Transcript PowerPoint
The Final Lecture (#40): Review
Chapters 1-10, Wednesday April 23rd
• Announcements
• Homework statistics
• Finish review of third exam
• Quiz (not necessarily in this order)
• Review Chapters 3 to 7
Reading:
Chapters 1-10 (pages 1 - 207)
Final: Wed. 30th, 5:30-7:30pm in here
Exam will be cumulative
Homework Statistics
14
Number of students
12
10
Mean = 81%
Median = 88%
8
6
4
2
0
20
40
60
Score (%)
80
100
Review of
Chapters 3 & 4
Classical and statistical probability
Classical probability:
•Consider all possible outcomes (simple events) of a process
(e.g. a game).
•Assign an equal probability to each outcome.
Let W = number of possible outcomes (ways)
Assign probability pi to the ith outcome
1
pi
W
&
1
i pi W W 1
Classical and statistical probability
Statistical probability:
•Probability determined by measurement (experiment).
•Measure frequency of occurrence.
•Not all outcomes necessarily have equal probability.
•Make N trials
•Suppose ith outcome occurs ni times
ni
pi lim
N N
Statistical fluctuations
Relative error (ni /ni ) ni1/ 2
Error: ni ni1/ 2 ,
N
0.0
-0.5
log a log N b
log( )
-1.0
-1.5
-2.0
-2.5
-3.0
N
1
0.5
10
0.15
100
0.04
1000 0.0132
10000 0.00356
100000 0.00145
0
1
1/ 2
a 0.516
2
3
log(N)
4
5
The axioms of probability theory
1. pi ≥ 0, i.e. pi is positive or zero
2. pi ≤ 1, i.e. pi is less than or equal to 1
3. For mutually exclusive events, probabilities add,
i.e.
p p1 p2 ........ pr
• In general, for r mutually exclusive events, the probability that one
of the r events occurs is given by:
• Compound events, (i + j): this means either event i occurs, or event
j occurs, or both.
• Mutually exclusive: events i and j are said to be mutually exclusive
if it is impossible for both outcomes (events) to occur in a single
trial.
Independent events
Example:
What is the probability of
rolling two sixes?
Classical probabilities:
p6
1
6
Two sixes:
p6,6 16 16
1
36
•Truly independent events always satisfy this property.
•In general, probability of occurrence of r independent
events is:
p p1 p2 ........ pr
Statistical distributions
6
7
8
9 10
ni
xi
Mean:
nx
x
,
i
N
i i
where N i ni
Statistical distributions
16
ni N
xi
Mean:
x i pi xi , where
ni
pi lim
N N
Statistical distributions
16
ni
xi
Standard
deviation
x
2
p x x
i
i
i
2
Statistical distributions
Gaussian distribution
(Bell curve)
2
1
x x
p( x )
exp
2
2
2
Statistical Mechanics – ideas and definitions
A quantum state, or microstate
• A unique configuration.
• To know that it is unique, we must specify it as
completely as possible...
Classical probability
• Cannot use statistical probability.
• Thus, we are forced to use classical probability.
An ensemble
• A collection of separate systems prepared in
precisely the same way.
Statistical Mechanics – ideas and definitions
The microcanonical ensemble:
Each system has same:
# of particles
Total energy
Volume
Shape
Magnetic field
Electric field
............
and so on....
These variables (parameters) specify the ‘macrostate’
of the ensemble. A macrostate is specified by ‘an
equation of state’. Many, many different microstates
might correspond to the same macrostate.
Ensembles and quantum states (microstates)
Volume V
10 particles, 36 cells
10
1
pi
36
16
3 10
Cell volume, V
Ensembles and quantum states (microstates)
Volume V
10 particles, 36 cells
10
1
pi
36
16
3 10
Cell volume, V
Entropy
Boltzmann hypothesis: the entropy of a system is related to
the probability of its being in a state.
1
p
W
S f n W W
S kB ln W
Rubber band model
d
N!
N!
W N , n
n ! n ! n ! N n !
Sterling’s approximation: ln(N!) = NlnN N
lnW N ln N n ln n N n ln N n
1 x 1 x 1 x 1 x
N
ln
ln
2 2 2 2
Chapters 5-7
•Canonical ensemble and Boltzmann probability
•The bridge to thermodynamics through Z
•Equipartition of energy & example quantum systems
•Identical particles and quantum statistics
•Spin and symmetry
•Density of states
•The Maxwell distribution
Review of main results from lecture 15
Canonical ensemble leads to Boltzmann distribution function:
exp Ei / k BT
exp Ei / k BT
pi
Z
j exp E j / kBT
Partition function:
Z j g j exp E j / kBT
Degeneracy:
gj
Entropy in the Canonical Ensemble
M systems
ni in state yi
M!
WM
n1 ! n2 !..ni !..
ni
S M k B M i
M
ni
ln
M
k B M i pi ln pi
Entropy per system:
S k B i pi ln pi
The bridge to thermodynamics through Z
Z exp E j / k BT ; js represent different configurations
j
F k BT ln Z
T ln Z
ln Z
F
S
k B ln Z T
k B
T V
T
V
T V
2 ln Z
2 ln Z
U TS F k B T ln Z T
k BT ln Z k BT
T
T
V
V
F
U
S
CV
T
T 2
T V
T V
T V
2
A single particle in a one-dimensional box
V(x)
V=∞
0
V=0
n x
n A sin
L
V=∞
0
2 2n 2
2
n
n
2
2mL
x=L
x
A single particle in a three-dimensional box
The three-dimensional, time-independent Schrödinger equation:
2
x, y, z V x, y , z x, y , z x, y , z
2
2m
2 2 2
x y z
2
2
2
2
n1 x n2 y n3 z
i x, y, z A sin
sin
sin
n1 , n2 , n3
L L L
2 2
2
n
n n2 n3 ,
2 1
2mL
2
2
ni 1,2,3...
Factorizing the partition function
Z trans e
n 21
e
n 22
e
n 23
n1 1 n2 1 n3 1
2
2
2mL
n 21 n 22 n 23
e
e
e
n1 1
n2 1
n3 1
n 21 n 22
n 23
e dn1 e dn2 e dn3
0
0
0
mk BT
L
2
2
3
3/ 2
V
3
D
T
mk B L
2 2
2
3/ 2
2
3/ 2
Equipartition theorem
Z trans e
n 21 n 22 n 23
e
n1 1 n2 1 n3 1
e
Z1 Z 2 Z 3
If the energy can be written as a sum of independent terms, then the
partition function can be written as a product of the partition functions
due to each contribution to the energy. Also,
F k BT ln Z1Z2 Z3 k BT ln Z1 ln Z2 ln Z3
free energy may be written as a sum. It is in this way that each degree
of freedom ends up contributing 1/2kB to the heat capacity.
Z N Z1 Z 2 Z 3
N
ln Z N N ln Z1Z 2 Z 3
F Nk BT ln Z1 ln Z2 ln Z3
Rotational energy levels for diatomic molecules
l = 0, 1, 2... is angular momentum quantum number
l
2
l l 1
2I
g l 2l 1
I = moment
of inertia
qR(K)
CO2
I2
HI
HCl
H2
0.56
0.053
9.4
15.3
88
Vibrational energy levels for diatomic molecules
n = 0, 1, 2... (harmonic quantum number)
n n 12
= natural
frequency of
vibration
qV(K)
I2
F2
HCl
H2
309
1280
4300
6330
CP (J.mol1.K1)
Specific heat at constant pressure for H2
CP = CV + nR
H2 boils
5R
2
Translation
9R
2
7R
2
Examples of degrees of freedom:
1
1
2
2
ELC C V L i 12 k BT 12 k BT
2
2
1
1
2
2
EHO k x m v 12 k BT 12 k BT
2
2
1
Etrans m v x2 v 2y v z2 23 k BT
2
1
2
2
Erot ,dia I x y 12 k BT 12 k BT
2
average, or r.m.s. value
Bosons
1
y 2,Bose x1, x2
i x1 j x2 i x2 j x1 y 2,Bose x2 , x1
2
y 3,Bose x1 , x2 , x2 i x1 j x2 k x3 i x2 j x1 k x3
i x2 j x3 k x1 i x3 j x2 k x1
i x1 j x1 k x2 i x1 j x3 k x2
• Wavefunction symmetric with respect to exchange. There are N! terms.
• Another way to describe an N particle system:
y i n1, n2 , n3 ,
Ei n11 n2 2 n3 3
• The set of numbers, ni, represent the occupation numbers associated
with each single-particle state with wavefunction i.
• For bosons, occupation numbers can be zero or ANY positive integer.
Fermions
1
y 2,Fermi x1, x2
i x1 j x2 i x2 j x1 y 2,Fermi x2 , x1
2
• Alternatively the N particle wavefunction can be written as the
determinant of a matrix, e.g.:
i ( x1 ) j ( x1 ) k ( x1 )
y 3,Fermi x1 , x2 , x3 i ( x2 ) j ( x2 ) k ( x2 )
i ( x3 ) j ( x3 ) k ( x3 )
• The determinant of such a matrix has certain crucial properties:
1. It changes sign if you switch any two labels, i.e. any two rows.
It is antisymmetric with respect to exchange
2. It is ZERO if any two columns are the same.
• Thus, you cannot put two Fermions in the same single-particle state!
Fermions
• As with bosons, there is another way to describe N particle system:
y i n1, n2 , n3 ,
Ei n11 n2 2 n3 3
• For Fermions, these occupation numbers can be ONLY zero or one.
2
0
Z Fermi e
/ k BT
e
2 / k BT
e
3 / k BT
Bosons
• For bosons, these occupation numbers can be zero or ANY positive
integer.
Ei n11 n2 2 n3 3
Z Bose 1 e / kBT 2e 2 / kBT e 3 / kBT e 4 / k BT
A more general expression for Z
• What if we divide by 2 (actually, 2!):
1 M i / k BT M j / k BT
Z2 e
e
2! i 1
j 1
12 e 21 / kBT 12 e 2 2 / kBT 12 e 2 3 / kBT
e 1 2 / k B T e
1 3 / k B T
2 3 / k B T
2 4 / k BT
e
e
e 1 4 / kBT
e
2 5 / k BT
• Terms due to double occupancy – under counted.
• Terms due to single occupancy – correctly counted.
SO: we fixed one problem, but created another. Which is worse?
•Consider the relative importance of these terms....
Dense versus dilute gases
Dilute: classical, particle-like
Dense: quantum, wave-like
D
•Either low-density,
high temperature or
high mass
•de Broglie wavelength
D
(mT )1/2
•Low probability of
multiple occupancy
•Either high-density,
low temperature or
low mass
•de Broglie wavelength
D (mT )1/2
•High probability of
multiple occupancy
A more general expression for Z
• Therefore, for N particles in a dilute gas:
ZN
and
Z1
N
N!
F Nk BT ln Z1 ln N 1
VERY IMPORTANT: this is completely incorrect if the gas is dense.
• If the gas is dense, then it matters whether the particles are bosonic
or fermionic, and we must fix the error associated with the doubly
occupied terms in the expression for the partition function.
• Problem 8 and Chapter 10.
Identical particles on a lattice
Localized → Distinguishable
Z N Z1
N
and
F Nk BT ln Z1
Delocalized → Indistinguishable
ZN
Z1
N
N!
and
F Nk BT ln Z1 ln N 1
Spin
3
2
5
2
Fermions :
1
2
Bosons :
0, , 2 , 3 ,....
,
,
1
2
}
Antisymmetric
3
4
12 y space spin
,....
Symmetric
Fermions:
x x x x
12 i x1 j x2 j x1 i x2
12
i
1
j
2
j
1
i
2
Particle (standing wave) in a box
n1 x n2 y n3 z
i A sin
sin
sin
L
L
L
x y z
2
2
2
n1 n2 n3
i
2 2 2
2m Lx Ly Lz
2
Lz
Ly
Lx
2
mk BT
Z Lx Ly Lz
2
2
Boltzmann probability: ni Npi N
e
3/ 2
i / k BT
Z
Density of states in k-space
n1
kx
Lx
kz
n2
ky
Ly
ky
kx
n3
kz
Lz
The Maxwell distribution
In 3D:
V/ 3 is the density of states in k space
2
Vk
D ( k ) 2 ; density of states per unit k interval
2
D(k)dk gives the # of states in the range k to k + dk
Number of occupied states in the range k to k + dk
D3 2 ( k ) / kBT
e ( k ) / k BT
f k dk N
D(k )dk N 2 k e
dk
Z
2
Distribution function f (k):
f k dk N
0
Maxwell speed distribution function
3
2
n(u)
D3 m 3 2 mu2 / 2 kBT
m 2 mu2 / 2 kBT
n u N 2 3 u e
4 N
ue
2 kT
2
u
Density of states in lower dimensions
In 2D:
A/ 2 is the density of states in k space
Ak
D(k )
; density of states per unit k interval
2
D(k)dk gives the # of states in the range k to k + dk
In 1D: L/ is the density of states per unit k interval
Density of states in energy
In 3D:
2
dk
Vk
dk
D( )d D(k ) d 2 d
2 d
d
k
d
k
If (k )
, then
2m
dk
m
2 2
2
Vkm Vm 2m
D( ) 2 2
2
2 2 3
1/ 2
Useful relations involving f (k)
g
k
All k
pk D(k ) pk dk = D(k )
0
e
( k ) / k BT
Z
0
0
0
dk 1
N gk pk ND(k ) pk dk = f (k )dk N
All k
A = D(k ) A(k )
0
e
( k ) / k BT
Z
dk
f (k ) A(k )dk
0
f (k )dk
0
The molecular speed distribution function
m
n u 4 N
2
kT
3/ 2
u 2 exp mu 2 / 2kT
n u
N
u / um
1/ 2
2kT
um
m
u / um
urms / um
1/ 2
8kT
u
m
1/ 2
urms
3kT
m
Molecular Flux
Flux: number of molecules striking a unit area of the container
walls per unit time.
2
dN (u) u n(u)
f u du
du d
dAdt
V
0
/2
0
sin q cos q dq
4
uf (u)du u n(u)du u e
2
uB =
0
f (u)du
0
0
0
4 mu 2 / 2 k BT
un(u)du u e
0
3 mu 2 / 2 k BT
du
du
0
uB u is the average molecular speed in the beam
The Maxwell velocity distribution function
1/ 2
m
N vx N
2 k BT
e
mv x2 / 2 k BT