Physics 213 Problem 4 Week 3
Two-State System: Spins
Consider a set of 7 spins which each point either up or down.
a) What is the total number of microstates for this 7-spin system?
Each spin, independent of the others, has two possible states. Thus, the number of microstates of
the system is Ω = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 27 = 128.
b) How many microstates have 3 spins up, 4 spins down?
Here we simply choose exactly 3 of the spins to be up, which leaves the other 4 spins down. The
number of ways to choose 3 things out of 7 when order of choice doesn't matter is given by the
binomial function (also called n-choose-m, or N-choose-Nup here):
It is useful to notice that we could as well have chosen the four spins to be down instead of the
three to be up. Common sense tells us that the number of ways to do this would still have to be
35. In general, choosing M things out of N is equivalent to choosing the N-M things one does
not want.
c) What is the probability in equilibrium that 3 spins are up, 4 down?
Since all of the microstates (particular arrangements of up and down for the 7 spins) are equally
likely, the probability is P(3) = 35/128 = 0.273.
d) The entropy σ is simply the (natural) logarithm of the number of states. What is the entropy
for the 3-up, 4-down configuration?
σ3up = ln (Ω3up) = ln (35) = 3.55
e) Plot σ for all 7 configurations (macrostates): (Notice that σ has only 4 different values.)
σ
4
3
2
1
0Nup
0 1 2 3 4 5 6 7
Solution
