This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 'd~ ~
~ P- s ~ P. ) + 1
.:: d) (5 points) In order to know the total number o f particles in the system we have
to sum over the density o f states (N)= L 1 s,sprns e xp ( € s ~ J1 ) + 1 We will call gs is the number o f independent spin states (2 for spin 1/2), and
perfonn the appropriate integral over phase space to sum over spatial quantum
states. Express <N> as a function o f gs, the volume V and an integral over cfp. ~f'~ . I t JC~\.-~-d ~£V-~\r~"\Q<;;' < ~;:>:= __ ~s l ~ J sr'~"'-s (?fpC £c.i.f 'l'/1- I f ; J 3 x J~
~1~~f t-.3 e) (5 points) What is the condition for (N y ) « 1 ? I f this low occupation number
is correct for most o f the states show that for non relativistic particles (N) = gsvex p( ~)( 2~~
or Jl = ; 3 Y =; gs vexp ( ~IOg(_n-J where n = (N)
gsnQ ~ )n Q i 3 =- = p'1..
2~ P","1 to"" P> 1-1 r~'L V This is the classical result (the gs factor comes from the spin degrees o f freedom and is
a}so present classically). This rigorous result justifies the Gibbs ansatz o f dividing by N t
the naIve partition function for a system o f N undistinguishable particles. In iliis derivation, you may waj:Oe:~ ~2:):: J 2KG...
View Full Document
- Spring '06