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Lecture 20. Continuous Spectrum, the Density of States
(Ch. 7), and Equipartition (Ch. 6)
Typically, it’s easier to work with the integrals rather than the sums. Thus, whenever
we consider an energy range which includes many levels (e.g., when kBT >> inter-level
spacing), and, especially, when we are dealing with continuous spectra, we’ll replace the
sum over a discrete set of energy levels with an integral over a continuum. When doing
this, we must replace the level degeneracy with a new variable, the density of states,
g(), which is defined as the number of states per unit energy integral:
P i  
  
di
exp   i 
Z
 k BT 
P   d  
g  
  
exp  
 d
Z
k
T
 B 
 i , di
The units of g(): (energy)-1
1 , d1
The Density of States
The density of states, g(), is the number
of states per unit energy
Page 279, Ch. 7
Let’s consider G() - the number of states per unit volume with
energy less than  (for a one- and two-dimensional cases, it will be
the number of states with energy less than  per unit length or area,
respectively). Then the density of states is its derivative:
The number of states per unit volume in the energy range  – +d  :
Since
 P  d  1
0
P   d  


the partition function is
dG 
d
g  d
1
exp   
Z
The probability that one of these states is occupied:
The probability that the particle has an energy between
 and +d  :
g   
1
g   exp     d 
Z
Z   g   exp    d 
0
The Density of States for a Single Free Particle in 1D
The energy spectrum for the translational motion of a molecule in free space is
continuous. We will use a trick that is common in quantum mechanics: assume that a
particle is in a large box (energy quantization) with zero potential energy (total energy=
kin. energy). At the end of the calculation, we’ll allow the size of the box to become
infinite, so that the separation of the levels tends to zero. For any macroscopic L, the
energy levels are very close to each other (E ~ 1/L2), and the
“continuous” description works well.
E4
E3
E2
E1

n 
x
L
k 
For a quantum particle confined in a 1D “box”, the stationary solutions
of Schrödinger equation are the standing waves with the wavelengths:

2 L 2

, n  1,2,...
n
kn
where
L
The momentum in each of these states:
k
pn2
h 2n 2 k 
En 


2m 8mL2
2m
k
2

pn 

h
n
n
L

the wave
number
hn
 k
2L

h
2
2
- the energy (the non-relativistic case)
In the “k” (i.e., momentum) space, the states are equidistantly positioned (in contrast to
the “energy” space).
The Density of States for a Single Free Particle in 1D (cont.)
The total number of states with the wave numbers < k is:
1
k

2 m
kL
G(k)
the
number
of

G k 
states

 
g 1D    
dG  
d

L  2 s  1

m
2
N k  
k
kL

 / L 
G   
L 2m

additional degeneracy of the
states (2s+1) because of spin
[e.g., for electrons 2s+1=2]
g()
1D

The Density of States for a Particle in 2D and 3D
For quantum particles confined in a 2D “box” (e.g., electrons in
FET):
n
n
ky
2D
k
kx 
kx

Lx
ky 
x
Lx
k  kx  k y
2
y
Ly
2
k 2  area 
1  k2
k2
A 2m
N k  

G k  
A G   
4  
4
4
4 2
Lx Ly
g()
A
2
s

1
m


# states within ¼ of
2D
g 2 D   
2
2
a circle of radius k
- does not depend on 

3D
kz
3/ 2
3
k 3  volume 
1  4 / 3  k
k 3V
V  2m 
N k  

G k  
G   
2
2
2 
2 



8
6

6

6





Lx Ly Lz
g()
g
kx
3D
V  2s  1  2m 
  
 2 
4 2


 1/ 2

ky
Thus, for 3D electrons (2s+1=2):
3D
3/ 2
g
3D
  
V
2 2
 2m 
 2 


3/ 2
 1/ 2
Partition Function of a Free Particle in Three
Dimensions
3/ 2
3/ 2 
 3D

2s  1  2m  1/ 2  V 2s  1  2m 
Z1  V  g  exp    d   g   
 
 2    exp    d
2  2 
2
4   
4
  0


0

Z1 – a reminder that this is the partition function for a single particle
The mean energy can be found without evaluating the integral:

E 

ln Z 

3/ 2

 exp    d
0

1/ 2

 exp    d
The numerator can be
integrated by parts:
0


3
3/ 2



exp


d


 1/ 2 exp    d
0

2 0
E 
3
k BT
2
(the integrand vanishes at both limits)
- in agreement with the equipartition theorem
The Partition Function of a Free Particle (cont.)
V  2m 
Z1 


4 2   2 


3/ 2 

 exp    d
0
 exp    d  k BT 
3/ 2
0
 mk T 
Z1  V  B 2 
 2  
(assume that s=0)


x exp  x  dx  k BT 
3/ 2
0
3/ 2
 2  2 

VQ  
 mkBT 
V

VQ
3/ 2
 h2 

 
 2m 
3/ 2

2
1  mkBT 

nQ 
 
2 
VQ  2  
3/ 2
- the quantum volume and density
For unit volume (V=1):
VQ  L3Q
LQ 
Z1 
h
2mkBT
1
 nQ
VQ
the quantum length, coincides [within a factor of
()1/2 ] with the de Broglie length of a particle
whose kinetic energy is kBT.
VQ is the volume of a box in which the ground state energy would be approximately
equal to the thermal energy, and only the lowest few quantum states would be
significantly populated. When V  VQ, the continuous approximation breaks down and
we must take quantization of the states into account. We also need to consider quantum
statistics (fermions and bosons)!
Partition Functions for Distinguishable Particles
There is nothing in the derivation of the partition function or the Boltzmann factor that
restricts it to a microsystem. However, it is often convenient to express the partition
function of a combined macrosystem in terms of the single-particle partition function.
We restrict ourselves to the case of non-interacting particles (the model of ideal gas).
F T ,V , N   kB T ln Z
Recall the analogy between  and Z: S U ,V , N   kB ln 
For a microcanonical ensemble, we know the answer for total: total  1  2  ...   N
How about the canonical ensemble?
Z total   exp   E1 s   E2 s 
For two non-interacting particles, Etotal =E1+E2 :
s
 Z total   exp    E1  s   exp    E2  s    exp    E1  s   exp    E2  s   Z1  Z 2
s
s1
all the states for the composite system
s2
distinguishable
non-interacting distinguishable particles
Z total  Z1  Z 2  ......  Z N
Example: two non-interacting distinguishable two-state particles, the states 0 and :
The partition function
for a single particle:
 0 
  
  
  exp  
  1  exp  

Z1  exp  
 k BT 
 k BT 
 k BT 
The states of this two-particle system are: (0,0), (,), (,0), (0,)
- distinguishable
2
  
  
 2  
  
  exp  
  1  exp  
  Z12
Z total   exp   i   1  2 exp  
1
 k BT 
 k BT 
 k BT  
 k BT 
4
Partition Functions for Indistinguishable Particles (Low Density Limit)
If the particles are indistinguishable, the equation
should be modified.
Recall the multiplicity for IG:
1
=
1 VN
N 
 ( the accessible momentum " volume" )
3N
N! h
If the states (,0) and (0,) of a system of two
classical particles are indistinguishable:
2
2
Z total  Z1  Z 2  ......  Z N
  
  
 2 
  exp  
  Z12
Ztotal   exp   i   1  exp  
1
 k BT 
 k BT 
 k BT 
3
1
(For the quantum indistinguishable particles, there would be three microstates if the
particles were bosons [(0,0), (0,), (,)] and one microstate [(0,)] if the particles were
fermions).
For two indistinguishable
particles:
Z total 
1
1
exp  E1 s  exp  E2 s   Z1  Z 2

2 s1 s2
2
Exceptions: the microstates where the particles are in the same state. If we consider
particles in the phase space whose volume is the product of accessible volumes in the
coordinate space and momentum space (e.g., an ideal gas), the states with the same
position and momentum are very rare (unless the density is extremely high)
For a low-density system of classical
indistinguishable particles:
Z total 
1 N
Z1
N!
Z and Thermodynamic Properties of an Ideal Gas
The partition function
for an ideal gas:
1 N
Z
Z1
N!
Z
1 N  mkBT 

V 
2 
N!
2




3N / 2
1  V 

N !  VQ 
N
Note that the derivation of Z for an ideal gas was much easier than that for : we are
not constrained by the energy conservation!
  V
ln Z  N ln V  ln VQ  ln N  1  N ln 
  NVQ


 
 n
  1  N ln  Q

 
  n
 
 
  1
 
  V  
  nQ  


F T ,V , N   k B T ln Z   NkB T ln
 1   NkB T ln    1


  NVQ  
  n  
F 

Nk BT
  Nk BT
P  
ln V 
V
V
  V T , N
- the “pressure” equation of state (“ideal gas” law)
  nQ 

  nQ  5 
F 
    nQ  

3/ 2






  
  NkB
S  
T
ln

1

Nk
ln

T
ln
T

1

Nk
    

B 
B ln 


 T    n  
T
  T V , N
  n 

  n  2


1  VQ
31 3
U 
ln Z  N
ln VQ  N
N
 Nk BT


VQ  
2 2
U  F  TS 
3
Nk BT
2
- the Sackur-Tetrode equation
- the “energy” equation of state
The Equipartition Theorem (continuous spectrum)
Equipartition Theorem: At temperature T, the average energy of any quadratic
degree of freedom is ½kBT.
We’ll consider a “one-dimensional”
system with just one degree of freedom:
E q   cq
E(q)
2
q – a continuous variable, but we “discretize”
it by considering small and countable q’s:





1
1
2
Z
exp


cq

q

exp  cq 2 dq


q q
q 
1 1
Z
q c
q
 E q 
2
Z   exp 

exp


cq


q
 k BT  q
The partition function for this system:

New variable:

 1
2
exp  x dx  exp  x dx     q

 
2
The average energy:
E 
 
q

c
1 Z
k T
 1
 1
      3 / 2    1  B
Z 
2
 2
 2

x  c q
exp(-x2)
x
Again, this result is valid only if the degree of freedom is “quadratic” (the limit of high
temperatures, when the spacing between the energy levels of an individual system is <<
kBT).
Equipartition Theorem and a Quantum Oscillator
Classical oscillator:
Quantum oscillator:
p 2 kx2
E

2m 2


E
k BT k BT

 k BT
2
2
1
2
 n   n     

k
m
 

exp






2
 



the equipartition function: Z   exp   n   exp     exp  n  
1  exp    
 2
0
0
ln Z  

2
1  exp      exp  2    ...
  ln 1  exp    
the average energy
of the oscillator:
E 
1


 exp     
1
ln Z 

   


2
1  exp    
 2 exp     1
the limit of high T:

    1 exp     1   
k BT
the limit of low T:

    1
k BT
E

2
1
1  1
E    
    k BT
2





(the ground state is by far
the most probable state)


the assumption of a continuous energy spectrum was violated
The Maxwell Speed Distribution (ideal gas)
The probability that a molecule has
an energy between  and +d is:
P d 
  
  
1
1
d 
d
g  exp  
g  exp  
Z1
nQ
 k BT 
 k BT 
3/ 2
1  2m 
2 1/ 2 3 / 2
g    2  2   1/ 2 
  nQ
4   

3D
g()
  

exp  
 k BT 


0
P  
2


 3 / 2 1/ 2 exp  
 

k
T
 B 
P()
g    C 
=
0

0
Now let’s look at the speed distribution for these particles:

1
2
  mv 2
The probability to find a particle with the speed between v and v+dv, irrespective of the
direction of its velocity, is the same as that finding it between  and +d where d =
mvdv:
Pvdv  P d  P mvdv
1/ 2
2
Pv    
 


m 3 / 2 v 2 exp   1  mv2    m 
 2
  2 k BT 
3/ 2
 mv 2 

4v exp  
 2 k BT 
2
Maxwell
distribution
Note that Planck’s constant has vanished from the equation – it is a classical result.
The Maxwell Speed Distribution (cont.)
 m 

Pv dv  
 2 k BT 
3/ 2
 mv2 
 4v 2 dv
exp  
 2 k BT 
vy
 4v 2 dv
The structure of this equation is transparent: the
Boltzmann factor is multiplied by the number of states
between v and v+dv. The constant can be found from
the normalization: 
3/ 2
 m
C  
 2 k BT
 Pv dv  1
0
" volume" v  v  dv 



v
vx
vz
P(v)
  
d
dN    NP d  N exp  
 k BT 
energy distribution, N – the total # of particles
 m 

dN v   NPv  dv  N 
2

k
T
B 

3/ 2
 mv2 
 dv
4v exp  
2
k
T
B 

2
v
speed distribution
dN  v x   NP  v x 
P(vx)
1/ 2
 m 
dv  N 

 2 k B T 
 mv x2 
exp  
 dv x
2
k
T
B


distribution for the projection of velocity, vx
vx
The Characteristic Values of Speed
 m 

Pv   
 2 k BT 
3/ 2
 mv 2 

4v exp  
 2 k BT 
2
P(v)
Because Maxwell distribution is skewed (not
symmetric in v), the root mean square speed is not
equal to the most probable speed:
The root-mean-square speed is proportional to
the square root of the average energy:
E
vmax
v
vrms
1
2
mvrms 
2
vrms 
2E
3k BT

m
m
v
The most probable speed:
2 k BT
 dPv 

0

v

max
 dv 
m
v  vmax
The average speed:

 m 
v   v  P  v  dv  

2

k
T
B


0
3
2 
 mv 2
0 4 v exp   2kBT
3
vmax : v : vrms  2 : 8 /  : 3  1 : 1.13 : 1.22

8k BT
dv


m

Problem (Maxwell distr.)
Consider a mixture of Hydrogen and Helium at T=300 K. Find the speed at which
the Maxwell distributions for these gases have the same value.
 m 

Pv, T , m   
 2 k BT 
 m1 


2

k
T
B 

3/ 2
3/ 2
 mv2 

4v exp  
 2 k BT 
2
 m1v 2   m2 
2
  

4v exp  
2
k
T
2

k
T
B 
B 


3/ 2
 m2 v 2 

4v exp  
2
k
T
B 

2
3
m1v 2 3
m2v 2
ln m1 
 ln m2 
2
2k BT 2
2k BT
3 m1
v2
m1  m2 
ln

2 m2 2k BT
m1
3 1.38 10 23  300  ln 2
m2

 1.6 km/s
m1  m2 
2 1.7 10 27
3k BT ln
v
Problem (final 2005, MB speed distribution)
Consider an ideal gas of atoms with mass m at temperature T.
(a) Using the Maxwell-Boltzmann distribution for the speed v, find the
corresponding distribution for the kinetic energy  (don’t forget to transform
dv into d).
(b) Find the most probable value of the kinetic energy.
(c) Does this value of energy correspond to the most probable value of speed?
Explain.
 m 
(a) dN v   NPv  dv  N 
 2 k T 
B 


dv
2
dN    NP 
d  v 
d
m

 m 

P   
 2 k BT 
(b)
P 
0

3/ 2
3/ 2
 mv2 
 dv
4v exp  
 2 k BT 
2

 m 
1 2

dv 
d   N 
2 m 
 2 k BT 
3/ 2
  1 2
8

exp  
d
m
k
T
2
m

 B 
  
4 2  1/ 2
 

exp
3/ 2
m
 k BT 
   1/ 2
    1 
1 1/ 2
   exp  
   
  0
 exp  
2
 k BT 
 k BT   k BT 
(c) the most probable value of speed
1  max

2 k BT
2 k BT
 dPv 

0

v

max
 dv 
m
v  vmax
the kin. energy that corresponds to the most probable value of speed
1
2
 max  k BT
doesn’t
correspond
  k BT