The nature of the question here raised is well illustrated by the special case in which the possible phases are restricted to two opposite phases. We may then conveniently discard the idea of phase, and regard the amplitudes as at random positive or negative. If all the signs are the same, the intensity is n2; if, on the other hand, there are as many positive as negative, the result is zero. But, although the intensity may range from 0 to n2, the smaller values are much more probable than the greater.
The simplest part of the problem relates to what is called in the theory of probabilities the “expectation” of intensity, that is, the mean intensity to be expected after a great number of trials, in each of which the phases are taken at random. The chance that all the vibrations arc positive is 2−n, and thus the expectation of intensity corresponding to this contingency is 2−n·n2. In like manner the expectation corresponding to the number of positive vibrations being (n − 1) is
2−n·n·(n − 2)2,
and so on. The whole expectation of intensity is thus
| 1 | { 1·n2 + n·(n − 2)2 + | n(n − 1) | (n − 4)2 |
| 2n | 1·2 |
| + | n (n − 1) (n − 2) | (n − 6)2 + ... } |
| 1·2·3 |
(1).
Now the sum of the (n + 1) terms of this series is simply n, as may be proved by comparison of coefficients of x2 in the equivalent forms
(ex + e−x)n = 2n (1 + ½ x2 + ... )n
| = enx + ne(n−2)x + | n (n − 1) | e(n−4)x + ... |
| 1·2 |