13 Quantum coins and dice

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) = 14.

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) = 12.

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) = 12. 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) = 13 and p(H|H) = 23.

The joint prob­a­bil­i­ties for equal out­comes are p(HH) = p(H|H) p(H) = (2/​3) × (1/​2) = 13 and p(TT) = 13. 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) = 13. 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 + 302 = 21 pos­sible out­comes. The prob­a­bility of casting a first 6 is 16. 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 27. The prob­a­bility of casting a total of 12 is thus (1/6) × (2/7) = 121. 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 = 121. (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) = 321 = 17.

When casting an (imag­i­nary) pair of FD dice, there are 302 = 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) = 215.

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.

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1. [↑] Misner, C.W., Thorne, K.S., and Wheeler, J.A. (1973). Grav­i­ta­tion, W.H. Freeman and Com­pany, p. 1215.