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4.2  The ideal Bose gas (density matrices) 213

       Wenow use the recursionformulain eqn (4.46) to count permutations
     of N elements for variouschoices ofcycle weights {z 1,...,z N }.Let us
     start with the simplest case, {z 1,...,z N } = {1,... , 1}, where each cycle
     length has the same weight, and every permutationhas unit weight.We
     expect to find that Y N = N!, and this is indeed what results fromthe
     recursionrelation, as wemay proveby induction.Itfollows, forthis case
     ofequal cycle weights, that the weight ofall permutations with a last-
     element cycle of length k is the same forall k (as is z k Y N−k /(N − k)!
     in eqn (4.46)). This is a nontrivial theorem.To illustrate it, let uscount
     last-element cycles in Fig. 4.16:
                               ⎧
                               ⎪in 6 cycles of length 4
                               ⎪
                             
 ⎪
                               ⎨
                  in Fig. 4.16,  in 6 cycles of length 3
                  element 4 is  ⎪in 6 cycles of length 2  .
                               ⎪
                               ⎪
                               ⎩
                                 in 6 cycles of length 1
     The element 4 is in noway special, and this implies that any element
     among {1,... ,N} is equally likely to be in a cycle of length {1,... ,N}.
     Asaconsequence, in a randompermutation (forexample generated with
     Alg.1.11 (ran-perm)), the probabilityofhaving a cycle of length k is
     ∝ 1/k.We need more cycles ofshorter length to come up with the same
     probability. Concretely, we find

                                ⎧
                                ⎪6 cycles of length 4
                                ⎪
                             
 ⎪
                                ⎨
                   in Fig. 4.16,  8cycles of length 3
                                                      .
                    there are   ⎪12cycles of length 2
                                ⎪
                                ⎪
                                ⎩
                                  24 cycles of length 1
     The number ofcycles of length k is indeed inversely proportional to k.
     Asa second application ofthe recursionrelation, let uscount permuta-
     tions containing only cycles of length 1 and 2.Now {z 1,z 2 ,z 3 ,... ,z N } =
     {1, 1, 0,..., 0} (every permutation has the same weight, under the con-
     ditionthatitcontains no cycles of length 3or longer). Wefind Y 0 =1
     and Y 1 =1,and fromeqn (4.46) the recursionrelation
                       Y N = Y N−1 +(N − 1) Y N−2 ,
     so that {Y 0 ,Y 1 ,Y 2 ,Y 3 ,Y 4 ,... } = {1, 1, 2, 4, 10,...}.Indeed, for N = 4,
     we find 10 such permutations in Fig. 4.16 ([1], [2], [3], [6], [7], [8], [15],
     [17], [22],and [24]).
       Inconclusion, wehave described in this subsectiona recursionformula
     forcounting permutations that lets ushandle arbitrary cycle weights.
     Weshall apply it, in Subsection4.2.3,to ideal bosons.

     4.2.3   Canonical partition function of ideal bosons

     In Subsection4.2.1, weexpressed the partitionfunction ofa bosonic
     system as a sum over diagonal and nondiagonal density matrices for
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