson system may be subjected add up to 1. Observe that this expression remains unchanged if we exchange |a> and |b>.
If, on the other hand, the distinction between the “left” boson and the “right” boson corresponds to something in the actual world — if, for instance, the “left” and the “right” boson are of different types so that “left” and “right” represent the respective types to which they belong — then the symbol |a,b> may be used, and we can write
<c,d|a,b> = <c|a> <d|b>
for the amplitude associated with the transition from |a,b> to |c,d>. In this case, however, the bosons are not completely described by the individual states |a> and |b>, inasmuch as these contain no information about the particle species to which each boson belongs.
In case the bosons are completely described by |a> and |b>, the corresponding transition amplitude is
(1/√2)(<c,d| + <d,c|) × (1/√2)(|a,b> + |b,a>).
The manipulation of this expression is straightforward:
(<c,d|a,b> + <c,d|b,a> + <d,c|a,b> + <d,c|b,a>)/2
= (<c|a> <d|b> + <c|b> <d|a> + <d|a> <c|b> + <d|b> <c|a>)/2
= <c|a> <d|b> + <d|a> <c|b>.
The transition probability is thus given by
p = |<c|a> <d|b> + <d|a> <c|b>|2.
This should remind you of the probability p = |A(N→E,S→W) + A(N→W,S→E)|2 we previously obtained. Just put <E|N><W|S> and <W|N><E|S> in place of A(N→E,S→W) and A(N→W,S→E), respectively.
Suppose, then, that an initial measurement indicated the presence of two indistinguishable bosons described, respectively, by |a> and |b>, and that the next (relevant) thing that can be deduced from an actual event or state of affairs is the presence of two bosons described, respectively, by |c> and |d>. Is the boson that was described by |a> the same as the boson that is now described by |c> (in which case the boson that was described by |b> is the same as the boson that is now described by |d>)? Or is the boson initially described by |a> the same as the boson now described by |d> (in which case the boson initially described by |b> is the same as the boson now described by |c>)? Which boson in the initial state of the two-boson system is identical with which boson in the final state?
Do these questions have answers?
Once again they don’t, but this time the reason is not simply that, if they had, the transition probability would be |<c|a><d|b>|2 + |<d|a><c|b>|2 rather than |<c|a><d|b> + <d|a><c|b>|2. Instead, we have put our finger on the reason both why these questions have no answers and why the transition probabilities come out the way they do.
This Quantum World 