Download Lecture 9 Introduction to Statistical Mechanics

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Lecture #9
Lecture 9
Objectives:
1. Be able to explain what statistical mechanics is and why it is important.
2. Basic tools of Statistical Mechanics
3. Classical Mechanics: Be able to write the Newtonian, Lagrangian, and Hamiltonian equations
4. Quantum Mechanics:
(a) Be able to answer the question: What is a wave function?
(b) Be able to write Schrödinger’s equation
(c) Be able to derive quantum operators
Introduction to Statistical Mechanics
1. What is Statistical Mechanics?
Thermodynamics deals with the macroscopic properties of matter. Thermodynamics is not
dependent on the fundamental nature of matter. That is, thermodynamics is valid whether
matter is made up of atoms and molecules or whether matter is made up of some kind of
continuum fabric.
Quantum mechanics deals with the behavior of matter on the microscopic scale. In quantum mechanics we calculate the energy and behavior of a few atoms or small molecules. Most
calculations are performed at T = 0 K. We can’t get the macroscopic properties (e.g., the
equation of state, etc). from Quantum mechanics.
Statistical mechanics is a bridge between quantum mechanics and thermodynamics. We
seek to use statistical mechanics to compute macroscopic properties from the quantum mechanical information about the atoms and molecules of interest.
Quantum
Mechanics
Statistical
Macroscopic
Thermodynamic
Properties
Mechanics
2. Why study statistical mechanics?
Consider an example: We have used equations of state (EOS). What are they? They give
you a method of calculating one of (p, V, T ) given any two of them. Usually EOS are cast in
some analytic functional form. How do you get EOS?
(a) Guess some functional form, P = f (V, T ). e.g., van der Waals made the guess that
P = RT /(V − b) − a/V 2 .
(b) Measure as much P V T data as you possibly can for a specific fluid.
(c) Regress the parameters in the EOS in some least squares fashion. E.g., find a and b the
the van der Waals EOS for the fluid.
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Lecture #9
Do you now have complete information? How about U, H, S, etc.?
What happens if you want to find the EOS of a different fluid? For example, you did the
above for propane, now you want the EOS for H2 O. In many cases you need to go back to
the beginning, i.e., guess a new functional form. This is because you can’t fit propane and
water with the same functional form.
Problems with the above process:
(a) Very time consuming and very expensive.
(b) Have to start all over again for many new fluids.
(c) Extrapolation outside the region of parameter regression is not valid. Extrapolation
often gives unphysical results. For example: What happens in the van der Waals EOS
if V = b?
(d) This process does not give you any physical insight about why the fluid behaves as it
does, or how to systematically improve the EOS.
(e) You don’t know anything about many of the other properties.
Is there some way we can get at the EOS from a fundamental approach?
If matter is truly composed of molecules, then it must be the interactions among these
molecules that give rise to the macroscopic observable behavior.
POSTULATE: If we can understand the microscopic interactions we can predict the macroscopic behavior. This is the goal of statistical mechanics.
Classical Mechanics
1. Newtonian Mechanics
F~ = m~a = m¨~q
where ~q = (x, y, z) is the position vector and the dots denote the time derivative,
2
¨~q = d ~q
dt2
For a conservative system the total energy is
E(~q, ~q˙ ) =
1X
mi |q~˙i |2 + U (~q)
2 i
where U is the potential energy. The force for a conservative system is the negative of the
gradient of the potential.
~
F~ = −∇U
In Cartesian coordinates
~ = ~ı ∂U + ~ ∂U + ~k ∂U
∇U
∂x
∂y
∂z
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Lecture #9
where ~ı, ~, ~k are the unit vectors in the x, y, z directions, respectively. Newton’s equations
of motion are most convenient in Cartesian coordinates. In spherical coordinates the three
~ are
components of ∇U
h
i
r
h
i
θ
h
i
~
∇U
~
∇U
~
∇U
φ
=
∂U
∂r
=
1 ∂U
r ∂θ
=
1 ∂U
r sin θ ∂φ
2. Lagrangian Mechanics. Lagrangian has the advantage over Newton’s equations in that they
are invariant under coordinate transform.
L=K −U
K is the kinetic energy, given in general by
2
K=
m~q˙
2
Legrange’s equations of motion are
d
dt
∂L
∂ q̇j
!
=
∂L
for j = 1, 2, 3
∂qj
where qj is the jth coordinate. The importance of this equation is that the equation is
the same whether the coordinates are in Cartesian or some other system. For example in
Cartesian coordinates
d ∂L
=
dt ∂ ẋ
d ∂L
=
dt ∂ ẏ
d ∂L
=
dt ∂ ż
∂L
∂x
∂L
∂y
∂L
∂z
(1)
For cylindrical coordinates
d ∂L
=
dt ∂ ṙ
d ∂L
=
dt ∂ θ̇
d ∂L
=
dt ∂ ż
∂L
∂r
∂L
∂θ
∂L
∂z
Example: Consider a particle moving in 2-d under a central potential U (r) = −a/r.
K=
m 2
(ṙ + r2 θ̇2 )
2
(2)
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Lecture #9
L=K −U =
m 2
(ṙ + r2 θ̇2 ) + a/r
2
d ∂L
=
dt ∂ ṙ
d ∂L
=
dt ∂ θ̇
∂L
∂r
∂L
∂θ
∂L
= mr̈
∂ ṙ
∂L
a
= mrθ̇ − 2
r
∂r
d ∂L
d 2 = m
r θ̇
dt ∂ θ̇
dt
= m(2rṙθ̇ + r2 θ̈)
∂L
= 0
∂θ
d
dt
This gives the coupled set of differential equations:
a
= 0
r2
rθ̈ + 2ṙθ̇ = 0
mr̈ − mrθ̇2 +
3. Hamiltonian Mechanics. Hamilton’s equations are more convenient for quantum mechanics
and statistical mechanics. We define the Hamiltonian, H, as
H =K +U
H is the total energy and has the property that
of motion. Hamilton’s equations of motion are
∂H
∂pj
!
= q˙j
∂H
∂qj
!
= −p˙j
where
pj =
∂L
∂ q̇j
!
is the momentum conjugate to qj .
dH
dt
= 0, i.e., H is constant, or is a constant
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Quantum Mechanics
At the level of everyday experience it appears that energy can take on any value. On the microscopic scale, energies are, in fact, quantized. Einstein’s paper on the photo-electric effect showed
that light had both wave-like and particle-like behavior. Quantum mechanics is the result of reconciling classical mechanics with these microscopic observations. Instead of specifying a position
and momentum, we must specify something that looks like a wave. The wave equation describes
the probability of finding a particle in space. The well-known Heisenberg uncertainty principle is a
result of the wave-particle dual nature of matter.
δpδq ≥ h
Note that there are at least two other expressions for the Heisenberg uncertainty relation:
δpδq ≥
h̄
2
δpδq ≥ h̄
where h̄ = h/(2π).
1. The fundamental premise of quantum mechanics is that matter can be described as a wave,
rather than as a collection of particles. The first person to write down the equation that
describes matter in terms of waves was Erwin Schrödinger, for whom the Schrödinger equation
is named. Ironically, Erwin Schrödinger was quoted to have said about quantum mechanics
“I don’t like it, and I’m sorry I ever had anything to do with it.” The wave function can be
found by solution of the Schrödinger equation,
Hψ = Eψ
where H is the Hamiltonian operator, and E is the energy of the system.
H=−
h2
∇2 + U (~q)
8π 2 m
~ ·∇
~ is the Laplacian. In Cartesian coordinates
where ∇2 = ∇
∇2 =
∂2
∂2
∂2
+
+
∂x2 ∂y 2 ∂z 2
In spherical polar coordinates we have
"
∂
1 ∂
r2
∇ = 2
r ∂r
∂r
2
1 ∂
∂
+
sin θ
sin θ ∂θ
∂θ
1
+
sin2 θ
∂2
∂φ2
!#
In cylindrical coordinates the Laplacian is
∇2 =
1 ∂
∂
r
r ∂r
∂r
+
1 ∂2
∂2
+ 2
2
2
r ∂θ
∂z
See Bird, Stewart & Lightfoot or any good math book for other coordinates.
Note that the Hamiltonian operator H is like the Hamiltonian, except that the kinetic energy
is replaced by the kinetic energy operator.
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Lecture #9
2. The wave function describes the position and momentum of a particle. The probability that a
Q
particle (or system) is located in the volume element (qi + δqi ) at time t is Ψ∗ (~q, t)Ψ(~q, t)d~q.
Where Ψ∗ (~q, t) is the complex conjugate of the wave function.
The wave function is normalized,
Z
Ψ∗ (~q, t)Ψ(~q, t)d~q = 1
That is, the particle must be found somewhere in the universe.
3. Eigenstates and Eigenvalues: Quantized Energy Levels.
In the language of linear algebra, the value of E in the Schrödinger equation is called the
eigenvalue, or correct value, of the equation. The wave function that correctly solves the
Schrödinger equation is called the eigenstate or eigenfunction.
A macroscopic example of energy eigenstates is the vibrational spectra of a molecule. If you
shine light on a molecule and tune that light you will find wave lengths that correspond to
very high absorbance, while close-by wave lengths will not be absorbed. Why? The specific
energy of the wave will excite the molecule. That energy is the eigenvalue.
4. Operators and Observables. If you want to get the total energy of a molecule you must use
the Hamiltonian operator.
HΨ = EΨ
∗
Z
Z
Z
∗
Ψ HΨ = Ψ
Z EΨ
Ψ∗ HΨd~q =
Ψ∗ EΨd~q
∗
Ψ HΨd~q = E
Z
Ψ∗ Ψd~q
Ψ∗ HΨd~q = E
In general, if you want to find an observable (i.e., some measurable property) you have to
find the quantum mechanical operator that corresponds to that observable, and use it in the
following equation:
hJi =
Z
Ψ∗ J Ψd~q
5. The wave function:
Let Ψ(~q, t) be the wave function, which is a function of time and position. Let us assume
that the form of Ψ is a product of a temporal and a spatial function,
Ψ(~q, t) = Aτ ψ
where A is the amplitude of the wave,
2πiEt
τ = exp −
h
2πipq
ψ = exp
h
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Lecture #9
Now, how do we find the momentum of the particle? Take the derivative and solve for p:
∂Ψ
∂q
= Aτ
t
∂ψ
∂q
= Aτ
t
2πipψ
2πip
=Ψ
h
h
hence,
h
pΨ =
2πi
∂Ψ
∂q
t
Likewise, to find the energy we take the derivative w.r.t. time,
∂Ψ
∂t
q
∂τ
= Aψ
∂t
q
=−
2πiEΨ
h
hence,
EΨ = −
h
2πi
∂Ψ
∂t
q
Note that in quantum mechanics mechanical variables are obtains by operating on the wave
function. Thus, mechanical variables are replaced by operators,
p ⇒ p̂ =
h
∇
2πi
E ⇒ Ê = −
h ∂
2πi ∂t
What is the kinetic energy operator?
1 2
1
h2
h2
1
p ⇒ K̂ =
p̂ · p̂ = 2 2 ∇ · ∇ = − 2 ∇2
K = m~v 2 =
2
2m
2m
8π i m
8π m
Now we can write down the Hamiltonian operator:
h
h2
H = K + U ⇒ ĤΨ = − 2 ∇2 Ψ + U Ψ = EΨ = −
8π m
2πi
∂Ψ
∂t
q
which is the well-known Schrödinger equation. Note that
Z
Ψ∗ ĤΨd~q =
Z
Ψ∗ EΨd~q = E
Z
Ψ∗ Ψd~q = E
Reason: Ψ is normalized, i.e., the probability that the particle is in the universe somewhere
must be one.