We refer to the statistics of identical bosons as Bose–Einstein (BE) statistics, to that of identical fermions as Fermi–Dirac (FD) statistics, and to that of distinguishable particles as Maxwell–Boltzman (MB) statistics.
As we all know, the probability with which a toss of a pair of MB coins results in two heads or two tails, pMB(HH) or pMB(TT), equals
p(H) × p(H) = p(T) × p(T) = 1/4.
There are two ways in which one coin can come up heads and one tails, each with probability 1/4, so that the odds for this to happen are given by the sum of the corresponding probabilities,
pMB(HT) + pMB(TH) = 1/2.
If we were using an (imaginary) pair of FD coins, there would be a single possible outcome — one heads and one tails — whose probability would therefore be equal to unity.
What about (imaginary) BE coins? The probability of tossing the first H is p(H) = 1/2. The probability p(H|H) of tossing a second H is twice the probability p(T|H) of subsequently tossing a T since the factor n+1 equals 2 in this case. And since the second toss yields either heads or tails, we have that p(T|H) + p(H|H) = 1, whence it follows that p(T|H) = 1/3 and p(H|H) = 2/3.
The joint probabilities for equal outcomes are p(HH) = p(H|H) p(H) = (2/3) × (1/2) = 1/3 and p(TT) = 1/3. And since now there is only one way of obtaining one heads and one tails — for in the (imaginary) case of indistinguishable coins nothing in the actual world corresponds to the distinction between “the first coin” and “the second coin” — the odds for this are 1 − (1/3) − (1/3) = 1/3. Observe that the possible outcomes are again equiprobable. Only in the case of MB coins there are four possible outcomes (HH, TT, HT, and TH), whereas in the case of BE coins there are three.
When casting a pair of MB dice, there are thirty-six possible outcomes, one of which yields a total of 12, two of which yield a total of 11, three of which yield a total of 10, and so on. The corresponding probabilities are therefore 1/36, 2/36 = 1/18, and 3/36 = 12, respectively.
When casting an (imaginary) pair of BE dice, there are 6 + 30/2 = 21 possible outcomes. The probability of casting a first 6 is 1/6. Since the probability of casting a second 6 is twice the probability of subsequently casting a different number, and since the probabilities of the six possible outcomes of the second cast add up to 1, the probability of a second 6 is 2/7. The probability of casting a total of 12 is thus (1/6) × (2/7) = 1/21. In all there are six ways of casting equal numbers, each with a probability of 1/21, and there are fifteen ways of casting different numbers, each with a probability of (1 − 6 × 1/21)/15 = 1/21. (Fifteen rather than thirty because, in the case of indistinguishable dice, the order in which they are cast does not enter the picture.)
Once again the possible outcomes are equiprobable. Only in the case of MB dice there are thirty-six, whereas in the case of BE dice there are twenty-one. One of these yields a total of 12, one yields a total of 11, two yield a total of 10, another two yield a total of 9, three yield a total of 8, and so on. The corresponding probabilities are therefore pBE(12) = pBE(11) = 1/21, pBE(10) = pBE(9) = 2/21, and pBE(8) = 3/21 = 1/7.
When casting an (imaginary) pair of FD dice, there are 30/2 = 15 possible outcomes. Since in each case the dice show different numbers, the possible outcomes are again equiprobable. None of them yields a total of 12, one yields a total of 11, one yields a total of 10, two yield a total of 9, another two yield a total of 8, and so on. The corresponding probabilities are therefore pBE(12) = 0, pBE(11) = pBE(10) = 1/15, and pBE(9) = pBE(8) = 2/15.
When casting two MB dice, the individual outcomes are uncorrelated. The probabilities assigned to the possible outcomes of one cast are independent of the outcome of another cast. When casting two BE or FD dice, on the other hand, the individual outcomes are correlated. For BE dice, the probability of obtaining a 6 increases with the number of dice already showing a 6, while for FD dice the probability of obtaining a 6 is zero unless no other die is showing a 6.
What mechanism or process could explain the correlations that obtain between the outcomes of measurements performed on indistinguishable particles? Misner et al.[1] have answered this question succinctly:
No acceptable explanation for the miraculous identity of particles of the same type has ever been put forward. That identity must be regarded, not as a triviality, but as a central mystery of physics.
1. [↑] Misner, C.W., Thorne, K.S., and Wheeler, J.A. (1973). Gravitation, W.H. Freeman and Company, p. 1215.
This Quantum World 