for
Statistical Physics of Fields
Updated July 2008
by
Mehran Kardar
Department of Physics
Massachusetts Institute of Technology
Cambridge, Massachusetts 02139, USA
, Table of Contents
1. Collective Behavior, From Particles to Fields . . . . . . . . . . . . . . . . 1
2. Statistical Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3. Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4. The Scaling Hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . 55
5. Perturbative Renormalization Group . . . . . . . . . . . . . . . . . . . 63
6. Lattice Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
7. Series Expansions . . . . . . . . . . . . . . . . . . . . . . . . . . 106
8. Beyond Spin Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 132
, Solutions to problems from chapter 1- Collective Behavior, From Particles to Fields
1. The binary alloy: A binary alloy (as in β brass) consists of NA atoms of type A, and
NB atoms of type B. The atoms form a simple cubic lattice, each interacting only with its
six nearest neighbors. Assume an attractive energy of −J (J > 0) between like neighbors
A − A and B − B, but a repulsive energy of +J for an A − B pair.
(a) What is the minimum energy configuration, or the state of the system at zero temper-
ature?
• The minimum energy configuration has as little A-B bonds as possible. Thus, at zero
temperature atoms A and B phase separate, e.g. as indicated below.
A B
(b) Estimate the total interaction energy assuming that the atoms are randomly distributed
among the N sites; i.e. each site is occupied independently with probabilities pA = NA /N
and pB = NB /N .
• In a mixed state, the average energy is obtained from
E = (number of bonds) × (average bond energy)
= 3N · −Jp2A − Jp2B + 2JpA pB
2
NA − NB
= −3JN .
N
(c) Estimate the mixing entropy of the alloy with the same approximation. Assume
NA , NB ≫ 1.
• From the number of ways of randomly mixing NA and NB particles, we obtain the
mixing entropy of
N!
S = kB ln .
NA !NB !
Using Stirling’s approximation for large N (ln N ! ≈ N ln N − N ), the above expression can
be written as
S ≈ kB (N ln N − NA ln NA − NB ln NB ) = −N kB (pA ln pA + pB ln pB ) .
1