Download Lecture 1: Review of Quantum Mechanics, Introduction to Statistical

Version: November 7, 2011.
MIT 3.20, Rickard Armiento, Lecture 1
Lecture 1: Review of Quantum Mechanics, Introduction to
Statistical Mechanics, Time Averages vs. State Averages, First
Postulate
• October 14, 2011, McQuarrie 1.1 - 2.1
Formalities
Lectures by: Rickard Armiento, armiento mit.edu, (617) 715-4275, room 13-5065.
• Next 4 weeks: fundamentals of statistical mechanics. This ends with a partial exam, (25%
of grade.
• Book: McQuarrie “Statistical Mechanics”, (alt. “Statistical Thermodynamics”; both versions work). Chapters 1-6 + 11.
Introduction to Statistical Mechanics
Thermodynamics = very general macroscopic theory.
Statistical mechanics connects microscopic models with thermodynamics.
Statistical Mechanics = study of macroscopic systems from an atomistic point of view. “Many, many particles + statistics → the macroscopic world”.
Example: heat capacity. Why is the temperature dependence of the
heat capacity of Ag shaped like this? Solid of N vibrating ions
⇒? thermodynamical quantities, Cv , etc., Cv grows with T and
approaches 3kB T .
Will find that entropy is connected to the “uncertainty” of microscopic states. Counting of states → we will frequently enter into
combinatorics.
Goal: understand and predict macroscopic phenomena starting
from models of microscopic interactions between the atoms/particles in the system + statistics.
1
Statistical Mechanics
Statistical Mechanics connects microscopic models with
thermodynamics.
• What is entropy and internal energy in “reality”?
• Can we understand the second + third law?
Many, many particles + statistics → the macroscopic
world.
Quantum Mechanics
Statistical Mechanics
Thermodynamics
Heat Capacity, Experimental vs. Model
MIT 3.20, Rickard Armiento, Lecture 1
Short Review of Quantum Mechanics
Review of Quantum Mechanics,1
• The state vector: |Ψi represents the state of the system.
• Wave-function formulation: Ψ(q̄, t) ← wave function
q̄ = {set of coordinates necessary}, e.g. q̄ = (r̄, σ̄) × N.
• Probability interpretation: probability that the system is
between (q̄, q̄ + dq̄): Ψ∗ (q̄, t)Ψ(q̄, t)dq̄.
• Probability that the system is in any state:
Z
Ψ∗ (q̄, t)Ψ(q̄, t)dq̄ = 1 (normalized wavefunctions).
all q̄
Review of Quantum Mechanics, 2
• Time dependent Schrödinger eq.:
ĤΨ(q̄, t) = i~
∂Ψ(q̄, t)
∂t
• Stationary state: Ψ(q̄) represents the system if the
probability density does not change with t.
• Ψ(q̄) is a solution to the time-independent Schrödinger
eq. (an eigenvalue eq.)
ĤΨ(q̄) = EΨ(q̄),
~ ∂ 2 ψ(x)
+ v(x)ψ(x) = ψ(x)
2m ∂x2
Equation + boundary conditions ⇒ set of solutions:
E.g., one particle, 1D:
−
Ψ1 , Ψ2 , ... and eigenvalues E1 , E2 , ...
2
MIT 3.20, Rickard Armiento, Lecture 1
ĤΨ(q̄) = EΨ(q̄) and boundary conditions ⇒
Ψ1 , Ψ2 , ... and eigenvalues E1 , E2 , ...
Degenerate states:
Ω(E) = # states with energy E (discrete) = degeneracy
Ω(E)dE = # states between E and E + dE. (continous) = density of states
Simple 1D Examples:
1. “Particle in a box”: Single particle in a 1d infinite well:
Ĥ = −
h̄2 ∂ 2
+ v(x);
2m ∂x2
r
Ĥψ = ψ
⇒
ψn =
v(x) =
2
nπx
sin(
),
a
a
0 if 0 < x < a
∞ otherwise
n =
h2 2
n , n = 1, 2, 3, ...
8ma2
a
2. Harmonic oscillator:
Ĥ = −
r
ψn =
mω 1
h̄π 2n n!
h̄2 ∂ 2
1
+ mω 2 x2 ,
2
2m ∂x
2
1/2
r
hn
Ĥψ = ψ
mω 2
mω
x e− h̄ x /2 ,
h̄
⇒
n = (n + 1/2)h̄ω, n = 0, 1, 2, ...
ħω
Other examples: Finite well, delta-function potential, linear potential.
A First Meeting with Degenerate Microstates
Degeneracy in typical 3D system: How degenerate are the states? Start with one particle in a
box.
3D infinite well: Ĥψ = ψ:
h̄2
Ĥ = −
2m
∂2
∂2
∂2
+
+
∂x2 ∂y 2 ∂z 2
⇒ nx ,ny ,nz =
+ v(x, y, z),
v(x, y, z) =
h2
1 2
(n2 + n2y + n2z ) =
(n + n2y + n2z ),
8ma2 x
M x
0 if 0 < x, y, z < a
∞ otherwise
nx , ny , nz = 1, 2, 3, ...
Density of energy levels: E.g., m = 10−22 g, a = 10 cm ⇒ 1/M = 10−41 J = 10−22 eV. Very,
very dense.
3
E = kx²/2
MIT 3.20, Rickard Armiento, Lecture 1
nx
5
4
3
2
1
Count the ways n2x + n2y + n2z give the same number. Degeneracy ω = one quadrant of a sphere:
R2 = n2x + n2y + n2z .
ω() = # states as a function of the energy,
r
1 2 3 4 5 ny
φ() = # states with energy ≤ Degeneracy in 3D well
25
9000
8000
7000
6000
5000
4000
3000
2000
1000
00
10
15
States, φ(²)
States ω(²)
20
10
5
00
4
2
M²
6
8
75
50
25
00 1 2 3 4 5
5
10
M²
15
20
25
Idea: φ ≈ Volume of sphere quadrant/(volume of one state):
1
φ() =
8
4πR3
3
π
π
= (M )3/2 =
6
6
8ma2 h̄2
3/2
.
# states between and + ∆:
π 3/2 ( + ∆)3/2 − 3/2 =
M
6
π
= f ( + ∆) ≈ f () + f 0 ()∆ + O(∆2 ) = M 3/2 1/2 ∆+O(∆2 ).
4
ω(, ∆) = φ( + ∆) − φ() =
Same numbers as before, one gas particle: m = 10−22 g, a =
10 cm, at room temperature: ∼ kB T with T = 300 K. The
number of states in an energy interval of 1% (∆ = 0.01) is
ω(, ∆) ∼ 1028 .
Many indistinguishable non-interacting particles: (see McQuarrie
1.3), Ω(E, ∆E) ∼ 10N , N ∼ 1023 .
4
Many Noninteracting Indistinguishable
Particles
Ω(E, ∆E) =
1
Γ(N + 1)Γ[(3N/2)]
2πma2
h2
3N/2−1
⇒ Ω(E, ∆E) ∼ 10N , N ∼ 1023
E3N/2−1 ∆E
MIT 3.20, Rickard Armiento, Lecture 1
Time Averages and State Averages
Many microscopic states ⇔ one macroscopic / thermodynamic state
(e.g., fixed E; V , N .) Our particles in a box can be in many, many
degenerate states ∼ 10N .
Microstates vs. Macroscopic state
Macroscopic,
"Thermo state"
E, V, N
Starting point of statistical mechanics: many macroscopic quantities we observe (e.g., p) are changing microscopic quantities that
has been time averaged.
The idea is valid both for classical physics and quantum mechanics. In classical physics each particle is described by position and
momentum, q̄ = {r̄i , p̄i }N
i .
Time average over classical microstates:
1
Ē = hE(t)i =
∆t
Z
t+∆t
E(r̄, p̄)dt.
t
Time average over quantum mechanical states:
Ē = hE(t)i =
1
∆t
Z
Microstates
Starting point of statistical mechanics: many
macroscopic quantities we observe (e.g., p) are changing
microscopic quantities that has been time averaged.
1st postulate: The ensemble average of a
thermodynamic property is equivalent to the
time-averaged macroscopic value of the property
measured for the real system.
Time averages
t+∆t
hΨ(q̄, t)|Ĥ|Ψ(q̄, t)idt.
Macroscopic,
"Thermo state"
t
E, V, N
p
Calculating these time averages is hard!
t
Ensemble of microstates: A copies of the system, all in the same
thermodynamical state, with N , V and E fixed. All the degenerate
Ω(E) states are present equal number of times = an “ensemble of
points in the phase space”.
1st postulate: The ensemble average of a thermodynamic property is equivalent to the time-averaged macroscopic value of the
property measured for the real system.
Microstates
Ensemble
State average over classical microstates:
Z
Ē = hE(t)i = E(r̄, p̄)P (r̄, p̄)dr̄dp̄.
A collection of points (volumes) in phase space.
P (r̄, p̄)dr̄dp̄ ∼ measure of time that the system spends in a state
between [r̄; p̄, r̄ + dr̄; p̄ + dp̄].
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MIT 3.20, Rickard Armiento, Lecture 1
State average over quantum mechanical states:
Ē = hE(t)i =
X
Ei Pi
i
Formally: Pi = probability to find the system in state i. (Informally: “probability of state i”.)
Example: sleepy students: N = 40 students falling asleep / waking up independently, probability
of being asleep = Psleep . How many are asleep?
n̄sleep

1st postulate

1
The ensemble = many copies of the lecture hall,
=
nsleep (t)dt =
= hn(t)i =


∆t ∆t
all possible combinations of sleeping students: 2N .
N
=


Z
2
X
nsleep
Pi
i
=
i=1
=
Group all states with same ni ;
Probability that exactly n students are sleeping = Pn
N
X
n=0
=
N
X
n=0
N!
nPn =
n!(N − n)!
N!
nP n (1 − Psleep )N −n = ... = N Psleep .
n!(N − n)! sleep
Given Pi → we can calculate everything! This will be a central concept of statistical mechanics.
Examples (glance ahead): Constant E, V, N → Pi constant. Constant N, V, T → Pi ∼ e−Ei /kT
Different boundary conditions → different Pi .
Homework suggestions
1.
2.
3.
4.
Think about sleepy students example, why is count of states with the same ni =
Exam problem 2001E2:3(a).
Exam problem 2004E2:1(a).
Exam problem 2007E2:3(a).
6
N!
n!(N −n)! .
MIT 3.20, Rickard Armiento, Lecture 2
Lecture 2: Microcanonical and Canonical Ensemble
• October 17, 2011, McQuarrie 2.1 - 2.2
Previous lecture
Ensemble of microstates: a collection of microstates that all are
in the same thermodynamical state.
Time averages
Macroscopic,
"Thermo state"
E, V, N
Postulate: The ensemble average of a thermodynamic property is
equivalent to the time-averaged macroscopic value of the property measured for the real system.
p
t
Microstates
State average over quantum mechanical states:
Ē = hE(t)i =
X
Ei Pi
Ensemble
i
Formally: Pi = probability to find the system in microstate i at a
specific instance in time. (Informally: “probability of state i”.)
The Microcanonical Ensemble
Isolated system, constant volume ⇒ E, N , and V fixed (both microscopically, i.e., in each microstate; and macroscopically).
Degenercay: Ω(E). Microcanonical ensemble contains all these degenerate states.
Ensemble average?
M̄ = hM (t)i =
X
Mi P i .
i
Where do we get Pi ?
Equal a Priori Probability:
Principle of equal a priori probability: Given an
isolated system in equilibrium, it is found with equal
probability in each of its accessible microstates.
Assume that all microstates are equally likely ⇒
7
A collection of points (volumes) in phase space.
MIT 3.20, Rickard Armiento, Lecture 2
State probability in microcanonical ensemble
Pi = 1/Ω(E) = constant.
The ergodic hypothesis was one (historical) motivation for this postulate.
The Ergodic Hypothesis
(Isolated, noninteracting particles)
A
B
For states degenerate in energy: over long time, we pass
through them equally much.
Equal a priori probability:
PA = P B
But of all possible degenerate states,
How many "look like" A? Like B?
Pspread out >> Pon left side
This specific hand:
2,598,960 : 1
This specific hand:
2,598,960:1
One pair
1.36 : 1
Royal straight flush
649,739 : 1
Microcanonical ensemble averages:
Ē =
X
Ei Pi =
i
1 X
EΩ
=E
Ei =
Ω(E)
Ω
i
M̄ =
X
p̄ =
X
Mi Pi =
i
i
1 X
Mi
Ω(E)
i
pi Pi =
1
Ω(E)
X
pi
i
Canonical Ensemble
System with constant N, V , but not thermally isolated. It is in thermal equilibrium with the
environment at temperature T (=“macroscopic temperature”, we do not yet know what the temperature is in terms of microstates, i.e. “in the microscopic world” where all microscopic degrees
of freedom are specified.)
Canonical ensemble C = all microstates corresponding to fixed N , V , T .
8
MIT 3.20, Rickard Armiento, Lecture 2
Ensemble average: sum the microcanonical way?:
M̄ =?
X
Mi P i
i
But, what are the {Pi } now?
System is not isolated ⇒ all accessible microstates generally not equally likely.
Cunning plan: construct a microcanonical ensemble that simulates a system in the canonical
ensemble + heat bath and use principle of equal a priori probabilities in the microcanonical
ensemble to derive probabilities Pi .
Step 1
1
1
2
3
3
2
...
1
...
1
2
3
...
1
2
3
2
1
A 2
N,V,T
...
3
... A
N,V,T
3
...
A
N,V,T
A
N,V,T
A
N,V,T
Abstract canonical ensemble,
all entries share the same
N, V, T.
All states NOT equally likely
𝒜
N,V,T
All possible states of heat bath =
microcanonical ensemble. Every
entry has the same total
E=𝓔 , N=𝒩 , V=𝒱.
All states equally likely
Simulated heat bath of 𝒜 systems,
1. Bring to thermal equilibrium at T
2. Isolate ⇒ E=𝓔 , N=𝒩 , V=𝒱.
Step 2
All possible states?
Every microstate of the
heat bath has the same total
E=𝓔 , N=𝒩 , V=𝒱.
1
1
2
3
2
3
1
...
1
...
2
1
3
A
3
2
A 2
N,V,T
...
The abstract canonical ensamble = all possible subsystem states.
Lay out 𝒜 subsystems in all ways that give E=𝓔 , N=𝒩 , V=𝒱.
= Combinatorial problem!
1
3
2
...
...
3
... A
A
A
1
2
3
...
A
A
Each microstate = same probability →
Subsystem probability ~ number of times it occur.
Create a very large “heat bath” system out of A microstates from the canonical ensemble, labeled
j = 1, 2, ... A . Bring the system to thermal equilibrium at T (using, e.g. an external heat bath),
and then isolate it. All possible microstates of this heat bath system = a microcanonical ensemble
9
MIT 3.20, Rickard Armiento, Lecture 2
D ⇒ 1) All microstates in D have the same energy value = E . 2) Principle of equal a priori
probabilities ⇒ finding the system in any of the possible microstates in D is equally likely.
Main idea: determine the probabilities by counting how all the subsystems within the equally
likely microstates in D distribute over the possible states:
Pi =
Number of subsystems in state i within all the microstates in D
Total number of subsystems in all the microstates in D
Average number of subsystems in state i over the microstates in D
=
Number of subsystems in one microstate of D
The canonical ensemble C contains every state a subsystem in D can be in. All possible microstates of D = every way systems in C can be mapped onto the labeled (thus distinguishable)
subsystems of D under the constraint that the total energy sums to E ⇒ a straightforward combinatorial problem!
Categorize entries in D according to the distribution of subsystems over quantum states: There
are ai subsystems in state i, a = {ai }. Count possible quantum states in i, labeled subsystems
in j.
X
X
X
X
Number of microstates: A =
1=
ai , total energy: E =
Ej =
ai Ei
j
Total number of subsystems
in state i
i
j
X
=
All distributions
=
X
i
ai in the
Microstates in D with
=
×
distribution
this distribution
W (a)ai (a).
a
What is W (a)?: Number of ways A labeled subsystems can be divided with a1 in one group, a2
in a second group, etc. ...
W (a) =
A!
A!
=
.
a1 !a2 !...
Πi ai !
Total number of subsystems in all systems in D:
X Number of subsystems X X
X
X
X
=
W (a)ai (a) =
W (a)
ai (a) = A
W (a).
in state i
i
i
a
a
P
⇒
Pi =
W (a)ai (a)
aP
A
a W (a)
=
i
Probability to find the system
.
in quantum state i
10
a
MIT 3.20, Rickard Armiento, Lecture 2
Finished? In principle we could calculate averages using this expression.
X
Mi P i
M̄ =
i∈All states
But dealing with Pi in this form is still hard.
Most probable distribution
Binomial distribution
W ({ai }) very peaked around a small set of {ai }. Larger and larger
heat bath ⇒ A → ∞ ⇒ more and more peaked (See McQuarrie
1.5)
Idea: find the most probable distribution a∗ that maximizes W ({ai })
over all {ai }. Assume it is enough to consider this distribution in
the limit A → ∞, ai → ∞.
P
W (a)ai (a)
1 W (a∗ )a∗i
1
a∗i
a
P
= {A → ∞} =
Pi =
=
A
A W (a∗ )
A
a W (a)
Purely
Pmathematics: find the {ai } that maximizes W (a) =
E = i ai Ei .
A!
Πi ai ! ,
a*
constraints: A =
P
ai , and
W (a) contains large factorials. Maximize ln W (a) instead, so we can use Stirling’s approximation, ln n! = n ln n − n.
Langrange multipliers (see McQuarrie 1.5) ⇒
(
)
X
X
∂
ln W (a) − α
ak − β
ak Ek = 0,
∂ai
k
i = 1, 2, ...
k
α and β are undetermined Lagrange multipliers. Insert Stirlings’ approximation:
⇒ − ln a∗i − α − 1 − βEi = 0
⇒
0
a∗i = e−α −βEi ,
α0 = α + 1.
1 −α−βEi
e
A
For a system in temperature equilibrium with its environment at thermodynamic temperature
T , the probability to find it in a state with energy Ei , falls off exponentially with that energy.
Pi =
Note: this is per individual energy. But a typical system has more energy levels at higher energies
⇒ likely to find the system in some intermediate state.
State averages:
M̄ =
X
i∈All states
Pi Mi =
X
P (Ei )Mi =
i
X
i
But, what are α and β? → story for next time.
11
e−α−βEi Mi
MIT 3.20, Rickard Armiento, Lecture 2
Suggested homework
•
•
•
•
McQuarrie 2.2
McQuarrie 2.1
McQuarrie 2.8
Exam problem: 2009E2 3(i).
12
MIT 3.20, Rickard Armiento, Lecture 3
Lecture 3: Identification of Beta and Entropy in the Canonical
Ensemble
• October 19, 2011, McQuarrie 2.2 - 2.3
The Story So Far
Ensemble: collection of microstates that appear the same macroscopically / thermodynamically.
P
Ensemble averages: thermodynamical quantities are given as: M̄ = i Mi Pi
Pi = probability of finding a system in microstate i of the ensemble.
Microcanonical Ensemble: E, N , and V fixed ⇒ Pi = 1/Degeneracy = 1/Ω(E) = constant.
Canonical Ensemble: N , V , and T fixed ⇒ Pi = (1/A )e−α−βEi , Ei = The energy of state
i; α, β are so far unknown Lagrange multipliers that may depend on N, V, T .
Identifying α
Just part of the normalization:
Pi =
1 −α−βEi
=
e
A
e−α
A
e−βEi
Normalize: The probability for the system to be in any state = 1, solve for α.
1=
1 −α X −βEi
e
e
A
⇒
eα =
i
1 X −βEi
e
A
e−βEi
Pi = P −βE .
i
ie
⇒
i
This denominator will occur very frequently in our expressions, so define:
The canonical partition function:
Q=
X
e−βEi =
i∈All states
X
Ω(Ei )e−βEi =
i∈All energies
X
Ω(E)e−βE
E
Q is a “weighted count of all microstates” (more on this later.)
Check: canonical ensemble with just one energy Ei = E is a microcanonical ensemble:
Pi =
e−βEi
e−βE
1
=P
=
.
−βE
Q
Ω(E)
Ω(E)e
E
13
MIT 3.20, Rickard Armiento, Lecture 3
The meaning and source of β
Take a system with a specific set of energy states {Ei }. The average energy is set by β:
P
−βEi
X
X Ei e−βEi
i Ei e
P −βE 0 = P
Ē =
Ei Pi =
−βE
i
i
ie
i0 e
i
i
Why is there a free parameter in the canonical ensemble probabilities that lets us choose the average energy?
Step 1
Recall the derivation: the simulated heat bath was first in thermal
equilibrium with its environment at T . Then, the system was isolated which resulted in a fixed total energy = E . This is the source
of the freedom. E is directly related to the average energy (and thus
β),
X
X
E =
ai Ei =
A Pi Ei = A Ē
i
1
1
2
3
1
Abstract canonical ensemble,
all entries share the same
N, V, T.
All states NOT equally likely
Simulated heat bath of 𝒜 systems,
1. Bring to thermal equilibrium at T
2. Isolate ⇒ E=𝓔 , N=𝒩 , V=𝒱.
Identifying β
Recall the “thermodynamic equation of state”:
dE = T dS−pdV ⇒
=T
N,T
∂S
∂V
−p
N,T
∂V
∂V
⇒
N,T
∂E
∂V
−T
N,T
∂p
∂T
= −p
V,N
Idea: compose the same relation from ensemble averages, and
β. Butwhat
identify
should be
∂p
∂ Ē
held constant? Try β. I.e., seek out relation between: 1) p̄, 2) ∂V
, and 3) ∂β
.
N,β
V,N
Energy from ensemble average: (derived in previous section)
P
{Measured energy} = Ē(N, V, β) =
)e−βEi (N,V )
.
Q(N, V, β)
i Ei (N, V
(1) Pressure from ensemble average: A system in state j: dEi = −pi dV is the work done on the
system for volume change dV , thus, we understand what pressure is for a single microstate:
∂Ei
pi = −
∂V N
Macroscopic pressure:
P ∂Ei p̄ =
X
pi Pi = −
i
∂V
N
e−βEi (N,V )
Q(N, V, β)
i
14
3
...
1
2
...
1
3
2
1
A 2
N,V,T
3
...
A
N,V,T
...
3
... A
N,V,T
A
N,V,T
All possible states of heat bath =
microcanonical ensemble. Every
entry has the same total
E=𝓔 , N=𝒩 , V=𝒱.
All states equally likely
i
∂E
∂V
2
A
N,V,T
𝒜
N,V,T
At this point it may seem logical to solve for β(E ) and then seek
E (T ). However, it turns out to be much better to directly seek β(T ).
3
2
...
MIT 3.20, Rickard Armiento, Lecture 3
(2) The energy derivative:
∂ Ē
∂V
n
= Calculus: f (x) =
N,β
g(x)
h(x)
⇒ f 0 (x) =
h(x)g 0 (x)−h0 (x)g(x)
h2 (x)
"
!
X
X ∂Ei 1
∂Ei
−βEi
−βEi
Q
e
−
Ei β
e
−
Q2
∂V N
∂V N
i
i
∂ Ē
∂V
o
=
!
!#
X ∂Ei X
−βEi
−βEi
−
β
e
Ei e
=
∂V N
i
i
= −p̄ + β(Ep) − β Ē p̄.
N,β
(3) The pressure derivative
∂p
⇒
= {Similar to energy derivative} = Ē p̄ − (Ep)
∂β N,V
Assembling (1-3):
⇒
∂ Ē
∂V
+β
N,β
∂ p̄
∂β
= −p̄
N,V
Compare ensemble averages vs. thermodynamics:
∂ Ē
∂V
+β
N,β
∂ p̄
∂β
= −p̄
⇐⇒
N,V
∂E
∂V
−T
N,T
∂p
∂T
= −p
N,V
Wrong sign? How do we fix this? Are the more than one solution?
∂ p̄
∂p
dT
dβ
β
= −T
⇒−
=
⇒
∂β N,V
∂T V,N
T
β
− ln T = ln β + C ⇒ − ln T − ln k = ln β ⇒ ln
β=
1
,
kT
e−Ei /(kT )
Pi = P −E /(kT ) ,
i
ie
1
= ln β
kT
k = constant.
What did we learn? In the derivation of the canonical ensemble, β appeared due to the freedom
created by the total energy of the simulated heath bath being left undetermined, E . We now have
the exact connection between the variable β and the thermodynamic temperature T . This defines
temperature in the “microscostate picture”.
15
MIT 3.20, Rickard Armiento, Lecture 3
What is k in β = 1/(kT )?
Canonical ensemble with pairs of systems: like the canonical ensemble, but every subsystem =
two separate systems in thermal equilibrium. Possible states A and B, energy state occupations
{ai }, {bi } ⇒
A ! B!
W (a, b) =
.
Πi ai ! Πi bi !
Most probable distribution:
A
B
e−βEi
e−βEi
Pij = P
= PiA PiB .
P
A
−βEi0
−βEiB0
i0 e
i0 e
From thermodynamics ⇒ the systems have the same T . From above ⇒ same β = 1/(kT ). ⇒
Two arbitrary systems in thermal equilibrium have the same k ⇒ k is a general numerical constant = kB . Take numerical value from any system, e.g., ideal gas;
kB = 1.38 · 10−23 J/K = 8.6 · 10−5 eV/K
Identifying the Entropy
Similar strategy as for β: use 2nd + 1st laws to express thermodynamical entropy in internal
energy change + work:
1st law
δQrev = T dS ⇒ dS = (1/T )δQrev =
= (1/T )(dE − δWrev )
dE = δQrev + dWrev
Corresponding quantities from canonical ensemble: Let the systems in the ensemble do some
reversible work: infinitesimal change of state (all system in the ensemble change by dV ) along
nice reversible path, in contact with heat bath so T = const.
X
X ∂Ei X
δ W̄rev = −p̄dV = −
Pi (pi )dV =
Pi
dV =
Pi dEi .
∂V N
i
i
i
P
Recall: partition function: Q = i e−βEi = Q(N, V, T ) = “weighted sum over all states the
system can occupy”. Microcanonical case {Ei } = 0 ⇒ Q = Ω(E0 ).
What can we construct that looks like dS? Q is a weighted count of the microstates. Entropy is
somehow connected to the number of microstates. But, entropy is additive and Q is multiplicative. Try: ln Q.
P
Define: f (β, {Ei }) = ln Q = ln( i e−βEi )
Total differential:
df =
∂f
∂β
X ∂f dβ +
dEi .
∂Ei β,Ej6=i
Ei
i
16
MIT 3.20, Rickard Armiento, Lecture 3
Note:
1.
2.
∂f
∂Ei
P
− i Ei e−βEi
∂f ∂Q
·
=
= −Ē
=
∂Q ∂β
Q
Ei
∂f ∂Q
βe−βEi
=
=−
= −βPi
∂q ∂Ei Ei6=j
Q
β,Ej6=i
∂f
∂β
This gives
df = −Ēdβ − β
X
Pi dEi ⇒
i
d(f + β Ē) = df + Ēdβ + βdĒ ⇒
⇒ d(f + β Ē) = β dĒ − δ W̄rev
Compare state averages and thermodynamics:
d(f + β Ē) = β dĒ − δ W̄rev
⇐⇒
dS =
1
(dE − δWrev )
T
We have found a way to define entropy on the microstate level of the theory, which is consistent
with thermodynamics!
dS = kB d(f + β Ē) = d(kB (f + β Ē)) ⇒
S = kB ln Q + Ē/T + constant
Set constant = 0. (Only ∆S is relevant in thermodynamics, constant related to third law.)
Entropy in canonical ensemble:
S = kB ln Q +
Ē
T
What did we learn? We can understand what entropy is on the microstate level! Its main component is: ln of a weighted count of the number of microstates. S is not a mechanistic entity, there
is no Si to be summed up for all states. S is a property of the whole probability distribution.
Its meaning is roughly “how large is the space of microstates this system may be in, given its
current thermodynamic properties”. (More on this next lecture.)
Entropy in the Microcanonical Ensemble
Q=
P
ie
−βEi
= Ei = E = Ω(E)e−βE ⇒
S = kB ln(Ω(E)e−βE ) +
Ē
T
⇒
Entropy in the microcanonical ensemble:
S = kB ln Ω,
(This is written on Boltzmann’s grave).
17
MIT 3.20, Rickard Armiento, Lecture 3
A Very General Definition of Entropy, Gibbs’ Entropy Formula
S = −kB
P
i Pi ln Pi .
(see McQuarrie problem 2.5, hint: just plug in Pi = e−βEi /Q.)
Suggested Homework
•
•
•
•
McQuarrie 2.5
Exam problem: 2006E2-1(a)
Exam problem: 2001E2-3(b)
Exam problem: 2001E2-5(a-d) (we have not talked about alloys yet, but just use the entropy expression for the microcanonical ensemble and combinatorics.)
18
MIT 3.20, Rickard Armiento, Lecture 4
Lecture 4: Entropy Discussion. Physical Insights from Statistical
Mechanics.
• October 21, 2011, parts from McQuarre 1.1 - 2.4
Past Lectures
Probabilities in the microcanonical ensemble: Pi = 1/Ω(E) = constant
Canonical partition function: Q =
P
ie
−Ei /(kB T )
=
P
E
Ω(E)e−E/(kB T )
Probabilities in the canonical ensemble: Pi = e−Ei /(kB T ) /Q
Entropy:
S=
kB ln Ω (Microcanonical)
(Canonical)
kB ln Q + Ē
T
= −kB
X
Pi ln Pi .
i
S is a property of the probability distribution of states, not a regular variable.
Interpretation of the Partition Function
Partition function = “how the probabilities are partitioned among
the microstates”.
Typical for probabilities:
Weight of one thing
P =
Summed weight of all things
Ensemble
⇐⇒
Pi =
e−Ei /(kB T )
,
Q
Q=
X
e−Ei /(kB T ) .
i
This interpretation fits well, since we got the sum in Q from normalizing the probabilities in the nominator. Hence, Q is the “sum of all
states multiplied by their weight” = “the size of the state space”.
Also note: the “size of the state space” is directly controlled by T .
Solving for Microstates
Solve ĤΨ = EΨ ⇒ energy levels with degeneracy:
Count like this: Ψj,k , where j = energy state (0, 1, ...) and k counts
the degenerate states for that energy k = 1, 2, ...Ω(Ej ).
E0 : {Ψ0,1 , Ψ0,2 , Ψ0,3 , ..., Ψ0,Ω(E0 ) }
E1 : {Ψ1,1 , Ψ1,2 , Ψ1,3 , ..., Ψ1,Ω(E1 ) }
E2 : {Ψ2,1 , Ψ2,2 , Ψ2,3 , ..., Ψ2,Ω(E2 ) }
19
A collection of points (volumes) in phase space.
MIT 3.20, Rickard Armiento, Lecture 4
... Probabilities Pi = Pj,k .
Example from lecture 1: one independent particle: ω ∼ 1028 .
23
Many non-interacting particles: Ω ∼ 1010 .
Physics is independent of E0
Q=
X
e−βEi = e−βE0
X
i
e−β(Ei −E0 ) = e−βE0
e−β∆Ei ;
(∆Ei ≥ 0); ⇒
i
i
Pi = Pj,k =
X
e−βE0 e−β∆Ej
e−β∆Ej
P
P
=
−β∆Ej
e−βE0 j,k e−β∆Ej
j,k e
This shows that the probabilities are independent of E0 , you can set your energy scale so that
E0 = 0. A partition function with only one energy E = 0:
X
Q=
e−β0 = Ω
j=[0]
3rd Law in Statistical Mechanics; S → 0 as T → 0 (β → ∞)
e−β∆Ej
−β∆Ej
j,k e
Pi = Pj,k = P
∆E0 = 0 (Groundstate); lim e−β∆E0 → 1.
β→∞
∆Ei > 0 (Excited state); lim e−β∆Ej → 0.
β→∞
⇒
lim P0 = lim P
β→∞
β→∞
e−β∆E0
1
=
−β∆E
j
Ω(E0 )
j,k e
e−β∆E0
=0
−β∆Ej
j,k e
lim Pi6=0 = lim P
β→∞
β→∞
No ground state degeneracy:
P0 = 1
Pi6=0 = 0
S = −kB
X
Pi ln Pi = 0.
Consistent with 3rd law.
i
Physics does not depend on absolute S, but the 3rd law sets S = 0 for T → 0, which for no
ground state degeneracy is consistent with our choice of ‘constant = 0’ when we derived entropy;
since ln 1 = 0.
20
MIT 3.20, Rickard Armiento, Lecture 4
With ground state degeneracy:
1
P0 = Ω(E
0)
Pi6=0 = 0
S = −kB
X
Pi ln Pi = kB ln Ω(E0 )
i
Inconsistent with 3rd law? But: the ground state degeneracy is generally small compared to the
huge numbers involved for larger T ⇒ 3rd law is valid in the macroscopic / thermodynamic
limit.
Typical example, Ω(E0 ) ∼ N , say 1023 ⇒ kB ln N = 1.38 · 10−23 ln(10)23 ∼ 10−21 J/K ≈ 0.
Infinite Temperature Limit, T → ∞, (β → 0)
e−β∆Ei → 1 for all states, All states are completely accessible.
Pi → constant, same for all states
S = Maximal, state space is as large as it can be.
This is a general observation: if all states are accessible and equally probable, S is maximal.
Entropy for Intermediate Temperatures 0 < T < ∞
Non-uniform distribution Pi .
Intermediate entropy between S = 0 and S = Smax (possibly ∞).
Probability distribution
Example: solid
Ordered
Liquid
Disordered
Eordered < Edisordered < Eliquid
T low: Pordered > Pdisordered , Pliquid . System resides in “few” states,
disorder energy-wise too expensive.
T intermediate: The many more energy levels available. Disordered
phase has higher energy but higher degeneracy. Pdisordered more
significant. Possibly other ordered states.
T high: Pliquid more significant. Ω(Eliq ) >> Ω(Eordered ). (More
about this later, 2nd law spontaneous processes).
21
MIT 3.20, Rickard Armiento, Lecture 4
Interpretation of Entropy
(Isolated, noninteracting particles)
The partition function ∼ size of state space. Temperature acts as a
tuning knob that limits the state space. ∆Sworld increase ⇒ “nature
wants to increase its state space” = heat transfers from one system
to another if this increases the total state space. Why? Purely because of the statistics involved. All states are “possible”, but larger
state space = more likely to find the system there. Huge numbers
involved ⇒ laws of thermodynamics. The balance is controlled
by the
in state space size to a change in internal energy;
response
∂S
1
∂U V,N = T .
A
B
Equal a priori probability:
PA = P B
But of all possible degenerate states,
How many "look like" A? Like B?
Pspread out >> Pon left side
This specific hand:
2,598,960 : 1
This specific hand:
2,598,960:1
One pair
1.36 : 1
Royal straight flush
649,739 : 1
Connection to disorder: There are almost always more disordered
states than ordered ones ⇒ disorder = high entropy, order = low
entropy, i.e., the increase of state space usually means the world
evolve from order → disorder.
Arrow of time, direction of memory: with reversible equations of
motions, why is time moving forward? ∆S > 0 and the idea that
disorder increases breaks time symmetry.
Entropy in Information Theory
Claude Shannon discussed the nature of “lost information” in communication signals.
Claude Shannon
“My greatest concern was what to call it. I thought of
calling it ‘information’, but the word was overly used, so I
decided to call it ‘uncertainty’. When I discussed it with
John von Neumann, he had a better idea. Von Neumann
told me, ‘You should call it entropy, for two reasons. In the
first place your uncertainty function has been used in
statistical mechanics under that name, so it already has a
name. In the second place, and more important, nobody
knows what entropy really is, so in a debate you will
always have the advantage.”
M. Tribus, E.C. McIrvine, “Energy and information”, Scientific American, 224 (1971).
22
MIT 3.20, Rickard Armiento, Lecture 4
Entropy in Information Theory
Entropy: a measure of uncertainty. Proportional to the
number of optimal binary questions needed to exactly
determine the state of a system.
For the complete system:
I
I
I
Only one state: S = 0, complete certainty, no
questions needed.
All probabilities equal: S=maximal uncertainty,
maximal number of questions needed.
Not all probabilities equal: S=intermediate, some
questions needed to determine the state.
Entropy = “uncertainty”: proportional to the number of optimal binary questions one needs to
ask to determine exactly which state a system is in.
Eight Sided Dice
“Dumb” questions
Probability
Pi
Probability
Pi
1/8
1/8
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
Is it face 1? probability 1/8 that the answer is yes, 7/8 that
it is no. Is it face 2? ...
Which side of the dice faces up?
Ask binary questions = questions that can be answered
with a yes or no.
Asking one such question = one bit of information.
23
Not optimal questions, average number needed =
1
(1 + 2 + 3 + 4 + 5 + 6 + 7 + 7) = 4.375
8
MIT 3.20, Rickard Armiento, Lecture 4
Unequal probabilities Pi
Probability
Pi
“Clever” questions
Probability
Pi
2/8
1/8
1/8
1 2 3 4 5 6
1 2 3 4 5 6 7 8
123456
12345678
Question 1
1234
12
5678
1
Question 2
12
34
56
3456
2
34
78
56
Question 3
1
2
3
4
5
6
7
8
3
Number of binary optimal questions: 18 (3 × 8) = 3
Can get this as: log2 (8) = 3
4
5
Average number of questions =
2 × 28 + 2 × 28 + 3 × 18 + 3 × 18 + 3 ×
6
1
8
= 2.5
Communication
Instead of a state, consider communication stream of bits.
General Case
What is the probability that the next bit is 0, 1?
Given a set of mutually exclusive possible states,
i = 1, 2, ..., each with probability Pi known beforehand,
I
I
Then:
I
The average number of optimal binary questions needed
to find the state is
X
I=−
Pi log2 Pi
i
Suggested homework
• 2002E2-2
• 2003E2-1a
• 2007E2-5(a-c)
24
I
Only 0:s in stream, S = 0, complete certainty.
Completely random: S=maximal uncertainty.
P0 = 0.6, P1 = 0.4, S=intermediate.
Some complicated function of previous bits (think:
message in english), S=intermediate.
I=−
P
i
Pi log2 Pi = bits of “information” in the message.
MIT 3.20, Rickard Armiento, Lecture 5
Lecture 5: Relations with Thermo. General Boundary Conditions
(the box)
• October 24, 2011, McQuarrie 2.4 - 3.2
What We Know
P
Ensemble averages: Mmeausured = M̄ = i Mi Pi
Pi = probability of finding a system in microstate i of the ensemble.
Probabilities in the microcanonical ensemble: Pi = 1/Ω(E) = constant
Entropy in the microcanonical ensemble: S = kB ln Ω(E)
P
P
Canonical partition function: Q = i e−Ei /(kB T ) = E Ω(E)e−E/(kB T )
Probabilities in the canonical ensemble: Pi = e−Ei /(kB T ) /Q
Entropy in the canonical ensemble: S = kB ln Q + Ē
T
Completing the Connection to Thermodynamics
Let T, V, N be constant: canonical ensemble.
Helmholtz Free energy for the canonical ensemble: the thermodynamic potential with the same
natural variables as the basis for the canonical ensemble:
n
o
Ē
A = E − T S = S = kB ln Q + Ē
= E − T kB ln Q + T ⇒
T
T
Fundamental relation in natural variables:
A(T, V, N ) = −kB T ln Q(T, V, N )
Connects the ‘microstate world’ (Q) and the macroscopic thermodynamical world (A). All other
relations follow from this one + thermodynamics:
Deriving equations of state:
∂A
S=−
∂T V,N
∂A
p=−
∂V T,N
∂A
µ=+
∂N T,V









∂ ln Q
∂T
S = kB ln Q + kB T
∂ ln Q
Eqns. of state ⇒ p = kB T

T,N

∂V


∂ ln Q



µ = −kB T

∂N T,V
25
V,N
MIT 3.20, Rickard Armiento, Lecture 5
It works equally well to just solve for the quantities:
E
Ē
A = E − T S ⇒ S = (E − A)/T = A = −kB T ln Q =
+ kB ln Q =
+ kB ln Q.
T
T
The energy:
E = A + T S = −kB T ln Q + T
kB ln Q + kB T
∂ ln Q
∂T
!
= kB T
2
V,N
∂ ln Q
∂T
V,N
Another useful formula for the energy:
)
( ∂Q
P
P
−βEi
−βEi
=
−
E
e
E
e
∂ ln Q
i
i
i
∂β
i
⇒ E=−
Ē =
⇒
.
d ln(f (x))
f 0 (x)
Q
∂β
V,N
dx
f (x) =
The partition function Q ⇒ everything (in the thermodynamical description of the system.)
Example
N distinguishable systems that can be in 2 energy states = 0, .
j = nj ,
⇒
nj = [0, 1],
E=
N
X
nj ,
ρ=
j=1
X
nj .
j
N

e−β
X
Q=
P
j
nj X
Y
{nj =1,0}N
j
j
=
{nj =1,0}N
j
e−βnj =
Y X
X
e−βnj = 
j nj =1,0
e−βn 
= (1+e−β )N
n=1,0
(General result we will see later, Q = q N for identical independent distinguishable systems.)
A = −kB T ln Q = −kB T ln(1 + e−β )N = −N kB T ln(1 + e−β ).
A
= −kB T ln(2) ≈ −0.69kB T
N
A
For T << 1, β >> 1 ⇒
= 0.
N
For T >> 1, β << 1 ⇒
The average energy is:
Ē = −
∂ ln Q
∂β
=−
V,N
∂N ln(1 + e−β )
∂β
=
V,N
n
d ln(f (x))
dx
=
f 0 (x)
f (x)
o
Entropy:
S = kB ln Q +
Ē
1 N
= kB N ln(1 + e−β ) +
.
T
T 1 + eβ
26
= −N
−e−β
N
=
−β
1+e
1 + eβ
MIT 3.20, Rickard Armiento, Lecture 5
Average occupation: note: this is a ‘microstate world’ question, and thus is not derived via
thermodynamics.
hρi =
X
Pi ρi =
i
(
X
P
j
nj =
Q
{nj =1,0}N
j
=−
−β
j nj )e
P
n
f 0 (x)
f (x)
=
d ln(f (x))
dx
o
=−
1 d
ln Q = Q above =
β d
N d
N −βe−β
N
= β
.
ln(1 + e−β ) = −
β d
β 1 + e−β
e +1
As expected: hρi · = Ē.
Microcanonical ensemble in large system limit
Energy levels of the system: m = 1...M,
Degeneracy: Ω(E, N ) =
m(E) = E/.
N!
(N − m)!m!


Stirling’s approximation
ln n! ≈ n ln n − n
=
Entropy: S = kB ln Ω(E, N ) =


Assumes n is large
kB (N ln N − N − (N − m) ln(N − m) + (N − m) − m ln m + m)
To get E as a function of temperature, we need to derive an expression for the temperature.
∂S
1
1
∂S
1
∂S
Temperature from thermodynamics:
= ⇒β=
=
=
∂E N
T
kB ∂E N
kB ∂m N
1 ∂
1
N −m
1
N
− m) + (N − m) −m ln m + m) = ln
(N ln N
= ln
−1
{z − N} |−(N − m) ln(N{z
}|
{z
} ∂m |
m
m
→0
→− ln m
→ln(N −m)
N
N
N
−1 ⇒ m=
⇒ E = m =
. (Same as above!)
β
m
1+e
1 + eβ
For N, m large, the microcanonical ensemble gives the same result as the canonical ensemble.
This is generally true (will be explored in a later lecture on ‘Fluctuations’.)
⇒ eβ =
Grand canonical Ensemble
Allow a system to exchange both energy and particles with its environment ⇒ µ, V , and T fixed.
What do we mean by µ for microstates? Same issue as with T for the canonical ensemble.
Derivation similar to canonical ensemble: Ensemble G contains all possible systems with the
same macroscopic values of µ, V , T . Lay out 1...A labeled subsystems in a huge energyparticle bath which at first is in particle and thermal equilibrium with its environment at T and
µ. Then isolate it. View the simulated bath as a microcanonical ensemble, D. Every system in
D has a distribution of subsystems over particle number N and state (N, j), counted as aN,j :
XX
aN,j = A
N
j
27
MIT 3.20, Rickard Armiento, Lecture 5
In how many ways can we take A systems from G and lay them out into the simulated heat +
particle bath system, in a way that fulfills:
XX
XX
aN,j EN,j = E ,
aN,j N = N .
N
j
j
N
A!
⇒W =
,
ΠN Πj aN,j !
P
W (a)aN,j
aP
PN,j =
A
a W (a)
(Exactly as for the canonical ensemble: number of ways to reorder all labeled subsystems: A !.
Reordering with a group represents identical states, so divide with aN0 ,0 !, aN0 ,1 !, ...)
In the limit of an infinitely large simulated heat bath, A → ∞, the only term in the sum over
distributions relevant is the one that maximies W (a), {a∗j }, given by:
a∗N,j = e−α e−βEN,j e−γN
PN,j (V, β, γ) =
a∗N,j
e−βEN,j (V ) e−γN
= P P −βE (V ) −γN
N,j
A
e
N
je
Grand canonical ensemble = the canonical ensemble but also with sums now also over all possible values of N . Identify β by suddenly making walls impenetrable ⇒ canonical ensemble
⇒ β = 1/(kB T ).
Identification of entropy and γ:
f (β, γ, {EN,j (V )}) = ln Ξ = ln
XX
N
df =
∂f
∂β
dβ +
γ,{EN,j }
∂f
∂γ
e−βEN,j (V ) e−γN .
j
X X ∂f dγ +
∂EN,j β,γ,E
β,{EN,j }
N
df = −Ēdβ − N̄ dγ − β
j
XX
N
|
dEN,j .
(M,k)6=(N,j)
PN,j dEN,j
j
{z
=−p̄dV
}
(For the last identification of −p̄dV , see derivation of canonical ensemble)
⇒
df = −Ēdβ − N̄ dγ + β p̄dV
Add d(β Ē) + d(γ N̄ ) to both sides, compare with thermodynamics:
d(f + β Ē + γ N̄ ) = βdĒ + β p̄dV + γdN̄ .
⇐⇒
T dS = dE + pdV − µdN
⇒
1) γ =
2) S =
−µ
kT
Ē N̄ µ
−
+ kB ln Ξ + constant (Third law ⇒ constant = 0).
T
T
28
MIT 3.20, Rickard Armiento, Lecture 5
Entropy in the grand canonical ensemble:
S=
Ē N̄ µ
−
+ kB ln Ξ.
T
T
The grand canonical partition function:
Ξ(V, T, µ) =
XX
e−βEN,j (V ) eβµN =
j
N
X
Q(N, V, T )eβµN .
N
Much more work to sum over N ? In many cases it can actually be harder to restrict N .
Which potential has the same natural variables as the grand canonical ensemble?
Φ(T, µ, V ) = E − T S − µN
(= −pV due to Eulers’s theorem E = T S − pV + µN )
Fundamental relation in natural variables:
Φ(T, µ, V ) = −kB T ln Ξ(T, µ, V ).
Represents the most straightforward connection between macroscopic (thermodynamical) quantities and the ‘microstate world’ represented by Ξ. A good starting point for deriving other
relations.
The Generalized “Box” Ensemble
Are there other ensembles? Yes, for example, the isothermal-isobaric ensemble of systems with
heat and volume exchange with the environment:
XX
⇒ ∆(N, T, p) =
Ω(N, V, E)e−E/kT e−pV /kT
E
V
Still others, other type of work (e.g., magnetic), all possible combinations, etc.
Generic ensemble, let Λ be the characteristic potential/function.
Pi =
e2i
,
Z
−βΛ = ln Z;
Z=
X
i
29
e2i
MIT 3.20, Rickard Armiento, Lecture 5
Reverse-engineering the proof above gives the following scheme:
Recipe for expressions for any ensemble with constant T :
1. Start from the definition of the thermodynamical potential with the right natural variables, e.g., E(S, Y ); dE = T dS − xdY ⇒ Λ(T, x) = E − T S + xY
2. Multiply by −β to get +S/kB on the right side ⇒ −βΛ(T, Y ) = S/kB − βE + βxY .
3. Everything on the right hand side except S/kB goes into the exponent of the partition
function: ⇒ 2i = −β(Ei − xYi )
4. The left hand side gives the thermodynamic connection: −βΛ = ln Z.
⇒ Pν =
e2i
e−β(Ei −xYi )
=
;
Z
Z
Z=
X
e2i =
i
X
eβ(Ei −xYi ) ;
Λ = −kB T ln Z
i
1. Canonical ensemble / Free energy
A(T, V, N ) = E − T S;
⇒ Pi =
e−βEi
,
Z
−βA = S/kB − βE
Z=Q=
X
⇒
e−βEi ,
2i = −βEi
⇒
A = −kB T ln Q
i
2. Grand canonical ensemble / Grand potential
Φ(T, V, µ) = E − T S − µN ;
⇒ Pi =
e−β(Ei −µNi )
,
Z
−βΦ = S/kB − βE + βµN
Z=
X
e−β(Ei −µNi ) =
i
XX
N
⇒
2i = −β(Ei − µNi )
e−β(EN,j −µN ) = Ξ,
⇒
Φ = −kB T ln Ξ
j
3. Isothermal-isobaric ensemble / Gibbs free energy
G(T, p) = E − T S + pV ;
⇒ Pi =
e−βEi
,
Z
Z=
−βG = S/kB − βE − βpV...
X
i
e−β(Ei +pVi ) =
XX
V
⇒
2i = −β(Ei + pV )
e−β(EV,j +pV ) = ∆,
⇒
G = −kB T ln ∆
j
If T is not constant? More complicated → since we cannot hold E constant easily when the
system exchanges energy as work. ⇒ e.g. the (H,p,N ) ensemble.
Suggested Homework
•
•
•
•
Exam problem 2001E2-3 (Note: (a) was already suggested for lecture 1, (b) for lecture 3)
Exam problem 2003E2-3
McQuarrie problem 2.11
McQuarrie problem 2.14
30
MIT 3.20, Rickard Armiento, Lecture 6
Lecture 6: Fluctuations
• October 26, 2011, McQuarrie 3.3
What We Have Covered So Far
P
Averages: Mmeasured = 1st postulate = M̄ = i Pi Mi
MicrocanonicalP
ensemble:
Pi = 1/Ω
Ω(N, V, E) = i = Degeneracy for energy E
Easiest connection to thermodynamics: S = kB ln Ω(N, V, E)
Canonical ensemble: Pi = e−Ei /(kB T ) /Q
P
P
Q(N, V, T ) = E Ω(E)e−E/(kB T ) = i e−Ei (V )/(kB T )
Easiest connection to thermodynamics:A = −k
B T ln Q(N, V, T )
∂ ln Q
Useful formula for the energy: Ē = − ∂β
V,N
Grand canonical ensemble: Pi = e−(Ei −µN )/(kB T ) /Ξ
P
P P
Ξ(V, T, µ) = N Q(N, V, T )eµN/(kB T ) = N j e−(EN,j (V )−µN )/(kB T )
Easiest connection to thermodynamics: Φ = −kB T ln Ξ(V, T, µ)
Fluctations
Different ensembles = different quantities fixed, others quantities
in balance with environment and allowed to fluctuate. How much
does a quantity in balance with the environment fluctuate? Does
measurements need to take this into account?
X
Mmeasured =? M̄ =
P i Mi
Pi
i*
i
For the energy:
Ē =
X
i
Pi Ei =
1 X
1 X ln Ω(E)−βE
E Ω(E)e−βE =
Ee
| {z } Q
Q
E
∼P (E)
E
How does the probabilities P (E) ∼ Ω(E)e−βE = eln Ω(E)−βE typically look? First lecture, one
independent particle:
ω(E)dE ∼ E 1/2 dE
⇒
P (E) ∼ e(1/2) ln E−βE
⇒
P (E) ∼ e(3N/2−1) ln E−βE
Many independent particles (ideal gas)
Ω(E)dE ∼ E 3N/2−1 dE
31
MIT 3.20, Rickard Armiento, Lecture 6
Ω(E)
e-ϐE
Ω(E)e-ϐE
≈Ce
-(E-E)2/(2σ2)
Looks similar to a Gaussian (but it is not exactly a Gaussian, view this as first term in an expansion)
−(E−Ē)2
1
−
2σ 2
P (E) =
e
(2πσ 2 )1/2
P (E) is approximately Gaussian-shaped. What is the variance σE ?


AB 6= ĀB̄ 
A+B =
2
but
= E 2 −2E Ē+Ē 2 =
= E 2 −2Ē 2 +Ē 2 ⇒
σE
= (E − Ē)2 = (E 2 − 2E Ē + Ē 2 ) =
Ā + B̄


AB̄ = ĀB̄
Variance of a fluctuating variable:
2
σE
1 X 2 −βEi
−
= E 2 − Ē =
Ei e
Q
2
i
1 X
Ei e−βEi
Q
!2
;
2
(σX
= X 2 − X̄ 2 )
i
Trick: the quadratic term can be rewritten as a derivative of QĒ:
!
1 X 2 −βEi
1 ∂ X
1 ∂(QĒ)
−βEi
Ei e
=−
Ei e
= −
=
Q
Q ∂β
Q ∂β
i
i
|
{z
}
QĒ
=−
n
o
P
Ē ∂Q Q ∂ Ē
∂ Ē
−βEi = −QĒ
=
−
E
e
−
= ∂Q
= Ē 2 −
i
i
∂β
Q ∂β
Q ∂β
∂β
32
MIT 3.20, Rickard Armiento, Lecture 6
Hence,
2
σE
= E 2 − Ē 2 = Ē 2 −
∂ Ē
∂ Ē
− Ē 2 = −
∂β
∂β
Typical result for variance of an extensive variable. Intensive mechanical variables are slightly
more complicated (See McQuarrie problem 3.18.) To connect to thermodynamics, lets change
the β derivative into a temperature derivative,
2
σE
=−
n
∂ Ē
∂ Ē ∂T
=−
·
= ∂T
∂β =
∂β
∂T ∂β
∂ 1
∂β kB β
o
∂ Ē
= − kB1β 2 = −kB T 2 = kB T 2
∂T
From thermodynamics
CV =
∂E
∂T
⇒
N,V
Remarkably simple connection between variance of energy fluctuations and the heat capacity:
2
σE
= kB T 2 CV
Relative root-mean-square of fluctuations of energy (in canonical ensemble)
σE
=
Ē
p
k B T 2 Cv
= relative deviation from average.
Ē
Order of magnitude → use ideal gas. (both E and Cv are extensive, i.e., scales with the system
size.)
Ē ∼ N kB T, Cv ∼ N kB ⇒
Typical fluctuations for a large system:
σE
1
∼ √ ≈0
Ē
N
In numbers: N ∼ 1023 ⇒
σE
Ē
∼ 10−11 = 0.000000001%.
• N = small → fluctuations more relevant.
• N = large (macroscopic) → probability distribution for E = practically a delta function.
Fluctuations of N , grand canonical ensemble
T, V, µ constant, E, p, N fluctuates. Grand canonical ensemble:
PN,j =
1 −βEN,j −γN
e
,
Ξ
Ξ=
XX
N
33
j
e−βEN,j −γN
MIT 3.20, Rickard Armiento, Lecture 6
The variance of the particle number:
2
σN
=
N2
1 X X 2 −βEN,j −γN
− N̄ =
N e
−
Ξ
1X
N e−βEi e−γN
Ξ
2
j
N
!2
i
Derivation almost exactly as for energy. Trick: rewrite the quadratic term as a derivative of QN̄ :


X
X
X
X
1
1 ∂ 
1 ∂(ΞN̄ )
=
N 2 e−βEN,j −γN = −
N e−βEN,j −γN  = −
Ξ
Ξ ∂γ
Ξ ∂γ
j
j
N
N
{z
}
|
ΞN̄
=−
n
o
P P
N̄ ∂Ξ Ξ ∂ N̄
∂ N̄
−βEN,j e−γN = −ΞN̄
=
−
N
e
−
= ∂Ξ
= N̄ 2 −
N
j
∂γ
Ξ ∂γ
Ξ ∂γ
∂γ
Hence,
2
σN
= N 2 −N̄ 2 = N̄ 2 −
∂ N̄
∂ N̄
∂ N̄ ∂µ n ∂µ
−N̄ 2 = −
=−
·
= ∂γ =
∂γ
∂γ
∂µ ∂γ
∂
∂γ
(−kB T γ) = −kB T
o
= kB T
From thermodynamics:
∂ N̄
∂µ
V,T
N2
=− 2
V
∂V
∂p
N,T
(McQuarrie prob. 3.26, hint: use Gibbs-Duhem equation, SdT − V dp + N dµ = 0 to get
⇒
2
σN
N̄ 2 kB T κ
=
;
V
1
κ=−
V
∂V
∂p
= compressibility.
N,T
Size? Take ideal gas:
1 ∂
κ=−
V ∂p
Relative fluctuations ∼
√1
N
N kB T
p
N kB T
1
= 2
=
p V
p
⇒
√
σN
1
N
=
=√ .
N
N
N
is typical in statistical mechanics for extensive quantities.
Systems with significant fluctuations
1) Small systems: some variable M̄ :
σM
M̄
∼
√1 .
N
2) Close to critical points; at the critical point the transition between liquid and gas becomes a second-order transition. Spontaneous fluctuations can become very large here. “Critical opalescence” discovered in 1869 → fluids gets “milky” near the critical
point due to large length-scale density fluctuations (on the order of
the wave-length of visible light). At the critical point, the Gibbs
free energy becomes very flat, and there is an absence of restoring
force for fluctuations. Good read: Chapter 10 of Callen, Thermodynamics and an Introduction to Thermostatistics.
34
∂µ
∂p
.)
∂ N̄
∂µ
MIT 3.20, Rickard Armiento, Lecture 6
Equivalence of ensembles
In the thermodynamic limit (N → ∞), away from critical points
fluctuations are exceedingly small ⇒ Results obtained in one ensemble are the same in any other!
Equivalence between microcanonical and canonical:
Q(N, V, T ) =
X
Ω(N, V, E)e
−βE
=
E
E = E ∗ = Ē
=
other terms zero prob.
∗
= Ω(N, V, E ∗ )e−βE = Ω(N, V, E)e−βE .
Calculate the entropy using this form:
∂ ln Q
E
S = kB ln Q + kB T
=
= kB ln Q +
∂T
T
V,N
E
= kB ln Ω
T
P
Same as in microcanonical ensemble! The sum E is “optional”
when the system is large. However, it is often easier to keep it.
= kB ln Ω − kB βE +
Re-examine example from last lecture
N distinguishable systems that can be in 2 energy states = 0, .
j = nj ,
nj = [0, 1],
⇒
E=
N
X
nj ,
ρ=
X
j=1
nj .
j
In the Canonical ensemble we found:
Q = ... = (1 + e−β )N
∂ ln Q
N
E(T ) = −
= ... =
∂β
1 + e/(kB T )
V,N
In the Microcanonical ensemble we found:
S = ... = kB (N ln N − N − (N − m) ln(N − m) + (N − m) − m ln m + m),
T (E) =
∂S
∂E
−1
= ... =
N
1
ln
N
−1
E
−1
⇒
E(T ) =
m = E/
N
1 + e/(kB T )
But note: the derivation of S used Stirling’s approximation! I.e. the large system limit.
Very typical that one ensemble gives you the same curve, but one as E(T ) and the other T (E).
Another example: Chemical potential in canonical vs. grand canonical ensemble
35
MIT 3.20, Rickard Armiento, Lecture 6
1) µ(N ) in canonical ensemble, Q(N, V, T ) ⇒
∂A
∂ ln Q
µ(N ) = dA = −SdT − pdV + µdN =
= A = −kB T ln Q = −kB T
∂N V,T
∂N V,T
2) N (µ) in grand canonical ensemble, Ξ(µ, V, T ) ⇒
∂ ln Ξ
∂Φ
= − Φ = −kB T ln Ξ = kB T
N (µ) = dΦ = −SdT − N dµ − pdV =
∂µ V,T
∂µ V,T
These should give the same curve for large systems.
Suggested homework
• Exam problem 2001E2-2(a-c) (also: after trying to solve it as in the lecture, make sure
to check the solutions to see another systematic path towards the solution; a “3-step algorithm”.)
• Exam problem 2002E2-1(a-b) for deriving the variance of an even more exotic variable.
• McQuarrie problem 3.18.
• McQuarrie problem 3.24.
• McQuarrie problem 3.26.
36
MIT 3.20, Rickard Armiento, Lecture 7
Lecture 7: Non-interacting particles, ideal gas, indistinguishable
particles, Boltzmann approximation, Ideal monoatomic gas,
Fermions and Bosons
• October 28, 2011, McQuarrie 4.1, 5.1-5.4
Things We Know
P
Averages: Mmeasured = 1st postulate = M̄ = i Pi Mi
MicrocanonicalP
ensemble:
Pi = 1/Ω
Ω(N, V, E) = i = Degeneracy for energy E
Easiest connection to thermodynamics: S = kB ln Ω(N, V, E)
Canonical ensemble: Pi = e−Ei /(kB T ) /Q
P
P
Q(N, V, T ) = E Ω(E)e−E/(kB T ) = i e−Ei (V )/(kB T )
Easiest connection to thermodynamics: A = −kB T ln Q(N, V, T )
Grand canonical ensemble: Pi = e(−Ei +µN )/(kB T ) /Ξ
P
P P
Ξ(V, T, µ) = N Q(N, V, T )eµN/(kB T ) = N i e−EN,i (V )/(kB T )+µN/(kB T )
Easiest connection to thermodynamics: Φ = −kB T ln Ξ(V, T, µ)
In large ‘thermodynamical’ systems → all ensembles give the same result. (So use the one
where the sums in the partition function is the easiest to do.)
Distinguishable Non-Interacting / Weakly Interacting Particles
Many-particle wave function: Ψ(q̄1 , q̄2 , q̄3 , ..., q̄N ).
Interacting particles → cannot easily divide Ψ into single-particle functions (such a division is
an “orbital theory”).
For no interaction / weak interaction:
Ĥ = Ĥ A + Ĥ B + Ĥ C + ...,
has individual eigensolutions: Ĥ A ψ A = A ψ A , ...
Let: ψ = ψ A ψ B ψ C ... then:
Ĥψ = (Ĥ A + Ĥ B + Ĥ C + ...)ψ A ψ B ψ C ... =
= ψ B ψ C H A ψ A + ψ A ψ C H B ψ B + ... =
= (A + B + C + ...)ψ A ψ B ψ C ... = (A + B + C + ...)ψ.
Hence, ψ = ψ A ψ B ψ C ... is an eigenfunction to Ĥ with eigenvalue: E = A + B + C + ...
B
C
Enumerated solutions to the eigenvalue equation: Ei = Ej,k,l,... = A
j + k + l + ...
37
⇒
MIT 3.20, Rickard Armiento, Lecture 7
We have reduced the many-particle problem to N single-particle ones.
Partition function of independent states:
Q=
X
e−βEi =
i
A
β
e−β(j +k +....) =
X
X
A
e−βj
X
j
j,k,...
B
eβk ... = qA · qB ...
k
and, if A = B = C ⇒ Q = q N .
The partition function is a product of independent terms. ln Q is a sum of independent terms.
(Check the example in lecture 5 again; using this makes getting the partition function very easy)
For distinguishable weakly interacting subsystems (think: independent labeled particles):
B
C
Ej,k,l,... = A
j + k + l + ...
Q(N, V, T ) = qA · qB ... = if all subsystems equal = q(V, T )N
Where {qA } are the partition functions for the individual subsystems.
Great simplification, but this only holds for distinguishable particles, e.g., localized in space and
it would be straightforward to label them e.g., by coordinate. Works for sites in lattices, segments
in polymer chains, etc. Does not work for identical particles in a box.
Hydrogen on a surface: hypothetical model system (not very realistic).
Hydrogen adsorbed on surface. We assume hydrogen atoms have no interaction with each other
or the surface and thus maintain their atomic energy levels.
Localized on surface = can be labeled → Distinguishable!
Each hydrogen can individually be exited into different electronic states. 1s, 2s, 2px , 2py , 2px ,...
States of the system i corresponds to different combinations of electronic states/excitations.
State 1: E1 = a1s + b2s + c1s + d2px + ....
State 2: E2 = a2s + b1s + c1s + d1s + ....
Q = qN ;
q=
X
e−βl = ‘mini’ partition functions for each particle.
l
Indistinguishable Particles, Fermions and Bosons
Non-interacting particles that are not localized / can not be labeled = indistinguishable.
Example:
state 1:
state 2:
A
2s
1s
B
1s
2s
C
3s
3s
D
4s
4s
38
a b c
MIT 3.20, Rickard Armiento, Lecture 7
Are state 1 and state 2 really different states? If ’A’, ’B’, ... refers to specific sites on a surface:
yes. If they are hydrogen atoms in a gas, then: no.
Can we just allow for all possible reorderings with a factor (1/N !) ? Indistinguishability in
quantum mechanics makes things more tricky.
Quantum mechanics must give the same physics when particles are relabeled:
|ψ(q̄1 , q̄2 , q̄3 , ..., q̄N )|2 = |ψ(q̄2 , q̄1 , q̄3 , ..., q̄N )|2 ⇒ ψ(q̄1 , q̄2 , q̄3 , ..., q̄N ) = ±ψ(q̄2 , q̄1 , q̄3 , ..., q̄N ).
The wave function is allowed to change sign when we relabel indistinguishable particles! Turns
out: + and - describe two different types of particles:
+ = Bosons = Integer spin particles,
− = Fermions = Half-integer spin particles
(Actually comes out from combining quantum mechanics and special relativity ⇒ Dirac Equation ⇒ particles have spin)
Two fermions cannot be in the same state: ψ(q̄, q̄) = −ψ(q̄, q̄) ⇒ ψ(q̄, q̄) = 0. No problem for
bosons. (This is what we see in nature!)
This is a type of “dependence” (=correlation) between the particles in that
P it regulates which
states that exists. I.e. which i we should count in partition function sums i .
Partition Function for Weakly Interacting Indistinguishable Particles
Independent particles → the proof that led up to this relation is still valid:
X
B
C
A
Q(N, V, T ) =
e−(l +m +n +...)/(kB T )
l,m,n,...
BUT, only as long as the sum over l, m, n, ... is restricted to only be over valid states, AND only
counts identical states once ⇒ we cannot easily divide it up into independent factors.
Example: two identical indistinguishable particles, each with two energy levels. Count the number of possible states (= terms in the i sum):
If we only take reordering into account:
1
1 X X
× Sum =
= 2 states
N!
2!
l=0,1 m=0,1
Fermions:
X
← 1 possible state.
(l,m)=[(0,1)]
Bosons:
X
← 3 possible states.
(l,m)=[(0,0),(1,0),(1,1)]
Hence, in general it is not enough to just consider all possible ways of reordering the states, since
it does not correctly handle the tricky terms where particles are in identical states.
39
MIT 3.20, Rickard Armiento, Lecture 7
Boltzmann Distribution
Realization: if there is a huge number of relevant different states → the number of terms
with “collision” of particles in the same state are relatively few, and thus including or missing those terms will not affect the sum significantly. This only works if number of particles <<
number of available states. Usually true if the temperature is normal → high, or density is relatively low.
In this case we can just divide with all ways of reordering the particles, and thus:
P P P
...e−β(l +m +n )
qN
Q(N, V, T ) ≈ l m n
=
.
N!
N!
The Boltzmann Distribution / Boltzmann statistics
Q(N, V, T ) =
qN
.
N!
Where q is the partition function of one particle. Valid for indistinguishable particles when
number of particles << number of available single-particle states.
Monoatomic Ideal Gas
N independent identical indistinguishable atoms in a box.
all available single particle
=
Recall first lecture: φ() =
states with energy ≤ ∼ kB T, T ∼ 300 K, m ∼ 10−22 g, a ∼ 10 cm
⇒
π
6
8ma2 h2
3/2
φ() ∼ 1030 >> 1023 = N.
⇒ single particle states >> number of particles.
We can use the Boltzmann distribution!
The degrees of freedom decouple (independent sum of Hamiltonians, same proof as for independent particles):
qatom = qtranslational · qelectronic · qnuclear .
Nuclear energy levels are far apart (∼ 1 MeV), atoms stay at nuclear ground state at normal
temperatures, we can set qnucl = 1.
Commonly electronic states have ∆ ∼ 1eV and thus e−β is small (∼ 10−17 ), and thus we can
set qelec = 1. This is not true for all atoms (e.g. halogen atoms). McQuarrie discusses some
cases.
Partition function for particle in a box
qatom ≈ qtrans =
X
e−βl ;
l = nx ,ny ,nz =
l
h2
(n2x + n2y + n2z ).
2
8ma
| {z }
=1/M
40
MIT 3.20, Rickard Armiento, Lecture 7
qtrans =
∞ X
∞ X
∞
X
−β(n2x +n2y +n2z )/M
e
=
nx =1 ny =1 nz =1
x, y, z independent
=
n = nx , ny , nz
∞
X
!3
−βn2 /M
e
=
n=1
β/M ∼ 10−21 ⇒ MANY states, terms slowly changing with n.
=
=
Take continous limit
s
!3 Z ∞
3 n
o
R
p
2πmkB T 3/2
πM
2
2
∞ −cx
−βn /M
π
=
V
=
e
dn = 0 e
dx = c =
β
h2
0
Alternative way, using the density of states ω() in the continuous limit from lecture 1:
Z ∞
Z ∞
X
π 3/2 1/2 −β
ω()e−β =
ω()e−β d = ω() from 1st lecture =
qtrans =
M e d = ... = same thing
4
0
0
Canonical partition function for the monoatomic ideal gas (with no electronic excitations)
qatom ≈ qtrans =
2πmkB T
h2
3/2
V
qN
1
Q = trans =
N!
N!
V
= 3,
Λ
2πmkB T
h2
Λ=
3N/2
h2
2πmkB T
1/2
VN
The length Λ is a “thermal De Broglie wavelength”. Criterion for Boltzmann statistics becomes
Λ3 /V is small, i.e., Λ < dimensions of the container. (See McQuarrie, end of section 5-1).
Thermodynamic connection:
A = −kB T ln Q ⇒ A = −kB T ln
=
qN
N!
= −N kB T ln q + kB T ln N ! =
Stirling’s approximation
= −N kB T ln q + N kB T ln N − N kB T ⇒
ln n! ≈ n ln n − n
"
#
2πmkB T 3/2 V
A = −N kB T ln
− N kB T.
h2
N
Step 1, the pressure:
dA = −SdT − pdV + µdN
⇒
p=−
41
∂A
∂V
N,T
∂ = N kB T
ln V ⇒
∂V N,T
MIT 3.20, Rickard Armiento, Lecture 7
Pressure for the ideal gas (equation of state 1)
p=
N kB T
V
Step 2, the energy:
A = E − T S, dA = −SdT − pdV + µdN ⇒
∂A
∂ ln Q
S=−
+ kB ln Q ⇒
= A = −kB T ln Q = kB T
∂T V,N
∂T
3N/2
2 ∂ ln Q
2 ∂ ln
T
=
E = A + T S = kB T
= kB T
∂T
∂T N,V
N,V
Energy in the ideal gas (2nd equation of state)
3
E = N kB T ;
2
The other equation of state valid for a monoatomic ideal gas!
Step 3, the chemical potential
µ=
= −kB T ln
"
∂A
∂N
"
2πmkB T
h2
= −kB T ln
T,p
3/2
V
N
2πmkB T
h2
#
=
n
V
N
=
#
3/2
kB T
p
+ kB T ln N +
V
o
"
= −kB T ln
|
N kB T
− kB T =
N
2πmkB T
h2
{z
µ0
3/2
#
kB T +kB T ln p = µ0 + kB T ln p.
}
Suggested homework
•
•
•
•
McQuarrie problem 4-12
Exam problem 2003E2-5
Exam problem 2001E2-4(a)
Go through the derivation for an ideal gas in the lecture, and check what changes if we had
set qelec = ωe1 + ωe2 e−β∆12 . (Check McQuarrie 5-3, if you need help.)
42
MIT 3.20, Rickard Armiento, Lecture 8
Lecture 8: Ideal diatomic gas. Rotational and vibrational DOF
• October 31, 2011, McQuarrie 6
Useful Formulas
P
Averages: Mmeasured = 1st postulate = M̄ = i Pi Mi
MicrocanonicalP
ensemble:
Pi = 1/Ω
Ω(N, V, E) = i = Degeneracy for energy E
Easiest connection to thermodynamics: S = kB ln Ω(N, V, E)
Canonical ensemble: Pi = e−Ei /(kB T ) /Q
P
P
Q(N, V, T ) = E Ω(E)e−E/(kB T ) = i e−Ei (V )/(kB T )
Easiest connection to thermodynamics: A = −kB T ln Q(N, V, T )
Grand canonical ensemble: Pi = e(−Ei +µN )/(kB T ) /Ξ
P
P P
Ξ(V, T, µ) = N Q(N, V, T )eµN/(kB T ) = N i e−EN,i (V )/(kB T )+µN/(kB T )
Easiest connection to thermodynamics: Φ = −kB T ln Ξ(V, T, µ)
In large ‘thermodynamical’ systems → all ensembles give the same result.
Independent distinguishable particles: Q = qA · qB · qC · ... = {if identical} = q N
Independent indistinguishable particles in high-temperature low-density limit: Q = q N /N !
Diatomic Degrees of Freedom
Diatomic Molecule
Assume the degrees of freedom are independent. Requires the BornOppenheimer approximation: electrons reach equilibrium on much
shorter timescale than ionic motion. Also assume amplitude and
timescale of vibrations are so small they do not interfere with rotations (= “rigid-rotor approximation”).
2) ROT
Ĥsingle particle = Ĥtrans +
Ĥrot
+Ĥvib + Ĥelec + Ĥnucl
|{z}
|
{z
}
Rigid-rotor
For now, assume
approx = fixed r
big energy spacing.
Q=
N
qsingle
particle
← Assume normal conditions, Boltzmann statistics ok.
N!
qsingle particle = qtrans · qrot · qvib · qelec · qnucl
| {z }
we can set = 1
43
3) VIB
1) TRANS
MIT 3.20, Rickard Armiento, Lecture 8
The Two-Body Problem
Two bodies in a potential that only depends on their distance. Energy in classical physics:
1
1
E = m1 ṙ21 + m2 ṙ22 + v(|r2 − r1 |)
2
2
Reorganize to get:
1
1 m1 m2 2
E = (m1 + m2 )Ṙ2 +
ṙ + v(|r|),
2 | {z }
2 m1 + m2
| {z }
m
R=
m1 r1 + m2 r2
,
m1 + m2
r = r2 − r1 .
µ
Hence, the problem is equivalent to one of a free translational motion of the center of mass, R,
using the sum of the masses, m = m1 + m2 , plus the independent movement, r, of one ‘virtual’
particle with mass µ in the potential v. In our quantum mechanical formulation:
(m +m2 )
Ĥ = Ĥtrans1
(µ)
(µ)
+ Ĥrot + Ĥvib
Translations
Easy! Same as we derived for monoatomic ideal gas, but m = m1 + m2 :
qtrans =
2π(m + m )k T 3/2
V
1
2 B
V = 3
2
h
Λ
3
Etrans = N kB T
2
⇒
Vibrations
Harmonic approximation (small vibrations) = harmonic oscillator.
First term in Taylor expansion of the potential:
∂v 1 ∂ 2 v +
v(z) =
v(0)
(r−re )+
(r−re )2 +...
|{z}
∂r r=re
2 ∂r2 r=re
| {z }
| {z }
set energy
scale so this = 0
F
= 0 at equilibrium
Harmonic oscillator with reduced mass:
Ĥvib = −
h̄2 ∂ 2
1
+ F r2 ;
2
2µ ∂r
2
reduced mass:µ =
m1 m2
m1 + m2
From: Wikipedia (Mark Somoza, 26 2006)
Harmonic oscillator solution (see lecture 1):
1
n = (n + ) h ν; n = 0, 1, 2...;
2 ↑
Note: not h̄
qvib =
X
n
−βn
e
=
∞
X
e
−β(n+1/2)hν
=e
1
ν=
2π
−βhν/2
n=0
∞ X
n=0
Introduce “vibrational temperature” Θv = hν/kB ⇒
44
s
e
F
.
µ
−βhν
n
=
Geometric series
P
∞
1
n
n=0 x = 1−x
=
e−βhν/2
.
1 − e−βhν
MIT 3.20, Rickard Armiento, Lecture 8
Vibrational partition function
qvib =
e−Θv /(2T )
1 − e−Θv /T
“Number of thermally available vibrational states.”
From: McQuarrie, Statistical Mechanics, Table 6-1.
vib
e−βn
Pn =
;
qvib
Pn>0 = 1 − P0 = e−Θv /T
N2 : Θv = 3374 K, T = 300 K ⇒ Pn>0 ≈ 1.3 · 10−5
K2 : Θv = 133 K, T = 300 K ⇒ Pn>0 ≈ 0.64.
Only first, or first few states are excited in room temperature.
The vibrational energy:
N
ln(N
!)]
2 ∂ ln(qvib /N !)
2 ∂[ln(qvib − Evib,ν = kB T
= N kB T
=
∂T
∂T
V,N
V,N
∂ Θv
Θν
−Θν /(2T )
−Θv /T
= N kB T
ln e
− ln 1 − e
⇒ Evib,ν = N kB
+ Θ /T
∂T V,N
2
e v −1
2
N2 , T = 300 K ⇒ Evib,ν /N ≈ 0.15 eV/molecule
K2 , T = 300 K ⇒ Evib,ν /N ≈ 0.03 eV/molecule
Compare: chemical bonds ∼ 1 eV.
45
MIT 3.20, Rickard Armiento, Lecture 8
Rotations
Hamiltonian for rigid rotor
h̄2
1 ∂
∂
1 ∂2
Ĥ = −
sin θ
+
2I sin θ ∂θ
∂θ
sin2 θ ∂φ2
(in spherical coordinates.)
where I = moment of intertia = µR02
⇒ J =
J(J + 1)h2
;
8π 2 I
J = 0, 1, 2, ...;
ω() = (2J + 1)
The energy scale is set by h2 /(8π 2 I). Typical values → 10−4 eV = microwave region. (Which
is how a microwave oven works.)
Note: ∼ 100 times as tight energy spacing as vibrational spectra.
Rotations in Unsymmetrical Molecule (e.g., NO)
Usual way of forming the partition function:
X
X
2
2
qrot =
ω()e−β =
(2J + 1)e−β(J+1)h /(8π I)
J
As for vibrations: define characteristic rotational temperature Θr =
qrot =
X
h2
8π 2 IkB
⇒
(2J + 1)e−(J+1)Θr /T
J
Will look at T > Θr and T < Θr separately:
1) If T >> Θr , then we can handle this sum similar to how we handled the translational freedom
of the monoatomic gas in lecture 7 = many, many terms which are very similar → continuous
limit:
0
Z ∞
Z ∞
T
0
J = J(J + 1)
−β(J+1)Θr /T
⇒ qrot =
(2J + 1)e
dJ = dJ
=
e−J Θr /T dJ 0 =
Θr
0
0
dJ 0 = 2J + 1
Sum → Integral can be seen as first term in a Euler-Maclaurin expansion (see McQuarrie 6-3),
which gives:
Rotational partition function for unsymmetrical diatomic molecule and T > Θr
!
T
1 Θr
1 Θr 2
qrot =
1+
+
+ ...
Θr
3 T
15 T
46
MIT 3.20, Rickard Armiento, Lecture 8
Erot,ν = kB T
2
∂ ln(qrot /N !)N
∂T
= N kB T + ...
V,N
2) On the other hand, if T is small or Θr is large, the terms drop off quickly → just cut the
sum after enough terms. (Mostly relevant at very low temperature, and for hydrogen at low
temperatures).
Rotational partition function for unsymmetrical diatomic molecule and T < Θr
qrot =
X
(2J + 1)e−(J+1)Θr /T = 1 + 3e−2Θr /T + 5e−6Θr /T + ...
J
Rotations in Symmetric Molecule (e.g., H2 )
Indistinguishability again; exchanging the two nuclei requires symmetric or antisymmetric wavefunction. Requires careful QM treatment (see McQuarrie 6-4,6-5). However, for high temperatures it works out to just a factor 1/2, roughly meaning that there are half as many states due to
inverted states being identical.
Rotational partition function for symmetric diatomic molecule and T > Θr
!
T
1 Θr
1 Θr 2
qrot =
1+
+
+ ...
2Θr
3 T
15 T
Probabilities to find the system in a state with J = J (for both
symmetric and unsymmetrical case):
PJ =
(2J + 1)e−J(J+1)Θr /T
→ Has an intermediate maximum, Jmax
qrot
Assembling the Full Partition Function
From: McQuarrie, Statistical Mechanics, Table 6-1.
(qtrans qrot qvib )N
Q=
;
N!
(Boltzmann statistics →
decoupled degrees of freedom)
Thermodynamic connection:
A = −kB T ln Q = −N kB T (ln qtrans +ln qrot +ln qvib )+kB T ln N !) = Atrans +Arot +Avib +kB T ln N !.
Equation of state from pressure:
p=−
∂A
∂V
N,T
N kB T
only qtrans depends
=
=
;
on volume
V
47
Same as monoatomic!
MIT 3.20, Rickard Armiento, Lecture 8
Energy:
E = kT 2
3
Etrans = N kB T ;
2
Erot =
∂ ln Q
∂T
= Etrans + Erot + Evib
N,V
T >> Θr : N kB T
;
otherwise: more complicated
Evib = N kB
Θv
Θv
+ Θ /T
2
e v −1
.
Heat capacity:
CV =
∂E
∂T
=
V,N
∂Etrans + Erot + Evib
∂T
= CV,trans + CV,rot + CV,vib
V,N
3
CV,trans = N kB ,
2
T >> Θr : N kB
otherwise: more complicated
2
eΘv /T
Θv
T >> Θv : N kB ← from taylor exp.: eΘv /T = 1 + Θv /T + ...
=
= N kB
T << Θv : 0 ← all molecules in ground state
T
(eΘv /T − 1)2
CV,rot =
CV,vib
If we would have 0 << Θr << Θv , this nicely splits up in ranges:
Simplified schematic of Cv(T)
3
3
T << Θr << Θv : N kB + 0 + 0 = N kB
2
2
5
3
0 << Θr << T << Θv : N kB + N kB + 0 = N kB
2
2
3
7
0 << Θr << Θv << T : N kB + N kB + N kB = N kB
2
2
Θr=1
Θv=1000
In reality, a few complications:
1. Θr is usually quite small (cf. McQuarrie table 6-1), (largest
for H2 ∼ 85 K).
2. Indistinguishability for H2 and other small molecules → heat
capacity splits depending on nucleus spin configuration.
3. The transition CV,rot for T ≈ Θr has a bit of structure (see
schematic).
Suggested Homework
•
•
•
•
Exam problem 2007E2-2
McQuarrie problem 6.8
McQuarrie problem 6.10
McQuarrie problem 6.22
48
Schematic of Cv,rot(T)
MIT 3.20, Rickard Armiento, Lecture 9
Lecture 9: Metals: phonons and electrons
• November 2, 2011, McQuarrie 4.2
A Collection of Formulas
P
Averages: Mmeasured = 1st postulate = M̄ = i Pi Mi
MicrocanonicalP
ensemble:
Pi = 1/Ω
Ω(N, V, E) = i = Degeneracy for energy E
Easiest connection to thermodynamics: S = kB ln Ω(N, V, E)
Canonical ensemble: Pi = e−Ei /(kB T ) /Q
P
P
Q(N, V, T ) = E Ω(E)e−E/(kB T ) = i e−Ei (V )/(kB T )
Easiest connection to thermodynamics: A = −kB T ln Q(N, V, T )
Grand canonical ensemble: Pi = e(−Ei +µN )/(kB T ) /Ξ
P
P P
Ξ(V, T, µ) = N Q(N, V, T )eµN/(kB T ) = N i e−EN,i (V )/(kB T )+µN/(kB T )
Easiest connection to thermodynamics: Φ = −kB T ln Ξ(V, T, µ)
In large ‘thermodynamical’ systems → all ensembles give the same result.
Independent distinguishable particles: Q = qA · qB · qC · ... = {if identical} = q N
Independent indistinguishable particles in high-temperature low-density limit: Q = q N /N !
Fermi-Dirac and Bose-Einstein statistics
What happens when we cannot use Boltzmann statistics for indistinguishable particles?, i.e.,
density is high, and Q 6= q N /(N !).
Recall lecture 7: two fermions cannot be in the same state, any number of bosons can. Can count
states manually (terms in the partition function sum over i) for two identical indistinguishable
particles with two states:
“Classical” Boltzmann statistics:
1
1 X X
× Sum =
= 2 states
N!
2!
l=0,1 m=0,1
Fermions:
X
← 1 possible state.
(l,m)=[(0,1)]
Bosons:
X
← 3 possible states.
(l,m)=[(0,0),(1,0),(1,1)]
Idea: reorganize the sum to be over number of particles in each specific state nk . In that case:
Fermions: nk = [0, 1],
Bosons: nk = [0, 1, ..., N ].
49
MIT 3.20, Rickard Armiento, Lecture 9
P
For this to work, the sum must be taken under the constraint that k nk = N . If we can do that,
we are just counting valid states for either fermions or bosons. We have:
X
X (i)
(i)
Ei =
k nk , Ni = N =
nk
k
⇒ Q(N, V, T ) =
X
e−βEi
i
k


X0


mean
Let
 X0

P
{nk }
= sum over all possible n
=
e−β k nk Ek
k

P

 {nk }
that fulfill k nk = N
← Tricky math.
Awkward condition on N . Lets assume N is large, use equivalence between ensembles, and try
to do this in the grand canonical ensemble instead:
Ξ(T, V, µ) =
∞ X
X
N =0
=
∞ X
X
0
P
e−β(Ei −µN ) = Ei − µN = k nk (k − µ) =
i
e−β
P
k
nk (k −µ)
∞ X Y
X
0
=
N =0 {nk }
N =0 {nk }
e−β(k −µ)
n k
k
Here we get the benefit of using the grand canonical ensemble. Since N → ∞, each nk actually
go over all possible values, independently.
nmax
nmax
nmax
3
2
1
⇒
Ξ=
X X X
...
Y
n1 =0 n2 =0 n3 =0
e−β(k −µ)
nk
k
Reorganizing the state
sum
1
1
1
# Particles in state 1
# state 2
# state 3
∑
N=0
I
∑
N=1
N=2
I
∑
n1=0 n2=0 n3=0
N=3
Pn1,max Pn2,max Pn3,max
contains no duplicate states; all
the terms are unique, correct, states with varying
number of particles = one term for each possible
state.
This has to be the same thing as: the sum over all N,
and then all the ways of correctly placing those N
particles among the states.
n1 =0
n2 =0
n3 =0
50
MIT 3.20, Rickard Armiento, Lecture 9
nmax
nmax
nmax
1
2
3
Ξ=
X X X
...
n1 =0 n2 =0 n3 =0
Y
e−β(k −µ)
nk
k


Similar step as in the example in lecture 5
a1 a2 + a1 b2 + a2 b1 + b1 b2 =
=
=


= (a1 + b1 )(a2 + b2 )
max
max
=
k
Y nX
−β(k −µ)
e
nk
=
k nk =0
k
Y nX
k
e−β(k −µ)
n
.
n=0
Examine the sum for the two different particle types:
Fermions:
nmax
k
⇒
=1
1 X
e−β(k −µ)
n
= 1 + e−β(k −µ) .
n=0
Bosons:
nmax
k
∞ X
=∞⇒
e
−β(k −µ)
n
P∞
1
1
= 1−x
=
−β(
k −µ)
But only when x < 1
1−e
n=0 x
=
n=0
n
(Note that the last step requires µ < k ∀k ⇒ µ < 0 . If there are terms with k ≤ µ (or even
k → µ in some limit) the occupation numbers of such terms diverge, i.e., pushing all bosons
into those states. This turns out to be an important property of boson systems that e.g. is relevant
for the phase transition giving a Bose-Einstein condensate.)
Grand canonical partition function for Fermions [+] and Bosons [-]:
Ξ=
Y
(1 ± λe−βk )±1 ;
λ = eβµ
k
Expression for Bosons: referred to as Bose-Einstein (BE) statistics.
Fermions: referred to as Fermi-Dirac statistics (FD).
Classical indistinguishable particles (see Lecture 7): Boltzmann or Maxwell-Boltzmann statistics (MB).
Find average occupation in state k:
n̄k =
P
1X
1 ∂Ξ
∂ ln Ξ
nk e−β k nk (k −µ) =
=
=
Ξ
Ξ ∂(−βk )
∂(−βk )
i
=
Split ln Ξ into terms,
only one term has k .
=
e−β(k −µ)
1
1
= β(−µ)
= −1 β
,
−β(
−µ)
k
λ e ±1
1±e
e
±1
FD:+, BE:−
High Temperature Limit of BE and FD Statistics
FD and BE are equal when eβ(−µ) is big, so, e.g., when λ is small. When does this happen?
At higher temperatures, states at higher energy becomes available. For this to work, the average
occupation of each state must go down, i.e.:
n̄k ↓
⇒
eβ(−µ) ↑
⇒
eβ(−µ) >> 1
51
(eventually)
MIT 3.20, Rickard Armiento, Lecture 9
Hence, this limit is the same one that we discussed in Lecture 7: at low densities and hightemperatures the number of available states for each particle are much larger than the number
of particles ⇒ MB statistics are valid. (Important: this works because the chemical potential
depends on temperature, otherwise we would have T → ∞, β → 0 and thus eβ(−µ) → 1.) A
formal derivation of the validity of MB statistics starts from here.
Occupation in FD, BE, and MB statistics:
n̄FD () =
1
eβ(−µ)
+1
,
n̄BE () =
1
eβ(−µ)
−1
,
n̄MB () =
Low Temperature Limit of FD and BE Statistics
Low temperature for FD statistics:
FD: n̄ =
1
eβ(−µ) + 1
=
=
1
eβ(−F )
Let µ(T → 0) = F
β→∞
+1
=
=
if > F : 0
if < F : 1
The free electron model of electrons in metals starts from this. In
T = 0 K the valence electrons fill all energy states up to the Fermi
level. At higher temperature electron levels are “smeared out” around
52
1
eβ(−µ)
= e−β(−µ) .
MIT 3.20, Rickard Armiento, Lecture 9
the Fermi level. Taking interactions into account → more complicated models.
Low temperature for BE statistics:
BE: n̄ =
1
eβ(−µ)
−1
,
β→∞
The higher states gets less and less occupied, pushing more and
more bosons into the ground state(s). If µ → 0 the occupation
of the lowest state can diverge (i.e., essentially all particles in the
ground state) even at a low but non-zero critical temperature Tc ,
giving a Bose-Einstein condensate. This is why the transition to a
BEC is seen as a phase transition. (See, e.g., McQuarrie section
10-4 if you want to know more.)
For massless bosons where particle number is not conserved, e.g.,
photons, and phonons (= lattice vibrations, more on this in Lecture
11) one normally have µ = 0. One possible argument goes, for a
particle A, in a chemical reaction at equilibrium, one have ∆µA =
0. If particles are freely created and destroyed:
nA = mA,
m, n are integers
⇒ ∆µ = mµ − nµ = (m − n)µ = 0
Since m − n can be nonzero, we must have µ = 0.
Suggested Homework
•
•
•
•
•
•
McQuarrie problem 4.6.
McQuarrie problem 4.8.
Exam problem 2007E2-1
Exam problem 2004E2-1
Exam problem 2005E2-1(b,c)
Exam problem 2001E2-4(b,c)
53
MIT 3.20, Rickard Armiento, Lecture 10
Lecture 10: Distinguishable particles, configurational entropy, ideal
solid solution
• November 2, 2011, McQuarrie 11-6
The Formulas
P
Averages: Mmeasured = 1st postulate = M̄ = i Pi Mi
MicrocanonicalP
ensemble:
Pi = 1/Ω
Ω(N, V, E) = i = Degeneracy for energy E
Easiest connection to thermodynamics: S = kB ln Ω(N, V, E)
Canonical ensemble: Pi = e−Ei /(kB T ) /Q
P
P
Q(N, V, T ) = E Ω(E)e−E/(kB T ) = i e−Ei (V )/(kB T )
Easiest connection to thermodynamics: A = −kB T ln Q(N, V, T )
Grand canonical ensemble: Pi = e(−Ei +µN )/(kB T ) /Ξ
P P
P
Ξ(V, T, µ) = N Q(N, V, T )eµN/(kB T ) = N i e−EN,i (V )/(kB T )+µN/(kB T )
Easiest connection to thermodynamics: Φ = −kB T ln Ξ(V, T, µ)
In large ‘thermodynamical’ systems → all ensembles give the same result.
Independent distinguishable particles: Q = qA · qB · qC · ... = {if identical} = q N
Independent indistinguishable particles in high-temperature low-density limit: Q = q N /N !
Ideal Solid Solution
Consider some arrangement of atoms on a lattice with M sites, where “something is different”
with N sites; could be vacancies (= “Schottky defects”), substitutions, etc. It does not matter if
the lattice is in 1D, 2D, 3D.
In general, different arrangements have different energies. First approximation, assume that the
total energy only depends on concentration x = N/M . For a fixed M , set E = N We want to calculate S, A, µ, .... If the number of special sites N are conserved, so is the energy
= microcanonical ensemble.
M!
M
Pick N special sites
Ω(E) =
=
=
.
from M sites
N !(M − N )!
N
Note: while lattice sites are distinguishable, when we choose N sites to be special, it does not
matter in which order we choose them.
Stirling’s approximation
M!
S = kB ln Ω = kB ln
= kB ln M !−ln N !−ln(M −N )! =
=
ln N ! = N ln N − N
N !(M − N )!
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MIT 3.20, Rickard Armiento, Lecture 10
− N ln N + − (M − N ) ln(M − N ) + = kB M ln M − M
N
(M
N) =
−
= kB M ln M − N ln N − M ln(M − N ) + N ln(M − N ) =
= kB (M − N ) ln M + N ln M − N ln N − M ln(M − N ) + N ln(M − N ) =
= −kB M
N
M
ln
N
M −N M −N
+
ln
= −kB M (x ln x + (1 − x) ln(1 − x)) ⇒
M
M
M
Ideal solid solution configurational entropy.
S = −kB M x ln x + (1 − x) ln(1 − x) ,
where x is the concentration
of vacancies, subsitutions, etc.
Free energy:
A = E − T S = N + kB T M x ln x + (1 − x) ln(1 − x)
Alternatively, it is also ok to change over to the canonical ensemble
and realize we have just one energy:
X
X
Q=
Ω(E)e−βE = Ω(E)e−βE
e−βEi =
i
E
⇒ A = −kB T ln Q = E − kB T ln Ω = same thing.
Find the concentration that minimizes the free energy:
Minimize energy per site:
A/M = f + kB T x ln x + (1 − x) ln(1 − x) ⇒
∂(A/M )
=0
∂f
⇒ e−/(kB T ) =
⇒
x
1−x
+ kB T ln
⇒
x
=0⇒
1−x
xmin =
e−/(kB T )
.
1 + e−/(kB T )
If the energy difference is large >> kB T :
⇒
xmin = e−/(kB T )
This gives the expected vacancy concentration in a crystal.
Chemical potential:
∂A
∂x ∂ µ=
= +kB T M
x
ln
x+(1−x)
ln(1−x)
⇒
∂N V,T
∂N ∂x V,T
|{z}
1/M
55
MIT 3.20, Rickard Armiento, Lecture 10
µ = + kB T ln
x
1−x
Ideal Solid Solution in Grand Canonical Ensemble
Think: substrate in particle equilibrium with gas phase at fixed gas pressure.
Recall: µ = µ0 (T ) + kB T ln(p). Hence, if gas pressure and temperature is constant, µ = const.
Grand canonical ensemble: µ, V, T = constant. Again, take E to be independent of arrangement,
but on number of particles. E = M .
Ξ=
XX
N
e−β(EN,j −µN ) =
j
M
X
Q(N, T, V )eβµN
N =0
For a specific N -term in this sum, there is only one energy, meaning Q(N, T, V ) = Ω(E)e−βE ⇒
Ξ=
M
X
N =0
e
βµN
M!
e−βN N !(M − N )!
=
M h
X
e−β(−µ)
iN
N =0
M!
N !(M − N )!
Use the binomial expansion:
(a + b)M =
M
X
N =0
M!
aN bM −N
N !(M − N )!
with a = e−β(−µ) , b = 1 ⇒
Ξ = [1 + e−β(−µ) ]M
Φ = −kB T ln Ξ = −kB T M ln(1 + e−β(−µ) )
dΦ = −SdT − pdV − N dµ ⇒
e−β(−µ)
1
∂Φ
N (µ) = −
=M
=
M
∂µ V,µ
1 + e−β(−µ)
eβ(−µ) + 1
Solve for µ(N ) ⇒
µ(N ) = + kB T ln
x
1−x
Deriving µ(N ) in canonical ensemble vs. N (µ) in the grand canonical ensemble give the same
relation, because of the equivalence of ensembles.
The equivalence of ensembles is a statement about the relation between properties at an equilibrium. At an equilibrium, the chemical potential is related to the defect concentration in a certain
way. If the system is thermodynamically big, it does not matter if we calculate that relation
through the canonical or grand canonical ensemble. This does not mean that the solution to a
problem posed for an open system has the same solution as a similar problem in a closed system.
What is the concentration of adsorbed atoms in a closed system? Since the number of adsorbed
atoms is constant, it is exactly what you set it to. What is the concentration of adsorbed atoms in
56
MIT 3.20, Rickard Armiento, Lecture 10
an open system with µ = 2.0? Use N (µ) above to find out; i.e., x = N (2.0)/M . But it does not
matter if you got N (µ) from the microcanonical, the canonical, or the grand canonical ensemble,
because they are the same (for thermodynamically big systems).
Here it happens to be valid for all N . The underlying reason is because we assume atoms do not
interact with each other (E is independent of configuration), and thus the effect of the boundaries
of the system are absent.
Mixing Entropy of Ideal Gases, Gibbs Paradox
Consider two containers of volume V with distinguishable ideal gases, which are brought together.
E
In general: S = kB ln Q +
⇒
T
3
N ,q
3
Use ideal gas, Q = qatom
atom = V /Λ , E = 2 N kB T :
V
3
3
N
S1 = kB ln qatom
+ N kB = N kB ln 3 + N kB
2
Λ
2
3
V
+ N kB
3
Λ
2
V
= 2N kB ln 3 + 3N kB
Λ
S2 = kB ln
⇒
Ssum
After equilibrium:
Stot = kB ln(qatom (2V ))2N + 3N kB = 2kB N ln
⇒
(2V )
+ 3N kB
Λ3
∆S = 2kB N ln 2
Entropy increased. But what if the particles really were the same = indistinguishable? Then we
N /(N !)
need to treat them as indistinguishable. If we use Boltzmann statistics: Q = qatom
S1 = kB ln
N
qatom
3
V
3
+ N kB = N kB ln 3 − kB ln N ! + N kB ≈
N!
2
Λ
2
V
3
3
V
5
− kB N ln N + kB N + N kB = N kB = N kB ln
+ N kB
3
3
Λ
2
2
NΛ
2
V
5
S2 = N kB ln
+ N kB
N Λ3 2
V
⇒ Ssum = 2N kB ln
+ 5N kB
N Λ3
After equilibrium:
≈ N kB ln
Stot = kB ln
(qatom (2V ))2N
2V
V
+3N kB ≈ 2kB N ln 3 −kB 2N ln(2N )+2kB N +3N kB = 2kB N ln
+5N kB
(2N )!
Λ
N Λ3
⇒ ∆S = 0
Interesting discussion → what if we did not know that the particles were different when we
mixed them?
57
MIT 3.20, Rickard Armiento, Lecture 10
Suggested Homework
•
•
•
•
Exam problem: 2000E2-3
Exam problem: 2001E2-3
Exam problem: 2002E2-3
Exam problem: 2001E2-5 ((a-d) was suggested already back on lecture 3, if you did it
then, you can now look back and understand the context better)
58
MIT 3.20, Rickard Armiento, Lecture 11
Lecture 11: Vibrations in Monoatomic Solid - Einstein and Debye
model
• November 7, 2011, McQuarrie 11.1 - 11.5
REMEMBER: EXAM, NOVEMBER 9, 9:30am, ROOM 50-340
Here We Go Again
P
Averages: Mmeasured = 1st postulate = M̄ = i Pi Mi
MicrocanonicalP
ensemble:
Pi = 1/Ω
Ω(N, V, E) = i = Degeneracy for energy E
Easiest connection to thermodynamics: S = kB ln Ω(N, V, E)
Canonical ensemble: Pi = e−Ei /(kB T ) /Q
P
P
Q(N, V, T ) = E Ω(E)e−E/(kB T ) = i e−Ei (V )/(kB T )
Easiest connection to thermodynamics: A = −kB T ln Q(N, V, T )
Grand canonical ensemble: Pi = e(−Ei +µN )/(kB T ) /Ξ
P
P P
Ξ(V, T, µ) = N Q(N, V, T )eµN/(kB T ) = N i e−EN,i (V )/(kB T )+µN/(kB T )
Easiest connection to thermodynamics: Φ = −kB T ln Ξ(V, T, µ)
In large ‘thermodynamical’ systems → all ensembles give the same result.
Independent distinguishable particles: Q = qA · qB · qC · ... = {if identical} = q N
Independent indistinguishable particles in high-temperature low-density limit: Q = q N /N !
Paramagnetism, Other Work Terms
Paramagnetic solid: internal dipoles. In an external field the dipoles align to minimize energy,
but the statistical motion of the dipoles due to temperature counteracts this.
Model: take N distinguishable non-interacting spins in magnetic field H. Each magnetic moment m is either parallel or anti-parallel with the field.
ni = ±1,
M=
N
X
ni m,
Etot = M H =
i=1
N
X
ni mH
i=1
The formalism of Zemansky, problem sets, old exam problems:
Magnetic work: δW = HdM, ⇒ dU = T dS − pdV + µdN + HdM
I.e., the work term is: δW = [intensive, field] · d[Extensive].
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MIT 3.20, Rickard Armiento, Lecture 11
This represents a choice of system where the mutual field energy (e.g., the energy stored in the
magnetic field) is included within the system. This makes U = energy from spins + field, and
thus E = E(T ) is independent of H and M . If the temperature is constant T = T0 , one can
set E(T0 ) = 0 making “the internal energy zero”. Analog example where a spring transfers
energy to a gas: if both the spring and the gas are inside the system we are merely redistributing
internal energy ⇒ dE = 0. In a paramagnetic system + field: the energy of the dipoles increase
as +HdM , but that energy is taken from the field −HdM ⇒ dE = 0.
Hence, U (S, V, N, M ) and H(T, V, N, M ) = U − T S are not helpful for studying the magnetization of the system ⇒ go instead to a more relevant thermodynamic potential Λ(T, V, N, H),
which has H instead of M as its natural variable.
Need thermo potential Λ(T, V, N, H):
Λ(T, V, N, H) = U − T S − HM
⇒
dΛ = −SdT − pdV + µdN − M dH
Corresponding partition function: (see Lecture 5)
Z(T, V, N, H) =
X
e−β(Ei −Mi H) =
i
X
e+βMi H =
i
Indep. spins
= (eβmH +e−βmH )N = (2 cosh(βmH))N
Z = zN
Λ = −kB T ln Z
Average magnetization: look at potential: dΛ = T dS − pdV + µdN − M dH
∂Λ
= N m tanh(βmH)
M̄ = −
∂H N,V,S
⇒
Alternatively, this also works:
M̄ =
1 X
1 X
1 ∂Z
∂ ln Z
Mi P i =
Mi eβ(Mi H) =
=
Z
Z
Z ∂(βH)
∂(βH)
i
i
1 ∂ M̄
Can now calculate susceptibility ( M̄
∂H ), etc.
Common alternative formalism in books, etc.
Other literature commonly do not include the field energy in the internal energy, giving instead
this for work term and 1st law:
δW = −M dH
⇒
dU = T dS − M dH
With this formalism one gets E = M H, the natural variable becomes H ⇒ U (T, N, V, H)
and A(T, N, V, H) (rather than our U (T, N, V, M ), etc.). Here, another Laplace transform is
not needed. One proceeds with the canonical ensemble, but, it is now Q(T, N, V, H) where the
internal energy in the exponent E = M H. Hence, the expression for this Q becomes exactly
the same as we had for our Z above! The calculation becomes identical, just
substitute: Z → Q,
∂A
Λ → A above, and note that we end up with the same derivative ∂H N,V,S = −M . For a
very helpful extended discussion, see Kittel ‘Elementary statistical physics’ Chapter 18. For an
explicit example, see, e.g., the solution manual to Chandler ’Introduction to Modern Statistical
Mechanics’, problem 3.18/3.19. Feel free to compare with our exam problem 2003E2-3.
60
MIT 3.20, Rickard Armiento, Lecture 11
Lattice Vibrations, Einstein Solid; independent potentials
Historic problem, one of the great successes of statistical mechanics:
• Dulong-Petit 1819: empirically CV ≈ 3N kB .
• Einstein 1907: Statistical mechanics (but before the formalism was developed) for lattice
vibrations ⇒ Dulong-Petit is the high temperature limit, ok agreement for intermediate
temperatures, wrong for low temperature limit.
• Debye Model 1912: Correct low temperature limit.
Einstein Solid
In 1D
N independent atoms vibrating. Harmonic approximation
v(z) = v(0) +
1 ∂2v 2
z + ...
2 ∂z 2
Exactly like diatomic molecule: assume N atoms that move independently in separable potentials ⇒ Single-particle Schrödinger equation:
r
h̄ ∂ 2
1 2
1
F
−
+ F z Ψ = Ψ , n = (n + )h̄ω; n = 0, 1, 2, ..., ω =
2
2m ∂z
2
2
m
Same procedure as for diatomic molecule:
qsingle particle =
∞
X
e
−β(n+1/2)h̄ω
=
n=0
Geometric series
eΘE /(2T )
= Θ /T
⇒
h̄ω
ΘE = kB
e E −1
For Q, should we divide by (N !)? No, Lattice sites are distinguishable; we could label each one.
Qsystem = q
N
=
eΘE /(2T )
eΘE /T − 1
!N
= 1D result.
In 3D: Idependent degrees of freedom ⇒
Q3D = Q31D =
eΘE /(2T )
eΘE /T − 1
!3N
Connection to thermodynamics (same as for diatomic ideal gas, except without N ! and the exponent is 3N :
ΘE
ΘE
∂ ln Q
Ē = kB T
= 3N kB
+ Θ /T
∂T
2
e E −1
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MIT 3.20, Rickard Armiento, Lecture 11
Heat capacity of Einstein Solid
!
2
Θ
eΘ/T
CV =
= 3 N kB
T
(eΘ/T − 1)2
V,N
↑
(Like ideal diatomic gas, but times 3 due to 3D)
∂E
∂T
Needs one fitting parameter:
ΘE =
hν
h̄ω
=
,
kB
kB
(= larger if crystal is stiffer.)
Limits:
T → high;
CV → 3N kB ;
T → 0;
CV → 3N kB
ΘE
T
2
Agrees with Dulong-Petit law
; classical mechanics
e−ΘE /T → 0;
To zero as e−ΘE /T .
Experiments: ∼ cT 3
Discrete energy levels used before quantum mechanics was developed. Historical impact on the adoption of both quantum theory and
statistical mechanics.
Examples: Diamond: ΘE ∼ 1320 K, Pb: ΘE ∼ 100 K.
F
∂F =-S
(∂T
)V,N
FCC
High temperature limit (similar to ideal diatomic gas): Cv = 3N kB T .
The energy levels starts to look continuous when ∆E of a new level
is << Etot → ‘classical limit’.
BCC
Ttransition
Stiffer crystal → more separated energy levels → higher Einstein temperature → lower heat
capacity.
Vibrations can be important in phase transformations!
Z
CV
CV
dS =
dT → ∆S =
dT ⇒
T
T
⇒ S(soft bond material) > S(stiff bond material).
(At T < ΘE , otherwise all materials CV = 3N kB .)
All else equal, bigger entropy = more stable (due to −T S term in free energy)
Often FCC crystal structure → BCC as temperature increase for metals. BCC is more open →
usually softer force constants.
62
T
MIT 3.20, Rickard Armiento, Lecture 11
Phonons
The Einstein model is wrong in the low temperature limit, because atoms do not vibrate independently.
j-1
j
zj-1
zj zj+1
j+1
Multivariable Taylor expansion around {zj = 0} : (generally valid as long as all {zj } are small)
N N
N N
N
X
1 X X ∂v 1 XX
∂v zj +
zi zj +... = v0 +
ki,j zi zj +...
v(z1 , z2 , z3 , ...) = v(0, 0, 0, ...)+
∂zj zj =0
2
∂zi ∂zj zi =0,zj =0
2
i=1 j=1
i=1 j=1
j=1
| {z }
= 0 since zj = 0 = equilibrium positions
We get back to independent harmonic oscillators if all cross-terms are = 0.
Mechanics: in a coupled system of linear forces one can remove the crossterms by changing to
“normal coordinates” = a linear combination of the Cartesian coordinates zj ⇒ in 3D: 3N − 6 ≈
3N independent harmonic oscillators! (but now all with their own masses / frequencies). (We
lose 6 degrees of freedom for uniform translations and rotations.)
Energies:
E=
3N
X
j=0
3N
X
1
h̄ωj (nj + ) = E0 +
h̄ωj nj
2
j=0
The system is composed by independent ‘virtual’ particles which occupy the states 1, 2, ..., 3N
with energies h̄ω1 , h̄ω2 , ...h̄ω3N . Only dependent on how many in each state = indistinguishable.
No restriction on nj = bosons. These are called phonons. Can use BE statistics formulas. Very
useful.
The Partition function:
Q=
X
P
e−β(
(i) 1
k (nk + 2 )h̄ω(k))
i
The missing piece is ω(k), we need knowledge of ω(k) to get properties from thermodynamics.
It is a function of lattice, symmetry, masses, bond types, etc...
Alternatively,
R ω(k) → phonon density of states g(ω) = density of states, or of frequency: g(ν).
Requires: g(ν)dν = 3N
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MIT 3.20, Rickard Armiento, Lecture 11
"Einstein model"
Monoatomic linear
chain w. neighbors
Debye model
ω(k)
ω(k)
ω(k)
ωD
"k"
k
k
Debye model
Einstein model
g(v)
Real solid
g(v)
g(v)
v
v
v
Debye Model
Idea: at low temperatures lowest frequencies are relevant. These should not be very dependent
on the nature of atoms = assume the crystal is continuous ⇒
⇒
ω(k) = ck
4πV ν 2
c3
g(ν) =
with c = velocity of sound. Taking all different wave-modes in 3D into account (transverse +
longitudinal) → average sound velocity c0 and:
g(ν) =
12πV
c30
Treat all phonon modes like this, but if so, we need to cut off g(ν) when it reaches 3N .
νD
Z
g(ν)dν = 3N
⇒
νD =
0
3N
4πV
1/3
c
Debye model
g(ν) =
h3 /(kB ΘD )3 9N ν 2 ,
0, otherwise
0 ≤ ν ≤ kB ΘD /h
,
Debye temperature: ΘD =
From the partition function we get:
CV = 9N kB
T
ΘD
3 Z
0
64
θD /T
x4 e x
dx
(ex − 1)2
hνD
kB
MIT 3.20, Rickard Armiento, Lecture 11
Examples:
Diamond: ΘD = 1860 K, Pb: ΘD = 88 K, Most metals ∼ 100 − 400 K (See McQuarrie, table
11-1)
Limits:
Dulong and Petit in high temperature limit: CV →
1
3
ΘD
T
Experimental ∼ T 3 law for low temperature limit: CV →
From Wikipedia
Suggested Homework
• Final Exam 2000-5(a,b)
• McQuarrie problem 11-2
• Problem Set 6, problem 5
65
3
12π 4
5 N kB
T
ΘD
3