763620SS STATISTICAL PHYSICS
Solutions 2
Autumn 2012
1. Continuous Random Walk
Consider a continuous one-dimensional random walk. Let w(si )dsi be the probability
that the length of the ith displacement is between si and si + dsi . Assume that the
displacements are independent on each other and obey the same distribution w(s).
a) P
Show that the probability of finding the total displacement (after N steps) x =
N
i=1 si between x and x + dx is
"
! #
Z Z
Z
N
X
P(x)dx =
· · · w(s1 )w(s2 ) · · · w(sN ) δ x −
si dx ds1 ds2 · · · dsN ,
i=1
where δ(x) is the Dirac delta function and the integrations go from −∞ to ∞.
Solution: Let’s take a set of N displacements si=1,...,N that sum to a fixed x. As the
displacements are independent of each other, the common probability is a straight
product of the probabilities for the individual displacements:
w(s1 )ds1 w(s2 )ds2 · · · w(sN )dsN .
But as we are not interested only the one set of N displacements si=1,...,N that sum
to a x, but all of them, we integrate (i.e. sum) over all values of si=1,...,N that sum
to x, formally:
Z ∞
Z ∞Z ∞
w(s1 )w(s2 ) · · · w(sN )ds1 ds2 dsN dx
...
P (x)dx =
−∞
−∞ −∞
{z
}
|
P
x=
Z
∞
Z
n
i=1 si
∞
=
Z
w(s1 )w(s2 ) · · · w(sN ) δ x −
...
−∞
−∞
"
∞
−∞
b) Show that
Z ∞
1
dke−ikx QN (k),
P(x) =
2π −∞
N
X
!
si
#
dx ds1 ds2 dsN .
i=1
Z
where
∞
Q(k) =
dseiks w(s).
−∞
Solution: Here, the key thing is to express the delta function by using the hint
given in the Ex. 2:
!
Z ∞
Z ∞
N
X
P
1
1
s
)
−ik(x− N
i
i=1
si =
δ x−
dk =
e
e−ikx eiksi eiks2 · · · eiksN dk .
2π
2π
−∞
−∞
i=1
1
Then, we can a bit make rearrangements in the P (x)dx
Z ∞
Z ∞
Z ∞
Z ∞
1
−ikx
iks1
iks2
eiksN w(sN )dsN
e
e w(s1 )ds1
e w(s2 )ds2 . . .
P (x)dx =
2π −∞
−∞
−∞
−∞
Z ∞
1
=
e−ikx QN (k)dk dx .
2π −∞
2. Continuous Random Walk - Discrete Steps
Assume that a particle is propagating along a discrete one-dimensional random walk with
the probability density w(s) = pδ(s−l)+qδ(s+l), where p (q) is the probability of taking
a step of length l to the right (left). Use the results of the previous problem to show that
the particle can be found only at locations
x = (2n − N )l , n = 0, 1, 2, . . . , N.
Show that the probability of finding the particle at those locations is
P (2n − N ) =
N!
pn q N −n .
n!(N − n)!
P
N!
n N −n
and the relation
(You might need the binomial expansion (x+y)N = N
i=1 n!(N −n)! x y
R
∞
1
−ikx
.)
δ(x) = 2π −∞ dke
Solution: First, we write the explicit expression for Q(k)
Z ∞
Z ∞
iks
ds eiks [pδ(s − l) + qδ(s + l)] = peisl + qe−isl .
ds e w(s) =
Q(k) =
−∞
−∞
Then the probability P (x)
1
P (x)dx =
2π
Z
∞
−ikx
e
−∞
pe
ikl
+ qe
−ikl N
1
dk dx =
2π
Z
∞
e
−∞
−ikx
N X
N
n=0
n
peikl
n
qe−ikl
N −n
dk dx
Z ∞
N X
N n N −n 1
=
p q
e−ikx eik(2n−N )l dk dx
n
2π
−∞
n=0
=
N
X
n=0
N!
pn q N −n δ(x − (2n − N )l)dx ,
n!(N − n)!
where we can read that the particle can be found only at the locations x = (2n − N )l,
!
n = 0, 1, 2, . . . , N , with the prob. P (2n − N ) = n!(NN−n)!
pn q N −n .
2
3. H-theorem - Approach to Equilibrium
Assume that Pr is the probability of finding the system in the microstate r. The transition
probability between any two microstates obeys Wsr = Wrs .
a) Write down the ”master equation” governing the time evolution of Pr .
Solution: The master equation means the equation for the time evolution of Pr , i.e,
dPr
. We interpret that the Wsr means the probability per time unit for transition
dt
from the state r to the state s. Thus Pr can increase by transitions from other
states to the state r and decrease by the transitions from the the state r to other
states, formally written:
X
X
X
dPr
=
Ps Wrs −
Pr Wsr =
Wrs (Ps − Pr ).
dt
s
s
s
b) Show that
d hln Pr i
≤ 0,
dt
where the equality holds when Pr = Ps for all r and s. Interpret this result!
Solution:. Now, hln Pr i means the expectation value for ln Pr , that is hln Pr i =
P
r Pr ln Pr . Our task is now to form the time evolution equation for it:
X
X dPr
dhln Pr i
1 dPr
dPr
=
ln Pr + Pr
(ln Pr + 1) .
=
dt
dt
P
dt
dt
r
r
r
Now,
P we can make estimates. First of all, for Pr ≤ 1, (ln Pr + 1) ≤ 1 and
d
r Pr = 0. These imply that
dt
P
dPr
r dt
=
X dPr
X dPr
dhln Pr i
=
(ln Pr + 1) ≤
= 0,
dt
dt
dt
r
r
that is, dhln Pr i /dt ≤ 0. And if Pr = Ps for all the states, we see that
X
dPr
=
Wrs (Ps − Pr ) = 0,
dt
s
which implies that also dhln Pr i /dt = 0.
Interpretation: If the probabilities Pr are not equal with each other (i.e. Pr 6= Ps for
all possible pairs s, r), then the system is in a state of higher order (H = hln Pr i is
higher, entropy S (∝ −H) is lower than in equilibrium). This state tends to change
so that H decreases, i.e. dhln Pr i /dt < 0, until the equilibrium is reached, where
Pr = Ps for all the state s and r. In other words, the entropy becomes maximized.
3
4. Spin System
Consider an isolated system of N very weakly interacting localized spin- 12 particles with
a magnetic moment µ pointing either parallel or anti-parallel to an applied magnetic field
B.
a) What is the total number of states Ω(E) lying in energy range between E and
E + δE, where δE is small compared to E but δE µB?
Solution: The energy of a 12 -spin in the magnetic field is E = −µ · B. Thus, if np
(na ) is the number of spins that are parallel (anti-parallel) with the magnetic field,
then we can form a set of equations for the total number particles N and the total
energy E:
(
(
E
+ N2
na = 2µB
N = na + np
.
⇒
E
np = − 2µB
+ N2
E = (na − np )µB
With fixed E, there number of micros states is
N!
N
N!
=
N (E) =
=
N
E
np !na !
np
! N2 −
− 2µB
2
E
2µB
.
!
The possible energy states are spaced by 2µB. By assuming that N is sufficiently
large, we can estimate that the total number of states lying in the energy range
[E, E +δE] is the number of microstates at the energy E multiplied with the number
of possible energy states within δE. In other words, we assume that N (E) changes
very slowly within δE. Then we write for the total number of states:
Ω(E) = N
2
N!
E
− 2µB
! N2 −
E
2µB
δE
.
! 2µB
b) Write down the expression for ln Ω(E). Use Stirling’s formula to simplify.
Solution:The Stirling’s formula is ln n! ≈ n ln n − n. By applying it we can write:
δE
N
E
N
E
ln Ω(E) = ln
+ N ln N −
−
ln
−
2µB
2
2µB
2
2µB
E
N
E
N
+
ln
+
.
−
2
2µB
2
2µB
c) Use Gaussian approximation for Ω(E) when |E| N µB.
Solution: The total number of states has maximum at E = 0 and around it
2
it behaves smoothly as Gaussian (∝ e−x ). Thus, we make Taylor expansion of
ln Ω(E) at E = 0:
ln Ω(E) = ln Ω(E = 0) +
dln Ω(E = 0)
1 d2 ln Ω(E = 0) 2
E+
E + O3
dE
2
dE 2
4
By forming the derivatives of the expression for ln Ω(E) in (b), we get that
δE
ln Ω(E = 0) = ln 2N + ln
2µB
dln Ω(E)
1
N
E
N
E
=
ln
−
− ln
+
dE
2µB
2
2µB
2
2µB
"
#
1
d2 ln Ω(E)
1
1
−N
=
− N
,
E
E
2
2
dE
(2µB)
− 2µB
+ 2µB
2
2
which implies that
δE
1
−
E 2,
2
2µB 2(µB) N
E2
δE N
2 exp − 2 2
.
Ω(E) =
2µB
2µ B N
ln Ω(E) = ln 2N + ln
5. Approach to Thermal Equilibrium
Assume that two systems with different values of β are brought into thermal contact.
Show that the system with higher value of β will absorb heat from the other until the
two β values are the same.
Solution: In a thermal contact, the total energy E (0) = E1 + E2 is fixed but the two subsystems can exchange energy. For the combined system, the total number of microstates
is
Ω(0) = Ω1 (E1 )Ω(E2 ) = Ω1 (E1 )Ω2 (E0 − E1 ).
In the non-equalibrium, we know that the entropy tends to increase (the second law of thermodynamics). Let’s study thus the entropy S = kB ln Ω(0) = kB ln Ω1 (E1 ) + kB ln Ω2 (E2 ):
d(kB ln Ω(0) )
dS
∂ ln Ω1 (E1 ) ∂E1 ∂ ln Ω1 (E2 ) ∂E2 ∂E1
=
= kB
+
dt
dt
∂E1
∂t
∂E2
∂E1 ∂t
∂E1
= kB (β1 − β2 )
≥ 0.
∂t
Thus, if β1 ≥ β2 then ∂E1 /∂t ≥ 0, i.e. the system with higher value of β will absorb heat
from the other system until the two β values are the same.
5