Statistical Thermod Mechanics Nassimi.pdf

Statistical Thermodynamics I,
23915
Ali Nassimi
[email protected]
Chemistry Department
Sharif University of Technology
July 14, 2021
1/1

Aim
Your most valuable asset is your learning ability.
This course is a practice in learning and specially improves your
deduction skills.
This course provides you with tools applicable in understanding many
natural, societal or nancial phenomena.
End of semester objective: You should become able to calculate any
equilibrium thermodynamic property of a system given intermolecular
interactions in that system.
2/1

Aim
Statistical mechanics is about deducing macroscopic properties of a
system from microscopic properties of the constituents of that same
system (molecules).
Equilibrium statistical mechanics aims at calculating equilibrium
properties of the system, e.g., energy, entropy and free energy.
Non-equilibrium statistical mechanics aims at calculating
non-equilibrium (transport) properties of the system, e.g., electrical
conduction, heat conduction, absorption spectra and viscosity.
This course will cover equilibrium statistical mechanics and leaves
non-equilibrium statistical mechanics to another course.
3/1

References
Statistical Mechanics by Donald A. McQuarrie
Statistical Thermodynamics by Donald A. McQuarrie Statistical Thermodynamics, Theories and Applications by G. A.
Parsafar (In Farsi)
4/1

Course structure
Midterm exam 16 Ordibehesht
11 AM
Ch. 1-4 35%
Final exam 12 Tir 3:30 pm Ch. 5 - 8 45%
Class presentation 15 Tir-22 Tir 20%
Raise your question and concern as it might be the question or
concern of your classmates.
Always remember that equations are the language of science but they
never do suce.
5/1

Topics
Preamble
Ensembles Classical and quantum statistics Monatomic and diatomic gases Classical statistical mechanics Polyatomic gases Chemical equilibrium Crystals
6/1

Classical Mechanics-Newton formalism
In the absence of external forces, motion will continue with a constant
speed.
~
F=m~a. If body A exert a force on body B, then B exerts the
same force in the opposite direction on A.
Two dimensional motion under coulombic attraction to a xed center. ~
F=k~r=r
3
, break down into components. mx=Fx=
Kx
(x
2
+y
2
)
3=2andmy=Fy=
Ky
(x
2
+y
2
)
3=2
Use polar coordinate systemx=rcosandy=rsinto derive.
mr
2_
= constant andmr=
k
r
2+
l
2
mr
3
7/1

Classical Mechanics-Lagrangian
Joseph-Louis Lagrange (1736-1813) was an Italian enlightenment era
mathematician and astronomer with signicant contributions to
analysis, number theory, and both classical and celestial mechanics.
Lagrange succeeded Euler as the director of mathematics at the
Prussian Academy of Sciences in Berlin, where he stayed for over
twenty years, producing volumes of work and winning several prizes of
the French Academy of Sciences.
Lagrange's treatise on analytical mechanics oered the most
comprehensive treatment of classical mechanics since Newton and
formed a basis for the development of mathematical physics in the
nineteenth century.
Lagrangian,LKU, whereK=
P
i
mi
2
_x
2
i
. Lagrangian dynamics:
d
dt
@L
@_qj
=
@L
@qj
, the form of this equation is
invariant under the change of coordinates.
Two dimensional motion under coulombic attraction to a xed center.
8/1

Classical Mechanics-Hamiltonian
Sir William Rowan Hamilton (1805{1865) was an Irish mathematician,
astronomer, and mathematical physicist, who made important
contributions to classical mechanics, optics, and algebra.
His studies of mechanical and optical systems led him to discover new
mathematical concepts and techniques.
His reformulation of Newtonian mechanics, now called Hamiltonian
mechanics has proven central to the modern study of classical eld
theories such as electromagnetism, and to the development of
quantum mechanics.
Momentum,pj=
@L
@_qj
. Hamiltonian,H=
P
j
pj_qjL. Kinetic energy,K=
P
j
aj(q) _q
2
j
. Total energy of the system,
H=K+V.
Hamilton's equations of motion,
@H
@pj
= _qj;
@H
@qj
=_pj.
9/1

Classical Mechanics-Liouvillian
Classical mechanics occurs in phase space which consists of one
dimension (axe) for each coordinate and each momenta.
The state of a system is determined by a single point in its phase
space.
Joseph Liouville (1809{1882) was a French mathematician. Liouville became a member of the Constituting Assembly in 1848.
However, after his defeat in the legislative elections in 1849, he turned
away from politics.
Every property, f, of the system is a function of coordinates, r, and
momenta, p.
Dynamics:
@f
@t
=ff;Hg Poisson bracket:ff;gg=
P
i
(
@f
@qi
@g
@pi

@f
@pi
@g
@qi
)
10/1

Quantum Mechanics-preamble
From the late 19th century people started to patch classical physics to
justify some observations including photoelectric eect and black body
radiation.
Quantum mechanics was formally formulated in the 1920s. Quantum mechanics lives in the Hilbert space where there is no hole.
A hole in a space occurs when the limit of a series cannot be found in
that space.
A vector, C, is denoted by a ketjCi. Complex conjugate of such a
vector,C
y
, is denoted by a brahCj.
bra * ket = bracket,hDjCi. ket * bra = operator,jCihDj. Operator acts on a vector to produce another vector Every observable in quantum mechanics is represented by an operator.
If we require that the expectation value of an operator^Ais real, then
^Amust be a Hermitian operator.
11/1

Quantum Mechanics-operators
Observable
Name
Observable
Symbol
Operator
Symbol
Operator Opera-
tion
Position r ^r Multiply byr
Momentum p ^p i~

^i
@
@x
+^j
@
@y
+
^
k
@
@z

Kinetic en-
ergy
T ^T
~
2
2m

@
2
@x
2+
@
2
@y
2+
@
2
@z
2

Potential
energy
V(r) ^V(r) Multiply by V(r)
Total energyE
^
H
~
2
2m

@
2
@x
2+
@
2
@y
2+
@
2
@z
2

+
V(r)
Angular lx
^lx i~

y
@
@z
z
@
@y

momentum ly
^ly i~

z
@
@x
x
@
@z

lz
^
lz i~

x
@
@y
y
@
@x

12/1

Quantum Mechanics
A density matrix is a matrix that describes the statistical state of a
system in quantum mechanics. The probability for any outcome of
any well-dened measurement upon a system can be calculated from
the density matrix for that system.
The extreme points in the set of density matrices are the pure states,
which can also be written as state vectors or wavefunctions. Density
matrices that are not pure states are mixed states.
Any mixed state can be represented as a convex combination of pure
states, and so density matrices are helpful for dealing with statistical
ensembles of dierent possible preparations of a quantum system, or
situations where a precise preparation is not known, as in quantum
statistical mechanics.
given a nite number of pointsx1;x2; : : : ;xnin a real vector space, a
convex combination of these points is a point of the form
1x1+2x2+ +nxnwhere the real numbersisatisfyi0
and1+2+ +n= 1.
13/1

Quantum Mechanics
Density matrix, ^, contains all information that can be known about a
system.
For a pure state, ^=j ih jin general ^=
P
i
cij iih ij Quantum Liouville equation determines quantum dynamics as
@^(t)
@t
=
i
~
[^(t);^H].
[^A;^B] =^A^B^B^A. f;g $
i
~
[;]. Formal solution: ^(t) =e
i
^
Ht=~
^(0)e
i
^
Ht=~
. Every linear operator has a matrix representation. Schrodinger equation for pure states:i~
@
@t
j i=^Hj i hxj i= (x;t) For a time independent Hamiltonian assume
(x;t) =(x)(t); =e
iEt=~
; ^H=E:14/1

Quantum Mechanics
Wave function or state function, has the important property that

(r;t) (r;t)dis the probability that the particle lies in the volume
elementdlocated atrat timet.
For the case of a single particle, the probability of nding it somewhere
is 1, we have the normalization condition
R
1
1

(r;t) (r;t)d= 1.
The wavefunction must also be single-valued, continuous, and nite. In any measurement of the observable associated with operator^A, the
only values that will ever be observed are the eigenvaluesa, which
satisfy the eigenvalue equation^A =a .
The values of dynamical variables can be quantized (although it is still
possible to have a continuum of eigenvalues).
If the system is in an eigenstate of^Awith eigenvaluea, then any
measurement of the quantityAwill yielda.
15/1

Quantum Mechanics
Although measurements must always yield an eigenvalue, the state
does not have to be an eigenstate of^Ainitially. An arbitrary state can
be expanded in the complete set of eigenvectors of^A(^A i=ai i)
as =
P
n
i
ci i
We only know that the measurement ofAwill yield one of the values
ai, but we don't know which one. However, we do know the
probability that eigenvalueaiwill occur{it is the absolute value
squared of the coecient,jcij
2
.
After measurement of A on yields some eigenvalueai, the
wavefunction immediately \collapses" into the corresponding
eigenstate i(in the case thataiis degenerate, then is projected
onto the degenerate subspace ofai). Thus, measurement aects the
state of the system.
16/1

Stern-Gerlach experiment and Spin
Magnetic moment is a quantity that represents the magnetic strength
and orientation of a magnet.
Loops of electric current (such as electromagnets), permanent
magnets, elementary particles (such as electrons), various molecules,
and many astronomical objects poses magnetic dipole moment.
The magnetic dipole moment of an object is readily dened in terms
of the torque that object experiences in a given magnetic eld.
=mB
The direction of the magnetic moment points from the south to north
pole of the magnet (inside the magnet).
A magnetic moment in an externally produced magnetic eld has a
potential energyU=mB.
17/1

Stern-Gerlach experiment and Spin
m=^x
@Uint
@Bx
^y
@Uint
@By
^z
@Uint
@Bz
.
m=IS,m=NIS.
m=
1
2
ZZZ
V
rjdV;
Since the particles creating the current (by rotating around the loop)
have charge and mass, both the magnetic moment and the angular
momentum increase with the rate of rotation. The ratio of the two is
called the gyromagnetic ratio orso that:m=L
The Stern{Gerlach experiment demonstrated that the spatial
orientation of angular momentum is quantized.
18/1

Stern-Gerlach experiment and Spin
Silver atoms were sent through a spatially varying magnetic eld,
which deected them before they struck a detector screen, such as a
glass slide.
19/1

Stern-Gerlach experiment and Spin
Particles with non-zero magnetic moment are deected, due to the
magnetic eld gradient, from a straight path.
The screen reveals discrete points of accumulation, rather than a
continuous distribution, owing to their quantized spin.
Spin is an intrinsic angular momentum of subatomic particles that is
closely analogous to the angular momentum of a classically spinning
object, but that takes only certain quantized values.
^
Sz z+=
~
2
z+
^
Sz z=
~
2
z
Only one component of a particle's spin can be measured at one time,
meaning that the measurement of the spin along the z-axis destroys
information about a particle's spin along the x and y axis.
Particles with half integral spin are called fermions while those with
integral spin are called bosons.
20/1

Quantum Mechanics
Thus an atomic-scale system was shown to have intrinsically quantum
properties.
This experiment was decisive in convincing physicists of the reality of
angular-momentum quantization in all atomic-scale systems.
Particle in a box:
(
U(x) = 0 0<x<a;
U(x) =1otherwise:
n=
h
2
8ma
2n
2
IfH=H1+H2then = 1 2andE=E1+E2.
21/1

Quantum Mechanics
Harmonic oscillator is one of the few quantum-mechanical systems for
which an exact, analytical solution is known:
U(x) =
1
2
kx
2
; ^H=
^p
2
2m
+
1
2
k^x
2
=
^p
2
2m
+
1
2
m!
2
^x
2
; n=
(n+
1
2
)~!; ! =
p
k=m.
n(x) =Nne

2
x
2
/2
Hn(x);n= 0;1;2;3; :::,=
p
m!/}, H0(y) = 1H1(y) = 2y
H2(y) = 4y
2
2
H3(y) = 8y
3
12y:
0= (

)
1=4
e
y
2
=2
1= (

)
1=4
p
2ye
y
2
=2
Rigid rotor:H=
~
2
2I

1
sin
@
@
(sin
@
@
) +
1
sin
2

@
2
@
2

; j=
J(J+1)~
2
2I
; ! j= 2j+ 1.
22/1

Particle in a box degeneracy
For a single particle in 2-D,E=
h
2
8ma
2(n
2
x+n
2
y),
Convince yourself that the number of states with energy smaller than
is the area of circle of radiusr
2
=
8ma
2
h
2in the rst quadrant.
23/1

Particle in a box degeneracy
In 3-D, number of states with energy betweenand+ ,
!(;) =

4
(
8ma
2
h
2)
3=2

0:5
+O(()
2
)
N non-interacting particle:E=
h
2
8ma
2
P
3N
j=1
n
2
j
.
We may dene a coordinate system in an n-D space which is
analogous to the spherical coordinate system dened for 3-dimensional
Euclidean space, in which the coordinates consist of a radial
coordinate r, and n-1 angular coordinates1; 2; ; n1, where the
angles1; 2; ; n2range over [0; ] radians andn1ranges over
[0;2) radians.
x1=rcos1;x2=rsin1cos2;x3=
rsin1sin2cos3; ;xn=rsin1 sinn1.
Vn=
R
Sphere
d=
R
a
0
hrdr
R

0
h1d1
R
2
0
hn1dn1=
R
a
0
Snr
n1
dr.
R
angles
dx1 dxn=r
n1
Sndr
24/1

Particle in a box degeneracy
In=
R
1
1

R
1
1
e
(x
2
1
+x
2
2
++x
2
n
)
dx1 dxn= (
R
1
1
e
x
2
dx)
n
=

n=2
.
In=
R
1
0
e
r
2
r
n1
Sndr=Sn(n=2)=2.
(x) =
R
1
0
e
t
t
x1
dt, show that (n+ 1) =n!, show that
(n+
1
2
) =
(2n)!
2
2n
n!
p
.
Vn=

n=2
(n=2+1)
a
n
!(E;E) =
1
(n+1)(3n=2)
(
2ma
2
h
2)
3n=2
E
3n=21
E
Partitionable Hamiltonian.H=H1+H2+ , = 1 2
Parity operator.
25/1

Thermodynamics
Universe:
'
&
$
%
'
&
$
%
System Environment
Zeroth law of thermodynamics, rst law of thermodynamics, Second
law of thermodynamics.
State functions vs. path functions dU=dq+dw=TdSpdV U =U(S;V) T= (
@U
@S
)V;p=(
@U
@V
)S.
26/1

Thermodynamics
Legendre transformation:F=F(x);s=
FG
x0
G(s) =Fsx.
Generally,G(s) =F
P
j
sjxj.
U(S;V);H(S;p) =?;A(T;V) =U(
@U
@S
)VS=
UTS;G(T;p) =?
A thermodynamic potential is a scalar quantity used to represent the
thermodynamic state of a system.
27/1

Thermodynamics
The concept of thermodynamic potentials was introduced by Pierre
Duhem in 1886. Josiah Willard Gibbs in his papers used the term
fundamental functions.
Internal energy U is the energy of conguration of a given system of
conservative forces and only has meaning with respect to a dened set
of references.
Expressions for all other thermodynamic energy potentials are
derivable via Legendre transforms from an expression for U.
For an open quantum systemdU=TdSpdV+dnwhere
= (
@U
@n
)S;V.
Grand potential or Landau free energy is dened by
G(T;V; )
def
=
def
=FN=UTSN
Fundamental equation:
dG=dUTdSSdTdNNd=PdVSdTNd
When the system is in thermodynamic equilibrium, Gis a minimum.
28/1

Thermodynamics
For homogeneous systems, one obtains =PV.
Internal energy (U) is the capacity to do work plus the capacity to
release heat.
Gibbs energy (G) is the capacity to do non-mechanical work. Enthalpy (H) is the capacity to do non-mechanical work plus the
capacity to release heat.
Helmholtz energy (F) is the capacity to do mechanical plus
non-mechanical work.
U is the energy added to the system, F is the total work done on
it, G is the non-mechanical work done on it, and H is the sum of
non-mechanical work done on the system and the heat given to it.
The principle of minimum energy follows from the rst and second
laws of thermodynamics.
When the entropy S and "external parameters" (e.g. volume) of a
closed system are held constant, the internal energy U decreases and
reaches a minimum value at equilibrium.
29/1

Thermodynamics
The following three statements are directly derivable from this
principle.
When the temperature T and external parameters of a closed system
are held constant, the Helmholtz free energy F decreases and reaches
a minimum value at equilibrium.
When the pressure p and external parameters of a closed system are
held constant, the enthalpy H decreases and reaches a minimum value
at equilibrium.
When the temperature T, pressure p and external parameters of a
closed system are held constant, the Gibbs free energy G decreases
and reaches a minimum value at equilibrium.
The variables that are held constant in this process are termed the
natural variables of that potential
30/1

Thermodynamics
If a thermodynamic potential can be determined as a function of its
natural variables, all of the thermodynamic properties of the system
can be found by taking partial derivatives of that potential with
respect to its natural variables and this is true for no other
combination of variables.
If there are D dimensions to the thermodynamic space, then there are
2
D
unique thermodynamic potentials.
U[j] =UjNj,F[j] =UTSjNj,H[j] =U+pVjNj
andG[j] =U+pVTSjNj.
Fundamental equations:dH=dA=dG= If the system has more external variables than just the volume that
can change, the fundamental thermodynamic relation generalizes to:
dU=T dS
X
i
Xidxi+
X
j
jdNj, Xiare the generalized forces
corresponding to the external variables xi.
Maxwell relations, e.g., using dA=-pdV-SdT to derive
(
@S
@V
)T= (
@p
@T
)V.
31/1

Thermodynamics
Extensive vs. intensive.
Fundamental relation can be written as,
dU
dV
=T(
dS
dV
)p, imposing
the constant temperature condition and using a Maxwell relation
yields (
@U
@V
)N;TT(
@p
@T
)N;V=p.
U=U(V,T), thusdU= [T(
@p
@T
)Vp]dV+CVdT. CV= (
@U
@T
)V=T(
@S
@T
)VandCp= (
@H
@T
)p=T(
@S
@T
)p. Thus,CpCv= [p+ (
@U
@V
)T](
@V
@T
)p Chemical potential,
j= (
@U
@Nj
)S;V;:::= (
@H
@Nj
)S;p;:::= (
@A
@Nj
)V;T;:::= (
@G
@Nj
)p;T;:::.
Homogeneous of degree n,f(x1; ; xN) =
n
f(x1; ;xN) Euler's theorem: If f is a homogeneous function of degree n,
nf(x1; ;xN) =x1
@f
@x1
+x2
@f
@x2
+ +xN
@f
@xN
n(n1)f(x1; ;xN) =
P
N
i;j=1
xixj(
@
2
f
@xi@xj
) G(T;p;Ni) =
P
j
Nj(
@G
@Nj
)T;p;=
P
j
Njj
32/1

Thermodynamics
Find expressions for other thermodynamic potentials.
Gibbs-Duhem equation:
P
j
Njdj= 0 For a chemical reactionaA+bBcC+dD, which can be
represented by
P
j
jAj= 0,dG=
P
j
jdNj= (
P
j
jj)d.
At equilibrium:
P
j
jj= 0 Phase equilibrium j=
0
j
+RTln
pj
p
0=
0
j
+RTlnp
0
j 0=RTln[
(P
0
c
)
c(P
0
d
)

d
(P
0
a
)
a(P
0
b
)

b
] =RTlnKp
33/1

Ensembles
An experimenter repeating an experiment under the same macroscopic
conditions is unable to control microscopic details, thus he might
expect a range of outcomes.
A large number of identical (on a macroscopic level) systems
constitute an ensemble.
An ensemble is a collection of systems sharing one or more
macroscopic characteristics but each being in a unique microstate.
The complete ensemble is specied by giving all systems or
microstates consistent with the common macroscopic characteristics
of the ensemble.
The system may be specied by N, V, E, or N, V, T, while the
ensemble hasAidentical systems.
There is an enormous number of microstates, O(10
N
), consistent with
the systems specications.
System properties depend on the microstate of the system. Time independent Schrodinger equation determines allowed energy
levelsEjand their corresponding degeneracies (Ej).
34/1

Canonical Ensemble: Denition
Ensemble average of the mechanical property B,B=
1
A
P
A
i=1
Bi
whereBi=
R

i
^B id.
Ergodic hypothesis states that time average of a mechanical property
equals ensemble average of the same thermodynamic property.
Make a large collection of systems each having walls impermeable to
matter but heat conducting. Bring this collection in contact with a
heat bath of temperature T. After equilibration isolate the ensemble.
Each system is specied by N, V, T while the ensemble is specied by
AN,AV, and.
Solve Schrodinger equation for the system specied by N V. Specify
statesE1;E2;E3; such thatEiEi+1. The number of systems
occupying each state, respectively, isa1;a2;a3; . Set of occupation
numbersa1;a2;a3; is called a distribution.
Energy time uncertainty relation: Et
~
2
The principle of equal a priori probabilities.
35/1

Canonical Ensemble: Averages
The number of ways the systems can take this distribution,
distribution multiplicity, isW(a1;a2; ) =
A!
iai!
where
P
i
ai=A
P
i
aiEi=.
aj=
P
a
ajW(a)
P
a
W(a)
Pj=
aj
A
=
P
a
ajW(a)
A
P
a
W(a)
M=
P
j
MjPj
What are unjustied assumptions in this treatment? Noting thatW(a) is a multinomial distribution, and letting
A ! 1 Pj=
aj
A
=
P
a
ajW(a)
A
P
a
W(a)
=
a

j
W(a

)
AW(a

)
=
a

j
A
@
@aj
W(a)
@
@aj
(A
P
i
ai)
@
@aj
(
P
i
aiEi) = 0
@
@aj
ln
A!
iai!

@
@aj
(A
P
i
ai)
@
@aj
(
P
i
aiEi) = 0
@
@aj
[AlnA A i(ailnaiai)] +(
P
i
ij) +(
P
i
ijEi) = 0
lna

j
++Ej= 0!a

j
=e
Ej. ThusPj=
a

j
A
=
e

e
E
j
A
. Using the normalization of probabilitiesPj=
e
E
j
P
i
e
E
i
.
36/1

Canonical Ensemble: Partition function
Canonical ensemble partition function,Q=
P
i
e
Ei, is a bridge
between quantum mechanical energy levels and thermodynamic
functions.
E=
P
j
Ej
e
E
j
Q
=
(@Q=@)N;V
Q
=(
@lnQ
@
)N;V
Adiabatic process, PV work onlydEj=pjdV;pj=(
@Ej
@V
)N
p=
P
j
pjPj=
P
j
(
@E
j
@V
)Ne
E
j
Q
=
1
Q
(
@Q
@V
)N;=
1

(
@lnQ
@V
)N;
Ensemble postulate of Gibbs.E=E p = p
(
@

E
@V
);N= (
@(
E
j
e
E
j
Q
)
@V
);N=p+EpEp
(
@p
@
)N;v=
@
@
(
@E
j
@V
)Ne
E
j
Q
=EpEp
(
@

E
@V
);N+(
@p
@
)N;V=p (
@E
@V
)T;NT(
@p
@T
)N;V=por (
@E
@V
)T;N+
1
T
(
@p
@(1=T)
)N;V=p
37/1

Canonical Ensemble: Value of
=
1
kT
To prove the universality of k, construct an ensemble composed of
systems A and B paired, with number of particles and volume,NA;VA
andNB;VB, respectively.
Figure:
38/1

Canonical Ensemble: Value of
The number of possible system distributions resulting in the
composite ensemble state ab,W(a;b) =
A!
jaj!
B!
kbk!
P
j
aj=A;
P
j
bj=B;
P
j
(ajEjA+bjEjB) =E
Pij=
ai
A

bj
B
=
P
ab
aiW(a;b)
A
P
ab
W(a;b)
P
ab
bjW(a;b)
B
P
ab
W(a;b)
Using maximum term method:Pij=
a

i
A
b

j
B
@
@al
lnW(a;b)1
@
@al
(A
P
j
aj)2
@
@al
(B
P
j
bj)
@
@al
(E
P
j
(ajEjA+bjEjB)) = 0
@
@bl
lnW(a;b)1
@
@bl
(A
P
j
aj)
2
@
@bl
(B
P
j
bj)
@
@bl
(E
P
j
(ajEjA+bjEjB)) = 0
Using the normalization conditionPij=
e
E
iA
QA
e
E
jB
QB
=PiAPjB
Thus two arbitrary systems in thermal contact have the same value of
. Since=
1
kT
they must have the same value of k.
k can be determined for any system including an ideal gas.
39/1

Canonical Ensemble: Value of
If the external parameters of the system remain constant then the
interaction is termed a purely thermal interaction. It is the
distribution of the systems in the ensemble over the various
microstates which is modied.
Suppose that the systemAis thermally insulated from its
environment. The systemAis still capable of interacting with its
environment via its external parameters. This type of interaction is
termed mechanical interaction, and any change in the average energy
of the system is attributed to work done on it by its surroundings.
On a microscopic level, the energy of the system changes because the
energies of the individual microstates are functions of the external
parameters. Thus, if the external parameters are changed then, in
general, the energies of all of the systems in the ensemble are modied
(since each is in a specic microstate).
Considerf(;E1;E2; ) = lnQ,
df= (
@f
@
)Ej
d+
P
k
(
@f
@Ek
);Ei6=k
dEk=Ed
P
k
PkdEk
40/1

Canonical Ensemble: interpretation of work and heat
d(f+

E) =(d

E
P
k
PkdEk)
Molecular interpretation of reversible work is a change in the energy of
levels without an accompanying change in the population of levels.
SincedE=
P
j
EjdPj+
P
j
PjdEj=qrev+wrev
qrev=
P
j
EjdPj
P
k
PkdEkis the ensemble average of the reversible work done by the
system.
d(f+E) =qrevis derivative of a state function. is an integrating factor ofqrev. According to the second law, integrating factor ofqrevis constant/T.
dS
k
=d(f+E). Thus
S=
E
T
+klnQ+constant=kln
P
j
e
Ej=kT
+
1
T
P
j
Eje
E
j
=kT
P
j
e
E
j
=kT+constant
41/1

Canonical Ensemble: Thermodynamic connection
Partition function in terms of levels:
Q(N;V;T) =
P
E
(N;V;E)e
E(N;V)=kT
By settingE0= 0, limT!0S= limT!0(kln
P
E
(N;V;E)e
E=kT
+
1
T
P
E
(N;V;E)Ee
E=kT
P
E
(N;V;E)e
E=kT) =kln (N;V;E0) which is very small. Thus
S=
E
T
+klnQ.
E=kT
2
(
@lnQ
@T
)N;Vandp=kT(
@lnQ
@V
)N;T S=kT(
@lnQ
@T
)N;V+klnQ A(N;V;T) =ETS=kTlnQ(N;V;T) To derive the second law
consider spontaneous process of going from state 1 to state 2 involves
removal of a constraint or barrier. Thus for an isolated system
2(N;V;E)1(N;V;E).
42/1

Second law
E.g., expansion of an ideal gas where
(E;E) =
1
(N+1)(3N=2)
(
2M
h
2)
3N=2
E
3N=21
V
N
E.
Or addition of a catalyst to a kinetically stable non-equilibrium
mixture.
Q2Q1=
P
E
(2(N;V;E)1(N;V;E))e
E(N;V)=kT
.
Thus, A=kTln(
Q2
Q1
)<0 for an spontaneous process, at constant
volume and temperature.
43/1

Grand Canonical Ensemble
Walls are heat conducting and permeable to the passage of molecules.
Each system is specied by V, T and.
For each value of N the system has energy statesfENj(V)g.aNjis the
number of systems in the ensemble with N particles and in energy
state j.
The number of ways to achieve any distribution,
W(faNjg) =
A!
NjaNj!
.
PNj=
aNj
A
=
P
a
aNjW(faNjg)
A
P
a
W(faNjg)
=
a

Nj
W(faNjg)

AW(faNjg)

Recourse to the maximum term method, W(a) should be maximized
subject to constraints
P
N
P
j
aNj=A,
P
N
P
j
aNjENj=Eand
P
N
P
j
aNjN=N.
a

NJ
=e

e
ENj(V)
e
N
PNj(V; ; ) =
a

Nj
A
=
e
E
Nj
(V)
e
N
P
N
P
j
e
E
Nj
(V)
e
N
(V; ; ) =
P
N
P
j
e
ENj(V)
e
N
44/1

Grand Canonical Ensemble
E(V; ; ) =
1

P
N
P
j
ENj(V)e
ENj(V)
e
N
=(
@ln
@
)V;
p(V; ; ) =
1

P
N
P
j
(
@ENj(V)
@V
)e
ENj(V)
e
N
=
1

(
@ln
@V
);
N(V; ; ) =
1

P
N
P
j
Ne
ENj(V)
e
N
=(
@ln
@
);V
Make the walls impermeable to molecules to derive a collection of
canonical ensembles. For all these ensembles=
1
kT
.
f(; ;fENj(V)g) = ln = ln
P
N
P
j
e
ENj(V)
e
N
df= (
@f
@
)
;fEjgd+ (
@f
@
)
;fEjgd+
P
N
P
j
(
@f
@ENj
)
;;fES6=jgdENj df=EdNd
P
N
P
j
PNjdENj
Ensemble average reversible work is
P
N
P
j
PNjdENj Assuming only PV workdf=EdNd+pdV d(f+E+N) =dE+pdV+dN
45/1

Grand Canonical Ensemble
d(f+E+N)

=d

E+ pdV+

d

N
TdS=dE+pdVdN
=

kT
;S=
E
T

N
T
+kln
Grand canonical partition function
(V;T; ) =
P
N
P
j
e
ENj(V)=kT
e
N=kT
Absolute activity,=e
=kT
.
The relative activity of a species i:ai=e

i

i
RT.
(V;T; ) =
P
1
N=0
Q(N;V;T)
N
. G=N=E+pVTS!S=
E
T
+
pV
T

N
T
. By equating statistical and thermodynamic entropies:
pV=kTln (V;T; ).
In cases where the constant N constraint make calculation of Q
awkward, one resorts to the calculation of .
46/1

Isothermal-Isobaric Ensemble
Walls are exible and heat permeable, while particles cannot pass
them (N,T,p).
Most chemical reactions are performed under these conditions. W(faVjg) =
A!
VjaVj!
shall be maximized constrained by
P
V
P
j
aVj=A,
P
V
P
j
aVjEVj=Eand
P
V
P
j
aVjV=V.
Every partition function can be constructed from (N;V;E)
multiplied by the appropriate exponential and summed over the
quantities which can pass through the walls of each system.
Isothermal-isobaric partition function
=
P
E
P
v
(N;V;E)e
E=kT
e
pV=kT
=
R
dV
V0
e
pV=kT
P
E
(N;V;E)e
E=kT
=
R
dV
V0
e
pV=kT
Q(N;V;T).
PVj=
e
E
j
PV

G(N;T;p) =kTln (N;T;p) dG=SdT+VdP+dN
47/1

Micro-canonical Ensemble
An isolated system, (N;V;E).
The whole of a grand canonical ensemble is a microcanonical system.
SGC= k(E+N+ ln )
=kln +k(
P
N;j
EN;j
e
E
Nj
(V)
e
N

+
P
N;j
N
e
E
Nj
(V)
e
N

)
= kln +k
P
N;j
(EN;j+N)
e
E
Nj
(V)
e
N

UsingPNj(V; ; ) =
a

Nj
A
=
e
E
Nj
(V)
e
N
P
N
P
j
e
E
Nj
(V)
e
N
SGC=kln k
P
N;j
(lna

Nj
+ ln lnA)
a

Nj
A
=

k
A
P
N;j
a

N;j
lna

N;j
+klnA
SMC=AS=klnW(a

Nj
)
48/1

Micro-canonical Ensemble
S=kln (N;V;E) for an spontaneous process
S=kln(2(N;V;E)=1(N;V;E))
Second law for a system at constant volume and energy. For an ideal gasS=Nkln[(
2mkT
h
2)
3=2Ve
5=2
N
] SincedS=
1
T
dE+
p
T
dV

T
dNwe have
p
T
= (
@S
@V
)N;E
pV=NkT, thusk=R=NA.
49/1

Fluctuation Theory: Canonical ensemble
Deviation of a mechanical property from its mean is called uctuation,
investigation of such deviations is called uctuation theory.
In the thermodynamic limit the possibility of observing any value
other than the mean value is extremely remote.
All ensembles are equivalent for all practical purposes.
2
E
=(EE)
2
=E
2
E
2
=
P
j
E
2
j
PjE
2
P
j
E
2
j
Pj=
1
Q
P
j
E
2
j
e
Ej=
1
Q
@
@
P
j
Eje
Ej=
1
Q
@
@
(EQ) =

@

E
@
E
@lnQ
@
=kT
2@

E
@T
+E
2

2
E
=kT
2
(
@E
@T
)N;V=kT
2
Cv
E

E
=
(kT
2
Cv)
1=2

E
For an ideal gas
E

E
=O(N
1=2
) thusEE

50/1

Fluctuation Theory: Canonical ensemble
P(E) =c(E)e
E=kT (
@lnP
@E
)E=E
= (
@ln
@E
)E=E
= 0 (
@
2
lnP
@E
2) = (
@
2
ln
@E
2) atE=E

;(
@
2
ln
@E
2) = (
@
@E
) =
1
kT
2
Cv
lnP(E) = lnP(E)
(E

E)
2
2kT
2
Cv
+ P(E) =P(E) exp [
(E

E)
2
2kT
2
Cv
]
2
p=p
2
p
2
p
2
=
P
j
p
2
j
e
E
j
Q
=
1
Q
P
j
(
@Ej
@V
)
2
e
Ej=
1
Q
[
1

@
@V
P
j
((
@Ej
@V
)e
Ej) +
1

P
j
(
@
2
Ej
@V
2)e
Ej] =
kT
Q
(
@
@V
(pQ)(
P
j
(
@pj
@V
)e
Ej=kT
)) =kT(
@p
@V
+ p
@lnQ
@V

@p
@V
)

2
p=kT(
@p
@V

@p
@V
)
Calculate
p
p
for an ideal gas?
51/1

Fluctuation Theory: Grand canonical ensemble

2
N
=N
2
N
2
=
P
N;j
N
2
PNjN
2
P
N;j
N
2
PNj=
1

P
N;j
N
2
e
ENje
N
=
1

@
@
P
N;j
Ne
ENje
N
=
1

@
@
(N) =
@

N
@
N
@ln
@
=kT(
@

N
@
) +N
2

2
N
=kT(
@

N
@
)
A is homogeneous of degree one,A(T; V; N) =A(T;V;N) According to Euler's theorem,N(
@A
@N
) +V(
@A
@V
) =A(1) Further,N
2
(
@
2
A
@N
2) + 2NV(
@
2
A
@N@V
) +V
2
(
@
2
A
@V
2) = 0 (2) (1)!NV(
@
2
A
@N@V
) +V
2
(
@
2
A
@V
2) = 0 (1)!N
2
(
@
2
A
@N
2) +NV(
@
2
A
@N@V
) = 0 N
2
(
@
2
A
@N
2) =V
2
(
@
2
A
@V
2) V
2
(
@p
@V
)T;N=N
2
(
@
@N
)T;V
52/1

Fluctuation Theory: Grand canonical ensemble
(
@
@N
)V;T=
V
2
N
2(
@p
@V
)N;T. Thus
2
N
=

N
2
kT
V
.
Isothermal compressibility,=
1
V
(
@V
@p
)N;T
N

N
= (
kT
V
)
1=2
Typical uctuations in statistical thermodynamics areO(N
1=2
).

=
N
N
= (
kT
V
)
1=2
P(N) =CQ(N;V;T)e
N
(
@lnP
@N
)N=N
= (
@lnQ
@N
)N=N
+= 0 (
@
2
lnP
@N
2)N=N
= (
@
2
lnQ
@N
2)N=N
=(
@
@N
)N=N
=
1
kT(@

N=@)V;T
P(N) =P(N) exp[
(N

N)
2
2kT(@N=@)V;T
]
53/1

Equivalence of various ensembles
Q(N;V;T) =
P
E
(N;V;E)e
E=kT
= (N;V;E)e
E=kT
.
A=kTlnQ=EkTln (N;V;E). ThusS=kln (N;V;E). =
P
N
Q(N;V;T)e
N
=
P
E;N
(N;V;E)e
E
e
N
=Q(N;V;T)e
N
= (N;V;E)e
E
e
N
kTln =kTlnQ+N=kTln E+N kTln =A+G=TSE+G=pV
54/1

Ideal gas
Q(N;V;T) =
P

e
E=kT
=
P
i;j;k;
e
(
a
i
+
b
j
+
c
k
+)=kT
=
P
i
e

a
i
=kT
P
j
e

b
j
=kT
P
k
e

c
k
=kT
=qaqbqc where molecular
partition functionq(V;T) =
P
i
e
i=kT
For N distinguishable identical particlesQ(N;V;T) = [q(v;T)]
N
,
e.g., sites in a solid crystal are distinguishable.
Partitioning of the molecular Hamiltonian,
H=Htrans+Hrota+Hvibr+Helect, lead to the division of molecular
partition function.qmolecula=qtranslationalqrotationalqvibrationqelectronic
The unrestricted sum for partition function is not valid for fermions as
it allows repeated indices, i.e., more than one particle in the same
energy state.
The unrestricted sum for partition function is not valid for Bosons as
it over-counts states with repeated indices, i.e., states where more
than one particle is in the same energy state.
55/1

Ideal gas
Number of quantum states with energy,
() =

6
(
8ma
2

h
2)
3=2
=

6
(
8m
h
2)
3=2
V
If the number of available molecular states is much greater than the
number of molecules in the system, two molecules in the same state is
a rare event.
Thus the only problem withQ(N;V;T) =q
N
is indistinguishability of
molecules, partition function can be corrected as:
Q(N;V;T) =q
N
=N!.
Valid when ()N!

6
(
12mkT
h
2)
3=2

N
V
Boltzmann statistics or classical limit become a better approximation
by increasing mass or temperature or decreasing density.
E=kT
2
(
@lnQ
@T
)N;V=NkT
2
(
@lnq
@T
) =N
P
j
j
e

j
=kT
q
=N =
P
j
j
e

j
=kT
q
Probability of a molecule being in the j'th energy statej=
e

j
=kT
q56/1

Ideal gas
For a non-interacting gas one can consider the container to be an
ensemble and each molecule to be a system.
Label each molecular state with 1;2;3; such thatii+1. Consider N as the total number of molecules andnias the number of
molecules in the state i.
Derive probability of state occupation similar to the last section by
maximizing ways of distributing molecules over energy states
W(fnjg) =
N!
ini!
subject to
P
i
ni=Nand
P
i
nii=E.
i=
ni
N
=
n

i
N
=
e

i
=kT
q
in the Boltzmann statistics.
Exact partition function for fermions is
Q(N;V;T) =
P

i6=j6=k6=
e
(
a
i
+
b
j
+
c
k
+)=kT
where * denotes that
i+j+ =E.
For bosonsQ(N;V;T) =
P

i;j;k;
e
(
a
i
+
b
j
+
c
k
+)=kT
where * also
denotes that any term with repeated indices is counted just once.
57/1

Fermi-Dirac and Bose-Einstein statistics
Ej=
P
k
knkandN=
P
k
nk.
Q(N;V;T) =
P
j
e
Ej=
P

fnkg
e

P
i
iniwhere * signies
P
k
nk=N.
Due to diculty of performing this sum, we turn to the grand
canonical partition function:
(V;T; ) =
P
1
N=0
e
N
Q(N;V;T) =
P
1
N=0

N
P

fnkg
e

P
i
ini=
P
1
N=0
P

fnkg

P
nie

P
j
jnj
=
P
1
N=0
P

fnkg

1
k=1
(e
k)
nk=
Pn
max
1
n1=0
Pn
max
2
n2=0

1
k=1
(e
k)
nk
(V;T; ) =
Pn
max
1
n1=0
(e
1
)
n1
Pn
max
2
n2=0
(e
2
)
n2
=

1
k=1
Pn
max
k
nk=0
(e
k)
nk
FD=
1
k=1
(1 +e
k) BE=
1
k=1
(
P
1
nk=0
(e
k)
nk) =
1
k=1
((1e
k)
1
); e
k<
1;e
(k)
<1;(k)<0! < 0.
58/1

Fermi-Dirac and Bose-Einstein statistics

FD
BE
=
1
k=1
(1e
k)
1
N=N=kT(
@ln
@
)V;T=(
@ln
@
)V;T=
P
k
e

k
1e

k
=
P
k
nk nk=
e

k
1e

k
E=
P
k
nkk=
P
k
ke

k
1e

k
pV=kTln =kT
P
k
ln[1e
k] In quantum statistics even non-interacting particles are not
independent (through symmetry requirement), thus molecular
partition function q is irrelevant.
In the limit of classical statistics nk!0. This requires!0
meaningN=V!0 for constant T orT! 1for constant N/V.
59/1

Fermi-Dirac and Bose-Einstein statistics
For smallnk=e
k!

N=q
nk

N
=
e

k
q
E!
P
j
je
j
=
E

N
=
P
j
je

j
P
j
e

j
pV=kT
P
k
ln[1e
k] = (kT)(
P
j
e
j) =kTq pV= ln =q=N
=e
q
=
P
1
N=0
(q)
N
N!
thusQ(N;V;T) =
q
N
N!
.
Thus Boltzmann statistics is valid in the limit of small.
60/1

Ideal monatomic gas
q(V;T) =qtransqelectqnucl
Translational energy
nx;ny;nz=
h
2
8ma
2(n
2
x+n
2
y+n
2
z)nx;ny;nz= 1;2;3;
qtrans=
P
1
nx;ny;nz=1
e
nx;ny;nz=
P
1
nx;ny;nz=1
e

h
2
8ma
2
(n
2
x
+n
2
y
+n
2
z
)
=
P
1
nx=1
e

h
2
n
2
x
8ma
2
P
1
ny=1
e

h
2
n
2
y
8ma
2
P
1
nz=1
e

h
2
n
2
z
8ma
2
= (
P
1
n=1
e

h
2
n
2
8ma
2
)
3
qtrans(V;T) = (
R
1
0
e

h
2
n
2
8ma
2
dn)
3
= (
2mkT
h
2)
3=2
V
qtrans=
R
1
0
!()e

dwhere!()d=

4
(
8ma
2
h
2)
3=2

1=2
d qtrans=

4
(
8ma
2
h
2)
3=2
R
1
0

1=2
e

d= (
2mkT
h
2)
3=2
V=
V

3 trans=kT
2
(
@lnqtrans
@T
) =
3
2
kT=
p
2
2m
is the thermal De Broglie wavelength of the particle.
61/1

Ideal monatomic gas
Condition for applicability of classical or Boltzmann statistics

3
=V1
qelect=
P
i
!eie
i=!e1+!e2e
e12
+
62/1

Ideal monatomic gas thermodynamic functions
k=0.695 cm
1
/deg-molecule, 1 eV=8065.73 cm
1
.
Nuclear levels are separated by millions of eV's. Nuclear states do not
contribute in thermodynamic change.
qnucl=
P
i
!nie
i=!n1+!n2e
n12
+ A=kTlnQ=NkTln[(
2mkT
h
2)
3=2Ve
N
]NkTln(!e1+!e2e
12
) E=kT
2
(
@lnQ
@T
)N;V=
3
2
NkT+
N!e2e12e

e12
qelect
p=kT(
@lnQ
@V
)N;T=
NkT
V
S=
3
2
Nk+Nkln[(
2mkT
h
2)
3=2Ve
N
] +Nkln(!e1+!e2e
12
) +
Nk!e212e

12
qelect
=Nkln[(
2mkT
h
2)
3=2Ve
5=2
N
] +Select, which is the
Sackur-Tetrode equation.
(T;p) =kT(
@lnQ
@N
)V;T=kTln
q
N
=kTln[(
2mkT
h
2)
3=2V
N
]
kTlnqeqn=kTln[(
2mkT
h
2)
3=2kT
p
]kTlnqeqn=
kTln[(
2mkT
h
2)
3=2
kT]kTlnqeqn+kTlnp=0(T) +kTlnp
0(T) =kTln[(
2mkT
h
2)
3=2
kT]kTlnqeqn
63/1

Atomic term symbols
Term symbol is an abbreviated description of the angular momentum
quantum numbers in a multi-electron atom.
Each energy level of an atom with a given electron conguration is
described by not only the electron conguration but also its own term
symbol, as the energy level also depends on the total angular
momentum including spin.
The usual atomic term symbols assume LS coupling (also known as
Russell-Saunders coupling or spin-orbit coupling). The ground state
term symbol is predicted by Hund's rules.
Hund's rule of maximum multiplicity states that the electron
conguration maximizing spin multiplicity is more stable.
Multielectron atomic Hamiltonian,
^H=
~
2
2m
P
j
r
2
j

P
j
Ze
2
rj
+
P
i<j
e
2
rij
+
P
j
(rj)ljsj=H0+Hee+Hso
Hsorepresent the interaction between the magnetic moment
associated with an electrons spin with the magnetic eld generated by
its own orbital motion.
64/1

Atomic term symbols
Russell-Saunders or L-S coupling: ForZ<40;^Hsocan be treated as
a small perturbation. Then [^H0+^Hee;^L] = [^H0+^Hee;^S] = 0. So S
and L are good quantum numbers.
^
L
2
=L(L+ 1)~
2

^
S
2
=S(S+ 1)~
2
L and S are the vector sums ofljandsjrespectively. For electrons,
L=l1+l2;l1+l21; ;jl1l2jand
S=s1+s2;s1+s21; ;js1s2j
Terms withL= 0;1;2; are denoted by S, P, D, ... When^Hsois taken into account only [^H;^J] = 0. Only^J=^L+^Sis
conserved. Eigenvalues of^J
2
= (^L+^S)
2
areJ(J+ 1)~
2
with
!j= 2J+ 1 corresponding to 2J+1 eigenvalues of^Jzwhich are
J~;(J1)~; ;J~
Allowed values of J areL+S;L+S1; ;jLSj. Term symbol is
written as
2S+1
LJ.65/1

Atomic term symbols
For heavier atoms j-j coupling is used. Total angular momenta for
each electron is dened asji=si+li. Total angular momentum, J, is
derived by coupling the j's. Term symbol is written as
2S+1
LJ.
For molecules, Greek letters are used to designate the component of
orbital angular momenta along the molecular axis.
For a given electron conguration: The combination of an S value and an L value is called a term, and
has a statistical weight (i.e., number of possible microstates) equal to
(2S+1)(2L+1);
A combination of S, L and J is called a level. A given level has a
statistical weight of (2J+1), which is the number of microstates
associated with this level in the corresponding term;
A combination of S, L, J and MJdetermines a single state.
66/1

Atomic term symbols
The product (2S+1)(2L+1) as the number of possible microstates
jS;mS;L;mLiwith given S and L is also the number of basis states in
the uncoupled representation, where S, mS, L, mLare good quantum
numbers whose corresponding operators mutually commute.
With given S and L, the eigenstatesjS;mS;L;mLiin this
representation span function space of dimension (2S+1)(2L+1).
In the coupled representation where total angular momentum is
treated, the associated microstates arejJ;MJ;S;Liand these states
span the function space with dimension of
Jmax=L+S
X
J=Jmin=jLSj
(2J+ 1) as
MJ=J;J1; :::J+ 1;J.
67/1

Adiabatic approximation
Adiabatic, Born-Oppenheimer approximation.
Adiabatic means not passing through and in thermodynamics refers to
a condition imposed on a system that prevents any passage of heat
into or out of the system.
In quantum dynamics, adiabatic refers to an inherent property of a
process, i.e., its tendency to occur without any change in quantum
state.
Ehrenfest showed that when the parameters of a system in a particular
quantum state are changed slowly, the system remains in the same
quantum state (adiabatic theorem).
68/1

Adiabatic approximation
Now, adiabatic implies that there are two sets of variables which
describe the system of interest and the system can be characterized by
the eigenstates dened at each xed value of one set of variables,
which change slowly compared to the other set.
In the rst step of the adiabatic approximation, the electronic
Schrodinger equation is solved, yielding the wave-function electronic
depending on electrons only.
During this solution the nuclei are xed in a certain conguration,
very often the equilibrium conguration.
In the second step of the BO approximation this function serves as a
potential in a Schrodinger equation containing only the nuclei.
The success of the BO approximation is due to the dierence between
nuclear and electronic masses.
69/1

Adiabatic approximation
Molecular Hamiltonian,
^H=
P
Nn
I=1
^P
2
I
2MI
+
P
Ne
i=1
^p
2
i
2me
+
P
i<j
e
2
j~ri~rjj
+
P
I<J
ZIZJe
2
j
~
RI
~
RJj

P
i;I
ZIe
2
j
~
RI~rij
=
^Tn+^Te+Ve+Vn+Vne
h
^Tn+^Te+Ve+Vn+Ven
i
(r;R) =E (r;R)
(r;R) =(r;R)(R); where(R) is a nuclear wave function and(r;R) is an electronic
wave function that depends parametrically on the nuclear positions.
The dierence in the nuclear and electronic mass also results in a
dierence in their momenta, i.e., nuclear momenta are greater, which
in turn causes the nuclear wave function (coordinate amplitude) to
change more steeply than the electronic wave function.
rI(R) rI(r;R)
70/1

Adiabatic approximation
^
Tn

(r;R)(R)

=
~
2
2
P
I
1
MI

(r;R)r
2
I
(R) + 2rI(r;R)
rI(R) +(R)r
2
I
(r;R)

~
2
2
P
I
1
MI
(r;R)r
2
I
(R)
h
^Tn+^Te+Ve+Vn+Ven
i
(r;R)(R) =E(r;R)(R)
[
^
Te+Ve(r)+Ven(R;r)](r;R)
(r;R)
=E
[
^
Tn+Vn(R)](R)
(R)
.
h
^Te+Ve(r) +Ven(r;R)
i
(r;R) ="(R)(r;R)
h
^Tn+Vn(R) +"i(R)
i
(R) =E(R).
The physical interpretation is that the electrons respond
instantaneously to the nuclear motion, therefore, it is sucient to
obtain a set of instantaneous electronic eigenvalues and eigenfunctions
at each nuclear conguration,R(hence the parametric dependence of
i(r;R) and"i(R) onR)
71/1

Distance dependent potential
h
^Tn+Vn(R) +"i(R)
i
(R;t) =i~
@
@t
(R;t).
"i(R) give rise to BO hypersurfaces. This equation describes the nuclear dynamics and vibronic states. Breaking down the motion of nuclei: In 2D consider two particles with a distance dependent potential.
E=
m1
2
( _x
2
1
+ _y
2
1
) +
m2
2
( _x
2
2
+ _y
2
2
) +U(x1x2;y1y2)
Center of mass and relative coordinates are dened as
X=
m1x1+m2x2
m1+m2
Y=
m1y1+m2y2
m1+m2
x12=x1x2y12=y1y2
x1=X+
m2
m1+m2
x12y1=Y+
m2
m1+m2
y12 x2=
X
m1
m1+m2
x12y2=Y
m1
m1+m2
y12
E=
m1+m2
2
(_X
2
+_Y
2
) +
m1m2
2(m1+m2)
( _x
2
12
+ _y
2
12
) +U(x12;y12) =
M
2
(_X
2
+_Y
2
) +

2
( _x
2
12
+ _y
2
12
) +U(x12;y12) =Ecm+Erel
72/1

Distance dependent potential
Thus the center of mass motion can be separated from relative motion
of a two particle system.
Mapping the relative motion into polar coordinates r andwhere
x12=rcosy12=rsin_x12= _rcosr_sin_y12=
_rsin+r
_
cos:
Erel=

2
( _r
2
+r
2_
2
) +U(r) =

2
_r
2
+U(r) +

2
r
2_
2
=Evib+Erot
Similarly in 3D consider two particles with a distance dependent
potential.
E=
m1
2
( _x
2
1
+ _y
2
1
+ _z
2
1
) +
m2
2
( _x
2
2
+ _y
2
2
+ _z
2
2
) +U(x1x2;y1y2;z1z2)
Center of mass and relative coordinates are dened as
X=
m1x1+m2x2
m1+m2
Y=
m1y1+m2y2
m1+m2
Z=
m1z1+m2z2
m1+m2
x12=
x1x2y12=y1y2z12=z1z2
x1=X+
m2
m1+m2
x12y1=Y+
m2
m1+m2
y12 z1=
Z+
m2
m1+m2
z12 x2=X
m1
m1+m2
x12y2=Y
m1
m1+m2
y12z2=
Z
m1
m1+m2
z12
73/1

Distance dependent potential
m1_x
2
1
+m2_x
2
2
2
=
(m1+m2)
2
_
X
2
+
m1m2
m1+m2
_x
2
12
E=
M
2
(
_
X
2
+
_
Y
2
+
_
Z
2
)+

2
( _x
2
12
+ _y
2
12
+ _z
2
12
)+U(x12;y12;z12) =Ecm+Erel
Finally transform the relative coordinates into spherical polar
coordinates.
_x12= _rsincos+r_coscosr_sinsin _y12=
_rsinsin+r_cossin+r_sincos _z12= _rcosr_sin
_x
2
12
+ _y
2
12
+ _z
2
12
= _r
2
+r
2_
2
+r
2_
2
sin
2

74/1

Diatomic molecule
Erel=

2
( _r
2
+r
2_
2
+r
2_
2
sin
2
) +U(r)
Evib=

2
_r
2
+U(r)Erot=

2
(r
2_

2
+r
2_

2
sin
2
) Hn=Htrans+HInt,n=trans+int,qn=qtransqint qtrans= [
2MkT
h
2]
3=2
V,Q(N;V;T) =
q
N
trans
q
N
int
N!
Relative motion of the two nuclei consists of rotary motion about the
center of mass and vibratory motion about the equilibrium
internuclear distancere.
Small amplitude of the vibratory motion allows treatment of the
rotary motion as the rotation of a rigid dumbbell.
U(r) =U(re) + (rre)(
dU
dr
)r=re+
1
2
(rre)
2
(
d
2
U
dr
2)r=re+ =
u(re) +
1
2
k(rre)
2
+
Rigid rotor-Harmonic oscillator approximation:Hrot;vib=Hrot+Hvib,
rot;vib=rot+vib,qrot;vib=qrotqvib75/1

Diatomic molecule
For a rigid rotor,
J=
~
2
J(J+1)
2I
J= 0;1;2; !J= 2J+ 1 I=r
2
e
For a harmonic oscillator
vib=h(n+ 1=2)n= 0;1;2; !n= 1 =
1
2
(
k

)
1=2
Selection rule for radiation induced rotational transition: 1) Possession
of permanent dipole moment. 2) J=1
=
j+1j
h
=
h
4
2
I
(J+ 1)J= 0;1;2; B=
h
8
2
Ic
, J(cm
1
) =BJ(J+ 1) Selection rule for radiation induced vibrational transition: 1) Change
of dipole moment by the respective vibration. 2) n=1
=
j+1j
h
=
1
2
(
k

)
1=2 Assume:H=Htrans+Hrot+Hvib+Helec+Hnucl; thus
=trans+rot+vib+elec+nucl.
76/1

Diatomic molecule: vibrational partition function
q=qtransqrotqvibqelecqnucl,Q(N;V;T) =
(qtransqrotqvibqelecqnucl)
N
N!
Zero of rotational energy taken as the energy ofJ= 0 state. Zero of vibrational energy taken as the bottom of the internuclear
potential well of ground electronic state.
Zero of the electronic energy is separated ground state atoms. qelec= e1e
De=kT
+ e2e
2=kT
+ qvib(T) =
P
1
n=0
e
n
=e
h=2
P
1
n=0
e
hn
=
e
h=2
1e
h
IfkTh qvib(T) =e
h=2
R
1
0
e
hn
dn=
kT
h
e
h=2
E=NkT
2dlnqv
dT
=Nk(
v
2
+
v
e
v=T
1
) in terms of vibrational
temperature.
CV;vib= (
@Ev
@T
)N=Nk(

T
)
2e
v=T
(e
v=T
1)
2
77/1

Diatomic molecule: vibrational partition function
fn=
e
h(n+1=2)
qvib
= (1e
h
)e
hn qrot(T) =
P
1
J=0
(2J+ 1)e

BJ(J+1) Characteristic temperature of rotationr=B=k
78/1

Diatomic molecule: rotational partition function
qrot(T) =
R
1
0
(2J+ 1)e
rJ(J+1)=T
dJ=
R
1
0
e
rJ(J+1)=T
dJ(J+ 1) =
T
r
=
8
2
IkT
h
2 rT
At small temperaturesqrot(T) = 1 + 3e
2r=T
+ 5e
6r=T
+ For intermediate temperatures Euler-MacLaurin summation formula:
P
b
n=a
f(n) =
R
b
a
f(n)dn+
1
2
ff(b) +f(a)g+
P
1
j=1
(1)
jBj
(2j)!
ff
(2j1)
(a)f
(2j1)
(b)g
The formula was discovered independently by Leonhard Euler and
Colin Maclaurin around 1735 (and later generalized as Darboux's
formula). Euler needed it to compute slowly converging innite series
while Maclaurin used it to calculate integrals.
Bernoulli numbers,B1= 1=6;B2= 1=30;B3= 1=42; Use to calculate
P
1
j=0
e
j
. Euler-MacLaurin formula is applied toqrot(T) with
f(J) = (2J+ 1)e

r
T
J(J+1)
;a= 0;b=1
79/1

Diatomic molecule: rotational partition function
qrot(T) =
T
r
f1 +
1
3
(
r
T
) +
1
15
(
r
T
)
2
+
4
315
(
r
T
)
3
+ g
lnqrot(T) = ln
T
r
+ lnf1 +
1
3
(
r
T
) +
1
15
(
r
T
)
2
+
4
315
(
r
T
)
3
+ g Erot=NkT
2
(
@lnqrot
@T
) =NkT+
Nj
N
=
(2J+1)e
rJ(J+1)=T
qrot(T)
.
Jmax= (
kT
2

B
)
1=2
1=2(
T
2r
)
1=2
. If the nuclei have integer spin they are Bosons and the molecular
wavefunction must be symmetric with respect to interchange of the
two nuclei.
If the nuclei have half odd integer spin they are Fermions and the
molecular wavefunction must be antisymmetric with respect to
interchange of the two nuclei.
For a homonuclear diatomic molecule symmetry requirement most be
considered.
If temperature is fairly large
qrot(T) =
T
r
f1 +
1
3
(
r
T
) +
1
15
(
r
T
)
2
+
4
315
(
r
T
)
3
+ g
80/1

Diatomic molecule: rotational partition function
Symmetry numberis the number of indistinguishable orientations of
a molecule.
Exclusive of the nuclear part
0
total
= trans rot vib elec
Consider the interchange of nuclei as rst inverting the molecule
followed by an inversion of only electrons.
Translational partition function is unaected by inversion. Vibrational partition function is unaected by inversion. Most molecules electronic ground state is
+
gwhich is symmetric
under both inverting the molecule and an inversion of only electrons.
Thus rotational wavefunction controls the symmetry of
0
total
Rigid rotor wavefunctions are the same functions as the angular
functions of the hydrogen atom.
81/1

Spherical harmonics
l= 0;ml= 0!Y0;0=
1
p
4
l= 1;ml= 1;0;1!
8
>
<
>
:
YPx
= (
3
4
)
1=2
sincos
YPy
= (
3
4
)
1=2
sinsin
YPz
= (
3
4
)
1=2
cos l= 2;ml= 2;1;0;1;2!
8
>
>
>
>
>
>
<
>
>
>
>
>
>
:
Yd
z
2
= (
5
16
)
1=2
(3 cos
2
1)
Ydxz
= (
15
4
)
1=2
(sincoscos)
Ydyz
= (
15
4
)
1=2
(sincossin)
Ydxy
= (
15
16
)
1=2
(sinsin 2)
Yd
x
2
y
2
= (
15
16
)
1=2
(sincos 2)
82/1

Diatomic molecule: rotational partition function
Eect of inversion on the orientation of the diatomic molecule is
(; )!(; +)

0
total
remains unchanged for even J and changes sign for odd J. Symmetric spin wavefunctions for spin-1/2 nuclei are,and
2
1=2
(+)
Antisymmetric spin wavefunctions for spin-1/2 nuclei is
2
1=2
()
Odd J levels have a statistical weight of 3 compared to a statistical
weight of 1 for even J levels.
Nuclei of spin I has 2I+1 spin states with eigenfunctions
1; 2; ; 2I+1
There are (2I+ 1)
2
nuclear wavefunctions. (2I+ 1)Iantisymmetric spin functions are
i(1)j(2)i(2)j(1);1i;j2I+ 1
Remaining (2I+ 1)
2
(2I+ 1)I= (2I+ 1)(I+ 1) are symmetric
nuclear functions.
83/1

Diatomic molecule: rotational partition function
For
+
gstates and integral spin
(2I+1)I antisymmetric nuclear spin functions couple with odd J (2I+1)(I+1) symmetric nuclear spin functions couple with even J For
+
gstates and half integral spin (2I+1)I antisymmetric nuclear spin functions couple with even J (2I+1)(I+1) symmetric nuclear spin functions couple with odd J These results apply to polyatomic linear molecules as well. Homonuclear diatomic molecule with integral spin:
qrot;nucl(T) = (I+ 1)(2I+ 1)
P
J even
(2J+ 1)e
rJ(J+1)=T
+I(2I+
1)
P
J odd
(2J+ 1)e
rJ(J+1)=T
Homonuclear diatomic molecule with half integer spin:
qrot;nucl(T) =I(2I+ 1)
P
J even
(2J+ 1)e
rJ(J+1)=T
+ (I+ 1)(2I+
1)
P
J odd
(2J+ 1)e
rJ(J+1)=T
84/1

Diatomic molecule: rotational partition function
If rT!
P
J odd
(2J+ 1)e
rJ(J+1)=T

P
J even
(2J+ 1)e
rJ(J+1)=T
1=2
P
J
(2J+ 1)e
rJ(J+1)=T

1=2
R
1
0
(2J+ 1)e
rJ(J+1)=T
dJ=
T
2r
qrot;nucl(T) =
(2I+1)
2
T
2r
=qrot(T)qnuclwhereqrot(T) =
T
2r
and
qnucl= (2I+ 1)
2
qrot(T)
8
2
IkT
h
2
1

P
1
J=0
(2J+ 1)e
BJ(J+1)
rT Spin isomers of hydrogen Hydrogen with opposite nuclear spins, singlet hydrogen, is called
para-Hydrogen.
Hydrogen with parallel nuclear spins, triplet, is called ortho-Hydrogen. The para form is more stable than the ortho form by 1.06 kJ/mol. ForH2qnuc,rot=
P
J even
(2J+ 1)e
rJ(J+1)=T
+ 3
P
J odd
(2J+ 1)e
rJ(J+1)=T
85/1

Diatomic molecule: rotational partition function
ForH2qnuc,rot=
P
J even
(2J+ 1)e
rJ(J+1)=T
+
e
1060J=molk=RT
3
P
J odd
(2J+ 1)e
rJ(J+1)=T
The conversion between ortho and para hydrogen in the absence of a
catalyst (e.g., Fe
III
or activated charcoal) is very slow.
Northo
Npara
=e
1060J=molk=RT3
P
J odd
(2J+1)e
rJ(J+1)=T
P
J even
(2J+1)e
rJ(J+1)=T Figure:
Wikipedia
86/1

Diatomic molecule: Thermodynamic functions
Predict theH2heat capacity at very low temperatures.
Try the same analysis onD2. Erot=NkT[1
r
3T

1
45
(
r
T
)
2
+ ] CV;rot=Nk[1 +
1
45
(
r
T
)
2
+ ] Srot=kT(
@lnqrot
@T
)N;V+klnqrot=Nk[1ln(
r
T
)
1
90
(
r
T
)
2
+ ] Harmonic oscillator-rigid rotor approximation:
q(T) = (
2MkT
h
2)
3=2
V
8
2
IkT
h
2e
h=2
(1e
h
)
1
!e1e
De=kT
lnq(T) =
3
2
ln(
2MkT
h
2) + lnV+ ln
8
2
IkT
h
2
h
2kT
ln(1e
h
) + ln!e1+De=kT
E
NkT
=
5
2
+
h
2kT
+
h=kT
e
h=kT
1

De
kT
Cv
Nk
=
5
2
+ (
h
kT
)
2e
h=kT
(e
h=kT
1)
2
S
Nk
= ln[
2MkT
h
2]
3=2Ve
5=2
N
+ln
8
2
IkTe
h
2+
h
kT
e
h=kT
1
ln(1e
h=kT
)+ln!e
87/1

Diatomic molecule: Thermodynamic functions
p=NkT(
@lnq
@V
)!pV=NkT
= (
@A
@N
)T;V=kT(
@lnQ
@N
)T;V=kTln
q
N
!

0
(T)
kT
=
ln[
2MkT
h
2]
3=2
ln
kT
P
0ln
8
2
IkT
h
2+
h
2kT
+ ln(1e
h=kT
)
De
kT
Centrifugal distortion eects, anharmonic eects can be included. state has zero total angular momentum. In other cases electronic
and rotational angular momentum must be coupled. I.e., the
electronic and rotational partition functions do not separate.
AtTrelectronic and rotational partition functions separate.
88/1

Classical statistical mechanics
q=
P
e
(energy)
!qclass
R

R
e
H(p;q)
dpdq
Available (eective) number of states vs. available (eective) volume
of phase space.
For a monatomic ideal gasH=
1
2m
(p
2
x+p
2
y+p
2
z). Thus
qclass
R

R
e

(p
2
x
+p
2
y
+p
2
z
)
2mdpxdqxdpydqydpzdqz=
V[
R
1
0
e
p
2
=2m
dp]
3
= (2mkT)
3=2
V
Compare withqtrans(V;T) = (
2mkT
h
2)
3=2
V
For a linear rigid rotorH=
1
2I
(p
2

+
p
2

sin
2

)
qrot
R
1
1
dp
R
1
1
dp
R
2
0
d
R

0
de

2I
(p
2

+
p
2

sin
2

)
= 8
2
IkT
Compare withqrot(T) =
T
r
=
8
2
IkT
h
2
89/1

Classical statistical mechanics
For a harmonic oscillatorH=
p
2
2
+
k
2
x
2
qvib
R
1
1
dp
R
1
1
dxe
(
p
2
2
+
k
2
x
2
)
=
kT

where=
1
2
(
k

)
1=2
Compare with the high temperature limit
qvib(T) =e
h=2
R
1
0
e
hn
dn=
kT
h
Conjecture:q=
1
h
s
R

R
e
H

s
j=1
dpjdqj A weighted area of h in phase space is equivalent to one weighted
quantum mechanical state.
This seem to be intimately related to space-momentum uncertainty
relation, xp
~
2
. And to the Wigner transform
Aw(Q;P) =
Z
dZhQ
Z
2
j^AjQ+
Z
2
ie
iPZ=~
.
90/1

Classical statistical mechanics
Q=
q
N
N!
=
1
N!

N
j=1
1
h
s
R

R
e
Hj
s
i=1
dpjidqji
Q=
1
N!h
sN
R

R
e

P
j
Hj

sN
i=1
dpidqi=
1
N!h
sN
R

R
e
H

sN
i=1
dpidqi
Conjecture:Q=
1
N!h
sN
R

R
e
H(p;q)
dpdq For monatomic gas
H(p;q) =
1
2m
P
N
j=1
(p
2
xj
+p
2
yj
+p
2
zj
) +U(x1;y1;z1; ;xN;yN;zN)
ThusQclas=
1
N!
(
2mkT
h
2)
3N=2
ZNwhere the classical conguration
integralZN=
R
V
e
U(x1;;zN)=kT
dx1 dzN
For a moleculeH=Hclass+Hquant!q=qclassqquantwhere
qclass=
1
h
s
R
e
Hclass(p;q)=kT
dp1dq1 dpsdqs
For entire systemH=Hclass+Hquant!Q=QclassQquant=
Qquant
h
sN
R
e
Hclass(p;q)=kT
dpclassdqclass
91/1

Phase space and the Liouville equation
Each system is represented by a phase point in the phase space and a
microcanonical ensemble is represented by a cloud of phase points in
the phase space.
Each phase point evolves according to
_qj=
@H
@pj
and _pj=
@H
@qj
j= 1;2; ;l=sN
The postulate of equal a priori probabilities states that the density of
points is uniform over the constant energy hyper surface.
Number of systems that have phase point in dpdq about the point p,
q at time t is f(p,q,t)dpdq.
R

R
f(p;q;t)dpdq=A
Ensemble average of a function of coordinate and momenta,

=
1
A
R

R
(p;q)f(p;q;t)dpdq
Gibbs postulate equates

with the corresponding thermodynamic
function.
92/1

Phase space and the Liouville equation
The number of phase points inside the volume element
p1 plq1 qlabout the pointp1; ;pl;q1; ;qlis
N=f(p1; ;pl;q1; ;ql;t)p1 plq1 ql
Remember divergence: The ow going through a dierential volume
per unit time is:
93/1

Phase space and the Liouville equation
Consider a hyper rectangular dierential volumep1 plq1 ql
around the pointp1; ;pl;q1; ;ql
Number of phase points entering through the face perpendicular to
theq1axis and located atq1isf_q1p1 plq2 ql
The number passing through the opposite face is
f(p1; ;pl;q1+q1; ;ql;t) _q1(p1; ;pl;q1+
q1;q2; ;ql)p1 plq2 ql=
(f+
@f
@q1
q1)( _q1+
@_q1
@q1
q1)p1 plq2 ql
net ow in theq1direction(
@f
@q1
_q1+f
@_q1
@q1
)p1 plq1 ql+
94/1

Phase space and the Liouville equation
Similarly, net ow in thep1direction is
(
@f
@p1
_p1+f
@_p1
@p1
)p1 plq1 ql+
Thus total ow of phase points is

P
l
j=1
(
@f
@qj
_qj+f
@_qj
@qj
+
@f
@pj
_pj+f
@_pj
@pj
)p1 plq1 ql=
d(N)
dt
@_qj
@qj
+
@_pj
@pj
= 0
@f
@t
=
P
l
j=1
(
@f
@qj
_qj+
@f
@pj
_pj) =
P
l
j=1
(
@f
@qj
@H
@pj

@f
@pj
@H
@qj
) In terms of Poisson bracket,fA;Bg=
P
l
j=1
(
@A
@qj
@B
@pj

@B
@qj
@A
@pj
),
@f
@t
+ff;Hg= 0
In terms of the Liouville operator,
@f
@t
+iLf= 0. Equivalent to Hamiltonian equations of motion. @f
@t
+
P
N
j=1
~
Pj
mj
rrj
f+
P
N
j=1
~
Fj rpj
f= 0.
95/1

Phase space and the Liouville equation
Liouville equation is equivalent to
df
dt
= 0.
Principle of the conservation of density in phase space: density in the
neighborhood of any moving phase point is a constant along the
trajectory of that point. Implies thatf(p;q;t) =f(p0;q0;t0).
p=p(p0;q0;t) andq=q(p0;q0;t). The existence and uniqueness theorem implies that no two trajectories
can pass through the same point, i.e. they never cross.
Conservation of extension in phase space:pq=p0q0or Jacobian
of (p,q) to (p0;q0) is unity.
In general if we are given two sets of momenta and their conjugate
coordinatedq1dq2 dq3ndp1 dp3n=dQ1 dQ3ndP1 dP3n
e.g.,dpxdpydpzdxdydz=dprdpdpdrdd. This is a reason for
using momenta and not velocities for describing classical systems.
96/1

Equipartition of energy
=
R

R
He
H
dq1dqsdp1dps
R

R
e
H
dq1dqsdp1dps
H(p1;p2; ;qs) =
P
m
j=1
ajp
2
j
+
P
n
j=1
bjq
2
j
+H(pm+1; ;ps;qn+1; ;qs)
=
R

R
(
P
m
j=1
ajp
2
j
+
P
n
j=1
bjq
2
j
+H(pm+1;;ps;qn+1;;qs)e
H
dq1dqsdp1dps
R

R
e
H
dq1dqsdp1dps
=
P
m
j=1
aj
R

R
p
2
j
e
H
dq1dqsdp1dps+
P
n
j=1
bj
R

R
q
2
j
e
H
dq1dqsdp1dps
R

R
e
H
dq1dqsdp1dps
+
R

R
H(pm+1;;ps;qn+1;;qs)e
H
dq1dqsdp1dps
R

R
e
H
dq1dqsdp1dps
Principle of the equipartition of energy state that each of the quadratic
terms contribute kT/2 to the energy and k/2 to the heat capacity.
E.g., for monatomic ideal gas
H=
p
2
x+p
2
y+p
2
z
2m
E= 3kT=2Cv= 3k=2
ajandbjcan be functions of the variables not involved in the
quadratic terms, i.e.,pm+1; ;ps;qn+1; ;qs.
E.g., for a linear rigid rotorH=
1
2I
(p
2

+
p
2

sin
2

);E=kT;Cv=k:
97/1

Equipartition of energy
Equipartition is a classical concept thus a small =kTbetween levels
is required for its validity.
For a diatomic moleculeCv=
5
2
Nk+
Nk(v=T)
2
e
v=T
(e
v=T
1)
2
The most general form of the equipartition theorem states that under
suitable assumptions, for a physical system with Hamiltonian energy
function H and degrees of freedom xn, the following equipartition
formula holds in thermal equilibrium for all indices m and n:
D
xm
@H
@xn
E
=mnkBT.
The general equipartition theorem is an extension of the virial
theorem, which states that
D
P
k
qk
@H
@qk
E
=
D
P
k
pk
@H
@pk
E
=
D
P
k
pk
dqk
dt
E
=
D
P
k
qk
dpk
dt
E
98/1

Ideal polyatomic gas
Start by applying the adiabatic or Born-Oppenheimer approximation
thus separating electronic and nuclear degrees of freedom.
ThenHnuc=Htrans+Hintnuc=trans+intqnuc=qtransqint
whereqtrans= [
2MkT
h
2]
3=2
V
Q(N;V;T) =
q
N
trans
q
N
int
N!
3n coordinates are required to specify the location of all atoms. 3
coordinates are to specify the center of mass while 2 or 3 (for linear or
non-linear molecule) are needed to specify its orientation. The
remaining 3n -5 or 3n-6 internal coordinates are needed for specifying
the relative position of the nuclei.
Use a rigid rotor-harmonic oscillator approximation to write
Q(N;V;T) =
(qtransqrotqvibqelecqnucl)
N
N!
Consider all atoms separated in their ground electronic state as the
zero of energy
99/1

Vibrational partition function
Potential energy of the molecule is a function of= 3n6 (3n5)
relative coordinate. At equilibrium conguration its gradient is zero.
U(~r) =U(~r0) +
P

i;j=1
1
2
(riri0
)(
@
2
U
@ri@rj
)r0
(rjrj0
) +
@
2
U
@ri@rj
are elements of a matrix called Hessian which at the pointr0
equalskij.
If the Hessian is positive denite at x, then U attains an isolated local
minimum at x. If the Hessian is negative denite at x, then U attains
an isolated local maximum at x. If the Hessian has both positive and
negative eigenvalues then x is a saddle point for U. Otherwise the test
is inconclusive.
Assuming molecular vibrations to be small we truncate this expansion
at the third term.
U(~r) =
P

i;j=1
kij
2
(riri0
)(rjrj0
).
The problem iscoupled harmonic oscillators.
100/1

Vibrational partition function
To practically deal with this problem assume all molecular motions as
vibrations and writeM
d
2
dt
2X=KXwhereX=
2
6
4
x1
.
.
.
x3n
3
7
5;M=
2
6
4
m1 0
0
.
.
.0
0 m3n
3
7
5;K=
2
6
4
k1;1 k1;3n
.
.
.
.
.
.
.
.
.
k3n;1 k3n;3n
3
7
5.
To solve this equation, we put it into complex form:
d
2
dt
2Z=M
1
KZ Try solutions of the formZ
(j)
=A
(j)
e
i!
(j)
t
. Yields eigenvalue equationM
1
KA
(j)
= (!
(j)
)
2
A
(j)
. M
1=2
KM
1=2
(M
1=2
A
(j)
) = (!
(j)
)
2
(M
1=2
A
(j)
) A real symmetric 3n3nmatrix has 3nreal eigenvalues and
correspondingly 3nreal orthogonal eigenvectors.101/1

Vibrational partition function
Thus there are 3neigenmodes characterized by eigenfrequency!
(j)
and eigenvectorM
1=2
A
(j)
, where (A
(k)
)
T
MA
(j)
=kj.
This is the origin of mass weighted coordinate in computational
chemistry.
Each molecular position can be described asQjM
1=2
A
(j)
Coordinate corresponding to the set of smallest frequencies represent
center of mass motion while the next set of small frequencies
represent rotational motion.
Hvib=
P

j=1
~
2
2
@
2
@Q
2
j
+
P

j=1
kj
2
Q
2
j
=
P

j=1
(nj+ 1=2)hjnj= 0;1;2; , wherej=
1
2
(kj)
1=2
qvib=

j=1
e

j
=2T
(1e

j
=T
)
, where j=
hj
kB
Evib=Nk
P

j=1
(j=2 +
je

j
=T
(1e

j
=T
)
) CV;vib=Nk
P

j=1
[(
j
T
)
2e

j
=T
(1e

j
=T
)
2
]
102/1

Rotational partition function
For linear moleculesj=
J(J+1)h
2
8
2
I
!J= 2J+ 1J= 0;1;2;
where moment of inertiaI=
P
n
j=1
mjd
2
j
qrot=
8
2
IkT
h
2=
T
r
Ixx=

X
j=1
mj[(yjycm)
2
+ (zjzcm)
2
]
Iyy=

X
j=1
mj[(xjxcm)
2
+ (zjzcm)
2
]
Izz=

X
j=1
mj[(xjxcm)
2
+ (yjycm)
2
]
AlsoIxz=
P

j=1
mj[(xjxcm)(zjzcm)]
103/1

Rotational partition function
There is a particular set of coordinates called the principal axes
passing through the center of mass of the body such that the inertia
matrix become diagonal.
Moments of inertia about these axesIXX;IYY;IZZare called the
principal moments of inertia. They are customarily denoted by
IA;IB;IC
principal axes often coincide with molecular axis of symmetry. Moments of inertia about principal axes are often found
experimentally and tabulated in terms of rotational constants as
A=
h
8IAc
B=
h
8IBc
C=
h
8ICc
in units ofcm
1
.
IA=IB=IC: Spherical top, e.g., CH4, CCl4.
IA=IB6=IC: Symmetric top, e.g., CH3Cl, NH3. IA6=IB6=IC: Asymmetric top, e.g., H2O, NO2.
104/1

Rotational partition function
For a spherical top:j=
J(J+1)~
2
2I
!J= (2J+ 1)
2
J= 0;1;2;
qrot=
1

R
1
0
(2J+ 1)
2
e
J(J+1)~
2
=2IkT
dJ Symmetry numberis the number of ways a molecule can be rotated
into itself. For H2O,= 2, for NH3= 3, for CH4= 12, for
C2H4= 4 and forC6H6= 12.
Symmetry number avoids over counting indistinguishable
congurations in phase space.
is the number of pure rotational elements in the point group of a
nonlinear molecule.
qrot=
1

R
1
0
4J
2
e
J
2
~
2
=2IkT
dJ=

1=2

(
8
2
IkT
h
2)
3=2 Any molecule with ann3-fold axis of symmetry is at least a
symmetric top.
105/1

Rotational partition function
For a symmetric topJK=
~
2
2
f
J(J+1)
IA
+K
2
(
1
IC

1
IA
)gJ=
0;1;2; ;K=J;J+ 1; ;J1;J!JK= 2J+ 1.
J is a measure of total rotational angular momentum of the molecule. K is component of rotational angular momentum along the axes C of
symmetric top.
qrot=
1

1
j=0
(2J+ 1)e
J(J+1)A
J
K=J
e
(CA)K
2
j=
~
2
2IjkT
: qrot=

1=2

(
8
2
IAkT
h
2)(
8
2
ICkT
h
2)
1=2
For a symmetric top:K=
p
2

2IA
+
(pp cos)
2
2IAsin
2

+
p
2

2Ic
Asymmetric top is the most common type of molecule.
Hamiltonian for an asymmetric top
H=
1
2IAsin
2

[(pp cos) cos psinsin ]
2
+
1
2IBsin
2

[(p
p cos) sin +psincos ]
2
+
1
2Ic
p
2

0 02 0 2are Euler angles.
106/1

Polyatomic thermodynamic functions
For asymmetric top
qrot=

1=2

(
8
2
IAkT
h
2)
1=2
(
8
2
IBkT
h
2)
1=2
(
8
2
ICkT
h
2)
1=2
=

1=2

(
T
3
ABC
)
1=2
Erot=
3
2
NkT CV;rot=
3
2
Nk Srot=Nkln[

1=2

(
T
3
e
3
ABC
)
1=2
]
Linear polyatomic molecule:
q= (
2MkT
h
2)
3=2
V
T
r
f

J=1
e

j
=2T
1e

j
=Tg!e1e
De=kT

A
NkT
=
ln[(
2MkT
h
2)
3=2Ve
N
]+ln(
T
r
)

j=1
[
j
2T
+ln(1e
j=T
)]+
De
kT
+ln!e1
E
NkT
=
3
2
+
2
2
+

j=1
[
j
2T
+
j=T
e

j
=T
1
]
De
kT
Cv
Nk
=
3
2
+
2
2
+

j=1
(j=T)
2e

j
T
(1e

j
=T
)
2
S
Nk
=
ln[(
2MkT
h
2)
3=2Ve
5=2
N
]+ln(
Te
r
)+

j=1
[
j=T
e

j
=T
1
ln(1e
j=T
)]+ln!e1
pV=NkT
107/1

Polyatomic thermodynamic functions
For nonlinear polyatomic molecules:
q= (
2MkT
h
2)
3=2
V

1=2

(
T
3
ABC
)
1=2
f

J=1
e

j
=2T
1e

j
=Tg!e1e
De=kT

A
NkT
= ln[(
2MkT
h
2)
3=2Ve
N
] + ln(

1=2

(
T
3
ABC
)
1=2
)

j=1
[
j
2T
+
ln(1e
j=T
)] +
De
kT
+ ln!e1
E
NkT
=
3
2
+
3
2
+

j=1
[
j
2T
+
j=T
e

j
=T
1
]
De
kT
Cv
Nk
=
3
2
+
3
2
+

j=1
(j=T)
2e

j
T
(1e

j
=T
)
2
S
Nk
= ln[(
2MkT
h
2)
3=2Ve
5=2
N
] + ln(

1=2
e
3=2

(
T
3
ABC
)
1=2
) +

j=1
[
j=T
e

j
=T
1
ln(1e
j=T
)] + ln!e1
pV=NkT
D0=Dej
1
2
hj
Residual entropy for CO and CH3D.
108/1

Hindered rotation
Rotation about a single bond.
U=
1
2
V0(1cos 3)
h
2
8
2
Ir
@
2

@
2+
1
2
V0(1cos 3) =
Iris eective moment of inertia.
At extremes of temperature this motion is a vibration or a rotation. Numerically solving Schrodinger equation for dierent values ofV0,
one can tabulateas a function ofV0=kT.
Partition function and thus thermodynamic properties can be derived
from tables of. Comparing calculated and experimental values of
thermodynamic functions one can deduceV0.
109/1

Chemical equilibrium

0
A
A(g) +
0
B
B(g)
0
cC(g) +
0
D
D(g).
=
NA
0
NA

0
A
=
NB
0
NB

0
B
=
NCNC
0

0
C
=
NDND
0

0
D
cC(g) +DD(g) +AA(g) +BB(g) = 0 dNj=jd dA=SdTpdV+
P
j
jdNj. At constant volume and
temperaturedA=
P
j
jdNj= (
P
j
jj)d.
At equilibrium
(
@A
@
)T;V= 0 =
P
j
jj=cC+DD+AA+BB.
Q(NA;NB;NC;ND;V;T) =
QA(NA;V;T)QB(NB;V;T)QC(NC;V;T)QD(ND;V;T) =
qA(V;T)
N
A
NA!
qB(V;T)
N
B
NB!
qC(V;T)
N
C
NC!
qD(V;T)
N
D
ND!
i=kT(
@lnQ
@Ni
)Nj6=i;V;T=kT(
@lnQi
@Ni
)Nj6=i;V;T=
kT(
@ln(q
N
i
i
=Ni!)
@Ni
)Nj6=i;V;T=kTln
qi(V;T)
Ni
110/1

Chemical equilibrium
Derive the equilibrium constant expression.
N

C
C
N

D
D
N

A
A
N

B
B
=
q

C
C
q

D
D
q

A
A
q

B
B
Kc(T) =

C
C

D
D

A
A

B
B
=
(qC=V)

C(qD=V)

D
(qA=V)

A(qB=V)

B
pj=jkTthusKp(T) =
p

C
C
p

D
D
p

A
A
p

B
B
= (kT)
C+DABKc(T)
E.g., Association of alkali metal vapor,
2Na(g)Na2(g)Kp(T) =
pdimer
p
2
monomer
= (kT)
1
(qNa
2
=V)
(qNa=V)
2
qNa(T;V) = (
2mNakT
h
2)
3=2
Vqelec(T)
qNa2
(T;V) = (
2mNa
2
kT
h
2)
3=2
V
8
2
IkT
2h
2
e
h=2
1e
h!1ee
De=kT
=
(
2mNa
2
kT
h
2)
3=2
V
T
2r
(1e
h
)
1
e
D0=kT
qNa
V
=?
qNa
2
V
=?
111/1

Chemical equilibrium
Isotopic exchange reaction,H2+D22HD. Born-Oppenheimer
approximation implies thatH2;D2;HDhave the same internuclear
potential, k andDe.
K(T) =Kp(T) =Kc(T) =

2
HD
H
2
D
2
=
p
2
HD
pH
2
pD
2
=
q
2
HD
qH
2
qD
2
=
(
2m
HD
kT
h
2
)
3
(
T

r;HD
)
2
(
e

;HD
=2T
1e

;HD
=T
)
2
e
2De=kT
(
2m
H
2
kT
h
2
)
3=2
(
2m
D
2
kT
h
2
)
3=2
(
T
2
4
r;H
2

r;D
2
)(
e

;H
2
=2T
1e

;H
2
=T
)(
e

;D
2
=2T
1e

;D
2
=T
)e
2De=kT
=
m
3
HD
(mH
2
mD
2
)
3=2
4r;H
2
r;D
2

2
r;HD
(1e

;H
2
=T
)(1e

;D
2
=T
)
(1e

;HD
=T
)
2
e
(2;HD;H
2
;D
2
)=2T
;HD
;H
2
=
HD
H
2
= (
H
2
HD
)1=2. Thusv;HD= (
3
4
)
1=2
v;H2
. For this reactionK(T) = 4(1:06)exp
77:7
T
.
112/1

Chemical equilibrium
For diatomic isotopic exchange reactions like
14
N2+
15
N22
14
N
15
N:
K(T) = 4(1 +

2
8M
2)e

2

M;vib=32M
2
T, where is the mass dierence
between isotopes and M is mass of the heavier isotope
CH4+DBrCH3D+HBr K(T) =
CH
3
DHBr
CH
4
DBr
=
qCH
3
DqHBr
qCH
4
qDBr
=
CH
4
DBr
CH
3
DHBr
(
MCH
3
DMHBr
MCH
4
MDBr
)
3=2IHBr
IDBr
(IAIBIC)
1=2
CH
3
D
(IAIBIC)
1=2
CH
4
qvib;CH
3
Dqvib;HBr
qvib;CH
4
qvib;DBr
qvib;CH
3
Dqvib;HBr
qvib;CH
4
qvib;DBr
exp[
P
j

CH
3
D
;j
+
HBr
;j

CH
4
;j

DBr
;j
2T
]
CH
4
DBr
CH
3
DHBr
=
121
31
= 4
113/1

Chemical equilibrium
Teller|Redlich product rule for isotopically substituted compounds:
(
M
0
M
)
3=2I
0
I
=
n
i=1
(
m
0
i
mi
)
3=2

3n5
j=1

0
j
j
for linear molecules and
(
M
0
M
)
3=2
(I
0
A
I
0
B
I
0
C
)
1=2
(IAIBIC)
1=2=
n
i=1
(
m
0
i
mi
)
3=2

3n6
j=1

0
j
j
Diatomic molecules:H2+I22HI
K(T) =
(qHI=V)
2
(qH
2
=V)(q
I
2
=V)
=
q
2
HI
qH
2
qI
2
=
(
m
2
HI
mH
2
mI
2
)
3=2
(
4r;H
2
r;I
2

2
r;HI
)
(1e

;H
2
=T
)(1e

;I
2
=T
)
(1e

;HI
=T
)
2
exp
(2D
HI
0
D
H
2
0
D
I
2
0
)
RT
Compare withd(lnK) =
H
R
d(
1
T
) Polyatomic molecules:
H2+
1
2
O2H2O K p(T) =
qH
2
O=V
(kT)
1=2
(qH
2
=V)(qO
2
=V)
1=2
qH
2
V
= (
2mH
2
kT
h
2)
3=2
(
T
2r;H
2
)(1e
;H
2
=T
)
1
e
D0;H
2
=RT
114/1

Chemical equilibrium
qO
2
V
= (
2mO
2
kT
h
2)
3=2
(
T
2r;O
2
)(1e
;O
2
=T
)
1
3e
D0;O
2
=RT
qH
2
O
V
= (
2mH
2
OkT
h
2)
3=2
1=2

(
T
3
A;H
2
OB;H
2
OC;H
2
O
)
3
j=1
(1
e
j;H
2
O=T
)
1
e
D0;H
2
O=RT
Restricted internal rotation: Ethylene-Ethane equilibrium C2H4+H2C2H6, one should nd the value ofV0 For the equilibrium
P
j
jj= 0 substitute(T;p) =0(T) +kTlnp
to get lnKp=
0
kT
=kTln(
q
N
) =kTln[(
q
V
)
V
N
] =kTln[(
q
V
)kT] +kTlnp 0(T) =kTln[(
q
V
)kT] depends on the unit of pressure. q(V;T) =qtrans(V;T)qrot(T)qvib(T)qelec(T)
115/1

Chemical equilibrium
Zero of energy conventions enters in calculation of
qelec(T) =!e1e
e1=kT
+!e2e
e2=kT
+ =
e
De=kT
(!e1+!e2e
12=kT
) =e
De=kT
q
0
elec
(T).
q(V;T) =qtrans(V;T)qrot(T)qvib(T)q
0
elec
(T)e
De=kT
=
qtrans(V;T)qrot(T)fj(1e
j=T
)
1
gq
0
elec
(T)e
(De1=2
P
j
hj)=kT
=
qtrans(V;T)qrot(T)q
0
vib
(T)q
0
elec
(T)e
D0=kT
q(V;T) =q
0
(V;T)e
D0=kT
=q
0
(V;T)e

0
0
=kT
partition as the product
of an internal partq
0
(V;T) and a scaling factor accounting for the
arbitrary zero of energy.

0
0
=kTln[(
q
0
V
)kT] +kTlnp
limT!0=
0
0 Convention: Energy of an element is zero at 0

K if it is in the physical
state characteristic of 25

C and 1 bar. For a molecule
0
0
represents
the energy of a molecule at 0

K relative to the elements, i.e., heat of
formation.
116/1

Chemical equilibrium
G
0
E
0
0
=RTln[(
q
0
V
)kT] whereE
0
0
is the standard free energy at
0

K.
E
0
0
=H
0
0
=G
0
0
(G
0
E
0
0
)=Tvaries slower with T compared to (G
0
E
0
0
).
RlnKp=
E
0
0
T
+ (
G
0
E
0
0
T
) S
0
298
= (
H298E
0
0
T
)(
G
0
298
E
0
0
T
)
117/1

Class presentations
Mr. Javadi Crystals, 11-1:11-3 15 Tir 11 am
Mr. Hadi Chemical Equilibrium 16 Tir 10:15 am
Mr. Moradi Crystals, 11-4:11-6 17 Tir 10:10 am
Ms. Mirzakhani Quantum Statistics, 10-1:10-4 19 Tir 2:45 pm
Mr. Sharbati Quantum Statistics, 10-5:10-7 20 Tir 2:45 pm
Ms. Madadi Simple theories of liquids 20 Tir 4 pm
Mr. Moham-
madvand
Polymers 22 Tir 3 pm
Mr. Hashemi Ideal systems in electric and mag-
netic elds
23 Tir 3 pm
Mr. Hajilou Imperfect gases 24 Tir 3:15 pm
Mr. Zamani Distribution functions in
monatomic liquids
24 Tir 4:30 pm
118/1

Table of physico-chemical constants
Quantity Symbol Value (SI units)
atomic mass constant mu= 1u 1:660538910
27
kg
Avogadro's number NA;L 6:022141710
23
Boltzmann constant k=R=NA 1:380650510
23
JK
1
Faraday constant F=NAe 96485.338 Cmol
-1
gas constant R 8.314472 JK
-1
mol
-1
,
0.08205 L atm mol
-1
K
-1
,
8.20573 m
3
atm mol
-1
K
-1
molar Planck constant NAh 3:9903110
10
J s mol
-1
electric constant (vacuum
permittivity)
"0= 1=(0c
2
) 8:85418781710
12
Fm
-1
119/1

Table of physico-chemical constants
Quantity Symbol Value (SI units)
magnetic constant (vacuum
permeability)
0 12:5663706110
7
NA
-2
Newtonian constant of gravitationG 6 :6740810
11
m
3
kg
-1
s
-2
Planck constant h 6 :62607004010
34
Js
reduced Planck constant~ 1:05457180010
34
Js
speed of light in vacuum c 299792458 m/s
electronic charge e 1 :6021910
19
C
electron mass me 9:1095610
31
Kg
120/1

Statistical Thermod Mechanics Nassimi.pdf

About This Presentation

Slide Content

Tags

Categories

Download

Quick Actions

Statistics

Related Slideshows

Statistical Thermod Mechanics Nassimi.pdf

About This Presentation

Slide Content

Slide 1

Slide 2

Slide 3

Slide 4

Slide 5

Slide 6

Slide 7

Slide 8

Slide 9

Slide 10

Slide 11

Slide 12

Slide 13

Slide 14

Slide 15

Slide 16

Slide 17

Slide 18

Slide 19

Slide 20

Slide 21

Slide 22

Slide 23

Slide 24

Slide 25

Slide 26

Slide 27

Slide 28

Slide 29

Slide 30

Slide 31

Slide 32

Slide 33

Slide 34

Slide 35

Slide 36

Slide 37

Slide 38

Slide 39

Slide 40

Slide 41

Slide 42

Slide 43

Slide 44

Slide 45

Slide 46

Slide 47

Slide 48

Slide 49

Slide 50

Slide 51

Slide 52

Slide 53

Slide 54

Slide 55

Slide 56

Slide 57

Slide 58

Slide 59

Slide 60

Slide 61

Slide 62

Slide 63

Slide 64

Slide 65

Slide 66

Slide 67

Slide 68

Slide 69

Slide 70

Slide 71

Slide 72

Slide 73

Slide 74

Slide 75

Slide 76

Slide 77