Summations Replaced with Integrals in Quantum Statistical Mechanics
Francesco R. Ruggeri
In statistical mechanics, involving quantum objects, one sometimes uses sums if a simple series sum rule exists. For example, in obtaining the Planck distribution 1/ (1-exp(hf/T)) one uses the sum of a geometric series. Otherwise, it seems one uses integrals, arguing that energy levels are very close together. For the Planck distribution with 1/ (1-exp(fon/T)) where n is 1,2,3…, one usually uses an integral, when a sum really applies. In (1), it is argued that such an integral does not yield the same result as the sum for this Planck distribution example. In particular, (1) considers fo as a very small number and shows explicitly that the summation depends on fo, while the integral does not. Given that this applies to the partition function, it carries over to all statistical quantities calculated from the partition function, as noted in (1). Here, we consider other examples. For example, if one tries to replace the geometric series sum used to obtain the Planck distribution, in the first place, with an integral, one immediately has a degeneracy issue. A one dimensional integral over dp (p=magnitude of momentum) yields one answer, while a three-dimensional result Integral 4*3.14 pp dp yields another. The geometric sum, however, does not include the degeneracies of p so it is not clear why one would use a 3-dimensional integral. Even if one uses a one dimensional integral, it does not yield the same result as the geometric sum. In fact, the geometric sum depends on the value of f/T, while the integral does not - it depends on T. Another example is the partition function for a single quantum particle in a box with infinite potential walls (e.g. a Fermi gas). Again an integral is used, but it seems that using p=po a where a is a small number proportional to 1/L, the length of the box and summing should yield an “a” dependent result, which the integral does not give.
•Sep 30, 2023•0 citations