Physics 7230 Spring 2004 Homework 2 Due Monday Feb 23
Do any four of the following five problems.
1. Density matrix. Consider a two-state system with energy eigenstates |0> and |1> and
pure states  1 and  2 .
it
,
1
1
1
and
it
,
2
2
2
  a 0  be
1
  a 0 be
1
2
2
where   E1  E0  . The coefficients are constant and normalized ( ai  bi  1 ), and
E0 and E1 are the energies of the eigenstates |0> and |1> respectively.
a) Show that both pure states satisfy the Schrodinger equation.
 b) Construct a density matrix from the two states using probabilities p1=p and p2=1-p.
Write down the density matrix in matrix form using the energy basis. Calculate the trace
of the density matrix and the trace of the square of the density matrix. Under what
conditions does the density matrix describe a pure state? Under what conditions does the
density matrix describe an equilibrium state?
c) For the equilibrium density matrix, relate p to . Show that “negative temperatures”
are really hotter than any positive temperature.
Solution:
a) H 1  0a1 0  b1eit 1 
the same goes for the other state.
b)


ˆ ( t)  p 1 1  (1 p)  2


a1 0  b1eit 1  0a1 0  b1eit 1

i t
 p a 2  (1 p) a 2
1
2
 2  
( pa*b  (1 p)a*b )eit
 1 1
2 2
( pa1b1*  (1 p)a2b2* )e it 

2
2

p b1  (1 p) b2

The density matrix is explicitly time dependent unless a1b1*=0 and a2b2*=0. If those
conditions are satisfied then the density matrix is built only of energy eigenstates so can
describe an equilibrium system.
ˆ (t)) 1 no matter what the values of p, a and b are.
Trace(



If the trace of 2 is unity then the density matrix describes a pure state. If the trace of 2
is less than unity then the density matrix describes a mixed state (or ensemble average).
The calculation is simplified if we choose (without loss of generality) a1 and a2 real and
i
b j  1 a 2j e j
ˆ (t)2 ) 1 2p(1 p)(a12  a22  2a12a22  2a1a2 1 a12 1 a22 cos(1  2 )) This
Trace(
positive but is less than unity in general. If p=0 or p=1 or if b2=b1 and a2=a1 then the trace
is equal to 1. In any of those cases the matrix is built from a single pure state.
c) If b1=0 and a2=0 then the density matrix looks like
p
0 
 in the energy basis. Since p must be proportional to exp(-)
0 1 p
ˆ  



 p 
1
so   ln 
. If p < 1/2 then >0. If p > 1/2 then <0. If p =1/2
1 exp()
1 p 
then =0. Negative temperatures just correspond to the excited state being more highly
occupied than the ground state. Since the energy is higher at negative temperature than
for any positive temperature negative temperatures are “hotter”. This nonequilibrium
 in a laser in which the higher occupation of the excited state is caused by
situation arises
pumping into the excited state and is called thermal inversion. Note that  is the more
natural representation of the temperature. The behavior depends smoothly on .
p
2. Isobaric ensemble: Consider the isobaric, isothermal ensemble partition function
Y( p,T,N) 



0
0
 exp(pV )Z(T,V,N)dV   exp(pV  F(T,V,N))dV
in which the volume has been traded in for the pressure.
a) Show that the logarithm of this partition function is related to the Gibbs free
energy by considering derivatives of ln(Y).
b) Show that your definition of G is consistent with the definition G=F+pV in the
thermodynamic limit by estimating the integral by expanding the argument of the
exponential about its maximum out to quadratic order and evaluating the
Gaussian. What constraints apply to F(T,V.N) for this to work?

 ln(Y ) 

 
 p T ,N
  V exp(pV  F(T,V,N))dV
0

 exp(pV  F(T,V,N))dV
  V
0


 ln( Y ) 

 
 N T ,N
F 
    exp(pV  F(T,V,N))dV
 
0 N p,T

 exp(pV  F(T,V,N))dV
F 
     
N p,T
0
F 
  
 exp(pV  F(T,V,N))dV
0   p,N


 ln( Y ) 

 
  T ,N

 exp(pV  F(T,V,N))dV
F 
  
 U

  p,T
0


Therefore ln(Y) has all the properties of the Gibbs Free Energy G=F+pV if we choose
G=-kBT ln(Y).
b) Now lets make sure both ensembles are exactly the same in the thermodynamic limit.
Evaluate the integral for Y by assuming the integral is dominated by the integrand near
its maximum. Expand argument of the exponential about its maximum at V  V .


 1  2F(T,V,N) 

2
Y   exp(pV  F(T,V,N))dV  exp(pV  F(T,V ,N))  exp 
(V
V
)
dV

V 2
V V
 2 

0
0

Note that the parameter p must be equal to the pressure as defined in the canonical
ensemble because the first derivative of the argument of the exponential is zero at the
maximum.
F(T,V,N) 
p  

 V
N.T ,V V
Therefore






1 
2
 F(T,V ,N)  pV  1 ln 2k TVK
G  k B T ln( Y)  F(T,V ,N)  pV  ln
 B T
2
2  F(T,V,N)  
2
 

V 2
V V 

.
The last term involves the isothermal compressibility which is positive by stability. This
gives the correct thermodynamic limit (G=F+pV) because the first two terms are of order
N and last term is only of order ln(N). Therefore the ln(N) term is negligible in the
thermodynamic limit.
3. Ideal gas with split ground state. Suppose the N-body Hamiltonian is a sum of
noninteracting atoms, but the ground state of each atom is split into two closely spaced
atomic energy levels 0 and 1 with degeneracies g0=1 and g1=3. Assume classical
behavior for the center of mass motion. Determine the N-body partition function. Show
the free energy, the entropy, the internal energy, and the heat capacity can each be written
as a sum of two terms, one related to the center-of-mass motion and one related to the
quantum levels. Comment on the high and low temperature limiting behaviors of the
quantum terms.
Since the Hamiltonian of the whole system is the sum of the Hamiltonia of the
individuals atoms
p2
H
 H electronic the partition function factorizes
2m
N
N
(Z cm Z elect ) N
1 V  N 0
Z
  3  e
1 3e  

N!
N!  


Therefore the free energy is
F  Fideal  Felect where Felect  N0  NkB T ln(1 3e  )

The ideal gas part comes from the center of mass motion. The internal energy, entropy
and heat capacity of the electronic terms are


3 
Uelectronic  N0 

 1 3e  



  
Selectronic  NkB ln(1 3e  )  NkB 

1 3e  


(3) 2 

Celectronic  NkB
1 3e  2 


At low temperature UN0 , S0, C0. At high temperature UN(0.+31)/4,
SkBln(4), and C0. The electronic entropy goes to zero at low temperature because
there is a single ground state. The electronic entropy goes to kBln(4) at high temperature
since there are 4 equally occupied states per atom. The energy goes to the weighted mean
of the two energies at high temperature since there three excited states for every ground
state.









4. Consider the vertical motion of a mass hanging on a spring. The unstretched length of
the spring is a and the spring constant is k. Include the force of gravity. The position y is
the vertical distance below the support point. The Hamiltonian is
p2 k
H
 (y  a) 2  mgy .
2m 2
Calculate the classical partition function. Calculate the internal energy, heat capacity, the
average position and the variance of the position. Explain the low temperature limits in
simple physical terms. Finally calculate the one-particle density defined by
n(z)   (z  y) .
Integrate the position over all values of height. If you are worried about the mass being
higher than the support we will see that is exponentially unlikely since mga>>kBT for a
reasonable mass on a reasonable spring.
p 2
k(ya )2
1   2m   2  mgy
k T
Z e
dp  e
dy  B e mgy
h 


mg
where y  a 
is the equilibrium position of the mass hanging from the spring.
k
k T 
F  kB T ln B  mgy
  
F 
 U  
 k B T  mgy equilibrium energy plus energy due to the equipartition theorem
  
with two squared degrees of freedom.
C  kB equipartition theorem with two squared degrees of freedom.

 k(y  y ) 2 
 y exp 2 dy

y  
 y hangs at the equilibrium point on average.
 k(y  y ) 2 
 exp 2 dy


 k(y  y ) 2 
y 2 exp
dy

2
k T
2


2

y  y  
 y 2  B in agreement with the equipartition
2
 k(y  y ) 
k
 exp 2 dy

theorem.

 k(y  y ) 2 
  (z  y)exp 2 dy
 k(y  y ) 2 
k
n(z)   (z  y)   

exp


 k(y  y ) 2 
2k B T
2


 exp 2 dy

The mass can be found in a Gaussian distribution about the equilibrium position. The
width of the Gaussian is proportional to the square root of the temperature according to
the equipartition theorem.







5. Determine the classical partition function for a simple rigid pendulum. The pendulum
has length l and mass m in gravitational field g. The Lagrangian is
2Ý
2
Ý)  ml   mglcos( ) .
L(,
2
Derive the Hamiltonian from the Lagrangian and calculate canonical partition function
(this can be evaluated in terms of Bessel functions). Allow the pendulum to move
through all angles. Calculate the internal energy and heat capacity vs. temperature. Plot
the heat capacity and compare the high temperature and low temperature limiting heat
capacities with the predictions of the equipartition theorem.
The momentum conjugate to the angle is
L 
Ýso the Hamiltonian is (note this is a Legendre transformation)
p   Ý ml2

2
Ý L  p  mglcos( )
H  p 
2ml2
The canonical partition function is
p2  2 
2ml2 kB T
1
Z   exp 2 dp  exp mglcos( )d 
I0 (mgl)
h
2ml  0
where I0 is a Bessel function of the second kind.
 2ml2 k T

B

F  kB T ln 
I
(

mgl)
0




F  k B T
I '(mgl)
U  
 mgl 0

I0 (mgl)
   2


I0 '(mgl) 2 
U 
1
2 I0 ''(mgl)

C    kB   mgl 
 
 I (mgl)  

2
T 
I
(

mgl)


0
0



At low temperature U-mgl +kBT (the pendulum hangs down and vibrates about
equilibrium state according to equipartion theorem with two squared degrees of freedom
so the heat capacity is kB. At high temperature U1/2 kBT so one squared degree of
freedom has been lost. Only the angular momentum contributes to the energy at high
temperature since the potential energy tends to zero (the pendulum is as likely to point up
as down). Note that the heat capacity increases slightly from kB at low temperature but
then drops to 1/2 kB at high temperature.
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