KOÇ UNIVERSITY
Phys 301
Statistical Mechanics
Instr: Alkan Kabakçıoğlu,
TA: Duygu Can
HW #1 SOLUTIONS
Q1. The macroscopic state of a system of N particles each of which can assume one of the three spin states {+,0,-}
is defined by the two quantities (s+ , s− ) which represent the excess number of “up” and “down” particles. The
excess number of remaining particles with zero spin is given by s0 = −(s+ + s− ). We are asked to find the probability distribution function PN (s+ , s− ) under the assumption that each microscopic state is equally probable. Let’s
use the generating function technique introduced in class to count the microstates consistent with a given macrostate:
0
N
(x + y + z)
=
N!
x nx y ny z nz ,
n
!n
!n
!
x
y
z
nx ,ny ,nz |
{z }
X
where
nx =
N
+ s+ ,
3
ny =
N
− (s− + s+ ),
3
nz =
N
+ s−
3
gN (s+ ,s− )
The total number of microstates is 3N , so we have
PN (s+ , s− ) =
gN (s+ , s− )
N!
= 3−N N
3N
( 3 + s+ )!( N3 + s− )!
N
3
.
− (s+ + s− ) !
Use Stirling’s approximation ln n! ' n ln n − n + 21 ln(2πn) and define σi ≡ si /N to get
N
ln PN ' −
(1 + 3σ+ ) ln(1 + 3σ+ ) + (1 + 3σ− ) ln(1 + 3σ− ) + (1 − 3σ+ − 3σ− ) ln(1 + 3σ+ − 3σ− )
3
2 2
4π N
3
3
ln
+
ln (1 + 3σ+ )(1 + 3σ− )(1 + 3σ+ − 3σ− )
+
2N
27
2N
which simplifies after using ln(1 + x) ' x − x2 /2 and ignore terms with O(N −2 ) to yield
1
4 2 2
2
2
ln PN ∼
+ σ+ σ− + σ−
− ln
π N
.
= −3N σ+
2
27
Exponentiate both sides and revert from σi ’s back to si ’s to obtain the desired result :
2
2
1
PN (s+ , s− ) = p
e−3(s+ +s+ s− +s− )/N
2
2
4π N /27